Ring Theory Volume I
This is volume 127 in PURE AND APPLIED MATHEMATICS H. Bass and S. Eilenberg, editors A list of titles in this series appears at the end of this volume.
Ring Theory Volume I
Louis H. Rowen Department of Mathematics and Computer Science Bar IIan University Ramat Can, Israel
ACADEMIC PRESS, INC. Harcourt Brace Jovanovich, Publishers
Boston San Diego New York Berkeley London Sydney Tokyo Toronto
Copyright 0 1988 by Academic Press, Inc. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher.
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Library of Congress Cataloging-in-Publication Data Rowen, Louis Halle. Ring theory. (Pure and applied mathematics; v. 127-128) Includes bibliographies and indexes. 1. Rings (Algebra) I. Title. 11. Series: Pure and applied mathematics (Academic Press) ; 127-128. 87-14536 QA3.P8 VOI. 127-128 510 s [512’.4] CQA2471 ISBN 0-12-599841-4 (v. 1) ISBN 0-12-599842-2(v. 2)
88899091 9 8 7 6 5 4 3 2 1 Printed in the United States of America
Contents ...
Foreword
Xlll
Introduction: An Overview of Ring Theory
xvii
Table of Principal Notation
xxiii
Chapter 0 General Fundamentals 0.0 Preliminary Foundations Monoids and Groups Rings and Modules Algebras Preorders and Posets Upper and Lower Bounds Lattices Modular Lattices Zorn’s Lemma Well-ordered Sets and Transfinite Induction Fields First Order Logic
1 1 2 5 6 6 7 8 10 11 12 12
0.1 Categories of Rings and Modules Monics and Epics Functors
13 15 16
0.2 Finitely Generated Modules, Simple Modules, and Noetherian and Artinian Modules Finitely Generated Modules
18 18
V
Contents
vi
Cyclic Modules Simple Rings and Modules Chain Conditions Noetherian and Artinian Modules 0.3 Abstract Dependence Algebraic and Transcendental Elements
19 19 21 22 23 25
Exercises
26
Chapter 1 Construction of Rings 1.1
Matrix Rings and Idempotents Matrices and Matrix Units Subrings of Matrix Rings Matrices Whose Entries Are Not Necessarily in Rings Idempotents and the Peirce Decomposition Idempotents and Simple Modules Primitive Idempotents Lifting Matrix Units and Idempotents
29 29 32 33 35 37 39 39
1.2 Polynomial Rings Monoid Rings Polynomial Rings Rings of Formal Power Series and of Laurent Series Digression: More General Constructions A Supplement: Ordered Groups
42 42 43 48 51 51
1.3 Free Modules and Rings Free Modules Independent Modules and Direct Sums Modules Over Division Rings Are Free Free Objects Free Rings and Algebras Free Commutative Ring Diagrams Universals Invariant Base Number Weakly Finite Rings A Supplement: The Free Group
53 54 54 55
1.4 Products and Sums Direct Products and Direct Sums
66 66
56
57 59 60 60 61 63 64
vii
Contents
Exact Sequences Split Exact Sequences Reduced Products Ultraproducts Products and Coproducts B Supplement: Free Products and Amalgamated Sums C Supplement: Categorical Properties of Modules: Abelian Categories 1.5
Endomorphism Rings and the Regular Representation Endomorphism Rings Endomorphisms as Matrices The Dual Base Adjunction of 1
67 68 69 71 73 76 78 81 81 84 86 87
1.6 Automorphisms, Derivations, and Skew Polynomial Rings Automorphisms Derivations, Commutators, and Lie Algebras Skew Polynomial Rings and Ore Extensions Principal Left Ideal Domains (PLID’s) Skew Polynomial Rings (Without Derivation) Over Fields Differential Polynomial Rings Over Fields The Weyl Algebra Skew Power Series and Skew Laurent Series Skew Group Rings
88 88 89 92 93 96 97 98 100 101
1.7 Tensor Products Tensor Products of Bimodules and of Algebras Properties of the Tensor Operation Tensors and Centralizing Extensions Tensor Products Over Fields Tensor Products and Bimodules
101 103 105 106 108 110
1.8 Direct Limits and Inverse Limits Direct Limits D Supplement: Projective Limits The Completion
111 112 115 116
1.9 Graded Rings and Modules Tensor Rings B Supplement: Constructing Free Products
118 121 124
viii
Contents
1.10 Central Localization (also, cf. 92.12.9ff.) Structure Passing From R to S-’R Examples of Central Localization Central Localization of Modules
129 132 134 134
Exercises
137
Chapter 2 Basic Structure Theory 2.1 Primitive Rings Jacobson’s Density Theorem Prime Rings and Minimal Left Ideals Finite-Ranked Transformations and the Socle Examples of Primitive Rings E Supplement: A Right Primitive Ring Which Is Not Primitive
150 151 153 154 158 159
2.2 The Chinese Remainder Theorem and Subdirect Products Subdirect Products Semiprime Rings
162 163 164
2.3 Modules with Composition Series and Artinian Rings The Jordan-Holder and Schreier Theorems Artinian Rings Split Semisimple Artinian Algebras Central Simple Algebras and Splitting Zariski Topology (for Finite Dimensional Algebras)
165 165 167 171 173 174
2.4 Completely Reducible Modules and the Socle
176
2.5 The Jacobson Radical Quasi-Invertibility Examples Idempotents and the Jacobson Radical Weak Nullstellensatz and the Jacobson Radical The Structure Theoretical Approach to Rings “Nakayama’s Lemma” The Radical of a Module F Supplement: Finer Results Concerning the Jacobson Radical Wedderburn’s Principal Theorem F Supplement: Amitsur’s Theorem and Graded Rings F Supplement:The Jacobson Radical of a Rng Galois Theory of Rings
179 180 182 183 184 186 188 189 189 192 194
196 198
Contents
ix
2.6 Nilradicals Reduced Rings Nilpotent Ideals and Nilradicals The Nilradical of Noetherian Rings Bounded Index Derivations and Nilradicals Nil Subsets G Supplement: Koethe’s Conjecture
199 20 1 202 205 206 207 207 209
2.7 Semiprimary Rings and Their Generalizations Chain Conditions on Principal Left Ideals Passing from R to eRe and Back Semiperfect Rings Structure of Idempotents of Rings H Supplement: Perfect Rings G Supplement: Left x-Regular Rings
21 1 212 215 217 219 22 1 224
2.8 Projective Modules (An Introduction) Projective Modules Projective Versus Free Hereditary Rings Dual Basis Lemma Flat Modules Schanuel’s Lemma and Finitely Presented Modules H Supplement: Projective Covers Categorical Description of Projective Covers
225 225 227 228 229 230 232 233 236
2.9 Indecomposable Modules and LE-Modules The Krull-Schmidt Decomposition Uniqueness of the Krull-Schmidt Decomposition Applications of the Krull-Schmidt Theorem to Semiperfect Rings Decompositions of Modules Over Noetherian Rings I Supplement: Levy’s Counterexample C Supplement: Representation Theory The Brauer-Thrall Conjectures Quivers and f.r.t. The Multiplicative Basis Theorem and the Classification Problem The Categorical Language
237 238 240 242 244 245 249 25 1 259 260 26 1
X
Contents
2.10 Injective Modules Divisible Modules Essential Extensions and the Injective Hull Criteria for Injectivity and Projectivity J Supplement: Krull-Schmidt Theory for Injective Modules
26 1 263 266 26 8
2.11 Exact Functors Flat Modules and Injectives Regular Rings Faithfully Flat Modules
270 273 276 278
2.12 The Prime Spectrum Localization and the Prime Spectrum Localizing at Central Prime Ideals The Rank of a Projective Module K Supplement: Hilbert’s Nullstellensatz and Generic Flatness Comparing Prime Ideals of Related Rings LO and the Prime Radical Height of Prime Ideals
280 282 283 285
2.13 Rings with Involution The Category of Rings with Involution Involutions of Rings with Minimal Left Ideals Simple Rings with Involution (*)-Radical and (*)-Semiprime
296 299 30 1 304 306
Exercises
308
269
286 29 1 294 296
Chapter 3 Rings of Fractions and Embedding Theorems 3.1 Classical Rings of Fractions Properties of Fractions Localizations of Modules
348 352 355
3.2 Goldie’s Theorems and Orders in Artinian Quotient Rings Annihilators ACC (Ann) Goldie Rings Goldie Rank and Uniform Dimension
356 356 357 358 36 1
xi
Contents
Annihilator Ideals in Semiprime Rings Digression: Orders in Semilocal Rings Embedding of Rings Universal Sentences and Embeddings Embeddings into Primitive and Semiprimitive Rings L Supplement: Embeddings into Matrix Rings L Supplement: Embedding into Division Rings L Supplement: Embedding into Artinian Rings Existentially Closed Rings
363 367 368 368 369 37 1 372 3 74 377
3.3 Localization of Nonsingular Rings and Their Modules Johnson’s Ring of Quotients of a Nonsingular Ring The Injective Hull of a Nonsingular Module Carrying Structure of R to ff
377 378 380 382
3.4 Noncommutative Localization Digression: The Maximal Ring of Quotients The Martindale-Amitsur Ring of Quotients INC Revisited C Supplement: Idempotent Filters and Serre Categories
3 84 385 386 389 39 1
3.5 Left Noetherian Rings 394 Constructing Left Noetherian Rings 395 The Reduced Rank 396 The Principal Ideal Theorem 399 Heights of Prime Ideals 400 Digression: Decomposition Theorems of Noetherian Rings 40 1 Artinian Properties of Noetherian Rings: Prelude to Jacobson’s Conjecture 403 Jacobson’s Conjecture 405 D Supplement: Noetherian Completion of a Ring 409 Krull Dimension for Noncommutative Left Noetherian Rings 414 Other Krull Dimensions 418 Critical Submodules 420 M Supplement: Basic Dimension 42 1 M Supplement: The Noncommutative Forster-Swan Theorem 425 J Supplement: Jategaonkar Theory 433 N Supplement: Gabriel Dimension 44 1 Exercises
443
Contents
XU
Chapter 4 Categorical Aspects of Module Theory 4.1 The Morita Theorems Categorical Notions Digression: Two Examples Morita Contexts and Morita’s Theorems Proof of Morita’s Theorem and Applications Digression: Simple Noetherian Rings
467 469 470 47 1 475 476
4.2 Adjoints Adjoint Pairs and Universals
477 479
Exercises
48 1
Appendix A. The Proof of Magnus’ Theorem and the Magnus-Witt Theorem (Theorem 1.3.38) Exercise
485
Appendix B. Normed Algebras and Banach Algebras
490
Exercises
493
The Basic Ring-Theoretic Notions and Their Characterizations
495
Major Ring- and Module-Theoretic Results Proved in Volume I (Theorems and Counterexamples; also cf. “Characterizations”)
499
References
507
Bibliography of Books Collections of Papers Bibliography of Articles Subject Index
490
507 510 512 531
Foreword (Obligatory) The theory of rings originally developed as a collection of brilliant insights for proving and improving diverse theorems in algebra, by extracting the essence of the theorems divested of superfluous hypotheses. This approach was startlingly successful and led to the search for solutions from “first principles”; moreover, there was a unifying structure theory due principally to Jacobson, and his book Structure of Rings (1956, rev. 1964) remains one of the principal references in the literature. (The other classic references are Herstein’s Carus Monograph Noncommutative Rings and, for the homological point of view, Cartan- Eilenberg’s book Homological Algebra.) However, as more and more “elementary” proofs found their way into the folklore, the structure of rings began to resemble a tower of Babel, where ring theorists speak such diverse languages (ring theory, module theory, category theory, etc.) that at times we cannot communicate effectively with one another. In the last few years there have been many very successful books in special topics, which are sometimes used as texts, notably Passman’s book The Algebraic Structure of Group Rings. However, any specialized book necessarily presents the subject vertically rather than horizontally; i.e., the selection of topics and proofs reflects the particular needs of the subject area instead of the broader picture, and the reader must often plow through technical results in order to arrive at theorems of more general interest. On the other hand, the basic theorems are often proved in an almost offhand fashion, to leave space for the author’s main objective. For example, any book concerned with finite dimensional algebras must deal with the Wedderburn- Artin theorem, that any finite dimensional simple algebra is isomorphic to a ring of matrices over a division algebra. One of several short, direct proofs is presented, although the reader thereby loses the opportunity of seeing Wedderburn’s theorem in one
xiii
xiv
Foreword
of two important broader contexts-the Jacobson structure theory and/or the Morita theory. The principal object of these two volumes is to present the mainstream of ring theory, being a source both of results and of proofs. The first volume deals with general structural results and could serve as an introductory text; the second focuses on the classes of rings that have occupied most attention in the literature. On the other hand, efforts were made to hold the material to readable size, leading to the following hierarchy: Main Text Supplementary material Appendices Exercises Digressions Chapter 1 deals mostly with constructions needed throughout the text but might well be skipped and referred to as needed. Thus a graduate course should probably start with Chapter 2, the structure theory (of both rings and modules), which is relatively easy to assimilate and lies at the heart of ring theory. From there one might proceed to Sections 3.1 and 3.2, in order to cover Goldie’s Theorem; then one might read Section 4.1 for Morita’s theorems. Parts which are needed less frequently are designated as “Supplements” and could be skipped on the first reading. In this manner one could progress quite far even in the first 150 pages of reading. The supplements are designed as a means of continuing a specific thread throughout the text. Thus “A Supplement” deals with ordered groups and their uses in ring theory, and occurs in &1.2,1.3; “B Supplement,”dealing with free products, occurs in R1.4,1.9. The two appendices are akin to supplements,but are put off until the end, in order not to disrupt the flow of the text. The exercises are designed as a secondary extension of the text, and often their “hints” are virtually complete proofs. One aim of the exercises is to present interesting theorems (such as the Popescu-Gabriel theorem) which, although important, are not needed later in the text proper. With one exception (Bass’ theorem on “big” projective modules in $5.1, which uses Kaplansky’s theorem that non-f.g. projective modules over a local ring are free), I do not know of any result in the text whose proof relies on an exercise. Also appearing as exercises are several interesting examples dispersed throughout the literature (many due to G. Bergman), so the reader is urged at least to peruse the exercises.
Foreword
xv
Parts of the text are labeled as “digressions.” These are paragraphs that do not bear directly on the remainder of the text and often are given scant explanation, in order not to distract the reader from the intended thrust of the exposition. The larger digressions point to a major area of recent interest. The shorter digressions are asides. Notwithstanding these various attempts to hold down the size of the main text, several worthy subjects have been slighted. The original intention was to give a fairly detailed account of the representation theory of Artin algebras, but lack of space required it to be shrunk to the rather meager sketch at the end of Section 2.9. Other areas have been similarly misrepresented. In contrast to the fundamental aspect of Volume I, the goal of Volume I1 is to give the flavor of the subjects of current research, although in most areas the results fall a few years behind latest work. The reader should also take note of the various indices, which are intended in part as aids to help organize the material. The issue of acknowledgments is very delicate, since so many people have graciously aided me in this project. Ami Braun, P. M. Cohn, Marie-Paule Malliavin, and Jean-Pierre Tignol have spotted many flaws in earlier versions of the manuscript. Bill Blair gave instrumental advice in bringing Volume I to final form. On the other hand, I am always indebted to my mentor Nathan Jacobson, as well as to S. A. Amitsur. Other colleagues and friends who provided useful help and advice include Ephraim Armendariz, Maurice Auslander, Miriam Cohen, S. Dahari, Dan Farkas, Ed Formanek, Larry Levy, John McConnell, Chris Robson, Shmuel Rosset, David Saltman, Lance Small, and Robert Snider. Thanks are also due to Tony Joseph and Rudolf Rentschler for pointing out the highlights of the theory of enveloping algebra. Small’s excellent compilations of reviews (Reoiews in Ring Theory, Amer. Math. SOC.(1980, 1984)) have been an invaluable tool. Finally, I would like to thank Klaus Peters of Academic Press for his efforts in bringing the work to fruition.
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Introduction: An Overview of Ring Theory
In the solar system of ring theory the Sun is certainly the semisimple Artinian ring, which can be defined most quickly as a finite direct product of matrix rings over division rings. Much of ring theory is involved in measuring how far a ring is from being semisimple Artinian, and we shall describe the principal techniques. The numbers in parentheses refer to locations in these two volumes.
The Structure Theory of Rings Any simple ring (with 1)having minimal nonzero left ideal is of the form M,,(D), by the celebrated result known as the Wedderburn- Artin theorem (2.1.25’). The general structure theory begins with Jacobson’s density theorem (2.1.6), which generalizes the essence of the Wedderburn- Artin theorem to the class of prirnitioe rings; using subdirect products (52.2) one can then pass to semiprimitive rings. The passage from semiprimitive rings to arbitrary rings leads one to the study of various radicals, which are intrinsically-defined ideals which when 0 make the ring much easier to analyze. Thus the radical is the “obstruction” in the structure theory. In the general structure theory there are several radicals, which we consider in order of decreasing size. The biggest, the Jacobson radical (52.5) denoted Jac(R), is 0 iff the ring is semiprimitive. Since Jac(R) is often nonzero, one
xvii
xviii
Introduction
considers various nilradicals (#2.6), so called because their elements are nilpotent. The largest nilradical is called the upper nilradical Nil(R) and the smallest nilradical is called the prime radical or lower nilradical and is the intersection of the prime ideals; a ring is semiprime iff its lower nilradical is 0. Even in the classical case of finite dimensional algebras over fields (for which semiprimitive rings become semisimple Artinian (42.3) and for which the radicals all coincide), the structure of the radical itself is extremely complicated; often the radical is considered an encumbrance to be removed as soon as possible. Thus the various structure theorems which enable us to “shrink” the radical take on special significance; most prominent is Amitsur’s theorem (2.5.23), which says that if Nil(R) = 0 then Jac(R[A]) = 0, where A is a commuting indeterminate over R. Other such results are considered in 42.6. Certain classes of rings are particularly amenable to the structure theory. If a ring R is Artinian then Jac(R) is nilpotent and R/Jac(R) is semisimple Artinian; furthermore, one can “lift” the idempotents from R/Jac(R) to R to obtain more explicit information about R ($2.7). There are other instances when Jac(R) is nil (42.5). One of the principal classes of noncommutative rings is Noetherian rings (43.5), for which Nil(R) is nilpotent. Fortunately for semiprime Noetherian rings one has Goldie’s theorem (43.2), which shows that the classical ring of fractions exists and is semisimple Artinian; this result makes available the Wedderburn- Artin theorem and thereby rounds out the WedderburnArtin-Noether-Jacobson-Levitzki-Amitsur-Goldie structure theory. Goldie’s theorem also applies to prime rings with polynomial identities (Chapter 6). Rings of fractions have been generalized to rather broad classes of rings, as described in Chapter 3. Since the structure theory revolves around primitive and prime rings, one is interested in the set of primitive ideals (of a ring) and the set of prime ideals. These sets are called the primitiue and prime spectra and have a geometry of considerable interest (42.12).Recent research has focused on the primitive and prime spectra of certain classes of rings. Since any ring is a homomorphic image of a free ring, the Cohn- Bergman school has conducted research in studying free rings in their own right. Such a project is of immense difficulty, for the very reason that the theory of free rings necessarily includes all of ring theory. Occasionally one wants to consider an extra bit of structure for a ring, the inuolution (42.13),which generalizes the transpose of matrices. The structure theory carries over fairly straightforwardly to rings with involution, with involutory analogues of the various structural notions.
Introduction
xix
The Structure Theory of Modules Semisimple Artinian rings are also characterized by the property that every module is a direct sum of simple modules (2.4.9), and, furthermore, there are only a finite number of isomorphism classes of simple modules (2.3.13). This raises the hope of studying a ring in terms of its modules, and also of studying a module in terms of simple modules. The “best” modules to study in this sense are Artinian Noetherian modules, for they have composition series ($2.3), and these are essentially “unique.” Unfortunately there is no obvious way of building a module from its simple submodules and homomorphic images, so one turns to building modules as direct sums of indecomposable modules, thereby leading us to the “Krull-Schmidt” theorem (2.9.17), which says that every module with composition series can be written “uniquely” as a direct sum of indecomposables. Unfortunately this leaves us with determining the indecomposables (52.9), which is a tremendous project even for Artinian rings and which largely falls outside the scope of this book. Thus one is led to try more devious methods of studying modules in terms of simple modules, such as the noncommutative Krull and Gabriel dimensions (53.5). The Krull dimension has become an indispensible tool in the study of Noetherian rings.
Category Theory and Homology When studying rings in terms of their modules, one soon is led to categories of modules and must face the question of when two rings have equivalent categories of modules. Fortunately there is a completely satisfactory answer of Morita (54.1). Every module is a homomorphic image of a free module, and perhaps one could learn more about modules by studying free modules. It turns out that a more natural notion from the categorical point of view is projective module (52.8), and indeed a ring R is semisimple Artinian iff every module is projective (2.11.7). The interplay between projective and free is very important, leading to the rank of a projective module (2.12.19) and the K O theory (55.1). The path of projective modules can take us to projective resolutions and homological dimension of modules (g5.1, 5.2). Homological dimension is most naturally described in terms of category theory, which lends itself to dualization; thus one also gets injective modules (52.10), which also play an important role in the more general theories of fractions (@3.3,3.4).Homology (and cohomology) lay emphasis on two important functors, YU,~ and &xt, which are derived respectively from the tensor functor and from the functor Xom. These latter two functors are an example of an adjoint pair (4.2).
xx
Introduction
Special Classes of Rings (Mostly Volume 11) Certain classes of rings are of special interest and merit intensive research. Classically the most important are finite dimensional algebras over a field; these fall inside the theory of Artinian rings, but much more can be said. The role of the radical is made explicit by Wedderburn’s principal theorem (2.5.37) when the base field is perfect; Wedderburn’s result has been recast into the cohomology of algebras (45.3). Much of the theory of finite dimensional algebras carries over to the more general realm of rings with polynomial identities (PI-rings, chapter 6). Deep results of representation theory can be obtained using the PI-theory by means of elementary arguments. Moreover, PI-theory is not tied to a base field, and so various generic techniques are available to enrich the PI-theory; relatively free PI-algebras have commanded considerable attention. It turns out that there is just enough commutatively in the PI-theory to enable one to obtain a satisfactory version of much of the theory of commutative algebras and, indeed, to build a noncommutative algebraic geometry (which however lies largely outside the realm of this book). In a different direction, commutative algebraic geometry leads us to consider a@ne algebras over a field, by definition finitely generated as algebras. Although not much can be said about affine algebras in general, affine algebras become more manageable when they have finite Gevand- Kirillov dimension (46.2); in particular, affine PI-algebras have finite Gelfand-Kirillov dimension and are a very successful arena for generalizing results from the commutative theory. Returning to the Wedderburn- Artin theorem, a very natural question is, “What can be said about the division ring D in M,(D)?” This remains one of the more troublesome questions of research today, because tantalizingly little is known about arbitrary division rings. When D is finite dimensional over a field however, a whole world opens up, comprising the theory of (finite dimensional) central simple algebras and the Brauer group (Chapter 7). Much recent research on division algebras is from the standpoint of noncommutative arithmetic and K-theory, but we deal mostly with the general structure theory of division algebras. Most of the exposition is given to understanding the algebraic content of several special cases of the amazing MerkurjevSuslin theorem. Our final chapter (8) is about those rings which arise most in representation theory, namely, group rings and enveloping algebras of Lie algebras. These rings in fact contain all the information of the respective representation theories, and therefore provide a key link from “pure” ring theory to the
Introduction
xxi
outside world. Both areas naturally lie on the border of algebra, drawing also on analysis and geometry. Nevertheless, the ring theory provides a guide to directions of inquiry; the question of central interest in enveloping algebra theory has been to determine the primitive ideals, for these correspond to the irreducible representations. Pure ring theory also yields a surprising amount of information, and much of the theory can be cast in the general framework of Noetherian rings (58.4).(The structural foundation is laid in 92.12.) Another topic in modern research is the Galois theory of rings. Jacobson developed a Galois theory to study extensions of division rings, in a similar manner to the Galois theory of extensions of fields. This has led to the study of fixed subrings under groups of automorphisms (end of 52.5), and more recently to Hopf algebras, which are treated in $8.4 as a simultaneous generalization of group rings, enveloping algebras, and algebraic groups. A word about current research-whereas the 1960s and early 1970s was the era of abstraction and beautiful general theories, the late 1970s and 1980s have displayed a decided return to specific examples. Thus considerable recent attention has turned to Weyl algebras and, more generally, rings of differential polynomials, and the theory of the enveloping algebra of sI(2, n) has produced many interesting examples and a few surprises.
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Table of Principal Notation Note: ! after page reference means the symbol is used differently in another part of the text.
1 l! 2 3 3 3 4 4 5 5 6 6 6 14 14 14 16 16 16 17 18
18! 19 19 22 29 29 30 32 43 43 44 44 48 51 51 53! 55 55 56 59 66
xxiii
Table of Principal Notation
67 67 70 73! 74 77 82 93 93 97 98 100 101! 103 104 112 112 115 117! 119 121 121 123 124 129! 129! 134
273 280 280 292 292 296 297 299 300 306 349 355 356 378 379! 388! 393 396 402 409 409!
41 1 41 5 42 1 424 425 426 421 429 442 47 1 471! 478 477! 49 1
0
General Fundamentals
50 Preliminary Foundations The object of this section is to review basic material, including the fundamental results from ring theory. The proofs are sketched, since they can be found in the standard texts on abstract algebra, such as Jacobson [SSB]. Two important conventions: N denotes the natural numbers including 0; also, given a function f:A + B we shall often use fa instead of the more standard notation f(a), and similarly fA denotes {fa:a E A}. Given sets A, I define A' = {functions f:I -,A}, which can be identified with the Cartesian product n { A :i E I } . A special case is I = { 1,. . .,n } ; in this case we denote A' as A("), which can be identified with A x x A (taken n times) under the bijection sendingf to ( f l , ...,fn ).
Monoids and Groups A monoid is a semigroup S which has a unit element 1, such that 1s = sl = s for each s in S. If a semigroup S lacks a unit element we can adjoin a formal element 1 to produce a monoid S' = S u { l } by stipulating 1s = sl = s for all s in S. In verifying that a monoid S is a group, one need only check that each element is left invertible, i.e., for each s there is s' such that s's = 1. (Indeed,
1
General Fundamentals
2
take a left inverse s" of s'. Then s" = ~"(s's)= (s"s')s = s proving ss' = 1 and s' is also the right inverse of s, i.e., s' = s-'.) Sym(n) denotes the group of permutations, i.e., of 1: 1 functions from { 1,. ..,n} to itself; Sym(n)is a submonoid of A A where A = { 1,. . .,n } . Although we have used the multiplicative notation for groups, we shall usually write abelian groups additively, with + for the group operation, 0 for the unit element, and - g for the inverse of g . Fundamental Theorem of Abelian groups: Eoery finitely generated abelian group is isomorphic to a finite direct sum of cyclic subgroups.
An element of a group is torsion if it has finite order. The set of torsion elements of an abelian group G is obviously a subgroup, and we say G is torsion-free if its torsion subgroup is 0. Since any cyclic group is isomorphic either to (Z, +) or some (Z/nZ, +) for some n in Z, we have Corollary: Eoery finitely generated abelian group is the direct sum of a torsion-fiee abelian group and a j n i t e abelian group.
Rings and Modules A ring is a set R together with operations +, (called addition and multiplication) and distinguished elements 0 and 1, which satisfy the following properties: (R, +, 0) is an abelian group. (R, .,1) is a monoid. a(b + c) = ab + acand (b + c)a = ba
+ cafor all a,b,cin R.
If ab = ba for all a, b in R, we say R is commutatiue, common examples being Z, Q, R, and C, as well as rings of polynomials over these rings; matrix rings (cf., $1.1) provide examples of noncommutative rings. The singleton (0) is a ring called the trioial ring. In all other rings 1 # 0 (since if 1 = 0 then any r = r 1 = r 0 = 0). A subring of a ring is a subset which is itself a ring having the same distinguished elements 0, 1; thus (0) is not a subring of any nontrivial ring. Unless explicitly stated otherwise, rings will be assumed to be nontrivial. Suppose R, T are rings. A ring homomorphism f: R + T is an additive group homomorphism satisfying f(rlr2) = (frl)(fr2) and f l = 1. By domain we mean a (nontrivial) ring in which each product of nonzero elements is nonzero; R is a division ring (or skew-field) if (R - {O},., 1) is a group, i.e., each nonzero element is invertible. A commutative domain is
.
3
$0 Preliminary Foundations
usually called an integral domain (or entire); a commutative division ring is called a field. Given a ring R we define a ( l e f )R-module to be an abelian group M (written additively), together with composition R x M + M called scalar multiplication, satisfying the following laws for all ri in R and xiin M: r(xl
+ x 2 ) = rxl + rx,;
( r l r 2 ) x= r l ( r 2 x ) ;
(rl
+ r 2 ) x = r l x + r,x;
l x = x.
In other words, every possible associative law and distributive law involving scalar multiplication holds. A right R-module M is an abelian group with scalar multiplication M x R + R satisfying the right-handed version of these laws. The motivating examples: (i) R is a left (or right) R-module, where the addition and scalar multiplication are taken from the given ring operations of R. (ii) The modules over a field F are precisely the vector spaces over F. Suppose M, N are R-modules. A module homomorphism f: M + N , also called a map, is a group homomorphism “preserving” scalar multiplication in the sense f ( r x ) = rfx for all r in R and x in M. In analogy to rings, a map is an isomorphism if it has an inverse which is also a map; clearly, this is the case iff the map is bijective. However, maps have additional properties. Write Hom,(M, N) for {maps from M to N}, made into an Abelian group under “pointwise” addition of maps, i.e., (f + g)x = f x + gx. The zero map sends every element of M to 0. Moreover, composition of functions provides a map from Hom(N, K ) x Hom(M, N) to Hom(M, K), which is bilinear in the sense (91 + 92)fl = S l f i
+92f2
and
Sl(f1
+ fi) = S l f l + 9 2 f 2
for any maps h: M + N and g i : N + K. A submodule of an R-module M is an additive subgroup N closed under the given scalar multiplication; in this case N is itself a module and we write N < M. A submodule N is proper if N < M. Viewing R as R-module as above, we say L is a lefi ideal of R if L I R; in other words, a left ideal is an additive subgroup L satisfying rx E L for all r in R and x in L. Note that for any x in a left R-module M we have Rx IM. The case M = R has special interest, because a left ideal L is proper iff 1 # L. In particular, for r E R we have Rr c R iff r has no left inverse. Right ideals analogously are defined as right submodules of R. A is a proper ideal of R (written A 4 R) if A is a proper left and right ideal of R. If f is a ring homomorphism or a module homomorphism, we define its kernel kerf to be the preimage of 0. Then rudimentary group theory shows kerf = 0 iff f is 1: 1. Consequently, f is an isomorphism iff f is onto with
General Fundamentals
4
kerf = 0. Note for a map f : M N that kerf is a submodule of M and is proper iff f is nonzero; on the other hand, i f f : R T is a ring homomorphism then kerf 4 R. Let us try now to characterize kernels structurally. First take R-modules N I M . Forming the abelian group M / N in the usual way, as {cosets of N } , we can define scalar multiplication by --f
--f
r(x + N ) = rx
+N
for r E R and x E M ,
thereby making M / N a left R-module called the quotient module (or residue module or factor module). There is a canonical map cp: M + M / N given by x + x + N , and N = ker cp. In this way we see every submodule of M is the kernel of a suitable map. Moreover, we have Noether's isomorphism theorems: Proposition 0.0.1: Suppose f: M + M' is a map of R-modules whose kernel contains a submodule N of M . Then there is a map f: M / N + M' given b y f ( x + N ) = f x , with kerf = (kerf )IN. In particular, i f f : M + M' is onto and kerf = N then f is an isomorphism. Proof(Sketch): These facts are standard for abelian groups, so one need merely check scalar multiplication is preserved. Q.E.D. Corollary 0.0.2: If M , , M , I M then MI and ( M , M , ) / M , x M l / ( M l n M,).
+
+ M, I M
and M , n M , I M ,
Proofi Define f : M , + ( M , + M , ) / M , by f x = x + M a . Clearly f is onto and kerf = {x E M , : x + M , = 0 } = M I n M , so apply the proposition. Q.E.D. Corollary 0.0.3: If K IN I M then ( M / K ) / ( N / K )x M I N .
Proofi The canonical map M + M / N yields an onto map M / K + M / N Q.E.D. with kernel N I K , so apply the proposition.
One also obtains similar results for rings. Given 14 R we define the quotient ring (also called factor ring or residue ring) R/I to have the usual additive group structure (of cosets) together with multiplication (r,
+ I)(r, + I) = r1r2 + 1.
This can be easily verified to be a ring, and there is a canonical ring homomorphism rp: R + R / I given by cpr = r + I .
$0 Preliminary Foundations
5
Proposition 0.0.4: Suppose f: R + T is a ring homomorphism whose kernel contains an ideal A. There is a ring homomorphism f: RIA + T given by f ( r + A) = f r , and kerf = (kerf )/A. If f is onto and kerf = A then is an isomorphism.
7
Corollary 0.0.5: If B E A are proper ideals of R then (A/B) 4 R/B and (R/B)/(AIB) !z R/A. If f:M + N is a map of R-modules then fM is a submodule of N; thus f is onto iff N / f M = 0, providing a useful test. For rings one sees for any homomorphism f:R + T that f R is a subring of T. Since fR is not an ideal of T we do not have the parallel test for rings, but anyway we shall find it useful at times to replace T by fR, thereby making f onto.
Algebras In the text C usually denotes a commutative ring. A C-algebra (or algebra over C) is a ringR which is also a C-module whose scalar multiplication satisfies the extra property c ( r l r 2 )= (crl)r2= r1(cr2)
for all c in C, and r l , r2 in R.
Any ring R is also a Z-algebra, by taking nr to be r + . * . + r (taken n times); the formal correspondence is given more formally in example 0.1.10 below. In general the theories of algebras and of rings are very similar. Indeed, if R is a C-algebra and A 4R is a ring then A .c R as C-module (since ca = c( l a ) = ( c l ) a E A for all c in C and a in A). Thus the ring RIA also has a natural C-module structure, with respect to which RIA is in fact a C-algebra. Put more succinctly, any ring homomorphic image of R is also naturally a C-algebra. Define the center of a ring, denoted Z(R), to be (z E R: rz = zr for all r in R}, clearly a subring of R. Under its ring operations R is an algebra over every subring of Z(R). Conversely if R is a C-algebra then there is a canonical ring homomorphism cp: C + Z(R) given by cpc = c l . (Proof c l E Z(R) since (c1)r = cr = c(r1) = r(c1); cp is a ring homomorphism because ( c l c , ) l = c l ( c 21 ) = c l (l(c21 ) = (cll)(c, l ) . ) In case C is a field we have ker cp 4 C so ker cp = 0, and we may identify C with a subring of R. Often it is easier to prove theorems about algebras over a field, which one then tries to generalize to algebras over arbitrary commutative rings. At times we shall need the following generalization of the center of a ring R. Suppose A c R. The centralizer of A in R, denoted C,(A), is
6
General Fundamentals
{ r E R:ra = ar for all a in A } , a subring of R. We say B centralizes A if B c CR(A).For example C,(R) = Z(R). Note that A G CR(CR(A)).
Proposition 0.0.6: Any maximal commutative subring C of R is its own centralizer. Proofi Let T be the centralizer of C in R. Then C G T. But for any a in T we see C and a generate a commutative subring C' of R; by maximality of C we Q.E.D. have C' = C so a E C ; hence T = C, as desired.
Preorders and Posets A preorder is a relation which is reflexive ( a I a ) and transitive (if a I b and b Ic then a I c). A preorder I on S is called a partial order (or PO for short) if Iis antisymmetric;i.e., a I b and b I a imply a = b. In this case (S, I)is called a poset. We write a < b when a I b with a # b. The following posets are of particular importance to us.
(i) Every set S has the trioial (or discrete) PO defined by declaring any two distinct elements are incomparable (Lea,a I b iff a = b). (ii) The power set B(A)of a set A is the set of subsets of A, ordered by set inclusion. (iii) If M is an R-module, define Y(M)= {submodules of M},partially ordered under I (so N , I N , if N1 is submodule of N,). When R is ambiguous we write 9 ( R M ) for Y(M). One can do the same for {ideals of a ring}.
Upper and Lower Bounds Suppose S is a set with preorder I.An upper bound for a subset S' of S is an element s in S for which s' I s for all s' in S'. Upper bounds need not exist. For example, no pair of distinct elements has an upper bound if I is the trivial PO. On the other hand, we call S directed (by I)if every pair of elements has an upper bound. An upper bound s of S' is called a supremum if s Is' for every upper bound s' of S'. Dually, we define lower bound and say S is directed from below if every pair of elements has a lower bound. A lower bound s of S' is called an injrnum if s 2 s' for every lower bound s' of S'. The supremum (resp. infimum) is denoted as v (resp. A ). The supremum and infimum of S (if they exist) are denoted, respectively, as 1 and 0. For any poset (S, I) we can define the dual poset (S, 2 )by reversing the inequality, i.e., now s1 2 s2 if previously s1 I s,. The technique of passing to
$0 Preliminary Foundations
7
the dual poset is extremely useful and often produces extra theorems with no extra work.
Lattices A lattice is a poset in which every pair of elements has both a supremum and an infimum; the lattice is complete if every subset has both a supremum and an infimum. Passing to the dual reverses v and A , so we see the dual of a (complete lattice is also a (complete) lattice. The following posets are in fact complete lattices:
(i) The power set Y(A),where v is set-theoretic union and A is intersection. (ii) 9 ( M )for an R-module M; here Mi = M iand A M i= M i . In fact, this important example is the raison d'itre for the study of lattice theory by ring theorists, and many purely lattice-theoretic results will have applications throughout the text (cf., exercises 3- 13).
n
v
This language enables us to state certain module-theoretic results more sharply. To see this we need a useful general result. Given posets A , B we say a function f:A -+ B is order-preseruing if a , < a, implies fa, < fa, for all a1,a2in A. If A, B are lattices, we say a function f:A -+ B is a lattice homomorphism if f(a, v a,) = fa, v fa, and f ( a , A a,) = fa, A fa, for all a,, a, in A. Every lattice homomorphism f:A -+ B is order-preserving, since if a, I a, we have fa, = f(a, v a,) = fa, v fa,. Proposition 0.0.7. Suppose f:A -+ B is a lattice bijection. Then f is a lattice isomorphism iff f and f are order-preserving.
-'
Proof: (a) as observed above. ( G )For any a1,a2in A we have fa, I f ( a l v a z ) and fa, I f ( a l v a,) so fa, v fa, I f(al v a,). Applying this argument to f - ' yields a, v a, =
f-'fa,
v
f-lfu,
I f-'(fa,
v fa,) I f-'f(a, v a,) = a, v a,
so equality holds at each stage. The proof for
A
is analogous.
Q.E.D.
We can apply this result to the lattice 9 ( M ) . Proposition 0.0.8: Iff: M N i s onto then there is a lattice isomorphismfrom 9 ( N ) to {submodules of M containing kerf}, given by N ' -+f-'N'. (The inverse correspondence is given by M ' -+ fM'). -+
Proof: Group theory yields us a 1: 1 correspondence from {subgroups of N} to {subgroups of M containing kerf}, and we wish to restrict this
8
General Fundamentals
to submodules. If N‘ IN then f -IN’ I M since f (rx) = r f x E N’ for all r in R and x in f -IN‘; if M‘ I M then clearly f M ‘ I f M = N. Thus we have an order isomorphism, which by proposition 0.0.7 is a lattice isomorphism. Q.E.D. The corresponding result for rings (proposition 0.0.10) can be obtained directly, but for the sake of variety one could make use of the following observation. Remark 0.0.9: (“Change of rings”) Suppose f:R + T is a ring homomorphism. Any T-module M can be viewed as R-module via the scalar multiplication defining rx to be ( f r ) x for r in R and x in M. Thus every submodule of M in T - d u d can be viewed in R - d u d , providing a lattice morphism from 9 ( , M ) to U ( , M ) . When f is onto this lattice morphism also is onto, since any R-submodule of M is also a T-submodule under the corresponding action. In particular, viewing T naturally as T-module we can identify {left ideals of T} = 9(,T) with 9 ( , T ) .
Proposition 0.0.10: Any onto ring homomorphism f: R + T induces a lattice isomorphism of {left ideals of R containing kerf} with {left ideals of T}; likewise for right ideals. In particular kerf Q R, and f induces a lattice isomorphism of {ideals of R containing kerf} and {ideals of T}; the inverse correspondences each are given by f - I . Proofi Proposition 0.0.8 and remark 0.0.9 yield the following composition of lattice isomorphisms: {R-submodulesof R containing kerf } + LQRT)+ 9(,T). The rest is clear.
Q.E.D.
Modular Lattices Since proposition 0.0.8 concerns lattices (of submodules) one is led to look for a more lattice-theoretic approach to modules, which stems from the following observation: Remark 0.0.11: In any lattice 9we have a v (b, A b,) Ia v b, for i = 1,2, so a v (b, A b,) I (a v b,) A (a v b,). In particular, if a I b we have a v (b A c) I b A (a v c).
9
80 Preliminary Foundations
In certain cases equality actually holds. The most important case is the lattice of submodules Y ( M ) , for if MI I M, and x, E M, n (MI + M 3 ) then x, = x 1 + x3 for x i in Mi so x3 = x, - x1 E M,, proving x, E M, + (M, n M 3 ) .We are led to the following definition: Definition 0.0.12: A lattice 9is modular if a v ( b A c) = b a s bin9.
A
(a v c) for all
9 ( M )is thus a modular lattice. A more symmetric condition in verifying the modularity of a lattice is given in exercise 2. Definition 0.0.13: A lattice 9 is complemented if for each a in Y there is a complement a' with a A a' = 0 and a v a' = 1. The lattice 9 ( S ) is complemented. On the other hand, our other major example of a lattice 9 ( M )is in general not complemented. Complements play an important role in module theory because of exercises 12 and 13, which anticipate key structural results in 42.4. Definition 0.0.14: A jilter of a lattice 9is a subset 9satisfying the following three conditions: (i) If a E F and b 2 a then b E 9 (ii) If a, b E F then a
A
bE9
(iii) 0 .$ 9 Filters have several important applications in rings and can be viewed as the "dual" of ideals, c.f., exercises 23J. It is important at times to have maximal filters; these are called ultrajlters, cf., remark 0.0.16 below. Example 0.0.15: (i) For any a E 9, % = { b E 9:b 2 a } is called the principal jilter generated by a. (ii) Suppose 9 = 9 ( S ) for some infinite set S. Then {complements of finite subsets of S ) is a filter, called the Frechet filter, or cojinite filter.
A direct way of obtaining a filter 9 of a lattice Y is by finding B E 9 satisfying properties (ii) and (iii) of definition 0.0.14, i.e., 0 $ B and if b, b' E a then b A b' 2 b" for some b" in B. Then { a E 9:a 2 b for some b in B } is a and a is called the base of the jilter 9.We shall define many filters filter 9, via their bases.
General Fundamentals
10
Zorn ’s Lemma A poset (S, I) is a chain if for all sl, s, in S we have s1 Is, or s, I s,; in this case Iis called a total order; for example, (Z, I)is a chain under the usual one sometimes finds that I induces a (total) order. Given a lattice (S, I), total order on a certain subset of S. For example, if A, E A, E A, E are subsets of A then {A,: i E N } is a chain in ( B ( A ) ,E). This situation is of great interest because of the following result, often called Zorn’s lemma. We say a poset (S, I)is inductive if every chain S’ in S has an upper bound in S. For example, (N, I) is not inductive since there exist chains not bounded from above, but (N, 2 ) is inductive since 0 is an upper bound. The real interval [0,1] is inductive with respect to I,but Q n is not inductive. An element s of a poset S is maximal if there is no element s‘ > s in S. Minimal elements are defined analogously. (Note that S may have many distinct maximal elements, the most extreme example being when the PO is trivial, in which case every element is both maximal and minimal.
[O,a]
(“Zorn’s lemma”): maximal element.
If (S, I)
is an inductive poset then S has at least one
The key application is as follows: Suppose A is a set and S c @ ( A ) such that for any chain {A,: i E I}in S we have A, E S. Then some subset of A is maximal in S. The most basic application of the maximal principle in ring theory is that every proper ideal of a ring R is contained in a maximal proper ideal, called a maximal ideal. Indeed, an ideal A is proper iff 1 $ A, so {proper ideals of R} is inductive. Likewise any (proper) left ideal is contained in a maximal left ideal. Incidentally this does not hold in general if we do not stipulate the existence of the element 1, which is the main reason we deal with rings with 1, cf., exercise 14. However, a ring need not have minimal nonzero left ideals (for example E), since the above argument has no analogue.
u
Remark 0.0.26:
Any filter is contained in an ultrafilter by the maximal
principle. Using remark 0.0.16 one sees that a filter 9 of a comeither a E 9or a‘ E 9. One plemented lattice is an ultrafilter iff for all a in 9, concludes that if a v b is in an ultrafilter 9 then a E 9 or b E 9? Remark 0.0.27:
Another important application of the maximal principle: {commutative subrings of R containing Z ( R ) } is inductive and thus has maximal members.
11
50 Preliminary Foundations
In other words, any ring R has maximal commutative subrings, and these are often useful in the study of R, in view of proposition 0.0.6. The maximal principle is proved by drawing from set theory and, in fact, is equivalent to the axiom of choice, which asserts that for any family {Si:i E I } of sets there is a suitable “choice” function f:I + Siwith f i E Si for each i, i.e., f “chooses” one element from each Si. At first blush this axiom seems obvious; however, the larger cardinality the index set I, the less credible the axiom becomes. P. J. Cohen proved that the axiom of choice is independent of the Zermelo-Fraenkel axioms of set theory, and today it is used freely by algebraists because the maximal principle is so powerful. To understand the connection we must bring in transfinite induction.
u
Well-Ordered Sets and Transjinite Induction Many definitions in general ring theory rely on transfinite induction. To understand this process requires some intimacy with the ordinals, and to this end we bring in some formalism from set theory. It is natural to build sets from the bottom up, starting with the empty set and then building sets whose elements themselves are sets. Thus we formally define the symbols
Z = i u {i} = {a, {a}}, and so forth. The axiom of regularity states for every set S # 0 there is s E S
0 = @,
i = 0 u (0) = {@},
with s n S = 0. This ensures that given two sets S,, S, we cannot have both S, E S, and S , E S , . (Indeed take S = {Sl,S,}.) Thus we can define an antisymmetric relation < on sets by whenever S, E S, or SI = S,. S, < S , Since the elements of a set S are themselves sets, we can view Ias an antisymmetric relation on the elements of S. S is called an ordinal if (S, I) is a chain. If a is an ordinal then a+ = a u { a } is also clearly an ordinal, called the successor of a. In particular 0 = 0, = 0+, 2 = i+, . . . are all ordinals. O n the other hand, there are ordinals which are not successors, the first of which is {E: n E N}; these are called limit ordinals. A chain is well-ordered if every nonempty subset has a minimal element. Every ordinal a is well-ordered under < as defined above; indeed if @ # S c a then any s E S with s n S = 0 is minimal in S. Thus we have the following generalization of mathematical induction: Principle of transfinite induction: Suppose a is an ordinal and S c a has the property for every ordinal a‘ < a that if {fl:fi < a ’ } G S then a’ E S. Then S = a.
12
General Fundamentals
The proof is rather easy, but the applications are wide-ranging; here are some set-theoretic implications we shall need (cf., exercise 15-20): (i) Every set can be put into 1:l correspondence with a suitable ordinal and thus is well-ordered under the corresponding total ordering. (Thus we shall often describe a set S as {s1,s2,. . .} even when S is uncountable); (ii) Zorn’s lemma, as stated above.
Fields In the structure theory of rings one often considers fields as “trivial” since they have no proper ideals # 0; in fact, most results from ring theory hardly require any knowledge of fields. Nevertheless, fields do play important roles in several key topics (such as division rings), and ideas from field theory provide guidelines for generalization to arbitrary rings. When appropriate we shall assume familiarity with the Galois theory of finite dimensional field extensions, including normal and separable extensions, and the algebraic closure of a field.
First Order Logic We shall need some tools from formal logic, in order to deal with certain important constructions, most notably reduced products and ultraproducts. The reader willing to accept these results (1.4.1lff) might well skip this account. On the other hand, some rigor is sacrificed. A j r s t order language 9 consists of constants (or object symbols) which are “designated elements” in the theory (such as 0, 1 and usually denoted as a, b,c,. ..) variables (or dummy symbols) denoting “variables” or “indeterminates” (such as x, y, z ) relative symbols of order n for n = 0, 1,. . . function symbols of order n for n = 0, 1,. . . connectives which are the symbols -I, A , v ,denoting, respectively, “not,” “and,” “or (inclusively),” quantijiers which are the symbols V (denoting “for all”) and 3 (denoting “there exists”)
A term is an expression defined inductively as follows: Every constant or variable is a term, and f ( t l , . . ., t,,) is a term for all terms t l , . ..,t,, and every n-ary function symbol f. An atomic formula is an expression without connectives.
$0.1 Categories of Rings and Modules
13
Formulas and the free oariables of a formula cp (denoted FV(cp)) are defined inductively in terms of rank. If cp is atomic then we say rank(cp)= 0 and FV(cp) is the set of variables used in writing cp. Inductively, if cpi are formulas of rank n, then is a formula of rank n, + 1 whose free variables a;e FV(cp,). cp,), (cp, v cp,), each is a formula of rank n, + n, + 1 whose free variables are FV(cp,) u FV(cp,). (Vx)cp, and (3x)cpl are formulas of rank n, + 1 for any x in FV(cp,), whose free variables are FV(cp,) - {x}. i cp,
(cp,
A
A jirst-order (or elementary) sentence of 8 is a formula without any free variables. A structure Y is a set S together with an assignment of the symbols in 9 to their interpretations in S. Constants are assigned to designated elements of S, relative symbols are assigned to relations on S, n-ary function symbols are assigned to functions S(”) --* S, and variables denote arbitrary elements of S. Let us define when a sentence cp “holds” in 9 Suppose S is the underlying set.
(i) The atomic sentence ( t , = t , ) holds in Y iff the terms t , and t , have the same assignment in S. The atomic sentence R ( t , , . . .,t,) holds in S iff the relation R is satisfied by the assignments of t , , ...,t,. We proceed inductively on rank, having just handled the rank 0 case. (ii) i cp holds (in 9) iff cp does not hold (in Y). (iii) cpl A 41, holds iff cp,, cp, both hold cpl v cp, holds iff cp, and/or cp2 hold (iv) (Vx)cp(x) holds iff cp(s) holds for all elements s in S; (3x)cp(x) holds iff cp(s) holds for some element s in S. Here ~ ( s denotes ) the formula obtained by replacing each occurrence of x by s and is easily seen to be a sentence.
Certain redundancies in the language 8 can be discarded when proving general assertions about the lower predicate calculus. In particular, the quantifier 3 and the connective v are superfluous, as the reader can readily check.
80.1 Categories of Rings and Modules The language of categories is useful, particularly in certain aspects of module theory. We presuppose a nodding acquaintance with this language; Jacobson [80B, Chapter 13 more than suffices for this purpose. In particular, the reader should know the definition of category, subcategory, (covariant)
14
General Fundamentals
functor, contravariant functor, natural transformation (of functors), and natural isomorphism of functors. Given a category W we write Hom,(A, B), or merely Hom(A, B), for the set of morphisms; 1,: A + A denotes the unit morphism. Our interest in categories will focus on R-dud, where R is a given ring; its objects are the R-modules, and its morphisms are the maps of R-modules. , objects are rings and whose morAnother category of note is W i ~ ywhose phisms are ring homomorphisms. R-Aud and W i ~ each y are subcategories of d’e,the category of abelian groups (written in additive notation); d’eis a subcategory of 92,the category of sets. Often the class of objects of a category is “too large” to be a set, for one can construct distinct objects for each ordinal. Such is the case with Wimg and R - A d ,cf., exercise 1.4.1. Accordingly, a category is called small when its class of objects is a set. The next example is quite useful. Example 0.1.1: Any set 1 with preorder I can be made into a small category whose objects are elements of 1, with Hom(i,j) either a singleton or 0 depending on whether or not i I j. Write E: for the morphism from i to j if i 5 j. Then E : = l i and E: = E ~ E ;for all i, j, k in I. Conversely, any small category in which JHom(A,B))I 1 for all objects A, B can be given a preorder (as a set) and can be described as above. If 1 is an arbitrary set then it can be given the trivial preorder, in which case the corresponding small category has Hom(i,j) empty unless i = j, in which case Hom(i, i) = l i . This trivial example is useful in understanding certain constructions of Chapter 1 (e.g., the coproduct). Definition 0.1.2: Given a category V we define the dual category WoP by Ob Vop = Ob W and Hornrpop(A, B) = Hom,(B, A) with composition in Cop given by g f = fg (where f E Homro,(A, B) and g E Homro,(B, C)). In other words, we reverse arrows and write things backwards. (In particular (VO”)”P = W.)
Any general theorem for all categories a fortiori holds for the dual categories; translating back to the original category yields a new theorem, the dual theorem obtained by switching all arrows. The main problem with this approach is that few theorems hold for all categories, and the dual of a wellknown category may be quite bizarre. Nevertheless, there are certain important examples for which the dual is well-known and useful. For example, the dual of example 0.1.1 corresponds to the reverse preorder on S, justifying the customary use of “dual” concerning posets and lattices.
50.1 Categories of Rings and Modules
15
Monics and Epics Recall a morphism f: A + B is monk if f g # f h for any g # h in Hom(C, A), for all objects C; dually f is epic if gf # hf for any g # h in Hom(B, C). Clearly the composition of monics (resp. epics) is monic (resp. epic). In any subcategory of Yet! each 1:l morphism is monic, and each onto morphism is epic; we would like to test the converse for R - A d and Wiltg. Proposition 0.1.3:
In R - A d , monics are 1: 1, and epics are onto.
Pro08 Suppose we are given f: M +N. If f is monic define g, h: kerf --+ M by taking g to be the identity and h = 0; then f g = f h = 0 implying g = h, so kerf = 0 and f is 1:l. If f is epic then define g, h: N + N / f M by g = 0 and hy = y + f M ;then gf = hf = 0, implying g = h so N / f M = 0, i.e., f M = N. Q.E.D. The story ends differently for W i ~ g . Example 0.1.4: A 1:l ring homomorphism which is epic and monic, but not onto. Consider the 1 :1 ring homomorphism f:Z + Q given by f n = n. For any morphism g : Q + R in W i ~ we g have g(mn-’) = (gm)(gn)-’ for all rn, n # 0 in Z, implying g is determined by its restriction to Z; it follows at once that if g # h then g f # hf, so f is epic. In Exercise 1 we see that all monics in Wilt9 are 1:l. Nevertheless, example 0.1.4could be called the tragedy of Wiltg. Perhaps rings categorically should be viewed in terms of the following example. Example 0.1.5: Suppose R is a ring. We form a category with only one object, denoted A, and formally define Hom(A,A) = R where the composition of morphisms is merely the ring multiplication. To differentiate the approach to rings and to modules, we designate ring homomorphisms as “homomorphisms;” a 1 :1 (ring) homomorphism is called an injection, and an onto homomorphism is a surjection. For modules we adopt, respectively, the more categorical terminology of map, monic, and epic. Nevertheless, since an ideal of R is merely a left and right submodule, we would like to introduce another category.
Definition 0.1.6: Suppose R, R’ are rings. An R-R’ bimodule is a left Rmodule M which is also a right R’-module satisfying the associativity
General Fundnmentals
16
condition (rx)r’ = r(xr‘)for all r in R, x in Myand r’ in R’. R-Mud’-R‘ is the category whose objects are R-R‘ bimodules and whose morphisms f: M + M’ are maps both in R-Modand d u d - R ‘ .
Remark 0.1.7: The R-R sub-bimodules of R are precisely the ideals of R.
Functors Certain functors arise continually in the study of R-Mod.Perhaps the most important are the “Hom” functors:
Example 0.1.8: (i) Hom(A,,-): % + Y e t is the functor sending an object A to Hom(A,,A) and sending a morphism f: A + B to the function f#: Hom(A,, A) + Hom(A,, B), defined by f# h = f h for each h in Hom(A,, A). Note Hom(A,,-) is covariant because for all h in Hom(A,,A) and each f:A +Band g:B + C we have
(Sf)#h = gfh = 9#f#h,
so (Sf)# = 9 # f #
(ii) Horn(-, A,): %‘ + Y t is the contraoariant functor sending A to Hom(A, A,) and sending f to the function f : Hom(B, A,) + Hom(A, A,) defined by f ’ h = hf for each h in Horn@, A,).
”
Other functors we need are the identity functor l,, and the “forgetful functors.” A subcategory W of 9 is full if Hom,(A,B) = Hom,(A,B) for any objects A, B in V. Given a full subcategory V of 9 we say the inclusion functor F: W + 9 has a retraction G:.9 + %fif G F = 1,.
Example 0.1.9: (i) .9 = at& and %‘ is the full subcategory of torsion-free abelian groups. Given A E d&define tor(A) = {torsion elements of A}. Then the functor G: .9-+ V given by GA = A/tor(A) is a retraction. This example has been abstracted to the torsion theory of modules, c.f., 53.3 and 53.4. (ii) Given a ring surjection f: R + T, remark 0.0.9 yields a functor F: T-Mud --* R-Mod. If M E T-Mud then viewing M as R-module we have (kerf)M = 0. On the other hand, if M E R-Mud and (kerf)M = 0 then we can reverse the procedure and view M as T-module by putting ( f r ) x to be rx. In this way F yields an isomorphism from T-Mud to the full subcategory W of R-Aod consisting of those modules M such that (ker j ) M = 0. There is a retraction G: R - A d + V given by GM = M/(ker f )M.
$0.1 Categories of Rings and Modules
17
Two categories V, 9 are isomorphic if there exist functors F : V + 9 and G :9 --* V with G F = 1% and F G = l a . This definition is very stringent, but is useful in identifying a pair of theories. Example 0.1.10: d&and Z - A u d are isomorphic categories. (Indeed, let F: Z - A u d + d&be the forgetful functor, and define G: d&+ Z - A d as follows: Given M E d& we view M as Z-module by introducing scalar multiplication nx = x + . * + x , the sum taken n times for n E N, and (- n)x = -(nx). Every group homomorphism then becomes a morphism in Z - A d , so we have the inverse morphism to F.) To identify categories of left and right modules, we need the notion of the opposite ring RoP. Definition 0.1.11: If R is a ring, RoPis the ring obtained by keeping the same additive structure but reversing the order of multiplication, (i.e., the product of rl and r2 in RoPis r2r1). RoPyields the dual category of the category obtained from R in example 0.1.5, thereby justifying the notation. Note (R"P)"P= R. Proposition 0.1.12:
R-Aod and Aud-RoPare isomorphic categories.
Proofi Given M in R-Aod, we define the scalar product of x E M and r E RoPto be rx; reversing the order of all products, we see this new scalar multiplication makes M a right R"P-module. Any morphism f: M , + M 2 in R - A d can be viewed at once as a morphism in Aod-RoP,so we thus have a functor R-Aud+.Mud-RoP. We have an analogous functor Aud-R"P+ (RoP)"P-Aod= R-Mud, and the composition of these two functors is the identity. Q.E.D.
Thus any general theorem about modules is equivalent to a corresponding theorem about right modules. Usually we want to weaken the notion of isomorphism of categories to equioalent categories, c.f., Jacobson [80B, p. 271; we shall see in Chapter 4 that categorical equivalence is a fundamental tool of module theory. We shall also need the following construction from time to time. Example 0.1.13: Suppose I is a small category and V is a category. Then there is a category V' whose objects are the functors from I to %? and whose
General Fundamentals
18
morphisms are the natural transformations, i.e., Hom(F, G) = {natural transformations from the functor F to the functor G}. The composition cq of two natural transformations q: F + G and G H is given by (cn), = 4‘,qA E Hom(FA, HA) for each A in Ob %?.
r:
--f
90.2 Finitely Generated Modules, Simple Modules, and Noetherian and Artinian Modules Returning to modules, we approach one of the nerve centers of the subject and look at the generation of modules by elements. Finitely generated (fag.)modules, also frequently called “finite” and “of finite type,” turn out to be much more tractable than arbitrary modules. In particular, we shall examine cyclic modules, leading us to simple modules. At the end we introduce the important classes of Artinian ’modules and Noetherian modules.
Finitely Generated Modules Given M in R d o d and AGR, SGM, define A S = ~ : , , a , s , : t ~ Na+A, , si E S}; if M E A u d - R we define SA analogously. For S = {s,: i E I } we often write Asi instead of AS. Usually A will be an additive subgroup of R, in which case AS is a subgroup of M for any set S; in fact As = {as:a E A}.
Proposition 0.2.1: Suppose SG M and M E R - A u ~If. L I R then L S I M. RS is the intersection of all submodules of M which contain S. If M E RMod-R’ then RSR’ is the intersection of all sub-bimodules of M containing S. Proofi LS is an additive subgroup of M, and for any r in R, r(LS) = (rL)S E LS proving LS 5 M. In particular RS IM. Now for any N I M with S G N we have RS E RN E N; since RS itself is a submodule containing each element s = 1s of S we get the second assertion. The last assertion is proved similarly. Q.E.D.
We say a module M is spanned by the subset S if M = RS. A module spanned by a finite set is called Jinitely generated, abbreviated as “f.g.” throughout the text. Definition 0.2.2: R - F ~ M uis~ the full subcategory of R - A d whose objects are the f.g. R-modules. Remark 0.2.3: If f: M + N is a map of modules and M is spanned by S then fM is spanned by fS (for if x = E l l s i then f x = c r i (fsi)). In par-
90.2 Finitely Generated, Simple, Noetherinn and Artininn Modules
19
ticular, if M is f.g. then f M is f.g. On the other hand, a submodule of an f.g. module need not be f.g. as seen in the next example. Example 0.2.4: Let Q [ A ] be the ring of polynomials in one commuting indeterminate over Q and let I = 3,Q[3,] and R = Z I c A. Then Id R but 1 4 R-Firnod. (Indeed, given any x l , . . .,x, in I , let the coefficient of 3, in xi be mi/ni and observe (2n, --*n,)-'3,E I Rxi.)
+
Cyclic Modules Of particular interest are modules spanned by a single element. Definition 0.2.5:
M is a cyclic R-module if M = Rx for some x in M.
R = R1 is a cyclic R-module, so remark 0.2.3 shows RIL is cyclic for every L < R. Conversely, every cyclic module has this form, as we shall see shortly. Dejnition 0.2.6: If M E R-Mud and S c M , define Ann,S (the left annihilator of S in R) to be {r E R: rs = 0}, a proper left ideal of R. If R is understood, we write Ann S for Ann, S; we also write Ann x for Ann{x}.
Lemma 0.2.7: If M E R-Mod, then for every x in M there is a map f,: R + M given by f,r = rx; kerf, = Annx, implying RIAnnx w Rx 5 M. Proof: Clearly f, is a map, and kerf, = {r E R: rx = 0} = Ann x, so Rx Q.E.D. f x R w R/ker f , = R/Annx.
=
Proposition 0.2.8: M E R-Mod is cyclic iff M w R / L for some left ideal L of R; in fact, for M = Rx then we can take L = Annx. Proof: As noted above, RIL is cyclic; the converse is lemma 0.2.7.
Q.E.D.
Simple Rings and Modules We shall turn now to a basic philosophy concerning arbitrary categories. One should like to examine objects by taking morphisms to objects whose structure we already know. Then the simplest objects would be those objects from which all morphisms are monic, motivating the following definition. Definition 0.2.9: A nonzero module M is simple if M has no proper nonzero submodules; a ringR is simple if R has no proper nonzero ideals. (Simple modules are called irreducible in the older literature.)
General Fundamentals
20
Remark 0.2.10: In WiBg or in R - ~ ? u d ,an object A is simple iff every nonkerf # A so kerf = 0. (e) Any zero morphism f: A + B is 1: 1. (Proof: (a) morphism A + A/Z is 1:1 so I = 0.)
In particular, we have the following categorical criterion for a module M to be simple: Every nonzero map from M is monic. An analogous criterion holds for W i ~ inf view of exercise 0.1.1, but the dual criterion only works for modules. Remark 0.2.11: An R-module N is simple iff every nonzero map f: M +N is onto. (Proof: (a) 0 # jM < N implies f M = N; (e) if 0 # N' < N then the injection N' + N is not onto.)
There are two immediate difficulties in trying to build a structure theory based on simple rings and/or modules: (i) There must be enough simples to yield general information about rings and modules. (ii) One needs some technique to study the simples. The first difficulty can be dealt with by means of maximal left ideals. Remark 0.2.12: (i) If L is a maximal left ideal of R then R/L is a simple R-module. (Immediate from the lattice correspondences pertaining to R as R-module.) (ii) If I is a maximal ideal of R then R/I is a simple ring.
Surprisingly this remark provides all the simple modules.
Lemma 0.2.13: If M is a simple R-module then M is cyclic. In fact Rx = M for eoery x # 0 in M. Proof: 0 # Rx I M so Rx = M. Proposition 0.2.14: mal left ideal of R.
Q.E.D.
M E R-Aud is simple iff M x R/L for a suitable maxi-
Proof: (=.) Write M = Rx and define cp: R + M by cpr = rx. Then cp is onto so M x Rlkercp, and kercp is a maximal submodule by proposiQ.E.D. tion 0.0.8 (.=) Reverse the argument
Before putting this idea aside, we note an important generalization.
50.2 Finitely Generated, Simple, Noetherian and Artinian Modules
21
Proposition 0.2.15: If M E R-.Fimod and N < M then N is contained in some maximal submodule M’. Consequently M J M ’ is a simple module in which the image of N is 0.
=I:=]
Proof;. Write M R x i , where all x i € M . For any chain M , IM , I... of proper submodules, some xi # Mj for all j (since otherwise all xi E Mj for large enough j, implying M = z R x i E Mj contrary to Mi proper). Thus xi 4 M j , which hence must be proper (and is clearly a submodule). Taking M I = N , we have proved {proper submodules of M which contain N } is inductive and thus by Zorn’s lemma contains maximal members, which must be maximal submodules of M. The rest is clear. Q.E.D.
u
It is difficult to study simple rings without imposing further restrictions because of the intrinsic complexity of ideals. Indeed, the smallest ideal of R containing a given element r is RrR = { = ril rri2: t E N, ril, ri2 E R } which cannot be described in the first-order theory of rings. On the other hand, the smallest left ideal containing r is Rr = {r’r:r’ E R } , which is much more amenable. When r is in the center then in fact RrR = Rr, and the situation is much easier to handle. In general, the classification of simple rings is an immense project, far from completion, but there is the following easy result concerning rings without proper left ideals.
c:
Proposition 0.2.16: (i) Rr = R iff r has a left inverse in R ; (ii) R is a division ring iff R has no proper nonzero left ideals; (iii) if R is simple then Z ( R ) is a Jield.
Proof;. (i) Rr = R iff 1 E Rr iff 1 = r’r for some r’ in R. (ii) (3) follows at once from (i). Conversely if 0 # r E R then Rr = R proving every element of R-{0} has a left inverse, so R-(0) is a multiplicative group. (iii) Rz = R for any z # 0 in Z ( R ) , implying z-l exists and is clearly in Z(R). Q.E.D.
In particular, simple commutative rings are fields.
Chain Conditions A poset ( S , I) satisfies the maximum (resp. minimum) condition if every nonempty subset has a maximal (resp. minimal) element. (S, I) satisfies the
General Fundamentals
22
ascending chain condition (abbreviated ACC) if there is no infinite chain s1 < s2 c s3 < *.., i.e., if every ascending chain is finite; dually (S, I) satisfies DCC if every descending chain is finite.
Proposition 0.2.17: A poset (S, I )satisfies the maximum condition iff it satisfies the minimum condition ifl it satisfies DCC. satisfies ACC. (S, I) Proofi We prove the first assertion; the second is its dual and follows by passing to the dual poset. First note that an ascending chain is finite iff it contains a maximal element. So if (S, I) satisfies the maximum condition then every ascending chain is finite, proving (S, I) satisfies ACC. Conversely, if (S, I) satisfies ACC then every subset s' is inductive (because every chain of S' is finite) and thus has a maximal element. Q.E.D.
Thus to verify a chain is well-ordered, we need only check that there is no infinite descending subchain. Also we see that if a lattice satisfies DCC then every subset is well-ordered.
Noetherian and Artinian Modules The point of studying chain conditions on lattices is in utilizing the lattice Y ( M )of submodules. Defiition 0.2.18: A module M is Noetherian if Y ( M ) satisfies ACC or, equivalently, if Y ( M ) satisfies the maximum condition. M is Artinian if Y ( M )satisfies DCC or equivalently the minimum condition.
Proposition 0.2.19: Suppose M E R-Aud and N I M. M is Noetherian iff N and M I N are Noetherian. M is Artinian iff N and M / N are Artinian. Proofi This is really a fact about ACC and DCC in modular lattices having 0 and 1, i.e., Y ( M ) satisfies ACC (resp. DCC) iff Y ( N ) and Y ( M / N ) satisfy ACC (resp. DCC). (*) ACC and DCC certainly pass to sublattices. But Y ( N )I Y ( M )in the obvious way, and Y ( M / N ) is isomorphic to the sublattice {submodules of M containing N ) by the correspondence M ' / N H M'. (.=) We do ACC; DCC is analogous. Suppose M I 5 M , I in M. Then we have M,nNIM,nNI...
and
(MI
+ N ) / N I( M , + N ) / N I
. . I
23
80.3 Abstract Dependence
in 9 ( N ) and Y ( M / N ) , respectively; so for some m we have Mi n N = Mi+1n N and (Mi + N ) / N = ( M i + , N ) / N for all i 2 m. Thus we need to show that if Mi n N = Mi+ n N and (Mi + N ) / N = (Mi+ + N ) / N then M i = M i + l . Suppose X E M ~ +Then ~ . X E M , + ~ + N = M , + N ;writing x = y i + x ‘ for yi in Mi and X ’ E N we have x ’ = x - y , ~ M ~ + ~ n N = M , n N so x E Mi as desired. Q.E.D.
,
+
,
Corollary 0.2.21: Every f.g. module over a left Noetherian ring R is Artinian) then M is Noetherian).
1:;:
Proofi We prove Noetherian; Artinian is analogous. Let N = 4, which is Noetherian by induction on t. But M / N =(N + N,)/N w N,/(NnN,) Q.E.D. is Noetherian so M is Noetherian by the proposition. We say R is left Noetherian (resp. left Artinian) if R is Noetherian (resp. Artician) as R-module. Corollary 0.2.21: Every f.g. module over a left Noetherian ring R is Noetherian. Every f.g. module over a lejl Artinian ring R is Artinian.
=xi=
Proofi We prove Noetherian. Write M is Noetherian, so M is Noetherian. Q.E.D.
Rx,. Each Rxiw R/Annxi
80.3 Abstract Dependence This discussion is motivated by the observation that modifying the proof in Herstein [64B, Ch.4, corollary 23 shows that all bases of a given vector space over a division ring have the same cardinality.
Definition 0.3.1: A strong dependence relation on a set A is a relation Edcp between elements and subsets of A , satisfying the following axioms for all S,S’
c A:
(i) (ii) (iii) (iv)
If s E S then s E~~~ S. If x Edep S then x Edep S, for some finite S , c S. If x edepS and s Edep S’ for all s in S then x Edep S’. If y Edep S u { x } and y $dep S then x Edep S u { y}.
The fourth axiom, called the (Steinitz) exchange axiom, can be tied in with another idea. Say a set S is independent if s $dep S - {s}for all s in S. Then the exchange axiom implies
General Fundamentals
24
(iv’) If x # d e P S and S is independent then S u { x } is independent. (Indeed, suppose S u {x} is dependent. Then some s Edep (S - {s})u {x} and by hypothesis s #dep S - {s}, so exchanging s and x yields x Edep (S - {s}) u {s} = s. Conditions (i) and (ii) are usually immediate in any candidate for being a strong dependence relation. However, (iii) and (iv) are trickier and may even fail. For example, one customarily says a subset S of a module M is independent if rss = 0 necessarily implies each rs = 0. This definition satisfies (i), (ii) and (iv) but not necessarily (iii). On the other hand, if we instead try to define x Edcp S by x E RS, then (iii) holds but (iv) may fail, e.g., R = M = Z, x = 2, y = 4 and S = 0.)Fortunately, these two definitions coincide when R is a division ring, so we get both (iii) and (iv). Another example of interest to us is {simple submodules of a given Rmodule M}. Then definition 1.3.5 below satisfies (i), (ii), and (iv), cf., argument of remark 1.3.8 below, and (iii) holds because N n 2 Ni# 0 implies N n N, = N (since N is simple) and thus N c N,.
xfinite
Theorem 0.3.2: Suppose 9’ is a set with a strong dependence relation Edep. Then any independent subset of 9’can be expanded to a suitable maximal independent subset B, and s Edep B for every s E Moreover, if B‘ is another maximal independent subset then IBI = IB‘J. Proof: The existence of maximal independent subsets follows from Zorn’s lemma and property (ii), and if s & p B then B u {s} is independent by (iv’), contrary to the maximality of B. Finally, we prove IBI 5 IB’I in two stages. First assume B is finite, i.e., B = {b, ,...,b,}. Then we claim there are (b’, ,...,bb} in B’ such that {b’, ,...,b;,bk+, ,..., b,} is independent, for each k In. Taking k = n this will prove {b’,,. . .,bb} is independent, so IB’I 2 n = (BI. The claim is proved by induction on k. For k = 0 this is clear, so suppose we have found b’,,...,&-, with {b;,...,b,-,,b, ,...,b,} maximal independent; we look for bl. Well b, Edep B’ and b, #dep {b’,,. ..,b:- I , b,, ,,. ..,b,}, so by (iii) we conclude there is some b‘ in B‘ with b’ &ep {b’,,...,blb,, . . .,b,}, and we take b’, to be this b’. {b;,. . .,4,b,, . . .,b,} is independent by (iv’), as desired. Next assume B is infinite. Write B‘ = { b’: i E I ’ } . For each i in I’ take a finite subset B, of B such that b: edepB,. Then bl edep B, for each b’ in B‘. Since b Edep B’ for each b in B, we have b Edep Bi for each b in B, proving Uie,,B, = B. Hence there are an infinite number of B,, proving
,,
,,
uisl,
u
,,
50.3 Abstract Dependence
25
IB’I is infinite, so then letting xo denote the cardinality of N we have IB’I = /B’Ixo 2 (Bil = I u Bil= IBI. We have proved IBI I (B’I in all cases; by symmetry IB’I 5 IBI, so (BI = IB‘I. Q.E.D.
xis,,
Remark 0.3.3: P. M. Cohn has observed that the conclusions of theorem 0.3.2 still are valid if we weaken (iii) by stipulating in its hypothesis that S is also independent. Let us call this a weak dependence relation.
Algebraic and Transcendental Elements One of the standard applications of the theory of abstract dependence is the study of transcendence bases in field theory. Since we shall rely on this theory when examining extensions of rings (first in 42.5 and more extensively in Chapter 7), let us review the fundamentals from the commutative theory. Let C dtf denote the category of algebras over a commutative ring C. Given R in C - d t g and r E R we write C [ r ] for ( c : = , c i r i : c Ei C , t E N). C [ r ] is a commutative subalgebra of R, and there is a surjection q,: C [ I ] --t C [ r ] given by qrl = r, where I is a commuting indeterminate over C ; the elements of ker qr are the polynomials satisjed by r. We say r is transcendental over C if ker qr = 0; otherwise r is algebraic over C, and we say r is integral over C iff r satisfies a monic polynomial. R is integral (resp. algebraic) ouer C if each element of R is integral (resp. algebraic) over C. When C is a field the notions, “algebraic” and “integral” coincide. More generally suppose the elements r l , . . .,r, of R commute with each other. Let CIIl, . .., I , ] denote the algebra of polynomials in the commuting indeterminates Al,.. ., I , over C. Writing C[rl, . . ., r , ] for the (commutative) C-subalgebra of R generated by r l , . . .,rt we have the canonical surjection q : C [ I , ,. . . , I , ] + C[r,,. . ., r , ] given by q I i = ri for 1 5 i I t ; we say r l , . . .,r, are algebraically independent (over C ) if ker q = 0. Example 0.3.4: We are ready to apply theorem 0.3.2 to the following situation: C c H are integral domains, whose respective fields of fractions are denoted F and K. Algebraic dependence over F (in K ) is a strong dependence relation, so any set of algebraically independent elements of K can be expanded to a maximal independent set, which is called a transcendence base of K over F. Moreover, the cardinality of a transcendence base of K over F is unique and is called the transcendence degree of K over F, or tr deg KIF. Note that elements of H are algebraically independent over C iff they are
General Fundamentals
26
algebraically independent over F, seen by “clearing denominators,” so we can also define tr deg HIC to be tr deg KIF. At times it is convenient to have the following generalization of “separable” for field extensions which are not necessarily algebraic; We say K 3 F is separable if either char(F) = 0 or char(F) = 0 and K is linear disjoint from F’Ip, the field obtained by adjoining p-th roots of all elements of F. In other words if al,. ..,a, E K are linearly independent over F then a,: . . .,a,P are also linearly independent over F. It follows that every subfield K O of K finitely generated over F is separably generated in the sense that K O is separably algebraic over a purely transcendental field extension of F. The standard proof, which can be found in Lang (65B, p.265) or Jacobson (80B, p.520) is an induction on the number of generators of K over F. In particular, any extension field of a perfect field is separable. We shall see more general uses of the word “separable” in 52.5 and then 55.3.
Exercises w.0 1. Given a set S with a preorder <, define an equivalence on S by stipulating a = b iff a < b and b Ia. Then the set of equivalence classes is a poset, where [a] I[b]
iff a I b. This observation is useful in generalizing results about posets. 1’. Define < on Z by a s b iff b divides a. Use this to define a PO on N via exercise 1, and identify this poset with Y(Z). 2. A lattice Y is modular iff Y has the property: If a 5 b and a A c = b A c and a v c = b v c then a = b. (Hint: (-=) Let a , = a v (b A c) and a, = b A (a v c). Then a , I a, by remark 0.0.11.To show a2 _< a , one needs only show a , A c 2 a, A c and a , v c 2 a, v c.) 3. We say b covers a if a < b and there is no x with a < x < b. Using exercise 2 show for any a, b in the modular lattice Y there is a lattice isomorphism from the sublattice {x E 9a < x < a v b to the sublattice {x E 4p:a A b < x < b}, given by x -P x A b, with inverse function y + a v y . (Hint: these are order-preserving). Conclude that if a v b covers a then b covers a A b. Many module theoretic results can be proved using lattice theory alone. In the next few exercises we sketch the development of this aspect of lattice theory, foreshadowing results to appear later in the text (esp. in Chapter 2). 4. A complete lattice Y is upper continuous if for every directed subset S of Y we have a A S) = v A s) for each a in 9, lower continuous is defined dually. Show the modular lattice Y ( M )is upper continuous but nor lower continuous. 5. An element a in a complete lattice Y is compact if for each directed set {b,:i E I} with b, 2 a we have some b, 2 a; Y is compactly generated if every element is a supremum of compact elements. Show every (complete) compactly generated lattice is upper continuous. On the other hand, the compact elements of Y ( M ) are the f.g. submodules of M, so Y ( M )is compactly generated.
(v
v
Exercises
27
Modular Lattices In exercises 6 through 13 assume 9 is a modular lattice. 6. The dual of Y is modular. Every interval of Y is a modular lattice, which is complemented if U is complemented. 7. Complements need not be unique in a complemented modular lattice. (Take sP(R‘z’).)However, if a < b in 9then no element can be a complement for both a and b. 8. If (a v b) A c = 0 then a A (b v c) = a A b. (Hint: a A ( b v c) I (a v b) A ( b v c ) = b v ((a v b) A c) = b.) 9. An element a of U is large (also called essential) if a A b # 0 for all b # 0. If b is maximal such that a A b = 0 then a v b is large. (Hint: use exercise 8.) 10. If 9 is upper continuous then for any a, b in Y with a A b = 0 there is c 2 b in Y such that a A c is large. An atom is a cover of 0. A lattice is atomic if for every a # 0 there is an atom a , I a. 11. Suppose Y is compactly generated. 9 is complemented iff every element is a
supremum of a suitable set of atoms. (Hint: (*) Any compact element c has some element b over which c is a cover, so c A b’ is an atom.) 12. The socle of 2, denoted soc(9), is defined as A{large elements of 9). The interval from 0 to soc(9) is a complemented lattice. (Hint: Use exercise 9.) soc(9) 2 V{atoms of U } ,equality holding if Y is compactly generated (by exercise 11). 13. Suppose Y is modular and complemented. Y satisfies ACC iff Y satisfies DCC. (Hint: Given an infinite chain a , > a , > . .. build an infinite ascending chain using complements and exercises 6 thru 9. The reverse direction follows from duality.)
Zorn’s Lemma 14. Let R be the set of polynomials (in one indeterminate 1)over Q having constant term 0. R satisfies all the ring axioms except the existence of the unit element 1. If H is an additive subgroup of Q then (HA R1)a R ; since Q has no maximal
+
15. 16.
17. 18.
19. 20.
additive subgroups, conclude R has no maximal ideals. Prove the validity of transfinite induction. (Hint: If S # a take a‘ minimal in a - S and prove the absurdity a’ E S since fl E S for all /? < a’.) If a # a’ are ordinals then either a < a’ or a’ < a. (Hint: Take a minimal counterexample a’.) Let B = {classof ordinals}. 0 is not a set (for otherwise 0 E 8,which is impossible.) Thus to show a class GR is not a set it suffices to find a 1:l function from 0 to GR. Any set S is in 1:1 correspondence with a suitable ordinal. (Hint: Otherwise define a 1: 1 function f:B + S as follows: fb = so for some so in S and, by transfinite induction, take fa to be some element of S - {ffl:fl E a } for each ordinal a.) Conclude that every set can be well-ordered with respect to suitable total order! For any set S , every function f:S + B ( S ) is not onto. (Hint: {sE S : s $ fs} is not in the image of f.)Consequently there is no 1: 1 function B ( S )+ S. Prove the maximal principle. (Hint: If ( S , I) were an inductive poset without maximal elements take an ordinal a of the same cardinality as B ( S ) and define j a + S by transfinite induction, taking fa‘ to be any element > sup{f/?: fl < a’} for each ordinal a’ < a. Then f is 1:1, contrary to exercise 19.)
General Fundamentals
28
Distributive Lattices and Boolean Rings A ( b v c) = (a A b) v ( a A c) for all a, b, c in Y.Show any distributive lattice satisfies the dual property a v ( b A c) = ( a v b) A (a v c). Y ( Z ) is a distributive lattice, but Y(,M) need not be distributive in general. 22. A lattice is Boolean if it is distributive and complemented. By a misnomer, Boolean lattices are usually called Boolean algebras; they are used in computer science and circuitry. Show any Boolean lattice Y has a ring structure given by ab = a A b and a + b = (a A b’) v (a’ A b). (Hint: The hard axiom to verify is associativity of addition, which is proved by showing the function (a,b,c) + ( a + b) + c is symmetric in a, b, and c. Indeed, a b = (a v b) A (a’ v b‘) so (a + b) + c = ( a A b A c’) v (a’ A b A c’) v (a’ A b’ A c) v ( a A b A c).) 23. An element e of a ring R is idempotent if e’ = e; we say R is a Boolean ring if every element is idempotent. Show every Boolean ring R is commutative, and r + r = 0 for all r in R. (Hint: r l r , = (rl + r$; then take r2 = 1.) Every Boolean lattice gives rise to a Boolean ring by exercise 22. 24. For any ring {idempotents of Z ( R ) }form a Boolean lattice under the partial order el < e, iEe,e, = e,. 25. Suppose Y is a Boolean lattice. Viewing Y as a ring as in exercise 22 show I Y as ring iff I is a filter of the dual lattice of 9. 26. The only Boolean integral domain is Z/2Z since e ( l - e) = 0. But every homomorphic image of a Boolean ringR is Boolean, so for A Q R it follows RIA is an integral domain iff RIA is a field.
21. A lattice 9 is distributive if a
+
+
1. Monics in 9 i ~ are q all 1 : l . (Hint: If f: R + R‘ is monk define a ring structure on
the Cartesian product R x R by componentwise operations and let T = { ( r l r r 2E) R x R: f r , = f r , } . Then in1 = fn, where ni: T+ R is the projection on the i component.) The proof of this exercise introduces two important constructions-the direct product and the pullback. 2. Given an object A of V one can define the constant functor FA:I + V given by F’i = A for all i E I and FAf = 1 , for all morphisms f. Any morphism g : A + B yields the natural transformation g: FA + Fs sending i to g. Now define the diagonal functor 6: V + given by 6A = FA and 6g = g. 3. Interpret example 0.1.13 when I is a set with the discrete PO, viewed as a small category as in example 0.1.1.
1. By comparing annihilators, show Z has nonisomorphic simple modules. 2. If f : M + N is epic and kerf and N are f.g. R-modules then M is also f.g. 3. If M is a f.g. R-module and A Q R is f.g. as left ideal then AM is f.g.
(Hint: If M =
1Rxi then AM = 1Axi.)
1
Constructions of Rings
There are several general constructions of rings and modules which play a central role in the theory of rings and which also provide many interesting examples and applications. In this chapter we consider these basic constructions, along with a little theory needed to shed light on them. The different constructions used are enumerated in the section headings.
$1 Matrix Rings and Idempotents It is fitting to start the main text with a discussion of matrices, since they are undoubtedly the most widely studied class of noncommutative rings. Although matrices are fundamental in the structure theory, we shall postpone most structural considerations until the next chapter, contenting ourselves with examining the elementary properties of matrices. Define the Kronecker delta 6, to be 0, unless i = j , in which case dii = 1.
Matrices and Matrix Units Definition 1.1.1: We define M,,(R), the ring of n x n matrices with entries in a given ring R, as follows: each matrix is written as (rij)where rij denotes the i-j entry for 1 Ii, j I n (with n presumed to be understood); addition and
29
Constructionsof Rings
30
multiplication are given according to the rules:
(ri,?))+ (r:)) = (ri,!’
+ rij’)
and
We delete the verification that M,,(R)is indeed a ring, analogous to the familiar special case when R is a field; moreover, towards the end of 51.2 we shall lay down general principles which at once imply M,(R) is a ring. We obtain a more explicit notation by defining the n x n matric unit eij to be the matrix whose i-j entry is 1, with all other entries 0. Thus (rij)= j = rijeij; addition is componentwise and multiplication is given according to the rule
I:,
(rieijNr2euu) = 4dr1r2)eiv
The set of n x n matric units is a base of M,(R) as R-module (with scalar multiplication given by r(rij)= (rrij)). Note that if a = (rij) we then see rijeuu= euiaej,, a very useful computation which pinpoints the entries of a. A scalar matrix is a matrix of the form reii.The set of scalar matrices is a subring of M,(R) which shall be identified with R under the isomorphism r --t x r e i i , and it is the centralizer of {eij: 1 5 1, j In}. (Indeed if a = (rij) commutes with each eij then ruieuj= aeiJ= eija = r p i , so matching entries we get rui = 0 = rjUunless u = i and u = j. For u = i and u = j we get rii = rjj for all i, j, implying a = rlreii.) In particular, Z(M,,(R)) = { zeii:z E Z(R)} and eii = 1 in M,,(R), leading us to an important internal characterization of matrix rings.
XI=
x:=l
I:=
1
Defnition 1.1.2: A set of n x n matric units of a ring T is a set {eij: 1 5 i, j I n} c T
such that
eii = 1 and eijeku = djkelufor all i, j, k, u.
Surprisingly, the existence of a set of n x n matric units makes any ring T an n x n matrix ring over a suitable ringR; the trick in the proof is to find some intrinsic description of R. Since the idea recurs, let us describe it briefly. If we already know that T = M,,(R) and if a = (rij)E T then clearly euiaeju;also it is useful to note the scalar matrix corresponding to riJ is rijei, = ei,aejj.
Proposition 1.1.3: T has a set of n x n matric units iff T % M,,(R) for a suitable ring R.
81 Matrix Rings and ldempotents
31
Pro08 (e) by definition of the eij.(a) Conversely suppose {eij: 1 Ii, j In} is a given set of matrix units of T. Let R = eulaelu:aE T}. R is a subring because it is closed under subtraction, and 1 = euu= x e u l l e l uE R and ( ~ e u l a e l u ) ( ~ e , l b e=u~l )e u l ( a e l l b ) eEl uR. It remains to define the euiae. = isomorphism cp: T + Mn(R). Given a in T put rij = JU eul(eliaejl)eluE R and define cpa = (rij).Clearly cp is an additive group homomorphism. Moreover,
{c:=
c:=,
I:=,
c:=,
r..e.. v v=
so cp(
n
C e .ae. e.. = 16 .e .ae..= e..ae.. n
u=l
U1
JU
lJ
u= 1
U1
UI
JJ
I1
JJ9
xi, rijeij I;, eiiaejj (Ceii)a( ejj) a, implying is 1 1; likewise, 1rijeij) (rij)so is onto. Finally the i-j term of cp(ab) is =
=
=
=
j=
cp
:
cp
eiiabejj= eiiU
n
n
k= 1
k= 1
1ekkbejj =
(eiiaekk)(ekkbejj)
implying cp(ab) = cpacpb, so cp is a ring isomorphism.
Q.E.D.
There are many close ties between the structure of a ring R and Mn(R), some of which we give now (also, cf., exercise 1). Proposition 1.1.4: Let Mat,, denote the full subcategory of B i ~ g whose objects are rings of n x n matrices. Then there is an isomorphism of categories F : 9 i B g + Aat, given by FR = Mn(R)where for any morphism f:R + R’ we dejine Ff: Mn(R)+ Mn(R’)by (Ff )(rij)= (frij).
Pro08 F f is indeed a ring homomorphism, since It follows at once that F is a functor, and its inverse G is given by GMn(R)= R where for any morphism g: Mn(R)+ M,(R’), Gg is given by the restriction to the scalar matrices. Q.E.D. For any subset S of R we let M J S ) denote the matrices whose entries lie in S. Proposition 1.1.5: There is a lattice isomorphism f: {idealsof R } + {ideals of M,(R)},gioen by f A = Mn(A). Proofi Suppose A 4R. Then Mn(A) is an additive subgroup of R, and the multiplication formula of matrices shows Mn(A)4M,(R). If B c A
32
Constructionsof Rings
then M,,(B) c M,,(A), so f is order-preserving, and it remains to find an order-preserving inverse. If I Q M,,(R) define g l = I n R (identifying R with the scalar matrices). Clearly, gf = 1, and to prove fg = 1 we must show that if a = (ri,) E I then each scalar matrix rijC:= e,,,, E I for every i,j. But Q.E.D. rijz:= euu= euiae,, E I, as desired.
c:=
There is a useful way to reduce the size of matrices, called partitioning, described as follows: Remark 2.1.6: Suppose n = mt. If R = M,(T) there is an isomorphism
f: M,,(T)+ M,,,(R) sending a = (a,,) to the t x t matrix (r,,,) where I,, = 1)+ Cy’r(Vaijeij. (Indeed f is clearly an isomorphism of
xYzt(,,-
additive groups; the straightforward but messy verification that f preserves multiplication is left to the reader. Note that f merely subdivides the matrix a into m 2 matrices, whence the name “partitioning.” The procedure can be done more generally, cf., exercise 3.
M,,(R) has extra structure obtained from the transpose t : M,,(R)+ M,,(R) defined by (rij)‘= (Iji). Rather than go into the structural implications now, we merely note that the transpose is an additive group homomorphism satisfying the properties (ab)‘= bra‘ and (ar), = a for all matrices a and b. The theory of matrices over fields should be familiar to the reader. There are ready extensions of the basic theorems to matrices over arbitrary commutative rings.
Subrings of Matrix Rings As we shall see later, there are large classes of rings which can be viewed as subrings of a matrix ring over a commutative ring. Here we consider examples of such rings. Example 1.1.7: Hamilton’s algebra W of real quaternions over R is defined as the 4-dimensional vector space (over R) having base l,i,j,k, subject to the conditions i 2 = j z = k 2 = - 1 and ij = -ji = k. One could, in fact, prove directly that multiplication is associative and distributive over addition, but it is quicker to display W as an 08-subalgebra of M,(C). Indeed,
taking A =
{(
-Y
!):x, y E C} x
-
c M,(C), where denotes complex con-
jugation, one sees A is a subring of M 2 ( C ) because it is closed under
41 Matrix Rings and Idempotents
33
multiplication, addition and subtraction. Hence A is an R-subalgebra and
so A is Hamilton’s quaternion algebra. This way of viewing
shows quickly that W is a division ring. Indeed,
xx zero element.
+ yy
W also
(-5 ;)(f -:)=
# 0 so we have constructed the inverse of each non-
Example 1.1.8: We give some general classes of subrings of M,(R)
(i) The ring of scalar matrices (isomorphic to R ) (ii) The ring of diagonal matrices (i.e., all i-j entries are 0 except for i = j ) (iii) The ring of (upper) triangular matrices (i.e., all i-j entries are 0 except for i Ij ) (iv) Those matrices (Iij)where rij = 0 unless i =j or i = 1 (v) Those triangular (rij)such that r l l = r Z 2= ... = r,, E Z ( R ) (vi) More generally, if T is a subring of the ring of diagonal matrices and A 4R , take the subring of triangular matrices consisting of each (rij)whose diagonal lies in T and whose off-diagonal entries each lie in A. (vii) Those matrices (rij)with rij = 0 unless nu-, < i, j s nu for suitable u (where 1 u I t and we are given 0 = no < n, < ... < n, = n. These matrices decompose into blocks of lengths nu - nu- .)
Matrices Whose Entries Are Not Necessarily in Rings One should observe that the ring structure of matrices did not require all the entries to lie in a ring. In particular, if we formally consider set of 2 x 2 matrices of the form
(: :)
(;
;),the
where a E A, b E B, c E C , d E D,
we see that addition of matrices require A , B, C, D, to be additive groups; multiplication merely requires various actions on the sets permitting us to define the product, with associativity and distributivity holding wherever possible. More explicitly, we have the following situation. Example 1.1.9: Suppose R, T are rings and M E R - A d - T . Then the mod-
ule operations make
(: ):
into a ring. (Indeed, one can define multi-
plication since if ri E R, x i E M , and si E T we have
34
Constructionsof Rings
and associativity and distributivity come from the given ring and module properties (exactly as usual for matrices). This example provides a wealth of interesting examples, several due to Small and Herstein, so we shall analyze its left ideal structure. Suppose Q is a left ideal of R M If X Y E Q then o z O T
( ). ( ) (7 :)=('
O)t
0 0 0 2 ')cQ
Thus Q = A
+ Le,,,
where A is an R-submodule of
(R ):
(viewed as
R-module by scalar multiplication, as usual) and L is a left ideal of T such that MLe,, G A. Conversely, any such set A + Le,, is a left ideal of
(u"
:)*
Analogously, the right ideals of
(R );
B is a right T-submodule of
+ Pel, where
, P is a right ideal of R, and
PMe,, z B. Consequently, the ideals of
have the form B
(f
F)
have the form
(i r)
where
P Q R, L Q T, and N is an R-T bimodule of M satisfying ML k N and PM E N. The reason we are interested in this example is that when R and T are suitably restricted, the left-right ideal structure is reflected in the bimodule structure of M, which can easily be arranged to be very asymmetric; in this manner one can often find a phenomenon occurring in the left ideal structure
of
(," );
but not in the right ideal structure. This theme will be pursued
35
$1 Matrix Rings and Idempotents
at times in the text; for the time being, the reader is invited to experiment with this example in the following two cases: (i) R E T and M = T; in particular when T is a field. (ii) T is a homomorphic image of R and M = T (viewed as R-module as in remark 0.0.9). Example 1.1.10: (Preview of the Morita Ring) Suppose R, T are rings with M E R-Mod-T and M‘ E T - A d - R . In order to define multiplication on
(L,y )
one needs compositions M x M’ + R and M’ x M + T which
we denote here respectively as ( , ) and [ , 1; i.e., for x E M and x’ E M’ we have (x, x’) E R and [x‘,x] E T. We define addition componentwise and multiplication in the following natural manner:
(a :j(:z): (
1112
= x;rz
+ (x,,x;) + six;
r1xz [x;,x,l
+ XlSZ + SlS2
).
Distributivity of multiplication over addition requires the forms ( , ) and [ , ] to be additive in each component, e.g., (xl + xz,x’) = (xl, x’) + (xz,x’), whereas associativity requires the following extra conditions: r(x, x’) = (rx, x’), (xs, x’) = (x, sx’), (x, x’)r = (x, x‘r) s[x’, x] = [sx’,
XI,
[x’r, x] = [x’, r x ] , [x’,x]s = [x‘, xs].
The Morita ring plays a key role in the structure of R-&ud, in addition to providing a huge class of interesting examples (and in fact yielding example 1.1.9 when we take M’ = 0).
Idempotents and the Peirce Decomposition An element e of R is idempotent if ez = e. Idempotents play an important role in ring theory and are closely linked to matrices since the matric units eii are idempotent. In the next few pages we shall study some general properties of idempotents and their impact on matrices and matric units. First note 0 and 1 are always idempotents and so are called trivial idempotents. Idempotents e and e‘ are orthogonal if ee‘ = e‘e = 0. If e is idempotent then 1 - e is an idempotent orthogonal to e. Lemma 1.1.11: Suppose e E R is idempotent.
(i) If L is a leji ideal of R then L n eR = eL. (ii) If A Q R then A n eRe = eAe.
36
Constructionsof Rings
Proof: (i) If er E L n eR then er = e(er)E eL, proving (C); (2) is obvious. (ii) As in (i). Q.E.D.
Proposition 1.1.12: Suppose e is an idempotent of a ring R. Then eRe is a ring with multiplicative unit e. Moreover, there is an onto lattice map $: {ideals of R} {ideals of eRe} given by A + eAe. Proof: Clearly eRe is a ring. To see $ is a lattice map note $ ( A + B ) = eAe + eBe = $ A + JIB $ ( A n 8) = e ( A n B)e = A n B n e R e = and ( A neRe) n ( B n eRe) = $ A n $B. If I 4 eRe then I = (eRe)I(eRe)= eRIRe = $(RIR), proving $ is onto. Q.E.D.
Thus we see that eRe and likewise ( 1 - e)R(1 - e) are rings whose ideal structures are at least as nice as that of R, and we are led to study R in terms of these rings, One useful link is the following straightforward result. Proposition 1.1.13: (“Peirce decomposition”) I f e is an idempotent of R then R = eRe 0 ( 1 - e)Re Q3 eR(l - e) 0 ( 1 - e)R(1 - e) as Abelian groups. Proof: For r in R take r , = ere, r2 = (1 - e)re, r3 = er(1 - e), and r4 = ( 1 - e)r(l - e); then ( r , r2) (r3 + r4) = re r(1 - e) = r. If r = r; + r; + r; rk is another such decomposition then erge = e((1 - e)Re)e = 0 and likewise erie = erbe = 0 so r; = erie = e(r; + r; r; + rb)e = ere = r,; Q.E.D. likewise r; = ( 1 - e)r;e = (1 - e)re = r2 and so forth.
+
+ +
+
+
Example 1.1.14:
(i) IfR=M,(T)ande=e,, theneRex Tand(1 - e ) R ( l - e ) x M , - , ( T ) , and the Peirce decomposition is the usual partitioning of the matrix. (ii) The Morita ring of example 1.1.10 has an idempotent e = and the Peirce decomposition partitions
G,F)
(: :),
into its components
(although of course this ring does not have a matric unit e12). Thus the Peirce decomposition enables us to analyze rings “resembling” matrix rings. (iii) If e E Z ( R ) then Re x eRe is a ring. In this situation we can refine the Peirce decomposition.
91 Matrix Rings and Idempotents
37
Proposition 1.1.14: R x R1 x x R, as rings ij,T there are pairwise orthogonal idempotents e, in Z ( R ) such that e, = 1 and Ri Re, for each i. Proofi (+) Identifying R with R, x ... x R, take e, = (0,..., 0,1,0, ..., 0) where "1" appears in the i position. Clearly ei E Z(R). (e) Let Ri = Re,, and define cp: R + Ri by r H (rel,. . .,re,). cp is a surjection since ( r ,e, ,. . .,r,e,) = cpf riei);Jl is an injection since if Jlr = 0 then each rei = 0 so r = rCe, = 0. Q.E.D.
1
n:=
Idempotents and Simple Modules Proposition 1.1.15: Suppose e is an idempotent of R. There is a functor F: R - A d + e R e - A o d given by FM = eM where for any map S:M + N , we take F f to be the restriction of f to eM. Moreover, if M is simple in R - A u d with eM # 0 then eM is simple in eRe-Aod (so F "preserves" simple modules not annihilated by e). Proofi The first assertion is immediate since (eRe)eM c e(ReM) 5 eM. To prove the second assertion note for any x in M with ex#O that (eRe)ex= e(Rex) = eM. Q.E.D. There is a converse for idempotents of matrices. Proposition 1.1.16: If M E R - A o d we can view M(") in M,,(R)-Aod by defining addition componentwise (i.e., (XI,.
.. ,x,)
+ (xi,. . . , x i ) = ( x l + x;, ..., x , + xh))
and multiplication by
Thus we get a functor G: R - A u ~ - + M , ( R ) - A by u ~GM = M("),where for any map f:M -+ N we define Gf:M(")-+ N(")componentwise. Proofi First note that the scalar multiplication we have defined is merely as n x 1 matrices, so matrix multiplication if we view the elements of M(") the straightforward verification that M(")E M , , ( R ) - A d is a special case of a well-known matrix argument. (This idea is generalized in exercise 3.) To see that G is a functor we note that G j is a map because putting h = Gf
Constructions of Rings
38
Theorem 1.1.17:
R - A u d and M , ( R ) - A u d are equivalent categories.
Proofi Identifying R with e,,Mn(R)ell let V = R - M u d and 9 = M,(R)A d . By proposition 1.1.15 and 1.1.16 we have functors F: 9 + V (taking e = el 1) and G :V + 9, so we must prove FG and GF are naturally isomorphic, respectively, to 1, and 1,. If M E R - A d then FGM = {(x,O,. . . ,O):x E M } so there is a natural transformation q: 1 + FG given by qHx = (x,O,. . .,O); each qH is clearly an isomorphism. Conversely, if M E M , ( R ) - A o ~then GFM = {((xl,O,...,O),.. ., (x,,O,. . . ,O)): ( x l , ...,x,) E M } so there is a natural isomorphism q': 1 + GF given by ~ t ( x l , . ,.x , ) = ((~1,0,. ..,O),...,(x,,O,.
. .,O)).
Q.E.D.
This is our first example of equivalent categories which are not isomorphic and will be fundamental to the study of equivalent categories (and the Morita theory). Once we know two categories are equivalent we can use the equivalence to transfer various categorical properties from one category to another, as illustrated in the next result. (Many more examples will be given in the treatment of Morita theory.) Corollary 1.1.18: M is simple in R - A u d iff M(")is simple in M , ( R ) - A o d .
M is simple iff each nonzero map f: M + N is monic. Since FG and GF are naturally isomorphic to the respective identities (notation as before), one sees easily that G preserves this categorical property, so M is simple in R - A u d iff M(")= GM is simple in M , ( R ) - A u d . Q.E.D.
Proofi
A more concrete proof of this result is given in exercise 4. The passage from M to M'") can be applied to reduce proofs about f.g. modules to the cyclic case, along the following lines: Remark l.Z.19: If M is spanned over R by the elements x l , . .., x n then M(") is cyclic in M , ( R ) - A u d , spanned by the element Jz = (xl,. ..,xn). (Indeed eijZ has x, in the i-th position.)
$1 Matrix Rings and Idempotents
39
Primitive Idempo tents An idempotent e is primitive if e cannot be written as the sum of two nontrivial orthogonal idempotents.
Remark 2.1.20: If e,,e, are orthogonal idempotents then Re, R(e, + e,) since el = el(el e,) and e2 = e2(el + e,).
+
+ Re, =
,
In particular, we see el is a primitive idempotent of M,,(F) for any field F (because e l , has rank 1). To understand this notion better, we look more closely at orthogonal idempotents. If el,e2 are idempotents with e2el = 0 then (1 - el)e2 is an idempotent orthogonal to e l , by an easy computation.
Proposition 1.1.21: An idempotent e of R is primitive iff there is no idempotent el # 0 such that Re, c Re. In fact, if Re, c Re then there is an idempotent e; such that Re, = Re', and e',, e - e', are orthogonal. Proof: We prove the second assertion, since the first is then an immediate consequence. Take e', =ee,ERe,cRe. Then e;e=e;, so (e;),=e;(ee,)= (e;e)el = e',el = ee: = eel = e', is idempotent. Also el = e: = (ele)el = ele', E Re', so Re, = Re',. Finally, e', = ee;e E eRe implying e - e', is an Q.E.D. idempotent orthogonal to e;.
Lifting Matrix Units and Idempotents Remark 1.2.22: If cp: M,,(R)+ T is a ring homomorphism then T x M,,(R') for a suitable ring R'. (Indeed {cpeij:1 < i,j I n} is a set of n x n matric units of T, so we are done by proposition 1.1.3.) Next we consider the converse: If R = T/A is an n x n matrix ring (i.e., has a set of n x n matric units) then is T a n n x n matrix ring? This question arises often in structure theory and can be handled in three stages. First note for any given set of matric units {eij:1 I i,j I n} that the eii are orthogonal idempotents. Thus we are led to ask (1) if every idempotent e of R has a preimage in T which also is idempotent, i.e., if e can be lifted to an idempotent of T ; (2) if a finite set of (pairwise) orthogonal idempotents can be lifted to orthogonal idempotents; and (3) if we finally can fill in a set of matric units. Surprisingly, the first question is hardest to handle, so we start by assuming it is solved.
40
C o n s ~ t i of o ~Rings
Definition 2.2.23: are satisfied:
A a T is idempotent-lifting if the followingtwo conditions
(i) 1 - a is invertible for all a in A ; (ii) every idempotent of T/A has the form x in T.
+ A for suitable x idempotent
Remark 2.2.24; If A satisfies (i) above then 0 is the only idempotent of A (for if a' = a E A then a = a(1 - a)(l - a)-' = 0).
Proposition 1.1.25: Suppose A 4 T is idempotent-lifting and R = T/A.
+
(i) If x, E T is idempotent and e E R is an idempotent orthogonal to x1 A then e can be lifted to an idempotent of T orthogonal to xi. (ii) Every countable set of (pairwise) orthogonal idempotents of R can be lifted to a set of orthogonal idempotents of T. (iii) Every set of n x n matric units of R can be lifted to a set of n x n matric units of T. Proofi
(i) Take X E T idempotent with e = x + A. Then xxl + A = e(xl + A ) = 0 so xxlEA, and we can form x'=(l -xx,)-'x(l -xxl) which is also idempotent and lies over e. Moreover, since x(l - xxl) = x - xx, = x(l - xl), we see x'xl = 0 so x2 = (1 - xl)x' is idempotent and lies over e, and x2x1 = 0 = X l X ' . (ii) Inductively, suppose we have lifted e l , . ..,e,- to orthogonal idempotents xl,. ..,xu- and want to lift e, to an idempotent xu orthogonal to x l , ...,xu-1.Let x', = xi and e = e,; by (i) we can lift e, to an idempotent xu orthogonal to x i . But for all i I u - 1 we have x;xi = xix: = xi, so x,xi = x,(x;xi) = 0 and xix, = (x,x;)x, = 0. (iii) Having lifted e l l , ...,en, to orthogonal idempotents x1,,. ..,xnn,we must also lift the eij to xij for i # j . Take bij in T with b, + A = eij. Define xil = xiibilxl, and a, = x l l - blixil. Then ai + A = e l , - e l i e i l = 0 so a, E A ; let d, = ( 1 - ai)-' and put xli = diblixii. Then xli lifts e l i and xlixil=dibrixiixil =diblixilxll= d i ( x l l -ai)xl, =d,(l -ai)xll = x l l . Hence ( x i , - xi1x1i)2 = x i - xiixilxli- xilxlixii+ (xilxli)'= xii - xilxli is an idempotent in A so is 0, implying xilxli= x i i . Then put xtj = xilxlj,a set of Q.E.D. matric units lifting eij.
,
c;:,'
The main tool for lifting idempotents involves an algebraicity condition.
81 Matrix Rings and Idempatents
41
Proposition 1.1.26: Suppose r E R and there exists a = x f = o z i r i for zi in Z ( R )such that r" = ar"". Then e = (ar)" is idempotent and, moreover, in any homomorphic image R in which 7 is idempotent we have 2 = E Proofi By induction we have r " = amrm+"for all m 2 1. Hence e2
= (ar)2"= an(anrn+n)= an r n -e.
If Fis idempotent then 2 = iinF2" = F" = E
Q.E.D.
To continue this idea to its conclusion we call an element r of R nilpotent if some power r' = 0; S c R is nil if every element is nilpotent (not necessarily with respect to the same t). If a' = 0 then (1 - a)(l
Remark 2.1.27: invertible.
+ a + ... + a ' - ' )
=1
so 1 - a is
Corollary 1.1.28: Every nil ideal A of a ring R is idempotent-lifting. In fact, i f 7 = r + A is idempotent in RIA then there is an idempotent e in the subring of R generated by r, satisfying Z = E Proof: Condition (i) follows from remark 1.1.27; to see (ii) we note that if x + A is idempotent then x 2 - x E A so 0 = ( x - x2)" = x" - nx"' + ... for suitable n; hence proposition 1.1.26 produces an idempotent lifting x + A. Surprisingly, one may not be able to lift an uncountable set of orthogonal idempotents modulo an idempotent-lifting ideal. Example 2.2.29: (Zelinsky [54]) A ring R with N Q R such that N z = 0 (so that N is idempotent-lifting), where RIN has an uncountable set of idempotents which cannot be lifted to an uncountable set of idempotents of R. Let F be any field and let R be a vector space over F with basis {bij,ei:i, j E I } given multiplication by the rules ez = e,, eiej = bij for i # j , e,bjk = 6,bik, bije, = hjkbi,, and b,b,, = 0. One checks readily R E F - d t g when multiplication is extended by distributivity to all of R. (To verify this the reader could use the criteria given towards the end of the next section.) Moreover, the subspace spanned by the b, is clearly an ideal N with N 2 = 0, and R = R / N has a set = {gi:i E I } of orthogonal idempotents. We shall show S lifts to an orthogonal set of idempotents {di:i E I } only if I is countable. Indeed, suppose we have the d i , and put x , = d , - e , ~ N since Zi=O. Let I ( i ) = { j E I : eixj = 0). Since any xi is a linear combination of a finite number of b,,
s
Constructions of Rings
42
we see by the multiplication table that for any j the set { i E I: eixj # 0} is i ) I. Also finite; hence for any infinite subset K of I we must have U i p K I (= define I’(i) = { j E I : xiej # 0 } which likewise is finite. But for i # j 0 = didj = (ei + xi)(ej+ x j ) = b,
+ eixj + xiej.
I f eixj = 0 then xiej = -bij # 0 so each I ( i ) E l’(i) is finite. Therefore I = UieKZ ( i ) is countable (seen by taking K infinite and countable), as desired. Eckstein [69] has a different method of obtaining the above results (and more) by means of semigroups.
81.2 Polynomial Rings This section features a very important construction, the monoid ring, which leads to huge classes of important examples. Of particular significance is the polynomial ring, a special case. Later on we shall see how to view the monoid ring in a more general setting. We start with a module construction over a ring R. Definition 1.2.1: Given a set S we form the R-module freely generated by S, denoted as RS, to be the set of formal expressions { rss:r, E R, almost all r, = 0}, endowed with addition and scalar multiplication defined “componentwise,” i.e.,
zseS
Cr,s
+ C r i s = C ( r s + r:)s
rCr,s = z(rr,)s,
where r E R. {s E S : r, # 0} is called the support of C r,s. For notational convenience, we identify each element s‘in S with the expression z e s r s s where r,, = 1 and all other rs are 0. Thus we view S as a subset of RS. Remark 1.2.2: RS is indeed a module. (Verifications are componentwise.) Later on we shall identify RS with the “free” module, but let us note that any element f = r,s of RS can now be viewed as the finite sum of those r,s for which s E supp(f); these r,s are called the terms of f, and r, is called the coeflcient of s (in R). Thus any element of RS is written uniquely as a linear combination of elements of S.
1
Monoid Rings We are interested in defining a ring structure on RS. To this end we first define multiplication on terms and extend it to arbitrary elements by distributivity. This is done easily when S is a monoid. Then we define the product
81.2 Polynomial Rings
43
of terms (r,s)(r$) = (r,r;)st
and more generally
where the inner sum is evaluated over all s, t in the respective supports such that st = u. We denote this ring structure as R [ S ] , called the monoid ring. When S is a group R [ S ] is called a group ring.
Proposition 1.2.3: The monoid ring R [ S ] is indeed a ring, whose unit element is lS, the unit element of S. Furthermore if R E C-dif" then R [ S ] is an algebra with scalar multiplication c r,s = E(cr,)s. There is an injection R R [ S ] given by r H rl,; Hence S centralizes R in R [ S ] .
1
We leave the straightforward but messy proof to the reader, noting that it is essentially the same proof the reader has had to plough through to show that the polynomials in one indeterminate over a field form a ring. There is one monoid of special interest. Assume throughout I is a wellordered set. Example 1.2.4: The word monoid on a well-ordered set 1. A word is a formal string w = (il ... i,,,); m is called the length, and max{i,: 1 I u I m} is called the height. Multiplication of words is formed by juxtaposition, i.e., (il * * im)(i',* . . ib)=(il * i,i; .. . ib), and is obviously associative. We get the word monoid by formally adjoining a unit element, written as ( ) and called the blank word. One may think of the blank word as having nothing in it and thus being of length 0, thereby yielding Remark 1.2.5: For any two words w l , w 2 we have length (w1w2)= length w1 + length w2 and height ( w l w 2 )= max(height wl,height w2). In case i repeats in a word, we write i' for i - . . i, taken t times.
Polynomial Rings Example 1.2.6: Suppose I is a singleton {A}. Then every word has the form A' for some i 2 0 (writing ' 1 for the blank word), and putting S = (2: i E N} we write R [ S ] as R I I ] = { C r = , r i A i : u ~ ~ , r i €the R } ,polynomial ring ouer R (in one commuting indeterminate), denoted R[A]. Note that A E Z ( R [ 1 ] ) .
44
Constructionsof Rings
When trying to generalize example 1.2.6 to form polynomials in several indeterminates, one has to decide whether or not one wants the indeterminates to commute with each other. We consider the cases separately. Example 2.2.7: Define X = {Xi:i E I}, where the Xi are formal symbols, called (noncommuting) indeterrninates. Clearly X and I are in 1:l correspondence, so we can well-order X by means of the well-ordering on I and form a word monoid S on X. The words (in the Xi) are called pure monomials; if C is a commutative ring then C[S] is written instead as C{X} = { c(i)Xil. .* Xim(i),where (i) denotes the word i, * im(i)of length m(i) (allowing repetitions), and qi)E C for each (i)}. The elements of C{X} are called noncommutative polynomials and C{X} is called the free C-algebra. In case I is finite, i.e., I = {il ,...,if}, we shall denote C{X} as C{Xl ,...,Xt}. In the current literature C{X} is usually denoted as C(X), cf. example 1.9.22below.
xfinite
To obtain commutative polynomials we could define inductively RCA,,. .., A f ] =(R[A,, . . .,Af- , ] ) [ A f ] , iterating example 1.2.6. However, from our point of view we should like to display this ring as a monoid ring preferably using words. To do this we examine words a little closer. Dejnition 1.2.8: A partial order I on words is defined inductively on the length, as follows: For arbitrary words w=(i, i,) and w'=(i; ih) which are not blank, w < w' if either i, c i', (in I) or i, = i; and (iz... i,) < (i; ib) (defined inductively), 1 . -
1 . .
For example, if I = {A, B, . ..,Z} with the alphabetic ordering then ACT < AT, but ACT and ACTION are not comparable. The reason we do not use the customary lexicographic (total) order is to make the following remark true: Remark 1.2.9: If w c W' and w,, w2 are arbitrary words then w l w c wlw' and wwz c w'wz. Nevertheless, I induces a total order on the set of words of a given length. A word w is basic if for each w' obtained by interchanging two letters, we have w I w'. For example, BEEN is basic since BEEN IEBEN, BEEN, EEBN, NEEB, BNEE, BENE.
Remark 1.2.10: (i, .*aim)is basic iff i, Ii, I ... Ii,; hence for every word w there is a basic word which we call Bw, obtained by rearranging its letters
81.2 Polynomial Rings
45
in increasing order. If w' is obtained by rearranging the letters of w then
Bw' = Bw. Definition 1.2.11: The symmetric word monoid on I is {Bw:w is a word on I } (including the blank word) under multiplication defined by w 1 w 2 =
B(w,WZ). The symmetric word monoid on I is a commutative monoid, since associativity and commutativity follow from remark 1.2.10: B(B(w,w2)w3)= B(w,w2w3)= B(wlB(w2w3))and B(w,w2)= B(w2w1)). Definition 2.2.22: Put 1= {Ai:i E 2 ) and let S be the symmetric word monoid on A. Then R [ S ] is denoted as R[A] and is called the polynomial ring over R on the commuting indeterminates ,Ii, i E 1. The words of S are called pure monomials. If 1= {A,,. ..,A,} we write R[1,,. . .,A,] instead of R[A]. To obtain further results we examine special kinds of monoid rings, introducing the notions of ordered monoid and filtration. Definition 1.2.13: A monoid S is ordered if S has a total ordering Isuch that whenever s < s' we have sls < sls' and ssl < s'sl for all s1 in S. A C-algebra A has a filtration by an ordered monoid S if A is the union of C-submodules {A(s):s E S} such that (i) A(s) =) A(s')
for all s < s' in S,
(ii) A ( s , ) A ( s 2 )s A ( s , s 2 ) (iii)
for all s1,s2 in S,
~ ( s= ) 0. seS
Note that if s1 < s; and s2 < s; then s1s2 < s;s2 < s;s;. Remark 1.2.13': The most important case in the literature is for S=(iZ, +), i.e.; A ( - l ) ? A ( O ) z A ( l ) z . . . . O f t e n A ( O ) = A ( - l ) = A ( - 2 ) = . . . = A, so we actually have a filtration over N. However, there are situations in which A(1) = A(2) = ... = 0, so we have instead a filtration by the negative integers. In the latter case we could write A, for A ( - n ) , and so ( i ) becomes A, G A, for m 5 n. At times this is used as the definition for filtration in the literature. Definition 2.2.14: The filtration of A by S is valuated if for each a # 0 the set {s E S: a E A ( s ) } has a maximal element, which we denote ua, thereby defining a function v : A - ( 0 ) S .
Constructions of Rings
46
In order to include 0 in the domain of u, we adjoin a formal element m to 0000 = m and sm = 00s = co > s for all s in S, and write S, = S u { m}. Putting VO = m we now have a function f: A + S,, where ua = 00 iff a = 0. The point of considering valuated filtrations lies in the following observation: (Sometimes this is called a “pseudo-valuation.”) S with
Remark 1.2.15: For any ai in A, u(ala2)2 (ual)(ua2). If u is a monoid homomorphism then for all nonzero a1,a2 in A, u(ala2) = (ua1)(ua2)E S implying ala2 # 0; thus A is a domain. We shall use this remark to obtain many examples of domains and start by applying it to monoid rings. Example 1.2.16: If R E C-dt’ and S is an ordered monoid then A = R [ S ] has the filtration given by A(s) = x ( R s ’ : s ’ 2 s), which is valuated since for a = r,s we see ua is the minimal s in the support of a; the corresponding r,s will also be called the lowest order term of a.
1
Proposition 1.2.17: If R is a domain and S is an ordered monoid then R [ S ] is a domain. Proofi Let risi denote the lowest order term of ai in A = R [ s ] for i = 1,2. Then ( r l ~ l ) ( r 2 ~ 2 ) = ( r l r 2 ) ( ~ 1 ~and 2 ) #is0 the lowest order term of ala2 since S is ordered. Thus u of remark 1.2.15 is a monoid homomorphism implying RCS] is a domain. Q.E.D. As we see, the idea of using the lowest order term to examine monoid rings (over ordered monoids) is very useful, and we shall also have occasion to use the highest order term.
Example 1.2.18: (Making the word monoid an ordered monoid). Define the following order on words: w <w’ if length ( w ) (length (w’), or if length w =length w’ and w is less than w’ according to definition 1.2.8. In other words, we sort words j r s t according to length and afterwards we order words of the same length. In view of remark 1.2.9 one sees easily that this ordering is total and satisfies the requirement of definition 1.2.13. Moreover, ordering the basic words in this manner, we also see the commutative word monoid also becomes an ordered monoid. Thus the noncommutative polynomial ring and commutative polynomial ring are monoid
51.2 Polynomial Rings
47
rings over ordered monoids, proving Proposition 1.2.19: If R is a domain then R{X} and R [ L ] are domains (Instant application of proposition 1.2.17.)
Some results relating R and R [ S ] do not require any additional assumptions on S. For any A E R we write A [ S ] in place of A S = { z f i n i t e a i s i : a, E A , si E S}. Proposition 1.2.20:
(i) I f 1 4 R then I [ S ] 4 R [ S ] and R [ S ] / I [ S ] M ( R / I ) [ S ] ; (ii) Z ( R [ S ] )E Z ( R ) [ S ] ,equality holding iff S is commutative.
Proof: (i) Define cp: R [ S ] + ( R / I ) [ S ] by c p ( 1 risi) = Fisi where Fi = ri + I . By inspection cp is a ring surjection and I [ S ] = ker cp 4R [ S ] . (ii) Write 2 = Z ( R ) . If I r i s i E Z ( R [ S ] ) then for all r in R , x ( r r i ) s i = ( r l ) I r i s i= ( C r i s i ) r l = I ( r i r ) s i implying rr, = rir for each i, so ri E Z and x r i s i E Z [ S ] . Conversely, for S commutative clearly S E Z ( R [ S ] ) so Z [ S ] G Z(R[S]). Q.E.D.
1
Example 1.2.21: The category ring. We can extend the construction of R [ S ] to an arbitrary semigroup S , but now R [ S ] may lack a unit element. There is one important situation where the unit element exists, and so R [ S ] is indeed a ring. Namely consider a category W with a finite number of objects, and let S be the disjoint union Hom(A, B). S is a semigroup whose multiplication is composition; precisely, given f:A + B and g : A’+B’ we define g f to be the composition when A’ = B and to be 0 otherwise. The unit element is l,, since for any f:A’ + B’ we have (IAEO 1,)f = 1,f = le.f = f‘. The ring R [ S ] is denoted R [ W ] and is especially useful when % is formed from a directed graph. (The objects are the vertices, and the arrows are the morphisms.)
UA,BEObY
xAEObw
To illustrate how wide-ranging this example is, let us display M,,(R) as a category algebra over R. Namely consider the category % having n objects v l , . . . ,v,, such that Hom(vj, I+)is always a singleton, which we denote as eij, with eijejk=eikfor all i,j, k. In R [ % ] we have eijeuU=djue,,,and ~ ~ eii= =1 , as noted above. Thus { e i j }are a set of n x n matric units for R[%] x M,,(R).
Constructions of Rings
48
Rings of Formal Power Series and of Laurent Series Often we are interested in embedding the monoid ring R[S] into a division ring. To this end we want to start with a larger module than RS. The obvious candidate is RS, the Cartesian product of copies of R indexed by S. Denoting a typical element of RS as an “S-tuple” (rJ, we can define R-module operations “componentwise,” i.e., (r,)
+ (r:) = (rs +
I:)
and
r(r,) = (rrs).
(Note RS IRS, equality holding iff S is finite.) If f = ( I , ) we define supp(f )= {s E S: rs # 0). We should like to define a ring multiplication as before, i.e., (r,)(r;)= ( r i ) where rf = ~ s r = u r , rUnfortunately, ~. this sum need no longer be finite, and so is meaningless in R. However, the sum is defined in one important special case. Suppose S has a given order and define R((S)) = { feRS: supp(f) is well-ordered (under the order of S)}. Clearly R((S))
s21 > s,, > since S, is well-ordered this process must terminate after a finite number of steps, proving S; is finite, yielding the claim. Thus we see that the proposed definition of the product does yield an element of RS, whose support we claim is well-ordered. Indeed, otherwise there is an infinite chain s10s20 > slls21> > ... with s l i E S1,s2i E S,. Take u - = ~ 0 and, inductively, given ui- choose ui such that sl,”,is minimal among {slu:u> u ~ - ~Then } . sluoI slulIslU2 I ... implying szuo > sZu1> s,,,~ > is an infinite descending chain, contrary to S2 being well-ordered. We now see the product yields a member of R((S)).To check associativity and distributivity we note that any term in the product involves only a finite - . a ;
49
51.2 Polynomial Rings
number of terms in the multiplicands and hence multiplication works just as in R [ S ] . Associativity and distributivity follow from the corresponding proQ.E.D. perties in RCS]. The rest of the proposition is straightforward. Corollary 1.2.23:
R((S))is a domain for any domain R and ordered monoid S.
Proof: Exactly as in the proof of proposition 1.2.17.
Q.E.D.
The ring R((S))is often useful because it has enough room to invert elements of R [ S ] . For example, the only invertible polynomials in FCA] (for a field F) are the nonzero constants; as we shall see presently, our new construction provides a method for inverting elements of F [ 4 . We shall discuss one case separately, since it is considerably easier but encompasses many examples. Let us say an ordered monoid S is strongly archimedian if given u in S we can find k in N with sk > u for all s > 1 in S. Proposition 1.2.24: Suppose R is a domain, S is an ordered monoid, and f E R((S))has lowest-order term rs. Then f is invertible in R((S))iff r-' E R and s-' E S . Proof: Suppose f-' exists in R((S)) and let r's' denote its lowest order term. Then the lowest order term of f f - ' is rr'ss'. But ff-' = 1 so rr'= 1 and ss' = 1. Likewise r'r = 1 and s's = 1 proving r' = r-' and s' = s-'. Conversely suppose r-' E R and s-' E S and let g = 1 - r - ' s - y . We shall conclude the proof by showing r- 's- 'f = (1 -g) is invertible, for then f-' = (1 - g)-'r-'s-'. For any u in S let r t ) denote the coefficient of u in g i (where go= 1); putting h j = ~ ~ = o g ~ i = ~ u , s ( ~ we ~ = have o r ~ (1 ~ -) gu) h j = h.(l - g) = 1 - gj+' = 1 - C u s S r ~ f " uAs . j -, 00 we shall see that the J hj "approach" (1 - g)-', so the natural candidate for (1 - 9)-' is h = C,,s(C,P"=o r!))u. We claim (1) for each u in S there exists some k(u) in N such that rg) = 0 for all i > k(u) (2) supp(h) satisfies DCC.
Indeed (1) permits us to define h, and (2) then shows h E R((S)),and a term-byterm check using (1) implies (1 - g)h = h( 1 -9) = 1. As we shall see, the proofs of (1) and (2) hinge on the fact that supp(g) = {s-'s':s' E supp(f)} - { l}, i.e., every element of supp(g) is greater than 1. At this point we could give the general proof, but present first an important special case which covers most applications.
Constructions of Rings
50
Special Case. S is strongly archimedian. Take k = k(u) such that wk > u for all w > 1 in S. Then wl. . . wi > (min{wl, ,w ~ }>) ~u for every i > k, so (1) is immediate; likewise if u = u1 2 u2 > in supp(h) then each uj E supp(hk) which is well-ordered (since hk E R((S))), so the chain must be finite, proving (2).
. ..
General Case. Let Q denote the set of finite sequences of elements of supp(g) and, given a sequence q = (sl,s2,. ..,s,), write val(q) = s1 s2 . . .s, E supp(q"), and length(q) = u. Then Q has a partial order given by q1 < q2 if either val(q,) < val(q2) or val(q,) = val(q2)with length(4,) > length (q2). Claim. Q has no infinite descending chain q1 > q2 > Otherwise pick a descending chain such that length(q,) is minimal; of all such possible descending chains we pick one with length(q,) minimal, and so on. Thus we have a chain q1 > q2 > * where (given q,, .. .,4,q, has minimal possible length. Write s, for the first element of q, and q: for the remainder of the sequence. Take uo = 0 and, inductively, given u j - l take uj > uj-l with s, minimal possible. We can do this since supp(g) is well-ordered, and have s,, 5 s,, 5 * * * , implying 41 > * * * > quI- > 4:, > qL2 > . . . is an infinite descending chain. Since q:, has smaller length than q,, we have a contradiction which yields the claim. If (1) were false we would have an infinite sequence i , < i , c in N with u in supp(g'j) for each j; taking qj to be a sequence with length(qj) = ij and val(qj) = o, we get q1 > q2 > q3 > ... contrary to the claim. If (2) were false we would have a chain val(ql) > val(q2) > val(q,) > ... for suitable qi in Q, so q1 > 4 , > q3 > ..- contrary to the claim. Q.E.D. a * . .
- a -
We now have the following immediate corollary: Theorem 1.2.24': r f G is an ordered group and D is a diuision ring then D[G] can be embedded in the division ringD((G)).
This is our first embedding of a ring into a division ring and is quite useful, cf., theorem 1.3.45 below. Embedding into a division ring is a recurring theme in ring theory. Remark 2.2.25: If w1 is a word of length k then for every nonblank word w clearly the length of wk+lexceeds k. This trivial observation shows that the
monoids of the next three examples are strongly archimedian.
Example 2.2.26: Define the monoid S= {A': iEZ} with multiplication given by A i A j = A i + j , ordered by A i c A j if i < j. S is strongly archimedian (as noted
51
$1.2 Polynomial Rings
above). R((S))is denoted R((1)) and called the ring of Laurent series ouer R because every element has the form f = CIP"_,riAi for suitable t in Z corresponding to the minimal element 2 of supp(f). For a field F, we see F((2))is a field containing F [ 4 , so that one could form the field of fractions K of F[1] inside F((I)),and at times we shall use this technique. Note that K c F((1)) since there are Taylor series expansions of functions which are not quotients of polynomials. (For example, take f = x n = o l l " / n !The formal derivative of f is f itself, implying f cannot be the quotient of two polynomials.)
Example 2.2.27: Take S = {Ai: i~ N}, a submonoid of the monoid of example 1.2.26. Now R((S))is denoted R [ [ 1 ] ]and called the ring of formal power series ouer R. Note that R c R[1] c R [ [ 2 ] ]c R((2)). Example 2.2.28: S is the (noncommutative) word monoid, with the strongly archimedian order of example 1.2.18. For any commutative ringC we denote C((S))as C { { X } } .
Digression: More General Constructions In all the examples described above the elements of S centralize R in R [ S ] . However, there are instances where we want to "skew" the multiplication somewhat. In this case we could simply define the product of two arbitrary terms, and extend it via distributivity, i.e.,
In order for us to have a ring the following conditions must hold for all terms
f , 9, h: ( f d h = f(9h);
( f + 9)h = f h + gh;
f ( 9 + h) = fg
+fh.
Also there must be an element 1 for which I f = f l = f for all terms f . Conversely, these conditions guarantee that we will have a ring structure. In $1.6 several important examples are given, although the verification of the ring structure will be done by means of the regular representation (discussed in $ 1 S ) .
A Supplement: Ordered Groups Theorem 1.2.24'is an important method of constructing division rings but to utilize it we must have examples of ordered groups. Thus we are led to a short excursion into group theory.
Constructions of Rings
52
Remark 1.2.30: Every ordered group is torsion free. (If g # 1 then g < 1 or > 1, so assume g < 1; then g" < 1 for all n in N, so in particular g" # 1).
g
The easiest torsion-free group is (Z, +) which obviously is ordered by the usual ordering of real numbers. Remark 1.2.31; (Z, + ) I is an ordered group for any set I. Indeed, we well-order I and compare a = (mi)with b = (ni)by saying a < b if mi < ni for the smallest i at which mi # n,.(This is merely the lexicographic-ordering.) In particular, every finitely generated torsion-free abelian group is ordered.
To study nonabelian groups it is convenient to find a more algebraic description of order. Using multiplicative notation we say an arbitrary group G is partially ordered by I if (G, I)is a poset in which a < b implies ga < gb and ag < bg for all g in G. As noted after definition 1.2.13,we see that if a c a' and b < b' then aa' < bb'. Consequently G has a submonoid P(G)= { g E G : g 2 1) called the positive cone. The following properties are straightforward, writing P-' for {a-':a E P } : (i) P n P-' = {I}. (ii) P is a normal submonoid of G (since for any a E P and g in G we have gag-'
> g1g-'
= 1).
(iii) If Iis a total order than P u P-' = G . Fortunately these above properties characterize positive cones.
Lemma 1.2.32: If P is a normal submonoid of G with P n P-' = { 1) then G has a partial order I for which P = P(G). If, moreover, P v P-' = G then G is ordered by <. Proofi Define I on G by a I b if a - ' b E P. If a Ib and b I c then a-'c = (a-'b)(b-'c) E P, so I is a preorder; moreover, if a I b and b Ia (so that a-'b = (b-'a)-' E P - ' ) we a-'b E PnP-' = (1) so that a = b, proving (G, I) is a poset. To see G is partially ordered by c we first note for a # b that ga # gb and ag # bg by cancellation. Now for a < b and arbitrary g in G we have ga < gb (since (ga)-'gb = a-'b E P) and ag < bg (since (ag)-'bg = g-'(a-'b)g E P ) . If, moreover, P u P-' = G then Iis a total order since either a-' b E P or a-'b E P-', in which case b-'a E P. Q.E.D.
$1.3 Free Modules and Rings
53
Using this algebraic criterion we are ready to consider certain classes of groups. Define the group commutator (y, h) = g-’h-’gh and, writing ( A ,B) for the subgroup of G generated by {(a,b): a E A, b E B}, define the lower central series G = Go 3 G, 3 G2 3 *.. where Gi = ( G , G i - , ) for i 2 1. Note that Gi- , / G i is a central subgroup of GIG,. Proposition 1.2.33: Suppose G = Go 3 G, 3 G2 3 is the lower central series for G. If each Gi-,/Gi is ordered, and if n,20Gi = ( 1 ) then G is ordered.
Proo) Let pi = {g E Gi:ij> in Gi = Gi/Gi+,}. Then en P;’ = 0 and P , u P f ’ = G i - G i + , . Let P = ( u c ) u { l } . Then P i s a normal monoid w i t h P n P - ’ = ( 1 ) a n d P u P - 1 = ( U i ( G i - G i + , ) ) u { l } =G,provingGis Q.E.D. ordered. There are interesting related results on ordered groups which we shall present in the exercises. Proposition 1.2.33 cannot be extended to solvable groups, by exercise 5.
51.3 Free Modules and Rings In this section, we shall study and construct free modules and rings. There is an extensive theory in Cohn [85B], which largely lies outside the scope of this book; in this section we cover only the most basic properties, leaving free products for 51.4 and 51.9. Intuitively, “free” objects in a category are constructed in the most general possible way, without any extra conditions, and thus “lie” above all other objects. It turns out that we have already constructed the free R-modules in 51.2, and we shall also use their properties to understand the nature of “free.” Then we construct the free monoid and the free ring; at the end we construct the free group, which is needed in several ring-theoretic constructions. Every free module has a base, and technical problems arise concerning the uniqueness of the cardinality of a base. This matter is addressed in the concept of IBN (invariant base number) in definition 1.3.24ff. In order to verify the theories we give a category-theoretic definition of free, which is put into context towards the end of the section when we introduce “universals,” which are applicable in a wide variety of situations. However, the reader should already be apprised that the definition is ill-suited to category theory, and in $2.8 we shall discuss a related notion, “projective,” which fits much better into a categoric approach.
Constructionsof Rings
54
Free Modules Since the module freely generated by a set has no superfluous conditions in its construction, we examine it to discover the nature of “free.” Definition 1.3.1: A base for an R-module M is a subset S of M such that r,s where the r, E R are almost all 0 and every element of M has the form are uniquely determined. A module with a base is called free. In order to prove S is a base it is enough to show S spans M and is independent, for then if r,s = rls then c ( r s - r:)s = 0 implying each r, - r: = 0 so r, = ri.
xseS c
Remark 1.3.2: S is a base of the module RS freely generated by S, as noted in definition 1.2.1; thus RS is free.
Proposition 1.3.3: Suppose a module F has a base S = {si:i E I}. For any module M und any { x i : i E I} c M there is a unique map f: F + M such that f s i = xi for all i in I.
c
c
Proofi Any such f must satisfy f( risi) = rif s i = rixi,proving uniqueness, and we can in fact use this formula to define f since S is a base. Clearly f is a map. Q.E.D. Example 1.3.4: (i) MJR) is free in R-Aud with base {eij:1 5 i,j In ) (which has n z elements). (ii) The ring of triangular matrices over R is free in R-.Mud with base {eij:1 I i I j I n } which has n(n + 1)/2 elements. (iii) R[1] is a free R-module, having base { 1’:i E N }. (iv) C{X}is a free C-module having base the words in the alphabet X. (v) R(”)is the free R-module. The concept of dependence of elements can be generalized t o modules, enabling us to understand free modules in terms of “direct sums.”
Independent Modules and Direct Sums Definition 1.3.5: A set { M i :i E I} of submodules of a module M is independent if Mi n Mj = 0 for each i in I; dependent means not independent. If
cj+i
$1.3 Free Modules and Rings
55
cisl
M = Mi and { M i :i E I } is independent we say M is an (internal) direct sum of the Mi and write M = @ M i . If each Mi z M we write M"' for
Oi
E I Mi. Before continuing we want to relate this definition to the prior definition of independence of a subset of a module (40.3).
Proposition 1.3.6: Suppose M E R-Jtod and S c M . S is independent ifl { R s :s E S > is an independent set of submodules with Ann, s = 0 for each s in S . Pro08 (a) If 0 # rs E Rs n x s , + s R s 'then we can write rs in two different ways in terms of the independent set S, which is impossible, so Rs n Rs' = 0. Likewise if r E Ann, s then rs = 0 = 0s so r = 0. (*) If c f i n i , e r = s ~0 then for each s we have r 2 s = - c s , + s r s , so rss E Q.E.D. Rs n f s Rs' = 0; hence rs E Ann s = 0, as desired.
cs,zs
c,,
oSEs
Corollary 1.3.7: If S is a base of M then M = Rs and R z Rs for each s via the map r + rs. In other words, every free module is a direct sum of copies of R . Remark 1.3.8: (i) Every dependent set of submodules can be shrunk to a finite dependent set. (Indeed if Mi n Mj # 0 then 0 # x i = xju for suitable t in N, x i in M i , and x j , in Mju;then {Mi, M j ,,..., M j , } are dependent.) It follows at once that {independent sets of submodules of M} is inductive, and thus each independent set of submodules can be expanded to a maximal independent set. In the same way, each independent set of elements of M is contained in a maximal independent set. (ii) If Y = { M i :i E I } is independent and N c M with N n Mi = 0 then Y u {N} is independent. (We must show Mi n ( N + Mj) = 0 for each i. Taking x j in Mj and y in N suppose xi = y + c j + i x j .Then y = xi - I j t i x j E N " I i E Mi, = 0, implying xi = E j t i x j E M i n C j + iMj = 0, as desired.)
cj+i
Xiel
cj+i
Modules Over Division Rings Are Free Modules over division rings are often called uector spaces since many techniques from linear algebra are applicable, largely because of the following easy results.
Constructiom of Rings
56
Remark 1.3.9: Suppose D is a division ring and M E D-Aod. Then Ann,x = 0 for all x in M. Moreover, if S c M and dx E DS E M with 0 # d E D then x = d - ’ d x E DS. Put another way, if x 4 DS then Dx n DS = 0.
Proposition 1.3.10: Every module M over a division ringD is free. In fact, the following statements are equivalent for any subset S of M : (i) S is a minimal generating set (of M ) . (ii) S is a maximal independent set. (iii) S is a base. Proofi Remark 1.3.8 (i) shows maximal independent sets exist, so if we can establish the equivalence we shall see they are all bases (so M is free).
(i)*(ii) If C d i s i = 0 with some di # 0 then siE cj+iDpj, contrary to minimality. (ii) => (i) For each x in M we have Dx n DS # 0, so x E DS, proving DS = M. Hence S spans M and is minimal since if S’cS then S q ! DS’ (by independence). (i), (ii) * (iii) is true by definition. (iii) =-(ii) S is independent, by definition. For any x E S we still have x E DS, so S u {x} is dependent. Q.E.D.
Free Objects We draw on proposition 1.3.3 to obtain a general definition of free, for any category whose objects are sets (perhaps with extra structure), i.e. there is a forgetful functor from % to %t.We shall call such a category concrete. Definition 1.3.11: A free object ( F ; X ) of % is an object F with a subset X = {Xi: i E I } whereby for every A in 0 6 % and every { a i :i E I } E A there is a unique morphism f:F A such that f X i = ai for all i in I. This definition is not suited to category theory since it deals with elements; in fact we shall see below that “freeness” is itself built from a functor G: % -,%d and is much more cumbersome (in terms of general category theory) than other notions (such as “projectivity”). Nonetheless, many of our results will focus on free objects, largely because of the following result.
Proposition 1.3.12: Suppose %? and I are given. (i) Any two free objects are canonically isomorphic.
$1.3 Free Modules and Rings
57
(ii) If a free object ( F ; X ) exists then every object of cardinality I II( is an image of F under a suitable morphism. Proofi
(i) Suppose (F;X)and (F’; X’)are free, where X’ = {Xi:i E I } . By definition, we have morphisms f:F + F‘ and g: F‘+ F such that f X i = Xi and gXi = X i for all i. Hence gf: F -+ F and fg: F’ + F’ with g f X i = X i and f g X ; = Xi for all i. But we already have 1,: F F with lFXi = X i so by the uniqueness in definition 1.3.11, gf = 1,; likewise, fg = l,,, proving f is an isomorphism. (ii) Index the elements of an arbitrary object A of cardinality 5 11) as {ai:i E I } so that all elements of A appear (possibly with duplication if IAl < 111); then defining f:F + A such that each f X i = a, we have fF = A Q.E.D. --f
We shall now show free objects exist in those other categories of interest to us.
Free Rings and Algebras If C is a commutative ring then C{X}is the free C-algebra. This is intuitively clear since for any C-algebra R and { ri :i E I } E R we can “specialize” X i H r,. We shall prove this fact from the previous module results, aided by the following tool: Dejinition 1.3.13: Suppose cp: R + T is a ring homomorphism, and M E R - A u d and N E T-Aod. An additive group homomorphism f: M + N is called a cp-map if f (rx)= (cpr)fxfor all r in R, x in M. cp-maps are also called semilinear maps. Remark 1.3.14: Any cp-map f : M -,N becomes a module map in the usual sense when N is viewed as R-module by remark 0.0.9 (defining ry as (cpr)yfor y in N). Thus for M free with base S, any function f: S --+ N can be extended uniquely to a cp-map f:M -+ N by putting f( r,s) = x(cpr,)fs.
1
This remark enables us to apply our module-theoretic results to rings, taking M = RCS].
Proposition 1.3.15: Suppose cp: R + T is a homomorphism of C-algebras. Also suppose S is a monoid and f: S -+ T is a monoid homomorphism such that cpR
Conshpetions of Rings
58
centralizes fS. Then there is a unique algebra homomorphism f: R [ S ] + T whose restriction to R is cp and whose restriction to S is f. Proofi By remark 1.3.14 we need check merely that f is a monoid homomorphism, which is true since f((rlsl)(r2s2))=f(rlr2s1s2)=cp(rlr2)f (s1s2)= Q.E.D. ~ 1 ~ 2 f s i f= s z( ~ i f s 1 ) ( ~ 2 f s=2 f) (risi)f(r2~2). If cpR c Z ( T )then the condition of the proposition is satisfied; in particular, this is the case if R = C is commutative, T E C-dt’, and cp: C + T is the canonical homomorphism c cl. This special case is worth restating explicitly: --f
Proposition 1.3.16: Suppose S is a monoid and T is a C-algebra. Any monoid homomorphism f: S -+ T can be extended uniquely t o an algebra homomorphism 3 C [ S ] -+ T. In view of proposition 1.3.16, we expect to find a free ring if we can locate the free monoid.
Proposition 1.3.17: Suppose X is an arbitrary set, and S is the word monoid built from X (i.e., X is the alphabet). Then ( S ; X ) is free in Moa. Moreover, (C{X};X) is free in C-dLy for any commutative ringC, and (Z{X};X) is free in W i ~ g . Proofi For any monoid V and {xi:i E I} z V, define f:S + V by f (Xi,- - * Xi,)= xi, xi,. Clearly, f is a morphism of monoids and is unique such that f X i = xi, proving (S;X) is free. Thus ( C { X } ; X )is free by proposition 1.3.16. Hence (Z{X};X) is free in Z - d f e , which is isomorphic to W i ~ g . Q.E.D. Corollary 1.3.18: Any C-algebra is isomorphic to C { X } / A for suitable A 4 C { X } ,where X is a suitably large set of noncommuting indeterminates.
One can use corollary 1.3.18 as one cornerstone of the study of C-algebras. Given any subset S of R, we write (S) for the ideal RSR, called the ideal generated by S; one views R/(S) as the ring formed by sending the elements of S to 0. Let us examine the situation more closely for R = C{X}. Writing xi for the image of X i in C { X } / ( S ) (where X = { X i : iE I } ) , we can write C{X}/(S) as C{xi: i E I } , or C{x} for short; and we say (xi:i E I } generate C { x }as an algebra (since no proper subalgebra of C{x} contains all the xi).
81.3 Free Modules and Rings
59
On the other hand, any polynomial f in C { X } involves only a finite number of indeterminates X i , say Xi,, . . .,Xi,; we denote the image of f in C { X } / ( S ) as f ( x i , ,. . . , x i , ) , which intuitively should be thought of as “substituting” x i for Xi. Then f ( X i , , . .. ,Xi,) E ( S ) iff f ( x i , ,. . . ,x i t ) = 0, so the elements of S can be thought of as relations among the x i . Thus defining a C-algebra as C { X } / ( S )is sometimes called “defining an algebra by generators and relations.” Such a construction is often very useful but involves the difficulty that perhaps ( S ) is much larger than one might expect and perhaps even ( S ) = C { X } . Sometimes this difficulty can be overcome by building a concrete example to go along with the abstract definition.
Free Commutative Ring When working with commutative rings one would like to have an analogue of corollary 1.3.18 using the free commutative ring, which we shall now identify.
Proposition 1.3.19: (i) The commutative word monoid is free in the category of commutative monoids. (ii) For any ring homomorphism f: R -, R’ and any set { z i :i E I } s Z(R’), there is a unique homomorphism f: R[1] -, R’ whose restriction to R is f , such that f;li = zi for all i. (Here 1 = {Ai: i E I } is a set of commuting indeterminates over R.)
Proofi (i) Let M be the commutative word monoid on the set I . Given a map f: I -,S where S is a commutative monoid, we can extend f to a monoid homomorphism f: M + Sgiven byf(i, ... i,)=(fi,)...f(i,), where i , <...
Corollary 1.3.20: For any set 1 of commuting indeterminates, (C[A],1)is free in the category of commutative C-algebras, and ( Z [ A ] ,A) is free in the category of commutative rings. Any commutative C-algebra has the form C [ A ] / Awhere A a C[A]. Let us exemplify these results. Using well-known properties of polynomials in one indeterminate over Z, one can easily verify Z[i] x Z[A]/(A2 + l),
60
Constructions of Rings
+ +
Z[w] x Z [ I ] / ( 1 2 I 1) for w a primitive cube root of 1, and Z[&] x Z [ I ] / ( A 2 - n) when n is not a square. The ease in handling commutative examples such as these is largely lacking in the noncommutative case.
Diagrams Before continuing our study of modules, let us develop some more abstract theory to enable us to see the border picture. One important new technique is the use of commutative diagrams, cf., Jacobson [80B, p. lo]. A dotted line is drawn to show there exists a morphism such that the ensuing diagram obtained by drawing a solid line is commutative. Thus A-B
f
means there is h: A -,C such that h = gf; we then say h completes this diagram. This notion enables us to put free objects in a general categorical context.
Universals Dejinition 1.3.21: Suppose we are given G:W -, 9 a functor of categories and X E O b ( 9 ) .A universal from X to G is an object U of %? together with a morphism u: X + G U such that for every A in Ob(%?)and every morphism f:X -+ GA there is a unique morphism f I : U + A such that f = (Gf’)u, i.e.,
Gf’completes the diagram X L G U
I
/
In case this definition appears dry, let us whet the reader’s appetite by observing that intuitively U is the “free” object of %? with respect to being built up from X. In particular we have
Example 1.3.22: Let %? be a concrete category and 9 = %Pet. For any set X the free object ( F ; X ) of W is the universal from X to the forgetful functor (where u is the identification of X as a subset of F). Thus the next result generalizes proposition 1.3.12(i).
$1.3 Free Modules and Rings
61
Proposition 1.3.23: (“Abstract nonsense”) Any two universals from X to G are isomorphic. Proof;. Let { U ;u: X + G U } and { U’;u’: X + G U ’ } be two universals. Applying the definition with A = U’ and f = u’ we get f’: U + U‘with u’ = ( G f ’ ) u ; interchanging the roles of U and U’ we get f”: U‘ + U with u = (Gf”)u’. Hence u = (Gf”)(Gf’)u = G ( f ” f ‘ ) u . On the other hand, taking the universal U and A = U and f = u, we get a unique f“’:U + U with u = (Gf”’)u; 1, and f’lf’ each fill this role so 1, = f’lf’. Analogously, l u r = f’f”,proving f’:U + U‘ is an isomorphism. Q.E.D. The reader should realize that there may not exist a universal from X to G . However, universals exist in many settings and help organize seemingly disparate constructions. Note: It is often easier to recognize that an object is a universal than to pinpoint the corresponding functor and category. (This is certainly the case with direct products, to be defined shortly.) Hence it is customary to write a diagram resembling that of definition 1.3.21 and to say a given construction “solves a universal mapping problem” and is therefore unique in this respect (up to isomorphism). It is a good exercise to write down each universal in precise terms. Aside: For any functor G :W + 9 the composite functor Hom(X,-): % + 92 is representable with representative (U,u) iff (U,u) is a universal from X to G , by Yoneda’s lemma, cf., Jacobson [80B, p. 46-47]. Consequently, the term “universal” has faded from category theory literature in light of the advance of representable functors. However, the concept is useful for us, since we shall come across many important examples of universals.
Invariant Base Number We shall find many occasions to deal with free modules and will want a workable “dimension,” usually called rank. The obvious definition of rank is the cardinality of a base, which leads us to the fundamental question of whether different bases can have different cardinalities. Definition 1.3.24: R has invariant base number (abbreviated I B N ) if R‘”) z R(”) implies m = n.
62
Constructiom of Rings
Lemma 1.3.25: R('")x R(") fi there is an rn x n matrix A and an n x rn matrix B, each with entries in R, such that A B = 1, (the identity m x m matrix) and B A = 1,. Proof:
Let x l , . ..,x, be a base of R('") and let y,,
. . .,y,
be a base of R(").
=cy=
(*) Let $: R(m)+R(")be an isomorphism. Put $xi a,,y, _ _ and $-'yi = bjkxk, SO X i = $-'$Xi = aijyj = xaij$-iyj = xj,kaijbjkXk for each i. from which one readilv concludes A B = 1;,.... likewise BA = 1.. aijyj, and ;': R(")--t (-)There are maps $: R'"'4 R'")given by $xi = R ( I )given by $'yj = bjkxk.Then $'$xi = xi for each i, so $'$ = 1 (since
$-'xy=
cy=
the two maps agree on a base). Likewise $'$ = 1 so $ and 1(/' are inverses, proving $ is an isomorphism. Q.E.D. Proposition 1.3.26: R lacks IBN iff there are m # n with an m x n matrix A and an n x m matrix B having entries in R such that A B = 1, and B A = 1,. Corollary 1.3.27:
(i) If R has IBN then M,(R) has IBN. (ii) If R lacks IBN and $: R 4T is a ring homomorphism then T lacks IBN. (iii) Every homomorphic image of a ring lacking IBN also lacks IBN. Proof: (i) Follows at once. (ii) If A , B are matrices over R with A B = 1, and BA = 1, for m # n then applying $I gives the corresponding equations for matrices over T. (iii) Is a special case of (ii). Q.E.D. This last result is very powerful. For example, any field has IBN, by proposition 1.3.26, so any nontrivial commutative ring C has IBN (since C/M is a field for any maximal ideal M). We can refine the method still further.
Definition 1.3.28: A trace map on a ring R is an additive group homomorphism t: R --t H for an abelian group H , satisfying t(rlr2) = t(r2rl) for all ri in R. Proposition 1.3.29: If R has a trace map t such that t l has no Z-torsion then R has IBN.
41.3 Free Modules and Rings
63
Proof: Suppose A = (aij) is an m x n matrix and B = (b,) is an n x m matrix with AB = 1, and BA = In,and suppose e = tl. Then
implying m = n as desired.
Q.E.D.
We shall see examples of trace maps later, particularly in the discussion of group rings. There are, indeed, rings lacking IBN (cf., example 1.3.33 below), and thus their simple homomorphic images lack IBN. O n the other hand, theorem 0.3.2 implies division rings do have IBN, and virtually every class of rings we shall study (Artinian rings, Noetherian rings, PI-rings, etc.) will be seen to have IBN.
Weakly Finite Rings There is a slightly stronger property than IBN which often is easier to verify.
Definition 1.3.30: A ring R is weakly n-finite if for all matrices A, B in M,(R) one has AB = 1 implies BA = 1. R is weakly finite (also called Dedekind finite or uon Neumann finite) if R is weakly n-finite for all n. Remark 1.3.31: If R is weakly finite then R has IBN, by proposition 1.3.26.
In the course of the text we shall see that many classes of rings are weakly finite, and thus satisfy IBN. At the moment let us note that IBN has the following connection to idempotents; further results will be collected in Chapter 3.
Proposition 1.3.32: (Jacobson).If R is not weakly 1-finite then R has an infinite set of orthogonal idempotents. More generally, if ab = 1 and ba # 1 in R then putting eij = b'( 1 - b a ) d we have eijeuu= djueiufor all i, j , u, u. proof: Note ajbj = ai-l(ab)bi-l = &1bj-1 = ... = 1 for all j . Also a(1 - ba) = 0 = (1 - ba)b. Thus .(1 - ba)ajb"(l - ba) = 0 unless j = u, in which case (1 - ba)ajhj(1 - ba) = (1 - ba)' = 1 - ba, so eijeuu= djubi(1 - ba)a"= djueiv.
Q.E.D.
Constructionsof
64
Rings
Example 1.3.33: Let us search for an example of a ring R lacking IBN. Of course it cannot be weakly finite, and proposition 1.3.32 leads us to look for some infinite analog of matrices. Let V be a countably infinite dimensional vector space over a field F, and let R = Hom,(V, V). ?hen R is a monoid under composition of functions and becomes a ring when we define f + g by (f + g)u = fu + go for all u in V. (This construction will be studied shortly in great detail.) Let B = { u i : i E N} be a base of V over F. We claim R is not weakly 1-finite. Indeed, since every map is determined by its action on B we need only find a 1: 1 function f: B + B which is not onto (for then f will have a left inverse which is not a right inverse). Define f to be the right shift,i.e., f u i = ui+ for each i. Actually R lacks IBN; in fact R") x R'" in R-.Mud, for all i , j in N-{O}. To prove this we need only show R x R"). Define CD: R + R(') by defining Of = (f,, fz)for each f in R, where the fi are given according to the following action on B:
,
f,ui = f u z i
and
fzui = f u z i -
for each i in N.
If (f,,f2)= (0,O)then f u i = 0 for all i, implying f = 0, so 0 is 1:1; conversely CD is onto since given (f,, fz)we recover f by putting f u z i = flui and f u z i - = f lui for all i. Clearly CD is a module map and is thus the desired isomorphism.
,
This example is also important in other aspects of ring theory.
A Supplement: The Free Group One other category which has free objects is %fi. We shall describe the free group in some detail because by ordering it we can utilize theorem 1.2.24' to embed the free algebra explicitly into a division ring; this construction is used in the class of Jategaonkar rings built in 62.1. Let us start with an explicit construction. Construction Z.334: (The free group) Take the free monoid M on the alphabet { X i , i E I}.We say two words h , , h z are equivalent if we can obtain h, from h, by successive steps of inserting or cancelling XiK or k;Xi.For example, X,YzXzY3 is equivalent to XlX4Y,Y3since first we can cancel YzX,and then insert X4Y4.It is easy to see the equivalence classes form a monoid (under the induced multiplication), which, in fact, is a group G since inverses are obtained by reversing the order and switching X i and k;. For example, the inverse of the class of X,X3X,Y5X4is the class of Y4X5Y6Y,Y,. Abusing notation slightly we shall write Xifor the equivalence class of Xi,
x:
81.3 Free Modules and Rings
65
x.
XI:' for the equivalence class of (This could be defined most concisely by means of universal algebra applied to monoids, cf., Jacobson [85B,p. 67)]. It is easy to see that G contains canonically the free monoid on {Xi: i EI}, and more information can be obtained by appealing to the basic structure theory of monoids: Proposition 1.3.35:
( G ;X ) is free in Ye#.
Proof;. Suppose H is any group with {hi:i E I } c H.Take the free monoid M on the alphabet {Xi, x : i E I } and define a monoid homomorphism f: M + H by f X i = hi and f Y;. = h i ' . Also define II/:M + G by $Xi = Xi and II/x = X;'. Clearly II/ is onto and ker II/ c kerf, so viewing Mfker II/ as G we have an induced monoid homomorphismf: Mfker II/ + H given by f X i = hi and fX;' = h;'. Since G and H are groups, we see f is a group homomorphism, and the uniqueness of f i s clear. Q.E.D.
Remark 2.3.36: Any element of the free group can be written uniquely in the form g;'...gY for suitable t, where the n, are nonzero integers and the gi E X with any two consecutive gi distinct. An analogous argument enables us to construct the free abelian group, which can be described explicitly as (Z, +)('); when I = { 1,. ..,n } we could rewrite this as the set of words g:' ...g>, ti E Z, where the group operation consists of adding the powers. Remark 2.3.37: Write G' for the commutator subgroup of G; if G is the free group then G/G'is the free abelian group since the condition of definition 1.3.11 clearly is satisfied. Our further study of the free group relies on two celebrated results of group theory whose proofs both involve an interesting interplay with Z{X}, thereby illustrating ring theoretic applications to group theory.
Theorem 1.3.38: Let G denote the free group on {Xi: i E I } written in multiplicative notation and write G =I G, 3 G2 =I for its lower central series.
n
G, = { l } (2) Magnus-Witt Theorem G,-,/G,is a free Abelian group, for each n. (1) Magnus' ITheorem
The proof of these theorems is rather lengthy, but involves an interesting interplay with the free ring Z{X},and is given in Appendix A. .
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66
Theorem 1.3.39: The free group can be ordered. Proof: Apply proposition 1.2.33 to theorem 1.3.38.
Q.E.D.
Theorem 1.3.40: I f G is the free group ordered as in theorem 1.3.39 and D is any division ring, then D((G)) is a division ring canonically containing D W . Proofi
By theorem 1.2.24'.
Q.E.D.
Corollary 1.3.41: The free algebra F { X } over a field F can be embedded in a division ring. There are other embeddings, as we shall see, but this embedding can be made explicit, as the reader can check by going through the proof.
51.4 Products and Sums The main theme of this section is direct products and direct sums, two of the most fundamental constructions for rings and modules. This is viewed best in a categorical setting, and exact sequences and split exact sequences are introduced here, leading at the end of this section to an excursion into the more abstract realm of abelian categories. On the other hand, direct products can be generalized using filters; the reduced product (and, more specifically, the ultraproduct) has turned out to be one of the more useful constructions in ring theory, because it can combine properties of disparate examples.
Direct Products and Direct Sums The direct product construction is very straightforward and probably familiar to the reader. Given a set of objects { A i :i E I } in a category %? we can often view the Cartesian product n i s , A i as an object of %,' where the relevant properties are defined componentwise. Rather than formalize this idea now, we present now the two examples of importance to us and defer a formal discussion until the treatment of reduced products. Definition 1.4.1:
n,,,
(i) The direct product of a set { R i : iE I} of rings, denoted R i , is the Cartesian product which is a ring endowed with componentwise operations (ri) + (ri) = (ri + ri) and (ri)(ri)= (rir:)where r i ,r; E Ri for all i.
67
51.4 Products and Sums
(ii) The direct product of a set {Mi:i E I } of R-modulcs, denoted nit, Mi, is the Cartesian product which is in R - d u d when endowed with componentwise operations (xi) + (xi) = (xi + x f ) and r(xi) = (rxi) where r E R and x i , x fE Mi for all i. The same sort of definition "works" in d&, '3+, d u ~and , 9 u a as is easily seen. The following observation is crucial for understanding how direct products work. Remark 1.4.2: Suppose {Ai: i E I } are objects in a category for which direct products have been defined (e.g., g i n g , R - d u d , etc.). For each j in I there Ai -+ Aj given by reading the j-th is an onto projection morphism nj: component i.e., nj(ai)= aj for each (ai) in A i . Moreover, given any object A and morphisms f i : A -+ Ai for each i in I, one has a unique morphism f : A -+ A i such that nif = 1;. for all i in I. (Indeed, such f can be defined by the rule fa = (La), the element whose i-th component is ha, and one sees easily that any such f must satisfy this rule and thus is uniquely determined.)
n,,,
n
n
We turn to the (external) direct sum of modules. The construction involves Mi in R - A u d by taking the direct product, as follows: Define pj:Mj p j x to have all components 0 except the j-th component, which is x. (Here x E Mj.) Then n j p j = l,, and nipi = 0 for all i # j. -+
ni,,
Definition 1.4.3: The direct sum @Mi of a set {Mi:i E I } of R-modules is the sum of the piMi in Mi. In other words, @Mi = {(xi)E Mi: almost all xi are 0). (Thus the direct sum and direct product of { M i :i E I } are the same iff I is finite.)
n
n
Remark 1.4.4: The notation @Mi is justified since @Mi is the internal direct sum of the piMi. (Indeed to see the piMi are independent, note that pix = pini(pix)for any x in Mi, so piMin
C pjMi = pini
j#i
j t i
since the nipj = 0.)
Exact Sequences To understand direct sums better we shall introduce sequences, one of the f' basic tools of category theory. A sequence of maps . . . M' -+ M 9 M" + . . . is exact at M if fM' = kerg; an exact sequence is a sequence exact at each of
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68
f
f
its modules. Thus 0 + M + N is exact iff f is monk, and M + N + 0 is exact iff f is epic. It turns out that exact sequences contain precisely the information one needs for proofs. The length of an exact sequence is the number of arrows. Remurk 1.4.5: The following information characterizes exact sequences of length m which start and end with 0. (i) m = 2. O + M + O is exact iff M x 0. (ii) m = 3. 0 + M 4 N + 0 is exact iff f: M + N is an isomorphism. (iii) m = 4. 0 + K % M N -,0 is exact iff g is monic, N = f M , and N x M / K (using g to view K as a submodule of M, and using f to induce the isomorphism). This is called a short exact sequence. (iv) m = 5. 0 - K --t L 4 M + N +Oisexact iff K x kerf and N x M/fL (using the given maps for the identifications).
We should observe that the information becomes weaker as the length increases. On the other hand, any exact sequence starting and ending with zeroes can be made longer by adding extra zeroes and arrows denoting the zero map, enabling us at times to apply results about exact sequences of longer length.
Split Exact Sequences Viewing a commutative diagram as a traffic flow, where the arrows denote one-way streets, one might sometimes want to go the “wrong way” on a oneway street to arrive at a given desired object. More formally, given g: M + N one might want to find h: N + M such that gh = 1, and/or hg = ly. These notions can be formulated most fruitfully for short exact sequences.
fM Definition 1.4.6: An exact sequence 0 + K + i s a m a p i : N + M s u c h t h a t g i = 1,.
N
+0
is split if there
oie,
Definition 1.4.7: If M = Mi we say each Mi is a summand of M and Mj is the complement of Mi. In the literature, summands often are called direct summands.
xj+i
Clearly if Mi is a summand of M then M is a direct sum of Mi and its complement. Thus the case of direct sums of two submodules is of special
51.4 Products and Sums
69
significance, and we shall now consider it closely. Note M = MI 0 M, iff MI and M , are summands of M and complements of each other. Remark 1.4.8: If M = M, 0 M , then any map f:MI -+ N can be extended to a map f: M + N by putting f ( x l x,) = f x , for all xi in M i(i.e., fM,= 0).
+
Lemma 1.4.9: Suppose 9: M -+ N and h: N -+ M are maps with gh Then g is epic; likewise h is monic, and M = ker g 0 hN.
=
1,.
Proofi If f g = f ' g then f = fl, = f g h = f ' g h = f', proving g is epic; likewise h is monic. For any x in M, g ( x - hgx) = gx - ghgx = gx - gx = 0, implying x - hgx E kerg, so M = kerg + hN. On the other hand, if x E (ker 9 ) n hN then writing x = hy we have 0 = gx = ghy = y, implying x = 0. Q.E.D.
Proposition 1.4.10: The following assertions are equivalent for an exact f M 3 N + 0: sequence 0 -+ K -+ (i) (ii) (iii) (iv)
The exact sequence is split. There is a map p : M + K with pf = 1,. fK is a summand of M. M x K 0 N ; in fact M = fK 0 iN for suitable i: N
+M
with gi = 1,.
Proof: (i) 3 (iv). By lemma 1.4.9, taking i: N + M with gi = l,, we have i monic and M = ker g 0 iN. But ker g = fK x K and iN x N. (ii) 3 (iv) is analogous, and (iv) (iii) is a fortiori. So it suffices to assume (iii), and prove (i) and (ii). Write M = fK 0 M'. Since ker g = fK,the restriction of g to M' is a map M' + N which is monic as well as epic and thus has an inverse i: N + M', yielding (i). Likewise, f induces an isomorphism K -+ fK with some inverse p : f K -+ K which extends by remark 1.4.8 to a map p: M + K Q.E.D. with pf = 1, yielding (ii).
In view of this result, we introduce the following terminology: A monic -+ N is split if there exists h: N + M with hf = 1,; an epic 9 : M -+ N is split if there exists h: N + M with gh = 1,.
f:M
Reduced Products We can generalize direct products by using filters. Since this method generates interesting examples and also provides a technique for shortening proofs, we shall examine it in some detail.
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Definition 1.4.11: Suppose 9 is a given filter of the power set Y(I),and we are given sets { A , :i E I}. The Jiltered product or reduced product of the A , with respect to % written H A i / % is the set of equivalence classes of the Cartesian product A , under the equivalence given by (a,) (a:) iff {i E I: a, = a ; } E 3? (This is seen to be an equivalence as a consequence of the definition of filter.) To simplify notation, we still write (a,) when technically we mean the equivalence class of (ai).
-
n
Note that the reduced product becomes the direct product when 9 is the trivial filter {I}. Of course, we want the reduced product to inherit the algebraic structure of the A , , as illustrated in the following straightforward observation. Remark 1.4.12:
fl
(i) Suppose the A , each are rings. Then A = A i / 9 is a ring under componentwise operations, i.e., 0 = (0), 1 = (I), (ai)(a:)= (a,a:), and (a,) - (a;) = (a,- a;);these are well-defined since filters are closed under finite intersections. (ii) If, moreover, Mi E Ai-Aod then M i ) / 9 E A - A u d by defining scalar multiplication @,)(xi)= @,xi).
(n
The verifications are easy and are examples of the following general considerations: Suppose in general each A , lies in a concrete category W. We want to see when the reduced product A = A , / 9 remains in W. It is convenient to assume W is axiomatizable in a first-order language 9, for then at least we can define a structure in 9 having A as the underlying set, as follows:
n
Any constant symbol is assigned to (ai) in A, where ai is the assignment in A , : Any n-ary relative symbol R is assigned to the relation on A defined by saying R((x,),( y , ) ,(z,),.. .) holds iff {i:R(xi,y i , z i , .. .) holds in A , } E 9; Any n-ary function symbol f is assigned to f:A(")+ A given by f ( ( x i ) , ( y,),(z,), . ..) = ( f(xi, y , , z,, . ..))). Again these definitions are well-defined because filters are closed under finite intersections. Proposition 1.4.13: Suppose cp is an elementary sentence whose only connecA , ) / S fi {i: cp holds in A i } E S tive is A . cp holds in
(n
Pro08 By definition the proposition holds for atomic formulas, so we argue by induction on the rank of cp (given in the definition of elementary sentences):
g1.4 Products and Sums
71
Case I. cp = cp, A cp,. cp holds in A iff cpl and cp, each holds in A , iff (by iff {i: cpl and induction) { i : cpl holds in A i } E B and { i : cp, holds in A i } E 9, cpz holds in A i } E 9 ? Case 11. cp = (3x)q1(x). cp holds in A iff cp,(s) holds for some s = (si) iff (inductively) { i :cpl(si)holds for some si}E E iff { i : ( 3 ~ ) 9 ~holds ( x ) in A i } E 97 Case 111. cp = (Vx)cp,(x). If { i : cp holds in A i } E B then cp holds in A, proved as in Case 11. Conversely, suppose cp holds in A and let J = { i : cp holds in Ai} and J’ = I - J . For each i in J ’ there is si in A i such that cpI(si)does not hold. Taking si arbitrarily in J and s = (sJ, nevertheless, cpl(s) holds; so letting J ” = { i : ‘pI(si)holds in A i ) we have J” E 9 by induction. Since J ” n J’ = 0 by choice of s we conclude J 2 J ” E 9. Q.E.D.
Since the axioms for rings and modules are atomic sentences (suitably quantified) we see that any reduced product of rings (resp. R-modules) is a ring (resp. R-module). Nevertheless, certain sentences do not carry over. For example, the direct product of domains is not a domain. The simplest sentence which does not carry over is the following:
Example 1.4.14: Let R , = 2/22 and R , = 2/32, Then R , and R , each satisfies the sentence (Vx)((2x = 0) v (3x = 0)) which nevertheless fails in R, x R,. A general class of sentences preserved under reduced products is given in exercise 6. Filters and reduced products can be described in purely ringtheoretic terms (cf., exercises 9 through 12) and this interplay enchances both ring theory and the study of filters.
Ultraproducts This technique becomes powerful when we restrict the particular filter under consideration. Recall that any filter can be embedded in a maximal filter, called an ultrafilter. A reduced product A i / 9 is called an ultraproduct when 9is an ultrafilter. Thus we know that ultraproducts are also plentiful, although it is usually near impossible to describe a given ultraproduct precisely. Nevertheless, they are important because of the following result.
n
n
Theorem 1.4.15: (ZoS’s theorem) Suppose A = A J Z where each A i is a structure in the same jirst-order language, and 4 is an ultrafilter. A jirst-order sentence cp holds in A iff {i E I :cp holds in A i } E F
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Proof: By induction on the rank of cp. Proposition 1.4.13has disposed of every case except cp = -I cpl, which holds in A iff cpl does not hold, iff { i E I: cpl holds in Ai} 4 9 ( b y induction), iff { i E I : i cp, holds in A i } E 9 (by remark 0.0.17). Q.E.D. Remark 2.4.26: The principal ultrafilters are rather dull. If 9is a principal ultrafilter on 9 ( I ) , generated by I, c I, then remark 0.0.17 implies I, is a suitable singleton {io};thus A i / 9 = Ai, and we have nothing new. However, every nonprincipal ultrafilter is interesting, as we see now.
n
Lemma 1.4.17: filter. Proof:
An ultrafilter 9 is nonprincipal iff 9 contains the cofinite
(e) 9does
not have any finite set, so is nonprincipal.
(*) We prove the contrapositive. If 9 does not contain the cofinite filter
and hence its complement is a finite set in then some cofinite set is not in 9, # I, n J E 9, so I, G J 9. Take a finite set I , minimal in 9If J E 9then by hypothesis, proving I, generates 9. Q.E.D. Proposition 1.4.18: Suppose 9 i s a nonprincipal ultrafilter on I , and cp is an elementary sentence holding in almost all A i . Then cp holds in the ultraproduct Ai/E
n
Proof: {i:cp holds in A i } €9by the lemma, so cp holds in A by t o i ’ theorem. Q.E.D. Example 1.4.19: Suppose Riis a ring of characteristic mi for each i E N where rn, < m, < m3 < If 9 is a nonprincipal ultrafilter on N then R i / 9 has characteristic 0. (Indeed for each m > 0 the statement m 1 # 0 holds in almost all Ri.
n
- . a .
.
Proposition 1.4.20: Suppose cp is a first-order sentence which is true for every ring of characteristic 0 in a given axiomatizable category. Then cp is true for all rings in this category having suitably large characteristic. Proofi Otherwise, there is a sequence rn, < m, < . * in N and rings Ri in this theory having characteristic m i in which cp fails. Then cp fails for their ultraproduct, which has characteristic 0. Q.E.D. This is usually called the “compactness argument” in logic and can be
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41.4 Products and Sums
applied to almost any “finiteness” condition which is not bounded; it shall be refined later on (cf., remark 2.12.37’ and its applications in 88.1 and 88.3). Also of importance is the following tool which often enables us to assume an algebra is finitely generated.
Proposition 1.4.21: Suppose % is a category defined in a first-order language, and % is closed under intersections. Any object A of % can be embedded in an ultraproduct of finitely generated subobjects of A. Proof: Index the finite subsets of A as { S , : i E I } and let Ai be the subobject of A generated by s,,i.e., A, = (){A’ E Ob%:Si s A’ G A } . Now put Ji = { j E I : S, c Aj}. Given i, i‘ in I we have Si u Sir = Sj for suitable j in I ; then j E Ji n J,. so Ji n .Ii. # 0. Thus (4: i E I } is a base for a filter contained in some ultrafilter 9% Put A^ = A,/% and define 9: A + A^ by ga = (ai) where a, = a for all i such that a E A i , and a, = 0 otherwise. (In other words, g is “almost” the diagonal map.) We claim g is the desired embedding. Indeed g is clearly a morphism. To see g is 1:1 suppose a # b in A. Then taking i such that S, is the pair {a,b} we see the i-th components of ga and gb are, Q.E.D. respectively, a and b for each j in Ji. Hence ga # gb.
n
Products and Coproducts We return now to direct products, using remark 1.4.2 to provide the following categorical characterization: Remark 1.4.22: Suppose A,: i E I are rings. Given any ring A and homomorphisms & : A + Ai for each i, there is a unique homomorphism f:A + Ai which simultaneously completes the following diagrams for each i:
n
A
One should bear in mind that the projection morphisms ni go from the direct product to the components. More generally we have A product of objects { A i :i E I } in a category % is an object denoted A i , together with morphisms ni: Ai + Ai such that for any object A and morphisms fi: A + Ai for each i there is a unique morphism f:A + A, with n,f = for each i in I.
n
Definition 1.4.23:
n
n
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74
Thus the direct product (in categories for which it is defined) is an example of a product in a category. To get back to direct sums we dualize the notion of product by reversing the arrows. Definition 1.4.24: A coproduct of objects { A i :i E I } in a category % is an object denoted A i , together with morphisms p i : Ai -+ A i , such that for any object A and morphisms fi: A i + A for each i there is a unique morphism f: Ai -+ A with f p i = fi for each i. In other words, f simultaneously completes the diagrams A
Example 1.4.25: (The direct sum as a coproduct) Notation as in definition 1.4.24, we see that @ M i together with the p j : Mi-,@ M i is the coproduct of the M i . (Indeed for any maps f;:Mi + M we define f: Mi + M by fx = c f i x i x , which makes sense since almost all nix = 0; clearly, f is uniquely determined such that 1;: = f p i for all i.)
0
Remark 1.4.26: We shall now view coproducts as universals, in a way which shall be generalized in exercise 1.8.7. Namely, given a category %? and a set I, we view I trivially as a small category (whose only morphisms are the identities i + i). Then the category %I given in example 0.1.13 can be described as follows: The objects are the functions from I to Ob(%),denoted as I-tuples ( A i )where Ai E Ob(%?);the morphisms $(Ai)-+ (Bi)are I-tuples (fi) where each f ; : A i-,Bi is a morphism. Let U = H A i E %? and u = ( p i ) , the morphism in %’ whose i-th component is the canonical morphism pi:Ai -+ A i . Then U and u define the universal from ( A i )to the “diagonal functor” 6: % %’ where 6 is given by 6 A = (A) and 6f = (f), i.e., each component of 6 A (resp. Sf) is A (resp. f).This displays the coproduct as a universal (if it exists), proving the coproduct is unique; by duality the product also is unique. -+
Surprisingly, coproducts of rings may be trivial. R , = 2/22 and R, = 2/32 have coproduct 0 since any ring R having homomorphisms pi:Ri+ R would satisfy 1 + 1 = pl(l 1) = 0 and 1 1 1 = p2(1 + 1 + 1) = 0, implying 1 = 0, so R = 0. In fact, the element 1 is precisely what ruins coproducts for Wiay; this is one of the few instances in which we might rather
+
+ +
gl.4 Products and Sums
75
have defined rings not to have 1, c.f., exercise 1.5.7. Nevertheless, we can construct a ring ‘‘like’’ a coproduct which is very useful in providing examples. Example 2.4.27: (Direct sums of rings) Suppose { R i :i E I } is an infinite set of rings and define the function p j : R j + Ri by taking p j r to have all components 0 except the j-th component, which is r. The p j satisfy all requirements of ring homomorphism except preserving 1; in fact, p j identifies R j with an ideal f i j = p j R j Q R i . Define the direct sum O R i of the R i to be the subring of Ri generated by all the R i . In other words,
n
n
fl
0Ri = {(m)+ xfinite ri: ri R i and rn E
E
H).
Since njpj = 1 and njpi = 0 for all i # j the direct sum satisfies the following universal property: If A: Ri + Tare ring homomorphisms, then there exists a unique ring homomorphism f:O R i -+ T such that 1; = fpi for all i. Indeed, define f by f((m) homomorphism we note
f((rn)
+ x f i n i , e r i=) m + x & n i r i ; to
check f is a
+ 1ri)((rn’) + 1r:) = f ( ( m m ’ )+ C(mri + m’ri + rir:)) = mm’ + x(mfinir; + mxniri + fiq(rir;)) = (rn + xJ;niri)(m’+ 1 f i q t - i ) = f((m)
+ Cri)f((m‘) + xri).
(The other verifications are instant.) This construction foreshadows the adjunction of 1, to be described in 41.5. See exercise 5 for an application. Despite our initial disappointment, coproducts of algebras are quite useful in several instances. We need an example for guidance. Example 1.4.28: Let X be a set of indeterminates and partition X = U X ( ~ ) where the i merely are superscripts. For any commutative ringC, the free C-algebra C { X } is the coproduct of the C ( X “ ’ } in C - d & where the p i : C { X ( ” } - + C { X }are the canonical inclusion maps. (Since C { X ( ” } are free C-algebras, the identity action on X(’’ produces pi.) To see this, suppose we are given g i : C { X “ ’ } -+ R for a C-algebra R ; we want to show there is a unique g : C { X ) + R satisfying J=gpi for each i. Indeed, g and gi must agree on X“’, so for each indeterminate X y ’ in X“’, we define the action of g on X y ’ by gXy’ = Six:’.These combine to give a set-theoretic map g : X + R which uniquely yields the desired homomorphism g : C { X } -,R .
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Constructions of Rings
Example 2.4.29: For one indeterminate Xi we see C{Xi} is a commutative polynomial algebra in one indeterminate. Thus C{X} = UiE,C{Xi}is a coproduct of free commutative C-algebras. This example merits further study, but first we note an important application. Theorem 1.4.30:
Coproducts exist in C-dl''.
Proofi Given C-algebras { R i :i E I } , write Ri = C{X")}/Ai for suitable disjoint sets of indeterminates X(') and ideals A i of C{X(i)}.Let X = X"), so that C{X} = IJC{X(i)} and let pi:C{X(i)}+ C{X} be the canonical inclusion map (c.f., example 1.4.28). Letting A be the ideal of C{X} generated by the p i A i , we put R = C{X}/A. The composite homomorphism C{X(')}+ C{X} + C{X}/A has kernel which contains A i , so we have an induced homomorphism pi:Ri + R. We claim that R together with the pi is the coproduct of the R i . Suppose we have another C-algebra R' with homomorphisms pi: Ri -+ R'. Then we have the homomorphism vi:C{Xci)}+ R ' given by the composite C{Xci)}+ Ri + R', which yields a unique homomorphism v: C{X} + R' such that vi = vpi for each i. Thus 0 = viAi = v ( p i A i )for each i, implying A C ker v. Hence v induces a homomorphism i? R+R' (since R = C { X } / A ) , and one has pi = Tpi. The uniqueness of T follows from the uniqueness of v, Q.E.D. so we have proved the claim.
u
Like many good results, this theorem raises more questions than it answers. When are the pi injections? More generally, how "big" is A compared to the piAi which generate it? We have seen in the worst case that the coproduct can be 0, so A can be all of C{X}, and now we intend to examine the best case, drawing our inspiration from example 1.4.28.
B Supplement: Free Products and Amalgamated Sums The coproduct is not of much interest to us when it is trivial; we much prefer when the coproduct contains each of its components canonically. Thus we are led to the following definition:
Definition 1.4.32: Suppose W is a concrete category with given objects A and { A i :i E Z}, where A is a subobject of each A i . An object B of W is a (free) amalgamated sum of the Ai over A if the following properties hold: (i) For each i there is a canonical 1 :1 morphism p i : Ai + B fixing A (k., via = pja for all i , j in I and all a in A).
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$1.4 Products and Sums
(ii) If piai = pjuj for i # j then a, = aj E A. (Intuitively Ai n A j = A.) (iii) If B' and pi: A , -,B' satisfy (i) and (ii) then there is a unique morphism g : B -+ B' satisfying g p , = pi for each i. B is called the free product of the A , if A has no proper subobject. (B should really be called the free sum, but this would contradict the current usage. In group theory the free amalgamated sum is called the free product with amalgamated subgroup.) We shall be most interested in the following situation:
Definition 2.4.32: Suppose W is a ring. W-ging is the category each of whose objects is a ring R together with a given homomorphism fR: W + R; a morphism in W-Bing is a ring morphism g : R + R' satisfying gfR = fR.. (This generalizes the notion of algebra.) Remark 2.4.32': Conditions (i) and (iii) of definition 1.4.31 show that any amalgamated sum over a ring W is a coproduct in W-Wi-y; this approach is most in accord with current practice. Notation as in example 1.4.28 the amalgamated sum of the C{X")} over C is C{X}in B h y , and this is the free product in C - d & Example 2.4.33:
Determining whether an amalgamated sum exists entails constructing a coproduct, which may not be an easy task. The next result shows condition (iii) of definition 1.4.31 is superfluous.
Proposition 1.4.34:
Suppose {Ai:i E I } are rings containing A. The A , have an amalgamated sum over A if there is some ring B satisfying ( i ) and ( i i ) of dejnition 1.4.32. Proof: Suppose v,: A, -+ Ai are the canonical homomorphisms into the coproduct, and let K be the ideal of Ai generated by {via - via: i, j E I and a E A}. We claim P = A i ) / K is the amalgamated sum along with vi: A i+ P given by the composite of vi with the canonical map A , + P. To see this we appeal to definition 1.4.31 using P and v, instead of B and pi.Of course there is a homomorphism g: A i+ B such that gvi = pifor each i. For each i, j E I and a in A we have g(via - vja) = p i a - pja = 0, implying g K = 0; thus g induces a homomorphism P + B and &Fi = pi, verifying condition (iii). Moreover, ker V, G ker pi = 0 so each vi is 1 : 1 and fixes A by definition, verifying (i). Finally if viai = ijaj then piai = %ai = fiaj = pjaj, Q.E.D. so the hypothesis on B implies ai = aj E A, proving (ii).
(ieIu
uic,
a:
Constructions of Rings
78
C Supplement: Categorical Properties of Modules: A belian Categories One of the main uses of category theory is to abstract the properties of the category R - A u d so that in effect one can utilize the module theory in varied settings. Although we shall not pursue this direction of inquiry, it is well worth pausing from time to time to note what happens in greater generality. This will occupy us for the remainder of the section. We say a category W is preadditive if Hom(A, B ) is an abelian group for any objects A, B in W, with respect to an operation + satisfying the following distributivity conditions: (91 + 92)f = 9lf
+ 92f
and
f(h1
+ h2) = f h l + fh2
for all morphisms f:A -,B, gi: B + B‘, and hi:A‘ + A. Another way of stating this requirement is that the composition Hom(B, C) x Hom(A, B ) -+ Hom(A,C) is bilinear for all A , B,C in ObW. (Unfortunately, the word “additive” was used for a less interesting definition of categories.) A functor 9:W -,9 is additive if W, 9 are preadditive categories and the map 9: Hom(A, B ) Hom(9A, 9 B ) is a group homomorphism for every A, B in Ob W. Example 0.1.8 is one of the classic examples of additive functors. We have seen R-Aud is a preadditive category. B h p is not preadditive, but every ring R can be made into a small preadditive category by means of example 0.1.5; conversely, every preadditive category having only one object can be viewed as a ring whose elements are the morphisms. This reinforces our philosophy of studying rings through their elements. There are other examples of preadditive categories which we shall encounter later. Many of the notions of R-Aud can be formulated categorically for an additive category W. A zero object of W is an object 0 such that Hom(A, 0) = (0) = Hom(0,A) for every A in ObW. Clearly, the zero object is unique up to isomorphism if it exists. If the 0 object exists then one can define a zero morphism 0: A + B by 0 = g 0 f where f: A -,0 and g:O -,B. --f
Definition 2.4.35: Given morphisms f i: B , --t A and f2: B2 -+A in a category 59, we say f i I f 2 if f, = f 2 h for some unique morphism h: B , -, B 2 ; f l and f2 are equivalent if h can be taken to be an isomorphism.
f2
Note in this definition that if f l is monic then h is monic; furthermore, if is monic then h is unique (since f, = f 2 h = f2h’ implies h = h’).
Definition 2.4.36: A kernel of a morphism f:A -+ B is a monic k: K -, A with f k = 0, which is the “largest” such in the sense of definition 1.4.35.
79
81.4 Products and Sums
Our goal is to find a class of categories in which all monics are kernels and then to use kernels (and their duals) to obtain a modular lattice generalizing the lattice of submodules LZ(M). The categories which we study turn out to satisfy an important property lacked by R - A d - t h e y are self-dual, where we say a category %‘is self-dual iff %‘ FZ WoP. In a self-dual category, the dual statement of a theorem is also a theorem. This observation is very useful and will be exploited in Chapter 5. Remark 1.4.37: Suppose iff gfl = 0
i.: Bi+ A
are monics and
f2 = kerg. fl
I fi
To proceed further we shall need the dual concept of kernel. The cokernel of a morphism f : A + B, denoted coker f , is an epic h: B --+ C with hf = 0 such that for every h’: B + C‘ with h ’ f = 0 there is a unique g: C + C‘ such that h’ = gh. For example if %‘= R-Mud then coker f = B / f A . There is a good criterion using cokernels for determining when a morphism is a kernel. Remark 1.4.38: If f is a kernel and coker f exists then f seen at once from the following diagram where f = ker g:
=
ker(coker f ) , as
Thus we are interested in categories with many cokernels. One of the first properties one learns about R-.Mod is that every submodule is a kernel. This can be expressed generally as follows. Remark 1.4.39: If f = f i g is monic where g: A + B is a cokernel and f,: B + C is a kernel, then f is equivalent to f l (and so is a kernel). (Indeed, we have f I f l so we must show f i I f. Write g = coker h. Thenfh = f i s h = 0 so h = 0. Hence l B h = 0 so there is 9’: B + A such that gg’ = 1, implying fg’ = f igg’ = f l , as desired.) Definition 1.4.40: A preadditive category V is abelian if %‘satisfies the additional properties:
(i) %‘is closed under finite products (i.e., if A , , . . . ,A, E Ob %‘then Ob 48.).
n
Ai E
Constructionsof Rings
80
(ii) Every morphism has a kernel and a cokernel. (iii) Every morphism f can be written as hg where g is a cokernel and h is a kernel. Clearly R-&d is abelian. Also we have seen that properties (ii) and (iii) show that all the previous remarks (and their duals) hold, and in particular every monic is a kernel. Often (iii) is written in the literature in the following equivalent form (cf., exercise 19): (iii’) Coker(ker f )is isomorphic to ker(coker f ) for every morphism f. Conditions (ii) and (iii) can be utilized in a particularly telling way. Define the image of a morphismf: A + B denoted im f,to be ker(cokerf). Thus f = imf iff f is a kernel. (Images can be defined in a more complicated fashion for arbitrary categories.) Let us view images more intuitively. Remark 1.4.41: Given f = hg where 9: A + C is a cokernel and h: C + B is a kernel we have coker h = coker f (since g is epic) so im f = ker(coker h) = h. Because of this, we write f A for C. Dually one has g = coker(ker f ).
Next we show (i) is self-dual by observing that a certain nice situation in R-Aud is not mere coincidence. Proposition 1.4.42: Suppose %? is a preadditive category with objects A , , .. .,A, which have a product A i , x i ) where I = { 1,. . .,t } . Then A i ,p i ) is the coproduct of A l , . ..,A, for suitable morphisms pi:Ai + A i which p j n j = 1 and nipj = 6 , (where aij:Aj + Ai is 0 unless i = j and satisfy 6ii = l A i . )
(n
zf=,
n
n
(n
Proof: Let A = A i . For each j we have a unique morphism pi: A j + A such that n i p j = 6, for 1 5 i It. Then n i x : = , p j n j = dijnj = ni for all i. Hence pjnj and 1, both complete the same diagram and are thus the same by the universality of the product. It is straightforward now to check ( A , p i )is the coproduct. Q.E.D.
I:=,
Corollary 1.4.43: Objects A , , ...,A, of a preadditive category have a product i f f they have a coproduct, in which case their product and coproduct are isomorphic. Proof: (+) was done; (e) follows by passing to the dual category, which is Q.E.D. also preadditive.
$1.5 Endomorphism Rings and the Regular Representation
81
Note that Corollary 1.4.43implies abelian categories are self-dual. On the other hand, R - A d is not self-dual, because the lattice Y ( M ) is upper continuous but not lower continuous (cf., exercise 0.0.4). The self-duality of abelian categories permits one to take shortcuts in proofs which are not available in R-Ad; yet much of the rich theory of R-Jtud is also available in arbitrary abelian categories, as can be seen in exercises 20ff.
51.5 Endomorphism Rings and the Regular Representation Although we have a fair number of rings at our disposal by now, we lack a unified framework in which to view them. The goal for this section is to introduce endomorphism rings and show that every ring is a subring of a suitable endomorphism ring, thereby providing a theorem parallel to Cayley’s theorem in group theory. The benefits are similar, enabling us to construct rings as subrings of known rings without having to verify associativity and distributivity. We shall also see how matrix rings are a special class of endomorphism rings. The underlying idea has its roots in abelian groups. In the foregoing discussion we always assume M # 0. Any module M can be viewed as an abelian group and thus as Z-module.
Endomorphism Rings Remark 1.5.1: Hom,(M, M) is a ring whose addition is sum of maps and whose multiplication is composition of maps. Moreover M E Hom,(M, M)Aud where the scalar multiplication f x is taken to be the action of f as a map. When M E R-&od we can transfer the elements of R to Hom,(M, M ) as follows: Definition 1.5.2: For M in R - d u d and r E R , define the left multiplication (also called lest homothety) map pr:M + M by prx = rx. Likewise, for M in A d - R define the right multiplication map p i : M + M by p i x = xr.
Remark 2.5.3: For every r in R , the left multiplication map pr is in Hom,(M, M ) , and there is a ring homomorphism p : R -,Hom,(M, M) given by p r = pr; moreover ker p = Ann, M. (Verifications are straightforward.)
82
Constructions of Rings
We should like p to be an injection, which is the case iff Ann, M = 0. Accordingly, we say M is faithful if Ann, M = 0. When M E R - A u d is faithful, we call p the ( l e f ) regular representation of R in Hom,(M, M). This can all be done from the right as well. Namely, if M E Aud-R define p’: R + Hom,(M, M) by p’r = p i . However p’ reverses the order of multiplication ( p ’ ( r , r 2 ) = (p’r,)(p’r,)). Such a map is called an anti-homomorphism. Note that in general a map f: R, + R, is a ring anti-homomorphism iff the induced map J R , + R;p is a homomorphism, so in order to avoid antihomomorphisms one must introduce opposite rings from time to time. For example we shall view the right (regular) representation as a homomorphism p ’ : R + Hom,(M, M)OP. The regular representation has many interesting applications.
Example 1.5.4: Viewing R in R-Mud we have Ann, R = 0 since Ann,{ 1 } = 0. Thus the regular representation identifies R with a subring of Hom,(R, R ) . The regular representation is one of the most important tools in ring theory, and we shall try now to hone it further to obtain many interesting applications. Proposition 1.5.5: If M E R - d u d then Hom,(M, M ) is the centralizer of pR in Hom,(M, M ) , where p: R + Hom,(M, M ) is the regular representation. Proof.- Suppose f E Hom,(M, M ) . Then f E HOM,(M, M ) iff f(rx) = rfx for all r in R, x in M , iff fp, = prf for all r in R. Q.E.D.
In particular, Hom,(M, M ) is a ring. Furthermore, one sees at once that M€R-&od-HOm,(M, MyP.Analogously, if M EA d - T then Hom,(M, M ) is a ring, the centralizer of p’T in Hom,(M, M); consequently, M E Hom,(M,M)-Mud-T. To unify these two situations we make the following definitions. Definition 1S.6: (i) If M is an R-module define End, M to be Hom,(M, M)OP,ie., multiplication in End, M is composition of maps in the reuerse order. (ii) If M is a right T-module define End MT to be Hom,(M, M).
We hope the asymmetry of this definition will not bother the reader, but it makes later assertions flow more smoothly.
83
$1.5 Endomorphism Rings and the Regular Representation
Proposition 1.5.7: Suppose M E R - A d - T is faithful as R-module. Then R E End M , by the (left) regular representation. Proof: By proposition 1.5.3, we have the injection p : R -P Hom,(M, M ) which, by hypothesis, centralizes p’T; so p R G End MT by proposition 1.5.5. Q.E.D. This method gives us many useful techniques and examples. Example 1.5.8: If R E C - d t y then taking M = R and T = C in proposition 1.5.7 (identifying C-Aud and A d - C ) , we have R G End Rc. However, End R, is defined using only the C-module structure of R. Thus, forgetting the original multiplication of R, we see that every possible multiplication on R which makes R a C-algebra produces a suitable subalgebra of End R,. This result can be used to construct the algebra structure by viewing R c End R , in the suitable way, just as Cayley’s theorem is used to construct finite groups. Then distributivity and associativity need not be verified since they are inherited from End Rc. Example 1.5.9: R z End R R for every ring R. (Indeed, it suffices to prove the regular representation p is onto. Take any f in End RR and put r = f 1. For all r’ in R we have
fr’
= f(1r’) = (f1)r’ =
rr’
= prr’,
proving f = p r . Likewise the right regular representation gives an isomorphism from R to (HornRR)OP= End, R. Remark 1.5.10: Taking M = R in proposition 1.5.5 we see the centralizer of p R in Hom,(R, R ) is HomR(R,R ) = p‘R. By symmetry we see the left and right representations of R are the centralizers of each other in Hom,(R, R).
When R is a C-algebra (cf., example 1.5.8), we conclude a fortiori that the left and right representations of R are the centralizers of each other in Hom,(R, R). Thus we see the right regular representation of R gives us a copy of R in End, R, whereas the left regular representation gives a copy of RoP, and these centralize each other. Example 1.5.11: If M E R - d u d is faithful then taking T = End, M we have R E End M,. An important part of the structure theory is the study of
R in terms of End MT, which is called the biendomorphism ring of M for obvious reasons. Example 1.5.12: Suppose M = R and T is a subring of R. Then R 5 EndR,. This is particularly useful when R is f.g. as T-module, as we shall now see.
Endomorphisms as Matrices Theorem 1.5.13: Suppose M E A d - T is spanned by xl,. . .,x,. Then End MT is a homomorphic image of a subring of M,(T). In fact, if {xl,. . .,x,,} is a base of M then End M T !z M,(T). Proof: Write R = { r = (a,) E M,,(T): there exists /3, in End MT with /?,xi = xiaijfor all j } . Note that the matrix r determines the action of /3, on the spanning set {xl,. . .,x,} and thus on all of M, so we have a function cp: R + End MT given by cpr = fir. Moreover, cp is onto since for any /3 in End MT we xiaij for suitable aij in T. have /3xj = To show R is a ring we note for r=(aij)and r’=(aij) that /3,-/3,.=/3,-,. and .3,/ = /3,/3,. since
=
i
i=l
xi(
k=l
aikaij) = /3,,,xj
for all xi.
This verification also shows cp is a homomorphism, so we have displayed End MT as a homomorphic image of R E M,(T), as desired,-If {xl,. . .,x,} is a base then (M; { xl,. ..,x,}) is free so every matrix r = (aij) defines an endo=l =0 morphism fir, implying R = M,(T); moreover, kercp = { ( a i j ) : ~ ~xiaij for all i } = 0, so cp is an isomorphism from M,,(T) to End M T . Q.E.D. Corollary 1.5.14: Suppose R is a ring with subringT, and as a right Tmodule R is free with a base of n elements. Then R E M,(T). Proofi
Example 1.5.12 shows R G End R T
= M,(T).
Q.E.D.
Corollary 1.5.15: Every jinite dimensional algebra over a jield is isomorphic to a subalgebra of a suitable matrix algebra.
81.5 Endomorphism Rings and the Regular Representation
85
Corollary 1.5.16: I f M E D-FimuA for a division ring D then k i d , M x M,(D) for some n. Proo)
Every module over a division ring is free, so apply the theorem.
These important corollaries begin to show us the power of theorem 1.5.13 even in the case when M is free. In fact, the relation between endomorphisms and matrices elevates matrix rings to perhaps the most prominent position in the structure theory of noncommutative rings, leading us to study endomorphisms in terms of matrix properties as in the following remark:
Remark 1.5.17: Suppose M has a base { x l ,..., x,} in A u d - T and (a,) E M,(T). Then xiaij:1 Ij I n} spans M iff (a,) is right invertible (since then we can recover the x i ) .
{I:=
When M # N then HOm,(M, N) is not a ring, but is nevertheless an abelian group which is also of considerable interest.
Remark 1.5.18: Suppose M E R - A u d and N E R-Am!-T. Then HOm,(M,N) becomes a right T-module with the scalar multiplication fa (for f:M -+ N and a E T ) given by ( f a ) x = ( f x ) a for all x in M . (The verifications are straightforward and left to the reader.) Likewise if M E A d - T then Hom,(M, N) E R - A u d with scalar multiplication rf given by (rf )x = r( f x ) . (Incidentally, this latter version “looks like” associativity.)
Finally, if M E R - A d - S and N E R - A d - T then simultaneously we can view Hom,(M, N) as S-module (as above) or as right T-module (as in remark 1.5.18), whereby, in fact, Hom,(M, N) is an S-T bimodule. We continue this brief digression to record a useful generalization of example 1.5.9. Proposition 1.5.19: Hom,(R, M) x M in R - A d where we oiew Hom,(R, M) as R-module via the action r f dejned by ( r f ) x = f ( x r ) for r in R and f :M+R.
Constructions of Rings
86
Proof: Define $: M + Hom,(R, M) by $ y = py where py is right multiplication by y . For all r, x in R we then have $(ry)x = p l y=~xry = p,,(xr)= (rp,,)x so $(ry) = r$y, proving $ is a map. $ is monic since y # 0 implies Q.E.D. 0 # y = py 1; $ is epic because f = p,l for any f:R -+ M.
Proposition 1.5.20: For any n in N and M in R - A d - T there is an isomorphism @:(End, M)'") + Hom,(M("), M) in Mud-T, sending (f l , . . .,f,,) to the map f dejned b y f ( X I , . . .,x,,) = &xi.
c?=
Proofi Write x i :M(")+ M for the projection onto the i-th component, and pi: M + M(") for the map p i x = (0,. . .,0, x,O,. . .,0) where x appears in the i-th component. Then @( fl, ...,f,,) = cl;:ni. Defining Y: HOm,(M'"', M) + (End, M)'") by Yf = (fpl,. . .,fp,,), we see @Yf = fpini = f pini = f and 'Pa(fl,..., f,,)= Y c l ; : n i = (fl ,..., fn). Thus @ = Y - ' and Y,@ are both isomorphisms. Q.E.D.
1
The Dual Base Taking M = R and T = H in proposition 1.5.20 and identifying End, R with R as in example 1.5.9, we see R(")z Hom,(R("),R) as abelian groups. This identification can be made more explicit. Definition 1.5.21: Suppose M E R - A d . Taking N = R E R-M0d-R in the natural way, we put M = Hom,(M, R), viewed in A u d - R as in remark 1.5.18. If M is free in R-Fiwzod with base { x l , .. .,x , } , define jZ in M for each i by 2icj"= I rjxj = ri (so that 2ixj = dij). We call . .,2,} the dual base of x l , . . .,x , . (An analogous construction exists for M in A d - R , in which case M E R-A~A.) Proposition 1.5.22: lf M is free in R-9imud with base { x l ,..., x,,} then . .,2,} is a base of M (so, in particular, M is free in A u A - R ) . Any map f:M - t R in M satisfies f ( ~r i x i~)= x~i i f x, i , so letting aj = f x j E R we have f = I; Zjaj = (because ( c j Z j a j ) ( rci x i )= C 2j( rixi)aj= riai = f rixi)). This proves that 21,.. .,i nspan M. Independence is straightforward; if z j r j = 0 then for each i we have Q.E.D. 0 = (1 i j r j ) x i= l(2jxi)rj= dijrj= r i .
Proof:
c
c (c cy= cj"= cj"=
There is an enlightening connection between M and M using category theory, given in exercises 2, 3. Note that for any map f:M -+ N we have (by example 0.1.8(ii)) a map f 3 + M given by f h = hf for any h: N + R. This
$1.5 Endomorphism Rings and the Regular Representation
87
map is especially important when R is a field and M is a finite dimensional vector space, for then the full force of linear algebra applies. In the literature, $1 is often designated as M * , and we shall also feel free to use the latter notation.
Adjunction of 1 Although we consider only rings with 1, there is an extensive literature on rings without 1 (i.e., all axioms are satisfied except for the existence of I), which forms a category g*zg whose morphisms cp satisfy all requirements of ring homomorphisms except we obviously need not have cpl = 1. Thus the forgetful functor (forgetting that 1 exists) enables us to view Britg as a subcategory of %zg, which is not full since the function p: Z --f Z x Z given by pn = (n,0) is a morphism in 9 ~ but g not in 9i*zf. The first question one may ask is, given a ring R , without 1, what is the universal from R , to the forgetful functor?
Definition 1.5.23: The ring Rb obtained by adjoining 1 formally to R , is the additive group Z0 R , together with multiplication defined by (ml>rI)(m29r2)= (mlm2,mlrZ + m2r1 (The trick here is to think of (m,r ) as m
+ r1r2).
+ r where m E Zand r E R.)
Proposition 1.5.24: The ring Rb is indeed a ring and, together with the canonical injection u: R , Rb given by ur = (0, r), is a universal from R , to the forgetful functor. --f
Proof;. We demonstrate Rb as a ring by means of the regular representa-Namely, for every (m,r) in ZO R , define (m,r) in End@ 0 R , ) by tion. (m,r)(m’, r’) = (mm’,mr’ + m’r + rr’) for all (m’, r’) in Z0 R,. We need to -) that Rb will be identified with a show ( m l , r l ) ( m 2 , r 2=) ( m l , r l ) ( m 2 , r 2so subring of End,@ 0 R,). Well __-
+ + r2r’) = (m,m2m’,m1(m2r’ + m‘r2 + r,r’) + m2m’rl + rl(m2r‘+ m’r2 + r2r’)) = ( m l m 2 , m l r 2+ m2r1 + rlr2)(m’,r’)
(m,,~l)(m2,r2)(m’= , r ‘(ml,r,)(m2m’,m2r’ ) m’r2
= (ml~rl)(m2,r2)(m’,r’)
as desired. Note that (1,O) is the multiplicative unit.
88
Constructions of Rings
To show universality, suppose R is a ring and f:R, + R is a morphism in Wq.We want to extend f to a ring homomorphism f’: Rb + R. Clearly, then, f’(1,O) = 1, so f’(m,r) = m + fr for all r in R,, and we see this in fact defines a homomorphism since f ’ ( ( m l > r m 2 , r 2 ) )= f ‘ ( m , m , , m , r ,
+ m2r1 + r l r 2 )
= m1m2
+ f(rnlr2 + m2rl + r1r2)
= m1m2
+ 4 f r 2 + (f7-1Im2 + fr1fr2
= (ml
+ fr1)(m2+ fr2)
= f’(rn1,r1)f’(m2,r2) Q.E.D.
The importance of the injection u is that it permits us to identify R, as an ideal of Rb. Thus rings without 1 should be viewed instead as ideals of rings. One problem with this construction is that if R, happens to be a ring (with 1) then Rb has a different 1. To correct this problem we call R, nondegenerate if Ann,, R , = 0. In this case the regular representation R, + End,, R, is an injection. Remark 1.5.25: Let T = hd,,R0. By proposition 1.5.24 the regular representation p: Ro + T extends to a homomorphism p‘: R b + T given by p’(m,r) = m + p i , and kerp’ = {(m,r): mr, + rr, = 0 for all ro in R,} = Ann,; R,. Taking R = Rb/kerp’ we have a ring injection R + T. Let ii: R, 4 R be the composite of the canonical homomorphisms u: R, + Rb and Rb + Rb/ker p’ = R. When R, is nondegenerate we have 0 = Ann,,, R, = R, n ker p’ = ker ii, so iiis an injection; we call R the reduced ring with 1 adjoined to R,. This construction has the advantage that if R , happens already to have an element 1 then iil = 1 because (1, - 1)R, = 0, implying U is a ring isomorphism.
a:
51.6 Automorphisms, Derivations, and Skew Polynomial Rings In this section we construct a noncommutative analogue of R[I] which still has many of its nice properties. In order to view the construction in context, we consider automorphisms and derivations of rings (leading to Lie algebras).
A utomorphisms Definition 1.6.1: An automorphism is an isomorphism Q: R + R. Write R“ for { I E R: or = r}, a subring of R called the fixed subring (under Q ) .
$1.6 Automorphisms, Derivations, and Skew Polynomial Rings
89
Remark 1.6.2: Any automorphism a of R restricts to an automorphism of Z(R). (Indeed, for any z in Z(R) and r in R, (az)r = a(za-'r) = o((a-'r)z) = r(az), proving az E Z(R), so a restricts to a homomorphism of Z(R), whose inverse, likewise, is the restriction of K'.) Automorphisms of fields give rise to classical Galois theory. O n the other hand, there are many automorphisms of R which in fact fix Z(R). For example, if a E R is invertible there is an automorphism (pa of R defined by qar = ma-' for all r in R; if z E Z(R) then qaz = aza-' = zaa-' = z. Automorphisms of the form q, for invertible a are called inner automorphisms. Definition 2.6.3: A unit of a ring R is an invertible element. Unit(R) = {units of R}, a multiplicative group; Inn Aut(R) = {inner automorphisms of R}.
Proposition 1.6.4: There is a group homomorphism @: Unit(R) + Aut(R) given by @a = q, (as dejined above). Then @(Unit(R))= Inn Aut(R) and ker@ = Unit(R) n Z(R), implying Inn Aut(R) x Unit(R)/(Unit(R)n Z(R)) as groups. Proofi For any a,b in Unit(R) and all r E R, qaqbr = $,(brb-') = abrb-la-' = qabrproving 0 is a homomorphism; ker@ = { a E R: ma-' = r for all r in R} = Unit(R) n Z(R). The last assertion is now clear. Q.E.D.
Derivations, Commutators, and Lie Algebras Definition 1.6.5: A derivation of a ringR is an additive group homomorphism d : R + R satisfying d ( r l r 2 )= (drl)r2 rldr2. It is important to note that this definition does not depend on the associativity of multiplication, and, in fact, we shall have occasion to deal with derivations of nonassociative algebras, cf., definition 1.6.6.
+
One motivating example is the usual derivative d on the polynomial ring R = W[A] given by d ( C f = ow J i ) = i w i A i - l . However, there is another fundamental class of derivations, for which we now aim. Given a ring R and ri E R, define the commutator [ r l , r 2 ] = rlr2 - r2r1.We shall rely heavily on the following computation:
Xi=,
[rl,r2r31= rlr2r3 - r2r3r1= [rl,r21r3 + r2[rl,r31. In particular, defining the map d,: R + R by d,r = [a, r], we see d, is a derivation, called the inner derivation given by a. Write [a, R] for {[a,r ] :r E R}.
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Constructions of Rings
Then d, = 0 iff [a,R] = 0 iff a E Z(R). Thus we have a map R -+ Deriv(R) given by a + d,, whose kernel is Z(R). To give this map algebraic significance we introduce a new structure.
Definition 1.6.6: A module M over a given commutative ring C is called a nonassociative algebra if M has binary multiplication satisfying the following properties for all c in C, x i in M : C(X,X,)
= (CXJX,
= x,(cx,).
In other words, nonassociative algebras share all properties of associative algebras except associativity of multiplication and the existence of the multiplicative unit 1.
Definition 1.6.7: A Lie algebra is a nonassociative algebra A where multiplication, written [a,a,] instead of ala2,satisfies the following laws for all a, in A : Calall [[ala2]a3]
+ [[a2a3]a1] + [[aJal]a2]
=0 = 0 (called the
Jacobi identity).
To avoid confusion, sometimes we write [a,,a,] in place of [ a , a,].
Remark 1.6.8: In any Lie algebra, 0 = [ a , + a 2 , a 1+ a,] = [alal] + Ca2a11 + Lala21 + [a2a21 = Ca2a11 + Ca1a21, implying Caza1l = -Ca,a,I for all a,. This is called the anticommutative law. The reason [ 3 is used for multiplication in a Lie algebra arises from the first of the following important examples. Example 2.6.9: Replacing the usual multiplication of an associative algebra A by the commutator [a,,a2] yields a nonassociative algebra, denoted A - . Clearly [a,a] = 0 for all a in A, and, in fact A - is a Lie algebra; the easy verification of the Jacobi identity is left to the reader. For this reason, the commutator is also called the Lie product. Of course, we want to deal with categories, so we need the morphisms.
Definition 1.6.10: If L , , L , are Lie algebras over C,a Lie homomorphism is aC-modulemapf: L , -+ L , satisfying theconditionf[a,a,] = [fa,,fa,] for all a , , a 2 in L,. C - Y his the category whose objects are Lie algebras over C and whose morphisms are Lie homomorphisms. If L E C-Y&, a Lie
41.6 Automorphisms, Derivations, and Skew Polynomial Rings
91
subalgebra of L is a C-submodule which is closed under the Lie multiplication of L (and thus is also in C-Yie, because the conditions of definition 1.6.7 follow a fortiori). When mentioning an associative algebra A in the context of C-LYie, we usually mean A - . For example, a Lie homomorphism f: R + T of (associative) C-algebras is a module map satisfying f [ r l , r 2 ] = [ f r 1 , f r 2 ] .
Example 2.6.12: If R = M J C ) for a commutative ring C, then M,,(C)- has the following Lie subalgebras: (i) the set of matrices of trace 0 (for tr[a, b] = tr(ab) - tr(ba) = 0). (ii) the set of skew-symmetric matrices. Proposition 1.6.12: Deriv(R) is a Lie subalgebra of (End,(R)-, for every algebra R. Proof:
(Note that we shall not use associativity of R.) If d1,d2E Deriv(R)
then
and, symmetrically,
so [ d l , d 2 ] is a derivation.
Q.E.D.
If L is a Lie algebra, define a Lie ideal I to be a submodule satisfying [ L l ] c I , i.e., [ax] E I for all a in L , x in I. It is straightforward then to verify that the quotient module L / I has the natural Lie multiplication [al + I , u2 + I ] = [ a l a 2 ]+ I, under which L / I becomes a Lie algebra; if 0: L + T is an onto Lie homomorphism then ker CD is a Lie ideal and L/ker CD z T as Lie algebras. Proposition 1.6.13: There is a Lie homomorphism CD: R- + Deriv(R) given by a + d, (the inner derivation corresponding to a).Writing Inn Der(R) for the set
Constructions of Rings
92
of inner derivations, we have O(R) = Inn Der(R) and ker O = Z(R), implying Inn Der(R) x R-/Z(R).
Skew Polynomial Rings and Ore Extensions We have noted the polynomial ring R[I] over a domain R satisfies deg(fg) = degf
+ degg
(1)
for all 0 # j,g E RCI], and deg 0 is undefined. In fact, (1) formally implies R[I] is a domain; hence we are led to ask under what general conditions R[I], viewed only as an R-module, can be endowed with multiplication making it a ring T graded by N which satisfies (1). Then T would automatically be a domain and might provide interesting new examples. We start by noting some necessary conditions obtained from polynomials of degree 0 and 1; where ri E R - ( 0 ) are arbitrary: (i) deg(r,r,) = deg(r,) + deg(r,) = 0 + 0 = 0,
so R
c T as domains.
so I r = (ar)I + 6r (ii) deg(Ar) = deg(l) + deg(r) = 1 + 0 = 1, for suitable maps a, 6: R + R with ker a = 0. (iii) a(rlrz)I+ 6(r1r2)= h 1 r 2 = (Ar1)r2 = ((ar,)I + 6rl)r2 = (m1)(mz)I (ar1)6r2 (8rI)rz;
+
+
matching components we get a(rlr2) = (ar1)(ar2)
and
6(r1r2) = (arl)6rz+ (6rl)rz.
In line with definition 1.6.5, we are led to the following definition. Definition 1.6.14: If a: R + R is a homomorphism, a a-derivation is a (E-module) map 6: R + R satisfying 6(rlrz) = (or1)&, (6r1)r2. Thus for a = 1, a a-derivation is merely a derivation. The properties described above in fact characterize T, as we shall now see:
+
Proposition 1.6.15: If the new ring structure on R [ I ] satisfies property (1) above, then for all r in R, I r = (ar)A 6r for a suitable injection cx R + R and suitable a-derivation 6 on R. Conversely, if a: R + R is an injection and 6 : R -,R is a a-derivation, then there is a unique new multiplication on R[I] (extending the R-module action) such that I r = (ar)l + 6r for all r in R ; moreover, this dejnes a ring T satisfying (1) and is thus a domain.
+
$1.6 Automorphisms, Derivations, and Skew Polynomial Rings
93
Proof;. The first assertion was prove above. For the converse we shall use the regular representation to obtain T as a subring of End, R [A]. Define 1in End, R[A] by the rule
Writing Ffor left multiplication by r define T = ( x ; = o c It iE: N, all ri E R } . If CJi = 0 then 0 = FiXi)A = riAi+ so each ri = 0; hence { 1': i E N} is a base of T as module over R = {F r E R).
1
(1
1
'
%=a
implying + 6r. Therefore, T is closed under multiplication and is thus a subring of End, RCA]; identifying R with R we see a t once that T satisfies (1) (and thus is a domain). Uniqueness of the multiplication follows Q.E.D. easily from (1). Definition 1.6.16: Suppose a: R -,R is an injection and 6 is a a-derivation. The Ore extension R[A;o,6] is the ring obtained by giving RCA] the new multiplication Ar = (ar)A + 6r, as in proposition 1.6.15. If 6 = 0 we write R[A;a] for R[A; a, S]; this is called a skew polynomial ring. R[A; l,S] is called a differential polynomial ring in the literature.
Note that an Ore extension of a domain is a domain, by proposition 1.6.15. Remark 1.6.17: A necessary and sufficient condition that each element of R[A;a,6] can be written in the form x i = o A i r iis that CJ is onto (i.e., a is an Airi, then clearly automorphism). (Indeed, if any rA can be written as t = 1, so rA = (ar,)A+ ( 6 r l )+ r o , proving ar, = r and thus a is onto. Conversely, if a is an automorphism then rA = Aa-lr - Ba-lr, and then one sees easily by induction on j that each term rAj can be written in the form Airi,and thus each element of R[A;a,6] can be written in this form.)
Ci=o
Principal Left Ideal Domains (PLID's) Let us now study the structure of skew polynomial rings more closely. Recall that F[A] is a PID (principal ideal domain) whenever F is a field, and this
Constructionsof Rings
94
property is used repeatedly in the commutative theory. The noncommutative analogue is a straightforward generalization, but we shall see later that the asymmetry of the construction of skew polynomial rings gives them some fascinating properties.
Definition 1.6.18: A principal left ideal domain (PLID) is a domain in which every left ideal is principal (i.e., of the form Rr). Every PLID satisfies the left Ore property that Rr, n Rr, # 0 for any nonzero elements r,, r, of R. (Indeed, we are done unless r2 4 Rr,, in which case Rr, + Rr, = Ra with a $ Rr,. Then a = x l r l x 2 r 2 for x i in R with x 2 # 0. But for some r # 0, we have Remark 1.6.19:
+
rl
= ra = r(xlrl
+ x2r2)= rxlrl + rx2r2
implying 0 # rxzr2 = (1 - r x l ) r l E Rr, n Rr,.) The left Ore property is of utmost importance in constructing rings of fractions, c.f., g3.1. The link between skew polynomial rings and PLIDs lies in the Euclidean algorithm.
Remark 1.6.20: Suppose 0 # f,g E R [ l ;c,S] where the leading coefficient of g is invertible in R. Then f = qg + p for suitable q, p in R[A;c,S] such that either p = 0 or degp < degg. (Indeed, let m, n be the respective degrees of f and g. We proceed inductively on m. If n > m then we are done with q = 0 and p = f.Otherwise, letting al", bA" be the respective leading terms of f,g, we note al" = (al"-"b-')bl", so putting h = f - al"-"bb-'g, we see deg h < m and, inductively, h = q'g + p for suitable q', p with p = 0 or deg p < n. Hence f = (al"-"b-' + q')g + p.) Of course, the hypothesis of remark 1.6.20 holds when R is a division ring, leading to the next result.
Proposition 1.6.21: Suppose D is a division ring. Then T = D [ l ; c , 6 ] is a PLID. Moreover, if Tf = Tg then f = dg for some d in D; in particular, deg f = deg g and f,g have the same number of (nonzero) terms. Proof: A left ideal L is generated by any element g in L of lowest degree (since i f f E L we write f = qg p implying p E L so p = 0). Now if Tf = Tg we see f = qg implies deg f 2 deg g; likewise, deg g 2 deg f; hence deg 4 = 0 so q E D as desired. Q.E.D.
+
51.6 Automorphisms, Derivations, and Skew Polynomial Rings
95
It is of interest to check when T is a principal right ideal domain (PRID). Proposition 1.6.22:
The following assertions are equivalent for T = D[A; a,d]:
(i) a is onto. (ii) T is a PRID. (iii) T is right Ore.
Proof: (i) 3 (ii) follows from remark 1.6.17 since the right-handed analogue of proposition 1.6.21 is applicable. (ii) (iii) is remark 1.6.19. (iii) =.(i). For any d in D we have some nonzero h E R T n dlT, so h = if = dAg for suitable f , g in T . Let f , g have leading terms adn and bA", respectively; then the leading term of h is = dabl"", so aa = dab, implying d = o(ab6')E aD,proving a is onto. Q.E.D. A minor modification of the Euclidean algorithm permits us to generalize proposition 1.6.21.
Remark 1.6.23: Suppose 0 # f , g E R[I;a,d] with deg f > degg, and or is invertible in R where r is the leading coefficient of g . Then we can write f = qg + p where either p = 0 or degp I degg. (Indeed, write g = r l n + Elid r i i i . The leading term of Ag is (ar)A"+',so apply remark 1.6.20 to f and lg.) Proposition 1.6.24: If R is a PLID with a(R R[A; a, d] is a PLID.
-
{O}) E Unit(R) then T =
Proofi Given a left ideal L of T let Lo = {leading coefficients of polynomials of L having minimal degree, n } . Then Lo < R so Lo = T r , for some r , in R . Take y E L of degree n having leading coefficient r,,. We claim L = Tg. Indeed take any j ' E L-Ty. Using remark 1.6.23 we may replace f by p and assume deg f = n so the leading coefficient of f is rr,, for some r in R; then f - rg is an element of L of degree < n so f - rg = 0, i.e., f = rg E Tg. Q.E.D. Of course this result only adds interest when a is not onto. However, it is important for two reasons: (i) the converse is true (cf., exercise 3) and the condition a(R - (0)) G Unit(R) permits an iterative construction which provides many interesting examples of Jategaonkar [69]; consequently, the
96
Constructions of Rings
condition is called Jategaonkar's condition. Presently we shall restrict our attention to special cases in order to introduce several important classes of rings.
Skew Polynomial Rings (Without Derivation) Over Fields First we consider the case 6 = 0, i.e., T = K [A; a ] .Clearly, TIi a T for i, and, more generally, TzA' a T for every z in Z ( T ) and all i. Surprisingly enough, every ideal has this form. Let K" denote {a E K : aa = a} Proposition 1.6.25: Suppose K is a jield having an endomorphism T = K [ 1 ; a ] and F = K".
CJ.
Let
(i) Z ( T )= F unless a is an automorphism of order n, in which case Z ( T ) = F[A"]. (ii) Every ideal of T has the form Tp1" for p in Z ( T ) . Proofi
1
(i) Obviously F[A"] E Z ( T ) .Conversely, supposef = a i I i E Z ( T )where ai E K . Then 0 = [ A , f ] = '&zi - ai)Ai+' proving aai = a, so a, E F for each i. Moreover, for every a in K we have 0 = [a,f] = l a i ( a - a'a)A', implying a = d a whenever a, # 0, i.e., oi = 1 so n divides i. Hence f E F[An] as desired. (ii) Suppose A a T and choose f # 0 in A of minimal degree t. By proposition 1.6.21 we have A = Tf and f has the same number of (nonzero) terms as any other polynomial in A of degree t . Replacing f by a;f' we may i+l assume a, = 1. Now [A, f ] = i(aai - a,)A has fewer terms than f and thus must be 0 by the above discussion. Hence each ai E F. Moreover, for every a in K we see (a'a)f - fa has degree It - 1 and is in A, so must be 0, implying (o'a)a, = a p ' a for each i. Hence a', = a'a whenever a, # 0, implying a' - i = 1 so n I ( t - i ) . Taking m to be the degree of the lowest order term of f we Q.E.D. conclude f~ F[A"]I" = Z(T)A".
1:
(The proof given above is redundant, for the proof of (ii) also implies (i). Indeed if f E Z ( T )then Tf U R and thus f = apl" for some a in K and p in F [ A n ] ,from which it readily follows a E F and m = 0. However, the argument of (ii) is also subtler than that of (i).) Example 1.6.26: Let K be a field with an endomorphism a which is not onto, e.g., K = Q(A) and a is given by aA = ,I2. (For a ( f ( I ) g ( A ) - ' )=
81.6 Automorphisms, Derivations, and Skew Polynomial Rings
97
f ( A 2 ) g ( 1 2 ) - 'one ; sees at once A I# O K , for if A = a( fg-') then g ( A 2 ) l = f (A2) contradicting the parities of degree.) K [ A ; a] is a PLID which is not a PRID by proposition 1.6.22; using proposition 1.6.25, we see the ideals only have the form (Ai), so, in particular, the ideals form a chain. Corollary 1.6.27: Suppose a is an automorphism of K of order n, and F = K". If p(A) # A is an irreducible polynomial of F [ l ] , then (p ( A n ) ) is a maximal ideal
of T = K [ A ;a] and T/( p(A")) is a simple ring. Any ideal containing ( ~ ( 2 " )has ) the form Tq(An)Am where q E F[A], so p(A") E Tq(l")l". If f is an irreducible factor of q(A") in F [ A ] then f I p(A"). If there were 9, h in F[A] such that gp hq = I , then g(l")p(A") h(l")q(l")= 1 contrary to the existence off. Hence, p , q are not relatively prime in F[AJ so p I q implying ( ~ ( 2 " )indeed ) is maximal. Q.E.D. Proof:
+
+
Example 1.6.28: (Cyclic algebras) Let R = K [ A ; o ] / ( l " - a ) where a is an automorphism of order n and a E F = K u. Since - a is obviously irreducible in F [ l ] , we see that R is a simple ring which we call the cyclic algebra ( K ,a, a). Let z be the canonical image of A in R and identify K with its image in R . Then R = Kz', and (2,: 0 I i < n} is a base for R as a vector space over K (for if aizi = 0 with some a, # 0 then ~ aiAi ~ E (A" ~ - ,a ) which ' is impossible by a degree comparison). Multiplication is given by the straightforward rule
c;:;
1::;
ifi+jn. In particular, Z" = a E F. Moreover, F = Z ( R ) (for if zyij aizi E Z ( R ) , then 0 = [z, aizi] = c(aai - ai)zi+ showing each a, E F; then 0 = [a, aizi] = c a i ( a - a'a)z' for every a in K , so taking a with aia # a we see a, = 0 for each i # 0.) Now n = [ K : F ] = [ R :K ] so [ R : F ] = n2.
c
',
1
Thus each cyclic algebra is a simple algebra of dimension n2 over its center for suitable n. This construction generalizes W = (C, a, - 1) where a is complex conjugation, and is the key to much of the theory of finite dimensional division algebras, which will be discussed in Chapter 7 in much greater detail.
Diflerential Polynomial Rings Over Fields Now we consider the opposite extreme, where (T = 1. Some of the methods are similar, but first we need to see how a derivation 6 "works" on a ring.
98
Constructions of Rings
Remark 1.6.29: ("Leibniz's formula") The following rule is seen by induction on n:
6"(ab)= i=O
(1) hia6"-'b
Leibniz's formula is used extensively in studying derivations. Our next result relies on the obvious consequence that if 6'a = 0 and 6 J b = 0 then 6"J- '(ab)= 0. Remark 1.6.30: Suppose 6 is a derivation of a ring R, and put Ri = { r E R : di+'r = 0 } and R , = U i c N R E i R . Then R , is a ring and satisfies RiRj E R i + j . In particular, Ro is a subring; if r E R , and r-' E R then r-' E R,. (Indeed, first note 61 = 6(1 1) = 61 61 implying 61 = 0 and R , is a subring. Then 0 = 6(rr-') = (6r)r-' r6r-l = r6r-I for any rin Ro which is invertible, implying 6r- = 0.)
-
'
+
+
Lemma 1.6.31: Notation as in remark 1.6.30. Suppose R = R , T = R [ I ; 1,6]. if A is any subset of T such that [ I ,A ] G A then
and
A n Ro[A] # 0.
(Note Ro[A] is the usual polynomial ring.) Proofi Of all the nonzero elements of A of minimal degree t, take f = riAi such that 6"rt = 0 for n minimal possible. (By hypothesis some hnrt = 0.) Then ~ ~ = , ( 6 r i )=A i[ I , f ] E A and 6"-'(6rt) = 0, so by assumption Q.E.D. we see [ L , f ] = 0, i.e., all 6ri = 0.
I:=,
The Weyl Algebra We are now going to define a ring which is very important both in the structure of enveloping algebras (58.3 below) and also in its own right. Example 1.6.32: Suppose C is a commutative ring. The Weyl algebra dl(C)= C { p , A } / ( A p - pL - 1) where p,A are indeterminates over C . In other words, we have & = 1 in A,(C). More generally, taking pl,... , p n ,A,,. .. , I , to be indeterminates over C , we define d , , ( C )= C { p , ,. , . , pn,I , , . . . ,A,,}/B where B = ([A,, pi] - 1, [ I , ,p j l , [Ai, Ajl,[pi,pjl for all i # .i). When C is understood, we merely write dnfor d n ( C ) .
9+
There is an alternate, more explicit description. Let W, = R[A; 1, 61 where R = C[p]and 6 is differentiation with respect to p. Note that L f - fL = Sj,
99
$1.6 Automorphisms, Derivations, and Skew Polynomial Rings
+
so, in particular, I p = pI 1, and we have a canonical isomorphism d,z W , . Proceeding inductively we put R ( n - l )= d n - , [ p , , ] and have d,, z R,,,- ,,[A,,;1 , 4 ] where 6, is partial differentiation with respect to p,,. a ,. . ., a for Because of this description, the notation C xi,. . .,x,,,
[
~
ax,
-1
ax,,
.d,,(C)has become widespread in the literature. (Thus x i replaces pi, and d / a x i replaces I i .) More generally, there has been increasing algebraic interest in differential operator rings, cf., recent work by K. Brown, J. McConnell, S. P. Smith, and T. Stafford. 1 so d,, = (d,,-,[A,,])[-p,,; 1,SJ On the other hand, -pA = A(-p) where 6; is partial differentiation with respect to A,,. (By convention take do= C ) .This is used to prove the following key result.
+
Proposition 1.6.33: If C is a commutative Q-algebra then every Lie ideal L of d,,intersects C nontrivially. Proof: In the notation of example 1.6.32 we have L n R ( , , - , , # 0 by lemma 1.6.31. But viewing instead d,,as (d,,-,[A,,])[-p,,; l,&] we apply lemma 1.6.31 to L n R , , - , , toget O # ( L n R ( , _ , , ) n d n - l [ A n ]= L n d n - l , a Lie ideal of d,,- Continuing inductively we conclude 0 # L n do= LnC. Q.E.D. Corollary 1.6.34: d , , ( F ) is a simple domain, for any Jield F of characteristic 0. More generally, if C is a commutative Q-algebra then every nonzero ideal B of .d,,(C)is generated by central elements, and d , , ( C ) / Bz d , , ( C / ( Bn C ) ) . Proof: Let B’ = B n C # 0, and write R for d,,(C).The canonical map R -,d , , ( C / B ’ ) ,obtained by taking coefficients modulo B’, clearly has kernel RB’, so R / R B ’ = d , , ( C / B ‘ ) .If RB’ c B then B / R B ’ corresponds to a nonzero ideal of d , , ( C / B ’ ) , which contains a nonzero element of C / B ’ = C / ( R B ‘n C ) x (C + R B ’ ) / R B ’ ,so B contains an element of C - RB’, conQ.E.D. tradiction. Hence RB‘ = B and R I B = d , , ( C / B ’ )as desired. Note d , ( F ) is also a PLID and PRID because of the symmetry given above and, thereby, is used in Chapter 3 as a source of interesting examples. More significantly, Weyl algebras are crucial to the theory of enveloping algebras of Lie algebras, cf., Chapter 8. Also cf., exercise 1.7.1. Let us return to the general setting.
Lemma 1.6.35: Suppose R = D[A; 1,6] where D is a division ring, and 0 # A a R . Then A = Ra for suitable a which centralizes D; if char(D) = 0 then 6 is inner.
Constructionof Rings
100
Proof: By proposition 1.6.21 we have A = Ra where a = 1' + x i 3 ; d i l i . Then for each d in D we have
ad = d1'
+ (t6d + d r - l d ) l ' - ' +
Now ad E A = Ra and deg(ad) = t = deg(a) so ad E Da; comparing coefficients of 1' proves ad = da. Now comparing coefficients of 1I-l shows t6d + d r - l d = dd,-,; so 6d = Q.E.D. [d, t-'d, - ' 1 if t {char@).
Proposition 1.6.36: Suppose 6 is a derivation on a division algebra D . (i) If 6 is inner given by [d,-] then there i s an isomorphism from the polynomial ring D[1] to D [ I ; 1,61 given by 1 -+ I + d , and eoery ideal is generated by a central element. (ii) I f 6 is not inner and char(D) = 0 then D[1; 1,6] is simple. Proof.. (i) [A - d , x d i A i ] = C ( [ l , d i ] A i - [ d , d i ] l ' ) = E(6diAi - 6diAi) = 0, so I - d acts like a commuting indeterminate, and we have the first assertion. Thus the second assertion can be checked in D[A]. Let A 4 D[A]. By the lemma we have A = D[A]a where a centralizes D and thus is in Z ( D [ I ] ) . Q.E.D. (ii) The contrapositive of the lemma.
Skew Power Series and Skew Laurent Series When 6 = 0 we can define a filtration on R[A;a] according to the lowest order monomial, as in example 1.2.26, which leads us to look for a generalization of power series rings and Laurent series in this case. Defnition 1.6.37: The skew power series ring R [ [ I ;a]] has the same additive structure as R[[A]] but with multiplication given by
(We assume 6:R + R is an injection.) Proposition 1.6.38: A = R [ [ I ; a]] is indeed a ring containing R [ I ; 03 canonically and has a filtration defined by putting A(t) = { riAi:ri = 0 for all i < t } . Consequently A is a domain if R is a domain.
,:I
Proof:
We prove A is a ring by using the regular representation as in the
41.7 Tensor Products
101
proof of proposition 1.6.15. The remainder of the proposition follows the discussion of filtrations in $1.2. Q.E.D.
Definition 1.6.39: The skew Laurant series ring R((A;a)),is the ring whose additive structure is that of R((A))= { x z m r i A i : mE Z , r i E R } but with multiplication given by (x,z,riAi)(C,",,r:Ai) = C,"=m+.(cI=",uDur:-u)Ai. There is an important subring denoted R[A, A - ' ; a] consisting of those elements having finite support, i.e., Z ; = m r i L iwhere m In are in h. Unfortunately R [ A , L - ' ; o ] is called a Laurant extension of R in the literature, but at least the notation is unambiguous. We write R[A,A-'] for RCA, A-'; 11.
Skew Group Rings It is easy to check that R[A, 1-'1 is isomorphic to the group ring R [ Z ] , so we are led to try to "skew" the group ring construction.
Definition 1.6.40: Suppose G is a group of automorphisms of R . Define the skew group ring R * G = rgg:r, E R } with multiplication given by the rule (rg)(sh)= (rsg)gh, where g h is the product in G, and sg denotes the element g(s) in R. We leave it to the reader to check that R * G is indeed a ring and that R * ( a ) z R [A,A- ;a] when D is an automorphism of R having infinite order. We shall refer to skew group rings only occasionally in the sequel.
{I,,,
'
61.7 Tensor Products The next construction to be considered in this chapter is the tensor product. This is an extremely important tool, some of whose uses will be outlined following the definition. We start with M in A d - R and N in R - M u d . Given an abelian group A we say II/:M x N -,A is a balanced map if satisfies the three conditions
+
Ill(& + X 2 , Y ) = W 1 , Y l ) + N X 2 7 Y l ) 4 + ( X , , Y l + Y 2 ) = W I ~ Y l+ ) W l l Y 2 ) +(Xlr7Yl) = ~ ( x 1 9 r Y l )
for all x i in M , y i in N , and r in R . This definition closely resembles that of bilinear form (cf., exercise 3). The tensor product turns out to be a universal for a suitable category involving balanced maps.
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Definition 1.7.1: The tensor product M Q N is the abelian group GIH, where G is freely generated (as Z-module) by the Cartesian product M x N, and H is the subgroup of G generated by all elements of the form
for xi in M, y i in N , and r in R. Write x Q y for the image of ( x ,y ) in M 6 N. If R is in doubt, we write M BRN in place of M Q N. Digression 1.7.2: Historically, the tensor product arose from a need to “multiply” two finite dimensional algebras A and B over a field F. Explicitly, given respective bases a,, .. .,a,,,and b l , . . .,b, of A and B over F, we could define A B B to be the vector space of dimension mn over F, whose base is written formally as {aiQ b,: 1 I i Im, 1 5 t In}. To define multiplication, one notes that multiplication in A and B are determined, respectively, by aiaj = aijkakand b,bu = &,,b, for suitable aijk,btUuin F ; then one could define
cF=
8 br)(aj @ bu)
(ai
=
k.u
aijkflruuak
0
and extend multiplication distributively to A Q B. The standard present-day treatment is much more general and elegant and still enables us to recover the above construction (cf., corollary 1.7.23 below). The approach is completely different. Whereas the x Q y span M Q N in Z.Mud they are not independent, so we cannot use them to define operations on M Q N. Instead, we rely almost exclusively on the following universal property.
Proposition 1.7.3:
There is a balanced map i,b: M x N -+ M Q N given by i,b(x,y)= x Q y. Moreover, if cp: M x N --* A is any balanced map there i s a unique group homomorphism Cp: M 8 N + A such that
cp(x B Y ) = cp(X,Y). Proof: The balanced map conditions are automatic from definition 1.7.1. To prove the other assertion, note cp extends naturally to a group homomorphism 9:G + A given by cpc(xi,yi)= cp(xi,y,) (notation as in definition
1
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51.7 Tensor Products
1.7.1);then H G ker cp yielding the desired Cp: G / H -+ A by Noether’s structure Q.E.D. theorem.
Corollary 1.7.4: Suppose f: M -+ M’ and g : N
-+ N ‘ are maps in A u d - R and R - A o d , respectively. Then there is a group homomorphism denoted f Q g : M Q N -+M’ Q N ‘ such that ( f Q g)(xQ y ) = f x Q gy.
Proof: Define cp: M x N -+ M ’ Q N ‘ by $(x,y ) = f x Q gy. Then cp is balanced; take f Q g to be 50. Q.E.D. Remark 1.7.5: Tensor products serve the following purposes: (i) keeping track of balanced maps; (ii) changing rings of scalars (for modules and algebras); (iii) providing a “multiplication” of algebras (over a common base ring); and (iv) providing universal constructions, such as the “tensor algebra”.
These uses of tensor product will occupy us for much of the book and could easily fill several volumes. For example, (iii) is the setting for the Brauer group and the ensuing theory of central simple algebras (Chapter 7) and Azumaya algebras (45.3). For the remainder of this section we shall introduce some of the main themes involving tensor products, largely postponing their very important categorical role until Chapter 4. Remark 1.7.6: If f:M -+ M ‘ and f ‘ :M’ .+ M “ and g : N N ‘ and g‘: N ’ -+ N “ are maps then f ‘fQ g’g = ( f ’ Q g’)( f Q 9). (Just check this on x Q y , which span M Q N . ) Consequently, there is a tensor functor from ( A d - R ) x ( R - A d ) to d&given by ( M ,N ) + M Q N and (f,g) -+ f Q g. (Note 1, Q 1, = l,@N.) -+
OR
Tensor Products of Bimodules and of Algebras Proposition 1.7.7: Suppose M E T - A d - R and N E R - A o d . Then M Q N E T - A d by the scalar multiplication a ( x Q y ) = a x @ y ; notation as in corollary 1.7.4, if f is also a map in T - A 0 d - R then f 0 g is a map in T - A d . Proof: First we redefine scalar multiplication in a manner which is obviously well-defined; namely, take the right multiplication map pa: M -+ M given by x + ax, and note ax Q y = (pa Q l N ) ( xQ y). Associativity is now
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Corollary 1.7.8 lf M E T - 4 u d - R then there is a functor M Q - from R - A o d to T - M d given by N + M 8 N and g + 18 g . Analogously, i f N E R - M o d - T there is a functor -8 N from J o d - R to Mud-T given by M + M Q N a n d f -+ f 8 1. Of course, we can always return to the original case by taking T = Z. For greater generality we may assume M and N are both bimodules.
Proposition 1.7.9: lf M E R-Mud-S and N R-Aud-T(as in proposition 1.7.7).
E
S-Aod-T then M Q N
E
A particular instance is when a ring T is viewed as a T-R-bimodule (for instance if R c T). Then TBR- is an important functor from R - M u d to T-Aod, which we shall study closely, and is called changing the ring of scalars from R to T. This technique is also applicable to algebras. Proposition 1.7.10: Suppose R and S are C-algebras. Then R Q S is a Calgebra with multiplication (rl 8 sl)(r2Q s2) = r l r 28 s1s2. Proofi Fix r 2 ,s2 and define f:R + R and g: S + S by right multiplication, i.e., fr = rr2 and gs = ss2. Then f Q g defines right multiplication by r2 8 s2, so the desired multiplication is well-defined. Associativity of multiplication follows from remark 1.7.6. Moreover, R Q S is a C-module, so it only remains to verify distributivity. To see this we work from the left, fixing a in R 8 S and defining II,:R x S -+R @ S by &(r, s) = a(r 8 s). Then is balanced and yields the map &: R Q S + R 8 S ; this is left multiplication by a and is thus distributive. Right distributivity is analogous. Q.E.D. Remark 1.7.11: If f:R + R' and g : T + T' are C-algebra homomorphisms then f Q g is also a C-algebra homomorphism. (This follows at once from proposition 1.7.7.)
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105
Properties of the Tensor Operation Having obtained the tensor product for modules and algebras, we are ready to ascertain some of its most fundamental properties. Proposition 1.7.12: Suppose Ri are rings for i = 0, 1, 2, 3 and Mi E R i &ud-Ri for 1 I i I3. Then (MI Q M , ) 8 M ,
%
,-
M , 0 ( M , Q M , ) in R,-Aod-R,
under an isomorphism sending ( x l Q x,) 0 x3 to X I 0 (
Q
~ 2 ~ 3 ) .
Proof: Fixing x in M , , define the balanced map t,hx: M , x M , + M , Q ( M , Q M , ) by $ J x 1 , x 2 )= x 1 Q ( x , Q x), yielding a group homomorphism $x:
MI Q M ,
+
MI Q ( M , 0 M3).
Now there is a balanced map cp: (MI Q M 2 ) x M , -+ M , Q ( M , Q M , ) given by cp( x l i 0 x Z i ,x ) = x l i Q x , ~ ) hence ; we have Cp: ( M , Q M , ) Q M, + MI Q ( M , Q M , ) sending (x, Q x,) Q x , to x 1 Q ( x , Q x3).Clearly (p is an R , - R , bimodule map, whose inverse can be constructed symmetrically, so (p is an isomorphism. Q.E.D.
1
1
Corollary 1.7.13: IfRiareC-algebrasthen(R, Q R , ) Q R , z R 1 0 ( R 2 Q R 3 ) canonically as C-algebras. Proposition 1.7.14: If M , N E C-Aud for C commutative then there is a Cmodule isomorphism M Q N + N Q M given by x Q y + y Q x ; this is an algebra isomorphism if M , N are C-algebras. Pro08
Define the balanced map $: M x N Q.E.D.
$ is the desired isomorphism.
+N Q M
by $(x, y ) = y Q x ;
One also has distributivity of the tensor product over direct sums (What else?). This is a special case of results in Chapter 4,but a short proof is available along the same lines as before. Ni) + @ ( M Q Ni) Proposition 1.7.15: There is an isomorphism M Q (eie, sending x 0( y i ) to (x 0y i ) . Proof: Define the balanced map t,h: M x (@N i ) + @ ( M Q N i ) by t,h(x,(yi))= ( x 0yi). Then $ is the desired map. To construct $-', let pi:Ni + @ Ni
106
Constructions of Rings
(aiE,
be the canonical injection and let f i = 1 @ pi:M 6 Ni + M @ Ni). Piecing together the fi yields a map f:@(M 0 Ni)-,M @ Ni)as in examQ.E.D. ple 1.4.25, and f ( x @ yi) = x @ ( yi),proving f and $ are inverses.
(0
Corollary 1.7.16: If F is a free W-module with base { x i :i E I } and R is a ring containing W, then R BW F is a free R-module with base { 1 6x i : i E I } .
Proofi View F as 0Wxi and R in R-AuA- W in the natural way.
Q.E.D.
Tensors and Centralizing Extensions One of the best ways of finding useful examples of tensor products is by means of the following notion. Definition 1.7.17: We say a ring T is a centralizing extension of R if T = CT(R)R,i.e., if T is generated by R and its centralizer (in T ) .T is a central extension if T = Z ( T ) R . Clearly central extensions are centralizing; centralizing extensions are also called extensions in the literature and are called liberal extensions in RobsonSmall [31]. Example 1.7.18: (i) For any monoid S the monoid ring R [ S ] is a centralizing extension of R. Likewise R [ I ] ,R [ [ I ] ] and , R((I))are all central extensions of R. (ii) MJR) is a centralizing extension of R (identifying R with the scalar matrices, whose centralizer contains the matric units).
Tensor products are also centralizing extensions, in a very special way. If R, R' are C-algebras there is a homomorphism f: R -+ R @ R' given by f r = r @ 1, and we write R @ 1 for f R . (We shall see below that f need not be injective in general.) We shall carry this notation for the next few pages.
Proposition 1.7.19: R Q R' is a centralizing extension of R @ 1, for any C-algebras R and R'. Moreover, if T is a centralizing extension of R and R' = CT(R) then there is an algebra surjection h: R Qc R' + T satisfying h(r @ r') = rr'.
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51.7 Tensor Products
Proofi Note ( 1 Q r')(r 0 1) = r Q r' = ( I Q 1)( 1 Q r'). Hence 1 Q R' centralizes R Q 1, and R Q R ' = ( I 0 R')(R 0 1). To prove the second assertion, we define the (onto) balanced map $: R x R' + T by $(r,r') = rr' and take h = 6:R 0 R ' + T. To check h is a ring homomorphism we note
h((r, Q r;)(r20 I ; ) ) = h(r1r20 r i r ; ) = r1r2r;r; = rlr',r2r> = h(r, Q r;)h(r2Q r;).
Q.E.D.
Corollary 1.7.20: Suppose T is a centralizing extension of a C-algebra R and R' = C,(R). If R ' contains a base B of T as R-module, then T x R Q R'. Proofi Let h:R 0 R ' + T be the surjection of proposition 1.7.19. If ri Q xi E ker h for x i E B then r i x i = 0, implying each ri = 0 by hypothesis; hence ker h = 0, proving h is an isomorphism. Q.E.D.
I:=
1
This somewhat technical result has several immediate applications.
Example 1.7.21: Assume R is a C-algebra, where C is an arbitrary commutative ring. Tensors below are taken over C . (i) The monoid ringR[S] has S as a base (as R-module), implying R[S] x R Q C[S]. (ii) MJR) has the eij as a base, implying MJR) x R 0 M,,(C). (iii) Let us partition M,,(R) into blocks of size n x n. Thus there are in2 blocks, and to the u-u block we define Euv = ~ ~ = l e ( , - l , ,$(,-, + il ) n + i for 1 Iu, u Im. There is an injection M,,(R) + M,,,(R) given by a + (". *
i.e., the block a is repeated m times along the diagonal and we write 0's elsewhere. Identifying M,,(R) with its image, we see easily that its centralizer contains E,,, and these E,, are a base for M,,,,(R) over M,,(R). Since the E,, are a set of m x m matric units we conclude that M,,,(R) x M,,(R)Q M,(C). In particular, M,,,(C) x M,(C) 0 M,,(C). (iv) The Weyl algebra d , , ( C )x d l ( C )Q ... 0 dl(C),cf., exercise 1. Example Z.7.2Z': Another useful example of tensor product is as follows: Suppose A 4 R. Then MIAM z (RIA) ORM for any R-module M. (Indeed, the map M 4(RIA) ORM given by x + 1 Q x has kernel containing AM, so we get a map MIAM + (R/A) ORM. Its inverse is constructed from the balanced map RIA x M .+ M/AM given by ( r + A,x) + rx + AM.) Connections of this kind between tensor products and changes of ring are very
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important because they give us the opportunity of obtaining general results about rings from properties of the tensor functor (corollary 1.7.8).See proposition 2.1 1.9ff and Ej4.2 to see how this idea unfolds. The following easy module-theoretic result is a useful converse to corollary 1.7.20. Proposition 1.7.22: T = R QwR'.
Suppose R, R' are rings canonically containing W,and let
(i) If R' is free in W-Mud with base B' containing 1 then the canonical maps R -+ T and R' -+ T (given by r + r 0 1 and r' + 1 0 r') are monic. Thus we may view R,R' c T in which R n R' = W. (ii) If R, R' are free in W-Mud with respective bases B, B' then T is free in W-Mod with base { b 0 6': b E B and b' E B'). Proof: (i) Corollary 1.7.16 implies that T is free as R-module with base { 1 0 b': b' E B } containing 1 0 1. Obviously, if 0 = r 0 1 = r(l 0 1) then r = 0; if r' = 1 wib: E R' and 0 = 1 0 r' = wi 0 b; then each wi = 0 so r' = 0. Thus R C_ T and R' C_ T canonically. If r 0 1 = 1 0 r' = 1 0 1 wibi then matching components shows r = w I E W, proving R n R' = W. (ii) If r' = wjb; and r E R, then writing rwj = wijbi in R, we have r 0 r' = rwj 0 bi = wijbi0 b;, proving the b Q 6' span T. To prove they are independent, suppose 0 = wijbi 8 bJ = Xi( wijbi)Q bJ for suitable wij in W; then each wijbi = 0 so each wij = 0, as desired. Q.E.D.
1
c
c
c
1
c
1
xi
Tensor Products Over Fields The classical theory of tensor products is for C a field, in which case the subject becomes quite straightforward. Corollary 1.7.23: Suppose C is a field, and T = R 0 R'. The canonical maps R R 0 1 and R' -+ 1 0 R' are isomorphisms; thus we may view R c T and R' c T, and T is a centralizing extension of R. Moreover, if 8,B' are respective bases of R, R' over C then { b 0 b':b E B, b' E B ' ) is a base of T over C. -+
Proof: By proposition 1.7.22(since R' has a base including 1).
Q.E.D.
Corollary 1.7.24: Viewed in R 0 R', we have Z(R) 0 Z(R') E Z(R 8 R'), equality holding i f C is a jield.
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109
Proof;. The first assertion is obvious. To obtain the reverse inclusion, expand a base B, of Z ( R ) over C to a base B of R (over C) and expand a base Bb of Z(R’) to a base B’ of R‘ (over C). If x c i j b i 0 bJ E Z ( R 0 R’) then 0 = [ C c i j b i@ b;,r @ 1 1 = x[cijbi,r] 0 b3 for each r in R, implying (for each j )0 = cijbi,r] for each r in R, so each cijbi E 2.Thus cij = 0 whenever bi # B,; likewise cij = 0 whenever bj $ Bb, implying cijbi @ bJ E Z ( R )@ Z(R’). Q.E.D.
xi
[Xi
1
We would like to match the ideal structure of R 0 R’ with those of R and R‘. Of course one should not be too optimistic. Indeed if K is a finite field extension of F then the canonical surjection K OFK+ K of proposition 1.7.19 sending a, 0 a, H a,a, is not 1 : 1 , by a dimension count. Thus K OFK is not a field and thus has zero-divisors. Our next example should be compared to theorem 2.5.36 below. Example 1.7.25: Suppose K is a purely inseparable field extension of a field F ; then K OFK has nonzero nilpotent elements. (Indeed, take a E K such that u p E F, where p = char(F). Then (a 0 1 - 1 @ a)p= u p@ 1 - 1 @ a” = 0.) Nevertheless, there are positive results to be had when C = Z ( R ) .
Lemma 1.7.26: Suppose R is simple and a, b E R are linearly independent over Z ( R ) . Then there are wil,wi2 in R, 1 Ii I k ( f o r suitable k ) such that Cf= wilawiZ# 0 and wi,bwi, = 0.
If=
Proof;. Otherwise there is a well-defined map f:RbR + R given by
for all ril,ri, in R, and f is clearly an R - R bimodule map. But RbR = R. Writing z = f l we have zr = ( f l ) r = f ( 1 r ) = f ( r 1 ) = rz, implying z E Z ( R ) ; Q.E.D. then a = f ( 1 b ) = zb, contrary to hypothesis. Theorem 1.7.27: Suppose C = Z ( R )and R is a simple C-algebra. Then every nonzero ideal A of R QcR‘ contains an element of the form 1 Q r’ # 0; in particular if R‘ is also simple then R QcR’ is simple. Proof: Choose 0 # a = rj 0 r; E A with u minimal possible. If u 2 2 then there are elements wil, wiz in R for 1 Ii Ik such that letting sj = wilrjwi2we have su- # 0 and su = 0. Then
110
Constructionsof Rings
contrary to minimality of u. Thus u = 1, and a = r l 0 r ; . Then 1 0 r; E ( R 0 l ) a ( R0 1) G A, proving the first assertion. If R‘ too is simple then 1 0 1 E (1 0 R’)(1 6 r ; ) ( l 0 R ’ ) G A proving Q.E.D. R 0 R‘ is simple. Corollary 1.7.28: If T = R W where R is simple and W is a Z(R)-algebra centralizing R then T M R @Z(R, W.
Proof: Otherwise the canonical epic cp: R 6 W -,R W given by cp(r 0 w ) = rw is not monic. But then there is some 0 # 1 0 w E ker cp which is absurd since 0 = cp(l0W ) = W . Q.E.D. Corollary 1.7.29: Suppose T is a finite dimensional subalgebra over a field F, and T = R OFR‘ where R is a simple subalgebra. Then R’ = CT(R). Proof: Let R” = CT(R). Obviously R’ c R”. But then T = RR” so T M R QFR” by corollary 1.7.28,yielding[ R : F ] [ R ” : F ]= [ T : F ]= [ R : F ] [ R ’ : F ] . Hence [ R “ :F ] = [ R ’ : F ] and thus R‘ = R“. Q.E.D.
In describing subsets of M 0 N it is convenient to write A 0 B for { c i a iQ b i :ai E A , bi E B}, where A c M and B c N. One must be careful not to view this as the tensor product of A and B even though the notation is the same. For example, the tensor product 22 0z(2/2Z) # 0; however, viewed inside 2 Bz(2/22) we have 2 2 0 (2/22) = 0 since any 2m 0 ii = m 0 2ii = m@O=O. Proposition 1.7.30: lf f:M + N is an epic R-module map and M‘ E A d - R then ker(1,. 0 f) = M’ 0 kerf (viewed in M’ QRM ) . Proof: Write K = M‘0 kerf. Then K I ker(1 0 f ) so 1 0 f induces a map f: ( M ’ 0 M ) / K + M‘ 0 N satisfyingf(x’ 63 x + K ) = x’ 0 fx. We shall show K = k e r ( l 8 f ) by proving f is an isomorphism, i.e., by constructing ?-I. Define the balanced map t,b: M‘ x N -P (M’ 0 N ) / K by $(x’,fx) = x’ 0 x + K. Then $: M‘ 0 N + (M’ 0 N ) / K satisfies q ( x ’ 0 fx) = x’ 6 x + K as desired. Q.E.D.
Tensor Products and Bimodules We have seen already that when transferring properties between rings R E R’ one often relies only on the R - R bimodule structure of R’. However,
$1.8 Direct Limits and Inverse Limits
111
bimodules are very complicated; for example, try describing the free R - R bimodule in terms of a base! At times the tensor product is a useful tool in studying bimodules, in view of the following fact: Proposition 1.7.31: Any R QaRoP-moduleM can be viewed as R - R bimodule under the scalar multiplications rx = ( r Q 1)x and xr‘ = ( 1 Q r‘)x, leading to an isomorphism of the categories R Qz R o P - A u d and R - A d - R , for any ring R. Proofi are
The two verifications of interest in proving M is an R - R bimodule
(rx)r‘= ((r Q 1)x)r’ = ( 1 Q r’)(r Q l ) x = ( r 0 1 ) ( 1 Q r‘)x = r(xr’). x(r’,r;) = (1 Qr’,r;)x
= ((1 Q r;)(l Q r’,))x = (1 Q r;)(xri) = (xr’,)r;.
Any R Q RoP-module map f:M + N becomes an R - R bimodule map, so we have a functor F : R Q R O P - A o d -P R - A u d - R . To obtain the inverse functor one takes an R - R bimodule M and defines a scalar multiplication as follows: The usual balanced map argument yields maps I,!Jx: R Q RoP-P M (for fixed x in M ) given by IL,(rQ r‘) = rxr‘; now allowing x to vary, define ( r Q r’)x = I,!Jx(rQ r’). To check M is a module over R Q RoP we observe
+
(rQr’)(x,+ x2) = r(xl + xz)r’ = rxlr’ + rx2r‘ = (rQr’)xl (rQr’)x2. ((rlQr\)(r2 Qr;))x = ( r l r 2Qr;r’,)x = rlr2xr;r’, = rl(r2xr;)r‘,= ( r , 0 r‘,)((r20 r;)x).
The other verifications are clear, and any R - R bimodule map becomes an R Q R’P-module map under this action, so we have the inverse functor to F, as desired. Q.E.D. Corollary 1.7.32: Let F = (R Q Rap)("), with base e l , . . .,en. Then F can be rieirI = (ri0 ri)ei. viewed as free R - R bimodule under the action Likewise, free R - R bimodules of infinite rank can be constructed.
XI=,
1
51.8 Direct Limits and Inverse Limits In $1.10 we shall explore localization over a central monoid. The proof that localization exists is cumbersome, and it actually can be obtained more easily via “direct limits,” a general construction which is useful also in many other
Constructions of Rings
112
settings. We shall also consider the dual construction, inverse limits, as well as completions as a special case.
Direct Limits As motivation, recall how the coproduct of {A,: i E I } in a category % is the “smallest” object A containing each A i “independently” (e.g., when %? = R-Aod). We would like to perform a similar construction when the A i are “glued together” in some way. An extreme case is when I = N and A l G A , C A, c ...; then one would take A = A,. Of course, A, G A , , really means there is a canonical injection cp: A i-,A i + 1, and there is also the problem of finding the suitable set in which to form the union A. To handle these two difficulties in a manner which also embraces the coproduct, we assume the index set 1 is given an arbitrary preorder I (i.e., < is reflexive and transitive).
u
Definition 1.8.2: A system (A,; (pi) is a set of objects {A,: i E I } in a category V, together with (pi in Hom(A,,Aj) whenever i 5 j, subject to the following two rules: whenever i 5 j Ik. (1) (p:(p{ = cp:
(2)
(pj =
lA,
for all i.
Examples 1.8.2: (i) if I has the trivial preorder then the cpi only exist for i = j since any i # j are incomparable. (ii) Any chain is a system, where I is given the usual order of ordinals, and cp; is the inclusion map. Definition 1.8.3: Suppose I is a preordered set, as before. The direct limit of the system (A,;cp{) is an object denoted 1 4A, (or, more precisely, 14(A,;q { )if necessary) together with morphisms pi:A, + & A, satisfying pi = p j q i for all i I j; for which given any family of morphisms gi:A, + A satisfying g, = gj(p{ for all i I j we have a unique g: lim A, + A such that gp, = g ifor each i, i.e., g completes the diagram
-
for each i Ij. Direct limits are defined by a universal property, and so are unique if they exist. In fact, they fit into a natural categorical setting, cf., exercise 4.
$1.8 Direct Limits and Inverse Limits
113
Example 1.8.4: We interpret the three examples of example 1.8.2 (when they exist). (i) If I has the trivial preorder then hr~ Ai is the coproduct (ii) If I is a chain then 1 4A i can be interpreted as A i .
u
Ai.
Thus we see the direct limit unifies several examples, but to obtain full benefit we want to prove that in these cases the limit must exist.
Proposition 1.8.5:
Direct limits always exist in R - A d .
Proof: Let ( M i ;qi) be a system of R-modules. Letting p j : Mj + @ M i be the injection into the j-th component, let N = c i s j ( p j q i- p i ) M i I@ M i . Each p j induces a map p j :M j + (@M i ) / N , and the constructions show ,iijq{= pi for all i Ij. We shall show &(Mi: pi) is ( @ M i ) / N together with the pi. Indeed given gi: Mi -+ M satisfying gjcpi = gi we first define the unique map 9’:@Mi + M satisfying g’pi = g i for all i. Then 0 = gJq .1!- g i = g’p.cp! J I - g‘pi = g ’ ( p j q i - pi) so N I ker 9’. Hence 9’ induces a map g : @ M i / N + M satisfying gpi = g i , as desired; g is unique since the p i M span @ M i / N . Q.E.D. On the other hand, we can also prove existence by restricting the systems under consideration.
Definition 1.8.6: A preordered set I is directed if each pair of elements has an upper bound. (It follows at once that each finite subset of I has an upper bound.) A directed system is a system indexed by a directed set. Theorem 1.8.7: Suppose the category %? is dejned in a Jirst-order language. Each directed system ( A i ;cpi) in %? has a direct limit. Proof: Let A’ be the disjoint union of the sets A i . Each element a of A’ then belongs to some suitable A i , and we indicate this by writing a“’ inon A’ by puting a ( i ) b”’ iff stead of a. Define the binary relation cpfa“’ = cp!b(j’ for suitable k 2 i, j. (Intuitively a and b “agree” far enough along I . ) Then clearly cpya“’ = cpj”b(j) for all m 2 k , from which it follows at once that is an equivalence, whose set of equivalence classes A ’ / - we denote as 2. It is easy to see 2 E ObV since each defining axiom involves only a finite number of elements and thus a finite number of indices, so there is an upper bound for these indices at which the appropriate verifications can be made. (For example, if %? = 9ing: then [O“’] = [O”’] and
-
-
-
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[l'"] = [ l ' j ) ] for each i, j since there is k 2 i, j and q10") = O(k)= qfO'"; moreover, given arbitrary elements [a'"] and [b'j)] in A take k 2 i , j and define [a(')]f [b")] = [qra(') f qfb("] and ab = [q:a")q;b(J)]and check easily that 2 is a ring.) The same argument gives us morphisms p i : Ai + 2 defined by pi& = [a'"], and p j q j = pi since q$") a") for all i I j . We claim ( & p i ) is the direct limit. Indeed, given gi: Ai + A satisfying g j q i = gi for i Ij , we want a unique g : A -+ A with gi = gpi for each i. As usual, uniqueness is obvious since 2 = p i A i . Existence of g is obtained by defining g[a'')] = gia('); this is well-defined since if [a")] = [b")] then taking qika(')= qjkb(j)we have gid" = gkqika")= g k q j k b ( j )= gjb(j).One sees easily that g is a morphism. Q.E.D.
-
u
Example 2.8.8: We return to the systems of example 1.8.2, whose direct limits were described in 1.8.4. (i) If I has the trivial preorder then I is not directed, and, indeed, we have seen the coproduct (which is the corresponding direct limit) need not exist in the category of nontrivial rings. Example 2.8.9: (A module is the direct limit of its f.g. submodules). Suppose M E R - A u d and {Si:iE I } are the finite subsets of M, viewed as a directed poset under set containment. Writing Mi for the submodule spanned by S i , one sees at once that M = lim Mi since any maps gi: Mi -+ N can be "pasted" together to a map g : M -+ N. The following result shall be generalized later (c.f., theorem 4.2.10): Proposition 1.8.10: If (4; ji) is a system in R-Aud with direct limit N and K ~ S - . d u d - R ,then ( K B R N i ; 1 O q ; ) is a system in S - d u d and K @IR N = hJ(K BRNi). 1 8 cpi) is a system in S - A d . If pi:4-+ N are the caProof: ( K OR4; nonical morphisms then we have morphisms 1 6p i :K 6Ni + K 8 N satisfying 1 8 pi = 1 @ p j q i = ( 1 @ pj)(l 6 qi)for i Ij , so it remains to verify the universality condition. Given gi: K 0 Ni -+ M satisfying gi = gj( 1 6 qi) we need g: K 0 N + M satisfying g(l 6p i ) = gi for each i. Fixing a in K define h i a : 4+ M by hiax = g i ( a 8x ) for x in N,.Clearly h, = hjaq{ for i Ij , so by hypothesis there is ha:N + M satisfying hapi = h, for each i. Note for any r in R that hi,,,x = hia(rx),from which it follows ha,x = h,(rx). Thus we have a balance map h: K x N + M given by h(a,x) = h,x, yielding a map g : K 8 N -+ M given by g(a 6x ) = h(a,x). Now g(l @ pi)@ @ x ) = g(a 8 pix) = ha(pix)= hiax = gi(a 6x ) for all a in
$1.8 Direct Limits and Inverse Limits
115
K and x in N , proving g(l 0 pi)= g,; uniqueness is seen by fixing a and Q.E.D. noting the h, are uniquely given.
D Supplement: Projective Limits The dual construction to direct limits is called “projective” or “inverse” limits. Whereas the definition is obtained formally by reversing arrows, the flavor is different. Direct limits involve “pasting together” objects, whereas inverse limits often involve a topological completion. Let us start with a category W and a preordered set I . An inverse system is a set { A , :i E I } of objects in %? together with morphisms (pi: A j A , for j satisfying each i I -+
(p: =
cpicpj
(pj =
1,;
whenever i I j I k. for all i.
Definition 1.8.11: The inverse limit (denoted @ I F and also called the projective limit) of the inverse system (A,;cpj) is an object @ A , of %, together A , --t A , with q i v j = vi for all i I j , which satisfy the with morphisms v,: lim t following property: For every A in Ob% and every set of morphisms gi:A + A i satisfying (pigj = g i for all i I j , there is a unique g: A -+ lim A , satisfying vig = g i for t each i in I . If I has the trivial preorder we have redefined the direct product. The usual argument shows that if the inverse limit exists it is unique up to isomorphism. We want to find situations in which it is guaranteed to exist; to obtain a good general condition we assume the inverse limit exists and see where it leads us. Suppose W is a concrete category with direct products, and suppose (A,;cpj) is an inverse system having an inverse limit. From the characterization of product, letting nj: A i-, A j denote the projection, we have a A i+ A isuch that n i f = vi for each i. For each unique morphism i < j we then have nif = vi = cp:vj = (p$njf;thus each (a,) in f(@ A,) c A i satisfies a, = viaj for all i 5 j . In particular, we have the following necessary nondegeneracy condition:
n
n
{ ( a , )E
n
fl A , : ai = q j a j for i I j } z 0
We shall call this the nondegeneracy condition (for inverse limits), and now show it often is sufficient as well as necessary in the important case
Constructionsof Rings
116
that % is a category defined in a first-order language. Suppose ( A i ; q j ) is an inverse system satisfying the nondegeneracy condition, so that L = {(ai)E A,: a, = qfaj for i I j} # 0.Also suppose L E %‘ and , let v,: L + Ai be the restriction of ni to L. If A E Ob %? and 9,:A + A , satisfy cpigj = gi then define g:A + L by ga = (gia).By definition we have v,ga = gia for all a in A, proving L = lim A , . This defines inverse limits in R-A’od, 9&g, and d8, and more generally for “universal algebra,” c.f., Jacobson [80B, Chapter 21.
n
The Completion We turn now to the most important special case of inverse limit. We shall carry it out first for abelian groups and then consider rings and modules as special cases. Example 1.8.12: Suppose we are given a set of subgroups { B i :i E I } of an abelian group G. Preordering I by putting i 5 j iff Bj E B,, we get an inverse system (G,;cpj) where Gi = G / B i and cpi: G j -P G, is the canonical map (viewing Gi as Gj/(Bi/Bj)).The nondegeneracy condition is satisfied by (g + B,) for each g in G, so we can form 6 = l@ G i , called the completion of G with respect to the Bi. We have a canonical group homomorphism $: G + given by $g = (g + Bi); $ is an injection iff Bi = 0.
+
n
n
Explicitly 6 = {(g, Bi) E G,: gj - g iE Bi for all j 2 i } . Thus $: G + 6 is onto iff for every such sequence (Gi + B,) there is g in G with g - giE Bi for each i ; in this case we say G is complete. Another description of the completion comes from topology. We say (g,) in G‘ is a Cauchy net if for each i there is some i’ for which gj - gi.E Bifor all j 2 i’. (Then for all j, j’ 2 i’ we have gj - gY = (gj- gi.)- (gj. - gi.)E Bi). {Cauchy nets} form an abelian group Cauchy(G) under component-wise addition, and are the set of convergent nets under the topology obtained by viewing {B,:i E I } as a neighborhood system of 0. There is a group homomorphism Cauchy(G) -+ obtained by “weeding out” the terms of a Cauchy net so that i‘becomes i for each i. The kernel N is the set of Cauchy nets which converge to 0; i.e., for each i there is some i’ for which gj E Bi for all j 2 i’. Thus 6 x Cauchy(G)/N, which is the topological completion of G.
niel
Digression 1.8.13: The topology on G (defined above) is Hausdorff iff Bi = 0. (Proof: (a) If g E Bi then 0 lies in the closure of {g}. (e) If
n
41.8 Direct Limits and Inverse Limits
I17
+
g # h~ G then g - h 4 Bi for some i, so g Bi and h proving the space is Hausdorff.)
+ Bi are disjoint open sets,
Example 1.8.14: Suppose R is a ring with filtration over (Z, +).Taking Bi= R ( i )we have Bj' Bi for all i i j so we can form R as above. In fact d is a ring with multiplication by (ri Bi)(ri + Bi) = (rir: + Bi), in the case that each Bi a R. There are two cases of interest:
+
+
(i) R ( i ) = 0 for all i > 0; then any Cauchy net (ri Bi) is equivalent to Bi), so R z R canonically, and R is complete. (ii) R has a filtration over N (i.e., R(i) = R for all i 5 0); then has a filtration over N, given by R ( n ) = {(r, B,): each ri E B,,}. Note this implies ri + B' = o for i I n (since B' 2 B,,). Consequently, R n k(n)= B,, = ~ ( n ) . Also if (r; + Bi) E k ( n ) then (ri Bi) - (rn+ Bi) E k ( n 1); since r,, E B, we see R(n) = ~ ( n ) R(n 1).
(r,,
+
+
+
+
,+
+
,
+
Suppose M = M, > M , > M , > ... in R-Mod. We can form the comis in fact an kpletion M of M with respect to the Mi. If M i= BiM then module under scalar multiplication (ri + Bi)(xi Mi) = (rixi Mi). In particular, we have the following special case of example 1.8.14(ii).
+
+
Definition 1.8.15: Suppose A a R. The A-adic completion of R is the completion of R with respect to the A-adic filtration (over N) given by R(0) = R and R ( i ) = A' for i > 0. Likewise the A-adic completion M of an R-module M is its completion with respect to the submodule Mi = A'M. This definition was motivated by the p-adic integers; here R = Zand A = pZ. The A-adic completion i? of R has the ideal A^ = {(ai A') E k : a , E A ) which contains $ A (recalling t,b: R + k is the canonical ring homomorphism given by t,br = ( r + A')) and plays the following key role:
+
Theorem 1.8.16: Notation as above, R I A idempotent-lifting ideal of R.
%
RIA^ canonically and
A^
is an
Proof: Composing t,b with the canonical map R + d/A^yields a homomorphism R + R / i whose kernel is A, yielding an injection $: R I A + RIA^. But II/ is onto, for if (r, + A') E R then (ri A') - ( r l A') E A^, SO (ri A') A^ = $(r, A). Thus R I A % R / i . Next we show 1 - a ^ is invertible for any element a = (a, + A') in 2. Note if (ai + A') E A^ then a , E A so a, E a , +A' G A
+
+
+
+ +
Constructionsof Rings
118
for each i, implying a: E A' and (1 - a , ) ( x { ~ i a=~ 1) - a; E 1 + A'. Thus a: we see (bi + A') is the desired inverse of 1 - 6. putting bi = It remains to show every idempotent e =(ri + A') + A^ of @A^ can be lifted to d . We want to write e = (ei+ A') + A^ where each ei + A' is idempotent in RIA'. Since r: - rl E A we can take el = r l , and we work inductively on i. Suppose we have found e l , . . .,ei for i 2 1. A'/A'+ is an idempotent-lifting ideal of RIA''' since (A'/A'+1)2 = 0. Viewing R/A' z ( R / A ' + ' ) / ( A ' / A ' + ' ) we can lift ei + A' to an idempotent e i + l + A'+' of RIA''', thereby completing the induction step. Q.E.D.
Ctli
Proposition 1.8.17: Let M be the A-adic completion of M , and f: M + M be the canonical map given by f x = (x + A'M). I f M = R x j then M = k f x j . In particular, if M is f . g . then fi is an f . g . d-module.
I:.=,
Proofi Given y^ = yi + A'M in auiE A' and zui E M . Then zui =
fi we write y i +
- yi =
xu= k
aUizUiwhere
ruijxjfor suitable ruij in R, so
Putting gj =(bij+ A') where bOj= 0 and, inductively, b , 1, = b, + = auiruij, we see gj E d and yi = Q.E.D. bijxj, implying y^ = bjjZj.
If=
81.9 Graded Rings and Modules In the last 20 years considerable interest has been attracted to the theory of graded rings and modules, primarily because any ring with filtration over Z has an "associated graded ring whose structure bears heavily on the structure of the original ring. This construction is deferred until definition 3.5.30, since it is most relevant to Noetherian rings; here we lay out the basic framework of the theory of graded rings and modules, including the category and its "free" objects. Then we construct the "tensor ring," an important source of examples of graded rings, and apply it to the free product construction. The reader interested in graded rings should pursue this subject in theorem 2.5.30 (the Jacobson radical), proposition 3.5.31 (the associated graded ring), definition 5.1.36 thru theorem 5.1.41 (the &,-theory), theorem 6.2.9 (the Golod-Shafarevich counterexample), and 58.3, 8.4 (enveloping algebras). Also, cf., exercises 2.6.1 1ff. Definition 1.9.1: Suppose S is a monoid. A C-algebra R is S - p a a d e d , or graded over S, if R = @ { R , : s E S } for C-submodules R , for which R,R,. E
51.9 Graded Rings and Modules
119
R,,,. Graded ring means graded Z-algebra (i.e., the R, are abelian groups). M, and rsxs,E Suppose R is S-graded. An R-module M is graded if M = M,,, for all r, in R, and x,. in M,.. R-%-&od is the category of graded Rmodules, whose morphisms are the maps f : M + N satisfying fM,c N, for each s. (These are called graded maps.) R-%-FLmod is the subcategory consisting of graded f.g. submodules. Likewise { S-graded C-algebras) is a subcategory of C - d t f whose morphisms are the algebra homomorphisms f : A + B satisfying fA, E B, for each s. If M is an S-graded module we call each M, a homogeneous component and x, for x, in M,, we define the call its elements homogeneous; writing x = Ifinit= support supp(x) = {s E S:x, # 0}, a finite set which is nonempty iff x # 0.
eSEs
Example 1.9.2: (i) A = R [ S ] is graded by S, where A , = Rs. (ii) A = M , , ( R ) is Z-graded, where A,=xj=i+uReij;thus AU=O for ( u J 2 n . (iii) Suppose is a primitive n-th root of 1. A = Z[[] is graded by Z/nZ, where A, = ['Z. (N, +) and (Z, +) are the grading monoids of greatest interest, largely in view of the following tie to filtrations.
Remark 1.9.3: If S is an ordered monoid then any S-graded algebra A has the filtration A(s) = Is,SsA,,, which is valuated by ua = A{supp(a)) for a # 0. Similarly, one can define the degree function dega = V{supp(a)}. Remark 1.9.4: In view of remark 1.2.15 the following conditions are equivalent when the grading monoid S of A is ordered: (i) A is a domain; (ii) u: A+S, is a monoid homomorphism; (iii) the product of nonzero homogeneous elements must be nonzero; (iv) deg(ab) = (deg a)(deg b) for all nonzero a, b in A (i.e., if one side is defined then both sides are defined and equal). It is easy to characterize the kernels in graded categories. If M is an Sgraded module and N < M , we write N, for N n M,. Clearly, N, I N ; we say N is graded if N, = N.
1
1
Remark 1.9.5: (i) The kernel of a graded map is a graded submodule. Conversely, if M is an S-graded module and N is a graded submodule then M / N is graded (by putting ( M I N ) , = M,/N,) and the canonical map M + M / N is graded. (The easy proofs are left for the reader.)
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Constructionsof Rings
(ii) The kernel of a graded homomorphism is a graded ideal; conversely, if R is an S-graded algebra and A 4R is graded then RIA is a graded algebra (as in (i)) and the canonical homomorphism R --* RIA is graded. Suppose g : M + N is a graded epic which splits (but not necessarily by a graded map). Then g does split by a graded map, i.e., there is graded f: N + M for which gf = 1,. (Indeed, we take ungraded f':N - r M for which g f ' = l,, and define f,:N,+M, to be the composition N,+N-+M+M, f' Remark 1.9.6:
where the outside arrows are the canonical maps. Then f = @fs: N + M is graded and gfx, = (gf'x), = x, for all homogeneous x,, implying gf = l,.) The category R-$+Aud turns out to be quite useful, especially for S = (N, +). If R is S-graded then R(")E R-Y+Fimod, viewing RP) as (R,)("). However, R(") is not necessarily free in R-%-9imud since its generators are in R?), whereas a graded module could be generated by homogeneous elements from other components. To rectify this situation we need a larger module. Given any S-graded module M we write M ( s ) for M with the new grade given by M(s),, = M,,,. Then &R(s) is a graded module which has the following "freeness" property. Remark 1.9.7: Suppose M = Rxi is S-graded. If x i = E x i s then each xis E M so M is generated by homogeneous elements. Moreover, if f,:R(')+
c;=,Rxis are epics then @fs:@R(')(s)+ @R(')(s) x (R('))(') is a free R-module.)
A4 is a graded epic. (Note
Finite grading sets are important largely because of the following link to Z and N. Remark 1.9.8: Suppose $: S
+ S' is a monoid surjection. Any S-graded algebra R can be S'-graded by defining R,, = R, for u in S'. This yields a functor from {S-graded C-algebras} to { S'-graded C-algebras}.
Applying this technique to example 1.9.2 (ii) yields a 2122grading of M,,(R) called the checkerboard grade: eijis in the 0 (resp. 1) component iff i + j is even (resp. odd). Example 1.9.9:
As an immediate consequence of proposition 1.7.15 we have the following tensor product connection:
Proposition 1.9.10 Suppose R is an S-graded ring. Then R, is a ring, and for any subring W of R, there is a functor R Ow- porn W-Aud to R-%-Aod.
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121
In case W = R,, this functor identijes R,-Aud with the full subcategory of R-Yer-Aud consisting of those graded modules M for which M = R M , . Note that we used the multiplicative notation for S ; in many applications S is written additively, with 0 as the neutral element, so we would write R, and M , instead of R, and M , .
-
Proof;. R, is closed under multiplication since 1 1 = 1. Given N in W-Aud we grade RQwN by (RQN),=R,QN, proving the first assertion. If W = R, then (R Q N), = R, 0 N sz N , so the functor M + M , (going the other direction) shows R,-Aud is categorically equivalent to the specified subQ.E.D. category of R-Ya-Aud.
Remark 1.9.11: Let us refine proposition 1.9.10. If R is S-graded then we can grade R, trivially (i.e., (R,),=O for all s # 1) and thereby have R,-Yer-Aud. These are merely the R,-modules, graded as a direct sum of submodules indexed over S . There is a functor F = R OR,-: R,-Yer-Aud + R-Yer-dud whereby we grade R OR,M by putting ( r 0 x), = ~ u v = s 0 r ux,. In the other direction, there is a functor G = R, OR-: R-Yt-Aud + R1-Yer-Aud whereby R, ORN is graded via (r Q x), = r 0 x,. G F is naturally isomorphic to the identity under the usual tensor product isomorphism. However, FG is not naturally isomorphic to the identify since the grade may be changed. We shall investigate this further in the proof of theorem 5.1.34.
Tensor Rings One useful type of graded ring arises from tensor products. Defnition 1.9.12: Suppose M is an R - R bimodule. Define the tensor power M @ " by M@O = R, M @ ' = M and, inductively, M @ ( " + l = ) M QR M @ " , viewed as R-bimodule in the natural way. Define the tensor ring T ( M ) = M @ " .When R is ambiguous we write T R ( M )for T ( M ) .
Clearly T ( M ) is an R - R bimodule; to show T ( M ) is a ring we need a multiplication. First we note M @ "is spanned by elements x,@(x,Q( ***Qx,)), which we write more simply as x, Q Q x,. By proposition 1.7.12 we have M 0 M + M wrn+") given by (x, 0 . * .0 x,,,) Q an isomorphism qmn: ( y , Q . . . Q y,) + x, Q * * * Q x,,, @ y , 8 . * - Q y , . Applying proposition 1.7.15 twice we have a map cp: T ( M )Q T ( M )+ T ( M ) @'
given by a,,, Q b, -,cp,,,,(a,,,Q 6,) for a,,, in M
and b, in M @",
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122
Now we have a multiplication in T ( M )by defining the product of a and b to be q(a @ b), and it is easy to verify the ring axioms. Summarizing, we have Proposition 1.9.13: T ( M )is an N-graded ring, with T ( M ) ,= R, T ( M ) ,= M , and, in general, T ( M ) ,= Ma”, multiplication given by (XI @
*
@ xm)(yl@ * @ y n )
= XI@ * * * @ X,
0~
1*.
’00Yn.
If R is a C-algebra then T ( M ) is also a C-algebra under multiplication c(x1 @ . . . @ x n ) = ( C l X 1 ) @ * . . @ X n .
The customary special case is when C is a commutative ring and M E C dud is viewed symmetrically as a bimodule. Then Tc(M) is called the tensor algebra of M . Tensor rings and tensor algebras are important as universals. To see this we first note there is a functor from { N-graded Calgebras} to C - d u d given by F T = Tl and Ff = f l where fl is the restrict i o n o f f : T + T ’ t o Tl. Proposition 1.9.14: Suppose C is a commutative ring and M E C - A o d . Then T = T,(M) together with the identijkation of M with Tl c T is a universalfrom M to the functor given above. Proof: We must show for every graded algebra A, each map f: M + Al extends to a unique homomorphism f’:T + A whose restriction to M is f. By induction on u one sees f’must satisfy f’(x1 @ ‘ . * * @ x u=) f ’ ( x , @ * * . O x u - , ) f x , =fx,...fx,
for all xl, ..., x u i n M .
(1)
Since the x1 8 ..-@ xu span T,,this shows f’must be unique, and this formula also motivates our definition of f ’. Namely put fi = f; define fu inductively on u by forming the balanced map $,,: T,,- x Tl + A , given by $,(y,x) = f u - l y f x ,and letf, = L1 0 Tl + A , . By induction on u, each f, is a map satisfying (l), so we can define f’:T + A by putting f ’ C y , = cfuyu where y, E T,. Noting f’(y,y,) = f,+,(y,y,) = f,y,f,y, by (I), where y, E T. and y , E T,, we conclude f ’ is indeed a homomorphism. Q.E.D.
6,:
If M=C(”’ then the tensor algebra T c ( M ) x C { X l..., , X,,}. Indeed, taking a base xl,. ..,x,, for M we define f:M + C{Xl,. . .,X,,}by Example 1.9.15:
81.9 Graded Rings and Modules
123
f x i = Xi and use the proposition to obtain a homomorphism f’: Tc(M)+ C{Xl,. . .,X,,} which is an isomorphism since the x i l 0 * 0 xi. form a base of T, over C.
This example is the motivating example in the study of tensor algebras, enabling us to obtain other constructions. Example 2.9.26: Let I be the ideal of Tc(M)generated by all x 0 x for x in M. Then I is graded, so T‘(M)/Z is an N-graded algebra called the exterior algebra of M , written as A ( M ) . Note for every x in M that x 2 = 0 in A ( M ) , and thus xx’ = - x ’ x for all x, x‘ in M . (Indeed, 0 = ( x + x ’ ) ~= x z + xx’ + x‘x + (x’)’ so xx’ + x‘x = 0.) This property can be used to characterize the exterior algebra as a suitable universal (cf., exercise 4)and leads to important applications in the theory of determinants (cf., Jacobson [85B,$7.23). Example 2.9.27: Let I be the (graded) ideal of Tc(M) generated by all x 0 y - y 0 x for x, y in M . Tc(M)/I is called the symmetric algebra of M and is commutative; commutative polynomial rings are obtained as a special case when M is free in C - A d . Example 2.9.28: (Upper triangular matrices as a tensor ring). Let R be a direct product of n copies of W, and let M = W ( ” - l )as W-bimodule, made into an R-bimodule as follows: Take a base of M over W and label it { e l 2 , e Z,..., 3 take a base of R over W and label it { e l l,..., en,,}, and define (weii)(w’ej,j+l) = dijww‘ej,j+l and ( w ’ e j . j + l ) ( w e i i )6i.j+1w’Wej,j+l~ =
In other words, the action from each side is matric multiplication. Then Mhasbase{ei,i+l~ei+l,i+2}sinceforj#iwehave -le i - l , i @ O = O . ei 1. i 0 ej,j + I = ei - 1, ieii 0 ej. j + 1 = e i - l , i O e i i e j , j +-
Ma
-
In this way we identify the u component of T,(M) as the u diagonal in the ring of upper triangular matrices, and T,(M) is easily seen to be isomorphic to the ring of upper triangular matrices over W.
r)
Example 2.9.29:
(,”
Suppose A, B are rings and M E A - A u d - B . The ring
can be viewed as a tensor ring by putting R = A x B and de-
fining left and right actions of R on M by (a,b)x = ax and x(a,b) = xb. (Verifications are as in the previous example.)
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Constructions of Rings
Example 1.9.20: Let T be the tensor ring built from the ring R and the free R-R bimodule (of corollary 1.7.32), writing Xi instead of ei. The elements of T are sums of monomials rlXi,r2XiZ...rfXi,rf+, where ri E R ; multiplication of monomials is by juxtaposition (and reduction by multiplying out adjacent elements of R). This important ring is called the free R-ring and is denoted R ( X ) ; note it is “more” noncommutative than the ring R { X }defined earlier, since R centralizes X in R { X } but not in R ( X ) . Remark 1.9.21: Cohn and Bergman defined a generalization which is very useful in various situations (cf., Rowen [80B]). Namely, if R E C-dlg define ( R ;C ) - a i ~ to g be the full subcategory of R-W&g consisting of R-rings which are also C-algebras naturally, e.g., if R c T then C 1 E Z ( R ) c Z ( T ) .Using C instead of Z in proposition 1.7.31 we get an isomorphism from R QcRoPMod to the full subcategory of R - A u d - R consisting of those bimodules M such that (cl)x = x(c1) for all c in C and x in M. The free bimodule in this category produces a tensor ring called R,(X) which is free in (R; C)-WRg. The form of the elements are the same as in example 1.9.20, except that we can move elements of C past the ri and Xi.
B Supplement: Constructing Free Products Suppose R , and R , are rings containing W canonically. Our twofold aim is to show the amalgamated sum of R, and R2 over W exists (when it does), and then to construct it explicitly enough to answer questions about it. The reason we consider only two rings at a time is for ease of notation; any finite number of rings could be handled by iteration, and infinite sets of rings could be treated using direct limits. To prove existence, we only need to find a common overring R‘ of R, and R2 with R, n R2 = W, by proposition 1.4.34. If, in fact, W is the center of the Ri then the obvious candidate is R , OwR,. Theorem 1.9.22: If C E Z ( R i )for i = 1,2 and R , is free as C-module with a base containing 1, then the free product of R, and R2 exists in C-dlg. Prooj; By proposition 1.7.22 there are the canonical injections R1+R, 0 1 and R , + 1 Q R, in R1 BCR,. Q.E.D.
Example 1.9.23: Suppose C c Z(R,) and R, = C { X } . Then R, Qc R , = R, 0 C{X} is not the free product, which is easily seen to be the C-algebra described in remark 1.9.21, as will become clear in the foregoing discussion.
81.9 Graded Rings and Modules
125
When W is not central we can still form Tl = R, Q w R 2 , which need not be an algebra, however. It is instructive to assume there is an overring R' as described above, and see how Tl relates to it. (Then we shall discard R' and find intrinsic conditions for the amalgamated sum to exist.) Define gl: T, -+R' by gl(rl 0 r 2 ) = r l r 2 (by means of a balanced map); likewise define inductively T + , = T Q w T l for ir 1 and g , + , : T + , + R by g i + l ( a i Q a , ) = (giai)(glal).The elements of T are the sums of terms of the form I (1' )
0 r y ) 8 r y ) 6 r 2( 2 ) ** r y ) Q r$)
where ry) E Rj for j = 1,2,
whose image in R ' is r y ) r y ) r \ 2 ) r y )...r ( i ) r ( i ) 1
2 '
To construct the amalgamated sum R of R, and R 2 we might consider trying the tensor ring of T, as W-bimodule. However, the ring multiplications of the R, do not "fit" with their images in the tensor ring; for example, elements of the form
... Q
,
@
1 Q r;''
and
.
*
Q
ry)
Q
,Q 1 Q rL2)Q
Q r;'? Q
. . . - . . . Q r U-lr" ( l ) (1) 6 1 6 1 Q r;') 0 . ..
*
*
- . . . Q rL1'
,Q rL2?,rL2)6 1 Q 1 Q . . .
are nonzero. To correct this problem we define K i to be the W - W submodule of T spanned by the elements of these two forms, and let ?;: = T / K , . Clearly K , E kerg,, inducing maps Si: -+ R'. We shall now appeal to the direct limit underlying the tensor ring construction, but applied to the ?;:. -+ 7;. obtained by tacking on For each i I j in N we have a map cp;: Q1 Q l . . . Q 1 at theend,e.g.,
ri2) r y ) )= r:" r?) ri2' Q ri2) 1 1 Q 1 1. - Each K , c ker yielding maps cp:: % + q. Since cpc jp; = cp:, we have a whose direct limit @ p i ) is to be our candidate for the amalsystem gamated sum. cp:(ry)
ry)
cpi,
(%;2)
Theorem 1.9.24: F is a ring under the multiplication given by the rule (piiZi)(pjh)= p i + j ( a i Qbj) for a, in T and bj in T j , and is the coproduct of R , and R2 in W-9i.ng..
Proofi By construction 7 is a W - W bimodule. To obtain the desired multiplication we take W- W bimodule maps TQq+F given by iii05 .+ p i + j ( a iQ bj). Fixing j and taking direct limits gives a map F Q 7 + T;
Constructions of Rings
126
now taking direct limits over j yields a map F 6 F + 7 given by
-
piEi 6p j b + pi+j(ai@ bj);this yields the desired multiplication. For j = 1, 2 we have maps i,hj:Rj T given by $lrl = pl(rl 6 1) and
$,r2 = pz(l 6r,); these are ring homomorphisms since, for example, $,(r1r’,)
= p1(r,r’, =
6 1) = p2(r14 6 1 6 16 1)
6 1 6r’, 6 1) = $lrl$lr’,.
It remains to prove the universal property for coproducts. Suppose
&: Rj + R’ were W-ring homomorphisms for j = 1,2. We could then create a W-bimodule map z: l+
R‘ given by
a:
as in the discussion preceding this theorem. This yields a map 7 + R’ satisfying & = g$j for j = 1,2, and g is a ring homomorphism, by inspection. Q.E.D. Having explicitly constructed the coproduct, we note that it is the amalgamated sum iff the following two properties hold:
(1) The canonical maps $j: Rj + are injections. (2) Viewing Rj c thereby we have R, n R, = W. Suppose R, is a free W-module having a base containing 1. By proposition 1.7.22 the Ri sit canonically in T, with R, n R, = W, so our only obstacle is in showing pl: +7 is monic. In general this can be a formidable obstacle, although it is overcome successfully by Cohn [SS], who passes to the modules Ri= Ri/W and uses homology to prove amalgamated sums always exist if W is von Neumann regular (as well as other results). We shall content ourselves here with several examples which are amenable to computation. Example 1.9.25: Suppose R,, R, each are free as W-modules having respective bases B,, B , each containing 1. We want to build a monoid resembling the word monoid. Namely, let B: = Bi - { l} and 93 = { l } u {words alternating in letters from B’, and B i } , i.e., a typical element s # 1 of 9 will be written in the form s = b j . . .b, where j = 1 or j = 2, and each b, E B’, for u odd and b, E B; for u even. The amalgamated sum R (of R , and R, over W) is to be the free W-module F having base 9,but to define multiplication we have to be a bit more subtle, using the right regular representation. Indeed, we shall embed R, and R, in End, F and take R to be the subring of End, F generated by their images. Thus we must define the suitable right multiplication of an
41.9 Graded Rings and Modules
127
element r of Ri (for i = 1 or i = 2) on a base element s = b j . . .b, of 9, and we then extend it by distributivity (taking l r = r). We proceed by induction on k - j, working in three cases for r: Case I . r E BI, where i = 1 or 2. If k $ i (mod 2) then the product of s and r is the string bj. * b,r. If k = i (mod 2) then the product is b j . ..b, - r' where r' is the product b,r taken in R,. (Note b , . ..b,- ,r' is defined by induction on k.) 9
Case 11. r E W. Then the product of s and r is b j . . .b,- ,(bkr).
I:=,
Case I l l . r = w,,bi,,for w, in W and b , in B, v B , . The product sr is now defined in terms of cases I and I1 as (sw,,)biu.
,
The reader should note the similarity between this construction and the free products of groups, especially in light of example 1.4.29. Techniques treating bases of free products are illustrated well in the following interesting theorem. Theorem 1.9.26 (Cohn). If R , ,R , are domains which are both algebras over a jield C then the free product R = R , R , is also a domain. Proof: (Martindale-Montgomery [83]) If Bi are bases of R i containing 1, then R has a base 9 = { l } v {words alternating in letters from B; and B ; } where BI = Bi- { 1). As before, we write a typical base element as s = a j . . . a , where j = 1 or j = 2, and each a,, E B; for u odd and a, E B; for u even. We would like to view 9as an ordered monoid in order to filtrate R as in 51.2, but this is impossible because 9 is not a monoid. Nevertheless we shall provide B with a fairly natural order, and see what we can do. Given s = a j . . . a , we say length (s)= k + 1 - j. Given also s' = a: * * . a: we say s < s' according to the following rules: First priority. length (s) < length (s'). Second priority. length (s) = length (s') and j > u (i.e., j = 2 and u = 1). Third priority. length (s) = length (s') and j = u and according to some specified ordering on the Bi we have a, < a: for the smallest number t for which a, # a;. We shall now study multiplication in R using this order on 9. For any r in R write length ( r )for the length of its highest order term. We view C E R via the identification c -+ cl. Suppose q,r E R . If q or r are in C then the product is obvious, and nonzero, so assume q and r are not in C , i.e., they each have length 2 1. We
Constructions of Rings
128
shall concentrate on the highest order terms of qr and their length. Our object is to show q is (right) regular, by which we mean qr # 0 for all r # 0 in R. Suppose S , I are the respective highest order terms for q,r and write s = o r a j - * . a , and t = f i b ; . - b , where O # a , P E C and the a,b are from B; u B;. Write n = length (q) + length (r) = (k + 1 - j ) + (u + 1 - u). Case I . a, and b, are from different B:. Then st = ufiaj.--a,b,... b, is the highest order term of qr so length (qr) = n and, in particular, qr # 0.
1
Case 11. q E R , . Write r = riti where ti are distinct strings starting with elements of B 2 ,and each ri E R , (although possibly ri E C). Then qr = C(qri)ti# 0 since each qri # 0. Case Ill. a, and b, are from the same B:. This is the interesting case. Say a,, b, E B; . Thus k is odd and u = 1. Also for convenience we assume j = 2. Then
length ( s ) = k - 1 and length ( t ) = u . Note that length ( s t ) s k + u - 2 because of the “collapsing” in the middle. Suppose momentarily that some (other) term of q of length k - 1 starts with an element of B ; , and let ;denote the highest order such term. Then ;ends with an element of B;; we see that i t has length k u - 1 and is the highest order term of qr, proving qr # 0. Thus assume all terms of q of length k - 1 start with an element of B ; . Now pick the string s’ = a; ... a ; - of length k - 2, of highest order such that s’ is the beginning of some term of q. Let q‘ be the sum of all terms of q starting with s‘, and write q’ = s’w. Length (w) = ( k - 1 ) - ( k - 2) = 1 so w E R , ; hence wr # 0 by case 11. Of course length (wr) 5 u. If length (wr) = u then taking the term h of wr having highest order, we see s’h is the highest order term of qr (and has length k + u - 2); thus qr # 0 unless length (wr) c u, so we assume this. Let r’ be the sum of all terms of r having length u. In order for wr to collapse, w must cancel the initial part of r - r‘; thus w is invertible in R , and r - r‘ = w-lr” with length (r”) < u. Continuing, we pick the string s“ of length k - 2 of next highest order (after s’)such that s“ is the beginning of some term of q. The sum of all terms starting with s“ is s”w’ for some w’ and provide the highest order term for qr (thereby proving qr # 0) unless u > length (w’r) = length (w’w-lr”); hence we must have w’w-l E C, so w’ E Cw. Continuing along the strings of length k - 2, we are done unless q = yw + z where yw contains all terms of length 2 k - 2, so that length (z) Ik - 3. Now one easily can organize the proof as follows: We claim that every q # 0 is regular, by induction on length (4). For q of length 0 this is obvious since q E C. In general, we have shown that if qr = 0 for r # 0 then
+
,
-=
51.10 Central Localization
q = qw-1
129
+ zw-’
has length Ik - 2 = length (q) - 1, so by induction is regular; however, q(wr) = qr = 0, implying wr = 0, which is impossible by Case 11. Thus every nonzero element is regular, proving R , R 2 is indeed a domain. Q.E.D. =y
$1.10 Central Localization (also, cf., 52.12.W.) One of the most fundamental problems in ring theory is to “invert” a subset S of R in the sense of finding a ring homomorphism 9:R + T such that s is invertible in T for all s in S. One would ideally want cp to be an injection, in which case a necessary condition is certainly that each s in S is regular in the sense that rs # 0 and sr # 0 for all r # 0 in R . Two familiar instances from commutative algebra, when R is an integral domain (so that every nonzero element is regular): (i) One can form the field of fractions F of R . (ii) Given a prime ideal P of R let S = R - P and let R , denote { s - l r : s E S , r E R } c F. R , is local, i.e., its unique maximal ideal is { s - l r : s E S , r E P}. Consequently, the process we are about to describe is called localization. Before proceeding, let us describe the situation more formally in terms of universals. Let 9 denote the category whose objects are pairs (R, S) where R is a ring and S c R ; morphisms f:(R,S ) + (R‘,S ‘ ) are ring homomorphisms f: R + R‘ for which f S E S’. Then 9 has the full category % consisting of,all ( R , S ) for which each element of S is invertible in R , and we can define the forgetful functor F:% --* 9 which “forgets” this condition on S. A universal from ( R , S ) to F is called a localization of R at S and is denoted S - ’ R if it exists. Definition 1.10.1:
Explicitly there is a given homomorphism v: R + S-’R such that vs is invertible for each s in S, and, moreover, for any ring homomorphism $: R + T such that s is invertible in T for all s in S we have a unique homomorphism completing the diagram
By abstract nonsense S-’R is unique up to isomorphism. There is a general construction of S-’R, but it is rather unwieldy and resists computation. (For
Constructiowi of Rings
130
example, it is difficult to determine even when v is 1:l.) However, when S G Z(R) we can mimic the construction from the commutative theory and bypass many of the difficulties. We shall see in corollary 1.10.18' that the construction actually can be achieved directly from the commutative theory, merely by using the tensor product. Consequently we shall perform the construction here for S c Z(R), called central localization, and defer noncommutative generalizations until Chapter 3. In the literature R, or R[S-'] are also written in place of S I R . We shall take advantage of this notational ambiguity by introducing different constructions by different notations and then showing they are the same. The first construction is by brute force.
-
-
Define an equivalence on S x R by (slrrl) (s2,r2)if s(slr2 - s2rl)= 0 for suitable s in S; letting s-lr denote the equivalence class of (s,r) define S ' R as S x R / - made into a ring under the operations Consrrucrion 2.20.2:
sT'r,
+ s;'r2
+
= (s1s2)-1(s2rl slr2)
-(s-'r) = s-'(-r) (sT1r1)bi1r2)= (sls2)-'(rlr2).
This construction is very intuitive, but the verifications that S ' R is a ring (done more generally in 43.1) can best be described as "tedious," especially is indeed an equivalence. (Soon we shall bypass these verificachecking tions.) Let us see S-'R has the desired universal property. Indeed define v: R + S-'R by vr = l-'r, obviously a homomorphism, and note ( v s ) - l = s-'l. Given $: R + T with $s invertible for each s in S, we have
-
$r$s = $(rs) = $(sr) = $s$r,
implying ($s)-'$r = ($r)($s)-l
for all r in R, s in S. If there exists $s satisfying $ = t,hsv then $s(s-'r)$s = $s(s-'r)$s(l-'s) = $r, implying $s(s-'r) = ($s)-'$r,
(1)
proving qSis unique. Conversely, we show $s exists by defining it via this equation; it is easy to show $s is a homomorphism with $ = &v, once we have shown $s is well-defined. But if s;'rl = then s(slr2 - s2r1)= 0 for some s in S and 0 = $(S(Slr2 - s z r d = IL(ssls2)(($Sz)-'J/r2- (ILs1)F1$r1)9
131
%l.lOCentral Localization
implying ($s2)-'$r2 = ($sl)-'$rl since $(sslsz) is invertible. Thus we have constructed S-'R explicitly. A quicker construction can be obtained using exercise 1, but it is instructive to consider a third construction using direct limits. Construction 1.10.3: For each s in S take R, = Hom,(Rs, R). Preordering S as usual, i.e., s I t iff s divides t, we define cp:: R, + R, (for s I t) by sending a map f:Rs + R to its restriction to Rt. Now (R,; cp:) is a directed system of abelian groups, whose direct limit we denote R,, together with the canonical maps p,: R, + Rs. Iff: Rs -,R we denote p, f as the element s - l r where r = fs; note s and r determine f since f(as) = ar for any a in R.
We want to verify in general that Rs serves as S-'R and to prove this directly so as to have a self-contained proof of the existence of K'R. Of course, we must define multiplication on R, but first we want some preliminary observations. (i) s-'r = (s's)-'(s'r)for all s' in S. (Indeed, writing s-lr = psf we have fs = r so f(s's) = s'(fs) = s'r and taking t = s's we have t-'s'r = prq:f= psf = s-'r.) (ii) s;'rl - si1r2= (s1s2)-'(s2rl- slr2), seen by passing to s1s2 using (i). (iii) ker p, = { f E Hom(Rs, R): s'(fs) = 0 for some s' in S}. (Proof ( 2 )Take t = s's and note 0 = (s's)-'(s'fs) = prcpff = p, f.( G) If p,f = 0, then by the proof of proposition 1.8.5 we can view everything in O R , and write f as a sum of terms cpifi - fi, where for notational convenience we wrote i instead of si,and sj 2 si. Let t be the product of s and all the si and sj appearing in these expressions. Then qifi(t)= f i ( t ) for all these i, j , so 0 = f(t). Writing t = s's we get 0 = s'fs as desired. (iv) sr'rl=si'r2 iff s(s,r2-s2rl)=0 for some s. (Indeed s;'rl -silt-,= - s1r2)so apply (iii).) (sls2)-1(s2r1 Now define multiplication by (s;'r1)(si'r2) = (s1s2)-'(rlr2). This is well-defined in view of (iv) and is distributive over addition by a straightforward verification; 1 1 is the multiplicative unit. At this stage we have verified all the conditions necessary to carry over the verification of construction 1.10.2, to define the homomorphism v : R + R, by vr = l-'r, and to show (1) holds, so R, is identified naturally with S ' R .
-'
This construction emphasizes the R,, which could be thought of as modules s-'R, and this vantage point will be used to good effect later. Remark 2.10.4: Notation as in (l), ker $, = S-'ker $. (Indeed, s-'r E ker $s iff ($s)-'$r = 0, iff +r = 0.
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Constructions of Rings
Proposition 1.10.5: If S G S' there is a canonical homomorphism cp: S-'R + (S')-'R given by s-'r + s-lr. If, moreover, vss is invertible for all s in S' then cp is an isomorphism. Proof: Take $: R + (S')-'R to be the canonical homomorphism and cp = t,bs. Under the additional hypothesis we can go in the opposite direction, starting with the canonical homomorphism $: R 4 S-'R; then cp-' = &.. Q.E.D.
In order to utilize central localization, we need to see first how structure passes from R to S-'R, then to note how S-'R may be in better form, and finally to show how one does not lose information when localizing wisely. Throughout S is a central submonoid of Z(R).
Structure Passing From R to S ' R We start by passing structure from R to S ' R , with special attention paid to ideals and the center. Given A c R write S-'A for {s-'a:s E S and a E A}, and s-'A for {s-'a: a E A ) . Remark 1.10.6: If L < R then S-'L I; S I R , by direct computation. Remark 1.10.7: For any finite set of elements xl, ...,x, of S-'L there is then some s such that each xi E s-'L. (Indeed, if xi = sf'ai take s = s1 xi = s-ysl ***Si-'Si+l.*.s,ai). "as,:
Note in remark 1.10.7 that we could multiply each xi by s to produce elements with denominator 1; this common but important technique is called clearing denominators, and often is used to pass from S-'R back to R. Write Y(R)for the lattice of proper left ideals of R, viewed as R-modules. If L c R and s E L n S then 1 = s-'s E S-'L, so S-'L = R. Thus every left ideal intersecting S "blows up", and we focus on gS(R)= ( L < R: L n S =
a}.
Proposition 1.10.8: There is an onto lattice homomorphism L?(S-'R) given by L + S-'L.
(D: Zs(R) +
Proof: Suppose L1,L,
$1.10 Central Localization
133
onto since if L' < S-'R we take L = { r E R : l - ' r € L ' } ;then L < R and Q.E.D. L' = S-'L. In general (D is not 1 :1, but it does act 1 : 1 on certain left ideals. We say a left ideal L is S-prime if sr E L implies r E L, whenever s E S and r E R. The importance of this concept will not be seen until we consider the prime spectrum in $2.1 1, but let us record some results here. Lemma 1.10.9: If L < R is S-prime and s-lr E S-'L then r E L. Proof;. Write s-lr = s;'xl where s1 E S and x , E L. Then s'(sx, for some s' in S so s'slr = s'sx, E L implying r E L. Q.E.D. Proposition 1.10.10: Suppose L,, L then L , I L. Proof;. If r E L, then 1-'r
E S-'L,
-= R and L is S-prime. I f I S-'L so r E L.
- slr) = 0
S-'L, I S-'L
Q.E.D.
Note that {S-prime left ideals of R } is not necessarily a sublattice of L?JR), since it need not be closed under sums. However, it is a lattice in its own right, isomorphic to Y ( S - ' R ) , as we see in the next result.
-
Theorem 1.10.11: The lattice Y s ( R )has a closure operation gioen by L= { r E R : Sr n L # @a>, under which the "closed" left ideals are precisely the S-prime left ideals. Thus {S-prime left ideals of R } is a lattice (where L , v L 2 = L , + L,) which is isomorphic to Ys(,(R) under 0 (of 1.10.8). Proofi If slrl E L and s,r, E L then sl!z(r, + r,) E L, from which one sees (for readily I, is a left ideal. Moreover, I , = L and L , n L, = Gn then sir E Li for suitable si in S so slszr E L , n L , and if r E Gn r E L , n L,). Clearly L = L iff L is S-prime. By proposition 1.10.10 the restriction (D' of 0 to S-prime left ideals of R is 1:l. Moreover, (DL = 0 L since if sr E L then 1-'r E S ' L . Thus @' is onto. In summary, we see {S-prime left ideals of R } is a poset under E and is naturally a lattice since any L , , L 2 have a sup (namely L, + __._ L,) and an inf (namely L, n L 2 ) . 0' is a lattice homomorphism since W ( L , + L,) = (D(Ll + L 2 )= (DL, + @L2 = 0'G + @'G. Q.E.D. Corollary 1.10.12: (D restricts to an onto lattice homomorphism from {Ideals of R disjoint from S } to {Ideals of S I R } . 0'restricts to a lattice isomorphism from {S-prime ideals of R } to {Ideals of S-'R}.
Constructionsof Rings
134
Proposition 1.10.13:
S-'Z(R) E Z(S-'R), equality holding if S is regular.
Proof: Suppose s1 E S and z1 E Z ( R ) .For any s-'r in S-'R we have [S;'Z~,S-'~]
=(s,s)-'[z,,~] =0
SO
S;'Z
E Z(S-'R).
Conversely, if S is regular and s;'rl E Z(S-'R), then for all r in R we have 0 = [sT'rl, l-'r] = s;'[r,,r] so [rl,r] = 0 proving rl E Z(R). Q.E.D.
Examples of Central Localization Example 2.10.14: If every ideal (resp. every left ideal) of R contains an element of S then S-'R is simple (resp. a division ring). This fact is used most often when S is all regular elements of R, in which case S-'R is the ring of central quotients of R. This coincides with the commutative ring of quotients when R is commutative. Example 1.20.25: Suppose S = { s i : i E N} for some fixed s in Z ( R ) . Then we write R [ s - ' ] for S ' R , and A [ s - ' ] for S-'A where A c R. For L < R note L [ s - ' ] z R [ s - ' ] iff L contains no power of s. Our final observations concern S-prime left ideals. Remark 2.10.26: By Zorn's lemma any left ideal disjoint from S is contained in a left ideal L maximal with respect to L n S = 125. L is then S-prime since L= {rER:srEL for suitable s in R} is a left ideal containing L , disjoint from S, so L = L. Thus we see S-prime left ideals exist in abundance, and for L as defined here S-'L is a maximal left ideal of S-'R. The same remark holds if we replace "left ideal" throughout by "ideal." These three examples are the basic tools for central localization and shall be examined closer in 52.1 1.
Central Localization of Modules Much of the previous discussion holds more generally for modules and when viewed in this way actually yields a tensor product connection between central localization and commutative localization, as we shall see. Definition 1.20.17: Suppose S is a submonoid of Z ( R ) and M E R-AoG!. For each s in S take M, = Hom,(Rs, M) and under the preorder s It iff s
135
$1.10 Central Localization
divides t, we define cp:: M, -+ Mt (for s I t ) by sending a map f:Rs + M to its restriction f:Rt-+M. Define S-'M =lim(M,; cp:), together with the canoni+ cal maps pi: M,+S-'M; given J Rs+M we denote p,f as the element s-'x where x = fs. We now make a series of observations analogous to those of construction 1.10.3 and omit the proofs when they are identical. First note if s is regular then Rs % R so M, % M as an abelian group. In general the following also hold for si in S and xi in M: (i) s-'x = (s's)-'(s'x) for all s' in S.
(ii) si'xl
- s;'xz = ( S ~ S ~ ) - ~-(SIXz). S~X~
(iii) ker ps = { f E Hom(Rs, M ) :s'(fx) = 0 for some s' in S}. (iv) si'xl = s;'xz iff s(s1x2 - szxl) = 0 for some s in S.
(v) There is a canonical map vs: M -+ Ms given by vsx = 1-'x, and ker vs = {x: S n Ann,x # @} (immediate from (ii) and (iv)). Thus the theory of central localization of modules parallels that of rings, with the simplification that we need not worry about ring multiplication. Indeed the following result holds: Proposition 1.10.18: -+
'
S-'R
BRM
!z S ' M , by an isomorphism s - l r 8 x
s- (rx).
Proofi There is a balanced bilinear map S-'R x M - + S - ' M given by (s-lr, x) 4 s-'(rx), which thus induces a map q :S-'R 0 M -P S-'M given by s-'r@x-+s-'(rx). On the other hand, cp-' is defined by the map s-'x --t s-'@ x, which is well-defined, since if si'xl = s;'xz then some s(s1x2 - SZX') = 0, so si' 0 X I = (ss1s2)-'ss2 0 X' = (ss1s2)-l0 S S Z X ' = (ss1s2)-l0 S S l X Z = s;'
Hence cp is the desired isomorphism.
0 x2.
Q.E.D.
Corollary 1.10.18': If S is a submonoid of Z x Z ( R ) then S - ' Z @ , R S-'R as Z-algebras. Proof: The map of proposition 1.10.18 respects multiplication.
%
Q.E.D.
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Constructions of Rings
Proposition 1.10.19: There is an onto lattice homomorphism 0:Y ( M )+ Y ( S - ' M ) given N -.+ S-'N for N < M. Proofi
Duplicate proof of proposition 1.10.8, writing N in place of L.
Instead of copying out all the previous results for M replacing R, we shall now formulate them in a categorical setting.
Theorem 1.10.20 There is a functor F : R - d o d -.+ S-'R-Aod given by F M = S-'M where for any map f: M + N we define Ff:S-'M + S-'N by F f (s-lx) = s - y x . This functor satisfies the following important properties: (i) ker(Ff) = S-'ker f for every map f. S-'N is exact. (ii) r f K A M 3 N is exact then S-'K -% S-'M (iii) F induces an onto lattice homomorphism Y ( M )-.+ Y ( S - ' M ) . (iv) Define N to be an S-prime submodule of M if sx E N implies x E N for each s in S. Then L ( M ) has a closure operation given by N = { x E M : s x E N for some s in S } ; {S-prime submodules of M } is a lattice isomorphic to Y ( S - ' M ) under F. Ff is the composition S-'M x S-'R 8 M S-'R Q N x S-'N. It follows at once F is a functor. Identify S-'M with S-'R Q M in what follows. Proofi
(i) Clearly S-'(ker f)= S-'R 8 kerf G ker(Ff ). Conversely if s-lx E ker(Ff) then s-yx = 0; hence there is s' in S with 0 = s'fx = f (s'x), so S I X E kerf implying s-'x E S-'(ker f ). (ii) ker(Fg) = S-'ker g = S-'( flu) = (Ff )(S-'K). (iii) Duplicate proof of proposition 1.10.8, where here Li < M . (iv) Duplicate proof of theorem 1.10.11. Q.E.D. One nice feature of the central localization functor is that even noncategorical properties are preserved.
Remark 1.10.21: If F is free in R - d d with base {xl,. ..,x,} then S-'F is free in S-'R-dod with base { l-'xl,. . . ,l-'xn}. (Either verify this directly or recall we have proved the more general result that tensor products preserve freeness.) Remark 1.10.22: If S E S' are submonoids of Z ( R ) and M E R-AUQ! then there is a canonical S-'R-module map S-'M + (S')-'M given by s-lx -.+
137
Exercises
sf'x; this is an isomorphism if 1 -IS' is invertible in S I R . (Clear, by applying proposition 1.10.5to tensor products.)
Proposition 1.10.23: If z, seZ(R),then ( R [ z - ' ] ) [ s - ' ] x R[(zs)-'] as rings and M [ z - ' ] [ s - ' ] z M [ ( z s ) - ' ] for any R-module M. Proof: Take S = {(zs)':i E N} and S' = {zisj:i,j E N}. Then S c S' and for Z ' S ~ ES' we have (zisj)-' = ( z ~ ) - ( ~ + j ) in z j sK~ ' R . Hence R [ z - ' ] [ s - ' ] x R [ ( z s ) - ' ] by proposition 1.10.5, and remark 1.10.22 gives the isomorphism M [ z - ' ] [ s - ' ] % M[(zs)-'].
Exercises $1.1 1. Suppose f: M , ( R ) -+ T' is a ring homomorphism. Then T' = M,(R') for some ring R', and f is given componentwise by a suitable homomorphism f R + R'. (Hint: Define the matric units eij = f e , . Now writing T' = M,(R') let f r be that r' such that f ( re,) = r'efi.) 2. If e , , e , are idempotents of R with e, E R e , and if e2 A = e l A for some idempotent-lifting ideal A then e l = e , e , . (Hint: e , ( l - e 2 )is idempotent.) 3. Write M,,,(A) for the set of rn x n matrices with entries in A, made componentwise into an abelian group if A is an abelian group. M,,JR) E M m ( R ) - A ~ d - M , , ( R ) where the scalar multiplication is taken to be multiplication of matrices; more generally, if N E R-Aod-T then M,,,(N) E M , ( R ) - A d - M , ( T ) via matrix multiplication. In particular, any matrix of M,+,(R) can be partitioned in the form
1
(: ):
1
+
+
where A E M,(R), B E M,,,(R), C E M,,,(R) and D E M J R ) and this
provides a special case of example 1.1.10.
4. If M E M,(R)-Aod and eiiM = 0 for some i then M = 0. Use this fact to prove corollary 1.1.18 directly. (Hint: Suppose M is simple in R-Aud and 0 # N I M("). Then 0 # eiiN I eiiM(")x M so eiiN x M ; hence N = M(").The converse is even
easier.)
Infinite Matrices 5. M , ( R ) is defined to be the set of infinite matrices which are row-finite, i.e., every
matrix has a countably infinite number of rows, but almost all entries of each row are 0. Show M , ( R ) is a ring whose opposite is isomorphic to the ring of columnfinite matrices. 6. Defining M , . , ( N ) to be the set of columns with entries in N , show there is a functor F: R-.Ai~nd+ M , ( R ) - A d given by N + M m , , ( N ) . 7 . Suppose W G R are rings and T = { w 1 + rijei,:w E W,rij E R } , the subring of M , ( R ) generated by W and the eij. Show Z(T) z Z(W).
-
Ifinits
Constructions of Rings
138
$1.2 1. (Incidence algebras) Suppose I is a finite poset, viewed as a small category %? as
in example 0.1.1, and F is a field; then F[V] is called an incidence algebra. Interpret the incidence algebras (in terms of matrices) arising from the following partial orders on { 1,. .. ,n}: (1) the usual total order; (2) i Ij iff i = 1 or i = j ; (3) i Ij iff i < j and i < i, for some fixed i,. Show an element is invertible iff each "diagonal" entry is invertible.
Ordered Algebraic Structures 2. Given g , , ...,g,, in G write S ( g , , . .., g n ) for the normal semigroup in G generated
by gl,. ..,g,,. For example S(g) is the set of products of conjugates of g. A normal submonoid P of G is extendible if for any finite g , , ...,g,, in G - { 1) one can find h,~{g,,g;'}for 1 ~ i ~ n s u c h t h S(h,, a t ...,h,,)[email protected] Pisextendiblethen P n P-' = { 1) (because for any g # 1 we have S(g) or S(g-') disjoint from P ) . A normal submonoid Po of G is extendible iff G has an order < with Po E P(G). (Extensive hint: (e) Given g i # l pick h i ~ ( g i , g ; ' }with h , < l , so S(hl,...,h,,) has only elements < 1. (a) Take a maximal extendible normal submonoid P 2 Po, by Zorn. It suffices to show P u P-' = G, or better yet, S(g) c P or S ( g - ' ) c P for any g # 1. Otherwise, P S ( g ) and P S ( g - ' ) are both normal submonoids =I P so thereareg,,, . . . , g , , and g,,, . . . , g ,,,such that theextendibilitycondition fails, but then {gl ..,gl,, g,,, . ..,g,,,,g} contradicts the extendibility condition for P.) 3. G is an ordered group iff { 1) is extendible. 4. G is an ordered group iff every finitely generated subgroup of G is ordered. In particular, every torsion-free abelian group is ordered. (Hint: Use exercise 3 This result, called Leoy's Theorem, can also be proved by ultraproducts cf., exercise 1.4.18.) 5. Take the group G = {x'y': i, j E Z} with multiplication given by the rule yx = x-'y (so y-'x = x-ly-' by taking inverses). Then (x) is a normal subgroup and G/(x) x (y), so the factors of the chain G 3 (x) 3 { 1) are torsion-free cyclic and thus ordered. However G is not ordered, since x E P o yxy-' = x - l E P.
,,.
$1.3 1. If a ring R is not a division ring then each simple module is not free. Conclude that
a ring is a division ring iff each module is free.
2. The free algebra C{X,,X,} in two indeterminates contains the free algebra C{X}
in a countable number of indeterminates. (Hint: Define an injection $:C{X} -+ C{Xl,X2} by $XI = XlX,.) The free group in two indeterminates contains the free group in a countable number of indeterminates. (Hint:X, -+ XiX,Xi.) 3. $ + R(") is epic iff gf is the identity for some map g: R(")+ R("). What is the situation for monics? 4. View M = R ( " ) as a right R-module with base {el,...,en},where ei=(O,...,0,1,0,...,0). Every n x n matrix A = (a,) over R corresponds to a unique map f:M + M given by fej = e,aij. Likewise, if the matrix B corresponds to g, show AB corresponds to fg. Conclude f is epic iff BA = 1 in M,,(R) for a suitable matrix B.
I:=,
139
Exercises
5. Using exercise 4 show R is weakly n-finite iff every epic j:R(")-+ R(") is an
isomorphism. 6. R is weakly n-finite iff R'"' x R(")6 M implies M = 0. (Hint: (*) by exercise 4. (e) If f:R(")-+ R(")is epic then apply exercise 3 to the exact sequence 0 + kerf -+ R ( " ) L R ( "-+) 0 to show R(")x R(")@ kerj. (If necessary refer to proposition 1.4.10.) 7. Define t: R + R / [ R , R ] by t l = 1 + [R, R ] . This is a trace map, called the Stallings trace map, and has the universal property that any trace map of R factors through it.
81.4 1. There is a 1:l function from {ordinals} to {rings}, proving {rings} is not a set.
(Hint: Define$ {ordinals} + {rings} by fa = Ea for suitable fi defined by transfinite induction such that if a, = a: then IZBII> IEa21.) 2. The following examples can be constructed easily using the direct product: (i) A ring which is not an integral domain but which has no nilpotent elements # O (ii) An infinite ring having prime characteristic p > 0. (iii) A ring in which for each k in N there exists r nilpotent with rk # 0.
n
3. If e = ( r i )E Ri and each ri = 0 or 1 then e is a central idempotent; e is minimal under the PO of exercise 0.0.24 iff all but one ri are 0. 4. There exists an infinite chain of nonisomorphic fields F , c F, c ...(Hint: Let 4 be a finite algebraic extension of 4- and count dimensions.) 5. (An example of nonisomorphic commutative rings R,, R, with injections R , 4R , and R, + R , ) Take fields F, c F, c ..*as in exercise 4 and let R , = OFzi-and R 2 = @FZi.The inclusion maps F 2 i - , + F2i give an injection R , + R , and, likewise, the inclusions F2i + F2i+ gives an injection R, + R,. But there is no isomorphism Jl: R, -+ R,. (Hint: let e be one of the minimal idempotents of exercise 3. Then Jle is minimal since II/ is an isomorphism, but this implies Jl: 4 + 4 for some j 2 i + 1. Applying the same argument to shows x 4 which is impossible.) 6. Suppose rp has the form (Qlxl)~~~(Q,xt)((Jll A ... A Jlu-,)+Jlu) where the Jli are atomic formulas. If { i E I: $ holds in A i } E 9then Jl holds in A = A i / S (Hint: As argued in proposition 1.4.13, one may assume cp is quantifier-free. Let J , = { i : Jl, A . . . A Jlu- each hold in A i } and 5, = { i : Jluholds in Ai}. Then J ; u 5, E 9; if J , E F then 5, 2 (J', u J2)n 5, E R)
,
,
,
I,-'
fl
,
Ultraproducts 7. The ultraproduct of division rings is a division ring.
(fl
7'. Suppose Ri z M,,( q )for each i in I, and F is an ultrafilter on I. Then Ri)/Fx M n ( n&/9). In particular, if each is a division ring then ( n R i ) / . F is n x n matrices over a division ring. 8. The ultraproduct of ordered groups is an ordered group. Conclude from proposition 1.4.21 that a group can be ordered iff each of its finitely generated subgroups is ordered.
Constructions of Rings
140
Reduced Products As Rings In exercises 9-12 suppose {R,:i E I} are rings and let
n,,,
R= R,. Write Z(r) = { i I:~r,SO} and for any A c R, define Z(A)= {Z(r):r E A} E B(I).Conversely, for any 9 G @ ( I ) , define 9(9) = {r E R: Z(r) E 9).
xJE9
9. For any set J c 1 there is a central idempotent e(J) with Z(e(J)) = J . (As in exercise 3). If 9is a filter of I then 9(9)= Re(J). 10. If 9 is a filter of 9 ( I ) then 9(9) Q R and Z ( Y ( 9 ) )= $ and R / Y ( 9 ) is isomorphic to the reduced product RJ9. (Hint: Z(r r’) 2 Z(r) nZ(r’) and Z(rr’) 2 2(r)uZ(r’)for r, r‘ in R, so 9(9) Q R with Z ( 9 ( 9 ) )= 9 by exercise 9.
fl
n
+
Define the obvious surjection $: R + R i / 9 and show ker 9 is the set of (ri) such that { i E I: ri = 0)E I, which is Y ( 9 ) . ) 11. 9()gives a lattice injection from {filters of B(I)}to {ideals of R}. This is a lattice isomorphism if each Ri is a field. However, if I = N and Ri x Mi@) for each i then {(ri):the rank of the r, is bounded} is an ideal of R not in the image of 9(). 12. The ultrafilters of a set I correspond to the maximal ideals of F’ where F is an arbitrary field.
Rings of Continuous Functions The following exercises were chosen to give a taste of Gillman and Jerison [60B], which is a study of the ringR of continuous real-
valued functions on a topological space X. (Gillman-Jerison denote this as C ( X ) . ) Typical elements of R are continuous f:X + R, and we write Z ( f ) for { x E X :f x = 0}, called the zero set off. A 2-set is a zero set of suitable f. 13. {Z-sets}is a sublattice of 9 J ( X ) . 14. Each 2-set is a countable intersection of open sets of
X. (Hint: Z ( f ) = X : I ~<~n-’~). I If X is compact then {Z-sets} is closed under countable intersections but not countable unions. (Hint: One may assume 0 If;.x I1 for each 1;: and each x in X, and note Z(1;:)= Z ( ~ j 7 2 ‘ ) J A 2-filter is a sublattice of {Z-sets} such that if 2, E 9and Z, c 2, then 2, E 3 If A 4 R define Z ( A ) = {Z(f): f~ A} and show it is a Z-filter. Also given a 2-filter 9 define 9(9) = {fE R:Z(f) E 9}. Show Y ( 9 )Q R, and R / Y ( 9 ) has no nilpotent elements # 0. If A is a maximal ideal of R then Z (A ) is an ultra-filter; if 9 is an ultra-Z-filter then 9(9)is a maximal ideal. In the same way, prime ideals of R correspond to prime filters (in the sense that I, u I, E 9implies II E 9or I, E 9). There is a 1:l correspondence from {ultra-2-filters} to the points of the StoneCech compactification of X. In this way, one can use geometry to study R, and, in fact, this example embodies many features of algebraic geometry. n n s N i X6
15.
16.
17.
18.
n
Abelian Categories 19. Conditions (iii) and (iii’)of definition 1.4.40 are equivalent. (Hint: Iff = hg where g is a cokernel and h is a kernel then kerf = ker g and coker f = coker h.) Assume throughout the remainder of this section’s exercises that the underlying category V is abelian.
Exercises
141
20. Any monic epic is an isomorphism. 21. (“Five lemma;” compare with proposition 2.1 1.15) In the following commutative diagram if f,g are epic and u,,u3 are isomorphisms then u2 is also an isomorphism:
(Hint: It suffices to show u, is monk epic; by duality it suffices to show u2 is monic. Suppose u,h = 0 for some h: C A , . Then 0 = gu,h = u 3 f h so fh = 0. Hence h = (kerf)h’forsomeh’:C+ A , so0 = u,h = u,(kerf)h’=(kerg)u,h‘soh’=O; hence h = 0.) 22. (Compare with exercise 1.8.1’) The pullback of morphisms f,:A , + A and f,: A , + A is a triple (P,p,, p,) where pi: P + Ai are morphisms such that f,pl = f2p2 and such that any commutative diagram -+
n
can be completed by a unique morphism P’+P. Show the pullback exists for any abelian category W; explicitly let p: P + A , A , be kernel of the morphism (f,7r,-f27r2):AIA , + A and let pi = n i p : P + A i . 23. In exercise 22 if f,,f2are monic then p1,p2 are m o n k Thus the pullback is the intersection of f l and f,. 24. Suppose A E Ob(W). Using definition 1.4.35, define a subobject of A to be an equivalence class of monics to A , and show ( 9 ( A ) ,I) is a poset. 25. ( 9 ( A ) ,I) of exercise 24 is a modular lattice. (Hint: Denoting subobjects of A by representatives fi: Ai -+ A show fi is the image of that morphism f: H A , -+ A given by 1;. = fpi for all i, where p i : Ai -+ Ai i t t h e canonical monic. To show V fi is indeed the sup, put f = imf.-Then f = f h for some epic h: Ai + A i so each fi = Jhpi implying each fi I f. On the other hand, if each fi 5 k for some monic k : A‘ + A then k ker(coker k ) so 0 = (coker k)fi = (coker k ) f p i implying (coker k)f = 0 and thusf I k. Write Ai for f ( Ai). Similarly, A fi is the kernel of the morphism f’:A -+ fiA given by f i = nif’ for all i, where f i = coker/;.. This shows ( Y ( A ) ,I)is a lattice. Write Ai for the domain of ker(Afi). To show ( 9 ( A ) ,4 ) is modular it suffices by exercise 0.0.2 to show for all monics fi: Ai A and h: B + A with fi I fi that if f, A h = f, A h and f, v h = f , v h then f, = f,. To see this one needs to calculate f, A f2 explicitly. Indeed under *
n
v
H
1
n
-+
Constructions of Rings
142
the notation of exercise 22, the pullback P is A, n A, and p = ker(f,n, - f2n2)yielding the exact sequence
Al n A , 2 A ,
n
A , &'A,
+ A,
where p' = finl - f2n2.
n
n
Note p' is epic. Taking u : A, -+ A, with f, = f i u yields a composition A, B A, 4 A, which together with the projection A, B -+ B yields a morphism u': A, B A, B. By hypothesis there are isomorphisms A, nB -+ A, nB and A, B + A, B so applying exercise 2 1
n
A, n B -
+
A 2 n B-A2
-+
fl
--)I
+
lu2+ B
to show u' is an isomorphism; hence u is an isomorphism.)
26. A subcategory 9 of V is dense if any object of V is isomorphic to a suitable object in 9. Show the natural transformations of % then are determined by the
natural transformations of 9. 27. A category I is skeletally small if it has a small dense subcategory. Show R-Simod is skeletally small, for the small dense subcategory is {R(")/M:n E N and M IR'")}. If I is skeletally small and V is abelian then %I (cf., example 0.1.13) is abelian, where q, q, is defined by (q, q2)i = q l i q,i for i in I. (Hint: componentwise verifications.)
+
+
+
91.5 0. If N is an R-module then End"") x MJEndN). (Hint: Modify the proof of theorem 1.5.13, writing N = Rx and sending (J,) E M,(EndN) to the map /I: N'")-+ N'")given by /I(x,, ...,xn)= (C:=, &,xi,... finxi).) 1. Define the idealizer 9(L)of a left ideal L of R to be { r E R: Lr E L } . Then Y(L)is the largest subring of R in which L is an ideal; also J ( L ) / Lx End(R/L), by sending an element r to its right multiplication map. 2. (The dual functor) Given f i n Hom,(M, N )define fi -+ M by taking f h to be the composition hf for any h: N -+ R. This gives a contravariant functor F:R-Aod -+ d 0 d - R given by FM = M and Ff = f, called the dual functor. Likewise, there is and the covariant functor GF: a contravariant dual functor G: Mod-R + R-Aou!, R-Aod -+ R - d o d is called the double dual. 3. (Ellenberg-MacLane)There is a natural transformation q from ihe identity functor of R-Aod to the double dual functor. (Hint: Defi_neqM: M -+ M by taking qMxto be the map (qMx)f = fx E R for f in M. Show ($(qyx))f = (qfi(gx))f for each f in M and x in M,proving &, = qag.) 4. If R is a division ring in exercise 3 then q restricts to a natural isomorphism in R-Simod. (Hint: Each f.g.R-module M is free; to show qM is 1: 1 show that if x # 0 there is f in M with fx # 0. A dimension count shows each qM is an isomorphism.) 5. Suppose K, M, N are finite dimeqsional vector spaces over a field F, and K M -8, N is exact. Then fi M 2 is exact. (Hint: Certainly kerf contains
,XI=,
4
4 4
Exercises
143
the dual of coker f so equality holds by a dimension count; likewise $3is dual of Mlkerg, implying $N and kerf have the same dimension.) In the terminology of $2.11 below, this says the dual functor is exact for F - ~ % z o ~ a; more general result is given in exercise 2.1 1.0. Rngs 6. Show the free object in Wng is {polynomials in Z{X}without constant terms}. Generalize this to C-algebras without 1. 7. Construct the product and coproduct in W n f . (Hint: Direct product and free product)
$1.6 1. In any domain if ab = 1 then ba = 1. (Hint: ba(ba - 1) = 0.) 2. If Ra = Rb in a domain R then b = ua for some unit u. 3. If T = R [A; a, 61 is a PLID then R is a PLID satisfying Jategaonkar’s condition. (Hint: specialize 1 4 0 to show R is a PLID. To verify Jategaonkar’s condition suppose 0 # r E R. Then Tr TJ. = Tb for some b in T; r E Tb implies b E R, and I E Tb implies ab is invertible. Specialize I + 0 to get b E Rr.) 4. If F is a field of characteristicp then Z ( d , ( F ) )x F [ A P , p P ] , so d l ( F )is not simple.
+
Derivations 5. If 6 is a derivation on R then 6(Z(R)) E Z ( R ) .(Hint: 0 = 6 [ z , r ] = [z,dr] + [6z,r].) 6. Using Leibniz’ rule show that if char(R) = p and 6 E Deriv(R) then 6 P E Deriv(R). 7. Suppose 6*a = 0. Then 6”(a”) = n!(da)”. 8. (Hirano-Yamakawa [84]) For any derivation 6 of R and for each positive integer k there are integers nki such that (ska)b= ~ : = , n k i 6 k - i ( a 6 i for b ) all a,b in R.
(Proof by induction). 9. (Hirano-Yamakawa [ 8 4 ] ) Let R = M,(F[A]) and 6 be defined by taking differen-
tiation at each component. Then 6 is nil on soc(R) but not on R. However, using exercise 8, one can easily prove the following result: Suppose the derivation 6 of R is nil on an ideal A. If d k S = 0 for some subset S of A for which Ann,S = 0, then 6 is nil on R. In particular, this holds if Ann, S = 0 for some finite subset S of A. Moreover, if dkA = 0 and Ann A = 0 then 63k-2R= 0. Suppose a is an automorphism of R. A a-ideal of R is an ideal A of R such that aA = A; we denote this by A Q (R, a). 10. (Goldie-Michler [ 7 4 ] ) The a-ideals are the kernels in the category of rings with automorphism a, where the morphisms preserve a. Explicitly, if A a (R, a) then a acts naturally on R I A ; conversely, if f: R + T is a ring homomorphism and R, T have automorphisms both denoted a and f ( a r ) = a ( f r ) for all r in R then ker f Q (R, a).
A ring is prime if the product of any two nonzero ideals is nonzero.
144
Constructions of Rings
11. (Pearson-Stephenson [77]) Suppose R has an automorphism a. A u-ideal P is called a-prime if whenever A 4 R and B 4 (R,a) with A B G P, one must have A E P or B E P. R is called a-prime iff 0 is a u-prime ideal. Show T = R[l;a] is a prime ring iff R is a-prime. (Hint: (-=) If A B = 0 and B 4 (R,a) then ( T A T ) ( T B T )E TABT = 0. (=.) If A',B' Q T with A'B' = 0 then take A = {leading coefficients of elements of A'} and likewise for B.) 12. (Cauchon [79]) Suppose T = R[l;a,6] where R is simple and a is an automorphism. If A 4 T then A contains a monic polynomial f of minimal degree, and then A = Tf.Furthermore, if we take a monk polynomial g of minimal degree > 0 such that T g Q T then A = Tzg" for suitable n and suitable z in Z ( T ) . (Hint: The first assertion follows easily from the Euclidean algorithm; note if f in A has minimal degree t then RfR contains a monic polynomial of degree t. To prove the second assertion, write f = hgn where h # Tg; then Th 4 T. It suffices to prove Th = Tz for some z in Z ( T ) .Write h = qg p , with deg(p ) < deg(g). Write n = deg h and m = deg(g). Note deg q = n - m. For all r in R one has gr - (a"r)g E T g implying gr = (a"r)g; likewise, hr = (a"r)h, and pr = (anr)p. Similarly, there are ri in R such that g l = (l + r , ) g and h l = (l r2)h; if (degp) < m - 1 then p l = (l r 2 ) p , so Tp 4 T, and by minimality of degg we get degp = 0, so p E R is invertible and p-'h is central. If degp = m - 1 then g = l ' p p' where deg(l') = 1 and degp' < degp. Taking a' = an-m one has l ' r = (a'r)l' for all r in R, so T x R[A';a'], and we are reduced to the case 6 = 0; in this case, TA 4 T so m = 1, and one can use the proof of proposition 1.6.25(ii).)Cauchon also describes the center.
+
+
+
+
$1.7
my=,
1. The Weyl algebra d n ( C )x d l ( C ) . (Hint: Take the natural homomorphism d l ( C ) + d n ( C )sending 1 and p of the i-th component to li and pi. This map sends a base to a base so is an isomorphism.) 2. Suppose A, B E C-df' and H is a commutative C-algebra. Then ( A aC H) @, ( B @CHI ( A @ c B ) &H. 3. Given M E R-&d write Mopfor the corresponding right RoP-module(i.e., same addition and opposite scalar multiplication. If M,N E R - d u d the bilinear forms B: M x N + R correspond to the balanced maps Mopx N + R (over R) and thus induce group homomorphisms Mop N + R. This explains the tie between tensor products and bilinear forms. 4. Suppose C is a commutative ring and A is an abelian group. If A E C - d ~ then d scalar multiplication induces a map p: C €3 A + A given by p(c 8 a) = ca. If, moreover, A E C-d!' then there is a map n:A €3 A + A given by .(al €3 a 2 )= a,a2,and E : C + A given by EC = cl. Show the following diagrams commute:
8 9 -
mR
-
Exercises
145
where $: C 6 A A 6 C is given by $(c 6a) = a 6 c. Conversely, show that any maps n , as ~ described above give rise to an algebra structure by defining ala2 = n(a, 6 a 2 ) and ca = X ( E C 6 a). Thus a C-algebra is given by the triple
( A ,n , 4 . 5. Writing nA:A
-
6 A A and cA: C + A for the maps of exercise 4, reprove that the tensor product A @ B of two algebras is an algebra, by taking =E~@Q, and nA to be the composite ( A 6 B ) 6 ( A 6B ) - ( A
-
6A ) 6( B @ B ) A 60 B,
where the first map is given by ( a , 6 b , ) 6 (a, 6b,) --t (a, 6a z )6 (b, 6 b 2 ) and the second map is n, 6 nB. Show this algebra structure agrees with that given in the text. 6. (Polynomials over a module) If M E R - A d define M [ L ] = { ~ x i L i : xE iM } , made into a module over the polynomial ring R [ 1 ] by putting riLi)( xjL’) = ~ u ( ~ ~ = , r i x u - iShow ) R U .this is a module by identifying M [ I ] with Z [ R ] B Z M as module over Z [ A ] mZR = R [ L ) in the natural way. 7 . If M E A G ~ - and R N E R - A G such ~ that R is a summand of N then the canonical map M M @ 1 E M @ N is monic. (Hint: Construct the left inverse map M 6 N M.) 8. Suppose fi: R i Ti are homomorphisms for i = 1,2. Then ker(f, 0 f,) = (kerf,) 6 R 2 + R , 6 kerf, viewed in R I 6 R , .
(c c
---
61.8 1. Define the pushout of morphisms f,:A .
-
-
A , and f 2 : A 0 A , to be the direct limit of the following system: i = {0,1,2} with PO defined by 0 < 1 and 0 < 2; the Ai are as given with pb = 1;: for i = 1,2 and cpi = lAi for all i. Thus the pushout is an object P together with morphisms p i : A i P for i = 1,2 satisfying plf, = p2f 2 . (Take p, = p,f, .) Show by example that the pushout need not exist in 9 h g . In R-Adshow P = ( A , $ A , $ A 2 ) / ( R( -a,f,a, 0) R( -a,O, f,a)) x ( A , @ A 2 ) / zR( -fla,f2a), the sums taken over all a in A , . Associate the pushout to the following diagram and universal:
-
1
+
I\ -
1’. (The pullback) Given morphisms 1;::Ai -P A , for i
= 1,2 define the pullback of j , and f2 to be the inverse limit of the system dual to that of exercise 1. Thus it is an object P together with morphisms p i : P Ai for i = 1,2 such that flpl = f 2 p 2 ,
Constructions of Rings
146
i.e., according to the diagram of exercise 1 with the arrows reversed. Show
P = { ( a o , a , , a 2 ) ~ n A i : a=of i a 1 = f 2 a 2 }
x { ( a , , a , ) ~ A lx A2:flaI = f 2 a 2 }
in R-Aod, a i m , , and d&. 2. If f:A + B is a map then cokf is a pushout of the system A + s B, and kerf is a
1
0
0
pullback of the inverse system A + j 3B.- In particular, cok is a direct limit and ker is an inverse limit. 3. (Reduced products as direct limits) Any filter f of subsets of a set I is partially ordered via set containment. Letting ng:Hi,, A + Ai denote the natural projection for T c S in E we get a system A i ; n s ) whose direct limit is A i / f together with the natural maps ps: Ai --* A i / S 4. Given a small category I define the inductive limit of a functor f: I -+W to be a universal from F (viewed as an object of U') to the diagonal functor 6: W + V', cf., example 0.1.13 and exercise 0.1.3. The direct limit is a special case via example 0.1.1, viewing any poset I as a small category. Thus the direct limit is a universal.
(n,,,n,,, nisS n
n
01.9 1. The Morita ring A =
(i r),
is graded by
and all other A , = 0. 2. Suppose A is an algebra over an algebraically closed field F, and let a E End, A. We define the characteristic subspace A , = { a E A : ( u- a)'a = 0 for suitable t } for a in F and recall by linear algebra that A = @ A , . If u is a ring endomorphism as well then A,A, E A,, (hint: (a - afl)(ab)= (a - a)aab aa(u - B)b for a in A, and b in A&, so A is graded by the multiplicative structure of F. If, moreover, a is an automorphism algebraic over F (viewing End, A as an F-algebra) then there are only a finite number of eigenvalues, since each eigenvalue is a root of the minimal polynomial, so there are only a finite number of nonzero components. If, instead, a is a derivation then A is graded by the additive structure of F (hint: show A,AF E A , + , by means of the equation (a - a - fl)(ab)= ((a - a)a)b a(a - fl)b). 3. Suppose A is graded. There is a lattice injection from {left ideals of A , } to {graded left ideals of A } given by L + A L (since ( A L ) , = L). This foreshadows a connection between the structures of Al and of A which will be of focal interest at times. Often A, has a special interpretation. For example, in exercise 2 if a is an automorphism which satisfies a separable polynomial then A , = A".
+
+
Exercises
147
Tensor Algebras 4. For every C-algebra A and every map f:M
+ A such that ( f x ) ’ = 0 for all x in M there is a unique algebra homomorphism f: A(M) + A extending f. Use this property to write A(M)as a suitable universal. 5. Suppose M is a free C-module with base indexed by I . Then A(M) x C { X } / A where A is the ideal generated by { X f , X i X j+ X j X i : i , jE I } . (Hint: Verify the universal property.) Furthermore, letting x i denote the canonical image of X i , one has a base B = { I } u { X i ; . . x i U : O< u < 00 and i , < ... < i,} of A(M) over C. The degree of the monomial x i l . . . x i uis defined to be u and is welldefined. Letting B, = { b E B: deg(b) is even} show A ( M )is Z/2Z-graded, where the 0-component is CB, = Z(A(M)).Thus A(M) is a free module over its center with base { I } u { x i :i E I } . 6. The symmetric algebra is a universal from a given module M to the forgetful functor (from {commutative C-algebras} to C - d ~ d . ) 6’. (Tensor rings as universals) Define the category W i ~ g - . @ i m ~whose d objects are pairs cp(R,M ) with M in R-dud-R, and whose morphisms are pairs (cp,f) where cp: R -+ R‘ is a ring homomorphism and f: M + M ’ is a p m a p on each side (i.e., q f ( r l x r 2 )= (cpr,)(fx)(cpr,).There is a functor F : {N-graded rings} + W i ~ g - B i m satisfying ~d FT = (To,TI).Taking u:(R, M ) + T R ( M )by sending R to the 0-component and M to the 1-component, show ( T R ( M ) , uis) the universal from (R, M ) to F. 7. Define the algebra of dual numbers (? over C to be free with base l , x and satisfying x 2 = 0. (This is the exterior algebra of a C-module_isomorphic to C.) If R E C-,&Y define a correspondence Deriv(R) + Aut(R €3 C) by sending 6 to the automorphism s^ of R @ ( ? given by s^(r,@l + r 2 @ x ) = r , @ 1 1 _ + ( r , + 6 r l ) 0 x ; inner derivations of R are sent to inner automorphisms of R €3C. 8. The free product R of any algebra over a field F with a polynomial ring F [ i ] in one in indeterminate is a prime ring. (Hint: Show r 1 h 2 # 0 for all nonzero r r2 in R by ordering a base.) This gives a useful procedure of creating examples of prime rings, as illustrated in exercise 3.2.9.
,,
Quadratic Forms and Clifford Algebras 9. A quadratic form on a vector space V is a map Q : V + F with Q ( a x ) = a 2 Q x for all x in V and such that the map B : V x V+F given by B ( x , y ) = Q ( x + y ) - Q x - Q y
is a bilinear form. Given a quadratic form Q on V, define the Clifford algebra C( V, Q ) = T ( V ) / I where T (V ) is the tensor algebra of V and I is the ideal generated by all x €3 x - Q(x).Then one has a canonical map $: V + C( V,Q ) ; writing X for $x one has X2 = Q x and X j j jE = B ( x ,y ) for x , y in V. In fact, C(V,Q ) is
+
universal with respect to this property: If A E F - d @ and f: V -+ A is a map such that ( f x ) ’ = Q x then there is a unique algebra homomorphism C( V , Q ) + A such that ?$ = f. Generalize this to arbitrary modules over commutative rings. 10. If { x i : i E I } i s a basefor V o v e r F t h e n E = { l } u { x i l x i 2 . . ~ x i r : t E N a n d i , < i, < ... < i , } span C( V,Q ) over F. (Easy induction on the degree of the monomial.) In particular, if [ V : F ] = n then [ C ( V , Q ) :F ] 2 2”. In fact, E is a base of C ( V , Q ) when B is nondegenerate, as seen in the next exercises.
Constructiolrr of Rings
148
11. Given a field F of characteristic # 2 and 0 #a, ~ E define F the quaternion algebra to be a four-dimensional algebra over F with elements x,y such that x2 =a, y2 = j,and x y = - y x ; then { I , x , y , x y } is a base. Note such an algebra must be
unique up to isomorphism in F-dt''. Q=(a, j)can be constructed as follows: Let K = F[1]/(,I2 - a ) , a commutative ring, and let a be the canonical image of 1 in K, so K = FLU].Then Q is the F-subalgebra of M 2 ( K )generated by the matrices 2a 0 and y = Note (o o) E K Q , implying K Q = M , ( K ) ; x = 0 -a a dimension count then shows M 2 ( K ) 7 z K @ Q , from which it follows Q is simple with center F. 12. (Explicit construction of the Clifford algebra.) Notation as above, suppose B is nondegenerate and char(F)#2. We construct an algebra A containing V in which x 2 = Qx for each x in V; moreover, if [ V : F ] = n then [ A : F ] = 2", implying C (V, Q ) 7z A and providing an explicit description of the Clifford algebras. Let { x i : 1 s i In } be an orthogonal base of V (c.f., Jacobson [74B,p. 3381). Since B is nondegenerate we have Q x i # O for each i. Let ai = Q x i . For each t < n we shall define inductively A ( x , , ...,x,), which will be the Clifford algebra on F x , ; it will follow A = A ( x , , . . .,x,).
(" )
(i .)A
c:=,
Case I . A ( x , ) = F [ 1 ] / ( L 2 - a,), identifying x 1 with the image of 1. Clearly, A ( x l ) = F F x , and x : = a, so [ A ( x , ) : F ] = 2 = 2' and we are done in this case. Note if a, is not square in F then l 2-aI is irreducible, so A ( x , ) is a field; otherwise, A ( x , ) z F x F by the Chinese Remainder Theorem.)
+
Case 11. A ( x l , x 2 ) is the quaternion algebra (a,,a2)2. (Indeed, taking yi with y:=aiandyly2= -y2y,,wecanidentifyxiwithyi,sinceforanyo=y1x, +y2x2 in V one has Q v = y:a, y:a2 = (yly, y2y2)'.)
+
+
Inductive Step. Suppose A ( x , , .. .,x,- 2 ) has been defined. Note for i < j It - 2 that xixj = - x j x i since (xi xj)2 = Q(xi xi) = ai aj = x: x f . Let 6 = ( x , . . . x , - ~= ) ~ f a l . . . a , and S =(a,-,6,a,-,6),,aquaternion algebra. Take y,- , , y , in S with y,- ly, = -y,y,- and y: = ai6. We define A ( x , , . .., x , ) = A ( x , , . . . ,x,- 2 ) BFS. Indeed, identify xi with (xl * * * x , -2 ) - 1 0 yi for i = t - 1 and i = t . Then x: = ai and xixj = -xixi for all i < j 5 t, as desired.
-,
+
+
+
+
,
$1.10 1. Suppose S is a submonoid of Z ( R ) ,and let 1= {As:s E S} be a commutative set of indeterminates. Then R , 7z R[1]/(ASs - 1: s E S), seen by verifying the universal
property.
2
Basic Structure Theory
This chapter contains the general results on rings and their modules which have come to be known as the “structure theory.” One of the main themes is Jacobson’s structure theory, via primitive rings and the Jacobson radical. This approach has been spectacularly successful in the 40 years since its inception. The approach is threefold: First one studies “primitive” rings, which are the building blocks, then one obtains the “semiprimitive rings,” and finally one studies the Jacobson radical J , the smallest ideal of a ring R such that R / J is semiprimitive. Since the inception of this theory, prime rings have come to the fore, especially with Goldie’s theorem, which characterizes left orders in simple Artinian rings. Although Goldie’s theorem and its consequences are not studied until Chapter 3, we shall also recast the structure theory in terms of prime rings; the radical associated with this theory is the “lower nilradical.” We shall also examine other nilradicals. Another side to the structure theory is the study of rings in terms of their modules. This can be done through the Jordan-Holder-Schreier theorem and the Krull-Schmidt theorem. New classes of rings arising in such a discussion are semiprimary, perfect, and seimiperfect rings, which are useful generalizations of Artinian rings. On the other hand, perhaps the best approach to
149
150
Basic Structure Theory
module theory is via projective modules and injective modules, thereby laying the foundation for homological algebra (Chapter 5). Sections 2.8-2.12 deal with this very important aspect of the structure theory. At the end of the chapter we consider involutions on rings. This extra piece of structure turns out to be very useful in studying those rings that possess it.
52.1 Primitive Rings Our object in this section is to become familiar with the class of “primitive” rings which is general enough to embrace a very large assortment of examples but is amenable to study by means of Jacobson’s “density theorem.” Our approach will be to develop the theory of primitive rings first and then describe some of the examples in detail. In the course of investigation we shall introduce some important tools of structure theory, including prime rings, minimal left ideals, and the socle. The definition of primitive ring itself is rather straightforward. Definition 2.1.1: A ring is primitive if it has a faithful simple module. Remark 2.1.2:
(i) Any simple ring R is primitive. Indeed every nonzero module M is faithful since Ann M 4R and is thus 0; hence R / L is faithful simple in R-.Mud for any maximal left ideal L. (ii) If D is a division ring and M is a right vector space over D then End MD is primitive (since M is faithful simple viewed as a module over End MD). Our principal objective is to relate primitive rings to End MD. First we locate D. Proposition 2.1.3: (“Schur’s lemma”) i f M E R - A o d is simple then D = End, M is a division ring.
By remarks 0.2.10,0.2.11 every nonzero element of D is an isomorphism and is thus invertible. Q.E.D. Proofi
If R has a faithful simple module M then by means of the regular representation (cf., example 1.5.11) we can view R c End M D where D = End, M. In fact, R takes up a lot of room, as we shall see now.
52.1 Primitive Rings
151
Jacobson's Density Theorem Definition 2.1.4: Suppose R is an arbitrary subring of End MD where M is a vector space over a division ring D, and view M E R - A u d by the given action of R on M. We say R is dense if for every n in N and every D-independent set { x , , . . . , x , } in M we have R X = M(") where x = ( x l , ..., x , ) E M("). (In other words, given y , , . . .,y, in A4 one can find r in R such that rxi = yi for 1 I i I n.) Remark 2.1.5: Every dense subring R of End MD is primitive. (Indeed, M is obviously faithful in R-~uQ! since f M = 0 implies f = 0; M is simple by the case n = 1 of definition 2.1.4.)
The term "density" stems from the following topological considerations: Given x E M and f E End MD define B(x;f )= { g E End M,: gx = f x } . The B(x;f )are a sub-base for a topology called the finite topology of End MD; R is dense in this topology iff R is a dense subring. The converse of remark 2.1.5 is the key to the subject. Theorem 2.1.6: (Jacobson's density theorem) Suppose R has a faithful simple module M and D = End, M. Then R is dense in End MD. Proof: We shall prove that R ( x , , . . , ,x,) = M(") for every D-independent set { x , , . . .,x , } in M. For n = 1 this is true since M is simple, so we proceed by induction on n. Claim: There is r in R such that rx, # 0 and rxi = 0 for all i < n. Otherwise r ( x l , ...,x,- 1) = O always implies rx, = 0 so we have a well-defined map f: M("- ')+ M given by f ( r ( x l , .. .,x, - ,)) = rx, (since R ( x , , . ..,x , - 1) = M("- by induction hypothesis). Then by proposition 1.5.20 there are d , , . . .,dn- in D such that xidi = f ( x , , . . . ,x,- ,) = x,, contrary to the assumed independence of x l ,. . .,x,. Having established the claim, we see by symmetry that for each j I n there exists rj#O with rjxj#O and rjxi=O for all i # j . Given arbitrary y,,. . . , y , in r k j . Then rxi = rjrjxi = M we find r; in R with r)(rj.xj)=y j and put r = Q.E.D. ririxi = yi for each i, as desired.
,
1;:;
,
An elegant proof of a generalization of the density theorem is given in exercise 2.4.1. One can also formulate these results in terms of matrices.
152
Basic Structure Theory
Proposition 2.1.7: If R is a dense subring of End MD then one of the following situations holds: (i) [ M : D ] = n for some n < co,in which case R x M,(D); (ii) For every n in N there is a subring of R having M,(D) as a homomorphic image. Proof: If [ M : D ] = n then R x End MD x M,(D). Otherwise, given n pick x,D and let R, = D-independent elements x l , . . .,x , of M , put V = { r E R: rV E V } . Then V E R , - A d so the regular representation gives us a homomorphism R, -+ End VDx M,(D) which is a surjection by the density theorem. Q.E.D. The structure of End MD is much less amenable when M is infinite dimensional over D;in this case End M D even lacks IBN (cf., example 1.3.33). One immediate consequence of Jacobson’s density theory is the following extremely important structure theorem.
Theorem 2.1.8: (Wedderburn-Artin) If R is primitive and satisjes DCC on left ideals then R x M,(D) for a suitable division ring D. Write R as dense subring of End M,, and let { x i :i E I } be a base of M over D. Let L, = Ann{xj:j Ii } . Then L, > L, > ... by density, contrary to hypothesis unless 111 is finite. But then we are done by proposition 2.1.7(i). Q.E.D. Proof:
The usual formulation of the Wedderburn-Artin theorem is “any simple ring satisfying DCC on left ideals has the form M,(D).” Since we are assuming 1 E R this statement can be improved (corollary 2.1.25’) and, furthermore, there are several short proof (exercises 8,9). Nevertheless this treatment shows how the density theorem gives us a firm grasp of primitive rings. There is a useful internal characterization of primitive rings. Define the core of a left ideal L (written core(L)) to be the sum of all those (two-sided) ideals of R contained in L. Thus the core is the unique largest ideal of R contained in L. Proposition 2.1.9: ( i ) Core(L) = Ann,(R/L). ( i i ) if L c R is maximal then R/Core(L) is primitive ring. ( i i i ) R is a primitive ring iff R has a maximal left ideal L whose core is 0. Proof: Let A = Core(R).
$2.1 Primitive Rings
153
(1) Ann,(R/L) is an ideal of R, and for any r in AnnR(R/L) we have r = r l E r R c L , implying Ann,(R/L) E A. Conversely, AR = A c L so A c AnnR(R/L). (ii) R/L is a simple module over R and is thus faithful simple over RIA. (iii) (F) follows from (ii); conversely, if M is a faithful simple R-module then M x R/L for some maximal left ideal L, implying 0 = Ann, M = Q.E.D. AnnR(R/L) = core(L) by (i).
Corollary 2.1.10: Every commutative primitive ring has no nonzero left ideals and is thus a field. There is a related criterion for primitivity which is sometimes easier to verify. We say a left ideal L is comaximal with all ideals if L + A = R for all 0 # A a R. Clearly, a maximal left ideal L has core 0 iff L is comaximal with all ideals, leading us to Proposition 2.1.11: all ideals.
R is a primitive ring iff R has a left ideal L comaximal with
Proofi (+) Take L maximal with core 0. (F) Take a maximal left ideal L‘r>L. Then L’ is comaximal with all ideals, Q.E.D. so core (L’) = 0.
Prime Rings and Minimal Left Ideals The density theorem relies heavily on the choice of faithful simple module, and one may ask whether all faithful simple R-modules need be isomorphic. In general the answer is “no” but an affirmative answer is available in one very important situation. We shall say a (nonzero) left ideal is minimal if it is minimal as a nonzero left ideal. Z is an example of a ring without minimal (left)ideals since if n E I 4h then 2nh c I . Primitive rings having a minimal left ideal are particularly well-behaved, and to elucidate the situation we introduce a more general class of rings. Definition 2.1.12: R is a prime ring if AB # 0 for all 0 # A, B Q R.
In generalizing definitions from elementary commutative ring theory, one often does best in replacing elements by ideals. We shall see that prime rings form the cornerstone of ring theory, being the “correct” noncommutative generalization of integral domain.
154
Basic Structure Tbeory
Proposition 2.1.13 The following conditions are equivalent: ( i ) R is prime; (ii) Ann, L = 0 for every nonzero left ideal L of R ; (iii)rl Rr, # 0 for all nonzero r1,r2 E R. Proof: (i)*(ii) Ann, L and LR are ideals of R with (Ann, L)(LR)=O, and 0 # L E LR,SO AnnRL= 0. (ii) (iii) if r2 # 0 then rl E Ann, Rr, implying rl = 0 by (ii). (iii) (i) if 0 # A, B Q R then taking nonzero a E A, b E B yields 0 # (aR)b E AB. Q.E.D. Remark 2.1.14: Every primitive ring is prime. (Indeed, if M E R-&d is faithful and simple and 0 # A,B Q R then 0 # BM I M so BM = M; implying 0 # AM = A(BM),so AB # 0.) The converse of remark 2.1.14is false even for commutative rings, by corollary 2.1.10.The “missing link” is minimal left ideals.
Proposition 2.1.15: Suppose R is prime and has a minimal left ideal L. Then L is faithful simple in R-Aod, implying R is primitive; moreover, every faithful simple R-module M is isomorphic to L, implying End, M x End, L. Proof: Ann, L = 0 by proposition 2.1.13, so L is faithful simple, and R is primitive. Suppose M E R - A u d is also faithful simple. Then LM # 0 so aOxO#O for some a, in L, x, in M. By remarks 0.2.10,ll the map L+M given by a + ax, is an isomorphism. Thus L z M and End, L x End, M. Q.E.D.
Finite-Ranked Transformations and The Socle To study minimal left ideals more closely, we view a primitive ring R as a dense subring of End M,, where M is faithful simple in R - A d and D = End, M. For any r in R define ran&) = [rM:D], viewing rM as a D-subspace of M. We are interested in the elements of finite rank, because they permit great flexibility in the study of R inside End MD. (We shall see below that “finite rank” is independent of the choice of M.) Remark 2.1.16: Suppose rl, r , in R have respective finite ranks t l , t , . Then rank(rl + 1,) I t l + t 2 and rank(rlr,) I min(t,,t,) by remark 0.2.3. The rank 1 elements play a special role.
52.1 Primitive Rings
155
Lemma 2.1.17: If rank(r) = t 2 1 then there are rank 1 elements r l , . . .,r, in Rr such that r = r,.
,
c:=,
Proof: Write rM = xiD, and for each i pick a, in R such that aixj = dijxi for 1 I j I t. Given x in M there are d , in D with rx = xjdj = a i x j d j = E : , , airx, so r=C:,, air. Taking ri=air we note riM=xiD so each ri has rank 1. Q.E.D.
c:,j=l
Lemma 2.1.18:
,
Suppose rank@) = t. Rr is a minimal left ideal of R iff t = 1.
Proof: (a) Take 0 # r , E Rr of rank 1. 0 # Rr, I Rr, so by hypothesis Rr, = Rr and r E Rr,; r has rank 1 by remark 2.1.16. (+) We prove Rr is minimal by showing rERa for every O f a E R r . Indeed, suppose a = r’r # 0 and pick x in M with 0 # a x = r‘rx. Let y = rx #O. Then rM = yD. Since M is simple there is rrrin R with r”(ax) = y. Take x’ arbitrary in M and write rx’ = yd’ for suitable d’ in D ; then rx’ = yd’
= (r”ax)d’ = r”r’(rX)d’ = r”r’yd‘ = rr’r’rxr = r”ax’
proving r = rrraE Ra.
Q.E.D.
To make use of these lemmas we must show that the existence of a minimal left ideal L implies the existence of finite-ranked elements. But proposition 2.1.15 shows M x L in R - A d , so we take M = L, and D = End, L. The crucial step is finding an idempotent in L. Lemma 2.1.19: Suppose L is a minimal lefi ideal of R and a E L with La # 0. Then La = L and Ann,a = 0, and L has an idempotent e with ea = a = ae. Proof: La # 0 and Ann,a are left ideals of R contained in L. Since L is minimal we get La = L and Ann, a = 0. Thus a E La so a = ea for some eE L. Then ea = e2a so (eZ- e) E Ann,a = 0, proving e is idempotent. Hence Q.E.D. Ann,e = 0 (as above) so a - ae E Ann,e = 0. Corollary 2.1.20: Suppose L is a minimal left ideal. If L z # 0 then L has a nonzero idempotent; in particular, this is the case if R is prime. Having produced an idempotent we are ready to utilize it in the structure theory. Proposition 2.1.21: Suppose R is any ring with idempotent e, and put L= Re.
156
Basic Structure Theory
Then there is an isomorphism from End, L to eRe preserving the right module action of L. If L is a minimal left ideal then eRe is a division ring, and, viewing L as module over the primitive ring R = R/Ann, L , we see all elements of L have rank 1.
Proof: Recall End, L = Hom,(L, L)OP.For any f in End, L we have f e = f ( e 2 )= efe E e L = eRe, so we can define $: End, L + eRe by $f = f e . Then $(fl + f 2 ) = (fl + f2)e = f i e + f i e = $fl + $f2 and $ ( f l f 2 ) = ( f z f 1 ) e = f 2 ( ( f 1 e ) e= ) ( f i e ) f 2 e= $fl$f2, so $ is a homomorphism, with ker$ = {f E End, L : f e = 0} = 0. Moreover, given ere in eRe we could define f i in End, L as right multiplication by re; then $.f, = f i e = ere, proving I(/ is onto and thus is an isomorphism. If L is a minimal left ideal then L is a faithful simple R-module and so eRe x End, L x EndE L is a division ring; eL= eRe so rank(e) = 1. Hence rank(re) = 1 for any re # 0. Q.E.D. We have prepared the ground for the definition of an important ideal, defined for an arbitrary ring R.
Definition 2.2.22: The socle of a ring R, denoted soc(R), is the sum of the minimal left ideals of R if R has minimal left ideals; otherwise soc(R) = 0. Remark 2.2.23: If L is a minimal left ideal of R then so is Lr for any r in R such that Lr # 0. (Indeed if 0 # ar E Lr for a in L then L = Ra so Lr = Rar.) Consequently soc(R) Q R.
Lemma 2.1.24: of R ) .
n
If R is prime and soc(R)#O then soc(R)= {nonzero ideals
Proofi Suppose 0 # A a R. For any minimal left ideal L of R we have 0 # A L I A n L I L; hence A n L = L and L c A , proving soc(R) E A. But soc(R)Q R by remark 2.1.23. Q.E.D.
Theorem 2.1.25: If R is a dense subring of End MD then soc(R)= {elements of R having finite rank}.
Proof: {elements of R having finite rank} is clearly an ideal A which by lemmas 2.1.17 and 2.1.18 is contained in soc(R). To prove s o c ( R ) c A we may assume soc(R)#O. Then A#O by corollary 2.1.20 and proposition 2.1.21, so soc(R)E A by lemma 2.1.24. Q.E.D.
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Corollary 2.1.25’: Any simple ring R with minimal left ideal L has the form M J D ) for a suitable division ring D. Proof: R is dense in End LD; the socle, being nonzero, must be the whole ring. In particular rank( 1)= n for some n, implying [ L : D ]= n, so R z M,,(D). Q.E.D.
Remark 2.1.26: If R is dense in End MD and soc(R) # 0 then soc(R) is also dense in End MD. (Indeed take any rank 1 idempotent e and take x E M with ex # 0. Now given independent x,,. . ., x , in M and arbitrary y , , ...,y, in A4 we take ril in R such that rilxi = x and rilxj = 0 for all j # i, and take ri2 in R such that ri2ex = y i ; then letting r = ri2eril E soc(R) we clearly have rxi = y i for 1 I i I t.) Actually, a much more general result (exercise 2) is available if we are willing to move to the category of rings without 1, noting the proof of the density theorem is still valid. We conclude this line of exploration by showing the socle of a prime ring is left-right symmetric.
Proposition 2.1.27: Suppose R is prime. Rr is a minimal left ideal of R iff rR i s a minimal right ideal. Proof: (a) Suppose 0 # r’ E rR. We need to prove r E r‘R. Take an idempotent e of Rr; then r E Re and r = r l e for some rI in R. Write r’ = ra = rlea. Then r,eaRr,ea # 0 so there is r2 in R with 0 # r,ear2rle E r’R. Since eRe is a division ring we can multiply ear2rle by some element to get e, implying r l e E r‘R;which is what we were trying to prove. (=) By left-right symmetry. Q.E.D.
Corollary 2.1.28: A prime ring R has minimal lejit ideals iff it has minimal right ideals, in which case soc(R)= x(minima1 right ideals of R). Let us pause to list the classes of rings described above, from general to specific: Prime -B Primitive
/ Prime with minimal left ideal l M n ( D )
Simple /
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Examples of Primitive Rings The class of primitive rings is fascinating since they vary greatly despite the unified framework imposed by the density theorem. We shall examine some examples in detail, starting with primitive rings with socle. The socle is very useful for constructing examples because of the following easy observation. Remark 2.2.29: If R is a subring of End MD containing soc(End MD)then R is dense in End MD. In particular, if R , is any subring of End MD then R , + soc(End MD) is a dense subring of End MD.(It is a ring since soc(End MD)is an ideal.)
This remark is very useful in constructing examples of primitive rings resembling given examples. For example, if C is an arbitrary integral domain, then taking an infinite dimensional right vector space M over the field of fractions F of C , we see C soc(End MF)is a primitive ring whose center is C. (Here we identify each element c of C with c 1, where 1 denotes the identity The following important example is also obtained transformation of End MF.) from this remark, cf., exercise 6.2.9’ below.
+
-
Example 2.2.30: Suppose F is a field and T is an infinite dimensional F-algebra. Viewing T E End TFby the regular representation, let R = T + soc(End TF). Then R E F-&’ and is a dense subalgebra of End TF.
Having seen how to construct interesting primitive rings with socle # 0 we turn to primitive rings having zero socle. In fact, we already have many examples and need merely to recognize them by means of the following result.
Proposition 2.1.31:
lf R is a domain having a minimal ldt ideal then R is a
division ring. Proofi There is a short structural proof using proposition 2.1.15 but we may as well use the following easy argument: Suppose L = R x is a minimal left ideal and 0 # a E R. Then 0 # Rax E R x so Rax = Rx and x E Rax, implying x = bax for some b in R. Then 1 = ba so we have found a left inverse for every nonzero element. Q.E.D.
Corollary 2.1.32: lf R is a simple domain which is not a division ring then R is primitive with socle 0. Proofi Simple implies primitive, so we apply the contrapositive of the proposition. Q.E.D.
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159
Example 2.1.33: The Weyl algebra over a field of characteristic 0 is a simple domain, so is primitive with socle 0. Example 2.1.34: For any field F the free algebra F{X1,X2} is not simple but, nevertheless, is primitive with socle 0. To see this, we take a right vector space M with countable base {ul, u 2 , . . .} over F and define the transformations rj in End MF by rlu,=O, r l u i + l = u i and r2ui=uiz+1. We see easily that F{X,, X 2 >is isomorphic to the subalgebra R of End MF generated by rl and r 2 , by the map XiH rj (because the “monomials” in rl and r2 are linearly independent). But M is obviously faithful in R-&ud and M is simple (because for any x in M we get some r;tx = uoa for suitable a # 0 implying u, E Rx, and clearly each ui E Ru, implying M = Ru,). Hence R is primitive. F{X1,X2} is a domain and thus has socle 0. Example 2.1.35: Suppose K is a field with an endomorphism a not of finite order. Then R = K [A;a] is primitive; indeed R(A - 1) is a maximal left ideal with core 0 since by proposition 1.6.25 all two sided ideals have the form RAi. On the other hand, R is a domain and thus has socle 0.
More examples are given in the exercises, as well as in the supplements here.
E Supplement: A Right Primitive Ring Which is Not Primitive Many basic concepts of ring theory are left-right symmetric, but not primitivity. While an undergraduate, Bergman [64] produced a right primitive ring which is not (left) primitive. His example relies on the asymmetry inherent in the skew polynomial construction. Later Jategaonkar found a general construction which has other applications, and we shall present Jategaonkar’s ring. (Also cf. Cohn [85B].) Example 2.1.36: (Jategaonkar’s counterexample) A Jategaonkar sequence of rings is a collection { R,: p I a> of rings for a given ordinal a satisfying the following conditions: (1) R , is a division ring D,and D c R, for each p. (2) For every successor ordinal y = 8’ there is an injection a:, R,
+
D
such that R , is the skew polynomial ring R,[I,; a,]. R, for every limit ordinal y. (3) R , =
To define a Jategaonkar sequence it suffices by transfinite induction to define
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the o,, but first we shall study the properties of Jategaonkar sequences, assuming that they exist. In the above notation we call R , a Jategaonkar ring if a is a limit ordinal. All proofs are straightforward by transfinite induction unless otherwise stated.
R, c R , for every i3 < y I a. Every nonzero element of R, can be written uniquely in standard form f= diIil.* * litfor suitable t, suitable di in D,and suitable ordinals il I i, I * * . I i, < a. In particular, we can define u ( f ) = {largest fl: I , appears in a monomial off}. If f E R, and fl > u(f) then I , = (o,f)-'I,f E R y f . Each R , is a PLID. (Proof by induction: For y a successor ordinal this follows from proposition 1.6.24. For y a limit ordinal and L< R , we take j < y with L n R, # 0 and, inductively, take L n R, = R,f for suitable f i n R,; for every B' > fl we have I,. E R y f by (iii), so, in fact, L = R , f . ) In particular, R, is a PLID. R,I, Q R , for each y (since I y I , = (D,I,)I, for p < y and I , I , E IyR,IZ, c R,I, for fl 2 y by (iii)). (vi) For each y < a there is a homomorphism R,+ R , obtained by specializing for I , + 0 for each fl > y. (sincexB, ,R,I, 4 Rmby (v)) (vii) Every nonzero left ideal L of R, contains a nonzero two-sided ideal of the form RI,. (Indeed, write L = R , f and take /? > u(f). Then I , E L by (iii) and RA, 4 R by (v).) (viii) R, is not primitive since every maximal left ideal has a nonzero core. '(1 + I,& E D for f;. in R , then each (ix) If fll < fl, < < fl, and fi = 0. (Otherwise, take such an equation with y minimal where y = max{u(fi): 1 I i I t } . If y > fl, then specializing y + 0 gives I 1( + I& E D with fi = giI, + 1;.But each u(f:) < y, contrary to hypo+ Isi)gi = 0, and by thesis unless each f: = 0; now, fi = g i I y , so induction on the degree we conclude each gi = 0, so fi = 0. This proves y Ifl,. If y
c:=
c(1
+
Thus Cl3:(1 + A )gi E D, so by induction on t, we see each g i = 0; si so, p v g , = 0, implying g , = 0, contrary to definition of d,.
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(x) R is right primitive. (Indeed, let A = c p < , ( l + 1,)R,. 1 $ A by (ix); A is comaximal with all ideals for if 0 # I 4 R then some 1, E I by (vii), implying 1 = (1 + 2,) - I , E A + I . Hence R is primitive by proposition 2.1.1 1.) In particular, R , is right primitive but not (left) primitive. R, has other interesting properties which shall be derived later. We turn now to the existence of Jategaonkar rings. This may seem to be a formidable task, but in fact can be handled quite easily by means of the “generic approach,” one of the themes of this book. The underlying idea, well illustrated in this example, is to use indeterminates to satisfy a given restriction in the most general possible manner. In the case at hand, we shall show that any given division ring Do is contained in a suitable Jategaonkar ring R,, for any ordinal R,. Let X = { X , : 1 I p < a } be a set of noncommuting indeterminates over D o . We shall build R , as a subring of a larger ring, with the X p taking the place of the A,. Since one large ring we know how to build from Do and X is the Laurent series ring D,((X)) of theorem 1.3.40, we shall take each R, to be a suitable subring. For y a limit ordinal we clearly want R, = {R,: fl < y}, so we shall henceforth assume y is a successor ordinal, say y = p’. Now we want R , = R,[X,; p,] for a suitable homomorphism p y :R, + D where D = R,; hence for all /? < y we must have X,X,X;’ = p y x BE D. The “generic” approach mentioned earlier leads us to use this condition to define D as follows: Let H be that subgroup of the free group % ( X ) generated by all X,X,X;’ for all fl < v. Then H inherits the ordering of % ( X ) and so Do((X)) has a division subring D = D,((H)) in view of theorem 1.2.24’. Let R, be the subring of Do((X)) generated by D and all X , for fl I y. Then there is an injection p,: R , + D given by pyf = X , f X ; ’ , and it remains to show that the X, act like indeterminates over R, or, more formally, that there is an isomorphism $ , : R , [ I ; p , ] + R , given by $ y ~ ~ = o Q i = c I = o l ; . X : yfor = 0, i.e., if each t . Clearly $, is a surjection, so it suffices to prove ker J/, ~ f = , 1 ; . X=~0 then each 1;. = 0. Assume, on the contrary, that ~ : , , 1 ; . X := 0, with each 1;. in R,, and f, # 0, and choose t minimal such. Then 0 = X , ( c f i X i ) X ; l = c ( p , f i ) X i , so we may assume each fi E D. Multiplying through by f;’ we now may assume f , = 1. But now X : XfZAfiX: = 0 = X : c : , ; ( p , f , ) X i (again conjugating by X , ) so‘ c:i,!,(1;. - p,f;,)Xt = 0; by minimality of t we see 1;: = p,f;: for each i.
u
+
+
Claim: If f E D and pyf = f then f E D o . (If we can prove this claim then
162
Basic Structure Theory
each J E D o , contrary to X , being an indeterminate over D o , so we conclude that ker $, = 0.)
Proof of Claim: Write f as a Laurent series x d h h where the dh E Do and h E H. Now every element of H is a string of XvXPX;’ or (XVXBX;’)-’ = XvXj’X, for fl < v ; so any h # 1 starts with some X,, and ends with some Xi:, and by inspection cannot commute with X,. If supp(f) # { l } then we can take h # { l } minimal in supp(f) (which is well-ordered). Then X,h < hX, or hX, < X,h. In the former case, we have X,hX;’ < h and X,hX;’ E supp(X, f X ; ’ ) = supp( f); in the latter case, Xi’hX, c h and Xi’hX, E supp(f). At any rate, we have contradicted the minimality of h, so indeed supp(f) = { l}, proving f E Do as claimed. Having established the claim, we conclude as noted above that ker $, = 0, so that R , = R,[X,;p,] as desired. Q.E.D.
82.2 The Chinese Remainder Theorem and Subdirect Products This short section provides a general method for simplifying the study of rings by breaking a ring into more manageable rings. Our motivation comes from a venerable result of number theory, often called the “Chinese Remainder Theorem:” Given m,, . . .,m,, n,,. ..,n, in N where m,, ... ,m, are pairwise relatively prime, one can find some n E N such that n = n, (modulo mi) for 1 I1 I t. We shall find a very easy proof of this fact by restating it in terms of the notions we previously encountered. Namely, given a collection ( A i :i E I } of ideals of R, we define the canonical homomorphism R RIA, by r + (r + Ai). The Chinese Remainder Theorem says that if n,, . . . ,n, are pairwise relatively prime then the canonical homomorphism Z + Z/niZ is onto. In fact, this result can be stated much more generally. We say ideals A,, .. . ,A, of R are comaximal if Ai + Aj = R whenever i # j.
+Hie,
fl:= ,
Proposition 2.2.1: (“Chinese Remainder Theorem”) Suppose A,, . . .,A, are RIA, comaximal ideals of R. Then the canonical homomorphism cp: R + is onto.
Hi=,
Proofi It suffices to show that (0,. . .,0, r + A,, 0,. . .,0) E cpR for each i, which is clearly the case if (0,. . .,0,1,0,. ..,0) E qR. By symmetry we need only show ( l , O , . .. ,0) E cpR, i.e., there exists r in R such that 1 - r E A , and r E A, n n A,. For each j > 1 write 1 = alj + aj for suitable aIj in A , and
s2.2 The Chinese Remainder Theorem and Subdirect Products
163
aj in Aj. Then
where a is a sum of terms from A, (since each term has some a I j as a multiplicand). Thus, letting r = a , ... a, we see 1 - r = a E A, and r E A, n . . . n A , , as desired. Q.E.D.
Subdirect Product Since proposition 2.2.1 gives a sufficient condition for cp to be onto, we now ask when cp is 1 : l . Proposition 2.2.2: If { A,: i E I}are ideals of R then the kernel of the canonical map 4p: R -+ RIA, is A,. In particular, if A,, . . . , A , are comaximal RIA,. then R / n ; = ,Ai x
n,,,n;=,nis,
Proof;. cpr = 0 iff r + A, = 0 (iff r E A,) for each i, proving the first assertion. The rest follows from Proposition 2.2.1. Q.E.D. Corollary 2.2.3: If A,, . . .,A, are distinct maximal ideals of R haoing intersection 0 then R s~ RIA,.
n:=,
Definition 2.2.3': An ideal P of R is prime if RIP is a prime ring. The following conditions are easily seen to be equivalent to this definition, cf. proposition 2.1.13: (i) If A , B a Rand AB E P then A c P o r B c P (ii) If A 2 P and B 3 P are ideals then AB $Z P (iii) If aRb c P then a E P or b E P.
A semiprime (resp. semiprimitioe) ideal is an intersection of prime (resp. primitive) ideals. Prime ideals are one of the keys to the theory of rings. Let us record a useful general result relating maximal ideals to prime ideals. Remark 2.2.4: Suppose A , , . . . ,A, are distinct maximal ideals of R and A, ... A, E P for some prime ideal P. Then P = Ai for some i. (Indeed, A, ... A,- G P or A, G P; by induction on t we conclude some A, c P, and thus A, = P since A, is maximal.)
,
This remark sheds further light on corollary 2.2.3. If A,, ... , A , are maximal ideals of R with Ai = 0, then any prime ideal P contains 0 = A, A,
n:=,
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and thus P = Ai for some i. Thus A l , . . . , A , are the only prime ideals of R. For future reference we digress a bit, with a module-theoretic version of corollary 2.2.3.
Proposition 2.2.5: If M has maximal submodules N,, where t is minimal, then M x Hi=,M / N i .
. ..,N,with nr=N, = 0
n:=,
+
Proofi Put Mi = M / N , . Define cp: M -, Mi by cpx = (x N , ,..., + N,). Then ker cp = Ni = 0 so it suffices to prove cp is epic. By symmetry it suffices to show for each element x in M there exists x’ in N , n n Nt with x N , = x‘ + N,. By hypothesis N2n ... n N, has some nonzero element y. Since M , is simple we have x + N , = r ( y + N,) for some Q.E.D. t i n R, so we take x’ = ry.
x
+
Definition 2.2.6: Given a collection { A i :i E I } of ideals of R, we say R is a subdirect product of { R I A , :i E I } if the canonical homomorphism cp: R -, RIAi is an injection.
Hie,
Remark 2.2.7: By proposition 2.2.2 we see R is a subdirect product of the RIAi iff Ai = 0. Thus we have an internal condition (in terms of ideals) for when R can be studied in terms of direct products of homomorphic images.
n
Subdirect products are much more general than direct products, as we see by the following examples. Z is a subdirect product of the finite fields Z / p Z for p prime, since pZ = 0. For any product of distinct primes m = p 1 * * * p t ,Z / m Z z Z/piZ. The polynomial ring F[A] is a subdirect product of the fields F [ A ] / ( p ) where p runs over the irreducible polynomials. For any product of distinct irreducible polynomials f = p l * * * pwe , have F [ A ] / ( f ) x F[A]/(pi).
n
n;=,
n:=,
Semiprime Rings Often “semi” means “subdirect product.” Definition 2.2.8: R is semiprime if R is a subdirect product of prime rings. Remark 2.2.9: By remark 2.2.7, R is semiprime iff
n(prime ideals of R} = 0.
Semiprime rings have a very useful property which gives them a central role in the structure theory.
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165
Remark 2.2.10: If L is a nonzero (or right) ideal in a semiprime ring R then L z # 0. (Indeed L $Z P for some prime P a R , so the image t # 0 in the prime ringR = R / P , implying L2 # 0 by proposition 2.1.13; hence L z # 0.)
In 82.6 we shall see that this property actually characterizes semiprime rings. For the time being we utilize it as follows: Proposition 2.2.11: Any minimal left ideal L of a semiprime ring contains a primitive idempotent. Proof: By corollary 2.1.20 L has a nonzero idempotent, which is primitive Q.E.D. by proposition 1.1.21.
52.3 Modules with Composition Series and Artinian Rings This section features the Jordan-Holder-Schreier theory of composition series. As an application, we determine the basic structure of left Artinian rings and also introduce central simple algebras. We shall rely on 50.2.17tT.
The Jordan- Holder and Schreier Theorems Let us say a chain M > M , > * . * > M, has length t. When verifying M is Artinian we shall often find it convenient to prove the stronger assertion that every chain in L ( M ) has length bounded by some t. This can usually be done directly, but we prove now that it is enough to examine one particular chain. Definition 2.3.1: A composition series of length t for M E R-Aud is a chain is of submodules M = M , > M I > ... > M, = 0 such that each Mi-,/Mi simple in R - A d ; each Mi- , / M i is called a factor of the composition series. Two composition series M = M , > M , > > M, = 0 and M = No > N, > ... > Nu = 0 are equivalent if u = t and there is a permutation 71 such that Nzi-,/Nzix M i - , / M i for 1 I i I t. (In other words, the two composition series have the same sets of factors, possibly permuted.) A rejnement of a chain of submodules is a new chain obtained by inserting additional submodules. Remark 2.3.2: A module M has a composition series iff M is both Noetherian and Artinian. ((3) follows at once from theorem 2.3.3 below. (e) M is Noetherian so has a maximal (proper) submodule M , ; M , has a maximal
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submodule M,, etc; the chain M > M, > M , > ... terminates since M is Artinian, thereby yielding a composition series.) Theorem 2.3.3: Suppose M E R - A d has a composition series % = (M = M , > M , > ..* > M,= 0). Then any chain of submodules of M has a refinement equivalent to %. (In particular, every chain has length < t.)
-
-
to denote equivalent chains, and let 9 = (M > N , > Proofi Write N , > .. .) be an arbitrary chain. We aim to prove 9 has a refinement %' %. We introduce the following general notation to permit manipulation of chains: picking Nu and letting 9'= (M > N , > > Nu)and 9"= (Nu> Nu+ > . * we write 9 = 9' + 9"and 9"= 9 - 9'.Also define %, = % ( M > M,) = ( M , > M , > > M, = 0), a composition series of length t - 1. The proof is best visualized by means of the following picture:
,
a)
The proof is by induction on t , the length of the composition series. If N , I MI then by induction we can refine 9'- (M > N , ) ( M , 2 N , ) = (M, 2 N, > N, > to a composition series 5%; W,, so 9 has the refinement (M > M,)+ %: ( M > MI) + = %, and we are done. Thus we may assume N , $ M , . Then N , M, = M since (N, M , ) / M , is a nonzero submodule of the simple module M / M , . By induction on t , we can refine the chain M1 > N, n M, 2 0 to a composition series %', %, and we can surely write U', = U; U;, where W; is a composition series for N , n M,and %; is a chain from M1 to N1 n MI whose factors are all simple. But M , / ( N , n MI) x ( M , + N , ) / N , = M / N , , so we can translate %> to a chain gofrom M to N , having the same (simple) factors. Denoting the length of %;' as rn, we see U; has length (t - 1) - m. Also N , / ( N , nM,)x ( N , M , ) / M , = M / M , is simple. Hence (N, > N1n M,) %\ is a composition series for N , having length t - m,which, by induction, is equivalent to some refinement 9, of 9 - (M > N , ) . Q0 9, is a composition series which is a refinement of 9, so we need merely show (9, 9,) %. To see this we examine the sets of factors, letting . . a )
-
-
+
+
+
+
+
+
-
+
+
-
$2.3 Modules with Composition Series and Artinian Rings
167
9( ) denote the set of factors of a chain.
+
5 ( g 0 9,) = 9(9,) u 9(9,) = T(%?‘J u 5 ( ( N l > N, n M,)
+
= 9(%>) { N , / N , n M,}
+ %;)
u 9(%; - %>)
= 9(%?:) u { N , / N , n M,}
Q.E.D. = S(%?,) u {MIMI} = F(%?). This theorem combines the “Jordan-Holder theorem” (stating the equivalence of any two composition series) with the “Schreier refinement theorem” (stating that any chain can be refined to a composition series). This very fundamental theorem enables us to define the length of M, written { ( M ) , to be the length of its composition series. (If M has no composition series we write [(M) = 00.)
Corollary 2.3.4:
Suppose [(M)
=n
< co.
(i) Every chain of submodules has length 5 n. (ii) P(M) = t ( N ) G ( M / N )for every N < M. (In particular, d ( N ) < [(M).)
+
Proof:
(i) is immediate; (ii) is clear when we refine the chain M > N. Q.E.D.
A related result to be used later connects composition series to primitive ideals, cf., remark 2.5.0 below. Remark 2.3.5:
If an f.g. module M has a composition series of length m then there are primitive ideals P,, .. . ,P,,, such that P, . * . P, M = 0. (Indeed, if M = Mo > M , > ... > M,,, = 0 is a composition series, take Pi = Ann,(Mi- , / M i ) ; thenP,M,-, E M,foreachisoP,,,..*P,MG M,,,=O.)
Artinian Rings Definition 2.3.6: A ring R is left Artinian (resp. right Artinian) if R is Artinian as left (resp. right) R-module. R is Artinian if R is left and right Artinian. It is easy to see that any homomorphic image of a left Artinian ring is left Artinian, because any chain of left ideals can be lifted up from a homomorphic image. We are ready to determine much of the structure of left Artinian rings.
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Example 2.3.7: Every finite dimensional algebra R over a field F is Artinian. (Indeed, if [R:F] = t and R > L, > L, > then viewing the Li as subspaces of R we have t=[R:F]>[L,:F]>[L,:F]> -**implying[Lj:F]=O for some j I t, so Lj = 0.) Remark 2.3.8: Every simple left Artinian ring has a minimal left ideal and thus by Wedderburn-Artin is of the form M,,(D) for suitable n and suitable division ring D. Conversely, R = M,,(D) is simple Artinian since we have the composition series of R-modules R > Ln-l > L n F 2>
> Lo = 0
where L, =
11 Rejj U
j=
(so that L,/L,-, x Reuuis a simple R-module); R is simple by proposition 1.1.5.)
Thus simple Artinian is just a different name for a ring of the form Mn(D). This characterization, resulting from work of Wedderburn, Artin, Hopkins, and Levitzki, is so fundamental that rings of the form M,,(D)are thereby called simple Artinian rings. Their module structure is very straightforward. Remark 2.3.8': Suppose e is a primitive idempotent of a simple Artinian ring R. Then Re is a simple R-module, and any simple R-module is isomorphic to Re, by proposition 2.1.15.
Theorem 2.3.9:
Suppose R is a left Artinian ring.
(i) Every prime ideal of R is maximal; in particular, i f R is prime then R is simple Artinian. (ii) There are only a finite number of distinct prime ideals PI,. . .,P,, and R/pi a direct product of simple Artinian rings. Rl fli= pi x
Proof: (i) Suppose P 4R is prime. Then RIP is prime Artinian and thus primitive by proposition 2.1.15. RIP is simple Artinian by theorem 2.1.8, so, in particular, P is a maximal ideal. (ii) Suppose Pl, P,, . . . are distinct prime (and thus maximal) ideals. Then the chain Pl
3
Pl n Pz 3 Pl n P2n P3 3
terminates at some Pl n n P,, implying for each prime ideal P that Pl n * - n P, E P. Hence P = pi for some i It by remark 2.2.4; the rest is proposition 2.2.2. Q.E.D.
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169
Armed with this result we shall now characterize a wider class of Artinian rings.
Theorem 2.3.10:
The following conditions are equivalent for a ring R:
(i) R is a Jinite direct product of simple Artinian rings. (ii) R = Soc(R). (iii) R is semiprime and leji Artinian. Proofi (i) * (ii) M,,(D)is the sum of its minimal left ideals, and this property passes to finite direct products. L,. Then 1 = ay,for suitable Lyiand uyiE (ii) * (iii) Write R = LYi;writing Li place of Lyiyields 1 E Liso R = Li. Putting Li = Lj gives a composition series R > 15,- > > .-.> L’, > 0 for R in R-Aod, proving R is left Artinian. To show R is semiprime, we note that 0 = Ann{ 1) 3 Ann Li.Since each Li is simple in R-Aod, we see each Ann L, is a primitive ideal; hence R is a subdirect product of the primitive (and thus prime) rings R/Ann L,. (iii) (i) By theorem 2.3.9, letting PI,.. . ,P, be the prime (and thus maximal) ,(R/4), and each R/C is simple Artinian. ideals of R, we have R w Q.E.D.
I:= ,
Eyer
n:=
n:=
A ring R satisfying the conditions of theorem 2.3.10 is called semisimple Artinian, or semisimple. Note that condition (i) is left-right symmetric, implying that the right-handed versions of (ii) and (iii) are also necessary and sufficient for a ring to be semisimple Artinian. Corollary 2.3.11: Suppose R is a semiprime ring in which every nonzero left ideal contains a minimal lefi ideal. If soc(R) is a sum of a finite number of minimal lejl ideals then R is semisimple Artinian. Proofi If soc(R) = Lifor minimal left ideals Li,then we have a composition series soc(R) = M,-l > ... > M , = 0 where each Mu= L, so L(soc(R))It . We shall use this fact to show soc(R) = Re for some idempotent e. Indeed, put e, = 0 and use the following inductive procedure: Given e, idempotent take a minimal left ideal L < R(l - e,) and write L = Ra, for a suitable idempotent a, of R (by proposition 2.2.11). Writing a, = r,(l - e,) we see aueu= 0 but (1 - e,)a, # 0; hence a: = (1 - e,)a, is an idempotent of L orthogonal to e,, so e,, = a: e, is idempotent in soc(R).
IT=
+
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170
The chain 0 < Re, < Re, < in soc(R) must have length It, so Re, = soc(R). If soc(R) # R then by hypothesis R ( l - e,) contains a minimal left Q.E.D. ideal not in the socle, which is absurd. Hence soc(R)= R. Corollary 2.3.12: Suppose a semiprime ring R has orthogonal primitive ideme, = 1. Then R is semisimple Artinian and each potents e l , . . .,e, with Re, is a minimal leji ideal.
I:=,
xi=,
Proofi Put A = soc(R) and L , = A n Re,. Then A = L , since any a = a C e i = ae, E L i . But L i , being in the socle, contains a minimal left ideal L ; = Re: where ei is idempotent (cf., proposition 2.2.1l), and proposition 1.1.21 shows Re, = L ; is minimal. Hence R = soc(R) is semisimple Artinian. Q.E.D.
1
Semisimple Artinian rings are so important in the structure theory that we shall summarize their basic properties in terms of idempotents.
Theorem 2.3.13: Write a semisimple Artinian ring R as Ri = M,,(Di) called the simple components of R.
ni=,R, for suitable
(i) Let z, = (0,. . .,0,l,O,...,0)where 1 appears in the i position. Then each zi is a central idempotent of R with z, = 1; A , = Rzi are the unique minimal ideals of R. Each A, is itself a ring with multiplicative unit z , , and Ri x A , as
,
rings. (ii) Any simple R-module M is isomorphic to Re for some primitive idempotent e of R. (iii) Every primitive idempotent e of R lies in some A,. Two primitive idempotents e, e' lie in the same A, iff Re x Re' as modules, iff eRe' # 0. (iv) There are precisely t isomorphic classes of simple R-modules, one for each simple component. Proof: (i) by proposition 1.1.14. (ii) M = A,M so some A,M # 0 implying A,M = M. Let 2 = Cj*,Ai. Then AA, = 0 so AM = AA,M = 0 and thus we view M as module over R I A x Ri. M is a simple &-module, by remark 0.0.9, so by remark 2.3.8' is isomorphic (in R,-Mud) to R,e for some primitive idempotent e of R,. Viewing e in A, we see M % Re in R-Mud. (iii) Write e = (el,.,.,e,) where e, E A,. The e, are orthogonal idempotents so e = e, for some i, and all other ej = 0. Clearly e, e' are in the same
42.3 Modules with Composition Series and Artinian Rings
171
component iff eRe' # 0 (since A iis simple as a ring). If e, e' E A i then Re x Re' as shown in (ii); conversely, i f f : Re --t Re' is an isomorphism and e E A i , then 0 # f e = f (zie)= zi f e E ziRe' implying zie' # 0, so e' E A'. (iv) Immediate from (ii) and (iii). Q.E.D. Let us disgress for a moment to show that the left-right symmetry of semisimple Artinian rings does not hold for left Artinian rings in general.
(r ,,L)
Example 2.3.14:
Suppose F is a field and K is a subfield, and let R =
. By example 1.1.9, the only proper left ideals of R are 0, Fe,,
,
,
+ Ke,, ,
Fe, and { a'e, + a'/3e, : a' E F } for each /3 in F, so R has length 2 in R - A u d and is left Artinian. On the other hand, for any K-subspace J of F, we see
(: i)
is a right ideal. Thus R is not right Artinian if [F:K] = 00.
Rather than enter here into the consequences of these basic results on Artinian rings, we close the discussion with a typical argument concerning an Artinian-like condition. Proposition 2.3.15: Suppose R has no infinite chain Rr 3 Rr2 3 Rr3 3 r in R. If a E R with Ann, a = 0 then a is invertible in R.
* *
for
Pro08 By hypothesis Ra' = Ra'" for some i, so ai E Ra'". Write ai = ba'+1. Then (1 - ba)ai = 0, implying 1 - ba = 0 (seen inductively on i ) ; hence ba = 1. But now (ab - 1)a = a(ba) - a = 0 so ab - 1 E Anna = 0 and ab = 1. Q.E.D.
Split Semisimple Artinian Algebras Definition 2.3.16: Suppose R is a finite dimensional F-algebra. R is split if Rx MJF) for suitable t, u in N . A field extension K of F is a splitting jield for R if R OFK is split as K-algebra.
n:=,
One of the prime uses of tensor products is in splitting algebras, largely because of the following few results. Note that if R is simple and split then R 2 M,,(F) for some n. Proposition 2.3.17: Any finite dimensional semisimple algebra R over an algebraically closed field F is split.
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172
n:=,
Proofi Write R = R, where each R, is simple Artinian. Then R, = Mnu(D,) where 0,is a division algebra finite dimensional over F. For any d in D,, we see F[d,] is a finite extension of F and is an integral domain (in fact a field); since d is algebraic over F we conclude d E F. This proves 0, = F for each u, so R x Hi=,MJF). Q.E.D.
Corollary 2.3.18: The algebraic closure K of F is a splitting field for a finite dimensional F-algebra R iffR BFK is semiprime. Proofi (=.) is a fortieri. (t) [R BFK: K] = [ R :F] so R BFK is Artinian; if semiprime then R BFK is semisimple and thus split. Q.E.D. The condition in corollary 2.3.18 does not always hold (cf., example 1.7.25), but an important positive result is available.
n:=,
Corollary 2.3.19: If R = R, where each R, is a simple finitedimensional F-algebra with center x F, then the algebraic closure K of F is a splitting field for R.
fl
Proofi By proposition 1.7.15 we have R BFK x R, QF K, each component of which is simple by theorem 1.7.27, so R Q K is split by corollary 2.3.18. Q.E.D. Remark 2.3.20: The general procedure for splitting R (if possible) is to write R = R, and take a field compositum F‘of the Z ( R , ) over F. If R OF x F’is still semiprime then we can split it by corollary 2.3.19.
n
Given a splitting field we can find a splitting field which is a finite extension of F, by the following.
Proposition 2.3.21: If K is a splitting field for R over F then K contains a splitting field which is finitely generated (as a ring) over F.
fl:=
Proofi Write R QFK = Mnu(K).Letting {el’: 1 I i, j Inu} be a set of matric units for R, we write e y = r l l Q a: for rll in R and a!; in K. Let K O be the subfield of K generated over F by all the a!:. Then K O is a finitely generated extension of F, and R BFKOcontains enough matric units Q.E.D. to be split (by a dimension check).
xk
$2.3 Modules with Composition Series and Artinian Rings
173
Central Simple Algebras and Splitting Definition 2.3.22: R is a central simple F-algebra if R is simple, F = Z(R), and [R: F] < 00. (In the literature, the “finite dimensionality” criterion is not always included in the definition.)
Every central simple algebra is clearly Artinian, but the theory of central simple algebras is considerably richer than that of simple Artinian rings. We shall develop some of the basic splitting results here, but leave the remainder of the theory for Chapter 7. Remark 2.3.23:
(i) If R is a prime algebra which is finite dimensional over its center then R is simple; indeed R is Artinian so apply theorem 2.3.9. (ii) Every subalgebra of a central simple division algebra is a division algebra. (Clear by (i) since it is a domain.)
Proposition 2.3.24: If R is a central simple F-algebra then the algebraic closure F of F is a splitting j e l d for R. Proof;. By theorem 1.7.27 R OFF’ is simple and thus split, by proposition 2.3.17. Q.E.D.
(See proposition 6.1.24 below for a related result.)
Corollary 2.3.25: If R is a central simple algebra then [R:F] = n 2 for suitable n, in particular, for n such that R OFF x M,,(F). Remark 2.3.26: Writing a central simple algebra R in the form M,(D) for D a division ring, we have the following information:
(i) D itself is a central simple F-algebra, with [D:F] = ( n / t ) 2 where n2 = [R:F]. (ii) R x M,,(F) BFDby example 1.7.21. (iii) ROP is also a central simple algebra, and R o P ~ M , , ( D o PM,,(F)OFDoP. )x Continuing this notation, we know by propositions 2.3.24 and 2.3.21 that R has a splitting field finite over F, but we shall improve this result in $7.0. We conclude this short discussion with a theorem so important that (according to Zelinsky) the master of central simple algebras, A. A. Albert, called it the most important theorem in the subject.
174
Basic Structure Theory
Theorem 2.3.27: If R is a central simple F-algebra and [ R : F ] = n then R &IF RoP x M,(F). Proof;. M J F ) z End,(R) contains both R and RoP via the right and left regular representations of R, respectively, cf., remark 1.510, thereby yielding a ring homomorphism cp: R BFRoP + M,(F) given by r 0 r' + rr'. But R RoP is simple by theorem 1.7.27 so cp is an injection; counting dimensions shows cp is an isomorphism. Q.E.D.
Zariski Topology (for Finite Dimensional Algebras) Definition 2.3.28: Suppose F is an infinite field, and let V = F("). Given a subset S of the polynomial ring FCA,,. . .,A,] define V ( S ) = {(al,.. .,a,) E V : f(al,. . .,a,) = 0 for all f in S}. Then the {US):S C F[A,, ...,A,]} constitute the closed sets of a topology on V called the Zariski topology. The Zariski topology of a finite dimensional algebra over an infinite field is its Zariski topology as a vector space. Write V(f)for V({f}).
n{
The open sets have the form V - V ( S )= V V(f):f E S } = V f E S}, so the sets V - V(f) form a base of the open sets.
u{ - V(f)):
Remark 2.3.29: V(f) # V for every 0 # f E FCA,, . . . , A , ] , by the fundamental theorem of algebra, cf., Jacobson [85B, theorem 2.193. It follows that any two nonempty open sets intersect nontrivially, seen by taking complements of V(fl) u V(fJ c V(flf2) # V. In particular, every open set is dense. This simple fact has far-reaching consequences. The idea in using the Zariski topology is to show various natural conditions are Zariski open and thus Zariski dense; hence any finite number of these conditions define a Zariski dense set. In this manner one can add on certain conditions at no expense. In what follows, R is a finite dimensional F-algebra, and wefixabaseb,, ..., b,. Definition 2.3.30: Let p,, . . .,p, be commuting indeterminates over R, and define R' = R [ p , , .. .,p,] and F' = F[p,, . ..,p , ] . The element x = bipi in R' is called the generic element of R (with respect to the base b , , . ..,b,). Viewing b,, .. .,b, as a base of R' over F' we can embed R' in M,(F') by mean of the regular representation; the characteristic polynomial of the matrix corresponding to x is called the generic characteristic polynomial px(A).
82.3 Modules with Composition Series and Artinian Rings
175
Remark 2.3.31: Because F’ is a unique factorization domain and p x is monic, the monk irreducible factors of p , in F’[A] are actually irreducible in F(pl, . . . , p , ) [A], so in particular the minimal polynomial m, of x over F ( p l , . . ., p , ) has coefficients in F’. This is called the generic reduced characteristic polynomial.
c
Definition 2.3.32: Suppose r = aibi E R, and m, = cf=oh(pl,. . . ,p,)ili where fi E F’ (so fd = 1). The reduced characteristic polynomial m, of r is Cf=,l;.(al,. . .,a,)Af. The reduced norm of r is ( - l)dfo(al,.. .,a,) and the reduced trace of r is -fl(al,. . .,a,). Remark 2.3.33’: These definitions are independent of the choice of base. (Indeed a change of base corresponds to conjugation of matrices.) Remark 2.3.34: (i) Suppose K is a field extension of F. The reduced norm (resp. trace) of r in R is the same as the reduced norm (resp. trace) of r Q 1 in R QFK, where we use the base bl Q 1,. . . ,b, 0 1. (ii) If R = M J F ) the generic reduced characteristic polynomial is easily seen to be the characteristic polynomial of x. Thus, in this case the reduced trace is the trace, and the reduced norm is the determinant, and these are independent of the choice of base.
Proposition 2.3.35: In the above notation every r in R has degree I d over F, and { r E R: the minimal polynomial of r has degree d } is a nonempty Zariski-dense open subset of R. Proofi As above let x be the generic element with respect to the base b , , ...,b,. Since x has degree d we have 1, x,. . . ,x d - l linearly independent so writing x k = hjkbkfor hjkin F’, we see the t x d matrix (hj,)has a nonzero d x d minor q in F’. Letting q1 = q, q z , . . . ,q,, denote the d x d minors in (hi,) we see that a suitable specialization of the pi H ai in F will give q(al,. . . ,a,) # 0 and thus the matrix (hjk(a1,..., a,)) has a nonzero d x d minor; letting r= a,b E R, we conclude that 1, r,. ..,r d P 1are linearly independent, so r has degree d. An arbitrary element r = aibi will have degree d iff 1, r, . . .,r d - l are independent, iff one of the minors ql(al,.. .,a,), . . .,qu(al,. . .,atj) is nonzero, i.e., iff (al,. ..,a,) belongs to the union of the Zariski open sets V - V(ql), Q.E.D. ..., V - V(q,,)where V - R.
1
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176
Proposition 2.3.36: An element r of R is invertible iff its reduced norm is nonzero. In particular {invertible elements of R } is Zariski open. Proofi Let p,(A) and m,(A) be the generic characteristic and generic reduced characteristic polynomials. Then m, divides p x , so specializing x H r we see rn,(A) divides p,(A). Let qr(A)be the minimal polynomial of r. Then q, I m,. But p , is the characteristic polynomial of the matrix corresponding to r under the regular representation, so q, and pr have the same irreducible factors; consequently, m, and p , have the same irreducible factors. Now r is invertible iff A{m,(A), iff Atp,(A), iff the reduced norm of r is Q.E.D. nonzero.
52.4 Completely Reducible Modules and the Socle In 52.1 we introduced the socle of a ring in order to study primitive rings. The careful reader will have noticed that a much more general definition was given in exercise 0.0.12, and we shall now examine (from scratch) how this definition applies to modules, leading to several important characterizations of semisimple Artinian rings. Definition 2.4.1: The socle of an R-module M, written soc(M), is the sum of the simple submodules of M (so soc(M) = 0 if M has no simple submodules). M is completely reducible (also called semisimple) if soc(M) = M.
In examining soc(M) we are led to the complement of a submodule N in M, which we recall is N' < M such that M = N 0 N' (where 0 denotes the internal direct sum throughout this section). Of course, the complement need not be unique, as evidenced by the example M = R'') and N = {(a,0):a E W}. We say M is complemented if every nonzero submodule has a complement. Modules are not necessarily complemented, so we introduce a related concept. Definition 2.4.2: A submodule L of M is large (also called essential) if L n N # 0 for all 0 # N < M. An essential complement of N I M is some N' I M such that N n N ' = 0 and N N' is large.
+
Thus every complement is an essential complement. Essential complements are very important in module theory because they exist, as we shall see now.
82.4 Completely Reducible Modules and the Socle
177
Lemma 2.4.3: Suppose A , B, C s M. (i) I f ( A + B ) n C # Oand A n C = 0 then B n ( A + C ) # 0. (ii) If A I C I A + B then C = A + ( B n C).
Pro08 Each result follows instantly from the observation if 0 # c = a then b = c - a E A + C. Q.E.D.
+b
Proposition 2.4.4: Every submodule of M has an essential complement. In fact, if N , Q IM with N n Q = 0 then N has an essential complement containing Q.
Proof: We prove the second assertion, since the first then follows for Q = 0. By Zorn’s lemma there is N‘ I M containing Q, maximal with respect to N n N ’ = O . We claim N + N’ is large. We must show for O # P c M that P n ( N + N’) # 0. This is obvious unless P $ N’ in which case N ’ + P > N ’ and so ( N ’ + P ) n N # O . Hence P n ( N ’ + N ) # O by lemma 2.4.3(i). Q.E.D. Proposition 2.4.5: If A 5 N I M and B is a complement (resp. essential complement) of A in M then B n N is a complement (resp.essential complement) of A in N.
Proof: The assertion about complements is a special case of lemma 2.4.3(ii) taking N = C and M = A B. To prove the assertion for essential complements suppose 0 # P I N ; then P I M so 0 # ( A + B)n P. Taking 0 # p = a + b we have b = p - a E N so P E A + (Bn N ) . Q.E.D.
+
When verifying a submodule N of M is large it is enough to show N n Rx # 0 for every 0 # x E M , since every nonzero module contains a nonzero cyclic submodule. This trivial observation will be used frequently, especially when we study large submodules in earnest in fi3.3, but let us make several easy observations. Proposition 2.4.5’: (i) Suppose N IN ’ I M . N is large in M iff N is large in N‘ and N’ is large in M . (ii) If M = M , + M 2 and Ni are large submodules of M ithen N 1 + N 2 is large in M.
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Proofi (i) Immediate from the definition. (ii) First consider the case N, n N2 = 0. If x = x, + x2 where x i E Mi we need r in R such that 0 # rx E N , + N,. This is trivial unless x1 # 0, in which case 0 # rlxl E N , for some r , in R. We are done unless r i x 2 # 0, in which case 0 # r2rlx2 E N2for some r , , and then we take r = r2r1. In general let M; be an essential complement of M, n M, in M,. Then M, M; is large in M, + M, by our special case; likewise N, + (N, n M;) is large in M, M; and so is large in M, + M, by (i). A fortiori N, + N , is large in M, + M , = M. Q.E.D.
+
+
Armed with these basic tools we shall now find the link between completely reducible modules and semisimple Artinian rings. Although the next two results have purely lattice-theoretic proofs (given in exercises 0.0.11 and 0.0.12),we shall present full module-theoretic proofs. Proposition 2.4.6: If M is complemented then M is completely reducible. Let M' be a complement of soc(M) in M. We shall show M' = 0 (so that soc(M) = M). Indeed, if there is 0 # x E M' we can take N I M' maximal with respect to x # N (by Zorn's lemma), and N has a complement N' in M' by proposition 2.4.5. We claim N' is simple, contrary to M' n soc M = 0. To prove the claim, suppose to the contrary that 0 < A c N'. Then A has a complement A' in N', and by choice of N we have x = y, + a = y, + a' for suitable yi in N, a in A, and a' in A'. Then y, - y, = a' - a E N n N' = 0, so a' = a E A n A' = 0 and thus x = y , E N, a contradiction. Thus the claim is proven, and we must conclude M' = 0 as desired. Q.E.D. Proofi
Theorem 2.4.7: Soc(M) =
{ large submodules of M}.
Proof: If 0 # S I M is simple and L I M is large then 0 # S n L E S implying S n L = S and so S E L. Hence soc(M) E L. Letting P = (-){large submodules of M }we thus have soc(M) E P. To prove the opposite inclusion we shall show P is complemented (and then we are done by proposition 2.4.6) Indeed if N I P then N has an essential complement N'; thus N N' is large, implying N I P I N N '. By lemma 2.4.3(ii) we see P = N @ (N' n P), so N has a complement in P, as desired. Q.E.D.
+
Theorem 2.4.8:
+
Thefollowing assertions are equivalent for M in R-.,4tud:
42.5 The Jacobson Radical
179
(i) M is completely reducible. (ii) M has no large proper submodules. (iii) M is complemented. (i) +(ii). M = soc(M) = (-){large submodules of M}, so M is its only large submodule. (ii) (iii) Every submodule has an essential complement, which is thus a complement. (iii) + (i) By proposition 2.4.6. Q.E.D. fro&
Since R is semisimple Artinian iff R is completely reducible in R - d o d , we have a new characterization of semisimple Artinian rings, with which we conclude this discussion.
Theorem 2.4.9
The following assertions are equivalent for a ring R :
(i) R is semisimple Artinian. (ii) R is completely reducible in R - A u d . (iii) Every R-module is completely reducible. (iv) Every R-module is complemented. (v) The only large left ideal of R is R itseg We already have (i) o (ii) and (iii) o (iv), and (iii) + (ii) is a fortiori. It remains to prove (ii) =. (iii). Write R = L, where Liare minimal left ideals, and suppose M E R - A d . For any x in M we have Rx = L i x . If Lix # 0 then a -+ a x gives an epic from Lito Lix, implying Lix is a simple module; hence x E soc(M) for each x in M, proving M is completely reducible. (ii) o (v) By theorem 2.4.8. Q.E.D. froofi
1
82.5 The Jacobson Radical We return to the general structure theory of rings. Having the Jacobson density theory at out disposal, it is natural to focus on primitive rings. This leads us to call an ideal P primitive if RIP is a primitive ring. Using subdirect products we can then study semiprimitive rings, i.e., rings R which are subdirect products of primitive rings, i.e., (){primitive ideals} = 0. We shall need an easy criterion for an ideal to be primitive.
P 4R is a primitive ideal iff P is the annihilator of a simple R-module M. (Proof (*) R / P has a faithful simple module M, which is then
Remark 2.5.0:
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180
a simple R-module with annihilator P, by remark 0.0.9; (-=) if P then M is a faithful simple RIP-module.)
= Ann,
M
The subject of this section is the obstruction to a ring being semiprimitive. The Jacobson radical Jac(R) = n{primitive ideals of R f . Clearly R is semiprimitive iff Jac(R) = 0, so we shall try to “remove” the Jacobson radical. Definition 2.5.2:
Proposition 2.5.1 ’: (i) If A a R and A 5 Jac(R) then Jac(R/A) = Jac(R)/A; in particular R/Jac(R) is semiprimitive. (ii) ff Jac(R/A) = 0 then A 2 Jac(R).
Proof: Write- for the image in I? = R / A . (i) If P is a primitive ideal of R then A E Jac(R) E P so primitive, implying P is a primitive ideal of R. Thus
Jac(l?) = =
RIP FZ R / P is
n{ P : P is a primitive ideal of R}
{ primitive ideals of R} = Jac(R)
(ii) If Jac(I?) = 0 then 0 = n { P :P is a primitive ideal containing A } = primitive ideals containing A} 2 Jac(R) implying Jac(R) 5 A. Q.E.D.
n{
Proposition 2.5.2: Jac(R) =
n{maximal leji ideals of R}.
Proofi ( G )By proposition 2.1.9 every maximal left ideal contains a primitive ideal and thus contains Jac(R). ( 2 )If P 4 R is primitive then for same simple M in R - A O J we have P = Ann, M = {Ann x : x E M } 2 {maximal left ideals of R ] since each Ann x is a maximal left ideal. Q.E.D.
n
n
Quasi-Invertibility In order to study Jac(R) as a set, we need a description in terms of elements. To this end let us say a E R is left quasi-invertible if 1 - a is left invertible in R, i.e., if 1 E R(l - a); a is quasi-invertible if 1 - a is invertible (from both sides). A subset of R is quasi-invertible if each element is quasi-invertible. (Compare with definition 1.1.23(i).)
181
$2.5 The Jacobson Radical
Lemma 2.5.3: Any left ideal L of left quasi-invertible elements is quasiinvertible. In fact if a E L and r ( l - a ) = 1 then 1 - r E L . Proofi 1 - r = -ra E L , so r = 1 - (1 - r) has a left inverse b. Hence the right and left inverses of r are equal, i.e., 1 - a = b is invertible. Q.E.D.
Proposition 2.5.4: Jac(R) is a quasi-invertible ideal of R which contains every quasi-invertible left ideal.
For any a E Jac(R) = n(maxima1 left ideals of R ) we cannot have 1 - a in a maximal left ideal, so R( 1 - a ) = R, proving a is left quasi-invertible. Hence Jac(R) is quasi-invertible by Lemma 2.5.3. Now suppose B is a quasi-invertible left ideal. If there were some maximal left ideal L @ B we would have B L = R, so b a = 1 for some b in B, a in L, and then a = 1 - b would be invertible (since b is quasi-invertible), contrary to L # R. Hence B is contained in every maximal left ideal, so B E Jac(R). Q.E.D. Proof:
+
+
In view of this result, Jac(R) is the same as what we would get from the right-handed analogue of definition 2.5.1. Indeed, calling this “right-handed” Jacobson radical J , we see J is quasi-invertible (by the right-handed version of proposition 2.5.4).Thus J CI Jac(R), and symmetrically Jac(R) E J .
Remark 2.5.4‘: Recall a left ideal L of R is nil if every element of L is nilpotent. Remark 1.1.27 applied to proposition 2.5.4 show Jac(R) contains every nil left (or right) ideal. The characterization given in 2.5.4 is also useful because invertibility (and thus quasi-invertibility) passes to homomorphic images and sometimes back again, as we shall see now.
Lemma 2.5.5: If J is a quasi-invertible ideal of R and r is an element of R whose canonical image in RIJ is invertible then r is invertible in R . Proof: Take r f in R such that 1 - r’r E J and 1 - rr’ E J . Then these elements are quasi-invertible, so r’r and rr‘ are invertible in R. It follows at once Q.E.D. that r is left and right invertible, so r is invertible.
Our final basic result concerning the Jacobson radical involves passing to homomorphic images.
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Proposition 2.5.6: (i) If cp: R + T is a ring surjection then cp(Jac(R))G Jac(T). (ii) If A 4R then (Jac(R) + A ) E Jac(R/A), equality holding if A G Jac(R).
Proofi Write J = Jac(R).
(i) If a E J then cp(1 - u)-’(l - cpa) = 1, implying cpa is left quasi-invertible; thus cpJ c Jac(T). (ii) The first assertion follows from (i); the second assertion is proposition 2.5.1’. Q.E.D.
Examples To give the reader the flavor of the Jacobson radical, we present some results concerning large classes of examples. Suppose 0 v {invertibleelements of R} is a division ring D.
Proposition 2.5.7: Then Jac(R) = 0.
Proofi Let J = Jac(R). Then J n D Q D implying J n D = 0. But if a E J then 1 - a is invertible so 1 - a E D and a E D n J = 0. Q.E.D. Corollary 2.5.8: Suppose A is a ring with a filtration by N such that A(0) is a division ring. Then Jac(R) = 0.
In particular, every skew polynomial ring over a division ring is semiprimitive. Let us now consider the opposite extreme.
Definition 2.5.9:
A ring R is local if { noninvertible elements of R} Q R.
(Some authors call these rings “quasi-local,” reserving the use of “local” for when the ring is Noetherian.) Proposition 2.5.10:
The following statements are equivalent:
(i) R is local. (ii) {noninvertibleelements of R} is a maximal left ideal of R. (iii) R has a unique maximal left ideal. (iv) Jac(R) is a maximal left ideal (and thus the unique maximal left ideal). (v) R/Jac(R) is a division ring.
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froofi Let J = {noninvertible elements of R}. If r 4 J then Rr = R, so r cannot be contained in any proper left ideal of R.
-
(i) (ii) (iii) is immediate from the above observation. (iii) =. (i) Let J' be the unique maximal left ideal of R. Obviously J' c J. But if r E J then Rr is a proper left ideal and thus contained in a maximal left ideal, which is J', proving J G J'. (iii) o (iv) by proposition 2.5.2. (iv) o (v) A ring is a division ring iff it has no nonzero left ideals. Q.E.D.
Example 2.5.21: If D is a division ring then the ring of formal power series D [ [ A ] ] is local (since the noninvertible elements are the power series having constant term 0, clearly an ideal). Remark 2.5.22: The easiest condition to verify that R is local is the following: If a + b = 1 then a or b is invertible. (For suppose R is not local. Then {noninvertible elements) is not closed under addition, so there are noninvertible elements x and y with u = x + y invertible. Hence xu-' + yu-' = 1, violating the condition.)
Idernpotents and the Jacobson Radical The Jacobson radical is particularly apt for dealing with idempotent-lifting since we now see that condition (i) of definition 1.1.23 is that every idempotent-lifting ideal is contained in the Jacobson radical. This connection will be examined closely in $2.7; here we collect a few basic facts. Remark 2.5.23: Jac(R) contains no nontrivial idempotent e # 0 by remark 1.1.24. It follows that a local ringR has no nontrivial idempotents. Proposition 2.5.14:
If e E R is idernpotent then Jac(eRe) = eJac(R)e.
froof: Write J = Jac(R). ( 2 ) If x E eJe and 1 =y( 1 -x) then e =ey(l -x)e= eye - eyxe = eye(e - x) since xe = x = ex, proving eye is the left quasiinverse of x in eRe, so eJe is left quasi-invertible in eRe. (E) Proposition 1.1.15 implies Jac(eRe)eM = 0 for all simple M in RQ.E.D. Aod, so Jac(eRe) = Jac(eRe)e E J. Corollary 2.5.15:
Jac(M,(R)) = M,(Jac(R)).
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Proof: This follows from corollary 1.1.18, but we now have a more elementary proof. Jac(M,,(R)) = M,,(J) for some J 4 R; identifying R with Q.E.D. eM,(R)e for e = el yields J = eJac(M,(R))e = Jac(R). This result leads us to the question of Jac(M,(R)), cf., exercise 1.1.5. The solution is given in exercises 2.8.29ff.
Weak Nullstellensatz and the Jacobson Radical One of the traditional aims of ring theory is to extract the essence of some well-known theorem and then find a straightforward generalization to noncommutative rings. One favorite such theorem has been Hilbert’s renowned Nullstellensatz, which is the statement, “Suppose F is an algebraically closed field, and consider polynomials f,f l , .. .,f, in the commutative polynomial ring R = F [ l l , . ..,An]. If f vanishes at all the common zeroes of f i , . ..,f,,then f” E Rfi for some m.”
In other words, for A =
RJ the conclusion of the Nullstellensatz is that
f + A is nilpotent in RIA. As in Kaplansky [70B, theorem 331, the hypotheses show f is in every maximal ideal of R containing A, so f + A E Jac(R/A).
Thus the key structural fact needed to complete the proof is that Jac(R/A) is nil. In other words, we want to show the Jacobson radical of every homomorphic image of R is nil. In general, one might say any ringR having this property satisfies the “weak Nullstellensatz.” Our next results give broad instances of when the Jacobson radical is nil, thereby providing the “weak Nullstellensatz” for left Artinian rings and for arbitrary algebras over suitably large fields. We shall return to the Nullstellensatz at definition 2.12.26ff. In case R is left Artinian the radical predates Jacobson, and we have the following result. Theorem 2.5.16:
Jac(R) is nilpotent for any left Artinian ring R.
Proofi Let J = Jac(R). Then J 2 J 2 2 J 3 2 ... so J‘ = J‘” for some t. Let N = J’;then N = N 2 . We claim N = 0. Otherwise N contains a nonzero left ideal L minimal with respect to L = N L . Then 0 # N u E L for some a in L , and N u = N 2 a = N ( N a ) implies L = N u (by assumption on L). Hence a = ra for some r in N. But then ( 1 - r)a = 0, contrary to r quasi-invertible. Thus 0 = N = J’ as claimed. Q.E.D.
Note that Jac(T) n R need not be quasi-invertible when R is a subring of
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T ; for example let T be a localization of Z,and R = Z. Nevertheless, there is a useful generalization of theorem 2.5.16:
Proposition 2.5.17: Suppose R is a subring of T. (i) If every element of R which is invertible in T is already invertible in R, then R n Jac(T) E Jac(R). (ii) If R is leji Artinian then R n Jac(T) G Jac(R) is nilpotent. Proofi (i) If r E R n Jac(T) then (1 - r)-l E T so (1 - I)-' E R, proving R n Jac(T )is quasi-invertible in R. (ii) If r E R is invertible in T then Ann, r = 0, implying r is invertible in R (by proposition 2.3.15). Now (i) implies R n Jac(T) c Jac(R) which is nilpotent by the theorem. Q.E.D.
Corollary 2.5.18: Suppose R is an algebra over a field F. Then every r in Jac(R) is nilpotent or transcendental. Proofi If r is not transcendental then F [ r ] is Artinian and r E F [ r ] nJac(R), a nil ideal by proposition 2.5.17(ii). Q.E.D.
Corollary 2.5.19: If R is an algebraic algebra over a jield then Jac(R) is nil.
Our further study of the radical of an algebra R over a field F requires the notion of the resolvent set of r in R, defined as { a E F:(r - a ) is invertible} (identifying F with a subfield of R). Write a(r) for the complement (in F ) of the resolvent set. Lemma 2.5.20: If r E R is algebraic (over a field F ) with minimal polynomial p then a(r) = { a E F:p(a) = O}. Proofi F[r] ss F [ A ] / ( p ) for some polynomial p = p ( l ) . We note a cyclical set of implications which implies all assertions are equivalent for arbitrary q(1) in F [ 1 ] : q(r) is invertible in R =. q(r) is not a zero divisor in F [ r ] =.p(1), q(1) are relatively prime => 1 = g(A)p(A) + h ( l ) q ( l )for suitable g ( l ) , h ( l ) in F [ l ] * 1 = h(r)q(r)+ q(r) is invertible in F [ r ] . Now taking q(A) = 1- a (which is irreducible) we see q(r) E a(r) iff 1 - a and p are not relatively prime, iff (1- a) I p , iff p(a) = 0. Q.E.D.
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Proposition 2.5.21: Suppose [F[r]:F ] 2 t. Then for any distinct a’,. ..,a, in the resolvent set of r we haue {(r - al)-’, .. .,(r - a,)-’} is F-independent.
Proofi
Suppose
If= P,(r - a,)- ’ = 0 for Piin F. Defining qi = n ( 2 - aj) j t i
and multiplying through by f l i = l ( r- a j ) yields &(r) = 0. On the other hand, deg(xfliqi)I t - 1, so, by hypothesis, xfliqi= 0. Thus for each u we get o = Cpiqi(a,,)= fiuq,(au)= B,,flj+,,(a,, - aJ, implying each B,, = 0 Q.E.D. for 1 I u I t.
Theorem 2.5.22: Suppose F is a field, R cardinal numbers). Then Jac(R) is nil.
E F-dt’,
and IF1 - 1 > [R:F] (as
Proofi If r E Jac(R) then ( 1 - ar)-’ exists for all a in F, implying r - a = -a(l - a-lr) is invertible for all 0 # a E F. If r were not algebraic then every finite subset of S = {(r - a)-1:a E F} would be F-independent by proposition 2.5.21, implying S is F-independent, contrary to hypothesis. Hence r is Q.E.D. algebraic and thus nilpotent by Corollary 2.5.18. Remark 2.5.22: It is very easy to construct a commutative algebra R of countable dimension over an arbitrary field F such that Jac(R) is not nilpotent; let C = F[Al,A2,.. .] and consider C/A where A = CA:.
xnsN
The Structure Theoretical Approach to Rings Let us outline a sequence of steps which one often follows in proving a theorem about a class of rings. The philosophy is to prove the theorem for a ring R in increasing generality, as in the following cases:
1. R is a division ring. 2. R is simple Artinian (matrices over a division ring). 3. R is primitive (use the density theorem to reduce to 2). 4. R is semiprimitive (reduce to 3 using subdirect products). 5. R has no nil ideals # 0. 6. R is semiprime. 7. General case. We shall encounter several instances of this method. Perhaps the most straightforward is the class of “commutativity theorems,” which enable one to conclude that rings are commutative under rather weak hypotheses.
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Although the list of people working in this area in the 1950s and 1960s reads like a "Who's Who in Algebra," interest has waned considerably since 1970, and, aside from a few technical questions, this area is largely of historical interest. Nevertheless, it is instructive in the use of structure theory and is included in a series of exercises. One of the most successful applications of the structure theory has been in the theory of polynomial identity rings, to be described in Chapter 6. When performing the reductions listed above, one can proceed immediately from 4 to 5 (or 6) when the Jacobson radical is nil (or nilpotent); several instances of this were just seen. Even when the Jacobson radical is not nil, there is a method of passing from 4 to 5, due to the following important theorem of Amitsur:
Theorem 2.5.23 (Amitsur): If R has no nonzero nil ideals then R[1] is semi primitive. Often we can transfer the conclusion of an assertion from R[A] to R, and in this way pass from 4 to 5. We now give a short direct proof of Amitsur's theorem; a second proof is given below in the context of graded rings.
ProoJ- Otherwise, let J = {nonzero elements of Jac(R[1]) having smallest degree}, and Jo = {leading coefficients of the element of J}. Clearly Jo u {0} is an ideal of R, which we claim is nil. Indeed suppose p E J. Then llp E Jac(R[A]), so there exists q in R[ll] such that (1 - 1p)q = 1. Then q = llpq + 1, which is the case m = 1 of the formula
which we verify now by induction on m. Indeed, assume (1) holds for m - 1. Then
= llmpmq
+
m-1
#?pi, i=O
as desired. Write p = C : = o r i l i . If x , r k x 2 = 0 for suitable x i in R then d e g ( x , p x , ) < k ; so, by definition of J , we have x , p x , = 0. We shall use this observation repeatedly. For example, write q = c : = o r i l l i for suitable ri in R and take m > t ; matching leading coefficients in (1) yields O = r r r : = r r - ' r k r i , so r,"-'pr;=O.
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Hence r?-’rir; = 0 for all i, so r?-’prir; = 0, implying rF-’p2r; = 0. Continuing this argument shows pmr;= 0, SO 0 = Impmr:Atand (1) becomes m- 1
C (2 + i1 2pi. i=O =O
f-1
q = Impm
As above, we now obtain pmr;- = 0; continuing in this way, finally yields 0 = rrrb = r? (since q = llpq + 1 implies r6 = 1). Thus we proved the leading coefficient of each element of Jo is nil, as desired. But now Jo u (0) s Q.E.D. Nil(R) = 0, proving J is empty, i.e., Jac(R[I]) = 0.
“Nakayama ’s Lemma” One of the basic tools in studying a non-semiprimitive ringR is the result Jac(R)M # M for every f.g. R-module M # 0. This is known as “Nakayama’s lemma,” although it is generally agreed that a better attribution would be Azumaya-Krull-Jacobson. Actually any of the following assertions shall be called “Nakayama’s lemma” in the sequel. Proposition 2.5.24: “Nakayama’s lemma.” Suppose 0 # M J = Jac(R).
E R-Fiwod
and
(i) M # J M . (ii) If N IM and M = N + J M then N = M. (iii) lf M J J M is spanned by x1 + J M , . ..,xk + J M then x l , . . .,xk span M. Proofi
c:=l
(i) Write M = Rxi with t minimal possible. If M = J M then x, = x a i x i for ai in J, so (1 - a,)x, = aixi; but 1 - a, is invertible so x, E Rxi, contradiction. (ii) M J N = J ( M / N )so M / N = 0 by (i), implying N = M. Q.E.D. (iii) Let N = 1:=1 Rxi. Then M = N + J M so N = M by (ii).
1:::
c:;:
There are several improvements of this result. First we record one due to Schelter, also, cf., exercise 12.
Proposition 2.5.25: r f M E R-Fimud and N < M then there is some primitive P Q R with N + P M # M. Proofi By proposition 0.2.15 there is some maximal submodule M‘ containing N. Then M J M ’ is simple so P = Ann, M J M ‘ is a primitive ideal Q.E.D. with PM c M‘; hence N + PM E M‘ # M.
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The Radical of a Module Definition 2.5.26: If M E R - A o d define Rad(M) = n{maximal submodules of M}, where we say Rad(M) = M if M has no maximal (proper) submodules.
Viewing R in R-.Mod we have Rad(R) = Jac(R) by proposition 2.5.2. Thus the radical of a module generalizes the Jacobson radical and is of considerable use in the technical study of f.g. modules, partly because of the following results. (However, our uses of Rad(M) will be confined to the exercises, cf., exercises 2.8.8ff.) Remark 2.5.27: Jac(R)M E Rad(M). (Indeed, if N is a maximal submodule then Jac(R) c Ann(M/N) so Jac(R)M E N, implying Jac(R)M c {maximal submodules}).
n
This improves Nakayama’s lemma because proposition 0.2.15 shows Rad(M) < M for every f.g. module M.
F Supplement: Finer Results Concerning the Jacobson Radical In this supplement we collect a wide range of lovely theorems about Jac(R), including Wedderburn’s principal theorem and several theorems of Amitsur. We start by comparing Jac(R) with Jac(T) when R is a subring of T. Our object is to compare Jacobson radicals of tensor extensions, but many results can be framed in the following general context. Definition 2.5.28: Suppose T E R - A d - R . An element x of T is R-normalizing if xR = Rx. (This parallels the group-theoretic notion of the normalizer of a subgroup.) T is R-normalizing if T is spanned (as R-module) by a set of normalizing elements; if this set is finite and R c T are rings we say T is a Jinite normalizing extension of R. (See exercises 13ff and 2.12.1Off.)
The obvious example of a normalizing extension is a centralizing extension; also the skew polynomial ring R[A;cr] is R-normalizing iff cr is onto. The next result is basic in studying normalizing extensions. Proposition 2.5.29: Suppose T is a Jinite normalizing extension of R spanned by R-normalizing elements a , . . a,,. If M E T-Mod is simple then M is a direct product of I n simple modules in R-Aod. ¶.
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,
Raix so M is an f.g. Proof: For any x # 0 in M we have M = Tx = R-module and thus has a maximal R-submodule by proposition 0.2.15. Let Ni = { x E M : aix E N for 1 I i I n}. Then ai(rx)E Raix G N for each x in N,, implying N,E R - A d . We claim each M / N i is simple in R-JIuLL. Indeed, if Ni < M' I M then aiM' is an R-submodule of M not contained in N so N + aiM' = M . For any element x in M we then have aix = y + a& for suitable y in N and x' in M', so y = ai(x - x'), implying x - x' E Ni and x E x' + Ni c M'. Therefore, M = M', proving the claim. N,= {x E M : a i x E N for each i } = {x E M : On the other hand, Tx c N } , a T-submodule of M which is thus 0. Hence M is a direct prodQ.E.D. uct of the simple modules M / N i by proposition 2.2.5.
n;=,
This result is originally due to Formanek-Jategaonkar [74], cf., exercise 15. Corollary 2.5.30: Suppose T is a Jinite normalizing extension of R. Then Jac(R) G Jac(T). More precisely, if P is a primitive ideal of T then P n R is a jinite intersection of primitive ideals of R ; if, furthermore, P n R E Spec(R) then P n R is a primitive ideal of R. (Aside: See also corollary 3.4.14)
Proof: Suppose P is a primitive ideal of T. Take a simple T-module M with P = Ann, M ; writing M = Ni for Ni simple in R-Aud and putting pi = Ann, 4, we have P, n* . * nP, = Ann, M = P nR. This proves the second assertion. If P n R is prime then P, * * P, c P n R implies some pi = P n R so P n R is primitive. The first assertion also follows, viewing Jac(R) as {primitive ideals}. Q.E.D.
fly=,
n
This raises interest in the reverse inequality: When is R n Jac(T) c Jac(R)? A positive result was given in proposition 2.5.17, but the following example should prevent over-expectations. Example 2.5.32: Fix a prime p and let S = { m E N : p # m } , writing the elements of S in ascending order as m1 = 1, m 2 , m 3 , etc. Let S ( i ) be the submonoid of N generated by ml, ...,mi (i.e., S ( i ) is the set of products of powers of m , , . . .,mi). Thus Rsci,= R,, a local ring, but each Rs(i)is semiprimitive (since each prime not dividing rn, * . .mi generates a maximal ideal and the intersection of these is 0).
u
Nevertheless, many positive results can be obtained by applying proposition 2.5.17(i).
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Lemma 2.5.32: The hypothesis of proposition 2.5.17(i) holds i f R is a summand of T either in R - A o d or in Mud-R. (Thus R n Jac(T) c Jac(R) in either case.) Proofi By symmetry we may assume R is summand of T in Aud-R. It suffices to show R n Tr = Rr, for then r invertible in T would imply Rr = R so r - l E R. We shall verify a slightly stronger result. Q.E.D.
Sublemma 2.5.32': If T is any right R-module of which R is a summand then R n T L = L for any left ideal L of R. Proof of sublemma: Write T = R 0 M. Clearly R n T L 2 L. Conversely, if r = x a i b i E R, where a, E T and b, E L, then writing a, = ri x i for ri in R and xi in M yields r = C r i b i + x x i b i ; matching components in R shows r = CribiE L. Q.E.D.
+
Actually sublemma 2.5.32' provides an easily verified criterion for the lattice of left ideals L ( R ) to be isomorphic to a sublattice of L(T), in case T is a ring. This will be one of the most-quoted results in later chapters. Before applying these results, we should point out that Heinecke-Robson [84] have proved the following general result: If T is a finite normalizing extension of R and R G W G T are rings then Jac(R) = R n Jac(W). Also see exercise 13. Proposition 2.5.33: If T is a Jinite normalizing extension of R which is free in R - A u d with a base containing 1, then Jac(R) = R n Jac(T). Proof: Corollary 2.5.30 and lemma 2.5.32.
Proposition 2.5.34:
Suppose R =
u
Q.E.D.
R,.
(i) If every element of R , invertible in R is already invertible in Ri then Jac(R) c Jac(R,); in particular, this holds i f R is free in Ri-Aud for each i. (ii) The reverse inclusion holds i f R is a Jinite normalizing extension of each R , .
u
Proof: (i) Any r in Jac(R) lies in some R,, so r E Jac(Ri) by proposition Q.E.D. 2.5.17(i). (ii) By corollary 2.5.30.
Theorem 2.5.35: If R E C-dt'' and H is an integral commutative C-algebra, then Jac(R 0 1) = (R 0 1) n Jac(R 0 H).
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Proofi Recall H is integral iff C [ h ] is an f.g. C-module for each h in H. By proposition 2.5.34 we may assume H is f.g. over C, so R 6 H is a finite centralizing extension of R 6 1 and we are done by proposition 2.5.33. Q.E.D.
One would like the stronger result that Jac(R 6 H ) = Jac(R) 6 H , but this has been seen to be false in example 1.7.25. There H was an inseparable field extension, and, in fact, the following useful result holds:
Theorem 2.5.36: Suppose R is an algebra over a jield F and H 2 F i s an algebraic j e l d extension. Then Jac(R) 6 H E Jac(R @I H ) , with equality if H is separable over F. (Tensors are over F.) Proof: Let { H i :i E I } be the finite extensions of F contained in H. Let J = Jac(R). If x = rj 6hj E J 6 H where each rj E J then some Hi contains h,, . . .,h, so x E J 6Hi E Jac(R 6H i ) and x is quasi-invertible, proving J 6 H E Jac(R 6H ) , which is the first assertion. Now assume H is separable over F, and let R = R/J. Since (R 0 H ) / (J 6 H ) x R6 H (by proposition 1.7.30) it suffices to prove Jac(R6 H ) = 0, i.e., we may assume J = 0. Moreover, any base of H over Hi is a base of R QFH over R & H i ; by proposition 2.5.34 we have Jac(R 6 H ) s Jac(R 6 Hi), so it suffices to prove each Jac(R 6 Hi) = 0, i.e., we may assume [ H : F ] < co. Letting K be the normal closure of H , we have just proved Jac(R @I H ) BHK G Jac(R @I H OHK) = Jac(R @ K), so it suffices to prove Jac(R @I K) = 0. K is Galois over F; letting C be the Galois group, we let G' = { 1 6cr: cr E C}, a group of automorphisms of R 6 K whose fixed subring is R. Suppose al,. ..,an is a base for K over F and z = ri 6 a, E Jac(R@K). For any given aj and cr in G, we have r i @ cr(uiaj)= a(z(1 6uj)) E Jac(R 6 K) by proposition 2.5.6; summing over all cr in G and letting ail be the trace TK,Fa,aj, we have aijriE R n Jac(R 6 K) = Jac(R) = 0 (using theorem 2.5.35). Cramer's rule yields dri = 0 for each i, where d = det(ai,) # 0. Thus each ri = 0, implying z = 0. This proves Q.E.D. Jac(R 6 K) = 0.
,
u
XI= XI=,
xi
Wedderburn's Principal Theorem We are ready for a celebrated theorem of Wedderburn, which explicitly describes finite dimensional algebras. Recall a field F is perfect if each algebraic extension of F is separable, e.g., if F is finite or char(F) = 0 or F is algebraically closed.
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$2.5 The Jacobson Radical
Theorem 2.5.37: Suppose R is a jinite dimensional algebra over a perfect jield F, and N = Jac(R). Then R = S 0 N (as a vector space over F ) where S is a subalgebra of R isomorphic to RIN. Proofi Write R for RIN. We shall use R repeatedly to analyze R.
n:=,
Case 1. N2 = 0 and R is split, i.e., R x M J F ) for suitable t and nu in N. Note N is idempotent-lifting by corollary 1.1.28. Let z, = (0,...,0,1,0,..., 0) in R, where 1 appears in the u position. Then z 1,..., z, are orthogonal idempotents of R which can be lifted to orthogonal idempotents e l , . . . , e l of R, and I e , = 1 (since 1 - C e , is an idempotent of N and thus 0.) Let R, = e,Re,. Then E,, = z,Rz, x M J F ) , so proposition 1.1.25 shows R, also has a set of matric units {e!;):1 I i , j Inu}. Let S, =.xi,jFety’x R,. Then S = X S , , is a subalgebra of R and S x S, x nRU x R. Since R = S/(N n S) we see N n S = 0; also [S + N: F] = [S:F] + [N:F] = [ R : F ] + [N:F] = [R:F], proving S + N = R.
fl
Case I!. N 2 = 0 (no assumption on E ) . This is the only place we shall use the fact F is perfect. Let m = [R:F], and let F‘ be the algebraic closure of F, which is separable by hypothesis. Then [R OFF’:F‘] = m and Jac(R 0 F’) = Jac(R) 0 F‘ = 0, so F’ is a splitting field for R, and R has a splitting field K which is f.g. over F, by proposition 2.3.21; letting R, = R OFK and N, = Jac(R,) = N 0 K, we have R, x S, 0 N, by case I, where S, x R,/N, is a subalgebra of R,. Let V be an F-subspace of R complementary to N, such that 1 E I/. Letting t = [V: F], we take a base { x , , .. .,x , } of I/ over F with x , = 1. Note m - t = [N:F] = [N,:K], so [S,:K] = t. Let { x , + , ,..., x,} be a base of N over F; then { x , , ...,x , } is a base of R over F and thus of R, over K (identifying R with R 0 1 c R,). Thus every element of S, has the .form y + I:=, aixi where y E N, and the a, E K; it follows readily (by counting dimensions) that S1 has a base { x u y,: 1 Iu It } where each y , E N, . Picking a base {a,, a 2 , .. .} of K over F with a, = 1, we define a balanced map R x K + R sending ( r , x a i a i ) to a,r; we thereby get an R - R bimodule map o:R 0 K + R satisfying o(r 0 a,ai) = alr. Let S = as,. Clearly S is spanned by the o ( x , + y,) = xu + o y , for 1 Iu It , which are independent because the xu are independent. Hence [S:F] = t. Moreover, using N = 0 we have
+
(xu + OY,)(X,
+ fly,) = xux, + (aY,)xu + X I P Y U = 4 ( x , + Y , ) ( X , + Y u ) ) E 6 = s
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for each u, u, implying S is a subalgebra of R. Clearly f x R so S n N = 0 and S 0 N = R. General case. Induction on [R:F]. By case I1 we may assume NZ # 0. Now write R for RIN’. By induction R = + lif for a suitable semisimple subalgebra x R / N x RIN. But the preimage S of also is a subalgebra and x S/(NZn S); since N is nilpotent we see N Zn S = Jac(S). By induction S = S , 0 (Nz n S) for some subalgebra S , x x RIN. Now
s
s
s
s
s
S1
+N =S,
+(N2nS)+ N = S + N = R
and N n S, = (N2n S) n S , = 0, proving R x S , 0 N as desired.
Q.E.D.
Definition 2.5.38: A finite dimensional algebra R over a field F is separable if R QFK is semisimple Artinian for every field extension K of F.
Wedderburn’s principal theorem can be obtained more generally for arbitrary separable algebras over a field, with virtually identical proof, cf., exercise 11; this is a true generalization by exercise 8. The main new ingredient needed here is the existence of separable splitting fields, proved by Koethe, cf., exercise 10. A more modern treatment of separable algebras including Wedderburn’s principal theorem is given in 55.3.
F Supplement: Amitsur ’s Theorem and Graded Rings We return now to Amitsur’s theorem (2.5.23) on when the Jacobson radical of R[I] is 0. Amitsur also proved a similar result for R(I), and, following a lovely idea of Bergman [73], we unify these results by means of a general theorem on graded rings.
Lemma 2.5.39: Suppose R is a (Z/nZ)-graded ring, and write R, for the homogeneous component corresponding to u for 0 Iu In- 1. If r r,E Jac(R) for ri E R, then each nr, E Jac(R).
=c:Ii
Proofi Let [ be a primitive n-th root of 1 in @. Then ZC-3 is a free Z-module (since the cyclotomic polynomial is monic) and thus putting R’ = R B nZ [ [ ] we have R’ is an f.g. free R-module by proposition 1.7.15. Let J = Jac(R’). Proposition 2.5.33 yields Jac(R) = R n J, so r E J . R‘ itself is Z/nZ-graded by putting R: = R, Q Z[r], and we can define an automorphism cr of R’ by putting a x a, = [‘a, (for I u,ER: arbitrary). Then aJ c J by proposition 2.5.6. Also, [ j k = z(ck)j = 0 whenever nfk, so
c;=,
$2.5 The Jacobson Radical
Hence nr,
EJ
195
n R = Jac(R), as desired.
Theorem 2.5.40: ideal of R.
Q.E.D.
(Bergman) If R is a Z-graded ring then Jac(R) is a graded
c
Proofi Let J = Jac(R). Given r in J write r = r, where the r, are the homogeneous components; then there is m such that r,=O for IuI > m. Take n 2 2m. Since the Z-grading induces a Z/nZ-grading (cf., remark 1.9.8),we have nr,EJ for each u. But using (n 1) instead of n we have each (n + l)r, E J,so r, E J , as desired. Q.E.D.
+
Lemma 2.5.41: nilpotent.
Suppose r E R and r1 is left quasi-invertible in R[1]. Then r is
,
Proofi Suppose (1 =:aili)(1 - r1) = 1. Matching powers of 1 shows a, = 1 and ai=ai- l r for each i 2 1, so inductively ai=ri. Thus rl+l =O. Q.E.D. Reproof of Amitsur's Theorem (2.5.23): We shall show for any ring R that Jac(R[I]) = N[1] for a suitable nil ideal N of R. Let J = Jac(R[1]) and N = J n R . Clearly "11 G J . Grade R[I] according to degree in 1. By theorem 2.5.40 if ril' E J then each riAi E J. But there is an automorphism D of R[1] given by a1 = 1 + 1; thus r i ( l + 1)' = a(ril')E oJ = J, so the constant term ri E J n R = N for each i. This proves J = N[1]. Finally, if r E N Q.E.D. then rI E J implying r is nilpotent by lemma 2.5.41.
c:=,
Theorem 2.5.42: (Another theorem of Amitsur) Suppose R is an algebra over a field F, and let F' be the field of rational fractions F(1). Then J a c ( R 6 F ' ) = N 0 F' for some nil ideal N of R. A, where A, = I"F(1") Proofi We claim F' can be Z/nZ-graded as To see this, first note A,A, = A,+, (subscripts modulo n). Also A o , . . .,A,are independent: I f l u g , = 0 with g,E F(1"), then clearing denominators we may assume each g , E F[1"] and thus has degree 0 (modulo n); the Pg, have different degrees for each u and cannot cancel, implying each lug, must already be 0, so each g , = 0, as desired. It remains to prove x : i A A U = F'. Clearly, F[1] cc::; A,, so we must show f E A, for f in F[1], and we may assume f is monic. Working in the splitting field K of the polynomial
c:::
c:::
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A" - 1 over F, we let A" - 1 =(A - 5') (A - c,,) and h = f ( A ) f ( h A ) - * * f(c,,A). It is easy to see h E F[A"], either by arguing combinatorically or by using Galois theory; details are left to the reader. But f divides h in KCA], so viewing F' c K(A) canonically we have f - ' h E K[A] n F' = F [ A ] c X A , . But h-' E F(A") = A,, so f-' = (f-'h)h-' E ( c A , ) A o = X A , , proving the claim. We are ready to apply the preceding theory. Write J = Jac(R @ F') and N = R nJ, viewing R as R @ 1 c R @ F. Clearly, N @ F' c J; to get the ri @ A?-' for suitreverse inclusion suppose x E J. We can write x = able t E N and f E F[A]. It suffices to show x ( l 6 f )E N 6F' since x = x(l@f)(l@f)-', so we may assume x = X : = , r i @ A i . For any n > t we have R 6 F' graded by ZfnZ with homogeneous components R @ nYF(A"), so lemma 2.5.39 implies nr,E J nR = N for each u. But using (n + 1) instead of n we also have (n + l)r, E N, so each ru E N, as desired. To see N is nil view F(A) in the Laurent series ring R((A)).For any r in N we have ( l - r A ) - ' = X , " = , r i A i ~ R @ F ' ,implying the powers of r generate a finite dimensional F-subspace of R, i.e., r is algebraic over F, and is nilpotent by corollary 2.5.18. Q.E.D
Xi=,
Corollary 2.5.43: Suppose R is an algebra over a j e l d F and F' = F(Ai:i E I) for any injnite set of commuting indeterminates. Then Jac(R 6F') is nil.
Proof: For [ I )finite this is an easy induction; for arbitrary I apply proposition 2.5.34(i). Q.E.D. See exercises 2.6.11 ff for results on graded rings.
F Supplement: The Jacobson Radical of a Rng Despite a determined attempt to impose 1 in ring theory, there are times when it is more convenient to deal with rings without 1. This is true in particular for certain "elementary" aspects of the theory. In this brief digression we shall 9 51.5). R, sketch Jacobson's original description of his radical for 9 ~ (cf., throughout will denote a ring without 1. We need a substitute for 1.
Definition 2.5.44: A left ideal L of R, is modular if there is e in R, such that r - re E L for all r in R,; e is called a left unit (modulo L). Remark 2.5.45: A modular left ideal L is proper iff e $ L (for, otherwise, r E re L E L for all r in R,, proving L= R,,). Consequently, Zorn's lemma shows any modular left ideal L # R, is contained in a maximal modular left ideal
+
52.5 The Jacobson Radical
197
(ZR,), with the same left unit. Thus one can define Jac(R,) = (){maximal modular left ideals of R,}. We can characterize Jac(R,) in terms of quasi-invertible elements, where now we say r E R , is left quasi-invertible if r + r’ - r’r = 0 for some r’ in R,. (If 1ER, this says (1 - r’)( 1- r)= 1, which is the definition we gave earlier.) Proposition 2.5.46: Jac(R,) is a quasi-invertible left ideal which contains every quasi-invertible left ideal.
Proofi Let J = Jac(R,). If a d is not quasi-invertible then L= {r-ra: rER,} is a modular left ideal, for a is a left unit modulo L; but L is contained in a maximal modular left ideal not containing a, contradiction. Thus J is quasiinvertible. Now let J’ be any quasi-invertible left ideal. We claim J’ is contained in every maximal modular left ideal L. Indeed, otherwise e E J ’ + L, where e is a left unit modulo L. Writing e = r + a for r in J’ and a in L, we see a = e - r E L. But r - re E L by hypothesis, so e - re E L. Take r‘ such that r + r’ - r’r = 0. Then e
= e - (r
+ r’ - r’r)e = (e - re) - r’(e - re) E L
contrary to hypothesis on e. Thus J’
c (){maximal modular left ideals} = J. Q.E.D.
It is now a straightforward matter to carry over the other characterizations of Jac(R) to B H ~ In , particular, Jac(R,) Q R, and is left-right symmetric. The reason behind such a project is to be able to focus on the radical itself as a rng; exercise 21ff presents Sasiada’s example of a simple radical rng. Having seen that modular ideals are useful, we would like to know that they are abundant enough. Proposition 2.5.47: If L , is a maximal modular left ideal and L, is any modular left ideal then L , n L , is modular.
Proof: This is obvious unless L , $Z L , ; thus L , + L , = R,. Let ei be a left unit modulo L, for i = 1,2. Writing e, = a, + b, for a, E L , and b, E L,, we claim e = a, + b , is a left unit modulo L , n L , . Indeed, for any r in R , we have r - re = r - r(a2 + b,) = r - re, and, likewise, r - re E L , , as desired.
+ r(a, - a z )E L ,
Q.E.D.
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Corollary 2.5.48: Any Jinite intersection of maximal modular left ideals is maximal. Proof: Induction applied to proposition 2.5.47.
Q.E.D.
Galois Theory of Rings With these modest beginnings, one can obtain some rather deep results concerning group actions on rings. Given a rng R, and a finite group G of automorphisms on R,, we write RE for { r E R,: or = r for all o in G}.Of course, R, may have 1, but we find it convenient not to require this. Write t(r) for or. We follow Puczylowski [84].
zaoG
Lemma 2.5.49: lf L is a maximal modular lefi ideal of R, then there is e in Ro for which IGJr- rt(e)E L for all r in R,.
naEG
Proof: Let e be a left unit modulo L' = oL, which is modular by corollary 2.5.48. Then r - re E L', so applying o and replacing r by o-lr shows r - roe E L'; summing over o yields lGlr - rt(e)E L'. Q.E.D.
Lemma 2.5.50: lf t(R,) c J = Jac(R,) then IGIR,
c J.
Proof: It suffices to prove IGIR, E L for every maximal modular left ideal L. Take e as in lemma 2.5.49. For all r in R, we have IGlr-rt(e)EL; but t(e)E J c L, so IG(r E L as desired. Q.E.D.
Theorem 2.5.51: (Martindale) IGIJac(RG)c Jac(R) whenever G is a Jinite group of automorphisms acting on a ring R. Proof: Let J=Jac(RG). Taking R,=RJ we have t(R,)=t(R)Jc R G J = J , a quasi-invertible left ideal of R,. Hence IGIR, is quasi-invertible by lemma 2.5.50 and is obviously a left ideal of R, implying [GIROc Jac(R) as desired. Q.E.D. A related theorem is the theorem of Bergman-Isaacs, that if G is a finite group of automorphisms on a rng R, and RE = 0 then (GIROhas a nontrivial nilpotent ideal. This follows at once from the following result, which also is amenable to Puczylowski's techniques:
Theorem 2.5.52: (Bergman-lsaacs) If G is a Jinite group of automorphisms acting on a rng R then there is n = n(G) such that (IGIAnn t(R))" = 0.
82.6 Nilradicals
199
Proofi Let A = [GIAnn t(R).We prove the result by means of a sequence of embeddings. S t e p 1 . A G Jac(R) by lemma 2.5.50, since t(Ann t(R)) = t(Ann t ( R ) R ) = (Ann t ( R ) ) t ( R )= 0. S t e p 2. A is nil by Amitsur's theorem (2.5.23) since A Jac(R[I]) by step 1.
c IGlAnnt(R[I])
G
Step 3. A is nilpotent. Indeed let X be a set of noncommuting indeterminates of the same cardinality as A and write X = {Xa:uE A}. Then A { { X } } c IGlAnn t ( R { { X } } )is nil by step 2. But in the formal power series ringR{{X}}, we see the element z , , u X a is nilpotent, i.e., (~,,,uX,)" = 0 for some n, and checking coefficients we see A" = 0. S t e p 4. A" = 0 where n depends only on G (not on R). Indeed, otherwise, for each m > 0 there would be a ring R, with G acting on R, such that (IGIAnnt(R,))"#O. But then taking R = H m E N R ,we see A=nlGIAnnt(R,) Q.E.D. so A" # 0 for each m,contrary to step 3.
Corollary 2.5.53:
Zf R G = 0 then IG1R is nilpotent. (Indeed t ( R ) c R G = 0.)
The embedding technique of this proof is obviously quite useful and probably has much unused potential. We return to it in exercises 2.12.20ff (Passman's "primitivity machine"). Further results on fixed rings under group actions are given in exercises 2.6.13ff and exercise 6.2.1. The results of this discussion assume their strongest form when IGl is invertible in R (e.g., char@) = 0); then Martindale's theorem says Jac(RG)E Jac(R), and Bergman-Isaacs says if Rg = O then Ro is nilpotent. On the other hand, one may want characteristic-free results. Then one must add an assump tion, such as G is outer on R, which classically means that the restriction of the automorphisms in G to Z ( R ) are all distinct. Such a theory will be needed in $7.2, where some of the classical results are presented; a sweeping theory of outer automorphisms due to Kharchenko is found in Montgomery [SOB].
82.6 Nilradicals For the purposes of this book we define a radical as the intersection of a certain class of ideals; for example, the Jacobson radical is the intersection of the primitive ideals. It is often difficult to reduce to the case Jac(R) = 0, and so it is useful to have smaller radicals to refine our study of rings, especially in
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light of theorems to be proved below on prime rings. In this section we consider radicals which are nil ideals. One candidate is the radical arising from prime ideals, called the “prime radical” or “lower nilradical;” another is the “upper nilradical,” “which is easier to describe, as the “largest” nil ideal, but which presents technical difficulties discussed in 2.6.33ff. The following observation holds if we fix a class W of rings and define Rad(R) = ( ) { A 4R:R/A E W}.
Remark 2.6.0: The proof of proposition 2.5.1’ goes through for the radical “Rad.” Thus if Rad(R) 3 A a R then Rad(R/A) = Rad(R)/ A; if Rad(R/A) = O then A 2 Rad(R). Although the situation for noncommutative rings is considerably more complicated than in the commutative case (for which every finite subset is nilpotent), nil ideals do satisfy certain pleasant properties.
Lemma 2.6.1:
If A 4 R is nil and S/A is a nil subset of RIA then S is nil.
Proofi Suppose s E S. Then some s” E A and is thus nilpotent, implying s is nilpotent. Q.E.D.
Proposition 2.6.2:
Any ring has a unique maximal nil ideal.
Proofi By Zorn’s lemma there exists some maximal nil ideal N. We claim N contains every nil ideal B. Indeed (B + N ) / N is nil, so B + N is nil by lemma 2.6.1, implying B + N = N , so B E N as desired. Q.E.D. Definition 2.6.3: Nil(R) is the unique maximal nil ideal of R. Remark 2.6.4: If A 4 R then (Nil@) + A ) / A c Nil(R/A), equality holding if A is nil. In particular, Nil(R/Nil(R)) = 0. In analogy to Jac(R) we would like to display Nil(R) as a radical.
Defiition 2.6.5: A ringR is strongly prime if R is prime with no nonzero nil ideals. P 4 R is strongly prime if RIP is strongly prime. Remark 2.6.6: (To be generalized shortly) if S is a submonoid of R and P 4 R is maximal with respect to PnS= 0 then P is a prime ideal. (Indeed, if Ai 3 P for i = 1,2 then there is si in Ai n S, so s1s2E A , A 2 n S implying A,A2 $ p.1
$2.6 Nilradicals
Proposition 2.6.7:
201
NiI(R) =
{strongly prime ideals of R}.
Proof: Let B = {strongly prime ideals of R}. If P E B then Nil(R/P) = 0, implying Nil(R) E P. Conversely, we shall show n { P E S}is a nil ideal and is thus Nil(R). To see this we shall show that for any a E R which is not nilpotent, there is some P in S not containing a. Indeed, S = {a’: i 2 0 ) is a submonoid of R, and by Zorn’s lemma there is some ideal P maximal with respect to P n S = 0. We claim P E 9 Indeed P is a prime ideal by remark 2.6.6. If R I P had a nonzero nil ideal B / P then B 3 P so B n S would contain some ak;but then some power of ak would be in P, contrary to P n S = 0.This proves the claim. Q.E.D. Note: An examination of the above proof shows that Nil(R) is also the intersection of those strongly prime ideals l?, satisfying the additional property that for each ringR, = RIP, there is some non-nilpotent element r, such that every nonzero ideal of R, contains a power of r,. This condition is somewhat technical, but is useful in implementing certain proofs.
Corollary 2.6.7’: rings.
Nil(R) = 0 iff R is a subdirect product of strongly prime
Two results motivate us especially to study Nil(R): (i) NiI(R) E Jac(R) by corollary 1.1.28. (ii) If NiI(R) = 0 then Jac(R[A]) = 0 by theorem 2.5.23.
On the other hand, we have seen in @2.5that Jac(R) is nil for several classes of rings.
Reduced Rings The structure theory developed so far applies very neatly to a special class of ring. Definition 2.6.8: A ring R is reduced if r 2 # 0 for every nonzero r in R.
Remark 2.6.9: In a reduced ring R if ab = 0 then bRa =O. (Indeed, (bRa)2= 0 so bRa = 0.) Consequently, a reduced ring R is prime iff R is a domain. In fact, a reduced ring is a semidirect product of domains, as we shall infer from the next result.
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ni
Remark 2.6.10: If Pl 3 P2 3 ... are prime ideals in a ringR then pi is also prime. (Indeed, if AB E pi and B $ 4 for some j then B $! 8 for all i > j, implying A c 8). Consequently, by Zorn's lemma, every prime ideal contains a minimal prime ideal (possibly 0).
n
n
Theorem 2.6.11: Suppose R is reduced. Then RIP is a domain for every minimal prime ideal P of R.
Proof.- Let S = R - P and S' = {sl * * * s k :Es S, i k E N}, the submonoid of R generated by S. To show RIP is a domain, we want to show that S is multiplicatively closed, i.e., S = S'. To see this we shall prove OqS'. Then by Zorn's lemma there is an ideal I maximal with respect to I n S' = 0,and such I is necessarily prime by remark 2.6.6. But I E P, so by minimality of P we conclude I = P, implying s' = s, as desired. So it remains to show 0 $ S'. Assume, on the contrary, s1 = 0 for si in S, where we take such a product with k minimal. Then k 2 2, and * * a s k
skR(s1*"sk-1) = 0 by remark 2.6.9. But skRs, contains some element s; in S (since P is prime), and S ; S ~ * * * S~~ s- ,~R s , . . . s ~=0, - ~ so we have a product of shorter length, contradiction. Q.E.D.
Corollary 2.6.12: (Andrunakievic-Rjabuhin) Every reduced ring R is a subdirect product of domains. Proof.- Nil(R) = 0, which implies (){strongly prime ideals} = 0, and thus Q.E.D. (-){minimal prime ideals} = 0. Another proof of Corollary 2.6.12, taken from Klein [SO], is given in exercise 3.
Nilpotent Ideals and Nilradicals Given a subset S of a ring define S' = S and, inductively, S" = S"'S. We say S is nilpotent of index n if S" = 0 and S"- # 0. Definition 2.6.13:
Whereas in a commutative ring every finite nil subset is nilpotent, this is hardly true in general, as evidenced by the following easy example: Example 2.6.14:
Let R = M , ( F ) and S = { e l z , e z l } ,
203
82.6 Nilradicals
This definition leads us to examine how far a nil ideal is from being nilpotent, with the hope at times of closing the gap. Accordingly, we define a nilradical to be a radical contained in Nil(R) which contains every nilpotent ideal of R. Nil(R) is obviously the largest possible nilradical and is therefore called the upper nilradical. On the other hand, we can define the prime radical of R, denoted prime rad(R), to be the intersection of all prime ideals of R. Proposition 2.6.15:
Prime rad(R) is a nilradical of R.
Proof: If P a R is prime and N a R is nilpotent then some N j = 0 c P implying N E P. Thus N c prime rad(R). On the other hand, every strongly prime ideal is prime, so prime rad(R) E (-{strongly prime ideals of R} = Nil(R) by proposition 2.6.7. Q.E.D. We shall show the prime radical is the smallest possible nilradical, and is therefore also called the lower nilradical. Let me first take the opportunity to state a personal preference for the upper nilradical, which is often much easier to use and which is certainly easier to derive than the lower nilradical. Nevertheless, the lower nilradical has played an important historical role in the proofs of many theorems. Our first step in discussing the prime radical is to generalize remark 2.6.6. A subset S of R-{0} is called an m-set if sl, s2 in S imply sirs2 E S for some r in R. Obviously, the complement of any prime ideal is an m-set.
Lemma 2.6.16: ( i ) The complement of an m-set contains a prime ideal; (ii) If R has no nonzero nilpotent ideals then R is semiprime. Proof: (i) Let S be an m-set. By Zorn’s lemma there exists P a R maximal with respect to P n S = 125; we claim P is prime. Indeed if P c Ai -4 R for i = 1,2 then each Ai contains some si in S, so S has an element slrs2 E A l A 2 implying A , A , $ P. (ii) For each r # 0 in R we shall build an rn-set containing r and be done by (i). Indeed, let s1 = r and inductively given si in S pick nonzero s i + l in siRsi.(If siRsi = 0 then RsiR would be a nilpotent ideal, contrary to hypothesis. We claim that S = {s1,s2,.. .} is the desired h-set. Indeed, we claim S n siRsj # 0 for all i,j, which is immediate for i = j. Thus we may assume i < j (for a parallel argument works if j < i). By induction on ( j- i ) there is some r‘ in R with si+lr’sjE S. Writing si+ = sirisi we get si(risir’)sj= si + r’sj E S, as desired. Q.E.D.
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Theorem 2.6.17: The following are equivalent for a ring R: ( i )R is semiprime; (ii)R has no nilpotent left ideals # 0; (iii) R has no nilpotent ideals # 0; (iu) if A U R and A' = 0 then A = 0. Proofi In view of remark 2.2.10 and lemma 2.6.16(ii) it only remains to prove (iv) +(iii). But if A" = 0 for A U R then (A"-')' = 0 so A"-' = 0, and Q.E.D. continuing by induction on n we see A = 0. This characterization of semiprime rings is so fundamental we shall use it without further reference. To continue our investigation we need to make the following definition by transfinite induction: N,(R) = 0, N , ( R ) = C(ni1potent ideals of R},
Na(R) = {r E R: r
+ N'(R)
Na(R) = Ua
E
N,(R/N'(R))}
for a = fl+,
for a a limit ordinal.
For example, N,(R)/N,(R) = N,(R/N,(R)). The Na(R) form an increasing chain which, once it stops growing, remains constant for all the succeeding ordinals. Thus, putting L ( R )= Na(R),where a is any ordinal of cardinality > IRI, we have L(R) = Nat(R)for all la'l > IRI. Remark 2.6.18: By transfinite induction applied to remark 2.6.0, every nilradical contains L(R).
Theorem 2.6.19: (Levitzki) L(R) is the prime radical of R (and is thus a nilradical contained in every nilradical). Proofi Let N be the prime radical of R. Then L(R) E N by remark 2.6.18. But R/L(R) has no nonzero nilpotent ideals, by definition of L(R), so by lemma 2.6.16,O =prime rad(R/L(R)) = N/L(R), proving N =L(R). Q.E.D. It is convenient to know at what ordinal a we already have Na(R) = L(R), i.e., how many steps are needed to reach the lower nilradical. Exercise 3 gives a general construction where we need arbitrarily large a to reach L(R). On the other hand, N,(R) = L(R)for large classes of rings (as we shall see), and, in fact, one can make the following observation: Remark 2.6.20: If N,(R) is nilpotent then N,(R) = L(R). (Indeed, for any nilpotent ideal N/N,(R) of R/N,(R) we see N is necessarily nilpotent, so N c N , ( R ) and N/N,(R) = 0, proving N = N,(R).)
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205
The Nilradical of Noetherian Rings We shall see now that the nilradicals of a left Noetherian ring coincide; in fact every nil left or right ideal is nilpotent. The proof uses the following fundamental connection between nil left and right ideals. If (ra)' = 0 then (ar)'" = a(ra)'r = 0. Consequently if Ra is nil then aR is nil. Remark 2.6.22:
Lemma 2.6.22: is nilpotent.
I f R satisfies ACC(two-sided ideals) then the lower nilradical
Proof: By remark 2.6.20 we need only show N , ( R ) is nilpotent. But by hypothesis {nilpotent ideals of R} has a maximal member, which is just N,(R). Q.E.D.
Theorem 2.6.23 (Levitzki): Every nil left or right ideal of a left Noetherian ring is nilpotent. Proof: Factoring out the lower nilradical (which we just saw is nilpotent), we could assume R is semiprime, so it suffices to prove the following stronger result.
Proposition 2.6.24: the form Ann, r.
Suppose R satisfies ACC on all chains of left ideals of
(i) Every nonzero nil right ideal T contains a nonzero nilpotent right ideal; (ii) If R is semiprime then R has no nonzero nil left or right ideals. .Proof: (i) Take 0 # a E T with Ann,a maximal possible. If r E R with (ar)" # 0 then Ann((ar)") = Ann a. (Indeed Ann(ab) 2 Ann a for all b in R, but Ann a was chosen maximally.) But ar E T so there is n 2 1 with (ar)" # 0 and (ar)"' = 0. Hence ar E Ann((ar)") = Ann a so ara = 0. This proves aRa = 0, so aR is nilpotent. (ii) Otherwise R has a nonzero nil left or right ideal T ; picking a # 0 in T we have Ra nil or aR nil (respectively), so aR is nil by remark 2.6.21. By (i) Q.E.D. R has a nilpotent right ideal, contrary to R being semiprime.
This result previews the Goldie theorems. Also a similar argument given in exercise 7 shows that under these hypotheses the lower nilradical contains every nil left or right ideal.
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Bounded Index Definition 2.6.25: A nil subset N of a ring R has bounded index It if a' = 0 for all a in N. N(R) = {a E R: Ra is nil of bounded index}.
Although N(R) will not be seen to be an ideal until further in this discussion, we can see the definition is left-right symmetric, by remark 2.6.21. N(R) is important to us because the following result links it to the prime radical; the elegant proof is due to A. Klein. Proposition 2.6.26: (i) lf L is a nonzero nil lejit ideal of R of bounded index n and a E L with a"-' # 0 then (Ra"-'R)' = 0. (ii) N,(R) C N(R) E Prime rad(R); in particular, R is semiprime iff N(R) = 0.
'
froofi (i) For r E R arbitrary writer' = ran- E L. Then r'a = 0 and (r')" = 0, so 0 = (r' + a)" = ai(r')n-i= (1 P)a"-'r' where r^ E aR is nilpotent, so 1 r^ is invertible and 0 = a"-%' = a n - ' r a n - ' , proving (Ra"-'R)? = 0, as desired. (ii) If a E N,(R) then Ra is nilpotent so a E N(R), proving the first inequality. We shall conclude by proving every nil left ideal L of bounded index is in the prime radical 1. Passing to R/l we may assume R is semiprime. But then L = 0 by (i). Q.E.D.
+
I::,'
+
Proposition 2.6.26(ii) provides a useful test for semiprime rings, which we shall need later. Theorem 2.6.27: (Amitsur) Given a ringR, let 1 be an infinite set of cardinality 2 IRI, and put R' = R', writing (ri) for the element of R' whose i-component is ri E R for all i in I. Identify R as a subring of R' under the "diagonal injection" r + (ri)with each ri = r. Then N(R) = Nil(R') n R. Consequently, N(R) 4 R and there is an injection R/N(R) + R'/Nil(R'). Proofi For any a E R n Nil(R') take y = (ri)E R' such that { r i :i E l } = R. Then (ya)' = 0 for some t, implying Ra is nil of bounded index I t ; thus a E N(R) proving R n Nil(R') 5 N(R). Conversely, if a E N(R) then Ra is nil of some bounded index t. Each component of R'a is in Ra so R'a is also nil of bounded index I t, implying R'a s Prime rad(R') by proposition 2.6.26. Therefore
N(R) E R n Prime rad(R') c R n Nil(R') c N(R),
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207
proving equality holds at every stage. Now the canonical homomorphism R -+R'/Nil(R') has kernel R n Nil(R') = N(R), yielding the last assertion. Q.E.D.
Derivations and Nilradicals For later reference let us see how derivations apply to nilradicals. Proposition 2.6.28: If R is a Q-algebra and 6 is u derivation then G(Nil(R))C Nil(R) and d(Prime rad(R)) G Prime rad(R). Proof:
Suppose a E Nil(R). For arbitrary r l i , r Z iin R we have
where each hUjhas the form rliar2'. But 6 ( r l i a r z i )= r1iaGr2i + rli(6a)r2i+ (Grli)ar2iE rli(du)rzi+ Nil(R), so applying the same idea to each G"(h,, h.,,) by iterating Leibniz' formula (1.6.29) we see E m( rli(6u)rZi)"+ Nil(R) 6"(( rliurZi)")
1
for suitable m > 0.
Now 1 r l i a r z i ~ N i 1 ( R so) ( ~ r l i a r 2 i ) n =for 0 some n, implying m ( ~ r l i ( b a ) r z i ) " c Nil(R) and is thus nilpotent. Hence 1 r l i ( 6 u ) r Zisi nilpotent. Since the r l i , rZi are arbitrary we see R(Gu)R is nil, proving G(Nil(R))E Nil(R). To show Prime rad(R) is invariant, we view Prime rad as the lower radical, built by transfinite induction (cf., theorem 2.6.19). The same argument as above shows that if a = p' then 6(NB(R))G NJR), yielding the desired result. Q.E.D.
Nil Subsets Often the question arises as to when a nil subset of a ring is nilpotent. Our approach to this subject will be through the structure theory, starting with simple Artinian rings and a theorem of Jacobson. Suppose M E R-Aod and S, N are respective subsets of R, M.If S'N = S i + 'N then S'N = SjN for all j > i. (Indeed, by induction on j we have SiN = SJ- ' N ; then SjN = S S j - 'N = SS'N = S'+ ' N = S'N.) Remark 2.6.29:
Call a subset S of R weakly closed if for each s l ,s2 in S we have some v in
Z (depending on s1 and s2) such that sls2 + vszsl E S. This encompasses the
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208
cases v = 0 (in which case S is a multiplicative semigroup), v = - 1 (Lie multiplication) and v = + 1 (Jordan multiplication). We shall call sls2+ vs2sl the weak product of s, and s2 in S. Proposition 2.6.30: (Jacobson) Suppose S is a nil weakly closed subset of R = Mn(D)where D is a division ring. Then S" = 0.
Take a minimal left ideal L of R. Then [ L : D ] = n viewing L in Aud-D.We claim S; = 0 for any nilpotent subset So of R. Indeed, otherwise S;L # 0 but some S',L = 0 by assumption, so remark 2.6.29 implies S F ' L c SbL for each i I n. Thus [L:D] > [ S o L : D ] > [ S i L : D ] > so [S;L: D] = 0, yielding S;L = 0 and proving the claim. Proof:
It follows {nilpotent subsets of S} has a maximal member So which is nonempty since any singleton is a nilpotent subset. Also S; = 0. Let V = SOLc L and rn = [ V : D ] , so 0 c rn c n. Put S , = {s E S:sV c V}. We work inductively on n. Since S, acts as a nil weakly closed set of transformations in End V, z M,(D) we have by induction STV = 0. On the other hand, S, acts on the quotient space L/V (over D) by the action s(a + V ) = sa + V, and [L/V:D] = n - rn c n, so S;-"L E V (again by induction), and hence S;L = S:S;-"L E STV = 0. Since So E S,, we have So = S, by the maximality of So, i.e., sSoL $ SOL for all s in S - So. We shall obtain a contradiction by assuming S # So. Let s1 0 s2 denote the "weak product" sls2 vszsl in S described above. If there exists s in S - So such that s So E So then for every element so in So and a in L we have s(soa) = (s 0 so)a - vso(sa)E SOL for suitable v in Z, proving s E S,, a contradiction. Thus we may assume s 0 So $L So for every s in S - S o . In particular, take sl in S - So and sol in So such that s2 = s1 0 sol 4 So; inductively, given si in S - So, take soi in So such that si+ = si0 soi # S o . Then sZn+ is a sum of ordinary products of length 2n + 1 of permutations of sl, sol,sO2,. . .,so, 2n; i.e., each term contains a consecutive string of elements of So of length n and is thus 0. Hence sZn+ = 0, contrary Q.E.D. to ~2,,+1 4 So.
+
0
,
,
,
Note that some restriction was necessary on the nil set S since { e 1 2 , e 2 , }is a nil subset of M 2 ( D )with e 1 2 e 2 ,= e l , idempotent. On the other hand, if S is indeed weakly closed then (in the above notation) L 3 S L 3 S2L3 3 S'L = 0 for some t I n; we can take a base of S'-'L (over D) and extend it to a base of S'-'L and then to S'-3L,etc. until we reach a base of L with respect to which S is in strictly lower triangular form. Thus we have a generalization of Engel's theorem from Lie algebra.
42.6 Nilradicals
209
We are ready to apply structure theory. Theorem 2.6.31: (Jacobson)Suppose R has a nilpotent ideal N such that RJN is semisimple Artinian. (In particular, this holds if R is leji Artinian.) Then there exists k such that Sk = 0 for every nil, weakly closed subset S of R.
n:=,
Proof: Write N = Jac(R). Then N' = 0 for some t and RIN x M,,(Di) for suitable ni. Let n = max{n,,. ..,n,}. Writing Si for the image of S in M,,,(Di)we have Sl = 0 for each i, by proposition 2.6.30, implying S" E N and S"' = 0. Q.E.D.
Theorem 2.6.31 hints that we may be interested in finding a criterion for a ring R to be a subring of a suitable left Artinian ring T ; then any nil, weakly closed subset S of R would a fortiori be a subset of T and thus nilpotent. It is also useful to know when multiplicative subsets of a ring are nil. The following result will be generalized later for P1-rings. Proposition 2.6.32: Suppose R is a jinite dimensional algebra over a field F. If S is a multiplicatively closed subset each of whose elements is a sum of nilpotent elements then S is nilpotent. Proof: By induction on n = [R:F]. Let R , = F + FS, the F-subalgebra generated by S. If R,, c R we are done by induction, so we may assume R, = R. In particular, FS is an ideal of R, which is maximal by a dimension count. If Jac(R) # 0 then FS/Jac(R) is nilpotent in R/Jac(R) by induction, so FS is nilpotent and again we are done. Thus we may assume Jac(R) = 0, so R is semisimple Artinian; checking the simple components, we may assume R is simple. Letting K be a splitting field of R, we can replace R by R @z(R) K and thus assume R x M,,(K). But every nilpotent element has trace 0, so every element of FS has trace 0, implying FS 4 R. Since R is simple, we have Q.E.D. FS = 0, SO S = 0. Remark 2.6.32':
S" = 0 by a dimension count, where n = [R: F ] .
G Supplement: Koethe's Conjecture Having seen that the unique largest quasi-invertible ideal contains every quasi-invertible left ideal, we are led to the following analog for nil ideals: Conjecture 2.6.33: (Koethe's Conjecture) The upper nilradical contains every nil ldt ideal.
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Although this conjecture has been verified for many different classes of rings, it is believed not to hold by most ring theorists, perhaps because it has withstood the efforts of several brilliant mathematicians. There are many equivalent formulations, which we shall discuss now. First note that Koethe's conjecture holds iff LR is nil for every nil left ideal L of R. Remark 2.6.34: If L is a nil left ideal of R and r E R then Lr is also a nil left ideal (for if a E L then ra E L so (ra)'" = 0 for some m, implying (ar)'"+' = a(ra)'"r= 0).
Theorem 2.6.35:
The following assertions are equivalent:
(i) Koethe's conjecture holds. (ii) The sum of two nil left ideals is necessarily nil. (iii) Nil(M,,(R)) = M,,(Nil(R))for all rings R and for all n. (iv) Jac(R[A]) = Nil(R)[1] for all rings R, where 1 is a commuting indeterminate.
Proof: Write N = Nil(R). (i) (ii) If L,, L, are nil left ideals then L , , L, G N so L, + L, E N is nil. (ii) * (iii) Nil(M,,(R))= M,,(A) for some A 4 R. But Ae, is nil, so A is nil. Thus Nil(M,,(R)) E M,,(N).It remains to show M,,(N)is nil. We use (ii), which implies inductively that any finite sum of nil left ideals is nil. M,,(N) = ,( Neil) and each Ne,, is nil (since aijeij)k= aijeij)a!,yI). (iv) * (i) Suppose L is a nil left ideal of R. Let R1 be the ring obtained by adjoining 1 formally to L (viewed as a ring without 1). A typical element of R1is then (m,a) where m E Z and a E L, which is nilpotent iff m = 0. Hence L = Nil(R,) so by (iv) we have L[A] = Jac(R,[1]). By Lemma 2.5.3 we see L[1] remains quasi-invertible when viewed as a left ideal of R[1], implying L[1] E Jac(R[A]) = (Nil R)[A], proving L E Nil@). Q.E.D.
,
cI=
XI=
(cI=,
(c
It remains to prove (iii) * (iv). This follows at once from the following result of Krempa-Amitsur.
Proposition 2.6.36: Let R be an arbitrary ring and N = Nil(R). r f M,,(N) is nil then every polynomial of N[1] having degree I n is quasi-invertible.
XI=,
Proofi We want to prove every polynomial p = 1 + a i l i is left invertible where a, E N. Since a, E Jac(R) we have (1 + a,) invertible; replacing p by (1 + a,)-'p we may assume a, = 0.Let q = 1 b,Ai be the inverse
I;,
82.7 Semiprimary Rings and Their Generalizations
21 1
of p in the ring of formal power series R[[A]], cf. proposition 1.2.27. Then each bj = C:= bj- ,ai, where we formally put b, = 1 and by = 0 for u < 0. Let A be the n x n matrix C::: e , + l , i+ a n + l - i e i n By . induction we see (O,... ,O,l)AJ= (bj-,,+ ..,bj- b,) for each j in N. But the matrix A" has entries only in N and so is nilpotent by hypothesis. Thus A' = 0 for some t, and hence bj = 0 for all j 2 t, thereby proving q E RCA]. Thus we proved every polynomial of "A] having degree In is left quasiinvertible; right quasi-invertibility is proved analogously. Q.E.D.
IT=,
82.7 Semiprimary Rings and Their Generalizations In this section we generalize left Artinian rings by isolating some of their properties, mostly in connection with chain conditions on certain classes of left or right ideals. As we shall see, these rings have certain easily discovered properties which cast strong light on the structure of Artinian rings themselves. Our starting point is the nilpotence of the Jacobson radical of a left Artinian ring (theorem 2.5.16).
Definition 2.7.1: R is a semilocal ring if R/Jac(R) is semisimple Artinian. A semilocal ring R is semiprimary if Jac(R) is nilpotent. Hopkins and Levitzki proved independently that left Artinian rings are left Noetherian. Their result can now be seen to be an easy application of semiprimary rings: Theorem 2.7.2: primary.
A ring R is left Artinian iff R is left Noetherian and semi-
Proofi Every left Artinian ring is semiprimary, so it remains to prove the following claim: Claim. A semiprimary ring R is left Noetherian iff R is left Artinian. Proof of Claim. Put J = Jac(R) and take n such that J " = 0.
We shall show if R is left Artinian or left Noetherian then R has a composition series in R - A d and thus is both left Artinian and left Noetherian. To build the composition series we start with R > J > J 2> . .. > J" = 0; clearly, it suffices to show each factor Mi = Ji-'/Ji has a composition series (where R = Jo). Since J M , = 0 we can view Mi in R-AoLLwhere R = R/J, so Mi is
212
Basic Structure Theory
a direct sum of simple R-modules, which can also be viewed as simple R-modules. If R is left Noetherian (resp. Artinian) then so is J i - ’ , viewed as R-submodule, and hence so is Mi. Thus Mi is a Jinite direct sum of simple submodules, and consequently has finite composition length, as desired. Q.E.D. The rest of this section deals with elementary properties of semiprimary rings, although results are obtained more generally for semiperfect rings. Certain key results, however, involve homological dimension, cf., exercises 5.2.40ffand Auslander [ 5 5 ] ; the groundwork is laid in exercises 16ff.One useful fact is that left ideals are “eventually idempotent,” cf., exercise 17.
Chain Conditions on Principal Lejl Ideals Semiprimary rings can actually be characterized in terms of a chain condition on principal left ideals; to see this we first need some preliminaries about Artinian rings.
Remark 2.7.3: Suppose R is semisimple Artinian and f: aR + bR is an isomorphism of right ideals of R (viewed as right modules) with fa = b. Then a = ub for some invertible element u of R. (Indeed, aR and bR have respective complements N1,N2 in R as right R-module. By the Jordan-Holder theorem there is an isomorphism 9: N, + N , , so h = f @ 9 : R + R is an isomorphism. Let u = h - l l . Then (h1)u = hu = 1 and u(h1) = h-’hl = 1, proving u is invertible; also a = h-’b = (h-’1)b = ub.) Remark 2.7.4: Suppose R is semisimple Artinian and Ra = Rb. Then a = ub for some invertible element u of R. (First assume a is an idempotent e. Then b E Re so be = b. Left multiplication by b thus gives an onto map f:eR + bR. Moreover, kerf = 0 since if 0 = f(er) = ber = br then 0 = Rbr = Rer. Remark 2.7.3 then yields e = ub as desired. In general, R a is a summand of R so there is an idempotent e with Re = Ra = Rb. Thus e = u,a = u2b by the special case, so a = u;’u2b.) When studying chains of principal left ideals one should realize that if Ra > Rb are principal left ideals then b = a‘a for suitable a’ in R, so Rb = Ra’a. Thus any chain of n principal left ideals can be put in the form Ra, > Ra,a, >
> Ru;*.u,.
In particular, if Ra, > ... > Ra;..a, is a chain in a homomorphic image ?l of R then Ra, > > Ra;..a,, leading to the following observation:
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213
Remark 2.7.5: Any chain condition on principal left ideals of R is transferred to all homomorphic images of R. (By "chain condition" we mean some restriction on the chains, such as ACC, DCC, or bounded length.)
The other key observation we need involves the Jacobson radical. Remark 2.7.6: Ifa E Jac(R) and Rr = Rar then r = 0. (For r = xar for some x in R, so (1 - xa)r = 0, implying r = 0 since xu is quasi-invertible.)
Theorem 2.7.7: R is semiprimary ifl every chain of principal left ideals of R has length Im for some m. More precisely, write J = Jac(R) and R = R/J. (i) If every chain of principal left ideals of R has length Im then J" = 0 and R has length s m as R-module (and is thus semisimple Artinian). (ii) If J " = 0 and R has composition length k then we can take m I k"". Proofi
(i) For any a,, ..., am in J we have R2Ra,2Ra,a,2...2Ram...a,20, so equality must hold at some stage, i.e., Ra,, ,ai .a, = Ra, * * a, for some i, so a,. . . a, = 0 by remark 2.7.6. It remains to show R has length Im. In view of remark 2.7.5 we may assume J = 0, i.e., R = R. Note that R has at most m orthogonal idempotents e l , e 2 ,... since otherwise R > R(l - e , ) > R ( l - e l - e,) > * * * is a chain of length > m. Thus soc(R) is a sum of at most m minimal left ideals. Moreover, every left ideal of R contains a minimal left ideal, since every minimal principal left ideal is a minimal left ideal (for if L < Ra then taking b E L one has Rb < Ra). Thus R is semisimple Artinian by corollary 2.3.1 1. (ii) This is the harder direction, and we prove it in a sequence of steps, the first two of which are applicable to arbitrary semilocal rings. 3.
Step 1. If L , 2 L, are principal left ideals with elements r l , r , of R with F, idempotent such that L2 = Rr,r,.
&=
= G,
then there are L , = Rr,, and
Proof of Step 1. Write L , = Ra. Then & = for some idempotent Zof R ; so S = Ua for some invertible ti by remark 2.7.4. Hence u is invertible in R lemma 2.5.5). Let r , = ua. Then L1= Rr, and 5 = ?t is idempotent. Write L, = Rb and b = rr, for suitable r in R. Then Rr, = & = fi so = 3 for some invertible u in R ; taking r2 = vr yields Rr2rl = Rub = Rb = L , , and =vb=c.
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wl
Step 2. We call a chain Rr, > Rr2rl > > Rrj...rl admissible if = r i - l for each i 2 1. In this case, putting L, = R r i . * * r lthere , is a sequence a,,. . . ,a, in J such that Li c Rai...al + Lj for each i I j.
-
* a -
Proof of Step 2. Let a, = r, - r j . . . r i for i I j . Then iii = Pi - 'J''"i = Pi - Ti = 0 so a, E J. Moreover, in evaluating a,... a, we have a sum of terms l r i r i - , - * . r l Thus . r i - - * rEl a i * * * a lR r j . . . r , , each ending in r j * * * rexcept so L, G Ra,...a, + L,.
+
Step 3. Every chain L, > * * * > L, for t 2 k" has an admissible subchain of length n - 1. Proof of Step 3. Clearly # R since if L, =Ru with ii invertible then u is invertible. Hence there are k" - 1 left ideals L , , L,, ... each having length 0, l , . .., or k - 1, implying there is a subchain Kl > ... > LI,, where L'; = . * * = LI, and s = k"-'. Write Lj = Rri...rl and b, = r , * . * r , .Then letting L: = Rbi we have L;' > * * . > Ly. Continue inductively using step 1. Conclusion of Proof. Step 2 shows we cannot have an admissible chain of length n, so step 3 shows there is no chain of length k"". Q.E.D. Note: With care one could lower the bound for m a bit; I think the sharpest bound is unknown.
Corollary 2.7.8: lf J " = 0 and R / J has (composition)lengthk then every chain of cyclic R-modules has length 5 k"". Proof.- Suppose M, > M2 > is a chain of cyclic modules; putting M = Rx, and letting x i f l = r i + , x i we have R > Rr, > Rr3r2 > Q.E.D. * . a .
One can translate results on chains of cyclic modules to chains of f.g. modules by means of remark 1.1.19 and the following observation: Remark 2.7.9: If R is semiprimary then M,(R) is semiprimary for each t . More explicitly, if J" = 0 and length(R/J) = k then M,(J)" = 0 and M,(R)/M,(J) % M,(R/J); moreover, length(M,(R/J)) = tk.
Corollary 2.7.10: If J" = 0 and R / J has length k then every chain of R-modules, each spanned by It elements, has length I (tk)"+'.
$2.7 Semiprimary Rings and Their Generalizations
215
Pro08 If MI > M , > is a chain of modules spanned by It elements then MY) > M!) > is a chain of cyclic M,(R)-modules, so apply corollary 2.7.8 to M,(R). Q.E.D. 1 . .
Example 2.7.11:
then
(9 ):
If M is a D - E bimodule where D and E are division rings
is semiprimary. For example
(0" ):
is semiprimary, but
neither left nor right Artinian.
Passing from R to eRe and Back It is easy to transfer many properties from R to eRe (and likewise to ( 1 - e)R(l - e)),as illustrated in the next observation. Lemma 2.7.12: The following properties pass from R to eRe, where e E R is idempotent: ( i ) left Artinian; (ii) left Noetherian; (iii) primitive; (io) simple Artinian; (o) semisimple Artinian; (oi)semilocal. Proof: (i) Suppose L 1 > L , > .-.are left ideals in eRe. Then RL, > R L , > are left ideals in R. (Note R L , # R L i + l since e R L , = (eRe)L, = Li # L i + l = eRLi + 1 .) (ii) Dual to (i). (iii) If M E R - d o d is faithful simple then eM clearly is faithful in (eRe)dod and is simple by proposition 1.1.15. (iv) By ( i ) and (iii). (v) If R x Ri where each Ri is simple Artinian then eRe z eR,e; so we are done by (iv). (vi) Write R for R/Jac(R) -which is semisimple Artinian. By proposition 2.5.14, eRe/Jac(eRe)z eRe is semisimple Artinian by (v). Q.E.D.
fli=
fl:=
We are also interested in passing from eRe and ( 1 - e)R(1 - e) to R. Lemma 2.7.13: Suppose e is an idempotent of R and N 4 R. N is nilpotent iff eNe and ( 1 - e)N(1 - e) are nilpotent. Proof; (-) is clear. Conversely assume (eNe)"' = 0 and ((1 -e)N ( 1 -e))"= 0; we shall show N"+" = 0. Indeed, suppose a, E N for 1 I i 5 m + n. Then
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216
putting t = m
+n
a,...a, = (e
+ (1 - e))al(e+ (1 - e ) ) a 2 . . . ( e+ (1 - e))a,(e+ (1 - e)).
So it suffices to prove e o u l e l a 2 e 2 ~ * ~=a r0efor , all ei E { e , 1 - e } . Either e, = e for at least m + 1 values of i, or e, = 1 - e for at least n + 1 values of i. Assuming the former, we can rewrite eOalel* * * a t eas , e o a b e a ; e . . - a ~ e a ~where + , , for 1 Iu Im each a: is a nonempty string of the form a i , + l e i u + l . * * a iEu N + l; hence e o a l e l ~ ~ ~ a , e , ~ e o a ~ ( e N = e )0,m a ~ + l as desired. Q.E.D.
Proposition 2.7.14: The following properties hold in R ifl they hold both in eRe and (1 - e)R( 1 - e), where e is idempotent: (i) semisimple Artinian in case R is semiprimitive (cf., example 2.7.1 1) (ii) semilocal (iii) semiprimary Proofi
(i) (a) is lemma 2.7.12(v). To prove (e) note R has a complete set of orthogonal primitive idempotents (putting together those from eRe and (1 - e)R( 1 - e)),so R is semisimple Artinian by corollary 2.3.12. (ii) Pass to R/Jac(R), recalling Jac(eRe) = eJac(R)e. (iii) By (ii) and lemma 2.7.13. Q.E.D. An immediate consequence of this result is the following basic fact about endomorphism rings. Corollary 2.7.15: If M is an f.g. module over a semisimple Artinian ring R then E = End, M is a semisimple Artinian.
Proofi Write M as a direct sum@:=1 Mi of simple R-modules. Take f E E. For some i we see f M i # 0 and thus is a simple submodule of M isomorphic to Mi (as in Schur's lemma); taking a complement M' of f M we define g to be 0 on M' and to be f-' on f M . Now (1 - g f ) M i = 0, proving f 4 Jac(E); since this is true for any f we conclude Jac(E) = 0. Letting e, be the projection of M onto Mi we know each eiEei is a division ring, so E is semisimple Artinian. Q.E.D. This fact and many of its generalizations (to be discussed later) can also be obtained as immediate consequences of the Morita theory of Chapter 4.
$2.7 Semiprimary Rings and Their Generalizations
217
Semiperfect Rings Much of the theory of Artinian rings comes from lifting idempotents and the information which they contain. The properties which we need are pinpointed in the next definition. Definition 2.7.16: A semilocal ring R is semiperfect if Jac(R) is idempotentlifting.
The usages “semiperfect” and “semiprimary” unfortunately do not fit our general meaning of “semi” given in 52.2, and it may be worth giving the mathematical etymology of these words. In definition 2.8.31 below we define a projective cover for a module. Bass [60] defined a ring R to be perfect if every module has a projective cover; R is semiperfect if every f.g. module has a projective cover. Our definitions of perfect (definition 2.7.3 1 below) and semiperfect are equivalent, respectively, to these definitions, although this shall only be seen in exercises 2.8.28 and 2.8.22. The terminology concerning “primary” and “semiprimary” is hazier. Jacobson [64B] used “semiprimary” where we use “semilocal,” and he used “primary” for a ring R such that R/Jac(R) is simple Artinian. As time passed, “semiprimary” took on the added connotation that Jac(R) is nilpotent. According to the usage of this book, we shall have the following kinds of rings in increasing generality: lejl Artinian, semiprimary, perfect, semiperfect, semilocal
As we shall see later in $2.7 and in 52.8, perfect rings are nice because of their various characterizations, which enable one to study them using a wide variety of techniques from ring and module theory. O n the other hand, most examples of perfect rings which one encounters are already Artinian. The class of semiperfect rings contains many non-Artinian examples, including all local domains (since there are no nontrivial idempotents to lift). Nevertheless, many important properties of Artinian rings are obtained in this setting, as we shall see now (through theorem 2.7.30) and later in theorem 2.9.1 8 (which relies on an added ingredient, the Krull-Schmidt theorem). Lemma 2.7.17: Suppose A Write R = R I A .
aR
--
and A
c Jac(R), and e E R is idempotent.
(i) r f L I Re and L + R(1 - e) = R then L = Re. (ii) If e is a primitive idempotent and A is idempotent-lifting then F is a primitive idempotent of R .
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(iii) If R is semiperfect and e is a primitive idempotent and L s Re with L $! Jac(R) then L = Re.
rl
= R then 1 E so L , has an Proofi (i) Let L , = L + R(l - e). If invertible element by lemma 2.5.5 and thus 1 E L,; hence e E L,e = L, implying L = Re. (ii) Otherwise, F = Fl + Z2 for orthogonal idempotents Zl, F2 # 0. Lift El to an idempotent e , orthogonal to (1 - e). Then ele = e = eel but e # el since 2 # el. Hence e = el + (e - e l ) is a sum of orthogonal idempotents, contrary to hypothesis. (iii) Put A = Jac(R), which by hypothesis, is idempotent-lifting. Then 0 = L I& and F is primitive by (ii), so = f& and L + R(l - e) = R; we conclude by applying (i). Q.E.D.
Lemma 2.7.18: Suppose e is a primitive idempotent of R and R is semiperfect; then eRe is a local ring.
R = R/Jac(R) is semisimple Artinian and Fis a primitive idempotent by lemma 2.7.17(ii), so Re is a minimal left ideal, implying & is a division ring by proposition 2.1.21. Q.E.D.
Proofi
Semiperfect rings have two key ingredients-one is lemma 2.7.18, and the other is the existence of a complete set of primitive idempotents, by which we mean orthogonal primitive idempotents e l , . . .,e, with ei = 1.
,
Remark 2.7.19: If A Q R is idempotent-lifting then any complete set { F l , . . .,F,} (in RIA) of primitive idempotents can be lifted to orthogonal idempotents e l , ...,e, of R by proposition 1.1.25; obviously, each ei is a primitive idempotent, and { e l ,...,et} is a complete set since 1 eii is an idempotent of A which is thus 0. In particular, every semiperfect ring has a complete set of primitive idempotents.
I:=,
Example 2.7.19’: (i) A commutative semiperfect ringC is merely a direct product of local rings. Indeed a complete set of primitive idempotents of C / J lifts up to C;but these are central since C is commutative, so provide the desired ring decomposition. (ii) If R is f.g. (as module) over a local Noetherian ring C then R is semilocal and (Jac R)k G (Jac C)R for suitable k. Indeed let J = Jac(C). Then for every simple R-module M one has 3M # M by Nakayama so JM = 0 implying J annihilates every simple R-module; thus J c Jac(R). But R/JR is f.g. as
82.7 Semiprimary Rings and Their Generalizations
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module over the Artinian ring C/J, so R / J R and thus R/Jac(R) are Artinian. Furthermore, Jac(R)/JR = Jac(R/JR) is nilpotent so (JacR)' c J R for some k. There is a rich theory in the literature in case C is complete; in this case R is semiperfect, and one should expect much of this theory to hold for arbitrary complete semiperfect rings (with respect to the Jac(R)-adic topology). Thus there are significant examples of semiperfect Noetherian rings which are not Artinian. Recapitulating, we have the following result: Proposition 2.7.20: A semiperfect ringR has a complete set of primitive idempotents e l , .. .,e, such that each eiRei is a local ring. This condition characterizes semiperfect rings, as we shall see in theorem 2.8.40. It is extremely useful when applied in conjunction with the KrullSchmidt theorem, cf., theorem 2.9.18 below. Proposition 2.7.21: If R is semiperfect and R/Jac(R) x M,,(D) then R x M,,(T) for some local ring T such that T/Jac(T) x D.(Rings having this property are sometimes called quasi-local.)
Proofi In view of proposition 1.1.25 we can lift a set of matric units of R/Jac(R) to a set of matric units of R, so R x M,,(T) for some T. But T x e , , R e l l is local by lemma 2.7.18, whose proof shows T/Jac(T) x D. Q.E.D. Example 2.7.22: A commutative semilocal ring which is not semiperfect. Let n be a product of distinct prime numbers p1 * * * p , and let q : Z + = Z/nZ be the canonical homomorphism. Let S = { m E Z:m is relatively prime to n } , so that the elements of are invertible. S-'Z is semilocal. Indeed cp extends to a surjection qs:S-'Z + given by qs(s-'r) = F-'F, and ker cps = nS-'Z is quasi-invertible, so ker cps G Jac(S-'Z); equality holds since the Chinese Remainder Theorem shows Z x ,(Z/p,Z) which is semisimple Artinian. Then has nontrivial idempotents for t > 1, but S-'Z c Q has no nontrivial idempotents and thus is not semiperfect.
z
s
z
n:=
Structure of Idempotents of Rings We shall now study the structure of the left module Re when e is idempotent and R is semilocal. This yields useful information for semiperfect R which is important even when R is Artinian.
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Remark 2.7.23: Suppose eER is idempotent. Any R-module map f: Re+ M is given by right multiplication by the element fe = efe E eM. In particular, f(Re) G ReM. (Indeed f(re) = f(re2) = (re)fe so fe = efe.)
Proposition 2.7.24: as additive groups.
If e, e' are idempotents of R then Hom(Re, Re') z eRe'
Proof: Define II/:eRe' + Hom(Re, Re') by letting $(ere') be right multi-
plication by ere'. Then II/ is a group homomorphism which is onto by remark 2.7.23; if ere' E ker II/ then 0 = e(ere') = ere' proving II/ is 1:1. Q.E.D.
Proposition 2.7.25: Suppose e1,e2 are idempotents of R. Then Re, z Re, ifl there are elements el, in e,Re, and e,, in e,Re, such that el2eZ1= el and ezlelz = e2.
Proof: (a) Let JRe, + Rez be the given isomorphism and g = f-'. Taking el, = elfel and ezl = e,ge, as in remark 2.7.23, we have el = gfe, = elel2eZl = el2eZ1and e, = fge, = e2eZ1el2= eZ1el2. (e) Define f:Re,+Re, by f(re,)=re,, and g:Re,+Re, by g(re,)=re,,; then gf(rel) = rel2eZ1= rel and fg(rez) = reZ1el2 = re, so f and g are inverses. Q.E.D. This result has several nice consequences.
Corollary 2.7.26: e,R z e,R.
Suppose e,,e2 are idempotents of R. Re, x Re, iff
Proof: The condition of proposition 2.7.25 is left-right symmetric. Q.E.D. Our other application involves the Jacobson radical.
Lemma 2.7.27:
Suppose e,e' are idempotents of R, and J = Jac(R). If
f: Re + Re' is- -a map then letting - denote the canonical image in I? = R / J we have a map f: Re + Re' given by f(Z) = f(re); if f is an isomorphism then f is an isomorphism in I?-M'od.
Proof: First note that if E = 0 then re E J and f(re) is a well-defined map. If f-' exists then isomorphism. Q.E.D.
hence implying f is an
= (re)fe E J ;
f-'= f-'
52.7 Semiprimary Rings and Their Generalizations
22 1
Proposition 2.7.28: Notation as in lemma 2.7.27. Re x Re’ in R - A u d iff % Re’ in R-Aod. Proof: (a) is the lemma. (-=) By proposition 2.7.25 there are elements a in eRe‘ and b in e‘Re with ab =F and ba = e’. Lemma 2.5.5 implies ab is invertible in eRe and ba is invertible in e‘Re‘. Hence the map fa: Re + Re’ given by f a x = xu is 1: 1 and onto, and thus is an isomorphism. Q.E.D. To appreciate these results, suppose R is semilocal and has a complete set of primitive idempotents e l , . . .,e,. (For example, this holds if R is semiperfect). Then by the Peirce decomposition R = @eiRej, and each eiRei is local. Thus much of the structure of R hinges on the eiRej;in particular, R is a direct product of local rings if the eiRej = 0 for all i # j . Definition 2.7.29: A set of orthogonal primitive idempotents { e l , ...,e k } is called basic if for every primitive idempotent e’ we have Re‘ x Re, for exactly e, is called a basic idempotent and eRe one e,, 1 I i I k. In this case e = is called a basic subring of R. R is a basic ring if 1 is a basic idempotent.
Theorem 2.7.30: Every semiperfect ringR has a basic idempotent e, and eRe/Jac(eRe) is a direct product of division rings. Take a complete set of primitive idempotents {el,...,e,}; reordering it we get some k I t such that for any j > k we have Rej x Rei for suitable 1 I i I k. Then { e l , .. .,ek}is a basic set. Indeed, for any primitive idempotent e‘,we have Re’ isomorphic to some F i n R for suitable j s t (since R is semisimple Artinian), so R e ‘ x R e j by proposition 2.7.28, and R e j x R e i for some Q.E.D. i I k. The remainder of the proposition is clear.
Proo$
Besides being much easier to handle, basic subrings are useful because R - A u d is equivalent (as a category) to T-Aud if T is a basic subring of R. This important result can be proved in a few lines, but we would like to save the proof to introduce Morita’s theorems, cf., example 4.1.11.
H Supplement: Perfect Rings Definition 2.7.31: R is right perfect if it satisfies the DCC on principal left ideals.
This definition is a very natural generalization of semiprimary rings, in view of theorem 2.7.7, but the use of “right” may seem strange (and for this
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reason Bjork calls these "left coperfect" in his papers). We can understand the terminology by turning to the history of perfect rings. They were brought to light by Kaplansky [SO], as part of a program to study rings satisfying abstract chain conditions, and various ad hoc results were obtained during the next ten years. The breakthrough came in Bass [60], who found many different characterizations, using varied tools from ring theory. Since then algebraists have tried to extend the theory of Artinian rings to this broader setting, but this project has slowed as it has become clear that many of the recent strides in the theory of Artinian rings (cf., 52.9) simply do not apply to non-Artinian perfect rings. In the next few pages we shall give several elementary characterizations of perfect rings, largely parallel to the previous results on semiprimary rings, so the proofs will be abbreviated; in the next section we shall give the important link to projective covers. The key characterization here is in terms of the Jacobson radical, and needs another definition.
Definition 2.7.32:
A subset S c R is right T-nilpotent (also called vanishing) it for every sequence s1,s2, ... in S there is suitable n with s ; * * s 1= 0. Theorem 2.7.33: R is right perfect iff R is semilocal with Jac(R) right Tnilpotent.
Proofi (a) Almost word-for-word the same as that of theorem 2.7.7(i) and is left for the reader. (e) Otherwise, there is an infinite chain of principal left ideals L , > = G = *. Pick an L, > ... ; starting far enough along we may assume infinite chain with having maximal possible length. We shall use Step 1 of with 5 idempotent, we can the proof of theorem 2.7.7(ii). Writing L , Then L;' = = ... for some j, so write L, = Lyr, with L;' > Lj' > = starting at j we may assume = L;' = Now, by hypothesis, -length length =length G;right multiplication by 6 gives an epic L;'+L2 which is thus an isomorphism. Writing L;' = Rr, we thus have r,r, for some invertible u,, by remark 2.7.3. O n the other hand, we could choose r2 such that = q (by Step l), so = m.Iterating this procedure yields a new chain L , > L, > such that Li = Rri...r, with = = for suitable invertible ui. Write 6 = F for e idempotent (recalling any nil ideal is idempotent lifting), and let a, = r, - ri * r,eu,'. Then ii, = 0 so a, E Jac(R) and hence there is n with 0 = a ; * * a , E r n * * * r+l r , , - . - r , e R .
=El
- a * .
. . a .
=w2
v, m.
223
82.7 Semiprimary Rings and Their Generalizations
Thus r;..r1 = r;-.r,er’ for some r’ in R. Letting si = e - r i * * * rfor l i2n we get ri. . . r , = (1 - s,)er’. But & = 0 so si E Jac(R) and Rr, * * rl = Rer‘ inQ.E.D. dependently of i; consequently, Li = Rer’ = Li+ for all i 2 n. Other characterizations can be obtained easily by bringing in idempotents. Lemma 2.7.34: Suppose e is an idempotent of R , and N 4 R. N is right Tnilpotent iff eNe and (1 - e)N(1 - e) are T-nilpotent. Proof;. (.=) Otherwise, we have an infinite sequence a l , a z ,... with each
a ; * . a , # 0. Thus we can choose e, in {e, 1 - e } for each i such that anen... a1el #O for each n. We may assume an infinite number of e, are e, but
then one easily builds a sequence from eNe which is not T-nilpotent (parallel to lemma 2.7.13), contrary to hypothesis. Q.E.D. Proposition 2.7.35: Suppose e is an idempotent. R is right perfect iff eRe and (1 - e)R(1 - e) are right perfect. I n particular, if R is right perfect then M,,(R) is right perfect. Proof;. Combine proposition 2.7.14(ii) with lemma 2.7.34.
Q.E.D.
There is an ascending chain condition discovered by Jonah [70]. Theorem 2.7.36: ring: (i) (ii) (iii) (iv) (v)
The following properties are satisjed by any right perfect
DCC on cyclic modules; DCC on modules spanned by I t elements; ACC on principal right ideals; ACC on cyclic submodules of any given right module; ACC on submodules spanned by It elements of any given right module.
Proof:
(i) As in corollary 217.8. (ii) Apply remark 1.1.19 to proposition 2.7.35, to reduce to (i). (iii), (iv), (v). We prove (v), which implies the other two properties. Suppose M E A o d - R and has a chain M , IM , IM , I ... of submodules of M , with each Mi spanned by It elements. Let N = Mi-< M , and let J = Jac(R). Then we write for the canonical image in N = N / N J , which can be viewed as a module over the semisimple Artinian ring R / J .
uzl
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If n is the (composition) length of RIJ then each Mihas length Itn as (R/J)-module, and thus the chain Mi Ia2IM 3 I..*must terminate. But, clearly, 15 = Mi so some Mk= 15, implying Mk + N J = N . To complete the proof we need to conclude Mk = N , which we get by applying the following variant of Nakayama’s lemma:
u
Lemma 2.7.37: If J is right T-nilpotent and N I M in M d - R with N + M J = M then N = M . Proofi Passing to MIN we may assume N=O. Inductively, we can construct a sequence al,az,... of J with Ma;..a1#O for each n, for if Ma,-,...a, f O then 0 # M J a , - , a, since M J = M . But this contradicts the presumed Q.E.D. T-nilpotence of J. The statements (i)-(v) are actually each equivalent to the ring being perfect, c.f., exercise 9. Moreover, Bjork has proved that theorem 2.7.36(ii) can be strengthened to the DCC on all f.g. modules, c.f., exercise 5 ; in fact, the DCC on cyclic modules formally implies the DCC on f.g. modules. A detailed account is given in Faith [76B]. Unfortunately, the notion of perfect is not left-right symmetric, as seen in the following example: Example 2.7.38: Let M be a right vector space over a field having a wellordered base x1,xz,x3,.. . and let R be the F-subalgebra of End MR generated by J = { f E soc(End MF): f x i E C j , i x j F ) . (This is an infinite analogue of triangular matrices.) Clearly, J 4 R and RIJ x F; moreover, J is T-nilpotent since given a, E J we have a, M E xiF for suitable n, and then, clearly, Jnal = 0. Hence R is right perfect. On the other hand, if we take ai = e i , i + l then a l . * * a= , e l , , + , # 0 for each n, so the left-handed version of T-nilpotence fails.
,
G Supplement: Lefr n-Regular Rings Our final generalization of left Artinian rings is a chain condition considered earlier in $2.3, namely that any chain Rr 2 Rr2 2 Rr3 2 ... terminates. This condition has acquired the unwieldy name “left n-regular,” but arises naturally in several contexts, including versions of “Fitting’s lemma,” c.f $2.9.
Lemma 2.7.39: If R is not left n-regular then there is a prime homomorphic image of R which is not left n-regular, so in particular RIPrime rad(R) is not left n-regular.
42.8 Projective Modules (An Introduction)
22s
Proof;. Suppose Ra > RaZ > Ra3 > By Zorn's lemma there is an ideal P maximal with respect to ai $ Rai+ + P for each i. We claim P is prime. Otherwise, there are ideals A, B x P with AB E P. By hypothesis a'ERa"' + A and d ~ R a j ++' B for suitable i,j. But then aiERai+j+'+ B as can be seen by iteration, so for suitable r,r' we have a'-ra'+'EA and u j - r ' a ' + j + ' ~ B ; so l ) ( a j - r ' a i + j + 1 ) E P implying a'+j E Rai+j+' + P, contrary to choice of P. Thus P is a prime ideal and contains the prime radical of R; since R I P is Q.E.D. not left n-regular we conclude RIPrime rad(R) is not left n-regular. * a * .
Proposition 2.7.40: left n-regular.
If Jac(R) = Prime rad(R) and R is semilocal then R i s
Proof;. Otherwise, by the lemma R/Jac(R) is not left n-regular, but R/Jac(R) Q.E.D. is semisimple Artinian, an obvious contradiction. Defining right n-regular analogously, one can prove every right n-regular ring is left n-regular, c.f., exercise 23; in particular, any left or right perfect right is left n-regular.
82.8 Projective Modules (An Introduction) In this section we introduce projective modules, a categorically defined notion which can often be used in place of free modules. Projective modules play an important role in much of the structure theory, and, in particular, they bear on the kinds of rings discussed in the last few sections (which is why they are introduced here).
Projective Modules Definition 2.8.1: An M-module P is projective if it satisfies the following lifting property: For any map f: P + N and any epic map h: M + N there is a map f P + M such that f = hf. We then say f lifts to f Clearly this definition could be formulated for any abelian category in terms of the diagram D
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where here the dotted arrow completing the diagram is not uniquely defined. Remark 2.8.2: Every free module P is projective. (Indeed, suppose { x i :i ~ l } is a base of P; select yi in M such that hyi = f x i and define f:P-, M such that ?xi = yi for all i. Then f x i = hfxi for each i, implying f = h?.) In fact, projective modules are “close” to being free, in view of the following basic characterization:
Proposition 2.8.3:
The following are equivalent for an R-module P:
(i) P is projective. (ii) P is a summand of a free module. (iii) Every epic h: M + P splits. (iv) If h: M + N is epic then the induced map Hom(P, M ) + Hom(P, N ) (given by g + hg) is onto. Proof: (i) o (iv) is a restatement of the definition. (i)+(iii) Take N = P and f = 1, in definition 2.8.1. Then h f = 1 so h splits. (iii)=. (ii) Take a free module M of large enough cardinality. (ii) 3(i) By remark 2.8.2 and the following observation: Q.E.D.
Lemma 2.8.4:
P =
BierPi is projective ifl each 6 is projectiue.
Proof: Let pi: 8 + P be the canonical monic, and ni:P + 8 be the projection for each i. Suppose h: M + N is epic. (-) Given f: -, N we have fni:P + N which lifts to some g : P --t M. Then f = ( f n i ) p i= hgpi, so f lifts to gpi. (-=) Given f P + N we lift each fpi to g,:q+M and then we have g: P+M such that gi = gpi for each i; hence fpi = hgpi for each i, proving f = hg. Q.E.D. Remark 2.8.4’: (“Eilenberg’s trick”) If P is projective then there is a free module F such that P @ F is free. (Indeed, if P @ Q = F is free then adding on another free, if necessary, we may assume F is not f.g. and so
F x F ( ~x) P @ Q + ... x P @ ( Q @ P)@(Q o P) + ..( x P Q3 F(”) x P Q3 F.)
This innocuous remark will be useful in the study of “large” projective modules, but one should note that F as constructed was not f.g. The condi-
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tion that P 0 F is free for some f.g. free F is stronger than P projective and will be studied in $5.1. Remark 2.8.5: Every f.g. projective module is a summand of an f.g. free module. In fact, if P = c y = lRxi then the proof (iii)*(ii) of proposition 2.8.3 shows P is a summand of R(“). Corollary 2.8.6: P E R - A o d is cyclic and projective iff P = Re for some idempotent e of R.
Pro08 (-) P is a summand of R by remark 2.8.5, so viewing P as a left ideal of R we have P = Re where e is the image of 1 under the projection R + P. (t) Re is a summand of R, so is projective. Q.E.D.
Projective Versus Free Example 2.8.7: Every minimal left ideal L of R = M,,(D) is cyclic and projective, but is not free if n > 1 since [ L : D ] = n but [ R : D ] = n2. Example 2.8.8: If m = n f = l U , E Z where the ui are powers of prime inteaiZ by the gers with a , , . . .,a, pairwise relatively prime, then Z/mZ w Chinese Remainder Theorem, yielding examples of projective modules which are not free when t > 1 (seen by counting the number of elements).
n:=
A deeper example of a nonfree projective module is given in exercise 1. Nevertheless, there are certain rings R over which all projective modules are in fact free, and we give three important instances. Example 2.8.9: Projective modules over local rings are free. This is obtained in exercise 2.9.2, but there is a straightforward proof for f.g. projective modules, which, anyway, is the case of greatest interest. In fact, we shall show now if P = Rxi with n minimal possible then (x,, ...,x,,} is a base of P. Let J = Jac(R) and let F be the free R-module with base { e l , .. . , e n } . Let n:F + P be the map given by xei = xi for 1 I i I n. We claim kern JF. Otherwise, there is x = riei E ker x with some ri 4 J ; we may as well assume r l $ J , so r l is invertible and 0 = nx = z r i n e i = c r i x i , implying x 1 = -1r‘ = 2 r 1- l r i x i , contrary to the minimality of n. Having proved the claim, we note n:F + P splits by proposition 2.8.3(iii); viewing P as a summand of F we have F = P @ kern s P + J F so F = P by “Nakayama’s lemma.”
c:=
228
Basic Structure Theory
Despite the categorical preference for projective modules, one often finds it easier to deal with free modules, and the second sentence of a proof involving a projective module P often is, “Take P’ such that P Q P’ is free.” For example, if R c T and P is a projective R-module, then T @R P is a projective T-module. (Proof If P = R(“) then T @I R(”) x ( T 6 R)‘”) x T(”)is free. In general, take P’ such that P Q3 P’ is free. Then (T 6P) Q3 (T 6 P ’ ) x T 6 (P Q3 P’) is free.) This is the key observation in utilizing example 2.8.9, since if T is local then T @ P is free; we shall pursue this approach further in the discussions of localizations. Example 2.8.10: Projective modules over the polynomial ring FCIZ,, . . .,A,] are free for any t. This is a famous theorem of Quillen and Suslin, which will be presented in $5.1. For the time being, we shall consider easy special cases of this result. For t = 0, every module over a division ring is free, so the first nontrivial case is t = 1, to be discussed now in a more general setting.
Hereditary Rings Definition 2.8.11: A ring R is (left) hereditary if every left ideal is projective. Example 2.8.12: Suppose R is a PLID. Then any left ideal has the form Rx for x in R, but there is a module isomorphism qx:R + Rx given by q x r = rx, so every left ideal is free in R - d u d . In particular, every PLID is hereditary. Example 2.8.13: The ring of upper triangular matrices over a division ring is hereditary, as seen easily from example 1.1.9. Likewise, the ring of lower triangular matrices over a division ring is hereditary. Triangular matrix rings are the motivating example in the study of hereditary rings and play an important role in their structure theory, cf., exercise 5.2.40, as well as in the construction of counterexamples. Exercise 5 gives a left but not right hereditary ring. The purpose of the present discussion is to describe the basic connection between hereditary rings and projective modules, which will lead to the homological description of hereditary rings in Chapter 5. However, it is difficult to study hereditary rings in full generality, and usually additional properties are imposed, such as left Noetherian or semiprimary or both. Some of the best examples of such research are Robson [72], FuelberthKuzmanovich [75], and Chatters-Jondrup [83]. A fruitful generalization of hereditary rings is PP-rings, c.f., exercise 3.2.20ff. Also see exercises 4ff.
$2.8 Projective Modules (An Introduction)
229
Theorem 2.8.14: Suppose R is hereditary. Every submodule of a free module F is isomorphic to a direct sum of lejl ideals, the number of summands being at most the cardinality of a base of F. Proof: Let {xi:i E I } be a base of F, and take a well-ordering on I. For any i in 1 let 7. = Oj
+ xi<
Corollary 2.8.15: is projective.
R is hereditary
+
every submodule of a projective module
Proofi (a) Any submodule of a projective module is obviously a submodule of a free module and is thus a direct sum of projectives, which by lemma 2.8.4 is itself projective. (0Every left ideal is a submodule of R and is thus projective. Q.E.D. Corollary 2.8.16: Every projective module over a PLID is free. In particular, fi F is a free module with base of cardinality a and M 5 F then M is free with base of cardinality Ia. Proposition 2.8.17: Suppose R is hereditary and P, Q are projective modules. If f E Hom(P, Q ) then kerf is a summand of P. In particular, if P has no nonzero summands then every nonzero map from P to Q is monic.
Proof.- Since R is hereditary f P < Q is projective, so P Q.E.D. P z f P 0 kerf.
fP
-+
-+
0 splits, i.e.,
Dual Basis Lemma So far we have no concrete way of determining projectivity in terms of elements. To rectify this we need to generalize another property of free modules. Remark 2.8.18: Suppose {ei:i E I } is a base of a free R-module F. Defining rjej)= ri, we see for any y in F that niy = 0 the projections x i :F + R by ni(1 for almost all i in I, and y = C(niy)ei.
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Basic Structure Theory
This elementary property actually characterizes projective modules, as we shall now see. Delinition 2.8.19: A dual base of a module P is a set of pairs {(xi,fi): i E I} where xi E P and fi:P -P R is a map satisfying the following condition for each x in P:f;x = 0 for almost all i in I, and x = ~,,,(f;x)x,. Obviously this equation shows each x E is a much more decisive result.
cRxi, so the xi span P, but there
Proposition 2.8.20: (“Dual basis lemma”) (i) Suppose P is projective and {xi:i E I} span P. Then there are fi: P + R such that {(xi,fi):i E I } is a dual base. (In particular, every projective module has a dual baie.) (ii) Conversely, if P has a dual base then P is projective. Proof: In each direction we assume { x i :i E I } spans P, and F is a free module with a base {ei:i E I } and projections ni: F -+ R as in remark 2.8.18. Define cp: F + P by cpei = xi for each i in I.
(i) Since P is projective we have $: P + F with 91)= 1,. Put fi = ni$ for each i in I. Then for every x in P we have
as desired. (ii) We shall show P i s a summand of F by finding $: P + F with cp$ = 1,. Indeed, define $ by $x = C(frx)eifor each x in P. Then cp$x = x(fix)cpei= c ( f i x ) x i= X . Q.E.D.
Flat Modules Let us note one more important property of free modules which is shared by projective modules. Delinition 2.8.21: An R-module M is $at if whenever f:N, in Mod-Rwe have f @ 1: N , QR M + N , BRM is monic.
N2 is monk
Remark 2.8.22; Obviously R is flat since N @lR R x N naturally.
231
$2.8 Projective Modules (An Introduction)
Proposition 2.8.23: M = @ M i is flat fi each Mi i s flat. Proof: Suppose f:N , + N2 is monic. Then using proposition 1.7.15 we can view f Q 1: N , Q M .+ N2 Q M as the composition
N,@
(@Mi) x @ ( N , 0 Mi).+ @(N2 0 Mi) % N2 Q ( @ M i ) ,
so we identify ker(f @ 1), as monic. Q.E.D.
@ ker(f 0 1
which is 0 iff each f 0
is
Corollary 2.8.23': Every projective module is flat. Every free module is a direct sum of copies of R and so is flat; a Q.E.D. projective module is a summand of a free module and thus is flat.
Proof:
Of course, nonflat modules exist; the Z-module M = 2/22 is seen to be nonflat when confronted with the monic Z.+ Q. Proposition 2.8.24: If 0 -,F, M is flat.
-,M .+ F2
.+
0 is exact with F , , F2 flat, then
Suppose f:N , + N 2 is monic. We want to show f @I 1, is monic. Looking inside N , Q M let K denote the submodule N , 0 F,. The composition Proof:
NiQ M
a N2 0 M
.+ N2
0 F2
has kernel containing K, so by proposition 1.7.30 we have a natural map N 1 6 F2
%
( N i Q M ) / K .+ N2 0 F2
which corresponds to f @ lF2.By hypothesis this is monic, so we conclude ker(fQ 1,) E K. But ker(f @ lF,) = 0 by hypothesis, so we conclude Q.E.D. ker( f Q 1,) = 0. Remark 2.8.25: If every f.g. submodule of M is flat then M is flat. (Indeed, otherwise there is f:N , .+ N 2 monic and some element z = xi Q y i # 0 in ker(f Q 1,); then M' = Ryi is not flat since 0 # z E k e r ( f 8 lM,).)
xi=,
Further study of flat modules requires the use of injectives and will be handled in $2.11.1Iff. Important homological properties of flat modules will be discussed in chapter 5.
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Basic Structure Theory
Schanuel's Lemma and Finitely Presented Modules One of the most important results concerning projective modules is the following fact, which lies at the heart of homology theory.
Proposition 2.8.26: (Schanuel's lemma) If 0 + K,+ 8 4 M with & projective for i = 1,2 then P, 0 K , x P20 K , .
-,0
are exact
Proof: Identify K i with kerg,. Lifting g 1 to a map h: P, + P2 satisfying g 1 = g2h, define f: P, 0 K , + P2 by f ( x ,y ) = hx - y . Note f is epic (for if x r E Pz there is x in PI with g , x = g 2 x rimplying g2(hx - x r )= g , x - g2x' = 0, so hx - x' E K , and x' = f ( x ,hx - xr)).Thus f is split so P, 0 K 2 x P2 0 kerf. Noting x E K , iff 0 = g , x = g,(hx) iff hx E K , , we see k e r f = { ( x ,y ) :hx = Q.E.D. y E K 2 } = { ( x , h x ) : xE K , } x K , .
The first among many applications of Schnauel's lemma is to f.g. modules. Definition 2.8.27: A module M is finitely presented if there is a mapf: F + M with F f.g. free, such that kerf is also f.g.
Equivalently, M is finitely presented iff there are f.g. free modules & with F, 5 F, --+ M + O exact. (Indeed, this condition clearly implies M x F,/hF2 is finitely presented; conversely, if M x F / K with K f.g. then taking an epic h: F2 + K with F, f.g. free we have F, 4F + M -,0 exact). Such a sequence will be called a j n i t e presentation of M . Examples 2.8.28:
(i) If R is left Noetherian then every f.g. R-module M is finitely presented. (Indeed take f: F + M epic with F f.g. free. Then kerf is f.g. since F is a Noetherian R-module.) (ii) Every f.g. projective module is finitely presented. (Indeed if K: R(')+ P is epic then R(') sz P 0 ker n so ker n is also an image of R(')and thus is f.g..
Proposition 2.8.29: If N is finitely presented and f . g . then kerf is f.g..
f:M
+N
is epic with M
Proofi First assume M is projective and take an exact sequence 0 --* K + P + N + 0 with P free and K, P f.g.. Schanuel's lemma shows kerf is a summand of M 0 K and is thus f.g..
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233
For general M take epic n:F + M with F f.g. free. Then ker( f n ) is f.g. by what we just saw. But kerf = n(ker fn). (Indeed, n(ker f n ) Ikerf; conversely, if nx E kerf then x E ker( fn)).Thus kerf is f.g.. Q.E.D.
H Supplement: Projective Covers Often when studying a module M we pass to a projective module P above it, i.e., there is an epic P + M. We shall now describe a situation for which a unique P “closest” to M can be found, which casts a strong light on idempotent-lifting, and perfect and semiperfect rings. Definition 2.8.30: A submodule M’ of M is small or superfluous (written M’ << M) if for any N < M we have N + M’ < M.
We usually verify the contrapositive, i.e., given N + M‘ = M we prove N = M. Small submodules play a hidden role in “Nakayama’s lemma,” which says Jac(R)M << M for every M in R-9imod. Moreover, small left ideals tie in with the Jacobson radical as follows: L << R iff L is contained in every maximal left ideal (seen by checking the definition), iff L IJac(R). Thus Jac(R) is a small left ideal which contains every small left ideal. Definition 2.8.32: A projective cover of a module M is an epic f: P such that P is projective and kerf << P.
-+
M
Part of the definition is easy to satisfy, since M is a homomorphic image of a suitable free module. Our next effort will be in analyzing the key condition that kerf is small, which leads to an improvement of Schanuel’s lemma in this case: Projective covers are unique up to isomorphism (proposition 2.8.34 below). Remark 2.8.32: If f: M -+ N is a map and M , IM with f M , = f M then M = M, + kerf. (Indeed, if x E M then f x = f x , for suitable x, in M , , and then x - xI E ker f.)
Lemma 2.8.33: Suppose M has a projective cover f:P -+ M. For any epic g : Q --t M with Q projective, we can write Q = P‘ 0 P” with P‘ = P and P’‘ I ker g, such that the restriction of g to P’ is a projective cover for M. Proofi Since Q is projective we can lift g: Q -+ M to h: Q + P such that g = f h . In particular, f ( h Q ) = gQ = M so hQ + kerf = P by remark 2.8.32,
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implying hQ = P since kerf is small. But P is projective so h splits, i.e., we can identify P with some summand P’ of Q and write Q = P @ P’’ where h is the projection onto P. Let K be the kernel of the restriction of g to P. Since gP = f h P = f h Q = M it remains to show K << P. Suppose N + K = P. Then g N = g P = M so hN = P as shown above. But hN = N so N = P, proving K is small. Q.E.D. Proposition 2.8.34: (Uniquenessof projective cover) l f fi: pi + M iis a projective cover of Mifor i = 1,2 and cp: M, + M 2 is an isomorphism then there is an isomorphism cp’: Pl + P2 with f 2 q ’ = cpf,. Proof: Apply lemma 2.8.33, taking f = f2, g = pf,, P = P2, and Q = Pl; we need only show the epic map h: Q 4 P obtained there is an isomorphism, for q f l = g = fh, and we would then take cp’ to be h. But P“ = kerh is small in Q (for P” Ikerg << PI = Q ) and P’ + P“ = Q, implying PI‘ = 0, as desired. Q.E.D. Example 2.8.35: A module need not have a projective cover. Indeed, suppose M is a cyclic but not projective module over a ringR which has no nonzero small left ideals. Then M has no projective cover (since, otherwise, taking Q = R in lemma 2.8.33 and letting f: R + M be the canonical onto map, we see there is a projective cover 8’: P‘ -+ M where P’ is a left ideal of R, but ker g’ K R implies ker g’ = 0 so M NN P’ is projective, contrary to assumption). In particular Z/2Z has no projective cover as a Z-module. This idea is carried further in exercise 20, in which it is shown for R semiprimitive that only the projective R-modules have projective covers.
Lemma 2.8.36: A cyclic module M has a projective cover iff M x Re/A for some idempotent e and left ideal A I Jac(R) n Re. In this case, the canonical map f: Re + M is a projective cover. Proofi (-=) kerf = Ae E Jac(R)Re << Re by “Nakayama’s lemma.” (=.) Let 9: R -+ M be the canonical onto map. Then R x P’ 8 P” under the notation of lemma 2.8.33 so P’ x Re for some idempotent e and M NN Q.E.D. Re/A where A << Re, so A E Jac(R).
Proposition 2.8.37: Suppose A 4 R and A E Jac(R). A is idempotent-lifting iff every summand of the R-module RIA has a projective cover. Proof: (=+) R = RIA has the same lattice of submodules in R-MULL or RIA-&& so any summand can be written as Re where Z E R is idempotent;
82.8 Projective Modules (An Introduction)
235
take e E R idempotent. Then x R e / A n R e = Re/Ae has a projective cover by the lemma. (e) Suppose ZE R = R / A is idempotent. The & is a summand of R and thus has a projective cover P which by the lemma can be written in the form Re/A' where e is an idempotent over 2. Q.E.D. We are ready for the key results of this discussion, tying projective covers in with idempotent-lifting and thus with semiperfect rings. Remark 2.8.38: Suppose there is an epic $: M + N with ker $ << M. Any projective cover f: P -+ N lifts to a projective cover g : P + M. (Indeed, f lifts to a map g : P + M with f = $ g ; then gP + ker $ = M by remark 2.8.32 so g is epic since ker $ is small. Moreover, ker g I kerf << P.) Lemma 2.8.39: (i) If K, << M and K , << M then K , + K , << M. (ii) If K i << Mi for 1 I i I n then K , 0 ... 0 K , << M I CT3 0 M.. (iii) If J : P;. + Mi are projective covers for 1 I i I n then (f,,. . .,fn): @:= 8 + @:= Mi is a projective cover.
,
,
Proof: (i) If K, + K, + N = M then K , + N = M so N = M. (ii) Inductively one may assume n = 2. Write M = M , 0 M,; by (i) it suffices to show K i << M for i = 1, 2. But if N + K, = M then ( N n M , ) + K , = M , so N n M , = M,,implying M , I N a n d N 2 N + K , = M. (iii) Clearly (f,,...fn) is epic, and its kernel is O k e r J , which is small by (ii). Q.E.D. Theorem 2.8.40: R is a semiperfect ring iff R has a complete set of orthogonal primitive idempotents e l , . . .,en such that each ei Re, is a local ring. Moreover, in this case every f.g. R-module has a projective cover. Proof: (*) is proposition 2.7.20. (-=) R is semilocal by proposition 2.7.14. Let J = Jac(R). We need to show J is idempotent-lifting, so in view of proposition 2.8.37 we need merely prove the last assertion, that every f.g. R-module M has a projective cover. M/JM is an f.g. module over the semisimple Artinian ring R = R / J , and el,.. .,Z, area complete set of orthogonal idempotents which are all primitive since q R e i x (eiRei)/Jac(eiRei)is a
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division ring. Thus MIJM x @finite&, for suitable u i , where each Reui z Re,,/Je,, is simple in R-Ad and thus in R - A d . Now M I J M has a projective cover by lemmas 2.8.36 and 2.8.39(iii), implying M has a projective Q.E.D. cover by remark 2.8.38. Note that theorem 2.8.40 shows for a semiperfect ring that every f.g. module has a projective cover. This condition characterizes semiperfect rings (exercise 22) and actually can be weakened slightly (exercise 23).
Categorical Description of Projective Covers The usual definition of projective covers is unsatisfactory from a categorical point of view since “smallness” of a submodule is not categorical. As an alternate, in view of lemma 2.8.33, we could say a given epic n:P + M is a projective cover of M if for every epic n‘: Q + M with Q projective there is a (split) epic f:Q + P satisfying nf = n‘. This criterion is indeed categorical but relies on our knowing all the projective R-modules. We are interested in a criterion which only depends on P. Let us say a map g: P + P lies above a map f:M -+ M if ng = fn. Since P is projective any map f:M + M has some g lying above f, seen by completing the diagram P
Remark 2.8.41: If gi lie above f;: for i = 1,2 then glgz lies above (Indeed, v 1 g z = (fin)gz = fi(ng2) = f1fi.J
f lfz.
This idea is useful in studying End, M because of the following result. Proposition 2.8.42: with P projective:
The following are equivalent for any given epic n:P + M
(i) n is a projective cover of M (i.e., kern << P ) . (ii) Given an epic f:M + M, every endomorphism (of P ) lying above f is epic. (iii) Given an isomorphism f:M -+ M , every endomorphism lying above f is an isomorphism. (iv) Every endomorphism lying above 1 is an isomorphism. (v) Every endomorphism lying above 1 is epic.
-
237
$2.9 Indecomposable Modules and LE-Modules
Proofi (i) (ii) Suppose g:P + P lies above f. Then ngP = f n P = fM = M = n P so P = g P ker a by remark 2.8.32. Hence g P = P. (ii)*(v)Takeg = lu. (v) + (i) Suppose N kern = P. We need t o prove N = P. Take some P‘ such that F = P 0 P‘ is free. Then F = N kern P’ so given a base x l , ..., xn of F we can write each xi = a i b, where a, E N a n d bi E kern P’. Define g: P + P by g( rixi) = riai, i.e., g is the restriction to P of the map F + P which sends xi to a,. Then g lies above l u so by hypothesis g is epic. But clearly N 2 g P = P proving N = P. (ii) * (iii) Suppose g lies above f. By hypothesis g is epic, so is split, since P is projective. Hence P = PI 0 kerg where Pl x g P x P. But if x E ker g then 0 = ngx = fnx so nx = 0 since f is monic; hence kerg Ikern and P = PI kern. We have already proved (ii) (v) e-(i) so ker n << P. Hence PI = P implying ker g = 0, i.e., g is an isomorphism. (iii) (iv) Take f = 1M . (iv) (v) A fortiori. Q.E.D.
+
+
1
1
+
+
+
+
+
52.9 Indecomposable Modules and LE-Modules We have seen in #2.4 that every completely reducible module can be written “uniquely” as a direct sum of simple modules. In this section we study modules having a weaker form of decomposition and apply this theory to modules having composition series and to semiperfect rings. The reason this section follows the section on projective modules is that one important module which we would like to decompose is the free R-module or, more specifically, R itself; in this case all the summands are obviously projective. Definition 2.9.1: An R-module M is indecomposable if M cannot be written as a direct sum of two proper submodules. A decomposition of M is a direct Mi = M for Mi indecomposable. M is completely decomposable sum (of Krull-Schmidt length I I I) if M has a decomposition as above.
OiE,
Our ultimate goal is to study the completely decomposable modules by means of the indecomposables. At first blush this task would seem near impossible, although amazing progress has been made in recent years. We defer the difficult project of determining the indecomposables and, for the time being, try to identify completely decomposable modules; one easy case is when M has a composition series.
Basic Structure Theory
238
The Krull- Schmidt Decomposition Proposition 2.9.2: (Existence of Krull-Schmidt decomposition) fi M has a composition series of length It then M is completely decomposable of Krull-Schmidt length It. Proofi (Induction on t ) If M is indecomposable we are done. So assume M = M , @ M , for proper submodules MI,M,. Then M > M , > 0 can be refined to a composition series of length It so M , and M , x M / M 1 each have composition series of respective lengths 5 tl,t, with t, + t , = t. By induction M,, M , are completely decomposable of length < t,, t,, respectively, so M = M , @ M , is completely decomposable of length < t, t , = t. Q.E.D.
+
Proposition 2.9.3: (i) Suppose M = M , @ M,, and let e, be the projection of M onto Mi for i = 1,2 (i.e., ei(xl,x,) = xi). Then e,,e, are orthogonal idempotents in End, M and el + e2 = 1. (ii) Conversely, if el,e2 are orthogonal idempotents in End, M with el + e, = 1 then Mi = e,M are submodules of M with M = M , @ M,.
Proofi Each step is straightforward; details are left to the reader.
Q.E.D.
Example 2.9.4: Take M = R so that R x End, R by the regular representation. R is a direct sum of n left ideals iff 1 is a sum of n orthogonal idempotents. In this way, we see proposition 2.9.3 is a rather sweeping observation which also has the following immediate application.
Corollary 2.9.5: M is indecomposable i g End, M has no nontrivial idempotents (which is true i f End, M is local, by remark 2.5.13). The criterion that End, M be local will turn out to be the key to this discussion. To see why, we need another fundamental result, called Fitting’s lemma.
Lemma 2.9.6: Suppose f f is an isomorphism.
E
End, M. If M is Noetherian and f is epic then
82.9 Indecomposable Modules and LE-Modules
239
Proof: kerf ~ k e r f ~ I . . . s o k e r f " = k e r f " "f o r s o m e n . B u t f o r a n y x in kerf we can write x = f "y for suitable y in M, so 0 = f x = f ""y implying 0 = f "y = x; hence kerf = 0 and f is monic. Q.E.D.
Proposition 2.9.7: ("Fitting's lemma") Suppose M has a composition series of length n, and f E End, M. Then M = M , @ M,, where MI = f "M and M, = kerf ". Each Mi is invariant underf; moreover, f restricts to an isomorphism f, : M , + M , and a nilpotent map f, : M , + M, . Proof: Note M 2 f M 2 f,M 2 ... and 0 I kerf I ker f 2 I ."; since M has1engthnweseeatoncef"M = f " + ' M = - . . a n d k e r f " = kerf"" = . . - . Since f"M = f"(f"M) we have M = f"M kerf" by remark 2.8.32. For x E f "M n kerf " writing x = f"y we have f 2"y = f "x = 0 so y E kerf 2 n = kerf" implying x = f"y = 0; this proves M = f"M @ kerf". Now fM,= M , so f, is an isomorphism by lemma 2.9.6; obviously f "M, = 0. Q.E.D.
+
Corollary 2.9.8: If M has a composition series and is indecomposable then End, M is a local ring.
Proofi Given f in End, M, we have either M, = M or M, = A4 in Fitting's lemma (because M is indecomposable), so f is either an isomorphism or is nilpotent. Thus we are done once we can prove the following result: Proposition 2.9.9: Suppose R is a ring each of whose elements is invertible or nilpotent. Then R is a local ring. Proof: We need to show if a , + a , = 1 then a , or a, is invertible, by remark 2.5.12. If a , is not invertible then a , is nilpotent by assumption, so Q.E.D. a, = 1 - a , is invertible by remark 1.1.27. Krull-Schmidt decompositions enable us to generalize corollary 2.7.15, as in the next interesting result.
Theorem 2.9.10: If M has a composition series of length n then EndRM is a semiprimary ring and Jac(End, M)" = 0. Write M = @:=, Mi where the Mi are indecomposable, and let ej E End, M be the projection onto Mi for 1 Ii I n. Then the e, are orthogonal primitive idempotents, and ei(EndRM ) e , x End,(e,M) = End, Mi is local by corollary 2.9.8. Hence End, M is semilocal, by proposition 2.7.14; Proof:
Basic Structure Theory
240
it remains to show J" = 0 where J = Jac(End, M ) . Suppose f E J . By Fitting's lemma we can write M = M 1 0 M 2 where M , = f "M and M 2 = kerf ";and f restricted to M1 has an inverse g, which we extend to an endomorphism of M by putting gx = 0 for all x in M , . Then (1 - g f ) M , = 0. But f E J implies 1 - gf is invertible. Hence 0 = M , = f "M, so we see J is nil of bounded index n. We shall actually prove the apparently stronger assertion that every nil (multiplicative) semigroup of End, M is nilpotent; then J k = 0 for some k > 0 and if J" # 0 we have M > J M > J2M > ... > J"M > 0 contrary to the Jordan-Holder theorem. (In fact, our assertion is equivalent to the nilpotence of J, by theorem 2.6.31) So assume S is a nil multiplicative subgroup of End, M. If J = 0 then we are done by theorem 2.6.31.Otherwise, End, M has a nilpotent ideal A # 0 by proposition 2.6.26, and 0 < A M < M so A M has length < n. S acts as a nil semigroup of endomorphisms on A M so by induction S'AM = 0 for some u > 0. Likewise, S " ( M / A M ) = 0 for some Q.E.D. u > 0, so S""M c S"AM = 0, as desired.
Uniqueness of the Krull- Schmidt Decomposition We turn to the uniqueness of the Krull-Schmidt decomposition, a fairly delicate issue which requires careful analysis.
Definition 2.9.11: M is an LE-module if End, M is local (i.e., M has a local endomorphism ring). We saw that every LE-module M is indecomposable, the converse holding when M has a composition series. In fact, LE-modules provide a fairly general situation in which it is possible to prove both the KrullSchmidt theorem and a generalization due to Azumaya. Following Beck [78] we obtain very precise information concerning decompositions of sums of LE-modules. Write J, for JaC(HOm,(M, M ) ) .
Lemma 2.9.12: Suppose f: M 1, + fg is an isomorphism iff 1,
and g: N 4 M are maps in R - A u d . gf is an isomorphism.
+N
+
Proofi (*) Noting f(1, + gf) = f + fgf = ( 1, + fg)f and, likewise, (1, + gf )g = g(1, + fg), one sees easily that 1 - g(1, + fg)-'f is the inverse of 1, + gf. The opposite direction is analogous. Q.E.D. Proposition 2.9.13: Suppose M, N E R-Aud and M is LE. Iff: M 4 N and g: N + M satisfy fg 4 J N then ghf is invertible for some h: N + N . In particular, f is split monic and g is split epic.
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Proofi By hypothesis, hfg is not quasi-invertible for some h: N + N, so 1, - hfg is not an isomorphism; lemma 2.9.12 implies 1, -ghf is not an isomorphism so ghf # J , . Since M is LE we conclude ghf is an isomorphism, as desired, and the second assertion follows from proposition 1.4.10. Q.E.D. To utilize this result fully we need an easy property of summands.
Proposition 2.9.14: Suppose M = K 0 N and eK is the projection of M onto K . M = L 0 N ( f o r L I M ) iff e, restricts to an isomorphism L + K . Proofi (t) Clearly L n N = 0. If x E M then e,x = e,y for some y in L , so x = y + ( x - y ) E L + N, proving M = L 0 N. (=) Let f be the restriction of e, to L. Then kerf = {a E L:e,a = 0) = L n N = 0. Note f is onto since for any x in K we can write x = a + b for Q.E.D. a in L and b in N, so then x = e K x = eKa + e,b = eKa.
Theorem 2.9.15: (“Exchange Property”) Suppose M = N 0 N’ = Oie1 Mi where each Mi is an LE-module. Then some Mi is isomorphic to a summand of N. Moreover, if N is indecomposable then we can pick i such that N x Mi and M = ( @ j + i Mj)0 N = Mi 0 N’.
Oiel
Proofi Let a: Mi + N be the projection onto N and let ai:Mi -+ N be the restriction of a to Mi; let ei:N + Mi be the restriction of the projection M + Mi. Pick x # 0 in N and write x = xu for xu in Mi”. Let f =1 aiueiUE Hom(N, N). Then x = ax = aimxu= aiueiux= f x so ( 1 - f )x = 0; hence f is not quasi-invertible and f #JN. Hence some aiei # J,. By proposition 2.9.13 this ai is split monic, so Mi x aiMi is a summand of N; likewise, e, is split epic. If N is indecomposable then ei and ai are both isomorphisms, so applying proposition 2.9.14 each time yields M = ( @ j + i Mj)0 N and M = Mi 0 N‘. Q.E.D.
BiclMi = OjeJ % are equiu-
Definition 2.9.16: Two decompositions M = alent if I = J and there is a 1:1 correspondence all i E 1.
6:I + J
with Mi x NOifor
Theorem 2.9.17: ( Wedderburn-Krull-Schmidt-Azumaya) If M is a direct sum of LE-modules then all decompositions of M are equivalent.
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Proofi Given an indecomposable summand K of M let Z(K) = { i E I:M, x K } and J ( K ) = { j E J:A$ x K}. Clearly it suffices to prove IZ(K)I = IJ(K)I for each K. By symmetry we need only show IZ(K)I IIJ(K)I; note by theorem 2.9.15 that I ( K ) # 0 iff J ( K ) # 0.
Case I. n = IJ(K)I is finite. For convenience write J ( K ) = { 1,. . .,n}. By M i , )Q N , . theorem 2.9.15 we have some i such that Mi x N , and M x ( Thus @ i , +Mi, i x MIN, x N,, we conclude by induction on n (having removed one module from each of I ( K ) and J ( K ) ) .
Oi,,
mj+,
+
Case II. IJ(K)I is infinite. We resort to the trick of theorem 0.3.2, suitably modified. Given j in J ( K ) write I ’ ( j ) = { i E I ( K ) :M = Mi @ Nj,)}. I ( K ) = U j s J ( K ) I ’ ( jby ) theorem 2.9.15, so it suffices to prove each I ’ ( j ) is finite since then II(K)I I IZ’(j)l I INJlJ(K)I = IJ(K)I. Pick x # 0 in A$,and write x = c x i where each xi E M i . Since M = @ M i we see almost all xi = 0. But if xi = 0 then x lies in the kernel of the restricso i # I’( j ) by proposition of the canonical projection M + Mi to 4, tion 2.9.14. Thus I ’ ( j ) is finite, as desired. Q.E.D.
xjeJ(K)
Applications of the Krull- Schmidt Theorem to Semiperfect Rings In 2.7.23-2.7.30 we initiated the study of a semilocal ring in terms of a complete set of primitive idempotents. The results become much sharper when we apply the finite case of theorem 2.9.17, usually called the Krull-Schmidt theorem. Note that these results are interesting even for Artinian rings.
Theorem 2.9.18:
Suppose R is semiperfect.
,
(i) There is an LE-decomposition R = @:= Re, where e l , . . .,e, is a complete set of primitive idempotents. (ii) Every summand of R is a jinite direct sum of submodules isomorphic to the Rei, and consequently every idempotent is a sum of orthogonal primitive idempotents (which can be expanded to a complete set of primitive idempotents). (iii) If f,,. ..,f,. is another complete set of primitive idempotents of R then t = t’, and there is some invertible u in R such that u-’eiu = f;: for 1 I i I t (where the fi have been rearranged such that Rei x Rf;:). Proof;. (i) is a restatement of proposition 2.7.20.
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52.9 Indecomposable Modules and LE-Modules
(ii) By example 2.9.4 we can write any summand L of R in the form Re for e idempotent. The exchange property shows L = @ Li where each Li x Rei for suitable i; but each Li is a summand of R so Li = Re: where clearly the ei are primitive and orthogonal. (iii) R = @Rei = @Rfi so by Krull-Schmidt, t = t' and the fi can be rearranged to yield an isomorphism i,hi: Re, -,Rfi. Then @i,hi: @Rei -, @Rfi is an isomorphism of R as R-module and is thus given by right multiplication by some invertible u in R. For ri in R we thus have 1;. = i,hi(riei)= rieiu for 1 Ii It. Then fiu-' = riei, so for 1 I j I t we get fiU-'e.u = r.e.e.u = &.r.e.u = &.f. 1 J J 1 1J 1 1 V 1
implying u-'e,u =
(I;=, fj)u-le,u = L,as desired.
Q.E.D.
The modules Rei of theorem 2.9.18 are projective, being summands of R , and are called the principal indecomposable modules of R , for obvious reasons. Corollary 2.9.19: Suppose R = M n ( T )where T is a local ring and {eij:1 I i, j In> is a set of matric units of R. lf {eij:1 Ii, j In } is another set of matric units of R then there is some invertible u in R with u-'eiju = eij for all i, j . Proofi From the theorem we have uo in R with uO1eiiuO= eli for 1 Ii I n, since { e l l ,..., en,,} is a complete set of primitive idempotents. Let u = -1 eilu0e;,and u' = eiluO e l i . Then
XI=
I
uu' = I e i l u o e ; l u ; l e l i = C e i l e l l e l i= C e i i = 1
and, likewise, u'u = 1, so u' = u - ' . Moreover, u-'e..u e..e. e' V = u'e..u V = e!11 u-'e 0 l i rj j 1 o 1 j = 41e;le;j = 4
as desired.
j
Q.E.D.
This has the following surprising consequence. Corollary 2.9.20: Suppose R = M n ( T )where T is a commutative local ring. Any ring injection f:R + R Jixing T is an inner automorphism. Proofi If {eij:1 Ii,j I n} is a set of matric units then so is { feij:1 I i, j I n}, so f e , = u-'eiju for some u in R, and the result follows at once. Q.E.D.
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We can also characterize the indecomposable projective modules. Proposition 2.9.21: Suppose R is semiperfect. An R-module P is indecomposable projective iff P is principal indecomposable. Consequently R has only a jinite number of indecomposable projective modules. Pro@ Let {el, ...,er} be a complete set of orthogonal primitive idempotents. (*) P is a summand of some free R-module, which by theorem 2.9.18 is a sum of copies of the Re,; the exchange property shows P FZ Re, for some i. The converse is clear. Consequently R has at most t indecomposable projectives. Q.E.D.
Applying theorem 2.9.18 to theorem 2.8.40 we see that a module M has a finite LE-decomposition iff End, M is semiperfect, and this has become a standard tool in generalizing the Krull-Schmidt theorem to more general classes of rings.
Decompositions of Modules Over Noetherian Rings Let us pause to see how these results fit into the general study of f.g. R-modules. When R is left Artinian then R is also left Noetherian, so every f.g. R-module is Artinian and Noetherian and thus has a composition series; hence this theory applies in its entirety. When R is left Noetherian but not Artinian, every f.g. R-module M is Noetherian and thus has a decomposition of finite length. (Indeed, let M1 be a maximal summand of M, and inductively let Mi+ be a maximal summand of M i ; writing N , + , for the complement of M i + 1 in Mi we have N , < N , 0 N , < ... in M , so M = N , 0 ... 0 Nr for some t.) This raises the hope of studying R - F i + t ~ ~ oind terms of the indecomposable modules. When R is a commutative principal ideal domain this has been done classically, including the “uniqueness” part of Krull-Schmidt (cf., Jacobson [85B; $3.93); however, in general, there are easy counterexamples:
,
Example 2.9.22: (Failure of Krull-Schmidt for Noetherian rings) If L,, L, are left ideals of R with L , + L , = R then L1 0 L, x R 0 ( L , nL,) since the epic L1 0 L , + R given by (xl,xz) + x1 - x, splits (because R is projective as R-module). If we can take L,, L , not principal then these two decomand positions are inequivalent. A concrete example: Take R = ZC-1 L , = (3,2 + G)and L, = (3,2 - G).
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Levy [83] has found a commutative Noetherian ring with a f.g. module having decompositions of different length (in addition to other properties), as a byproduct of his characterization of the indecomposable Z[G]-modules where G is a finite cyclic group of square-free order. (Osofsky [70] had produced a non-Noetherian example.) We present Levy’s example below. O n the other hand, Evans [73] showed that f.g. modules over a local Noetherian ring can be “cancelled,” in the sense that M 0 N , FZ M 0 N2 implies N , z N2 cf., exercises 15, 16. In exercise 14 the idea of example 2.9.22 is used to produce a local Noetherian ring over which Krull-Schmidt fails. However, there are positive results for complete semiperfect rings, cf., Swan [60], Evans [73], and Rowen [87]. At times one can make do without the Noetherian hypothesis; although Krull-Schmidt can fail for cyclic modules over semiprimary rings (by exercise 2.7.22), Bjork [71] obtained an LE-decomposition for finitely presented modules, cf., Rowen [86].
I Supplement: Levy’s Counterexample We shall now present the highlights of Levy’s example, which relies on nothing more than the Chinese Remainder Theorem. Example 2.9.23: (Levy [83]) Fix n 2 2 and let p , , . . . , p 2 , be distinct odd prime numbers; but P, = puZ for 1 Iu s 2n. Let R = {(tl,. . . ,t n ) E Z(“): ti = t i + l (modulo p 2 i - 1 p 2 i for ) 1 I i 5 n - l}. Clearly R is a ring and is a Z-submodule of Z@); hence R is a Noetherian Z-module and is thus a (commutative) Noetherian ring.
) M ( a )= { ( m , , . ..,m,) E Z(”): Imi= For any a =(a1,. ..,a2,) E Z(’”define r n , + , ( m ~ d p , ~ - ,if) a 2 i - 1# P 2 i - l , and a 2 ~ m=i m,+,(modp,,) if a z i #P2,}. Then M ( a ) is a f.g. Z-module as seen above, and is thus a f.g. R-module; in particular, M ( 1 ) = R and M ( 0 ) = Z‘”). The module M ( 0 )0 M ( 1 ) = Z‘”)0 R has peculiar behavior, to be documented below. First we describe how to construct certain elements in M ( a ) .
,,
Method 1. Given mi as the i-th component, we shall find m , , . . .,mimi+ . . ,m, such that ( m , , . . .,m,)E M ( a ) . Clearly, if we can find suitable and m i _ l then we can continue inductively in each direction. But each of these can be found by means of the Chinese Remainder Theorem. This method becomes trivial for R = M ( 1 ) ; a, = b, = 1 so m i + l = mi k p 2 , - 1p2iand mi- = mi + k’p2,- 3p2i-2for arbitrary k, k’ in Z.In particular,
,,.
+
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(0,. ..,0,ri,O,. . .,0) E R where ri = p Z i - 3 p Z i - 2 p 2 i - lp2i appears in position i. Method 2. By the Chinese Remainder Theorem there is c in Z with c u,(mod pu)for 1 Iu c 2n - 2, and thus (1, c, c2, ...,c"- ') E M ( a ) .
=
Using these methods of generating elements, we can make a series of straightforward observations with interesting conclusions. Write a = (ul,. .., u2,) and a' = (a;, ...,u;,) in Z('"), (i) Any R-module map f: M ( a ) -P M ( a ' ) can be described as componentviewing , M ( a ) , M ( a ' ) in Z("). wise multiplication by some (zl,. ..,z,) in Z'") (Indeed, define fi: Z -+ Z as the hoped-for map obtained by restricting to the i-th position, i.e., if f(m1, . . .,m,)= (mi,. . .,mL) then fimi = mi. By method 1, one can extend any mi to an element of M ( a ) for which mi is in position i; to prove fi is a map we need merely show 1;. is well-defined, which follows if mi = 0 implies mi = 0. Suppose, otherwise, i.e., mi = 0 and mi # 0. Then for ri = P Z ~ - J P Z ~ - ~ P Z ~we - ~have P~~ 0 = f(0) = f((0,. . .,0,r i ,0,. . .,O)(m,, .. .,m,)) = (0,. ..,ri,0,.. .,O)(m;, ...,mb) = (0,. . . ,r,mi,O,.
..,O),
which is false since rim: # 0. Now, clearly, f = (fi,. . .,f,)and each fi: Z + Z is given as multiplication by zi = fil.) (ii) Suppose a = (ul,. . .,u2,) where for each i 5 n - 1 we have u Z i - 4 PZi- or uZi4 P Z i .Then M ( a ) is indecomposable. (Indeed, if f E End, M ( a ) is idempotent then f is given by componentwise multiplication by suitable (zl,...,z,) where z ; = zi for each i ; hence each zi E (0, l}. Using method 2 there is (1, c, . ..,c"-') in M ( a )where c = u,(modp,) for 1 I u I 2n - 2; then f(1, c, .. .,cn- 1 = ( Z l , C Z Z , . . . , c"-lz,) E M ( a ) . If u2i-l # PZi- then
,
u2i-lci-1zi= cizi+l (modp,,-,);
since c = u Z i - f O(mod p Z i - 1) we get ci(zi- z i +1) E PZi- so zi - z i + E P2i-1 and zi = z i + l (since zi - z i + l E (0, l } and 1 4 PZiw1). Likewise, if = Zn, SO uZi4 PZi then zi = z ~ + thus ~ ; by hypothesis we see z1 = z2 = f = 0 or f = 1. Hence End, M ( a ) has no nontrivial idempotents, proving M ( a ) is indecomposable.) (iii) Define aa' = (u,ui,. ..,u2,u;,) where a = (ul,. .., u Z n ) and a' = (u", .. .,u;,); also define supp(a) = {u: a, $ P,} E { 1,. . ., 2 n } . If each component of a and a' is 0 or 1 and if supp(a) u supp(a') = {l,. . . , 2 n } and aa' = (O,O, .. . ,0) then M ( a ) QI M ( a ' ) x M ( a a ' ) @ R. (Indeed, take c = a,
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(modp,) for 1 I u I2n, and then take c’ such that both c’ = al(modp,) for 1 5 u 5 2n and c’ = l(modp) for all other primes p dividing c. Then c and c‘ are relatively prime, so c’d’ - cd = 1 for suitable integers d, d’. For u E supp(a) we have c’ = 0 and c = 1 = dmodp,). For u E supp(a’) we have c = 0 and c’ = 1 = &(mod p,).
Thus c‘w - dw’ = - w’(mod p,) for each u E supp(a), and -cw w’(mod p,) for each u E supp(a’). Define I,:M ( a ) @ M ( a ’ ) + M ( 0 ) 0 R by
I,(u,u’) = (d’u + d d , +~c’u’) ~ = ((d’o, + d U ; , d ‘ V Z where u = ( u , , . . ., u,)
E M(a)
I, is given by the matrix
+ d’w’ =
+ dub,. . .), (cvl + c‘u;, ...)),
and u‘ = ( u i , . . .,uL) E M ( a ’ ) . In other words,
(: t,)
in SL(2, Z), whose inverse
M ( 0 )@ R M ( a ) 0 M(a’), where I,-’(w, w’) = (c’w - dw’, gives us I,-’: -CW + d’w’).). (iv) Take a = (l,O, 1,0,. ..) and a’ = (0, l,O, 1,...). Then M ( a ) and M ( a ’ ) are indecomposable by (ii). On the other hand, (iii) yields --f
M ( a ) 0 M ( a ’ ) z M ( 0 )@ R = Z(”) @ R = Z 0 * * * @ Z @ R,
so we have decompositions of lengths both 2 and n + 1. (v) We can find more decompositions of Z(“) 0 R by means of the following simple observation, where M = M ( a ) : If azi- E PZi- and a,, E PZifor some i then End, M has an idempotent f corresponding to componentwise multiplication by (1,. . .,1,0,. ..,0)where 1 appears i times. But f also acts analogously on R; clearly f M is indecomposable in f R - A u d if f M is indecomposable in R-Mud, which we can check by means of (ii). Likewise for (1 - f ) M and (1 - f ) R . One readily concludes the following result: Let G ( a ) be the graph with vertices (1,. . .,n } and with an edge connecting i and i + 1 iff aZi- 4 PZi- or aZi4 PZi. If t is the number of connected components of G ( a )then M ( a )is a direct sum of t indecomposable modules. Now let a = (l,O,l,O,..., 1,0,0,0,...) where 1 appears n - t times for I I t I n, and a’ = (0, l,O, 1,.. .,0,1,1,1,. . .). Then M ( a ) @ M ( a ’ ) = Z(”) @ R by (iii). The edges of G ( a ) connect the vertices 1,. .., n - t + 1, and the remaining vertices n - t + 2,. . .,n are isolated points; hence G ( a ) has t connected components. M ( a ) is a sum of t indecomposables (including t - 1 copies of Z); M ( a ’ ) is indecomposable. Thus Z(,) @ R has a decomposition into t + 1 indecomposables for any t between 1 and n. We shall now derive other interesting properties by generalizing (iii).
’
,
’
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(vi) Given a and a‘ define e = (e,..,,e2,) where e, = 1 ifu E supp(a)u supp(a‘) and e, = 0 otherwise. Then M ( a )8 M ( a ’ ) x M ( a a ’ ) 0 M(e). (Indeed, as in (iii) take c,c’ relatively prime such that c = a,(modp,) and c’ = a:(modp,) for each u in supp(a) u supp(a’). Let q = n { p , : u E supp(a) n supp(a’)}. For any j 2 1 there are integers d j , d; with (c’)’qd> - cjdj = 1. Note that if pulauthen q = 0 or c’ = O(mod p,) so dj E c-j(mod p,); if p , I a, and p , I a: then qd; = (c’)-j(mod p,). Define I,:M ( a )0 M ( a ’ ) -P M ( a a ’ ) 0 M ( e ) by I,(u, u’) = (w, w ’ ) where w = ( w l , . . . ,w,) and w’ = ( w i , . . .,w:) with wj = qd:+ -juj + d,+ - juJ and w’, = cn+ 1 -juj + ( c r y + -’us. To see that w E M ( a a ) note for p , in supp(a) n 1 supp(a’) that
’
where i = [$(p + l)]; a similar verification shows w’ E M ( e ) . Moreover, I,is given by componentwise multiplication by matrices in SL(2,Z),so it is easy to compute I,-’:M ( a a ’ ) 0 M ( e ) + M ( a ) 0 M ( a ’ ) ; namely, +(w, w’)= (u, u’) where uj = ( c ’ ) ” + ’ - j w j - d n + l - j w J and u‘.J = - c n + ’ - j w j + q d ; + l - j w ; , proving I,is the desired isomorphism.) (vii) If each entry of a is f 1 then M ( a ) is a projective module, because M ( a ) 0 M ( a ) x M ( a 2 )@ R = M ( 2 ) 0 R = R @ R. (viii) If M ( a ) x M ( a ’ ) then for each i I n - 1 we have ei = f 1 such that a p~2 i - 1) ~ and - aii = ~ ~ a , ~ ( mpZi). o d (Indeed, by (i) any isoaii- = ~ ~ (mod morphism f:M ( a )+ M ( a ’ ) is given by componentwise multiplication by (zl,.. .,z,) in 27”; since f-’ exists we see zi is invertible and is thus k 1. If p , divides both a, and a: then clearly a: = 0 = a, 5 -a,(modp,), so we are done unless p , does not divide a, or a:; by symmetry, we may assume pula:. Using method 1 we can construct (ml,. . .,m,) in M ( a ) with mi = 1 and mi+,=a,(modp,), where i=[(u+ 1)/2]. Writing (mi ,..., rn:)=f(m, ,..., m,) we have zi+,aU= z i + l m i + l = mi,’
= a:m:
= a:zimi = a:z,(modp,),
implying a: = Eia,(modp,) where ei = zi+ ‘zi.) (ix) “Direct sum cancellation” fails in R-Aou!, i.e., 0 M(2) = M ( 0 )8 M ( 2 ) x M ( 0 )0 R by (vi), but (viii) shows M(2) x M ( 2 ) = R . (x) Nonisomorphic modules can have isomorphic “powers.” Indeed, if supp(a) = { 1,. .. ,2n} and t = l(p, - 1) then a: = l(mod p,) for each u implying M(a)(‘)x M(2)(‘-l ) 0 M ( a ‘ )x M ( l ) ( ‘ by ) iterating (vi), so all these M ( a ) have the same power. In fact, by (vii) and (viii) we have at least 2“-’
n::
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nonisomorphic projective modules M such that M 0 M x R 0 R. Still more properties of the M ( a ) can be found in Levy’s paper.
C Supplement: Representation Theory The object here is to describe in a nutshell some of the most important and exciting module theory research of recent years. Originally I intended to go into this research in considerable detail, but as the other material expanded it became clear that an adequate treatment would require more space than this book permits. Nevertheless, some mention should be made of recent advances in the subject. The main underlying problem is the classification of the f.g. indecomposable modules of a left Artinian ring, for then by the Krull-Schmidt theorem we would “know” all the modules. Of course, one could consider this problem for even wider classes. Nevertheless, as we shall see, even the Artinian question is exceedingly complex. When R is semisimple Artinian every indecomposable module is simple, by theorem 2.4.9, and there is precisely one isomorphism class of simple module for each irreducible component, by remark 2.3.8’. Thus we have only a finite number of nonisomorphic simple modules. Definition 2.9.24: R has j n i t e representation type (f.r.t) if it has only a finite number of f.g. indecomposable modules (up to isomorphism). Thus every semisimple Artinian ring has f.r.t., and, moreover, we know how to determine all its indecomposables. (Other examples are given in exercises 20,22.) Hence by Maschke’s theorem (8.1.10 below) every group algebra of a finite group over a field F of characteristic 0 has f.r.t. When char(F) # 0 the following sort of example arises. Example 2.9.25: Let G = (2/22)(2)= { l,g, h, g h ) and let F be an algebraically closed field of characteristic 2. Then F[G] is commutative of dimension 4 over F. The homomorphism F[G] --t F given by g H 1 and h H 1 has kernel N = Fa + Fb + Fc where a = g - 1 = g + 1, b = h + 1, and c = ab = g h + g + h + 1; clearly N 2 = Fc and N 3 = 0, so F [ G ] is local with radical N . Fc is the unique minimal ideal, so any ideal of dimension 2 over F is indecomposable. Any such ideal has the form Mep = F(aa + pb) Fc for suitable a, p in F, and we shall prove there are IF1 of these by showing Mas z Me.s. iff C 1 a ’ = fl-’j. Indeed (e) is clear for then F(aa + fib) = F(a‘a + j’b). To prove (-) suppose there is an isomorphism cp: Mas + Mars..
+
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Then cp(aa + Pb) = y(a'a + P'b) for some y in F, so
+ Pb)) = ay(a'a + P'b) = P'yc acpc = cp(b(aa + Pb)) = by(a'a + P'b) = a'yc,
Pcpc = cp(a(aa
so P-lP'yc
= a-la'yc
and
implying P-'j?' = a-la' as desired.
Let us explore this example some more. The class of indecomposables constructed above all have dimension 2 over F. There is only one indecomposable of dimension 1 (namely Fc), and the question arises as to how many indecomposables there are of dimension > 2. Are there any at all? To answer this question we turn to a functor. Defniton 2.9.26: Suppose R is a finite dimensional algebra over a field F. A Nazarova-Roiter functor with number m is an exact functor q: F[A]-Aud + R - A u d (cf., definition 2.1 1.2 below) such that (i) If M = F[A]/(A - a ) for a in F then [ q M : F] = m. (ii) If q M , x q M , then M, x M,. (iii) If M is an indecomposable F[A]-module then qM is indecomposable.
Proposition 2.9.27: Suppose F is algebraically closed. If there is a NazarovaRoiter functor q with number m then for each n there are I FI nonisomorphic indecomposable R-modules of dimension mn over F. Proofi Let Ma;"= F[A]/((A - a)"), which clearly is an indecomposable + Ma;,+ given F[A]-module, and is simple iff n = 1. There is a monic Main by x + ((A - a))" H (A - a)x + ((A - a))"" and thus an exact sequence 0 + Ma;,,+ Ma;"+1 + Mail + 0. Let Na;"= ?Ma;, which by (iii) is indecomposable. Then [Nu;,:F] = m by (i), so by induction applied to the exact sequence 0 + Nu;"+ No;"+,+ Na;,-+0,we see [Nu;,:F] = mn for all n. But the Na;"are all nonisomorphic Q.E.D. by (ii), so are a set of IF1 indecomposables of dimension mn.
Example 2.9.28: A Nazarova-Roiter functor with number 2, for example, 2.9.25. Given M E F[A]-Aod define q M = M ( , ) as F-vector space, made into an F[G]-module under the action g ( x , , x , ) = ( x , , x , + x , ) and h ( x l , x Z ) = (xl, Ax, + x , ) for (xl, x,) in M(,). VeriJication: Exactness is clear since it holds as F-vector space, and (i) is also
$2.9 Indecomposable Modules and LEModules
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clear. To check (ii) and (iii) first let a = g + 1 and b = h + 1; then a(x,,x,) = ( 0 , ~ and ~ ) b(x,,x,) = (O,Ax,). (ii) If q : q M , q M , is an isomorphism then for any x in M we have cp(0, x) = cp(a(x, 0)) = acp(x, 0) = (0, y) for suitable y in M,, so define the map +: M, + M , by +x = y. Clearly is monic, and we could obtain analogously from cp-’, so it remains to show A+x = +(Ax). But by definition if +x = y then --f
+
+-’
(0, AY) = b d x , 0)) = 440, Ax) = (0, +(Ax))
proving A+x = +(Ax) as desired. (iii) Suppose M is indecomposable and qM = N’O N ” for N ‘ , N” in R - A u d . Letting n be the vector space projection onto the first component we make the following series of claims:
+
1. M = nN’ 0 nN” as F-vector spaces. (Indeed, clearly M = nN’ nN“. But if (x, y’) E N ’ and (x, y”) E N” then (0, x) = a(x, y ’ ) = a(x, y ” ) E N ‘ n N ” = 0 so x = 0.) 2. nN’ is an F[A]-module. (Indeed, suppose x E nN‘. Then Ax = x‘ x” for suitable x’ E nN‘ and x” E K N “ . Take (x,y) E N‘, (x’,y’) in N‘ and (XI’, y”) in N“. Then a(x”, y ” ) = (0, x”) = (0, Ax x’) = b(x, y ) a(x’, y ’ ) E N “ nN ‘ = 0 proving x“ = 0, and thus Ax E nN’ as desired. Likewise, RN” is an
+
+
+
F[A]-module. 3. By (1) and (2) we must have nN‘ = 0 or nN” = 0. Thus we may assume nN” = 0. Then N “ = 0. (Indeed, if (0, x) E N “ then taking (x,y’) in N ’ we have (0, x) = a(x, y ’ ) E N “ n N ’ = 0 so x = 0.) This proves cpM is indecomposable, as desired.
The Brauer- Thrall Conjectures This example leads one to the famous Brauer-Thrall conjecture, which we formulate for any left Artinian ring R, letting length (M) denote the composition length of an f.g. module M. BT I. If R does not have f.r.t. then there are f.g. indecomposables of arbitrarily large length. BT 11. If R does not have f.r.t. then there are arbitrarily large n for which R has infinitely many f.g. indecomposables of length n.
BT I was proved rather startingly for finite dimensional algebras over fields by Roiter [68], by means of certain categories of vector spaces. Auslander [74] proved BT I, as formulated here, by a completely different method.
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Auslander's techniques are a triumph for abstract category theory, since his proof was framed in an abelian category with all the properties of R - A d , but which mysteriously provides stronger results than R - A o d itself. His methods continue to produce important new results, and there is some controversy as to whether these are due principally to the power of his tools or to his genius. We shall present some of his methods in down-to-earth terms following Ringel [80] and Pierce [82B], and then indicate how this is all abstracted. Accordingly, we work in R-9imud at all times. The starting point is the Harada-Sai lemma:
Proposition 2.9.29: (Harada-Sai) Suppose M = M,, ...,Mt are indecomposable modules of length I n, and t = 2"'. If f;.:Mi --* M i + , are nonisomorphisms for 1 I i < t then f,- ...f l M has length I n - m; in particular, for m = n we have f,-
,
,
fl
= 0.
Proof;. Induction on m, the assertion being obvious for m = 1 (since f;. is noninvertible). Let u = 2"-' = t/2, and g = f,-, ...fl and h = f,-,**.f,+,. We want to prove hf,gM, has length I n - m; by induction hypothesis gM, and hM,+ each have length 5 n + 1 - m, so we are done unless we have both the following:
,
n
+ 1 - m = length gM, = length fugMl =
n
+ 1 - m = length hM,,,
=length hf,M,
= length =
Note (1) implies the restriction of hf, = 5-
f,-, fugM,.
...length h f , . . . f , M , .
,
,,
M,, x g M , @ kerhf, I
,
and
(2)
f, to g M , is monic, yielding
0 = gM, n ker hf, = f u g M , n ker h.
On the other hand, since f u g M , I f , M , 5 hf,M, = hM,+ so remark 2.8.32 yields
(1)
(3)
(2) implies hf,gM, =
M u + , = f,gM, 0 kerh.
Since Mu and Mu+ are indecomposable we see hf, and h are monic, and f u g M , = Mu+ Consequently, f, is monic and epic and thus an isomorphism, contradiction. Q.E.D.
,.
The Harada-Sai lemma is actually a generalization of Fitting's lemma. In order to utilize this result we need some way to study nonisomorphisms M N for N indecomposable, noting M is a direct sum of indecomposables. Accordingly we have --f
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Definition 2.9.30: An almost split (or Auslander-Reiten) sequence is a non-split short exact sequence 0 + K + M + N + 0 where K , N are f.g. LE-modules such that if h: N‘ + N is not split epic then h lifts to a map g: N’ + M, i.e., yielding the diagram “
Note that N cannot be projective, for then the sequence would be split. The definition is closely tied in with Ext’ as will become clear in 55.2, but we shall require only the following straightforward facts, based on exercise 1.4.21, the “five lemma” (cf., proposition 2.1 1.15 below for a more concrete treatment).
Lemma 2.9.31:
Suppose we have the commutative diagram 0-K-M
i
N-0
with K and N LE-modules, and the rows are non-split. Then g and h are isomorphisms.
Pro08 By the “five lemma” it is enough to show g is an isomorphism. Thus assume g is not invertible in End K. Then 1 - g is invertible in End K, and also, clearly, (1 - h)M C k e r j = K ; but (1 - g)-’(l - h)i = 1, so i is split monic, contrary to hypothesis. Q.E.D. Remark 2.9.32’:
The dual of lemma 2.9.3 1 is proved in the same way.
Remark 2.9.32: To use the lemma recall that if
is commutative then h(kerj) E kerj‘. In this way we can obtain g to apply lemma 2.9.31 .
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Proposition 2.9.33: The definition of “almost split” is self-dual, in the sense that a nonsplit sequence with LE-modules at the end is almost split iff it satisjes the condition obtained by reversing all the arrows in dejnition 2.9.30.
Proofi By symmetry it is enough to show any almost split sequence 0 -+ K 4 M L N -+ 0 satisfies the dual property that any h‘:K -+ K‘ not split monic extends to a map g’: M -,K ’ with h’ = g‘i. To see this we draw the diagram i f O-K-M-N-0
where P is the “pushout.” This is described in exercise 1.8.1’ and in 5.0(7) below: P = ( K ’ @ M ) / A where A = {(h’x,ix):xE K ) I K’O M , and p’,p are the maps induced by the canonical monics K ‘ -+ K ’@ M and M -, K ’ O M . The composite K ’ Q M + M L N has kernel K ’ O K I A , and thus induces an epic h: P + N with kernel p‘K‘, thereby yielding the above diagram. If h is split then p‘ is split monic, and we have the desired map M 3 P + K f extending h’. But if h is not split then by definition 2.9.30 we have g : P -+ M lifting h. Then gp: M + M is an isomorphism by lemma 2.9.31 set up using remark 2.9.32. Hence (gp)-’gp’ extends h’, as desired. Q.E.D. Proposition 2.9.34: Any almost split sequence ending in N is unique in the following sense: lf O-+K‘-+M’-*N-+O is almost split then there is the commutative diagram 0-K‘-M‘-
f’
N-0
By definition there is h’: M‘ -+ M such that f ’ = fh’. Let g’ be the restriction of h’ to K’. Then g’K G K by remark 2.9.32. Likewise, there is h”: M -,M‘ with f = f‘h“, and h” restricts to g‘‘: K -+ K‘. Letting h = h’h“ and g = g’g’’ in lemma 2.9.3 1 we see g: K -+ K is an isomorphism so g” is split monic. But K’ is indecomposable so g” is an isomorphism, implying h“ is an isomorphism by the “five lemma.” The assertion follows, where the vertical Q.E.D. arrows are g” and h”. Proofi
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Remark 2.9.34’: Dualizing proposition 2.9.34 by means of proposition 2.9.33, we thus see that any almost split sequence originating in a given indecomposable is also “unique.” Proposition 2.9.35: Suppose 0 -+ K -+ M -+ f N + 0 is nonsplit exact, and K , N are f.g. LE-modules. The sequence is almost split ifff: M -+ N is minimal in Horn(-, N ) in the sense that if f = f’f”where f’:M’ -+ N is not split, then there is n:M ’-+ M such that fn = f‘ and nj“ = 1,. (In other words, the factorization f’f”is really fnf“ = fl, and is thus “trivial.”)
Proof: (a) If f = f‘f”and f’ is not split epic then by definition f ’ = f g for suitable g : M‘ -+ M . Thus gf” is an isomorphism on M by lemma 2.9.31 (set up means of remark 2.9.32). But f = f’f”= f g f ” so taking n = (gf”)-’g we have fn = fgf”(gf”)-’g = f g = f’and nf” = 1,. (e) Suppose we have h: N ‘ --+ N not split epic. Consider M 0 N ‘ with the canonical monics p , : M -+ M 0 N ’ and p,: N ’ -+ M 0 N ’ , and define f ’:M 0 N ’ N by f ’(x, y ) = f x + hy, which is epic since f is epic. Then f = f ’pl. Furthermore, f ’ is not split for if 1 = jf’ then, since End N is local, we would see by remark 2.5.10 that either f or h is split epic, contrary to assumption. Hence the hypothesis yields some n:M 0 N ’ M such that fn = f ‘and npl = l y . Letting g = np,: N ’ + M we see f g = fnp, = f’p, = h, as desired. Q.E.D. --f
-+
Definition 2.9.36: A class of R-modules satisfies the T-condition if any chain M , -+ M, -+M , -+ of nonisomorphisms is eventually 0. (This is called “Noetherian” in the literature, but really resembles definition 2.7.32.) We say R has almost split sequences if for any f.g. indecomposable, nonprojective module N there is an almost split sequence 0 -+ K
-+
M
-+
N +O.
Theorem 2.9.37: Suppose { f.g. indecomposable modules} satisjies the Tcondition, and all f.g. indecomposable R-modules are LE. Then R has almost split sequences. Proof: Given N f.g. indecomposable nonprojective, we want to find an epic f:M -+ N with kerf indecomposable, such that f is minimal in the sense of proposition 2.9.35. First we need a trick for assuring kerf will be indecomposable: If kerf = K 0 K‘ with K indecomposable then we could
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replace f by the map f: M / K ’ + N given by f ( x + K’) = f x ; clearly kerf= (kerf ) / K ’ x K. Let f, = f, and assume K1 = kerf, is indecomposable. Using the trick, we see if f, is not minimal then there is a “nontrivial” factorization f, = f,f y where f,:M, + N (for some M , ) and ker fi is indecomposable. Inductively, given nonminimal fi: Mi + N with K i = ker fi indecomposable, we likewise obtainf;.+,,Ki+, =kerJ;.+,,andf;’:Mi+Mi+,;f;.+, isnot splitand f;’is not an isomorphism. Let gi be the restriction of f to K i . By the “five lemma” the g i are not isomorphisms, so some g,- “*gl = 0 by hypothesis. But then there is a well-defined map i : N + M, given by i( fix) = ...fyx for all x in M,, and, clearly, f , i = l N , contrary to f, not split. Q.E.D.
fr- ,
Corollary 2.9.38: Suppose R is left Artinian and every f.g. indecomposable R-module has length c n for some n. Then R has almost split sequences. Proofi
Proposition 2.9.29 provides the T-condition.
Q.E.D.
Auslander-Reiten [75] actually prove any Artin algebra (=finite dimensional algebra over commutative Artinian ring) has almost split sequences; in fact, they construct almost split sequences using the “duality” and “trace” functors. Recent work by Auslander and others has shown how almost split sequences arise naturally in integral group rings, the geometry of rings and other situations, and almost split sequences have become a subject of interest in their own right. Given an f.g. indecomposable N we can construct a possibly new indecomposable K at the other end of the almost split sequence 0 + K + M + N -,0; also we have the indecomposable summands of M.This leads us to ask how many indecomposables can be found in this manner. We shall see now there are enough to prove BT I and more. Dewtion 2.9.39: We say an indecomposable M is adjacent to N if there is an almost split sequence 0 + K + M’ + N + 0 with M a summand of M’. We can form a graph whose vertices are the f.g. indecomposables, drawing directed edges between adjacent modules. Define So= {simple R-modules} and inductively 4+,= {f.g. indecomposable R-modules adjacent to a member of 4 } .Thus we are constructing the connected components of the simple modules. Of course, ISol= number of simple components of R/(Jac(R)) for R semilocal, and thus by proposition 2.9.34 we see each -16, is finite.
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Almost split sequences only permit the study of nonprojective indecomposables. What about the projective indecomposables? Actually these are the principal projectives studied in proposition 2.9.21; there are only a finite number of these, and their connected components are finite by the Harada-Sai lemma coupled with remark 2.9.34'. We shall say an f.g. indecomposable is admissible if it is not connected to a projective indecomposable. To proceed further we need the trace module T ( M , N ) = f M : f E Hom(M, N ) } .
I{
Theorem 2.9.40: (Positive solution of B T I ) Suppose R is lejt Artinian and every f.g. indecomposable R-module has length I t. Then every f.g. indecomposable R-module is in 4 for some n I 2', and, consequently, R has f.r.t. Proof: Given nonprojective indecomposables K , N we define T,(K,N ) = fi f;.. f , K , summed over all chains of nonisomorphisms K = M , + f" M. = N where the Mi are all indecomposable}. Note if n 2 2' MI + ..*+ then T J K , N ) = 0 by the Harada-Sai lemma. Thus our result will follow from
{I
Urns,,%.
Lemma 2.9.41: Suppose K is f.g. indecomposable and K 4 Then T,,(K, N ) = N for euery admissible simple image N of K . (Note K has maximal submodules and thus simple images since K is f.g.) Proof of lemma: We prove more generally the following assertion for any admissible indecomposables K and N : Define Y,(N) = {modules isomorphic to N } and 4,, ( N ) = {f.g. indecomposables adjacent to a member of 4 ( N ) . ] Claim. If K $ UrnsnJ$,(N) then T , ( K ,N ) = T ( K ,N ) . This does generalize the lemma, for if N is a simple image of K then T ( K ,N ) = N # 0. We prove the claim by induction. For n = 0 the hypothesis K 4 9 , ( N ) means K x N, and thus T,(K, N ) = T ( K ,N ) by definition. So assume the claim holds for n - 1, and we prove it for n. Take an almost s N + 0. Write M = @:= Mi for M iindecomsplit sequence 0 + K ' + M + posable, and let xi:M + Mi be the canonical projections and p i :Mi + M be the canonical rnonics. Of course, each fpi is not an isomorphism, since, otherwise, fpi( f p i ) - ' = l,, contrary to f not split. Clearly, K $ 4, ( M i ) for each i, so by induction, T,- ( K ,M i ) = T ( K ,Mi). Suppose hE T ( K ,N ) . Since K is indecomposable, h is not split, so by definition 2.9.30, there is g:K + M such that h = fg. But g = @qg and each xig E T ( K ,M i ) = T. l(K, Mi); hence h = f g E T,(K,M i ) as desired. Q.E.D. ~
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Basic Structure Theory
Implicit in this proof is the construction of nonisomorphisms f:Mi -,N, which cannot be factored as f = f If" unless f ' is split epic or f " is split monic; such maps f are called irreducible. We saw, in general, that irreducible maps arise from an almost split sequence 0 + K + M + N + 0 (provided it exists), and thus there are only a finite number of irreducible maps (up to natural isomorphism) ending in a given indecomposable N.Dually, there are only a finite number of irreducible maps beginning with a given indecomposable. Auslander-Reiten's explicit description of almost split sequences readily leads to
Lemma 2.9.42: If R is an Artin algebra there is m = m(R) such that for any irreducible f:M + N with M, N f . g . indecomposable we have m-' length(N) I length(M) I m length(N). The proof of the lemma is given in Ringel [80, p. 1121. The contrapositive of the Harada-Sai lemma says that if there is a chain of length t = 2" of nonisomorphisms between indecomposable modules Mi whose composition is nonzero then some Mi has length > n. On the other hand, if we stipulate that the chain is of irreducible maps and ends in an indecomposable of length Id, then length(Mi) 2 t"d by lemma 2.9.42. Thus for each nontrivial chain of irreducible maps of length t, we have an indecomposable of bounded length > n. Using this information it is a simple matter to prove
Sesqui-BT (Smald): If there are an infinite number of (nonisomorphic) indecomposables of length I d then there are an injinite number of indecomposables of length d' for an infinite number of d' > d. Proofi Clearly, we need find only one such d', for then we can repeat the procedure taking d' in place of d. Take { K :i E N} admissible indecomposable of length I d, and take an irreducible map Mil + Ni. Throwing out duplications, we may assume the Mil are all nonisomorphic and admissible. Proceeding inductively we build chain of irreducible maps of length 2d + 1, where for each j the Mi, are nonisomorphic; each has an indecomposable of bounded length > d, so there are an infinite number having some length d' > d. Q.E.D.
Using sesqui-BT I, we see'that to prove BT I1 it is enough to show there is some d for which there are an infinite number of indecomposables of length I d, but this seems to be difficult to demonstrate in general; see Ringel
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[81,@2.5]for positive results. Nazarova-Roiter [71] proved BT I1 for finite dimensional algebras over an algebraically closed field F by proving the existence of a Nazarova-Roiter functor and then appealing to proposition 2.9.26; Ringel extended this to the case F is a perfect field.
Quivers andf.r.t. Possibly the most important development in the theory of finite dimensional algebras was the introduction of the quiver r = T(R) of an algebra R, which is defined as the directed graph whose vertices correspond to the idempotents of R, and which has an edge from e, to ej iff Hom(Rei,Rej)# 0; see exercise 2.7.11 for equivalent conditions. To secure an undirected graph which will contain this information, we define the separated quiver rs = Ts(R), whose vertices are taken from the disjoint union of two copies of r-thus we denote vertices as ordered pairs (e,,O)or (ei,1) where e, E r; by definition, an (undirected) edge exists in rsfrom (e,,0) to (ej, l), whenever there is a directed edge in r from e, to ej. Gabriel [72] proved the following striking result: “Gabriel’s theorem”: Suppose R is a Jinite dimensional algebra over an algebraically closed field, and Jac(R)’ = 0. R has f.r.t. iff P ( R ) is a disjoint union of Dynkin diagrams of types A , , D,, E , , E , , or E , . Dynkin diagrams had become well-known through the classification of semisimple Lie algebras, cf., Jacobson [62B], but their entry into the representation theory of associative algebras was a profound surprise. The solution to this sublime mystery was given in a beautiful reworking of the proof in terms of quadratic forms, by Bernstein-Gelfand-Ponomarev [73]. This approach connects the theory of finite dimensional algebra to the theory of the Weyl group and opened the gateway to the theory of quivers. Any quiver can be viewed as a small category whose objects are the vertices and whose morphisms are the paths between edges; composition is the juxtaposition of paths. The category algebra of example 1.2.21 is then available, now called the “quiver algebra.” Quiver algebras take on a key role in the structure theory, since one sees easily that every algebra is a homomorphic image of a quiver algebra in the natural way. A similar route leads one naturally to a representation theory of quivers, where a representation p of r consists of a finite dimensional vector space p(u) for each vertex u, together with a vector space map p(u) t p(w) for each arrow u + w. The vector (dim p(u)) is called the dimension vector of p, and plays a fundamental role, cf., Kac [84], Ringel [84B], and Schofield [86].
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The Multiplicative Basis Theorem and the ClassiJication Problem Using the tools described above there now exists a systematic approach to the classification problem! The machinery hinges on algebras which are “minimally” of infinite representation type and the extended Dynkin diagrams. Note that certain initial reductions can be made by assuming the algebra cannot be written as a direct product of two algebras (i.e., there are no nontrivial central idempotents) and that the algebra can be replaced by its basic algebra, cf., 62.7; thus R is its own basic algebra. We shall henceforth make these two assumptions implicitly, as well as assume F is algebraically closed. Besides Gabriel’s theorem, there are certain fundamental old facts about f.r.t. algebras: (i) The lattice of ideals is distributive (Jans [SO]). (ii) If e, f a r e primitive idempotents of R then the algebra eRe z F [ L ] / ( L ” ) (called a truncated polynomial algebra), and eRf is either cyclic as eRemodule or is cyclic as rightfRf-module (This is due to Jans and Kupisch). Building on these results led to the following startling result: MultiplicativeBasis Theorem: (Bautista-Gabriel-Roiter-Salmeron [SS]) Any F-algebra of f.r.t. has a base B = { b l , .. .,b,} over F which is multiplicatively closed (i.e,,bibj E B u {0} for all bi in B ) ! Actually the multiplicative basis theorem holds also for algebras which are “minimally” of infinite representation type and thus can be used as a tool to prove BT 11. Examples of algebras with multiplicative bases include M,(F) and group algebras, which, in view of example 2.9.25, need not have f.r.t. Nonetheless, the multiplicative basis theorem is certainly a deep, striking result with powerful repercussions throughout the theory. (The reader should note that the multiplicative basis theorem cannot be generalized to F not algebraically closed. Indeed any division algebra with a multiplicative basis must needs be a group algebra, and thus must be F itself, since any other group algebra has a nontrivial augmentation ideal!) Several caveats before the reader delves into the literature: (i) Duality is a key feature of the subject. Since the dual of projective is injective, to be treated in the next section, one should certainly know about injective modules. (ii) Much of the theory is developed via the homology of abelian categories.
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The necessary background could be obtained by “abstracting” Chapter 5. Basic tools are the “five lemma,” the pushout and pullback, projective and injective resolutions, and Ext. In particular, rings of global dimension 2 play a key role in the theory.
The Categorical Language The bulk of the research about representation type is in the language of abelian categories. In order to translate into this language one tries to take properties which are most arrow-theoretic. Let W denote any additive category and 9 - M u d the category of additive functors Wop + d&. (The morphisms of W-Mudare the natural transformations.) The obvious example of such 9 is a ring R, viewed as a small category. However, there are other categories which are “skeletally small,” such as W = R - T i ~ u d .In fact, we “think” of W as a ring but apply it as some category of modules. A morphism f:A + A is idempotent if f’ = f. Let W be a category in which each idempotent morphism is split. We say an object P of R-.,Mod is f.g. projectiue if P (as a functor Bop+ d&) has the form Horn(-, A) for some object A of 9. Yoneda’s lemma thus says A H Horn(-,A) gives an equivalence from 9 to a full subcategory of W-.Mud. An object M of R - A u d is finitely generated (f.g.) if there is an exact sequence P + M -+ 0 for some f.g. projective P; M is finitely presented if there is an exact sequence PI 4P2 -+ M + 0. For the reader’s convenience let us give a concrete example of B. Let C be a “good” commutative ring. (Often C is a field; Auslander usually takes C to be Artinian, although sometimes C is a complete local Noetherian domain.) R is a C-algebra which is f.g. free as C-module, and W often is {f.g. R-modules which are free as C-modules}.
52.10 Injective Modules In category theory one often is led to dualize important concepts with the hope that the new notions will also turn out to be useful. This strategy has been quite successful at times. (The reader should be warned, however, that any strategy followed blindly can lead to disaster, and many pages in the literature have been wasted in this manner.) The concept dual to projective modules is in fact one of the most important in module theory. Defiition 2.10.1: An R-module E is injective if it satisfies the following property dual to the lifting property. For any map f:N + E and any monk h:
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N + M there is a map j M + E such that f = f h , i.e., f completes the following diagram (not uniquely): h
0-N-M
In this case we say f extends f. Remark 2.10.2: E is injective iff for every monic h: N + M the map h‘: Hom(M, E) + Hom(N, E) given by g H gh is onto. Remark 2.10.3: Any map f:R + E is given by right multiplication by an element of E (namely fl). In particular, if E is injective and L is a left ideal of R then any map f:L + E is given by right multiplication by some element of E (seen by first extending f to a map f R + E). This property actually characterizes injective modules, as we now see.
Proposition 2.10.3’:
(“Baer’s criterion”) E is injective ifl for euery left ideal L of R, each map g : L+ E can be extended to a map 6: R + E. Proofi (a) a fortiori. (I) Suppose f:N + E and N 5 M; we want to extend f to M. Let 9 = {(N’,f’): N 5 N ‘ I M and f is the restriction of f’to N}. We make 9 a poset by putting (N’,f‘) I(N”,f”) if N’ I N” and f‘ is the restriction off” to N’. Clearly Y is then inductive, and thus has some maximal (No,fo). We shall conclude by proving No = M. Otherwise, there is x in M - N o . Let L = { I E R: rx E No} I R, and define g : L + E by ga = fo(ax).By remark 2.10.3, g is given by right multiplication by some x’ in E. Define f,:No + Rx + E by f , ( y + rx) = foy + rx’. Then f, is well-defined, for if y , + r l x = y 2 + r2x E No + Rx then (r2 - r l ) x = y 2 - y , E No so ( r , - r 2 ) E L and fl(Y1
+ r1x) - f l ( Y 2 + r2x) = (fob + r1x’) - ( f o Y 2 + r2x’) = fo(Y1
- Y 2 ) + (rl
= fo(Y1
- Y 2 ) + gb-1 - 1 2 )
= fo(Y1
- Y z + (r1 -
- r2)x‘
Thus f,is a map extending fo, contrary to the choice of fo.
= f00 = 0.
Q.E.D.
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52.1 Injective Modules
Certain properties of projective modules (e.g., proposition 2.8.3 with “free” replaced by “projective,” and lemma 2.8.4) could have been formulated and proved in any abelian category %? satisfying the axiom that for each object C there is an epic f P -+ C with P projective. (In fact, one can take P to be free but we want to stay in a categorical framework.) We shall say such a category has enough projectives. Then the dual results would apply automatically to injectives, provided %? has enough injectives in the sense that for any object C there is a monic g : C + E with E injective. Although we shall reprove these results (2.10.13 through 2.10.15) from scratch, we see that the key must be to show every module is contained in an injective module. This is surprisingly difficult to show, but the rewards are great, since we shall end up with a theory stronger than the dual of #2.8.
Divisible Modules In order to study injective modules further, it is convenient to introduce a related elementary concept. Definition 2.10.4: An element x in M is divisible by r in R if rx‘ = x for some x’ in R . M is divisible by r if each element of M is divisible by r, i.e., if left multiplication by r is an epic from M to M . Finally M is a divisible module if M is divisible by every regular element of R . Proposition 2.10.5:
Every injective module E is divisible.
Proof: Suppose r E R is regular, and x E E . There is a map f Rr + E given by f ( r ’ r ) = r‘x, which is well-defined since rlr = r2r implies r l = r 2 . But f is given by right multiplication by some x’ in E, so x = fr = rx’ proving divisibility. Q.E.D.
There is a partial converse. Proposition 2.10.5‘:
Any divisible module over a PLID is injective.
Proof: (By Baer’s criterion). We shall extend a given map f:L + M to all of R , where L < R. By hypothesis L = Ra for some a in R. If a = 0 then f = 0 and we are done. If a # 0 then a is regular so fa is divisible by a, and there is some x such that ax = fa. Fixing x, define R + M by fr = rx. Then f ( r a )= Q.E.D. rux = rfu = f (ru)for all r in R, as desired.
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The advantage of moving back and forth from injective to divisible is that divisibility passes easily to other modules:
Remark 2.10.6: (i) Every homomorphic image of a divisible module is divisible. (ii) Every direct product of divisible modules is divisible. (iii) Every direct sum of divisible modules is divisible. (Verifications are immediate from the definition.) Example 2.10.7: If R is an integral domain then its field of fractions Q is divisible as R-module, so Q/R is also a divisible R-module. If R is a PID then Q/R is injective by proposition 2.10.5’; in particular, Q/Z is an injective Z-module. Example 2.10.8: Any Z-module M can be embedded in an injective Zmodule. Indeed, write M = F / K where F is a free Z-module with some base {bi:i E I}. Then taking F‘ to be the free Q-module with base {bi:i E I} (ie., F‘ = F B nQ) and viewing F c F’, naturally we have F’ divisible so F ‘ / K is divisible (and thus injective) as Z-module, and M c F ‘ / K , naturally.
To proceed further we need a basic fact relating homomorphisms and tensor products. Suppose A E R-Aud-S, B E s-Mud-T, and C E R-Aud- W. Using remark 1.5.18’ we have two very natural ways of forming a T - W bimodule. (i) HOmR(A, c)E S - J h d - W SO Homs(B, HOmR(A, c))E T-.dhd-w. (ii) A 0 s B E R-Aod-T SO HOmR(A @ B, C) E T-&ud- W.
In fact, these two constructions can be identified: Proposition 2.10.9: isomorphism
(“Adjoint isomorphism”) Notation as above, there is an
0:HOmR(A
B, c)+ Homs(B, HOmR(A,c))
in T - J h d - W , given as follows: For f in Hom(A BSB, C) and b in B, define A + C by fba = f(a @ b). Then @f is defined by (@f)b = fb.
fb:
Pro08 It is easy to check 0 is a T- W bimodule map, using remark 1.5.18’. To construct 0-l suppose g:B+Hom(A,C). Define 5 : A x B + C by $(a, b) = (gb)a. This is balanced, so 5 induces a map A 8 E + C which we Q.E.D. denote Jlg. Clearly, IJ and 0 are inverses.
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The phenomenon described above will be framed quite generally when we study adjoint functors; this proposition is used throughout the theories of projective and injective modules. Our current interest is to prove the following result: Theorem 2.10.10: Every R-module M can be embedded in an injective R-module. Proof: By example 2.10.8, M can be embedded into an injective Z-module G, which as an abelian group is a Z - Z bimodule. Viewing R E Z - d u d - R naturally, let E=Hom,(R,G)ER-dud by remark 1.5.18’. We have a series of monks M z End,(R, M) + End,(R, M) -+ End,(R, G) = E, so it remains to prove the following fact: Proposition 2.10.11: If G is injective in Z - A u d then E = Hom,(R,G) is injective in R - d u d . Actually, we shall prove the following more general result: Proposition 2.10.12: If G is injective in Z - A u d and F is a Jlat right R-module then E = Hom,(F, G ) is injective in R - A o d (viewing F in Z - d u d as an abelian group). Proof: Suppose h: N -+ M is monic in R - d ~ dBy . remark 2.10.2 we should show the canonical map h’: Hom(M, E ) -+ Hom(N, E ) is epic. But 1 0 h: F 0, N -+ F 0, M is monic since F is flat, and we have
HOm,(M, E ) = H0mR(M,Hom,(F, G)) z Hom,(F
O RM, G).
But Hom,(F 0 M, G) + Hom,(F 0 N, G) is epic by the Z-injectivity of G; we easily translate this to h# being epic. Q.E.D. This argument can be given more precisely in terms of “exact functors,” which we shall describe later, We now have “enough” injectives to dualize the projective theory and more. Lemma 2.10.13: Every summand of an injective module is injective. Proof: Suppose E= E,@E2 is injective, and let n:E-E, be the projection.
Any map f : L - E , when composed with the inclusion map E,+E is given by right multiplication by some x in E. But then f is given by right multipliQ.E.D. cation by nx, proving E , is injective by Baer’s criterion.
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Proposition 2.10.14: E is injective ifl every monic h: E + M splits. Proof;. (3)Taking N = E and f = 1, we have as in definition 2.10.1 proving h is split monic.
j M + E with
1, = f h ,
(e) Let E' be an injective module containing E. Then the inclusion map E + E' splits, proving E is a summand of E' and by the lemma is injective.
Q.E.D.
n
Remark 2.10.15: I f {Ei:i E Z } are injective then Ei is injective. (Indeed, given a monic h: N+M and a map f: N + n Ei write f =(A) with fi: N+Ei, i.e., f x = ( f i x ) ; extending each fi to M + Ei let f =
i:
(i).)
Essential Extensions and the Injective Hull We turn now to the dual notion of projective cover, called the injective hull, which turns out to be much more important (because it always exists!) We continue the discussion of large submodules, initiated in 2.4.2ff. Definition 2.10.16: I f N is a large submodule of M we say M is an essential extension of N , or M is essential over N. Remark 2.10.17: Suppose N < N' c M . M is essential over N, iff M is essential over N' and N' is essential over N. (Immediate from definition 2.4.2.) Remark 2.10.18: If N ' is an essential complement of N in M then M I N ' is an essential extension of N. (Indeed, N x ( N + N ' ) / N ' I MI". To prove N is large in MI", suppose N' I K I M with ( K I N ' )n N = 0. Then K n ( N + N ' ) I N ' so ( K n N ) n ( N + N ' ) I N n N ' = O ; by hypothesis, N + N ' is large in M so K n N =0, implying K = N' by maximality of N ' , so K/N'=O.)
Lemma 2.10.19: (Extension lemma) Suppose f: M+E is monic and E is injective. Zf N is an essential extension of M and we take j N 4E extending f, then falso is monic. Proof;. Mnkerf=O, so kerf=O since N is essential over M.
Theorem 2.10.20:
The following are equivalent for modules M I E:
(i) E is a maximal essential extension of M. (ii) E is injective and is essential over M . (iii) E is a minimal injective containing M.
Q.E.D.
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267
Moreover, given any module M one can find E satisfying these equivalent properties. Proofi (i)*(ii) Take E ’ 2 E injective and let N be an essential complement of E in E’. Then E ’ / N is essential over ( E + N ) / N x E, so the hypothesis implies equality holds, i.e., E‘ = E 0 N so E is injective, yielding (ii). Next we show there exists E satisfying (i). Take E’ injective containing M . {essential extensions of M contained in E ’ } is inductive and non-empty (since M is essential over M ) and so has a maximal element E. To see E satisfies (i) suppose N 2 E is essential over M . By the extension lemma, the natural inclusion map E + E’ extends to a monic i : N + E’. Thus we may assume N IE‘, so E = N by choice of E. Hence E satisfies (i) and thus (ii), so E also is injective. (iii) * (i) Take E , maximal in {essential extensions of M contained in E } . We just saw E, satisfies (i) and is injective, so E , = E by hypothesis. (ii)*(iii) Suppose M I E, IE with E, injective. Then the inclusion E , IE splits so E = E , @ N for some N IE; since N n M = 0 and E is essential, Q.E.D. we conclude N = 0, i.e., E = E,. Definition 2.10.21: The module E of theorem 2.10.20 is called an injective hull or injective envelope of M .
Property (ii) shows us the injective hull is the dual of the projective cover. However, whereas arbitrary projective covers do not necessarily exist (unless the ring is perfect), the injective hull of M always exists, and its use pervades ring theory. We shall now show the injective hull is unique. Proposition 2.10.22: The injective hull E of M is unique up to isomorphism. In fact, if E’ is another injective module containing M then the inclusion map i: M + E’ extends t o a monic E + E’.
Proofi The second assertion follows from lemma 2.10.19; the first assertion Q.E.D. then follows from theorem 2.10.20 (iii) applied to E’. Remark 2.10.23: Writing E ( M ) for the injective hull of M , one has the following properties:
(i) E ( N ) = E ( M ) for every large submodule N of M . (ii) E ( M ) = M iff M is injective. (iii) E ( M , 0 M , ) = E ( M , ) 0 E(M,).
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(Proofs: (i) E ( M ) is an essential extension of N, so use theorem 2.10.20(ii), which also yields (ii); to see (iii) note E ( M , ) @ E ( M , ) is an essential extension of M ,Q3 M , and is injective.)
Criteria for Injectivity and Projectivity There are several useful equivalent conditions for a module to be injective or projective. The first one is important in the study of quotient modules and rings, and improves Baer's criterion.
Remark 2.10.24: Suppose L I M and f: L + N is an R-module map. If L' is an essential complement of L in M then f extends to a map L 0 L' -t N by h 1 , x z ) = fx,. Proposition 2.10.25: E is injective iff for every large left ideal L of R, every map f: L + E extends to a map R -t E.
f
froofi Apply remark 2.10.24 to Baer's criterion.
Q.E.D.
Proposition 2.10.26: E is injective if every short exact sequence 0 -t E -t M + C -,0 with C cyclic, splits.
froofi Otherwise, let E' be the injective hull of E and let E"/E be a cyclic submodule of E'IE. By hypothesis, the inclusion map E+E" splits, contrary to E' being essential over E. Q.E.D. Our final result along these lines links projective modules to injective modules.
Proposition 2.10.27: A module P is projective iff for each epic h: E -t N with E injective, every map f: P -+ N can be lifted t o a map f P + E (such that f = hf).
Proofi (a) is by definition. Conversely, suppose h: M -t N is epic; we want to show any map f: P - t N lifts to M. Embed M in an injective module E. Then N x M/ker h is a submodule of Elker h; we may view f:P + E/ker h, which by hypothesis we can lift to P E. But, in fact, j P E M ; indeed, if x E ~ P then hx E h j P = f P E N so hx = hx' for some x' in M , implying x - x' E ker h E M and thus x E M . Therefore we have lifted f. Q.E.D.
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269
J Supplement: Krull- Schmidt Theory for Injective Modules Injective modules have a very pleasant decomposition theory. Our first proposition is reminiscent of the Shroder-Bernstein theorem of set theory. Lemma 2.10.28: Suppose M = K @ N and there is a monic f:M -+ K . Then { f ' N :i E N is an independent set of isomorphic submodules of M . Proofi We shall show by induction that N , f N , . . .,f " N are independent for each n. This is clear for n = 1. Given the induction hypothesis for n note f N , .. . ,f N are independent since f is monic; also "
+
n+ 1
N n x f'NINnfM=O, i=l
implying N , f N , .. . ,f
"
+
' N are independent, as desired.
Q.E.D.
Proposition 2.10.29: lf E l , E , are injective and each isomorphic to a submodule of the other then El x E , . Proofi We use repeatedly the fact that an injective submodule is a summand. El = E ; @ N where E , x E ; , and there is a monic f: El + E ; . Thus { f ' N :i E N} is independent. Let A = @ ,: f ' N IE ; ; A = fA @ N . E ; is injective and thus contains an injective hull B of fA. Hence the injective module E O N is an injective hull of f A @ N = A . Since A % fA we have B @ N x B. Writing E ; = B' 0 B we get
El = E ; @ N = ( B ' @ B ) @ N
=B'@(B@
N ) % B ' Q B = E; x E,. Q.E.D.
Next we have the exchange property, which is treated most easily by means of a useful notion. Definition 2.10.30: A submodule N of M is closed if N has no proper essential extension in M.
Lemma 2.10.31: Any closed submodule N of an injective module E is injective. Proofi By theorem 2.10.20 it suffices to show N is maximal essential. But by lemma 2.10.19, any essential extension of N is isomorphic to a submodule Q.E.D. of E and thus equals N by hypothesis.
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Basic Structure Theory
Proposition 2.10.32: If E = El 0 E2 = E ; @ E; then there are decompositions El = M , @ N , and E , = M , 0 N , such that M , @ M , x E; and N, 0 N, x E;.
Proofi Let ni: E + Ei be the usual projections for i = 1,2. Let Mi = El n E', and M , = n,E;. M , is closed in El, for if M , is large in N IEl then clearly E', is large in N + E ; so N c E', and thus N E El n E; = M I . Hence M , is injective by lemma 2.10.31, so the exact sequence 0 + M 1 + E', + M z + 0 splits, proving E', x M , 0 M , . Thus for i = 1, 2 we see Mi is injective and so has a complement 4in Ei. Since M , = (1 - n , ) E ; IE; + El we have
+
Let K = E; n ( N , + N,). If we can show K = 0 then N , 0 N 2 x (E; N , + N,)/E; = E / E ; x EL as desired. But n 2 K In 2 E i n n,(N, N , ) = M,nN,=O, and so K I E , n E ; = M , . Hence n 1 K I M l n n l ( N , + N 2 ) I MI n N , = 0, as desired. Q.E.D.
+
Theorem 2.10.33: Suppose E is a finite direct sum of indecomposable injectives El @ ... @ En. Then every decomposition of E is equivalent to this decomposition. Proofi Use proposition 2.10.32 in the argument of theorem 2.9.17.
Q.E.D.
82.1 1 Exact Functors In this section we continue the study of projective and injective modules by tying them in with exact sequences, using the very important idea of exactness of functors. This leads us to characterize various kinds of rings and modules in terms of exact sequences,especially using flat modules. The motivation springs from some easy observations about R-Aod; the interested reader will have no difficulty in seeing how to generalize to arbitrary abelian categories. A functor F is additive if for all maps f,g: M + N we have F ( f + g ) = F f + F g . Then if M = M , @ M , we must have F M = F M , @ F M , . Indeed, if ni: M + Mi are the projections and pi:M i+ M are the canonical monics for i = 1, 2 satisfying nipi = l,, and plnl + p2n2 = l,, then FniFpi = lFM,and FplFnl + Fp2Fn2 = 1 F M . We have proved
271
82.1 1 Exact Functors
Remark 2.11.1: Any additive functor F: R-&ud + W-Aud preserves split f M 3 N + 0 is split exact then exact sequences, in the sense that if 0 + K + so is O + FK-% F M S F N +O.
What happens when the given sequence merely is exact? To handle this situation we formulate the following definition.
-
Definition 2.11.2: A functor F:R-&od + W-Aud is exact if whenever f M 4 N is exact we have FK FS F M 3 F N is exact. K+ Piecing together this definition, we see an exact functor “preserves” the exactness of every exact sequence. However, it is convenient to focus on short exact sequences.
-
Proposition 2.11.3: A functor F is exact Ifs whenever 0 + K is exact then 0 + F K Ff F M FN + 0 is exact. Proof:
s
+M
3N
+0
(=.) from the above discussion. (e) One can cut a given exact
1M sequence K +
3N
into two sequences: f O+ker f -,K +fK -0 0 + kerg + M
4 gM + 0
Applying F to each of these sequences shows ( F f ) F K = F( flu) = F(kerg) = ker(Fg) as desired. Q.E.D. This criterion is useful because it usually is not satisfied! This paradox will become clear when we introduce another definition:
-
f M 3 N exact Definition 2.11.4: F is a left exact functor if 0 + K -, FS F M Fg\ FN is exact; F is right exact if K M 3 N +0 implies 0 + F K 3 exact implies F K Ff F M - Fg F N + 0 is exact. s M 3 N + o exact implies A contravariant functor F is left exact if K + Ff o + F N 3 F M -+ F K is exact. Viewing Hom(A,-) and Horn(-, A) as functors from R - d u d to Z - d u d we have the following important examples. Proposition 2.11.5: Hom(A,-) is a left exact functor and Horn(-,A) left exact contravariant functor for any R-module A.
is a
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Basic Structure Theory
Proofi We show the contravariant functor F = Horn(-, A ) is left exact; the other verification is analogous. Suppose K J M 4 N + 0 is exact. If h E FN = Hom(N, A ) with 0 = (Fg)h = hg then h = 0 since g is epic, proving o -+ FN Fg, F M is exact. TO see FN Fg, F M - FfF K is exact we first note (Ff ) ( F d = F ( g f ) = 0. It remains to show that if h: M + A with (Ff ) h = 0 then h = h’g for some h’: N + A. To obtain h’ write any x in N as gy for suitable y in M , and define h’x to be hy. To see h’ is indeed well-defined suppose gy’ = gy = x; then y - y’ E kerg = fK so h ( y - y’) E hfK = ( ( F f ) h ) K = 0, implying hy = hy’ as desired. Q.E.D. Corollary 2.11.6: Hom(P,-) is an exact functor iff P is a projective module. Horn(-, E ) is an exact contravariant functor iff E is an injectioe module.
Proofi Proposition 2.8.3 and remark 2.10.2, coupled with proposition 2.1 1.5. Q.E.D. We can use these results to characterize semisimple Artinian rings in terms of exact functors. Remark 2.11.7: Recall a ringR is semisimple Artinian iff every short exact sequence in R-Aud splits. It follows at once that R is semisimple Artinian iff every R-module is projective, iff every R-module is injective. Proposition 2.11.8: R is semisimple Artinian ifl for every ring W,every additive functor F : R-Aod + W - h d is exact.
Proofi (*) Every short exact sequence in R - A d splits, implying F is exact by remark 2.1 1.1 and proposition 2.1 1.3. (-=) Take W = Z. By hypothesis, for every R-module P, Hom(P,-) is exact so P is projective; hence R is semisimple Artinian by remark 2.1 1.7. Q.E.D. Using Horn(-,E) one could prove the analogous assertion for contravariant functors. One can also define functors by means of the tensor product. Proposition 2.11.9: For A E R-.Mod-T define A 0 -: T-Aod + R - d u d sending M to A OTM ; deJine -@ A : A d - R + Aod-T by M + M g RA.
273
$2.1 1 Exact Functors
Then A 0 - and -Q A are right exact functors, where one sends f: M respectively, to 1 0 f : A O M + A @ N and f'Q 1 : M Q A + N Q A.
+N,
Proof: These are clearly functors, and it remains to prove right exactness. Suppose K + f M 3 N + 0 is exact; we aim to show A 0 K SAQ M %A 0 N + 0 is right exact. (The other verification is analogous.) Clearly, ( 1 0 g ) ( A Q M ) = A Q gM = A Q N. Likewise, ( 1 0 f)(A 0 K) = A 0 fK = A 0 kerg, which. is ker(1 0 g) by proposition 1.7.30. Q.E.D. Corollary 2.11.10: a .flat R-module.
The functor - 0 A of' proposition 2.1 1.9 is exact iff A is
Proof: The missing part of exactness is precisely the definition of flatness. Q.E.D. Remark 2.11.10': Let us analyze this in a particular case. Let A 4 R and F: R - A d + ( R / A ) - A d be given by M + M / A M . By example 1.7.21' this functor is given by tensoring, and thus is right exact. On the other hand, for F to be exact we must have RIA flat as right R-module. Thus we see projective, injective, and flat each play a role in making a suitable functor exact.
Flat Modules and Injectives We shall continue the study of flat modules by relating them to injectives. The key here is the character module M # = Hom,(M, Q/Z). Besides being injective in Z-Jtiod, Q/Z satisfies the following useful property: Remark 2.11.11: If G is an arbitrary abelian group and g E G then there is a group homomorphism cp: G + Q/Z with cpg # 0. (Indeed, by injectivity of O/Z in Z - A o d it suffices to take G = Zg. If g has finite order t define cp: G + O/Zby cp(ng)= n/t + Z;if g has infinite order define cp by cp(ng) = n/2 + Z).
1M Lemma 2.11.12: A sequence K -, K # is exact.
3 N of maps is exact iff N # 3 M#C
Proof: (-) The contravariant functor Horn,(-, injective.
Q/Z) is exact since Q/Z is
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274
(e) Take any k E K. For every cp: N + Q/Z we have cp(gfk)= (f ’g’cp)k = 0; by remark 2.1 1.1 1 we conclude gfk = 0. Thus f K E ker g. Now given cp: M/fK + Q/Z we define Jl: M + Q/Z by putting Jlx = cp(x + f K ) . Then 0 = Jlf = f ’ J l so Jl = g’p = p g for some p in N’, implying Jl(kerg) = 0. This proves cp((kerg)/fK) = 0 for any such cp, implying (kerg)/fK = 0 by remark 2.11.11, and kerg = f K . Q.E.D.
Theorem 2.11.13: An R-module F is pat i f fF’ is injective in .Mod-R. Proof:
(a)By
proposition 2.10.12. (e) Any monk h: N + M in A u d - R induces an epic HomR(M,F’) + HomR(N,F’) which by the adjoint isomorphism (2.10.9) translates to an epic (M@ F)’ + (N 0 F)’, which in turn provides the monic h @ 1: N @ F + M @ F by lemma 2.1 1.12. Q.E.D. We are finally ready for a useful intrinsic characterization of flatness.
Theorem 2.11.14 The following conditions are equiualent for an R-module F: (i) F is pat. (ii) For any right ideal I of R, the canonical map Jl: I @ F + R @ F is nwnic. (iii) For any right ideal I which is f.g. (as right R-module), the canonical map 10 F + R 0 F is monic. (iv) For any right ideal I the map cp: I 6 F -,I F giuen by cp(a @ x) = ax is monic (of course, cp is always epic.) (v) As in (iu),for any f.g. right ideal I. (i) 3(ii) by definition. (ii) * (i) ( R @ F)’ + ( I @ F)’ is epic so the adjoint isomorphism shows HOmR(R, F’) + HomR(l,F’) is epic, i.e., F’ is injective by Baer’s criterion, so F is flat. (ii) * (iii) is a fortiori; conversely, (iii) (ii) since any right ideal is a direct limit of f.g. right ideals, enabling us to use proposition 1.8.10. (ii) (iv) R BRF sz F = RF so we consider the commutative diagram Proof:
-
-
which shows $ is monic iff cp is monic. (iii) o (v) As in (ii) o (iv). Q.E.D.
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275
Condition (v) is sometimes called the “flat test” and can be used to provide yet another criterion for flatness, by turning to a technique of great interest in the sequel.
Proposition 2.11.15: (“The 5 lemma”)Suppose we have the following commutative diagram of module maps in which the rows are exact:
f
A-B-C-D-E
9
h
i
(i) If a is epic and and 6 are monic then y is monic. (ii) If p and 6 are epic and E is monic then y is epic. (iii) If every vertical arrow but y is an isomorphism then y is also an isomorphism. The technique of proof is called diagram-chasing, and the proof actually shows what is needed in the hypothesis.
Proo$
(i) To show y is monic, suppose yc = 0. Then 0 = h’yc = 6hc; 6 monic implies c E ker h = gB. Writing c = gb we have 0 = yc = ygb = g’pb so pb = f’a’ for some a’ in A’; writing a’ = aa yields pfa = f’aa = f’a‘ = pb, so f a = b and c = gb = gfa = 0. (ii) To show y is epic suppose c’ E C‘. Then h’c‘ = 6d for some d , and 0 = j‘h‘c’ = j’6d = Ejd, implying j d = 0. Hence d = hc for some c. Now h’(c’ - yc) = h’c’ - h’yc = 6d - 6hc = 0 so c’ - yc = g‘b‘ for some b’. But b’ = pb for suitable b in B,so y(c + gb) = yc + g’pb = yc + g‘b’ = c’, proving c‘ E
yc.
(iii) Combine (i) and (ii).
Q.E.D.
Corollary 2.11.16: Suppose we have the following commutative diagram in which the rows are exact, and p is an isomorphism. K-M-N-0
Then cp is epic iff I) is monic. Proof: (a) Take A = K, B = M , C = N , and D = E = 0 in proposition 2.1 1.15(i). (-=)Take A = B = 0, C = K , D = M and E = N in (ii). Q.E.D.
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Proposition 2.11.17: Suppose F is p a t and 0 + K The following assertions are then equivalent:
+F
--*
M + 0 is exact.
(i) M is p a t . (ii) I K = K n I F for every right ideal I of R. (iii) IK = K n IF for every f.g. right ideal I of R. Proof.- Apply Corollary 2.1 1.15 to the diagram I@K-I@F-l@M-O
0-
K
n IF -IF
-1M
noting that 1,9 monic is the flat test, and cp is epic iff I K = K n IF.
Q.E.D.
We continue the study of flat modules in exercises 4R one useful result (exercise 8) is that every finitely presented flat module is projective.
Regular Rings Recall R is semisimple Artinian iff every R-module is projective, iff every R-module is injective. This raises the question, “When is every R-module flat?” The answer is quite easy. Definition 2.11.18: R is a (von Neumann) regular ring if for each r in R there exists s such that rsr = r.
Note this definition is left-right symmetric, so we could interchange ‘‘left’’ and “right” in what follows. Remark 2.11.29: (i) Notation as above, rs and sr are idempotents of R. Moreover, Rr = Rsr (since Rr = Rrsr c Rsr c Rr). Consequently, every principal left ideal is generated by an idempotent. (ii) Since Jac(R) contains no nontrivial idempotents, ( i ) implies Jac(R) = 0.
The connection with flat modules is in the next result. Proposition 2.11.20: The following conditions are equivalent for a ring R: (i) R is regular. (ii) Every principal right ideal is generated by an idempotent.
$2.1 1 Exact Functors
277
(iii) Every f.g. right ideal is generated by an idempotent. (iv) Every f.y. right ideal is a summand of R. (v) Every R-module is pat.
Proofi (i) * (ii) by remark 2.1 1.19 (right-hand version). (ii) 3 (iii) It suffices to show that if e l R and e , R are principal right ideals then I = e l R e , R is principal; by hypothesis, we may assume e l and e , are idempotent. I = e , R (e, - e,e,)R; write (e, - e,e,)R = e R for e idempotent. Then e,e E e l ( e z - e , e , ) R = 0; so e = (e - e l ) e E (e - e , ) R and e l = e - (e - e l ) E (e - e , ) R , proving I = ( e - e , ) R , as desired. (iii) + (iv) Obvious. (iv) (v) By theorem 2.1 l.l4(v);if I is an f.g. right ideal then taking a complement I‘of I in R we have for any module F:
+
+
-
( I @ F ) @ ( I ’ @ F ) x ( I @ 1’)@ F
FZ
R @ F x F = I F @ I‘F,
so the restriction I @ F + IF is monic. (v) +(i) Take F = R, K = Rr, and I = rR in proposition 2.11.17; then Q.E.D. r E K n I = K n IF = I K = rRr.
Corollary 2.11.21:
R is semisimple Artinian iff R is Noetherian and regular.
Proof: (=.) R is regular by (v) and is Noetherian by Levitzki-Hopkins. (e) Every right ideal is f.g. and thus a summand of R, by (iv), implying R has a composition series in A u d - R . Q.E.D. The next example of a regular ring is more typical.
Example 2.11.22: If an R-module M is semisimple then End, M is regular. (Indeed, given f:M + M let M, be a complement of kerf. The restriction f:MI +f M is an isomorphism whose inverse f:f M + M, can be extended to M as in remark 2.10.24. Clearly f g j = f.) The abstract theory of noncommutative regular rings has been a disappointment. For example, the following two properties of commutative regular rings were hoped to hold in general but are known now to have counterexamples. 1. Is every prime regular ring primitive? (No, cf., example 8.1.40) 2. If all prime images of a semiprime ringR are regular then is R regular? (No, cf., exercise 16.)
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Some of the basic properties of regular rings are given in exercises 17-32, and regular quotient rings are discussed in 53.3. Further information can be found in Goodearl’s books and Jacobson [64B].
Faithfully Flat Modules The definition of flatness raises the converse question of whether a sequence which becomes exact by tensoring with M must itself be exact. More precisely, we have the following definition: Definition 2.11.23: A flat module M is faithfully flat if the following condif N 3 N” is exact whenever tion is satisfied for N, N‘, N” in Aod-R:N‘ + Nt@M%N@M88‘-N’@Misexact.
Clearly R is faithfully flat; to get more examples we need the following useful criterion:
Proposition 2.11.24: R-module F :
The following conditions are equivalent for a flat
(i) F is faithfully Jut. (ii) N 6 F # 0 for euery right module N # 0. (iii) N 6F # 0 for eoery cyclic right module N # 0. (iv) F # IF for eoery right ideal I. (v) F # IF for eoery maximal right ideal I. Proofi (ii) o (iii) since every nonzero right module has a nonzero cyclic submodule. (iv) o (v) since every right ideal is contained in a maximal right ideal. (i) => (ii) Take N‘ = N” = 0 in the definition. f e (ii) (i) We wish to prove N’ + N 4 N“ is exact, given N‘ 6 F + N 6 F + N “ @ F is exact. Note gf 6 1 = 0 so g f N ’ @ M = 0, implying gfN’ = 0 by (ii), i.e., gf = 0. Thus f N ’ c kerg and we have ((kerg)/fN’)@ F x ker(g 63 lF)/(fN’@ F) = 0, implying (ker g)/fN’ = 0, i.e., ker g = f N ’ . (iii) cs (iv) Write N = R/I: then N @ F x ( R @ F ) / ( I 0 F ) x F / I F , so N 6 F # 0 iff F # IF. Q.E.D.
Note condition (v) is similar to “Nakayama’s lemma,” but stronger. Corollary 2.11.25: fully pat.
Eoery nonzero f.g. J u t module ooer a local ring is faith-
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Jac(R) is the unique maximal right ideal, so apply Nakayama's lemma. Q.E.D. froof:
Corollary 2.11.26: A direct sum F of a flat module and a faithfully flat module is faithfully p a t . Proof:
F is flat, and by proposition 2.1 1.24(v) is faithfully flat.
Corollary 2.11.27:
Q.E.D.
Every free module F is faithfully flat.
Inspection of the definition enables us to assume F = R(").We then conclude by corollary 2.1 1.26 and induction on n, since R is obviously a faithfully flat R-module. Q.E.D. Proof:
Although faithfully flat modules do not occupy the same dominating position in the noncommutative literature that they enjoy in commutative algebra, the following special instance will be needed.
Definition 2.11.28: A projective R-module P is faithfully projective if A P # P for every A a R. In view of corollary 2.8.23' and proposition 2.11.24, we see a projective module P is faithfully flat iff IP # P for every proper right ideal I of R, and definition 2.1 1.28 is a slightly weaker requirement on P (which is the same if R is commutative). This notion is related to progenerator, described in W.1, as seen by the following result which also presages some of the techniques of Morita theory. Proposition 2.11.29: Suppose R E T are rings and T is faithfully projective as R-module. Then R is a summand of T , under the natural injection i : R + T. (Thus sublemma 2.5.32' is applicable.) Proof: By remark 2.8.3' we have F = T 0 R @ )is free as left module for some cardinal a. Let A = { f 1: f E Hom(F,R)}. We shall view R E T E F naturally and now note F is actually an R - R bimodule, since T and R(') are R - R bimodules. Given r in R define f, to be right multiplication by r in R or in F according to context. Now we see A is an ideal of R, since
(f1)r = (f,f ) l E A
and
r(f1)
= fr = (ff,)l E
A
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1
for all f: F + R and all r in R. O n the other hand, writing 1 = liei where { e i :i E I } is a canonical base of F and letting ni:F + R be the canonical projection given by niej = d,, we see ri E ail E A and thus 1 E Tn A F = AT, so A = R by hypothesis. Thus 1 E A so 1 = f l for suitable f: F + R. Restricting f to a map f T + R we see f is an epic split by the inclusion map i: R 3 T, Q.E.D. so iR is a summand, as desired. This proof is quite remarkable for, although rather straightforward, it seems to defy categorization, relying on free bimodules of large size.
92.12 The Prime Spectrum In this section we introduce Spec(R), one of the main themes in ring theory. After establishing some basic properties we turn to the use of prime ideals in studying the rank of a projective module. Then we look at various noncommutative versions of the Nullstellensatz, and finally compare Spec of ring extensions. Definition 2.12.1: Define the prime spectrum Spec(R) = {proper prime ideals of R}, partially ordered under set inclusion. Given A E R define Y ( A )= {P E Spec(R): A E P } . Jacobson [64B] introduced the primitive spectrum, defined analogously as (primitive ideals of R}, but the prime ideals were catapulted into prominence by Goldie's theorems (c.f., 53.2 below). Later we shall relate the primitive and prime spectrums in certain cases; first we consider the following important topology. Remark 2.22.2: There is topology on Spec(R) having { Y ( A ) :A 5 R} as the closed sets, as a consequence of the following facts: (i) 0 = 9({ l}) and Spec(R) = Y({O}) (ii) Y ( A ~= ) Y( A ~ ) (iii) Y ( A , ) u 9 ( A 2 )= Y ( A ,RA,) (iv) If A, E A, then 9 ( A 1 )2 Y ( A , )
0
u
In the opposite direction define 3: Spec(R) --+ {Ideals of R} by Y ( Y )= { P E Y }for any Y in Spec(R). Write Y ( A )for Y(Y(A)). Remark 2.12.3: (i) f ( A ) is a semiprime ideal of R and A G f ( A ) , equality holding iff A is a semiprime ideal of R.
52.12 The Prime Spectrum
28 1
(ii) If A E B E Y ( A )then Y ( A )= Y ( B ) .Thus Y ( ) is determined by the semiprime ideals. (iii) Y(A)=Y(RAR) by (ii); consequently, for A i d R we have
Partially ordering the ideals of R by means of reverse inclusion, i.e., A, < A , iff A, c A,, we see Y ( ) is an order-preserving function from {ideals of R} to {closed sets of Spec(R)},whose restriction to { semiprime ideals of R} is 1:1 and onto. Thus the topological properties of Spec(R) take an algebraic significance. It is natural to ask which topological spaces can be of the form Spec(R), and we can make the following reduction to R semiprime: Remark 2.12.4: Let R = R/Prime rad(R), with - denoting the canonical homomorphic image. Then Y ( A )+ Y(2)defines a homeomorphism from Spec(R) to Spec(R). (Indeed, P p gives a 1: 1 correspondence of the prime ideals; moreover, A G P iff 2 G p.) --f
Certain standard topological concepts have significance in this discussion. Proposition 2.12.5: ( i ) Spec(R) satisjies the jinite intersection property. ( i i ) Every point in Spec(R) is closed iff every prime ideal of R is maximal. (iii) lf R =R, x R, then Spec(R,) and Spec(R,) are clopen subsets of R. (iu) Spec(R) is connected iff RIPrime rad(R) has no central idempotents. (v) For R commutative the clopen subsets of R are in 1 :1 correspondence to { Y ( e ) :e idempotent in R}. (vi) The closure of a set { p i : i E I } is Y ( n i , ,pi); in particular, { p i : i E I } is dense ifs pi = Prime Rad(R). Proof: (i) Suppose Ai a R with Qr = ni,,Y(Ai)= Y ( l A i ) .Then X A , is not contained in any maximal ideal so 1.1 Ai implying 1 AiUfor suitable i , , . . . ,i t ; thus Y ( A i U= ) Y ( Ai ) = fzr. (ii) {P}E Spec(R) is closed iff no other prime ideal contains P, so P is maximal. (iii) Writing n:R+R1 for the canonical projection let e = n l EZ(R). Then Y ( e )u Y (1 - e ) = Y(eR(1 - e))= Y ( R e (1 - e))= Y ( 0 )= Spec(R) and Y ( e )n Y ( l - e ) = Y ( R e + R ( l -e))=.Y(R)=Qr; thus Y(e)=Spec(R,) and Y(l - e ) = Spec(R,) are clopen. (iv) We may assume Prime rad(R) = 0 by remark 2.12.4. (-) By (iii). (t) Suppose Spec(R) has clopen sets % and 9,= Spec(R) - %, and let Ai = $(q). Then A, n A, = $(% u 9,= )0 and A, + A, = R (since otherwise
nr=,
1
EX:=,
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A, + A, would be contained in a proper maximal ideal P E 9, n 9, = 0.) By the Chinese Remainder Theorem R x RIA, x RIA,. (v) By (iv), since idempotents lift up the prime radical. (vi) Y ( pi) is closed by definition; on the other hand, if each @Y(A) for suitable A then A c pi for each i and thus A c_ pi. This proves the Q.E.D. first assertion, and the second is an instant application.
nis,
nie,
Many results of ring theory depend on selecting the correct set of primes having intersection 0, i.e., a dense subset of Spec(R).
Corollary 2.12.5’: Suppose R is prime and { p i : i E I } is dense (in Spec(R)). Then for any 0 # r E R we have { p i : r 4 p i } is dense.
n
Proof.’ Let A = { pi: r # pi}. Then rA E Q.E.D. so A = 0 since R is prime.
n{ pi:
r E pi} nA =
n{ p i :
i E I } = 0,
Remark 2.12.6: If f: R, + R, is a ring homomorphism for which R, is a centralizing extension of fR, then f - l : Spec(R,) + Spec(R,) is continuous.
Proposition 2.12.7 (“Prime avoidance”): If A, B d R and P,, . . .,P,ESpec(R) with A - B E pi then either A E B or A E pl. for some i.
u:=,
,
Proof.’ To make the argument symmetric we define P,+ = B (not necessarily prime) and write A E uiI pi. We shall prove by induction on t that A c pi for some i. For t = 0 this is clear so assume the result is true for t - 1. We are done unless for each j there is aj E A - u{pi: i # j } ; then each aj E 4. Note a 2 R a 3 R . . * a , R a l + , ~ so P l there is some a in a 2 R a 3 R ~ ~ - a l R a l + , - PThen ,. j , contrary to a, + a E A. for each Q.E.D. a E (7;: pj so a, a 4
+
There is one more fundamental result about prime ideals which will be of frequent use. Recall by remark 2.6.10 that minimal prime ideals exist.
Proposition 2.12.8: Suppose Pl ,.. .,Pn are incomparable prime ideals of R for n > 1 with I= pi = 0. Then {PI,.. .,P , } contains the set of minimal prime ideals of R.
n
Proof.’
For any prime ideal P we have Pl
Pn= 0 E P, so some pi E P. Q.E.D.
Localization and the Prime Spectrum Let us turn now to the effect of central localization on the prime spectrum.
283
52.12 The Prime Spectrum
Remark 2.12.9: Suppose S is a submonoid of Z(R) and P is a prime ideal Then P is S-prime. (Indeed, if sr E P then sRr c P so of R with P n S = 0. r E P ) Consequently, if s-'r E S-'P then r E P, by lemma 1.10.9.) Proposition 2.12.9': The correspondence (D of proposition 1.10.8 induces a homeomorphism from { P E Spec(R): P n S = @} (with the topology induced from Spec(R))to Spec(S-'R). Proof;. Remark 2.12.9 shows that theorem 1.10.11 can be used to show (D acts 1:1 on { P E Spec(R):P n S = a}.On the other hand, if v: R + S-'R is given by vr = 1-'r then for any P' in spec(S-'R) we have v-'P' E Spec(R) and clearly @ ( v - 'P') = P', yielding an (onto) 1:1 correspondence. The homeomorphism follows at once from the following observation:
Lemma 2.12.10: S-'AcS-'P. Proof;. UER
Suppose P E Spec(R) and P n S = 0.Then A E P iff
(a) is
obvious. (c=) If a E A then 1-'a Q.E.D.
E
S-'A E S-'P implying
Define V ( a ) ={PESpec(R):a$P} = Spec(R)-Y({a}), an open set. Clearly, the U ( a )are a base of Spec(R) since any open set has the form R - 9 ( A ) = U{U(a):aEA}. Corollary 2.12.11: Spec(R[s-'1) is homeomorphic to the open set V(s) of Spec(R), for any s E Z ( R ) . Remark 2.12.12: If S is a submonoid of Z(R) and A 4 R with A n S = 0 then S-'R/S-'A x S-'R, writing for the canonical image in RIA. (Indeed, the surjection S-'R H S-'R given by s-'r H F ' T has kernel {s-'r:s'r E A for some s' in S } = S-'A.)
Localizing at Central Prime Ideals Definition 2.12.13: Given P E Spec(Z(R)) let S = Z(R) - P ; write R, for S-'R, and for any R-module M write M, for S-'M. (This notation may be confusing initially but is unambiguous since 0 E P, and we certainly would not invert 0 in a ring.) If A E R write A, for S-'A. The next result shows how localization gets its name.
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Example 2.12.14: If C is a commutative ring and P E Spec(C) then Cp is a local ring whose unique maximal ideal is P,, and C,/Pp z field of fractions of C / P . Indeed, let S = C- P. Then P is maximal with respect to PnS = 0, implying by proposition 2.12.9' that Pp is the unique maximal ideal of C,, so C, is local; the last assertion is remark 2.12.12.
Proposition 2.12.15:
Let 2 = Z(R).
(i) I f M E R-Aud then there is a monk M + H{MP: P maximal ideal of Z(R)} given by x + (l-lx). (ii) lf M is (f.g.) projective in 2-Aod then Mp is free in Z,-Mud for every P in Spec(2). Proof: (i) Suppose XEM.Then l#Ann,x, which is thus contained in a maximal ideal P of 2; hence sx # 0 for each s in 2 - P so 1-'x # 0 in M,. (ii) Mp z 2, 8 M is projective over the local ring 2, and is thus free (cf., example 2.8.9 and the discussion following it). Q.E.D. We should like to examine now what happens to free modules under central localization; the key to the puzzle lies with finitely presented modules.
Remark 2.22.26: (relying heavily on proposition 1.10.5) If M is a finitely presented module and S is a submonoid of Z(R) then S-'M is a finitely presented S-'R-module. (Indeed, if f:F + M is epic with F f.g. free and kerf is f.g. then f extends to an epic S-'F + S-'M and ker f = S-'(ker f ) is f.g. in S-'R-Aod;since S-'F is f.g. free we see S-'M is finitely presented.)
Proposition 2.12.17:
Suppose M is Jinitely presented and S-'M is free in S-'R-Aod. Then there is s in S such that M[s-'] is already free in R[s-']Aod, with a base of the same size as the base of S-'M over S-'R.
Proofi Pick a base Z - ~ X ~ , . . . , Z - ' Xof, S-'M as S-'R-module, where ZGS and xiE M. Although not necessarily free, M[z-'] is spanned by z-'xl,. . ., z-'x, and is finitely presented in R[z-']-Mod by remark 2.12.16. Let F be a free R[z-']-module with base e l , ..., em and define f : F + M[z-'] by f e i = z-'xi. Then kerf is f.g. by proposition 2.8.29 so we can write kerf = R[z-']yj for suitable yj in F. But S-'ker f = 0 by proposition k10.5 so there are sj in S with sjyj=O. Writing s=sl * * * s , zyields s kerf =0, so the induced map R[s-'](") + M[s-'] is an isomorphism, i.e., M[s-'] is free. Q.E.D.
I:.=,
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285
Corollary 2.12.18: If M is f.g. projective over a commutative ringC then for every P in Spec(C) there is s in C - P such that MCs-'] is a free C [ s - ' ] module. Proofi
M is finitely presented and M p is free over C,.
Q.E.D.
The Rank of a Projective Module Define the rank of a free module to be the cardinality of its base. This is welldefined when R has the invariant base number, which is the case if R is commutative (c.f., corollary 1.3.27). Proposition 2.12.19: Suppose M is an f.g. projective module over a commutative ring C. Then there is a continuous function f: Spec(C) + N (under the discrete topology) given by fP = rank M p (as C,-module). Proofi We must show for any n in N that f -'(n) is open. Suppose PEf -'(n), i.e., rank M p = n and take S E C - P such that rank M [ s - ' ] = n, using corollary 2.12.18. We claim rank MQ = n for all Q in the open set U(s),i.e., whenever s $ Q. Indeed, taking S = C - Q we have
MQ = S-'M = S - ' ( M [ s - ' ] ) = ~ - ' ( R [ s - ' ] ' " ' )x (S-'R[s-'])'"' = (S-lR)'"' = R'p"',
so rank MQ = n, as desired. Thus we have shown f -'(n) is a union of open sets. Q.E.D. Corollary 2.12.20: values in N.
Notation as above, f takes on only a Jinite number of
Proof: { f -'(n): n E N} is an open cover of Spec(C) and thus has a finite Q.E.D. subcover by the FIP (proposition 2.12.5(i)).
Definition 2.12.21: A projective module M over a commutative ring C has constant rank n if rank M p = n for all P in Spec(C).
Theorem 2.12.22: Suppose M is an f.g. projective module over a commutative ring C. Then C has orthogonal idempotents el,. . . ,e, such that each Mi = eiM is projective of constant rank over Ci = eiC; also M z Mi and C x = Ci canonically.
n:
fl:=,
286
Basic Structure Theory
Proofi By corollary 2.12.20, Spec(C) is covered by a finite number of disjoint clopen sets, each corresponding to the ranks of localizations of M ; by proposition 2.12.5 these correspond to orthogonal idempotents of C, and the Q.E.D. rest of the assertion is clear.
Corollary 2.12.23: Notation as in the theorem, if C has no nontrivial idempotents then M has constant rank. Remark 2.12.24: If M is projective of constant rankn then for any maximal ideal P 4 C we have M I P M free over the field C / P of dimension n. (Indeed, the canonical map M + M / P M yields an epic M p M I P M ; any base of M p gets sent to a base of M I P M by Nakayama’s lemma.) --f
Definition 2.12.24’: A projective module of constant rank 1 is called invertible.
Invertible projective modules have a deep connection with algebra and geometry, since they are the natural categorical generalization of ideals of a PZD (which are isomorphic to the ring itself), cf., exercises 7, 8.
K Supplement: Hilbert h Nullstellensatz and Generic Flatness For commutative algebras R = F[a,, . . .,a,], Hilbert’s Nullstellensatz shows the Spec topology coincides with the Zariski topology defined in 42.3. On the other hand, in Ej2.5 we introduced the “weak Nullstellensatz,” that Jac(R/A) is nil for every A 4R and noted that this property generalizes Hilbert’s Nullstellensatz. Now we shall describe a different generalization and see how these properties are interrelated. These results are at the foundation of the noncommutative versions of the Hilbert Nullstellensatz, and thus are relevant to theories of noncommutative algebraic geometry. We start with a technical definition. We say a prime ideal A is a G-ideal if A # r ) {PESpec(R):A c P } . An integral domain C is a G-domain if 0 is a G-ideal. Remark 2.22.25: An integral domain C is a G-domain iff C[s-’] is a field forsome0 # sinC.(Indeed(*)isseenbytakingO # S E n ( 0 # P E Spec(C)j, Every since then CCs-’] has no proper maximal ideals so must be simple. (e) nonzero prime ideal contains a power of s and thus s itself.) Definition 2.12.26: A ring R is Jacobson if every prime ideal is the intersection of primitive ideals. An algebra R over a field F satisfies the
$2.12 The Prime Spectrum
287
Nullstellensatz if End, M is algebraic over F for every simple module M # 0.
The Nullstellensatz can be studied in terms of two auxiliary notions: A C-algebra R satisfies the strong Nullstellensatz if Ann, M is a G-ideal of C for every simple R-module M # 0; R satisfies the maximal Nullstellensatz if for every simple R-module M # 0 we have Ann, M is a maximal ideal of C such that R/Ann, M satisfies the Nullstellensatz over the field C/Ann, M. An alternative formulation of the maximal Nullstellensatz is, “If P is a primitive ideal of R then P n C is a maximal ideal of C and R I P satisfies the Nullstellensatz over the field C/P.” This is seen by writing P = Ann, M for M simple. Remark 2.12.26‘: If R is Jacobson then Spec(R) produces the same topology on R as the primitive spectrum. Thus the Jacobson property is very significant and is the focus of this discussion. Note that P E Spec(R) is the intersection of primitive ideals iff Jac(R/P) = 0, and it follows easily that R is Jacobson iff Jac(R/A) = Prime rad(R/A) for every A 4 R. In most classes of rings of interest to us it will turn out Nil(R/A) = Prime rad(R/A), cf., theorem 2.6.23 for example. Thus the key to the discussion is verifying the “weak Nullstellensatz,” that Jac(R/A) is nil.
To avoid repetition of the word “Nullstellensatz” we write Null, SN, and MN respectively for the Nullstellensatz, strong Nullstellensatz, and maximal Nullstellensatz. Note by definition that M N implies SN since every maximal ideal is a G-ideal. On the other hand, MN and Null become the same if C is a field, but SN is vacuous for fields. Remark 2.12.27: Null is quite powerful when F is algebraically closed, since then for any simple R-module M we see the division ring End, A4 is algebraic over the algebraically closed field F, implying End, M = F. When R is primitive we can take M to be faithful; then each element of Z(R) yields an endomorphism of M and thus belongs to F, proving Z(R) = F. Example 2.12.28: Suppose R is an F-algebra and IF1 - 1 > [R:FJ as cardinal numbers. Then R satisfies both the weak Nullstellensatz and Null. Indeed, the hypotheses pass to RIA for each A 4 R so we get the weak Nullstellensatz by theorem 2.5.22. On the other hand, if M is a simple R-module then D = End, M is a division algebra over F and M is a right vector space over D ; since M is cyclic we have [R: F ] 2 [ M : F ] = [ M : D ] [ D : F ] so [ D : F ] < IF1 - 1. Consequently, for any d in D,
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{(d - a)-': a E F} is F-dependent; proposition 2.5.21 implies d is algebraic over F, proving Null. Remark 2.12.29: (i) If C is semiprimitive (commutative) and not a field then C is not a G-domain (since n{maximal ideals} = 0). (ii) If R is a C-algebra and M is a simple R-module then Ann, M E Spec(C), analogously to remark 2.1.14.
Proposition 2.12.30: Suppose R is an algebra over a field F. If R[A] satisjes SN over F[A] then R satisfies Null. Proof: Suppose M is a simple R-module. We want to show any endomorphism f : M + M is algebraic over F. View M as an R[A]-module by the action riAi)x = l r i f i x for x in M. This action restricts to the given action over R, so M is a simple R[A]-module, implying P = Ann,[,, M is a prime G-ideal of F[A]. But F[A] is semiprimitive so P # 0; taking 0 # p(A) E P Q.E.D. we have 0 = p(l)M = p ( f ) M as desired.
(1
Duflo's approach stems from the following easy observation. Suppose R is an algebra over an integral domain C whose field of fractions is F, and M is a simple R-module which is faithful over C. Then M is naturally an F QcR-module. If M is free over S-'C (under this action) for a submonoid S of C then S-'C = F.
Lemma 2.12.31:
Proof: Given 0 # c E C let c^: M + M denote the left multiplication map by c. By Schur's lemma c^ is invertible in End, M , so we have the action of F Q R on M given by (c-' @ r)x = c^-'(rx) where r E R and x E M; this is clearly well-defined and makes M an F @ R-module. Now suppose M is free over S-'C. Take a base xlr.. .,x,. If z E S-'C then zixi for zi in S-'C. Then x1 = zzixi; matching coefficients of z-'xl = Q.E.D. x1 shows zzl = 1, so z is invertible, proving S-'C = F. As a special case we see if M is free over C[s-'] for some 0 # s E C then C[s-'] = F so C is a G-domain. This leads us to the following definition. Definition 2.12.32: R satisfies generic jlatness over C if for every simple R-module M there is s # 0 in C such that M[s-'] is free over C[s-'1.
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289
The usual definition of generic flatness requires the condition to hold for all f.g. R-modules, but we shall only require M simple (so that lemma 2.12.31 is applicable). Several trivial observations should be noted. Remark 2.12.33:
(i) In verifying generic flatness we may assume M is faithful over C, for if cM = 0 then M [ c - ' ] = 0 is free over C[c-']. In particular, C is a domain by remark 2.12.29(ii). (ii) Any commutative Noetherian algebra satisfies generic flatness, by proposition 2.12.17. We shall find another important example in 56.3. Lemma 2.12.33': l f RIPR satisfies generic flatness as CIP-algebra for each P in Spec(C) then R satisjes S N . Pro08 If M is a simple R-module then P = Ann, M E Spec(C) and PR 5 Ann, M; since M as RIPR-module is simple and (C/P)-faithful, we see P is a Q.E.D. G-ideal of C by the comment after lemma 2.12.31. In case C is Jacobson, we can conclude more. Remark 2.12.33": Suppose R satisfies SN over a Jacobson ringC. Then Ann, M is maximal in C for every simple R-module M. (Indeed, P = Ann, M is a G-ideal and is prime by remark 2.12.29(ii).Thus P is the intersection of primitive ideals of C, so Pis itself a primitive ideal. Hence C / P is commutative primitive and thus a field.) Remark 2.12.34: Suppose R is a C-algebra and view C E R; suppose A d R. Then any of the hypotheses Null, SN, MN, and generic flatness pass from R as C-algebra to RIA as C / ( C n A)-algebra. Indeed, any simple RIA-module M is also simple as R-module under the "change-of-rings" action, and End, M z End,,, M canonically. Lemma 2.12.35: Suppose R satisfies S N over a Jacobson ringC, and (R/PR)[I] satisfies Null over C / P for every maximal ideal P of C. Then the weak Nullstellensatz holds. Proofi In view of remark 2.12.34, we need only show Jac(R) is nil; by lemma 2.5.41 it suffices to show R[A](1 - rA) = R [ I ] for all r in Jac(R). If R[I](l - rI) < R[A] take a maximal left ideal L of R[A] containing 1 - rA
Basic Structure Theory
290
and let M = R[L]/L.By remark 2.12.33" Ann, M is a maximal ideal P of C. But 1 4 L so left multiplication by I defines a nonzero map 0 : M + M which is invertible by Schur's lemma. By hypothesis, 0 - l is algebraic so B = p(a-') for some pin (C/P)[A]. On the other hand, letting xo = 1 + L E M we see (1 - lr)xo = 0 so h x o = xo,implying lirix0 = (lr)ixo= xo for all i. Thus K i x 0 = rixo for all i, implying gx0 = p(a-')x0 = p(r)x, so 0 = (1 - rl)xo = xo - rdxO= (1 - rp(r))xo,
contrary to 1 - rp(r) being invertible (since r E Jac(R)).
Q.E.D.
Theorem 2.12.36: Suppose R is an algebra over a commutative Jacobson ring C. Consider the condition (*) RIPR satisfies generic Jatness over C I P for all P in Spec(C).
Then R satisfies both the weak Nullstellensatz and M N , provided R[A1,A2] satisfies (*) as C[L,]-algebra.
Proofi R[1,,A2]satisfies SN over C[1,] by lemma 2.12.33'. Also R satisfies SN over C, seen by remark 2.12.34 by specializing A,,1, + 0. For any maximal ideal P of C we see (R/PR)[1,,1,] satisfies SN over (C/P)[A,],implying (R/PR)[1,] satisfies Null by proposition 2.12.30. Therefore the weak Nullstellensatz holds by lemma 2.12.35. To verify MN suppose M is a simple R-module. Then P = Ann, M is maximal by remark 2.12.33", and R/Ann, M is a homomorphic image of (R/PR)[I,]and thus satisfies Null. Q.E.D. (Note: If we assume (*) for R we can weaken the hypothesis on R[1,,1,] by restricting our attention to those prime ideals of C[1,] whose intersection with C is maximal.) Corollary 2.12.37: (Commutative Nullstellensatz) If R is a commutative algebra which is finitely generated as an algebra over a field F, then R is Jacobson and satisfies Null over R. Proofi
R[ll, 1 4 is commutative Noetherian and thus satisfies (*). Q.E.D.
Remark 2.22.37': The commutative Nullstellensatz enables us to sharpen use of ultraproducts described in theorem 1.4.15ff.Suppose we want to prove
92.12 The Prime Spectrum
29 1
an elementary sentence cp about an integral domain C of characteristic 0 which is not a field, knowing (i) cp holds in characteristic p for almost all p > 0, and (ii) cp passes to subrings, i.e., cp is universal. In view of theorem 1.4.15 and proposition 1.4.21, we may assume C is finitely generated as Z-algebra. By the commutative Nullstellensatz, C is semiprimitive so it has an infinite set (5:i E N} of maximal ideal having intersection 0. By hypothesis, cp holds in each C / e , and often we can conclude by applying proposition 1.4.20. This technique is called passing to characteristic p and is particularly useful for group algebras and enveloping algebras, cf., Chapter 8. The point of generic flatness is that it can often be verified more readily than the Nullstellensatz properties, and at this stage one could already prove significant generic flatness results, such as Irving [79, theorem 21. However, we shall postpone the climax of this discussion until $8.4 and McConnell’s “vector generic flatness.” Incidentally when the “generic flatness” condition is dropped, some of the results fall flat on their faces, cf., exercise 5.
Comparing Prime Ideals of Related Rings Suppose throughout R c R’ are rings. One of the main techniques of computing Spec(R’) is by comparing it to Spec(R). To do this we must have a way of passing prime ideals back and forth from R to R’. This procedure is so basic that it has been investigated very carefully in the literature, leading to striking results by Heinecke-Robson [83]. Unfortunately, a presentation in this generality would obscure the simplicity of the proofs in special cases, so we shall settle on the compromise, studying centralizing extensions and treating normalizing extensions in exercises 10ff.
1
Remark 2.12.38: Suppose R‘ = Rri where each ri E C,,(R). If A 4 R then AR‘ = Ari = x r i A = R‘A (since A = R A = A R ) ; in particular AR’ 4 R‘.
1
In view of remark 2.12.38we say R all A 4 R. (Thus R‘AR’ = AR‘.)
c R’ is ideal compatible if AR‘ Q R’ for
Proposition 2.12.39: If R c R‘ is ideal-compatible there is a map Spec(R’) + Spec(R) given by P’ + P‘ n R. (Likewise, if P’ a R’ is semiprime then P’ n R is a semiprime ideal of R.)
Pro08 Suppose A B G P ’ n R for ideals A , B of R . Then A R ’ B R ‘ sABR’C P‘ Q.E.D. so AR‘ c P’ or BR‘ G P’, proving A c P’ n R or B G P’ n R .
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292
Our main concern is how much information is lost by this map, especially concerning chains of prime ideals. This leads us to study the following concepts: Definition 2.12.40: Given Pl E P, in Spec(R),define the following possible situations:
(i) LO(Pl) = “lying over” means there is P’ in Spec(R’)with P’ n R = PI. (ii) GU(P,,P,) = “going up” means given any P‘, in Spec(R’) lying over P we have P ; lying over Pz with P‘, E P i . (iii) INC(Pl) = “incomparability” means that one cannot have P’, c P;’ in Spec(R’)each lying over Pl. Similarly GU(-, P) means CU(Pl,P) for all Pl E P in Spec(R), and LO, GU, and INC mean, respectively, LO(P), GU(-,P), and INC(P) for all P in Spec(R). A’
Given A‘ Q R‘ and B Q R we say an ideal B’ of R’ is (A’, B)-maximal if c B’ and B’ is maximal such that B’n R c B.
Lemma 2.12.41: If A ’ n R G B then (A’,B)-maximal ideals exist. Furthermore any such ideal is a prime (resp. semiprime) ideal of R’ if B is a prime (resp. semiprime) ideal of R. Prooji The first assertion is by Zorn’s lemma. If P’ is (A’, B)-maximal and B’,, B; 3 P‘ with B ; B ; E P’ then (B’, n R)(B; n R) G P’n R E B; if B is prime (resp. semiprime with B’, = B ; ) then B’, n R G El or B; n R E B,, so B’, = P’ or B; = P’ by maximality, proving the second assertion. Q.E.D. Remark 2.12.42: If R n R’PR’ = P then LO(P) holds. (Indeed, take P’ to be (R’PR‘,P ) = maximal; then P c P’ n R E P so P‘ lies over P).
The key to both LO and GU is the following result: Proposition 2.12.43: In order t o prove LO(P) and GU(P,, P ) hold in R’ over R it suflces to prove for each P’ in Spec(R’) with P’ n R c P that the following weakened version of LO holds inR‘ = R‘/P’over R = (R + P ‘ ) / P ’ : There is 2 a for which 0 # A‘nR E i’.
Proofi To prove LO(P) we take P‘ in Spec(R’) to be (0,P)-maximal, and claim P c P’. Otherwise, # 0 in = R’/P’ so, by hypothesis, there is an R E P; then A‘ n R c P contrary to choice ideal A’ 2 P’ such that 0 #
zn
293
82.12 The Prime Spectrum
of P'. Thus we established the claim and LO(P). The same argument shows if PI G P and Pi lies over PI then any ( P i , P)-maximal ideal of R' lies over P, Q.E.D. proving GU(Pl, P). Remark 2.12.44: If R sz R' is ideal-compatible then G U implies LO. (Indeed, given P in Spec(R) take Pi to be (0, P)-maximal. Then P', E Spec(R') so applying GU(P', n R, P) to Pi produces a prime ideal of R' lying over P.) Let us now verify LO and G U in several instances. Proposition 2.12.45: Suppose R E R' have a common ideal A. Then LO(P) and GU(Pl, P) hold under the following circumstances:
(i) A $i P. (ii) R c R' is ideal-compatible and P is minimal prime over A suitable ideal B = R n R'BR' such that A $Z B.
+ B for
a
Proofi We verify the hypotheses of proposition 2.12.43. (i) If P'
2' n R
=
R c P then 0 # 2 4 R'
A' E p.
= R'/P'.
Let
A' = A X # 0. Then
(ii) Passing to RIB and R'IR'BR' one may assume B = 0. Note that if P' n R c P then A $i P' n R, since, otherwise, we contradict the minimality of P over A; hence we conclude as in (i). Q.E.D. Remark 2.12.45': To apply (ii) we need instances when B = R n R'BR'; this is clearly the case if B = 0 or if B is a prime ideal not containing A (by (i)). Actually a stronger incomparability result is also available. Proposition 2.12.46: Suppose R c R' have a common ideal A, and A $Z P for some P in Spec(R). Then INC holds; in fact, if P',, P; are primes of R' and P i n R G P; n R = P then P', E P;. Proofi AP',
5
P', n A C
Pi n R E P; but A
$Z P; so P', E Pi.
Q.E.D.
Remark 2.12.47: If R' is a centralizing extension of R and R' is prime then Ann, a = 0 for every R-normalizing element a # 0 of R'. (Indeed, let A = Ann,a; then AR'a = C,.(R)AaR = 0 implying A = 0 since R' is prime.)
Theorem 2.12.48: tensions.
(Bergman) LO and G U hold for all finite centralizing ex-
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294
Proof: (Robson-Small [81]) in view of proposition 2.12.43 we may assume R’ is prime,and need only show for each 0 # P E Spec(R) there is A’ Q R‘ for which 0 # A ‘ n R E P. Note R is prime by proposition 2.12.39. Write Ra, with each a, E C,,(R) and a , = 1. We expand a, to a maximal R‘ = independent set of the a, which by reordering the a, we may assume is { a , , . . .,a,,,}.Then T = Ra, = aiR is free as left or right R-module by remark 2.12.47. Let Bi = { r E R:ra, E T} and B = B, # 0. Then PB # 0 and PBR’ a R’. Let A‘ = PBR’ E P T G T; sublemma 2.5.32’ shows Q.E.D. A‘ n R E P T n R = P, and A’ n R 1 PB # 0, as desired.
xy’,
n;=,
So far the only known proofs of INC for centralizing extension involve the “central closure,” a construction which we shall not encounter until $3.4; consequently, we postpone the proof of INC until theorem 3.4.13. We can push theorem 2.12.48further by means of the following observation.
Proposition 2.12.49: Suppose R‘ is the direct limit of subrings Ri 3 R each satisfying LO, where the cpj of example 1.8.2 are the inclusion maps. Then R‘ 3 R satisfies LO. Proofi By remark 2.12.42 we are done unless R n R’PR’ 3 P for some P in Spec(R). But then writing r = xkrLlpkrL24 P for p i in P we see this equation holds in some R,, implying R n R,PR, 3 P contrary to LO(P) in R,. Q.E.D. Corollary 2.12.50: If R is an algebra ouer a field F then R and G U over R for any jield K 2 F.
mFK satisjes LO
Proof: By proposition 2.12.49 we may assume K is finitely generated. Then K is a finite algebraic extension of some purely transcendental field extension K O of F; by theorem 2.12.48 it suffices to show R K O satisfies LO and G U over R. By induction on the size of the transcendence base we may assume K O = F(I), i.e., we need to show R(I) I> R satisfies LO and GU. But LO is clear by sublemma 2.5.32‘,and G U follows then from proposition 2.12.43. Q.E.D.
LO and the Prime Radical One instant application is to the prime radical. Remark 2.12.51: If R’ is an ideal-compatible extension of R then Prime
52.12 The Prime Spectrum
295
rad(R) G R n Prime rad(R'), equality holding if LO is satisfied. (Indeed, let N' = Prime rad(R'). Then R n N ' = n { R n P ' : P ' E S p e c ( R ) } 2 n{PESpec(R)}, equality holding if we have LO.) This modest observation gives us the analogue to corollary 2.5.36. Theorem 2.12.52: Suppose K is a Jield extension of F, and let N = Prime rad(R) and N' = Prime rad(R QFK). Identifying R with R 8 1 we have N = R n N ' ; furthermore, N' = NK i f K is a separable Jield extension of F. Proof: The first assertion follows by applying remark 2.12.51 to corollary 2.12.50. To prove the second assertion note we have proved N' 2 NK, so factoring out NK we may assume N = 0, i.e., R is semiprime. We want to show R OFK has no nilpotent ideal A # 0. Otherwise, A has some nonzero element in RK, for some finite extension K, of F, so we may replace K by K, and assume K is finitely generated. Letting KObe a subfield generated by a transcendence base of K over F, we have RK, z R 8 K O semiprime; replacing R, F by RK, and KO, we may assume K is algebraic over F, so K is a finite separable extension. Let K be the normal closure of K. Then R 8 K is a finite centralizing extension of R 0 K , so by remark 2.12.51 it suffices to prove R Q K semiprime. At this point we can copy out the proof of theorem 2.5.36 since the prime Q.E.D. radical is invariant under automorphisms.
Corollary 2.12.53: If R is an algebra over a Jield F of characteristic 0, then Prime rad(R OFK) = Prime rad(R) Q K for any Jield extension K of F.
Unfortunately if R' is a finite normalizing extension of R and P E Spec(R') then P n R need not be a prime ideal of R, so one might be tempted to throw up one's hands in despair. However, P n R turns out to be a finite intersection of prime ideals of R, and corresponding weakened versions of LO and GU can be proved. INC also holds but is quite difficult. Definitive results were obtained by Heinecke-Robson [8 13. Passman [81] showed how to transfer many of these questions to the primitive spectrum, and we shall use his approach in exercises 10ff. The results for normalizing extensions were generalized to the case where R' is merely a subring of a finite normalizing extension; this is called an intermediate normalizing extension in HeineckeRobson [84a].
2%
Basic Structure Theory
Height of Prime Ideals Deytnition 2.12.54: If P is a prime ideal in R we say P has height 0 if P is minimal; inductively, height (P) = 1 + max{height(Q): Q c P i n Spec(R)). In other words, height (P) is the largest length of a chain Po c PI c ... c P in Spec(R). (The height is sometimes called the “rank.”) Although prime ideals need not have finite height, in general, the height is a very useful measure of a prime ideal.
Example 2.12.55: If S is a submonoid of R with P n S = 0 then the height of P in R equals the height of S-’P in S-’R, by proposition 2.12.9’. We bring in the height here because of its connection to LO, GU, and INC.
Remark 2.12.56: If R C R is ideal-compatible then LO(P) holds for every minimal prime ideal P of R. (Indeed, if P‘ is (0, P)-maximal then R n P‘ c P implies R n P‘ = P since P is minimal prime.)
Proposition 2.12.57: Suppose R c R‘ is ideal-compatible, and suppose P E Spec(R) of height n. (i) ’1 INC holds then height(P’) I n for every P‘ in Spec(R’) lying over P. (ii) If GU holds then there is some P’ lying over P with height(P’) 2 n. (iii) ’1 INC and GU hold there is some P’ lying over P with height(P’) = n.
Proof.(i) If Pb c ... c Pi = P‘ in Spec(R) then Pb n R c * . * c P; n R = P in Spec(R), so t I n. (ii) Suppose Po c ... c P, in Spec(R), then there is Pb lying over Po and inductively we build a chain Pb c -.. c PL by applying GU(pi- pi) to PiThus Pi has height 2 n. (iii) Combine (i) and (ii). Q.E.D.
82.13 Rings with Involution Definition 2.13.2: An involution (*) of a ring is an anti-automorphism of order 2, i.e., (*) is a group automorphism of order 2 of the additive group ( R , +) satisfying (r1r2)* = rfr: for all ri in R. We write (R,*) to denote the ring R with involution (*).
297
52.13 Rings with Involution
The motivating example of an involution is the transpose ( t ) on matrices, which is inextricably bound in with the structure of matrices; as we shall see, involutions often have a profound impact on the structure of rings, especially of simple finite-dimensional algebras. Our object here is to present certain structural generalities which make it easier to work with involutions: In particular, we deal with the structure of matrix rings with involution. There are certain special kinds of elements which arise from involutions. Definition 2.13.2: Suppose (R, *) is a ring with involution. An element r is symmetric if r* = r; r is antisymmetric (also called skew-symmetric)if r* = -r. Write ( R ,*)+ for {symmetric elements of R } and ( R , *)- for {antisymmetric elements of R ) K = ( R , * ) - and S = ( R , * ) + are subgroups of R (under r-r* r+r* + . K is in fact a Lie addition)and R = K 0 Sif f E R, sincer = 2 2 subring of R (i.e., [ K , K] c K ) , whereas S is a Jordan subring of R, by which we mean S has the following properties: Remark 2.13.3:
~
1 E S (since 1* = 1*1 s1s2s1 E S
if
= (1*1)* =
1** = 1).
sI,s2E S .
Digression 2.13.4: Since the structure of rings with involution interrelates deeply with the theory of Jordan rings, let us pause to note that this definition of Jordan subring matches the “quadratic” Jordan theory, which handles all characteristics including characteristic 2. The following formal argument shows that the ‘‘linear’’ Jordan product sls2 s2sl E S for all sl, s2 in S :
+
+ l)s,(s, + 1 ) E S . Thus sls2s, + (sIs2+ s2s1) + s2 E S, implying slsz + szsl E S since the other s1
+ 1E S
so
(sl
two terms are in S . Incidentally, the characteristic 2 case is usually more delicate than characteristic # 2 and often requires different methods to handle it since then K = S . The Lie and Jordan structures of K and S have been tied to the structure of R by some lovely theorems of Herstein and others, cf., exercises 5ff as well as Herstein [69B], Herstein [76B], Martindale-Miers [SS], and Benkart [76]. We can build new involutions by means of inner automorphisms.
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298
Remark 2.13.5: Suppose u E Unit(R). If (*) is an involution and u* = Lu then R has another involution ( J ) given by r J = (u-'ru)*. Indeed
(rJ)J= (u-'(u-'ru)*u)* = (u-lu*r*(~-')*u)* = (u-1ur*U-lu)* = I** = r; (rlr21J = (u-lrlrZu)* = (u-lrluu-'r2u)* = r ; r i .
Note uJ = u*. Let p,,: R + R be the group isomorphism given by left multiplication by u, i.e., p,,r = ur. Write K = (R,*)-, K' = ( R , J ) - , S = (R,*)+, S' = (R, J)'. If u E S then pus = S' and p,,K = K'. If u E K then pus = K ' and p,,K = S'. (This is all clear since (ur)-' = (u-'uru)* = (ru)* = u*r*.) Heuristically remark 2.13.5 gives the same type of involution if u E S and gives a different type if u E K (since the role of symmetric and antisymmetric have been reversed). We illustrate this idea in the following very important example.
Example 2.13.6: Suppose R = M,(C) where C is a commutative ring with 3. Then K = (R,t)- and S = (R,t)+ are free C-modules of respective dimensions in(. - 1) and $n(n l), having respective bases {eij - eji: i c j } and {eij+ eji: i I j}. If n = 2rn is even then viewing an n x n matrix partitioned as rn x rn
+
matrices in the form
(: ;)
we can take u =
(y
-3
EK
(where I is the
identity m x rn matrix), and form a new involution, as in remark 2.13.5, called the canonical symplectic involution (s). Explicitly (s) is then given by the formula
One can describe explicitly S = (R, s)+ and K = (R, s)-, as follows: S
=
I(:
:):a'
= d, b' =
-b, and c' = - c
I
,
a free C-module of dimension in(n - 1) having a base comprising all i , j I rn and ei,,,+ - ej,m+iand elements of the forms eij + em+j , , , + i for 1 I em+, for 1 Ii < j I m. K={(:
f;):a'=-d,b'=b,andc'=c
I
,
299
$2.13 Rings with Involution
+
a free C-module of dimension 3n(n 1) having a base comprising all element of the forms eij - e m + j , m +for i 1 I i , j I m and e i , m + + j e m + j , iand e m + , , + e m + j ,for i 1 I i 5 j I m.
The Category of Rings with Involution It is time for us to view involutions categorically. Namely, we consider the category %‘i+zy-(*) whose objects are rings with involution (R,*); the morphisms must respect this structure, so we define a homomorphism ( R , *) + (T,J) (where T is a ring with involution J) to be a ring homomorphism f:R .+ T satisfying f ( r * ) = ( f r ) Jfor all r in R . Homomorphisms of rings with involutions are called (*)-homomorphisms for short and lead us to the following definition. Definition 2.23.7.- A is an ideal of ( R , *), written A 4( R ,*), if A is an ideal of R satisfying A* E A. If A 4( R ,*) we also say A is a (*)-ideal. Remark 2.23.8 If f:( R ,*) + (T, J) is a homomorphism then kerf 4( R , *). (Indeed if f r = 0 then f r * = ( f # = 0.) Conversely, if A 4( R , * ) then R I A has an involution (*) given by ( r + A)* = r* + A, and the canonical homomorphism R -,R / A is in fact a (*)-homomorphism. Remark 2.23.9: (i) If A 4( R , * ) then A* = A since A* E A = A** E A * . (ii) If A 4 R then A* a R so A + A*, A n A * , AA*, and A*A are ideals of ( R , *). (iii) If A 4(R, *) then Ann, A 4( R , *). Now that we have the “right” definition for ideal we can define (*)analogues of the standard ring-theoretic concepts. Definition 2.23.20: ( R , *) is simple if 0 is the only proper (*)-ideal. ( R ,*) is prime if AB # 0 for all nonzero (*)-ideals A, B. ( R , *) is semiprime if A 2 # 0 for every (*)-ideal A # 0 of R . We often use the prefix “(*)-” to emphasize the role of the involution, as in (*)-ideal, (*)-simple,(*)-prime, etc. The following example shows the (*)-structure theory is a true generalization of the noninvolutory theory. Example 2.23.22: If R is any ring then R 0 RoPhas the exchange involution o given by ( r l , r 2 ) 0= ( r 2 , r l ) .If R is simple (resp. prime, semiprime) then so
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is (R 0 ROP, o), because ideals of (R @ ROP, 0 ) all have the form A 0 AoP for A 4 R. (Indeed, if (al, at) E B q (R @ ROP, 0 ) then (al, 0) E B and so (0,al) = (a19 0)" E B.1 Note that (R @ ROP, 0)' = {( r): I r, E R} x R as a Jordan ring, whereas (R 0 RoP,0)- = {(r, - r):r E R} x R as a Lie ring. Remark .2.13.12: Suppose R has an involution (*), and A 4R with A n A* = 0. Then (*) induces an anti-isomorphism from RIA to RIA*, so RIA* x (R/A)OP.Thus there is a natural injection (R, *) -+ (R/A 0 R/A*,O) given by r --t (r + A,r + A*). In view of example 2.13.11, the center of a (*)-prime (resp. (*)-simple) ring with involution need not be prime (resp. simple). To rectify this we need a (*)-center. Defnjrion 2.13.13: Z(R, *) = { z E Z(R): z* = z } . We say (R, *) has the first kind if Z(R) = Z(R, *); otherwise, (R, *) has the second kind. Remark 2.3.14: If z E Z(R) and z* = & z then Rz a (R, *). In particular, if (R, *) is simple (resp. prime) then Z(R, *) is a field (resp. a domain). Remark 2.13.15: If (R, *) has the second kind then (*) induces an automorphism of order 2 on Z ( R ) over Z(R, *). Remark 2.13.16: If 2 = Z(R, *) and H is a 2-algebra then involution of R BZH.
*01
is an
The following (*)-version of central localization will be useful to us.
Proposition 2.13.17: If S is a submonoid of Z = Z(R, *) then 1 0 (*) is an involution of S-'Z BZR x S ' R , which we thereby denote as (S-'R, *). The canonical map R --t S-'R given by r --t l - ' r is a (*)-homomorphism; if S is regular this is an injection and Z(S-'R,*) = S-'Z. Finally, if (R,*) is prime then (S-'R, *) is also prime. Proofi The first two assertions are immediate. S-'Z c Z(S-'R, *) by proposition 1.10.13. If S is regular and s - l r E Z(S-'R, *) then r E Z ( R ) by proposition 1.10.13 and s - l r * = ( s - l r ) * = s - l r implies r = r*; so r E 2, proving S-'Z = Z(S-'R, *).
52.13 Rings with Involution
301
Finally suppose ( R ,*) is prime. Then S is regular. Furthermore, R has a prime ideal P with P n P* = 0; thus S-'P n S-'P* = 0 (seen by clearing denominators), and S-'P E Spec(S-'R), proving (S-'R, *) is prime. Q.E.D.
Involutions of Rings with Minimal Left Ideals Our tour of the structure theory begins with R semiprime with minimal left ideals, where we can garner surprisingly strong results.
Lemma 2.13.18 Suppose a semiprime ring R has a minimal left ideal L = Re, with e idempotent. If R has an involution (*) and L*L # 0 then e = eb'e*be for suitable b, b' in R. Proof: e*Re = L*L # 0 so 0 # (Re*Re)2= Re*ReRe*Re so there is b' in R such that 0 # eb'e*Re, which is a right ideal of the division ringeRe (cf., Q.E.D. proposition 2.1.21), so e E eb'e*Re.
Proposition 2.13.19: Suppose R is a semiprime ring with an involution (*) and soc(R) # 0. Then a minimal nonzero le) ideal L of R can be chosen satisfying one of the following conditions: (i) L has a symmetric nonzero idempotent, or (ii) End, L is a held and a*a = 0 for all a in L , or (iii) L*L = 0.
Proof: Case I. There is some minimal left ideal L with La*a # 0 for some a in L. Let s = a*a E (R,*)+.Lemma 2.1.19 says L has an idempotent e with es = se = s, and Ann,s = 0. But e*es = e*s = (se)* = s* = s so (e*e)2- e*e E Ann, s = 0, proving e*e is a symmetric idempotent in L, yielding (i). Case ZI. La*a = 0 for every minimal left ideal L and for all a in L. We claim some minimal left ideal L has the following property: Property P. a*a = 0 for all a in L. Note that if L lacks property P then LL* = 0. Indeed, for any a in L with a*a # 0, we have Ra*a = L so LL* = L(Ra*a)* = La*aR = 0. To prove the claim, take any minimal left ideal L of R and a # 0 in L . Then aR is a minimal right ideal of R by proposition 2.1.27 (whose proof
Basic Structure Theory
302
only requires R is semiprime), so Ra* = (aR)* is a minimal left ideal. We are done unless each Ra* lacks property P, so O = Ra*(Ra*)* = Ra*aR, proving a*a = 0 for all a in L, as claimed. Taking L with property P let e # 0 be idempotent in L (cf., proposition 2.2.11) For any r in R we have e*r*re = (re)*re = 0, so e*r*e + e*re = e*(r* + l)(r + l ) e - er*re - e*e = 0, proving e*r*e = -e*re. Suppose (iii) does not hold. We shall show End, L = eRe is a field, thereby yielding (ii). By lemma 2.13.18, we have e = eb’e*be for suitable b, b‘ in R ; thus for all r,, r2 in R we have
e*ber,er,e = -(e*rfe*b*e)r,e = -e*rf(e*b*er,e) = e*rfe*r:e*be = -e*b*er,er,e
= e*ber,er,e,
yielding e*be[er,e,er,e] = 0 so, 0 = eb’e*be[er,e, er,e] = e[er,e, er,e] = [er,e, er,e]. Thus eRe is commutative and is thus a field.
Q.E.D.
In order to understand this result we turn to the following slight generalization of bilinear forms:
Definition 2.13.20: Suppose V is a right vector space over a division ring D. A map (,): V x V+D is a sesquilinear form if (Ciui,Cjwjdj) ( u i , wj)dj for all u i , wj in V and d j in D.(,) is nondegenerate if (V, u ) # 0 for all u # 0. (,) is alternating if (u, u ) = 0 for all u in V ; (,) is Hermitian with respect to a given involution (*) on D if (ul,u2) = ( u 2 , u 1 ) * for all ui in V.
=ci,j
Again, this definition of alternating form was formulated to include characteristic 2, and readily implies ( u l , u z ) = - (uz,ul) since (v1,uz)
+ (VZ,Vl)
= (V1
+ UZ,Ul + u 2 ) - (V1,VI)
- ( ~ , , V , ) = 0.
In case D is a field fixed by (*) we note that a Hermitian form becomes the more familiar symmetric bilinear form. To bring in involutions we introduce the following set-up. R is a ring with involution, L is a minimal left ideal containing an idempotent e, and D = eRe = eL. We say (,):L x L + D is (*)-compatibleif (,) is nondegenerate and (rxl,xz) = (xl,r*x2) for all r in R and x,, x, in L. Theorem 2.13.21: Suppose R is a semiprime ring with involution (*), and soc(R) # 0. Then there is a minimal nonzero left ideal L satisfying one of
303
g2.13 Rings with Involution
the following properties (viewing L as right vector space over D = End, L): (1) L has a (*)-compatible Hermitian form which is not alternating. (2) L has a (*)-compatible alternating form and D is a field. (3) L*L = 0.
In fact, we may choose L in (l), (2) such that for any x,, x2 in L there is r in R with rx = x1(x2,x) for all x in L. Proof;. We shall show the conditions of proposition 2.13.19 imply the respective conditions here. If (i) holds then taking L = Re with e* = e idempotent we define the nondegenerate Hermitian form (,) by (ale,a 2 e ) = ea:a2e. This is (*)-compatible so (1) holds; moreover, taking r = qea.2, we have rx = a,eatx = a,e(a,e, x) for all x in L, yielding the last assertion. (iii) and (3) are the same. Finally suppose (ii) holds, but not (iii). Then we have L = Re with a*a = 0 for all a in L, and e = eb'e*be for suitable b, b' in R , by lemma 2.13.18. Define (a,e,a2e) = eb'e*a:a,e, an alternating (*)compatible form, which is nondegenerate because if 0 = (Re, a e ) = eb'e*Rae then 0 = eb'e*beRae = eRae so (Rae), = 0, implying Rae = 0 and ae = 0. Thus (3) holds; moreover, taking r = a,eb'e*af we have rx = a,e(a,e,x) for Q.E.D. all x in L, as desired. Example 2.23.22: Suppose R = M,(F). If (*) is the transpose involution then take e = e l , and L = Re,, and (x,,x,) = tr(x:x,) E F for all xi in L. Thus theorem 2.13.21(1)holds where (,) is the trace bilinear form. The matric unit r = eij corresponds to the transformation x --t e i l ( e j l , x ) . If (*) is the canonical symplectic involution then n = 2m for some m. Again take e = e , , and L = Re,,, but now define (x,,x,) = tr(e1,,,,+,x~x2)for xi in L. To show (2) holds note for x = aieil that
I;= x* = 2 aiem+,,i+m - 2 rn
n
i=1
i=m+l
+ I:=,+,
aiem+l,i-m
so x*x = - aiai+mem+l,l ai-maiem+l.l= 0 for all x in L . Here the transformation x -P eil(ej,,x) corresponds to left multiplication by ei,j+mfor j < m, and by -ei,j+mfor j > m. In some sense these are the only possible involutions when F is algebraically closed, as we shall see in the upcoming discussion.
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Basic Structure Theory
Simple Rings with Involution To apply the preceding results as fully as possible, we need more general results about (*)-simplerings.
Lemma 2.13.23: (R, *) is simple iff R has a maximal ideal A with A n A* = 0.
Proofi (a) If A a R is maximal then A n A* a ( R , *) so A n A* = 0. (F) Suppose B 4 (R, *). We claim B = 0. If B c A then B = B* c A* so B E A n A* = 0. So assume B $ A. Then A + B = R so A* + B = (A* + B ) ( A + B ) c A*A + B = B implying A* c B. Hence A c B implying A = B Q.E.D. since A is maximal. Contradiction. Proposition 2.13.24: If ( R , * ) is simple then either R itself is simple or R x ( R , 0 RYP, 0 ) for a simple homomorphic image R1 of R . Proof: Apply lemma 2.13.23 to remark 2.13.12 and the Chinese Remainder Theorem. Q.E.D. Corollary 2.13.25: If (R, *) is simple of the Jirst kind then R is simple. Corollary 2.13.26: If ( R ,*) has the second kind and R is simple then taking H = Z ( R ) and 2 = Z ( R , *) in remark 2.13.16 yields ( R @,H,*) x ( R @ RoP,0). Proofi Let R' = R BZH . Then A = Ann(1,O) is a maximal ideal of R' with A n A* = 0; since R'IA x R we can conclude using lemma 2.13.23. Q.E.D.
In view of proposition 2.13.24 and corollary 2.13.26 we may continue our study of simple (R, *) by assuming R is simple and (*) is of the first kind, for, otherwise, we have example 2.13.11. Definition 2.13.27: If R is simple Artinian we say an involution (*) on R has orthogonal type if some minimal left ideal L has a nondegenerate (*)compatible Hermitian form; (*) has symplectic type if some minimal left ideal L has a nondegenerate (*)-compatible alternating form.
The key is that every involution of a simple Artinian ring has symplectic or orthogonal type by theorem 2.13.21. We need to quote a standard fact.
$2.13 Rings with Involution
305
Proposition 2.13.28: Suppose V is a j n i t e dimensional vector space over a j e l d F, having a nondegenerate bilinear form (,). (i) If (,) is alternating then n is even and writing n = 2m we have a base v l , ..., v, of V such that ( v ~ , v , , , + ~ ) =f o~r all 1 1 i 1 m and ( v i , v j ) = O whenever Ii - j l # m. (ii) If (,) is symmetric then there is an orthogonal base v l , . . .,on of V, i.e., ( v i , v j ) = 0 for all j # i. Proof: Jacobson [85B, $6.2 and $6.31.
Q.E.D.
We are ready for a decisive result concerning matrices.
Theorem 2.13.29: Suppose R = M,,(F) has two involutions (*) and ( J ) of the jirst kind, where F is a j e l d of characteristic # 2. These involutions have the same type iff there is a finite dimensional j e l d extension H of F such that, extending the involutions naturally to M,,(H) x M,(F) OFH we have ( M J H ) ,*) x (M,,(H),J).Furthermore, i f (*) and ( J ) are of symplectic type then we can take H = F. Proof: Step 1. We first show that if (*) and (J) both have symplectic type then (M,(F),*) x (M,,(F),J).Write for (*) and (*J for (J).By definition, for u = 1,2 there is a minimal left ideal L, with a nondegenerate (*,)-compatible alternating form (,),, with respect to which we have a base v'i"), .. .,of) as in proposition 2.13.28(i). Write rG' for the element of R satisfying riy)x = v I u ) ( v y ) , x ) , for all x in L. Let m = n/2. For u fixed the r:;) are independent over F. Indeed, if j = aijr!T' = 0 then for each x in L and each i I m we have (dropping the subscripts and superscripts u)
I:,
o=
(
-v
.
i aijrijx) i (vi,u,+i)aij(vj,x)
m + i , i .j = 1
=
j= 1
=
c aijuj,x),
( j 1 1
implying aijvj = 0 (since, (,) is nondegenerate), so aij = 0 for all i I m and all j ; the analogous argument shows aij = 0 for all m c i 5 n. By dimension count we now see {r!;): 1 I i, j I n } are a base of R as vector space over F, so there is a vector space isomorphism cp: R --+ R given by cp(Caijrlj))= Cu..r!?. 11 V In fact, cp is a ring homomorphism and thus an isomorphism; to see this it suffices to show cp(r:j)rki))= rf)r;i), which is clear
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Basic Structure Theory
since they have the same action on each x in L, . A similar argument shows cp is a (*)-isomorphism, as desired. Step 2. Assume now that (*) and (J)each have orthogonal type. For u = 1,2 there is a minimal left ideal L, with a nondegenerate (*)-compatible symmetric form (,) with respect to which we have a base oy), ..., v r ) as in proposition 2.13.28(ii). Since (,)" are nondegenerate, each (or), vy') is a nonzero element of F, which we call 1:.' Let H be the field extension of F of dimension 2 22" obtained by adjoining each square root yy' of pi"). Replacing F by H and v y ) by ( y y ) ) - ' u y ) , we may assume each ( u p ) , $ ) ) = 1, and choosing r;;) as in step 1 we see by an analogous argument that (M,,(H),*) z (Mn(H),J ) . Steps 1 and 2 constitute a proof of (a). Now let S = (M,,(F),*)+.Taking ( J ) to be the canonical symplectic involution we see [ S : F ] = i n ( . - 1) if (*) has symplectic type; taking (J) to be the transpose we see [ S : F ] = [(M,,(F),*)+:F ] = $n(n + 1 ) if (*) has orthogonal type. Thus the type is distinguished by the dimension of S over F, proving (e). Q.E.D.
Corollary 2.13.30: Hypotheses as in theorem 2.13.29. If F is algebraically closed then (M,,(F),*) x (M,(F),J)ifl(*) and ( J ) have the same type. We see the subdivision of involutions of the first kind into orthogonal and symplectic types is decisive in the analysis of matrix rings with involution. This will be extended later to more general situations including finite dimensional central simple algebras.
(*)-Radical and (*)-Semiprime Definition 2.23.31.- If (R,*) is a ring with involution and M is an R-module define Ann(,,,, M = { r E R: rM = 0 = r * M } . M is (*)-faithful if Ann(,,,, M = 0. (R, *) is primitive if R has a simple, (*)-faithful module. Proposition 2.13.32: The following assertions about (R, *) are equioalent: ( i ) (R, *) is primitive. ( i i ) Ann(,,*) RIL = 0 for some maximal left ideal L of R. ( i i i ) R has a primitive ideal P with P n P* = 0.
Proofi (i) *(iii) Take P = Ann, M where M is simple and (*)-faithful. (iii) a (ii) The primitive ring RIP has a faithful simple module M x R/L where L is a maximal left ideal of R. Then Ann(R/L) E P so Ann(,,,, R/L E Q.E.D. P n P* = 0. (ii) =. (i) Take M = R / L .
$2.13 Rings with Involution
307
There does not seem to be a good (*)-analog for the density theorem in general, although there is an excellent version for (*)-primitive rings having minimal left ideals (cf., exercises 3,4). The next step in the structure theory is noting that Jac(R) is a (*)-ideal; indeed, an ideal A is left quasi-invertible iff A * is right quasi-invertible.
Corollary 2.13.33: If (R, *) is a ring with involution then (R/Jac(R), *) is a subdirect product of (*)-primitive rings. Proof: Let J = Jac(R). If P is primitive ideal of R then J = J* E P* so J E P n P*. Thus n { P n P*: P is a primitive ideal of R} = J, so R is a subdirect product of the RIP n P*, each of which is primitive by proposition 2.1 3.32(iii). Q.E.D. If A is a nil ideal of R then A* is also a nil ideal. Thus Nil(R) is a (*)-ideal, and Amitsur’s theorem is applicable without modification for rings with involution. However, we turn to prime and semiprime rings.
Proposition 2.13.34: A ring R with involution (*) is (*)-semiprime iff R is semiprime. Proof: (e) a fortiori. (a) Suppose A 4 R and A 2 = 0. Then ( A n A * ) 2 = 0 implying A n A * =O. Hence (A + A * ) 2 = A 2 + ( A * ) 2+AA* +A*A = ( A * ) 2 = ( A 2 ) * = 0 so 0 = A + A* 2 A, as desired. Q.E.D. In particularly, any (*)-prime ring is semiprime. But, in fact, we can characterize (*)-prime rings.
Proposition 2.13.35: A ring R with involution (*) is (*)-primeiff R has a prime ideal P with P n P * = 0. Proofi (e) If A, B 4(R,*) with AB = 0 then AB E P so A E P or B E P. Assume A E P; then A = A* E P * so A E P n P* = 0. (a) By Zorn’s lemma there is an ideal P of R maximal with respect to P n P* = 0. We claim P E Spec(R). Indeed, suppose A 2 P, B 2 P and ABGP. Then ( A n A * ) ( B n B * ) ( A n A * ) E P n P * = O implying A n A * = O or B n B * = 0. By maximality of P we get A = P or B = P, as desired. Q.E.D.
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Basic Structure Theory
Exercises g2.I 1. Suppose 1 is a simple R-module with D = ld,M, and x1,...,X , E M are arbitrary. If V is a finite D-subspace of M and x 1 # V then there exist elements d, = 1, d2,.. .,d, in D such that for any x in M there is r in Ann, V such that rx, = xdi for 1 I i I n. (Hint: Enlarge a base of V to a base of V + x,D containing x, and apply the density theorem to this set.) This formulation of the density theorem is amenable to computation. 2. Suppose M is a faithful simple R-module with D = End, M. If L < R with LM # 0 then (as a ring without 1) L is dense in End MD. In particular, this holds if L R. 3. Suppose R, is a rng. If 0 # L < R, then L n Ann ’L 4 L. If R, is prime (resp. primitive) then so is the rng L/L n Ann’L. (Hint: M / { x E M: L x = 0 ) is faithful and simple.) 4. Let I be the ideal of F { X l , X 2 }generated by XlX2 - X 2 X , - X,. If char(F) = 0 then F { X,, X2}/Iis primitive with 0 socle. 5. Let R = End MD where M is a right vector space over D. Then R/soc(R) is a prime ring. (Hint: If a, b E R do not have finite rank then find r such that arb does not have finite rank.) 6. An example of a primitive ring R such that R/soc(R) is not prime. Let R be the subring of T = End MD generated by soc(T) and a transformation a given by ax, = xi if i is even and ax, = 0 if i is odd. Then aR(1 - a ) E soc(R) but a, 1 - a 4 SOC(R). 7. Display the Weyl algebra explicitly as a dense subring of End MD. 8. (RielTel’s short proof of the Wedderburn-Artin theorem). Viewing any left ideal L of R as R-module put T =End, L and R’ = End L T . For any x, y in L and f in R’, note (fx)y = f ( x y ) because right multiplication by y is an element of T. Now assume R is simple. Viewing R E R’ by the regular representation, conclude L is a left ideal of R’ so 1 E R = LR I R’ proving R‘ = R is simple. To prove the Wedderburn-Artin theorem note that if L is a minimal left ideal then T is a division ring and 0 # soc R’4 R’ implies [ L : T ] = n for some n < co, implying R x M,(T). 8’. If R is a prime ring with a left ideal L satisfying LR = R then R x End LT where T = End, L. (Hint: Same proof as above.) 9. Perhaps the slickest proof of the Wedderburn-Artin theorem: Suppose R is simple with minimal left ideal L. Write 1 = airi for a, E L, ri E R with n minimal. Then there is an epic JI: L(”)-+ R given by ( x , , ...,x,) + xiri; show ker $ = 0 because n is minimal. Hence R FZ End, R x End,L ),( x M,(End,L) by exercise 1.5.0.) 10. Prove directly that any domain R with a minimal left ideal R a is a division ring. (Hint: a E Rra for any r in R.)
Primitivity of Free Products Following Lichtman [78] we shall show that the free product of two nontrivial algebras over a field is primitive provided one of them has dimension 2 3. This hypothesis is needed because of the following example:
Exercises
309
+
2 F[Z/2Z], i.e., R, = F Fa, where a! = 1. Then R I U R , x F I G ] where G is the group ( a , , a , : a : = a: = 1). Show a = a1a2has infinite period, and ( a ) G of index 2; furthermore, R , U R , is not primitive. (Hint: Embed G in GL(2,Q) by a, - + e l , e,, and a, + 2 e 1 2 + $ e 2 1 . Since (a) has index 2, one can then embed F[G] into M,(F[(a)]) by the regular representation. But no nontrivial group ring is simple, so using the density theorem show that F[G] cannot be primitive. A faster conclusion would be via Kaplansky’s theorem on polynomial identities.)
11. Suppose R , z R ,
+
In exercises 13-16 we carry the following set-up: C is a commutative domain and R , , R , are C-algebras which are free as C-modules having bases each containing 1. For k = 1,2 denote the base of R, as { 1 ) u Bk. Enumerate B , as {e,: i E I}, and 8, as { X :j E J } . Let W = {words alternating in the e, and A}; then R , U R , exists and is a free C-module with base W u { 1). Assume lBll 2 IB21. Let f i = {words of W starting with an element of B,}. For example, f i e l E and ezfle, E “w;. Let Y = {words of of even length}, and let T be the C-subspace of R , U R , spanned by f 12. T is a subalgebra of R I U R , isomorphic to the free C-algebra (generated by the eiA, which act as indeterminates). 13. A n T # O for every nonzero ideal A of R , U R , . (Hint: Let O # a E A . Multiplying on the left by el and on the right by f,,if necessary, one may assume some word in supp(a) of maximal length n is in T. Writing euel = a, ~ a U k and e k f.fl = p, c a u k f , for suitable a,,/3. in C, show modulo T that e.(e,f,) = a.f, and e,,fte,(e,f,) = a,e,ftf, = aJ,e,. Thus any term of a’ = (e,fl)”a(e,fl)”+l not in T is congruent to ( e l j l ~ e l ( e l j lfor ) ~ suitable i, j with j > 0. Then 0 # a’e,f, a,/?,a’E T n A. 14. Let R = R I U R,. For each 0 # A 4 R choose an element s, in T n A, and let S = {sA: A Q R } . R is primitive if either of the following holds: (i) R is countable
+
+
and !B,122.(ii) I B , J 2 R .(Hint: (i) S is countable so enumerating S as {s,,~,,..,} define r , = 1 + sne,j,(eJ,)” + f,s,e,(f,e,)“, and let L = Enm= Rr,. Matching coefficients of words ending in e,f,(e,j,)”, one sees l$L. But each nonzero ideal A contains some s,, so 1 E Rr, A c L A, implying R is primitive.) (ii) B , has some subset which can be listed as {e,:sES}, taking r,= 1 +f,se,+ seJ, and L = x S E S R r Sshow , as in (i) that L is comaximal with every ideal and thus R is primitive. 15. If C is a field and 1 B , 1 ~ 2then R is primitive, without any restriction on R. (Hint: It suffices to find S as in exercise 14 with S countable or IS1 5 IB11, since then one can conclude as in exercise 14(i) or (ii), respectively. Suppose s = aLwkE A n T where wk E V are words and akE C. Replacing s by (e,fl)”+ls(e,fl)”)”C1 = Cakukwhere u k = (e,fl)”+’w,(e2f,)”+’ we see u , , ..., uk are algebraically independent. But then some higher commutator of the uk is in A n T, and there are at most lBIl of these if (Ellis infinite. For example, if s = alwl + azw2 + a3w3then
+
+
16. For a > 2 show example 2.1.36 is not a unique factorization domain. (Hint: pro-
perty (iii).)
Basic Structure Theory
310
,
17. Suppose R has a sequence of idempotents e l , e,,e,, ... such that each e l + E e,Re,. Then there is a primitive ideal P such that each ei # P. (Hint: Take a maximal left ideal Lcontaining R ( l - e,) and let P = core(,!,).)
1
$2.2 1. A ring R is subdirectly irreducible if n{nonzero ideals of R} 2 0 . Prove every ring is a subdirect product of subdirectly irreducible rings. (Hint: Given r in R take A, R maximal with respect to r # A,. Then n { A r : rE R} = 0 and R / A , is sub-
directly irreducible.)
f,,.. .,f; of pairwise relatively prime polynomials then C[A]/(f) = C[A]/(f,). 3. If A, E are comaximal ideals of R then Amand B” are comaximal, for any m, n in N. 4. Suppose A,, .. . ,A, are comaximal ideals of R. For any r l , . . .,r, in R and any m , , ... ,m, in N there is some r in R such that r - ri E A? for 1 < i < 1. (Hint: Apply exercise 3 to proposition 2.2.1.) This is a useful “approximation” result.
n:=,
2. Suppose C is a commutative ring and f E C[A]. If f is a product
$2.3 1. (Lattice version of Jordan-Holder theorem) A lattice L with 0, 1 has a composition chain a, = 0 c a , c a c ... c a, = 1 if each a, covers a,- If L is a modular lattice
,.
2.
with a composition chain then every chain can be modified to a composition chain of the same length. Moreover, defining h(a) as the length of a composition chain ending in a, one has h(a) + h(b) = h(a A b) + h(a v b). (Hint: Use exercise 0.0.3.) If M is an R-module of length n then every R[A]-submodule N of M[A] is spanned by at most n elements. Here we view M[A] naturally as module over R[A], as in exercise 1.7.6. (Hint: Viewing the elements of M[d] as polynomials, let Ni = {leading coefficients of elements of N having degree
xi=,
3. 4.
5.
6.
Exercises
31 1
7. Suppose R is a ring not necessarily with 1. If R is semiprime Artinian then R has 1 after all. (Hint: If R is prime then it is primitive, so use the density theorem. In
general, show R has only a finite number of prime (and thus maximal) ideals.
$2.4 1. (Jacobson's density theorem for completely reducible modules) Suppose M E RMud is completely reducible and r = End, M. For every f in End MT and any x , , . . .,xn in M there is r in R with rxi = f x i for 1 5 i 5 n. (Hint: For n = 1 write
M = R x , B N and let I [ E Tbe the projection from M to Rx,; then fx, =f(xx,)= n f x , so fx, E Rx,. For arbitrary n define f = (f,. ..,f):M(") M(") and note for T' = End, M(") that f~ End MT.. By the previous case there is r in R with r ( x l , ... , x n )= f ( x , , .. . ,x,) = ( f x , , . . . ,fxn).) Note that this short proof (found in Lang [65B,p. 4441) yields Jacobson's original theorem as a special case. 2. soc(M) is a completely reducible module which contains every completely reducible submodule of M. 3. If M is a completely reducible R-module then R/Ann, M is semiprimitive. (Hint: Take the intersection of the annihilators of the simple components of M.) Conversely, if R/Ann, M is semisimple Artinian then M is a completely reducible R-module.
$2.5 Characterizations of Jac(R)
-
+ r2 - rlr2 = 1 -(1 - rl) (1 - r2).Then o is associative and r o 0 = 0 o r = r. Jac(R) is a group under 0 , and r + 1 - r defines a group injection Jac(R) + Unit(R). 2. Jac(R) = { r E R: r + a is invertible whenever a is invertible}. (Hint: (r + a)-' = (1 + a-'r)-'a-'.) 3. Jac(R) = (-){cores of maximal left ideals of R}. (Hint: proposition 2.1.9.) 4. A (Jordan) inner ideal is an additive subgroup A of R such that ara E A for all a in A and r in R. (Examples: (i) any left or right ideal is inner; (ii) the intersection of inner ideals is inner.) Show Jac(R) = (-){coresof maximal inner ideals of R}, where the core of an inner ideal is the sum of all ideals contained in it. (Hint: ( 2 )Every maximal left ideal is contained in a maximal inner ideal by Zorn's lemma, and their cores must be equal. ( G ) Otherwise, Jac(R) $ core(/) for some maximal inner ideal I, so 1 + Jac(R) is an inner ideal containing 1 and is thus R. Then r + a = 1 for suitable r in Jac(R) and a in I; hence a is invertible implying 1 = R.) Note that we could define R to be inner primitioe if R has a maximal inner ideal whose core is 0. As just shown, all primitive rings are inner primitive, but the converse is false (since "inner primitive" is left-right symmetric). These ideas have been used with astounding success by E. Zelmanov in studying Jordan algebras (cf., Jacobson [81B]). 1. Define the circle operation o on R by r , o r2 = r ,
Basic Structure Theory
312
5. (For those who have been following Banach algebras) Suppose A is a Banach algebra over F and let J = {z E A:(az)" + 0 for each a in A} and J' =
{z E A:(za)" + 0 for each a in A}. Then J is a quasi-invertible left ideal. (Hint: Take a E F with ltll > 1. Then there is n such that Il(aaz)"ll < 1 for all m > n; hence Il(az)"ll < lal-" so 1 + az + (az)' + ... is dominated by Elal-" and thus converges to (1 - az)-'.) On the other hand, if z E A and 1 - az is invertible for all a in F then the Spectral Radius Formula (Rudin [66B, p. 3551) implies limn-.mllz"II1/" = 0, i.e., z is generalized nilpotent in the standard terminology; consequently Jac(A) is a generalized nil ideal. Conclude Jac(A)= J and symmetrically Jac(A) = J'. Thus Jac(A) contains every generalized nil left (resp. right) ideal of A. 6. Let E be the infinite dimensional exterior algebra in xI,x2, ... over a field F. Then N = E E x i is a nil ideal, and E / N x F; thus E is local. Furthermore, ( N X , ) ~= 0 for all xi in E. On the other hand, if a(resp. 6) is the endomorphism (resp. derivation) of E over F given by x i w xi+l then E [ I ; a ] and E [ I ; 6 ] are prime. ELI; 01 is semiprimitive, whereas Jac(E[I; 63) = ELI; 6 ] N . (Hint: The assertions for c are easy; for example, if 0 # a, b E ELI; a] and no xi appears in a for i > n then al"b # 0, seen by checking the "leading" term in the xi. To prove E [ I ; 61 is prime first note any ideal B # 0 contains a monomial x, * x,. But then x2...x2,= 6 r ( ~ 1 ~ . . ~ , ) ~ , +and 2 . thus ~ ~ ~for 2 tany , i there is suitable i' > i such that B contains xi + lxi + * . .xi,. Conclude AB # 0 for all ideals A, B # 0, for if xl " ' x uE A then x1 .'.xu.E AB.) This example is taken from Bergen-MontgomeryPassman [ 8 7 ] , which contains further ramifications. 7. If Ji = Jac(Ri) for each i in I and 9 is an ultrafilter of I then J a c ( n R i / F ) = 4/9. (Hint: ( 2 )by quasi-invertibility. ( G )If ( a i )$ JJ9then { i : ai 4 J } E 9 so 1 - (ai)(bi)is not invertible for some (bi).)
n
n
Separable Algebras 8. If F is perfect and R is a semisimple Artinian F-algebra then R is separable.
(Hint: To show R & K is semiprimitive one may assume K is finitely generated; using theorem 2.5.42 one may assume K is separable and algebraic.) 9. Suppose D is a division algebra of characteristic p with center F, and d E D - F with d PE F. Then there exists a in D with dad-' = a + 1; in particular, F(a) is not a purely inseparable extension of F. (Extensive hint: Let 6 denote the inner derivation a -+ [d, a ] . Then 6P = 0 for 6Pa = [dp,a ] = 0. Choose b with 0 # [d, b]. Take n minimal with 8"'1b = 0 and put c = (6"-'b)(6"b)-'. Then 6c = 1, so putting a = cd conclude [d, a ] = 6a = 6(cd) = d as desired.) 10. (Koethe-Noether-Jacobson theorem) Suppose D is a noncommutative division algebra which is algebraic over its center F. Then D contains an element not in F which is separably algebraic over F. (Hint: Otherwise, each element of D - F is purely inseparable over F, so we have d in D - F with dP E F. But then exercise 9 yields a contradiction.) Conclude that if [ D : F ] is finite then D has a separable splitting field. (Hint: Split off a piece of D and use induction on [ D : F ] . ) 11. Prove the following improved version of Wedderburn's principal theorem: Let R be a finite dimensional algebra over a field F, and N = Jac(R). If R/N is separable
Exercises
313
as an F-algebra then R x S @ N as R-modules, where S is a subalgebra of R isomorphic to R/N.
Normalizing Extensions (also, cf., exercises 2.12.1off) 12. Strong version of Nakayama's lemma (Robson-Small [81], Resco [82]) Suppose M is an f.g. normalizing R - R bimodule and L c R. If ML = M then there is a in L with M(l - a) = 0. Consequently, ML < M if M is faithful; in particular, this holds for f.g. normalizing extensions, so the hypothesis of proposition 2.5.17 also holds. (Hint: Write M = Rx, where xiR = Rx, and formally x,+ = 0. Then xj = 1 xiaijfor aij in L, so it suffices to prove the following: 12'. Suppose M is an R - R bimodule and L < R and there are xl,. .. ,xt in M with xiR = Rx, for 1 5 i 5 t satisfying equations xi = I xiaijfor 1 2 j 5 t, suitable
c::
,
c:=
aijin L . Then for each n I t there is a,, in L with (CisnRx,)(l- an)E ~ , , , , x , L . (Hint: induction on n. For n = 1 note x l ( l - a l l ) = Ci>,,xiaij. In general, xn(1- 0,,,,)(1- an- ,) = x,aij(l - a,- ,) + xiaij(l - a,,- 1) c C i ZxiL n by b ,bi in L. induction. Write r = (1 - ann)(l- a,- E 1 + L and xnr = ~ , ~ , , x ,for Then x,(r - b,,) E Ci>,xiL so take a,, = 1 - (1 - a,,- l)(r - b,,).) 13. If T is a finite normalizing extension of R then Jac(R) = R n Jac(T). (Hint: exercise 12.) 14. Suppose T is a finite normalizing extension of R and M E T-Aod. Then any large R-submodule N of M contains a large T-submodule of M. (Hint: Write T = C R a , where Rai = a,R and Mi = {x E M:a,x E N),a large R-submodule. Then T ( Mi) is large and T( Mi) = 1 a,( (7 M i ) IN.) 15. (Formanek-Jategaonkar [74]) Suppose T is a finite normalizing extension of R and M E T - A d . (i) M is Noetherian iff M is Noetherian as R-module. (ii) M has a composition series iff M has a composition series as R-module. (Hint: (i) (e) trivial. (a) By Noetherian induction, assume M/K is Noetherian as R-module for every K < M in T - A d . Assume M has an R-submodule N which is not f.g., take N maximal such. Then N is large as R-submodule since N K is not f.g. for any essential complement K of N, so N contains a large T-submodule N o . But N/No is f.g.; also No is Noetherian in T-Aod and thus f.g. in R-Aod. Hence N is f.g., a contradiction. (ii) By (i) and proposition 2.5.29.) 16. If R E T E M,,(R) then Jac(T)" c M,,(Jac(R)).Hint: Every simple M,,(R)-module has the form M(")by corollary 1.1.18, so it suffices to prove Jac(T)"M(")= 0. But M(")has composition length s n in R - A d and thus in T-Aod.) 16'. If R c T are rings and T x R'") as R-module then Jac(T)" E Jac(R)T; in particular, if Jac(R) = 0 then Jac(T) is nilpotent. (Hint: View T G M,,(R) by the left regular representation, so any element of Jac(T)" can be viewed as a matrix (rij) with rij in Jac(R). Writing 1 = C rixi one has r = l r = ririjxjE Jac(R)T.) 17. Suppose S = C i ERa, , where the a, are R-normalizing elements of S, and, moreover, S # I, l Ra,. , (This can be arranged if I is finite.) If S is a simple ring then there are maximal ideals { Mij:i, j E I} of R such that Mij = 0; in particular, if I is finite then R is a finite direct product of simple rings. (Hint: Embed Ra, in an R - R sub-bimodule M of S maximal with respect to a, $ M, and let M i j = ( r ~ R : a , r a j ~ M } Q The R . Mij are maximal ideals since if M i j c B Q R then
xi<,,
xi,,,
n
+
c
n,,
xi+
Basic Structure Theory
314
aiBaj + M = S so a,aj E aibaj + M and a,(l - b)aj E M, implying 1 E B. Moreover, S( (-)Mi,)S c M # S, implying (-) Mij = 0. Use the Chinese Remainder Theorem at the end.) 18. Define the Brown-McCoy radical B M ( R ) = (){maximal ideals of R}. If A Q B M ( R ) then B M ( R / A ) = B M ( R ) / A .Moreover, if a E B M ( R ) then 1 E R(l - a)R. (Hint: Otherwise, 1 - a E M for some maximal M a R; but a E M, so 1 E M.) We say A Q R has the BM property if a E R(l - a)R for all a in A. Then B M ( R ) has the EM property and contains every ideal A having the BM property. (Hint: If A $Z M for maximal M Q R then 1 E A + M so 1 - a E M for some a in A, implying a E R(l - a)R E M , so 1 E M.) Despite its superficial similarity to the Jacobson radical, the Brown-McCoy radical has not been very useful because the condition 1 E R(l - a)R is much more unwiedly than quasi-invertibility. 18’.Hypothesis as in exercise 17, show B M ( R ) c BM(S). We shall see later that B M ( R ) = R n B M ( S ) when I is finite. 19. N Q R is strictly nil if N BCH is a nil ideal of R BCH for every commutative C-algebra H. Show N is strictly nil iff N[A,, A,, ...] is a nil ideal in the polynomial ring R [ I , , A,, ...] in countably many indeterminates. Prove the following (easier) variant of Amitsur’s theorem: If R has no strictly nil ideals # 0 then R [A,, A,, . ..] is semiprimitive. (cf. exercise 2.6.6.)
Simple Radical Rng The object of the next few exercises is to present Sasiada’s example of a simple ring without 1 which is its own Jacobson radical. We follow Sasiada-Cohn [67] via Herstein [69B].
20. The monoid S of words in two noncommuting indeterminates X and Y is
ordered lexicographically, yielding a domain F((S)) whenever F is a field; we denote this F{{X, Y}}. Using proposition 1.2.24 show F{{X, Y} is a local ring whose Jacobson radical J is the set of elements having constant term 0. 21. Notation as in exercise 20. Suppose one can show X # ( X - Y X * Y ) . Let 1 2 (X - YX’Y) be an ideal of J (in 41~9) maximal with respect to X # I, and let A be the image of (X) in J/I. Show A is the unique minimal ideal of J / l and A Z # 0, so A would be a (nontrivial) simple radical ring without 1. Thus it remains to show X (X - YX’Y), which is the object of the next three exercises. 22. Let r = X - YX’Y. If C X + dY = XI=, a,rb, in F{{X, Y}} with b , # J then there are bl in J with CX + dY = cr(1 + b ; ) + C I = 2 a i r b ;(Hint: . Write b, = /?wwhere w has constant term 1, and b, = Pi b;‘ where b;’ E J. Then C a i r b i = (/?al /?,a,)rw+ air(b, - /?,w).Thus we may assume b , - 1 E J and bi E J for i 2 2. Write b , = 1 + b;X + h l Y and bi = b;X + h,Y for i > 1. Comparing terms ending in X yields c = a,(l + rb;) + a,rbi. But (1 + rb;)-’ = 1 + Z(-rb’,)j; multiplying through and solving for a, yields the desired result. 23. Suppose dY = airbi with n minimal. Then each bi E J . (Hint: Exercise 22 with c = 0.) 24. Show ( r ) has no monomial and, in particular, X # ( r ) . (Hint: Otherwise, take a monomial c = airbi with c chosen such that n is minimal possible and take such c of minimal degree. If each b, E J then one could match terms ending in the last letter of c and cancel the last letter, producing an example of smaller degree. Thus some bi 6 J. Assume b , J . Note c must end in X by exercise 23. Writing
+
xi,,
+
xi,,
xi,,
cy=,
+
+
Exercises
315
+
XI=,
b',) + a,rb; by exercise 22. This is the case the following claim: For any rn 5 n there are monomials p l , . . .,pm # 1 and elements d,, ...,d,, in F { { X , Y}} satisfying c = ciX one has c , X = c,r(l
rn
= 0 for
c
m
1
n
c l p l . ~ ~ p m X = c lpI".pj-lrdj+ j= 1
ajrdj
withsomed,EJ.
(1)
j=m+ 1
Assume (1) has been obtained, and suppose di E J for all j > m. Taking i minimal with di 4 J and writing dj = aj + d;X + dy Y for aj in F (so a, = 0 for j > m),one matches parts ending in X to get n
1 pl...pj-.lrdi + C 1 m
aic,pl ...pi-,(1
+ f ) = c,
j=m+l
j=
ajrd;,
(2)
where f = c j , i a ; ' a j p i . ' - p j - l E J . Since ai + ( a i l - rdj) is invertible, one can now express c , p , " ' p i - as a combination of < n terms, contrary to minimality of n. Thus dj# J for some j > m.But now the claim can be proved by induction on rn, by applying exercise 22 to (1) in the inductive step. (The computation is rather intricate, cf., Herstein [69B, p. 129-1301 where this notation F ( { X , Y } } , J , r , m replaces his A , A', u, r, respectively.) But now the above argument shows this is impossible for m = n, in view of exercise 22.
Commutativity Theorems The following exercises comprise the classical literature on commutativity theorems, based on the theorem of Wedderburn that every finite division ring is commutative (theorem 7.1.1 1). 25. (Jacobson) If D is an algebraic division algebra over a finite field F then D is commutative. (Hint: Otherwise, exercise 9 yields elements d,a with dad-' = a 1. Then d,a generate a finite division algebra D , over F, which by Wedderburn's
+
theorem is commutative; contradiction.)
26. (Jacobson) Suppose R is a ring in which for every element r there is n(r) > 1 with ,.W = r. Then R is commutative; in fact, R is a subdirect product of fields. (Extensive hint: Obviously R has no nonzero nilpotent elements. Jac(R) = 0 since if r E Jac(R) then r"(r)-' is an idempotent and thus 0, implying r = 0. Note the hypothesis goes over to subrings and homomorphic images; thus we may assume R is primitive. If R is not a division ring then Mn(D)is a homomorphic image of a subring of R for some n > 1; this is impossible since M J D ) has nilpotent elements. Thus R is a division ring of finite characteristic since 2" = 2 for some n > 1. R is algebraic over its characteristic subfield and is thus commutative.) Say a ring R is radicalover a subring A if each r in R satisfies rn(r)E A for suitable n(r) 2 1. 27. Suppose a field K is a radical over a subfield F. Every element a of K - F which is separable over F has the form a = (1 - w)(w' - a)-'for w,w' roots of 1 lying in the normal closure of F,(a) over F,, where F, is the characteristic subfield. (Hint: Let L be the splitting field of F(a) over F, and let 0 # 1 be an automorphism of L with b = cra # a. Then ba-' and ( b + l)(a + l)-' are roots of 1; write b = wa and b + 1 = o ' ( a + 1) = wa 1 and solve for a.)
+
316
Basic Structure Theory
28. (Kaplansky) Suppose K is a field radical over F but not purely inseparable
over F. Then K is algebraic over its characteristic subfield F, and char(K) > 0. (Hint: Take a E K separable over F. The normal closure of F,(a) has only a finite number of roots of 1, so { a + n:n E N} is finite and char(F,) # 0. For any a in F we have a + a and a algebraic over F,, so F is algebraic over F,,.) 29. (Kaplansky-Herstein) If Jac(R) = 0 and R is radical over Z(R) then R is commutative. (Hint: For a division ring this is true by exercises 25 and 28. For R primitive use the density theorem to reach a contradiction unless R is a division ring.) The next main result (exercise 3 1) relies on the Cartan-Brauer-Hua theorem (exercise 7.1.23): If D is a division ring and A is a division subring invariant under every inner automorphism of D then A = D or A E Z(D).
30. Define the hypercenter Hyp(R) = { a E R: for each r in R there is n = n(a,r) such that ar" = rna}. Show Hyp(R) is a subring containing Z ( R ) which is invariant under every automorphism of R. 31. If D is a noncommutative division ring then Hyp(D) = Z(D). (Hint: Other-
wise Hyp(D) = D by Cartan-Brauer-Hua. Suppose a, b E D generate a division subring D,. Every element of D1 has a power commuting with a and b, so D, is commutative by exercise 29.) 32. (Faith) If a division ring D is radical over a proper subring A then D is commutative and A is a field. (Hint: If a E A then some a-' E A implying A is a division ring. Suppose x E D - A. Then xa'"x-' = a , and (1 + x)a'"(l + x)-' = a, for suitable m in N and a, in A. Then am- a, = (a, - al)x implying urn= a, and a, = a,. Thus xu'" = a?. Likewise, for any a' in A, x + a' commutes with a power of a, so a' commutes with a power of a. Hence A = Hyp(A) is commutative, implying A E Hyp(D) = Z(D), so apply exercise 29.) 33. (Faith) Suppose R is radical over A. If R is primitive then A is primitive (dense in the same ring End MD). If Jac(R) = 0 then Jac(A) = 0. 34. (Herstein [75]) If Jac(R) = 0 then the hypercenter Hyp(R) = Z(R). (Hint: Assume R is primitive and not a division ring; use density.) Herstein actually proves Hyp(R)=Z(R) when R has no nonzero nil ideals but the argument is very intricate. 35. (Herstein [76]; Anan'in-Zjabke [74]) If Jac(R) = 0 and suitable powers of r l and rz commute for any r 1 , r 2in R then R is commutative. (Hint: First assume R is a noncommutative division ring. Let C, = { r E R: some power of u commutes with r}. Then C, is a division ring over which R is radical, so R = C, for each a. Hence R = Hyp(R) = Z(R). Pass to primitive and semiprimitive in the usual way.) This is known for rings with no nonzero nil ideals, but is much harder to prove. 36. Let Z = Z(R). If for every r in R there is some r' in Z [ r ] with r'r2 - r E Z then every nilpotent element is central. (Hint: r"(r'r')' - r'r' E Z for some r", so r"r'r3 - r E Z ; continue in this manner.) 37. If R is a ring in which every nilpotent element is central then every idempotent is central. (Hint: (ere - er)' = 0 so ere - er E Z(R)and 0 = (ere - er)(l - e) = -er + ere so er = ere; analogously ere = re.)
317
Exercises
Nakayama [ S S ] proved the following theorem: Suppose C is a finitely generated field, F 2 C is not algebraic over its characteristic subfield, and K is a field extension of F which is not purely inseparable. If for each x in K there is x' in C [ x ] and a number n ( x ) 2 1 such that ~ " ( ~ ) (+1 x'x) E F then K = F. (Compare with Jacobson [64B,p. 183ff.l) Using this theorem Nakayama [59] proved the sweeping result that if Cis a finitely generated ring and if for each r in R there is r' in C [ r ] such that r'r' - r E Z then R is commutative. The bulk of the argument is contained in the following exercise which combines material of Herstein, Martindale, Nakayama, and Utumi: 38. Suppose R E C - d f f and for each r in R there is r' in C [ r ] such that r'r' - r E Z ( R ) . If C is finitely generated and every field which is a homomorphic image of C is perfect, then R is commutative. (Hint: By exercise 2.2.1 one may assume R has a unique minimal ideal A. Let Z = Z ( R ) and J = Jac(R). If J = 0 then R is primitive and thus a division ring (for otherwise a homomorphic image of a subring of
R is matrices, contrary to exercise 36); by exercise 10 we have a splitting field separable over Z, which must be Z by Nakayama's theorem on fields, so R is itself a field. Thus one may assume J#O, so A E J . If aEA and z = a ' a 2 - a E Z then z$zA so z A c A and 0 = zA = (a'a - 1)aA; thus A' = 0 so A E Z by exercise 36. Also every idempotent is central and thus trivial. If r # 0 is not invertible in R then z = r'r' - r is nonzero (for, otherwise, r'r is a a nonzero idempotent so r' = r-'.) Consequently A E Rr. Let B = Ann, A. If Ann,r # 0 then it contains A so r E B. Also [ R , R ] c B since RA E A E 2.Let R = R / B . R is a field since if ra # 0 for a E A then Rra = A and a E Rra. Claim B E 2.Otherwise, some r = [ b , x ] # 0. Let z = r'r' - r E Z n B ; z # 0 since rA = 0. Also Ann(r'r - 1) = 0 since r'r - 1# B . Given a # 0 in A,write a =yr for y in R . For all n note 0 = a(r'r - 1)b" = yzb" = yb"z so yb"r E Ann(r'r - 1) = 0. Since y [ b " , x ] = y ( b " - ' [ b , x ] + [ b " - ' , x ] b ) = y [ b " - ' , x ] b , show inductively y [ b " , x ] = 0 for all n 2 2. Taking b'b' - b E Z conclude 0 = y[b'b',x] = y [ b , x ] = yr = a, contradiction. Now take f E C[LJ of minimal degree with z = f(r)r' - r E 2,where r 4 2. Pick x such that: O # [ r , x l ~ B s ZLettingg . be the formal derivative of / ( l ) L 2 - A then 0 = [z,x ] = g(r)[r, x ] so g(r) E B and g(F) = 0. Thus F is algebraic over the field of fractions of Repeating the procedure one may assume the derivative of g is 0. Writing g = h ( 1 p ) one has h(rP) E B, and proceeding by induction show F is purely inseparable over Thus F E so r E C + B E 2,a contradiction, proving R = Z.)
c.
c.
$2.6 1. If R is a reduced ring and S E R then Ann, S a R and R /Ann, S is a reduced ring. 2. (Klein [ 8 0 ] ) Prove every reduced ring is a subdirect product of domains, by proving for each x that if P is an ideal disjoint from { x ' : i > 0 ) maximal with
318
Basic Structure Theory
respect to RIP reduced then RIP is a domain. (Hint: Otherwise, P is not a prime ideal; take A , B 3 P with A B G P, and using exercise 1 conclude A and B each contain powers of x.) 3. (Amitsur [71]) For each ordinal a there is a ring R, with N,+ ,(R) 3 N,(R). In fact, the R, are defined inductively on a, with the inductive step described as follows: Given a ring T let M , ( T ) be the ring of row-finite matrices of exercise 1.1.5 and view T c M,(T) via scalar matrices (i.e., a -+ a 1). Let T’ be the subring of M , ( T ) generated by its socle (i.e., T = { a 1 + ~finilcaijeif:a,aij E T} of M,(T), and T ’= { a 1 + ~ f i n i t e a i j eaij i , E: Prime rad(T) for i < j } . Thus T c T c T. Given R, = T ‘we will take R,+ = T’. Taking Ro to be any field, this defines R for successor ordinals; for limit ordinals R, is defined as the direct limit of the {R,:p c a}. The desired properties of R, are proved by examining the inductive step. Given B Q T define B’ = {b 1 bUeif:b, b, E B } Q T’; if B E Prime rad(T) then B’ Q T ’and, in fact, N,(R,)’ c N,(R,+) for each y. Writing N, for Prime rad(R,), prove N, = N,(R,) and R,/N, = R o . (Extensive hint: When a = p’ prove, in fact, N, = N,(R,) = Nb and R,/N, x R,/N, by induction, as follows: Define $: R, -+ R,/N, by @(r 1 + x r i j e i j )= r N,. R,/N, x Ro so N, E ker @.But any matrix in ker $ is strictly upper triangular modulo Nb so the left ideal it generates is nilpotent modulo Nb. N, = N,(R,) by induction so N; E N,(R,) and ker$ E N,(R,) E N,; hence equality holds.) 4. If e is an idempotent of R and N = Nil(R) then eNe = eRe n N E Nil(eRe). Show equality need not hold. 5. If L is a quasi-invertible left ideal of R then M,,(L) is quasi-invertible. (Hint: View L G Jac(R).) By adjoining 1, show Koethe’s conjecture is equivalent to the assertion, “If every quasi-invertible left ideal of R is nil then the same holds for Mn(R).” 5’. Show Koethe’s conjecture holds for algebras over uncountable fields. (Hint: theorems 2.5.22 and 2.6.35(iii).) 6. The following result provides an easier substitute for Amitsur’s theorem (2.5.23) and also relates to Koethe’s conjecture. Let R‘ = R[I,, a,, .. .] the polynomial ring in an infinite number of commuting indeterminates. Jac(R’) is nil. (Hint: If a = j ( A , , ...,A,)€ Jac(R‘) then (1 -aA,+l)-l exists but formally is xlPo,oailf+l, implying a” = 0 for some n.) Define Absnil(R) = R nJac(R‘), a nil ideal of R. If Absnil(R) = 0 then Jac(R’) = 0; thus Jac(R’) = (Absnil(R))[d,, a,, .. .].Develop a radical theory for this interesting nilradical. 7. Hypotheses as in proposition 2.6.24, show any nil left or right ideal L of R is contained in the lower nilradical N. (Hint: Otherwise, take a E L - N with Ann, a maximal possible. But Ann(ara) 3 Anna for each r in R so aRa E N, proving a E N, contradiction.) In particular, N = Nil(R). 8. (The Levitzki radical) A subset S of a ring is locally nilpotent if every finite subset of S is nilpotent. Prove there is a unique maximally locally nilpotent ideal of R, called Lev(R),and this contains every locally nilpotent left ideal L of R. (Hint: For the first assertion note if N Q R is locally nilpotent and (S + N)/N is locally nilpotent in RIN then S is locally nilpotent. For the second assertion show every finite set { a 2 r l ,...,a,,,r,,,:aiE L } is nilpotent.)
-
-
. + xfinile
-
+
a
Exercises
319
9. An ideal A of R is locally nilpotent i f f A 8 z T s Nil(R@,T) for every T in Z(R)-dt’ (Hint: (t) Take T = Z{X} and consider a, @ xi).
,
Graded Rings 10. Suppose G is a group, written multiplicatively and g,, ...,gnr E G for n, t E N. Writing hi = g, . .. gi, if { h , , . ..,h,,} take on at most n distinct values then there exist 0 I u,, 5 u1 I ... Iu, 2 n with gu,+,~~~gU,+, = 1 for 0 5 j I t - 1. (Hint:
+
t 1 of the hi are equal, i.e., h,, = h,, = ... = h,,.) 11. Suppose R is graded by a multiplicative group G with unit 1, and JsuppR(= n. R, denotes the identity component of R. (i) If A is a graded subring (without 1)
of R with A, = 0 then A” = 0. (ii) If A < R, and A‘ = 0 then (RA)”’ = 0. (iii) If R has no nilpotent graded ideals then R, is semiprime and (Rr), # 0 for every r # 0 in R. (Hint: (i) Show A,, ...A,” = 0 for every g,.. ..,gn in G. This is certainly true unless g, . . . yi E supp R for each i, in which case exercise 10 is applicable with t = 1. (ii) Note (RA), = A and use exercise 10. (iii) By (ii) and (i).) 12. If R is graded by a multiplicative monoid S then L + RL gives a lattice injection from Y(R,) to {graded left ideals of R}; in particular, if R satisfies ACC (resp. DCC) then so does R,. Conversely, suppose the set-up of exercise 11 (iii) holds and R, is semisimple Artinian. Then R is Artinian; in fact, each R, is f.g. in R,-J&d. (Hint: 1 is a sum of orthogonal primitive idempotents ei in R,, so it suffices to show eiRgej eiR,ei in eiRei-&d whenever eiR,ej # 0. Exercise 1l(iii) shows eiR,ejRhei # 0 where h = g-I; taking xh in ejRheisuch that x,,R,ej # 0 we have a nonzero right ideal of the division ring ejR,ej so xhR,ej = ejR,ej and xhyg= ej for some y,. Conclude eiRsej = eiR,eiy,.)
Group Actions 13. If G is a finite group of automorphisms on a ringR then IGI Prime rad(RG)E
Prime rad(R). (Hint: Apply the embeddings in the proof of theorem 2.5.52 to theorem 2.5.51.) In particular, if R has no IGI-torsion and is semiprime then RGis semiprime. 14. Suppose G is a finite group of automorphisms of R, with 1GI-l E R. Then e = [GI-’ x g E C gis an idempotent of the skew group ringR * G, and e(R * G)ezRG. In view of lemma 2.7.12, this exercise shows that information passed from R to R * G can land in RG. M. Cohen found a Morita context (c.f., $4.1 below) involving RG and R * G, and recently the results have been put into the very general framework of Hopf algebras (c.f., $8.4 below). The standard text is Montgomery [80B] although there are several more recent developments, c.f., Passman [83], Montgomery [84], and BlattnerCohen-Montgomery [86]. The next two exercises present two standard counterexamples. 15. (A prime ring without 1 having solvable group of automorphisms whose fixed subring is 0) Let F be a field of characteristic p # 0, containing a primitive n-th root [ of 1. Let W be the free associative F-algebra F { X,, X2} and R’ = M2(W ) .
320
Basic Structure Theory
Let w(u,u) = ("XuE W and define H as the submonoid generated by
(i 3,
an abelian multiplicative subgroup of R'. Letting G be the multiplicative subgroup of R' generated by H and
we see H is normal in G and G/H is cyclic.
Let R be the subring of R' consisting of those matrices whose entries each have constant term 0. Then conjugation by G in R' induces a group of automorphisms of R, and RH c Wel2 so R G = 0.
A({
16. Let R = M 2 ( F )for a field F of characteristic 2, and let G =
y):o<m
a cyclic group of order p, which acts on R by conjugation. Then
which is not even semiprime, although R is simple. Note that this example and the previous example show there is something "wrong" with inner automorphisms, and many of the positive results involve groups of automorphisms which are not even restrictions of inner automorphisms (suitably defined), c.f., the papers of Kharchenko and also Montgomery [80B]. 17. If a is an invertible element of R then C,(a) is the fixed subring of the inner automorphism determined by a. This fact is used by M. Cohen [78] to link the structure of R and C,(a) after determining the structure of fixed subrings under an algebraic automorphism.
82.7 1. The following assertions are equivalent for a module M over a left Artinian ring R: (i) M is f.g.; (ii) M is Artinian; (iii) M is Noetherian. (Hint: (ii) => (i) Induction on the smallest t such that J'M =O where J =Jac(R). J ' - 'M is f.g. by induction, and
M / J ' - 'M is an Artinian R/J-module and is thus f.g.)
2. The ring of upper triangular matrices over a field is a semiprimary ring in which central idempotents of R/Jac(R) lift to idempotents which cannot be central.
Perfect Rings
nz
3. In example 2.7.38 if [ M : F ] is uncountable then Jac(R)' # 0 since the transformation taking x, to x1 is in each Jac(R)1 (where o is the first infinite ordinal). 4. If R is (right) perfect then Jac(R) is locally nilpotent. (Hint: If S is a finite subset of Jac(R) which is not nilpotent then for some s, in S one has Sks, # 0 for each k in M. Continue in this way to build a non-T-nilpotent sequence). 5. (Bjork) The following assertions are equivalent: (i) R is right perfect. (ii) R has no infinite set of orthogonal idempotents, and every R-module contains a simple submodule. (iii) Every R-module satisfies DCC (f.g. submodules). Hint: (iii) =$ (ii) follows instantly from the DCC on cyclic submodules. (ii) * (i) If a,, a 2 , ... E J = Jac(R) with a ; . . a , # 0 for each n then take L < R maximal with respect to
Exercises
321
a;..a, $ L for each n. Take a simple submodule L , / L of R/L. Some a ; ~ ~ a l E L 1 , so L , = Ran+ ...a, L, implying a;..a, E L as in remark 2.7.6. This con-
,
-
+
tradiction proves J is T-nilpotent; prove R/J is semisimple Artinian by corollary 2.3.1 1 and proposition 2.2.1 1) (i) (iii) the hardest direction. Take N < M maximal with respect to satisfying DCC (f.g. submodules), and assume N < M. We want a contradiction. There is x E M - N with J x c N (for, otherwise, one could define a nonvanishing sequence in J by stipulating inductively ai+,(ai...a,x)$N. There is a primitive idempotent e with ex$N; we may assume M = N Rex. Let M, > M, > ... be an arbitrary chain of f.g. submodules; to show it terminates one may assume each Mi $ N. The strategy is to build f.g. N, 2 N n Mi with N, > N, > ... and show this sequence must terminate. In fact putting No = N, Mo = M, and xo = ex, we shall find 4 5 ii-n Mi and xi in eN, - such that Mi = Ni + Rwi where wi = =, xi. Indeed, given yi in Mi < Mi one inductively has yj-, in 4- and rl in R with yi=yj- +riwi- This implies r,exo $ N so rie 4 J . Hence e E Rrie (lemma 2.7.17) so e = ebirie for some bi. Take xi = ebiyj- E e&- Then wi = ebiyi E eMi. Write Mi = z R x i j and, as before, xij = x i rijwi- where x i E N i - Define N,= R(x; - rijxi) c 4-n Mi and note Mi<4+XjRrijwi< Mi.) This argument can be modified to show DCC (cyclic submodules) implies DCC (f.g. submodules), as shown in Faith [76B].) 6. (Converse to theorem 2.7.35, Jonah [70]) If every right R-module satisfies the ACC on cyclic right submodules then R is right perfect. (Hint: Show for any sequence r l , r 2 ,... in R then there is i < j with rj...riR = rj...ri+lR,seen as follows: Let Si = riR and define an equivalence on the disjoint union S of the Si by putting si s, if there is j > max(i,u) with r j " . r i + , s i = rj...ru+,su (where s i ~ Sand i s, E S,). Then M = S/- is a right module and is a special case of the direct limit; the canonical image of each Si is a submodule M i . Since M, 5 M, < ... one has Mi = Mi+ for some i, yielding r j . ..ri + ri E r j . . .ri + 2(ri + R) as desired. Now go down the structure theory. If each ri E J then remark 2.7.6 shows J is T-nilpotent so assume J = 0. An easy application of the density theorem shows any primitive image of R is simple Artinian, so, in particular, (-){maximal ideals of R} = 0. Enumerate the maximal ideals Ai taking Ai P A , n ... n A i - and ri E Ai A,,; one may assume 1 E ri + Ai for each i, so that r j * * . r i R# r j * . . r i + , R for all j > i, a contradiction, unless a finite number of maximal ideals has intersection 0. Conclude by the Chinese Remainder Theorem.) 7. Every left Noetherian right perfect ring is left artinian. (Hint: Special case of exercise 8). 8. The following assertions are equivalent for a right perfect ring R with J = Jac(R): (i) R is left Artinian; (ii) R / J 2 is left Artinian; (3)J/J2ER-9imod. (Hint: (iii)-(i) If J = J 2 + Ra, then J" = J"+' + xRai,nil...aim,so by the proof of theorem 2.7.2 it suffices to show J is nilpotent. This is obvious unless J > J 2 > 53 > ... , in which case for each n some ai, * ..ain$ .In+', contrary to T-nilpotence.)
+
,,
xi
,
+
,
,.
,.
,
x;:),
,.
~
,
,
-
-
,
,
,
,
nil:
,
Semiperfect Rings 9. Suppose R is semiperfect, and J = Jac(R). An idempotent e is primitive iff Re/Je is simple in R - A u d iff J e is the unique maximal submodule of Re. (Hint: If N is
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Basic Structure Theory
a maximal submodule of Re then N = Le where L = R(l - e) + N is a maximal left ideal of R.) 9‘. If A a R then Hom(Re/Ae, R) x eAnn’ A in Mud-R for any idempotent e. (Hint: Given r in Ann’ A define cpr: Re + R by right multiplication by er, and note cpr yields a map RelAe + R.) 10. Suppose R is semilocal and J = Jac(R), and e, e‘areidempotents of R with R e x Re’. Then any mapf: Re+Re’ satisfies f(Re) EJe’. (Hint: in lemma 2.7.27 show f=O.) 11. Hypotheses as in exercise 10, the following are equivalent: (i) Hom(Re, Re’) # 0; (ii) eRe’ # 0; (iii) eJe’ # 0. (Hint: If 0 # f E Hom(Re, Re’) then 0 # fe E eJe’.) This exercise has important applications to representation theory. 12. (Zaks [67]) Suppose R is semiperfect and has complete set of idempotents el,...,en such that e,Re, is a division ring for each i. Then R = R ‘ + J where R‘=E{e,Re,: Re, x Re,} is semisimple Artinian and J = Jac(R) = E{e,Re,: Re, x Re,}. (Hint: Although this turns out to be a special case of Wedderburn’s principal theorem, we sketch Zak’s easy direct proof. First note that if Rei x Re, then eiJej = 0, thus implying every map f:Re, + Re, is an isomorphism. Thus putting R, = x{eiRej:Re, x Re, x Re,}, a simple ring, one concludes R’ x n R , is semisimple artinian. To see J a R it suffices to show e,rej E J implies eirejr’ek E J; but this is clear unless Re, x Re,, in which case right multiplication by rejr’ek gives an isomorphism from Re, to Re,, implying Rei x Rej, contradiction. The rest is clear.) 13. Suppose R is a semiperfect ring with J = Jac(R), and e is a primitive idempotent. An R-module M with composition series has a composition factor isomorphic to the simple module RelJe iff eM # 0. 14. “Block decomposition.” Suppose R is a semiperfect ring. We can then find a finite set of orthogonal central idempotents ul,. .. ,u, such that R x R , , with each Rui indecomposable as a ring (i.e., having no nontrivial central idempotents). Calling these Ru, blocks, show that each primitive idempotent of R belongs to a suitable block. Moreover, we say two primitive idempotents e, e’ are linked if there is a sequence of primitive idempotents e, = e, e2,...,em- em= e‘ such that for each j rn - 1 the modules Re, and Re,, have a common composition factor. Show if e E Ru, then Ru, = z{Re’: e’ is linked to e}, proving the blocks are uniquely determined. (Hint: The first assertions are immediate. Let [Re] = {Re’: e’ is linked to e}. If e and e” are not linked then [Re] n [Re“] = 0 from which we see easily each [Re] is an ideal and R = x[Re] summed over a suitable subset of a basic set of idempotents. Hence each [Re] is a block. If e and e‘ are linked then by induction on the length of the linkage they lie in the same block, using exercise 13. 15. A Loewy chain of submodules of M is a collection { M # : j< a } for some ordinal a such that M, = O,M, 5 MB whenever y < 8, MB= My for every limit ordinal j,and MB/My= soc(M/M,) whenever j = y’. M is Loewy if M has a Loewy chain with Ma = M for suitable a, and a is called the Loewy length of M. Note this generalizes composition series. Show that every f.g. module over a right perfect ring has a Loewy chain. Suppose R is semilocal. An R-module M has Loewy length n iff J ” M = 0 where J = Jac(R). (Hint: (e) 0 < J”-’M < ... < J M < M is a Loewy series; (-) note J soc(M) = 0 and proceed inductively.)
cy
323
Exercises
Structure of Semiprimary Rings 16. If R is semiperfect with Jac(R) nil then for all L < R and all j > k one has
17. 18. 19.
20.
21.
L' = LJ + ( L n Jac(R))'. (Hint: Let Lo = L n Jac(R). Then L = Lo + Re for a suitable idempotent e, by corollary 1.1.28, and an easy induction shows Lk= ReL + Lt for all k.) If R is semiprimary and L < R then L' = L"' where Jac(R)' = 0. (Hint: Exercise 16.) If e is a primitive idempotent of a semiprimary ring R then Re > J e > J'e > ... is a composition series. Given R let R' = Ri/$ where each Ri = R and 9 is an ultrafilter. If R is semilocal then R ' is semilocal (cf., exercise 2.5.7 and 2.3.6). If R is semiperfect then R' is semiperfect. Notation as in exercise 19, suppose R is not left Artinian. Then R' is not left Noetherian. Indeed, if L , > L , > L , > ... are left ideals of R then taking I, = Li-,)/F where L, = 0 for u I 0 show I , < I , < ... . This gives examples of semiperfect rings which are not left Noetherian. Analogously, if R is not right perfect then we can take each Lito be principal, in which case lk is principal and R' fails the ACC on principal left ideals. Notation as in exercise 20. M = R'/ll has an onto map which is not 1:1. (Note R' is semiperfect but fails ACC on principal left ideals.)
n
(n
Fitting's Lemma and mRegular Rings 22. (Bjork) A semiprimary ringR with a cyclic module M having an onto map which is not 1 : l . Let K be a field and D be a division ring containing the free
C
algebra K { X , Y } . The field of fractions F of K [ X ] can be viewed in D. Then R
=
(0" ):
is semiprimary; taking L =
module spanned by
(i y),
: 1 ,K Y X ' we see
:'L)
is a cyclic
and right multiplication by X is the desired map.
23. (Dischinger [76]) R is right n-regular iff R is left n-regular. (Hint: (a) It suffices to prove if a = a2b and b = b'c then a E Ra'. Note ac = a3b2c = a z and abc = a'b'c = a. Thus (c - a)' = c(c - a), implying ab(c - a)' = 0 and b2(c - a)' = b(c - a). But (c - a)k= (c - a)'+'d for some d , so b(c - a)'d = bk(c- a)'+'d = b(c - a). Hence 0 = ab(c - a)'d = ab(c - a) so aba = a. But 0 = ab(c - a)d = ab2(c- a)'d = ab2(c- a ) so a = aba = ab'ca = (ab'a)a E Ra'.)
$2.8 1. (Kaplansky) Let C = {continuous real-valued functions on the interval [0, l]}, a commutative subring of R[o*ll, and P = {f E C:f vanishes at a neighborhood
of 0). Then P and p,x
1 C . For each n in N - {0}define pn in P by p,,x = 0 for x In
1 1 for x > -. Then { p,,: n E N } spans P. Find a dual base, thereby n n
=x --
proving P is projective.
Basic Structure Theory
324
2. Q is not projective in Z - d o d . (Hint: the only map Q Z is 0.) 3. The following assertions are equivalent: (1) R is semisimple Artinian. (2) Every
R-module is projective. (3) Every cyclic R-module is projective. (Hint for (3)=4 1): Suppose L < R. Then the epic R + R/L splits, so L is a summand of R.) 4. A ringR is semihereditary if every f.g. left ideal of R is projective. In this case show every f.g. submodule of a projective module is projective; every f.g. projective module is isomorphic to a finite direct sum of copies of left ideals of R. 5. The ring R =
(:
z)
is left hereditary but not right hereditary. (Hint: If L is a
left ideal then L = e , , L @ e,,L; e , , L is a Q-submodule of Q @ Q and e z 2 L is a Z-submodule of Z. On the other hand, the right module R e , , acts like Q as a Zmodule and is not projective by exercise 2.) A complete characterization of hereditary rings of the form of example 1.1.9 is given in Goodearl [76B, theorem 4.71.
A KrulJ domain is an integral domain C whose field of fractions F has a set of discrete valuations { u i :i E I} for which the intersection of the valuation rings is C and such that for any 0 # a in F one has u p =0 for almost all i. This is a useful generalization of PID in commutative ring theory and is studied in Bourbaki [72B, Chapter 51. Whereas Krull domains arise in the study of centers of hereditary rings, cf., Bergman [71a], imposition of additional hypotheses restricts the center further, cf., Robson-Small [74] and Chatters-Jondrup [83]. This makes the following example interesting. 6. Page [84]. An example of a left hereditary, left Noetherian subring R of M , ( K ) for a suitable field K, whose center is an arbitrarily given Krull domain C. Indeed,
let F be the field of fractions of C; by Bourbaki [72B, Chapter 7, exercise 31cl there is a PID H with field of fractions F(1),such that F nH = C. Let K be the algebraic closure of F(1).Take f in K such that f " = 1,where n = 2 if char(F) =0 and n = p + 1 if char(F) = p. Letting 1'= 1 + f one has 1algebraic of degree n over F(A'), and F ( L ) n F ( A ' )= F in K. Then R = F ( l ) e l l + F(1,A')e12 Hezz 5 M , ( K ) is left hereditary and left Noetherian since 1is algebraic over F(A'), but Z ( R ) = H.
+
Isomorphisms (also, cf., exercises 2.1 1.12ff) 7. Suppose P is an f.g. projective module over a commutative ring C, and M E C-Mud. Then Endc M @ End,P z End,(M €3 P). (Hint: Define the balanced map End, M x Endc P + Endc(M €3 P) by sending ( f ,g) to f @ g. This defines the desired map, which is an isomorphism for P = C, and thus for P f.g. free, and finally for P a summand of an f.g. free C-module.) 7'. Suppose P is an f.g. projective R-module and M E R - d u d - T and N E d u d - T . Then HomT(M,N) @ l RP z Hom,(Hom,(P, M), N). (Hint: Define the balanced map Hom(M, N ) x P + Hom(Hom(P, M),N) by (f, x) + fx where fJg) = f ( g x ) . This induces a map from the tensor product which is an isomorphism for P f.g. free, and thus for P tg. projective.)
Exercises
325
The Radical of a Module cf., Definition 2.5.26. 8. If N I R a d ( M ) then Rad(M/N)= Rad(M)/N; in particular, Rad(M/Rad M)=O. On the other hand, if Rad(M/N) = 0 then Rad M 2 N. (Hint: As in proposition 2.5.1'.) 9. Rad N I Rad M for all N 2 M. (Hint: If M' is a maximal submodule of M use lemma 2.4.3(ii) to show M' n N is either N itself or a maximal submodule.) 10. If M' << M then M' is contained in every maximal submodule of M, so M' 5 Rad(M) (which we recall is the intersection of maximal submodules of M). 11. Rad(M)=x{small submodules of M}. (Hint: Prove for X E Mif Rx is not a small submodule of M then there exists a maximal submodule M" of M not containing x; indeed, if Rx + N = M for N < M then one can take M" 2 N maximal with (2) is by exercise 10.) respect to x $ M". This proves ( c); Rad(M) need not be small, cf. exercise 17. 12. Rad(M, @ M,) = Rad(M,) @ Rad(M,) for any M,,M, in R - A d . (Hint: ( G ) Let N=Rad(M,@M,). N I N ' O M 2 for every maximal submodule N' of MI, so NIRad(M,)@M,; likewise N < M,@Rad(M,). ( 2 )by exercise 9. A similar argument shows Rad(@ Mi) = @(Rad(M,)). 13. If F is free in R-&& then Rad(F) = Jac(R)F by exercise 12. 14. Rad(P) = Jac(R)P for every projective module P. (Hint: Write J = Jac(R) and P @ P' = F free; then apply exercises 12 and 13.) 15. Rad(P) # P for every nonzero projective module P. (Hint: Otherwise, P = Rad(P) = Jac(R)P. Take P @ P' = F free and let n:F + P be the projection. Let { y , :i E I} be a base of F and note P E Jac(R)F. Hence any x # 0 can be written as x a , y , for a, in Jac(R), so x = nx = xa,(ny,). Writing ny, as x j b i j y . for b, in Jac(R) yields 0 = x - x = a,y, - Cj,,a,bijyj = Cia,(6, - h i j ) y j so C i a i ( 6 , - bij) = 0 for each i. Hence a ( l - (b,)) = 0 where a = (a,); since there are a finite number of b, the matrix (bij)is quasi-invertible implying a = 0.)
xi
xi(
T-Nilpotence 16. Suppose M is a free right R-module with countable base xI,x2,.... Given rl,rz, ... in R let yi = xi -xi + ri and let N = yiR I M. Then { y , , y , ,. ..} is an indepen-
,
1
dent set of elements. Also xLE N iff r,r,- ... rk = 0 for some n > k; thus N = M iff for each k we have suitable n = n(k) > k with r, . .. rk = 0. 17. The following are equivalent for a right ideal J of a ringR: (i) J is T-nilpotent. (ii) M J # M for every M # O in A d - R . (iii) MJ<<M for every M # O in &od-R. (iv) M J << M where M is a free R-module with countable base. (Hint: (i) * (ii). If M J = M then one could inductively find a 1 , a 2 ... , with Ma;..a, # 0 for all n; (ii) =s(iii) If N I M then (M/N)J # M/N; (iv) =. (i) By exercise 16.) 18. If every right R-module has a maximal proper submodule then Jac(R) is Tnilpotent. (Hint: Use (ii) (i) in exercise 17.) 19. (Strooker [ 6 6 ] )We say a map f: M --* N is a couer if kerf << M. Suppose A 4 R and N is an f.g. RIA-module. If fi M + N is a cover then M is f.g., and if N is RIA-projective then M/AM x N. (Hint: let q : M+M/AM be the canonical map. Lift an epic (RIA)'") -+ N to a map g: R(")-+ M such that the appropriate square
-
Basic Structure Theory
326
commutes; conclude g is epic. If N is RIA-projective then viewing N as a summand of M/AM show equality holds by moving around the square.)
Projective Covers 20. If Jac(R) = 0 and M has a projective cover then M is itself projective (by exercises 1 1 and 14.) 21. Using lemma 2.8.39 reprove any finite set of orthogonal idempotents can be lifted orthogonally. 22. R is semiperfect iff every tg. R-module has a projective cover. (Hint: (-=) Every summand of R = R/Jac(R) has a projective cover, so J is idempotent-lifting. R is semisimple Artinian by exercises 3 and 20.) 23. R is semiperfect iff every simple R-module has a projective cover. (Hint: It suffices to prove every f.g. module M has a projective cover; since Rad(M) << M one can pass to M/Rad(M) which is a finite subdirect product of simple modules; conclude using proposition 2.2.5.) 24. Suppose P E R - A ois~projective and fEEnd, P. Then fEJac(End, P ) iff fP<
+
+
+
+
Characterizations of Right Perfect Rings 27. R is right perfect iff R/Jac(R) is semisimple Artinian and each right R-module has a proper maximal submodule. (Hint: exercises 17, 18.) 28. R is right perfect iff every right module has a (right) projective cover, in which case any projective right module is isomorphic to Oie, eiR for suitable primitive idempotents ei of R. (Extensive hint: (a) Suppose M E R - A oand ~ J = Jac(R). M/MJ = OiE,e,R/eiJ for primitive idempotents e,. Let P = Oi,,eiR. Exercise 15 shows P J c P and MJ<<M.Thus P + M / M J is a projective cover, which then lifts to a projective cover P+ M. In case M is already projective then P x M.
327
Exercises
(-=) R is semiperfect so it suffices to show any nonzero right module M # 0 contains a maximal submodule. Let f: P + M be a projective cover. By exercise 15
there is a maximal submodule N of P,and fN is maximal in M.) Other characterizations are given in exercise 2.9.10,2.11.7.
Amitsur-Ware-Zelmanowitz Proof of Characterization of Jac(M,(R)) We say a set { L i :i E I} of left ideals of R is right-vanishing if for any sequence a l , a 2 , ... where auELiwwith iu distinct we have a , ...a, = O for suitable n. Ware-Zelmanowitz [70] proved for any projective module P that any f:P+P is in Jac(End, P) iff P has a dual basis {(xi,gi):igI} such that {gifP:i E I } is a vanishing set of left ideals contained in Jac(R), iff every dual basis satisfies this property. As an application they reproved the theorem of Sexauer-Warnock [67] characterizing Jac(M,(R)) since this is Jac(End, F) for F countable free. We shall give this special case of their proof. Accordingly, we say a base {xl,x2,...} of F is WZ (for Ware-Zelmanowitz) ouer a map f : F -+ F if { n i f F :i E I} is a vanishing set of left ideals contained in Jac(R), where x i : F + R is given by nixj = 6,. 29. Write E = End, F and J = Jac(E) where F is countable free. The following are equivalent for f in E: (i) f E J ; (ii) F has a base WZ over f; (iii) every base of F is WZ over f. (Hint: Recall from exercise 24 that f E J iff f F << F. (ii) =. (i) Let { x , , ~ , ,...} be the given dual base WZ over f, and put Li = nJF. Then f x E Lixifor each x in F, so it suffices to show Lixi << F. Suppose, on the contrary, N + c L i x i = P with N # P. Then there is x in P with 0 # X E F = PIN, so 2 = x i s , , a i Z i for ai in Li and suitable finite I‘E I; for each j in I’ we have Z j = C i E , . . , j ) a i j Xfor i some finite I ” ( j ) c I . If j E I ” ( j ) then Xj=c((l -aJ1aij)ii summed over i E I ” ( j ) - { j } ; thus we may assume j # I”(j ) . Continuing in this way, for each n write Z as a sum of terms of the form ailai,* . . a i n i i nwhere the aiu are form distinct L,; thus one could build an infinite sequence which contradicts the right-vanishing of the L i . (i)=.(iii)Given any base { x l , x 2 ...} , let Li = nijF G Jac(R) by exercises 10, 14. It remains to show the Li are right-vanishing. We aim to show for any sequence u l , a z ,... with aiELi that a , * . . a , = O for suitable n. Write ai=nifwi where wi€F. Define $: F + F by the following inductive procedure. Take m , = 1 and m, > 1 such that f w I ~ x { R x j < : m Z } and put $x, = x 2 and $xj=O for all j with m , < j < m , ; inductively, given mi take m i + , > m i + 1 such that f w u ~ x { R x j :< m i + , } for each u < mi and put $xj = 0 for all mi < j < mi+ and $xm, = bixi+ where bi = a m , + , ~ ~ ~ a m , +Putting l - , . b, = 1 let N = ~ , ~ l R ( x i - b i - l a m , x i + ,Then ). alxz = nlfw,xz = $ j w , so x1 = (x, - a,x2) + alx2 E N $fF. Inductively, each x i E N + $ f F so N = F since ~ j isF small; by exercise 16 some product a , . . . a, is 0.) 30. An infinite matrix (rij)EJac(M,(R)), iff each rijEJac(R)and the left ideals generated by the columns of (rij)are right vanishing. 31. Extend exercise 29 to free modules of arbitrary cardinality. (Hint: (ii) =-(i) is the same; (i)-(iii) is obtained by defining $ to send all the “extra” base elements to 0.) 32. Prove the Ware-Zelmanowitz theorem as quoted above, for arbitrary projective modules P.(Hint: (ii) (i) is almost exactly the same as before, using the gi of the
c
1
,
,
+
-
-
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Basic Structure Theory
dual basis in place of n,;(i) (iii) seems to require writing P 0 P' = F free and defining JI on F using the same trick of "spacing out" the base elements.)
82.9 Countable Modules and LE-Decompositions A module is countable if it is spanned by a countable set of elements. (In particular, every f.g. module is countable.)
Oie,, Mi and N 5 M where N is spanned by { x j :j E J}.Then there is some set I' G I with 11'1 I IJI such that N I OiC,, M i . (Hint: Each xi is in a finite sum of the M i . ) (Kaplansky-Walker) If M is a direct sum of countable submodules then so is every summand. (Extensive hint: Say M is dsc if M is a direct sum of countable submodules. Write M = OiE, Mi with each Mi countable. Suppose M = M ' $ M", and consider all pairs (N', N") satisfying N' < M', N" < M" with N' and N" each dsc, and such that N' $ N" is a direct sum of M i . Order such pairs by putting (N', N") c (P',P") if N' and N" are respective summands of P' and P" whose complements are also dsc. Then Zorn's lemma gives a maximal pair (N',N"); write N' $ N" = @ i s , a Mi. Replacing I by I - I' one may assume N ' = N" = 0. Let n l r z2 denote the respective projections of M onto M' and M". Now take any Mi # 0 and let Q1 = M i . Inductively, given Qj countable, we take N ; = nlQj and N ; = n2Qj,which are countable, and let Qj+ be a countable sum of Mi which contains N ; $ NY. Then Qj= N ; $ NY with each module dsc, contrary to hypothesis.) Suppose M = M' 8 M" and N is a summand of M with an LE-decomposition N = @= 4. Then there are respective summands L', L" of M', M" such that M = L' 0 L" $ N. (Hint: By induction and proposition 2.9.14 one may assume N is an LE-module.) (Crawley-J0nsson-Warfield theorem) If M = Oi, Mi is an LE-decomposition with each Mi countable then every summand M' of M is a direct sum of modules isomorphic to suitable M i . (Hint: M' is a direct sum of countable submodules by exercise 2, so we may assume M' is countable. Let { x 1, x 2 , . ..} be a countable spanning set for M'. Inductively one wants to find independent submodules N , , N2 ,... of M' such that N, $...$ N, is a summand of M' containing x1 ...,x, for each n, and, moreover, such that each A$ is a finite direct sum of submodules isomorphic to various M i . (For then M' = @,?. So take No = 0 and given N l , . . ., N,- take N b - , such that Nu)@ N , - , = M'. Then x, = a, b, where a, E @: Nu and b, E N b - 1. Surely b, E @ie,(n)Mi for some finite subset I(n) of I. Let S, = Mi, a finite direct sum of LE-modules, and write M" for a complement of M'. Then exercise 3 permits us to write M = L' $ L" $ S, for suitable respective summands L',L" of M ' , M". Put N, = M' n(S, $ I,").Then b, E N, so x, E @:= Nu. Moreover M' = N, $ L'. Let N" be a complement of L" in M". Then S, x M/(L' @ I,") e N, @ N". But S, has a finite LE-decomposition so N, is a sum of various Mi.) Every projective module P over a local ringR is free. (Hint: Write P $ P' = Oia,R, an LE-decomposition since End, R x R is local, and apply exercise 4.)
1. Suppose M =
2.
u
um , =,
3.
4.
,
5.)
(0::;
,
5.
,
+
329
Exercises
6. Every projective module P over a semiperfect ringR has an LE-decomposition P = I pi where pi is a direct sum of copies of Re, and { e l , .. .,e,} is a basic
Of=
set of idempotents. (Hint: This is true for R and thus for any free module, so apply exercise 4.)
Complement Summands 7. A decomposition M
Oie
= Mi complements summands if for every summand N of M there is J c I such that M = Mi) Q N. The decomposition complements maximal summands if the above property merely holds for every summand N whose complement is indecomposable. In fact, every decomposition into LE-modules complements maximal summands (with J a singleton). Thus every finite decomposition into LE-modules complements summands (by induction). If the decomposition Mi complements summands (resp. maximal summands) then for every subset I' of I the decomposition Mi also complements summands (resp. maximal summands). If M has a decomposition which complements summands and N is any summand of M then N has a decomposition which complements summands (by proposition 2.9.14.) 8. Suppose M = Oi, Mi complements maximal summands. If Mi x Mj for some i # j in I then End Mi is a local ring. (Extensive hint: We may assume i = 1 and j = 2; by exercise 7 we may assume M = M, 8 M , x N(2)where N = M,. To show End N is local we claim f g = 1 implies f or g is a unit. Indeed, for j + g = 1 put M ' = { ( f x , - g x ) : x ~ N } and M " = { ( x , x ) : x ~ N } x N .Clearly M = M ' Q M" so N(2)= Mi Q M' for i = 1 or i = 2 implying the restriction of the canonical projection M -P M2-i to M' is an isomorphism, and f or g is thus invertible.) 9. The following are equivalent: (i) R is semiperfect. (ii) Every f.g. projective R-module has a decomposition which complements summands. (iii) R'2) has a decomposition which complements summands. (Hint: (i) =- (ii) R has such a decomposition by exercise 7; hence so does every f.g. free module and every summand. (iii) =- (i) by exercises 7, 8, and theorem 2.8.40.) Also the right-handed version holds by symmetry. 10. The following assertions are equivalent: (i) R is right perfect. (ii) Every projective right R-module has a decomposition which complements summands. (iii) The free R-module F = R'"' has a decomposition which complements summands. (Hint: (i) 3 (ii) By exercise 2.8.28 any projective right R-module P has a decomposition @eiR for primitive idempotents e,; this is LE. Since each summand is also projective apply theorem 2.9.15. (iii) =- (i) R is semiperfect by exercise 9. It remains to show Jac(R) is 7'-nilpotent, for which one should appeal to exercise 2.8.16. Namely, let x,, x 2 , ... be a base of F and let r , , r 2 , ... be any sequence in Jac(R). Put yj = xj - xi+ lrj, and let e , , . .. ,e, be a complete set of orthogonal primitive idempotents of R. Then {x1,y2,x3,y4,.. .} is a base of F so @m,= y2,R is a summand of F. Likewise, {yl,x2,y3,x4,. ..} is a base of F so F = @ , ~ = , ( @ ~ = , ~ ~ 2 j e , R Q y , j ~ ,by e i hypothesis, R)); F = N, Q N,Q(@,~=,Y,~R) where N1 is a suitable sum of xZjeiRand N2 is a suitable sum of y Z j - ,eiR.
(OiGJ
Oiel
OiSl,
+
,
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Basic Structure Theory
,
Let nj:F -+ xjR be the projection map. Applying 7 r l j - shows x,j-1R=n,j~,N,+~,j-ly2j-2R=n2j-lN2+~2j-lr2j-zR.
,
,
,
,
Smallness shows xzj- R = n Z j - N , for all j , so N , = B y z j - R. Hence @,?= yjR is summand of F, yielding f: F + F with fyi = xi for each i. Write fxj = ~ : ( ~ )xuaju. , Then xj = fyj = f(xj - xj+ ,rj) = ~ x u ( u j-u aj+,,"rj) so aju - aj+l,urj =aj,. Thus ajjrj-,...r1 =aj-,,jrj-z...rl = . ' . = a l j r l . Take n > n ( l ) . Then a , , + , , , , r , , ~ ~=(a,,,,~r, l)r,,-,...rl = a l n r l -r,,-,.*.r, = -r,,-,...r, so 0 = (a,,+,,,,r,, + 1)r"- .. 'r, so r,,- ...r l = 0.) 11. R is a local ring if€ the canonical decomposition R',) = R 0 R complements summands. (Hint: (-=) by exercise 8.) 12. (Cancellation). If M is an LE-module and M @ N x M @ N' then N x N'. Hint: Let M = M @ N,n , , ~ be , the natural projections from M to M and N',respectively. Then n1 + n2 = 1 so n, or n2 is an isomorphism.) 13. (Beck [78]) Extend theorem 2.9.15 to the following result (with similar proof): If M = N' @ N" = Mi where N' is an LE-module then there is some i such that Mi = M i @ MY for suitable Mi, MY with M = ( @ j + i Mj) Q MI @ N' = MY @ N".
,
,
Beck [78] has many more interesting results. Motivated by theorem 2.9.15 we say = is an LE-decomposition if each Mi is an LE-ring. Then each LEdecomposition complements maximal summands; Beck shows that it need not complement (arbitrary) summands by characterizing exactly when an LE-decomposition complements summands.
M
Bi,,Mi
14. (Swan)Failure of Krull-Schmidt for an f.g. module over a local Noetherian ring R. Let F be any field and let A = F[x,y]/(x3 - xz y'). Then A is a commutative domain in which the images of x and y generate a maximal ideal P. Localizing at S = A - P produces a commutative local Noetherian domain R = S-'A whose maximal ideal we call J. Let K be the field of fractions of R and let z = y-'x E K. Let T = R[z] c K. T is not local, having maximal ideals M, = J T(z - 1) and M, = J T ( z 1). As in example 2.9.22, one has M, @ M, x T @ (M,n M,) as T-modules and thus as R-modules. These modules are all torsion-free and indecomposable. But if T x Mi over R then one would have T x Mi over T, which is impossible since M,, M, require two generators as T-module. Note that the trick was to expand R to a nonlocal ring T, in which it is easier to violate uniqueness
+
+
+
+
of the Krull-Schmidt decomposition. 15. We say 1 is in the stable range of a ringR if whenever Ra
+
Rb = R one has a rb invertible for some r in R. Show any semilocal ring R has 1 in its stable range. (Hint: Passing to R/Jac(R) one may assume R is semisimple Artinian.) 16. (Evans [73]) Suppose 1 is in the stable range of E = End, M for an R-module M. If N,, N, are R-modules and N, @ M x N', @ M then N, z N', . In particular, this holds if R is semiperfect and M is an f.g. projective R-module. (Hint: Suppose f:N, @ M -+ N', @ M is an isomorphism with inverse g. Write f = ( J j ) and g = (gij) where f,,:N, -+ N',, f21: N, M ,flz: M -+ N',,and f,,: M -+ M, and likewise for g. Then in E one sees g2,fZ2 gZ1flz= 1 implying Efz2 + Eg2,f1, =
+
+
331
Exercises
E. Thus f z z
+ h g 2 , f , 2 = 1 for some h: M -+
M. Let g'
=
(:it,
TI2).
Then g ' j
is an isomorphism so g' is an isomorphism. Diagonalizing shows gl, - g12hgZ1 is an isomorphism from N, to N', .) 17. (Fuchs [71]) An R-module M satisfies the substitution property if whenever K = M 0 N , = M, @ N 2 with M x M, then K = M' @ N , = M' @ N, for some M'. Note by proposition 2.9.14 that necessarily M' x M, and also the substitution property implies cancellation (described in exercise 12). However Z - d o d satisfies cancellation by Cohn [56], whereas Z fails the substitution property in Z"'. (Hint: 2"' = Z(m, n) Z(p, k ) if mk - np = 1. But if Z"' = Z(u, u) + Z(0,l) = Z(u, u ) + Z(p, k ) then u = f 1 so k + pu = f 1. Thus taking k f & l(modp) yields a counterexample.) Fuchs also shows R satisfies the substitution property iff 1 lies in the stable range of R.)
+
Finite Representation Type (f.r.t.) 18. Any homomorphic image of a ring with f.r.t. has f.r.t. 19. If R is a semiperfect PLID then every indecomposable is principal indecomposa-
ble; thus any semiperfect PLID has f.r.t. 20. If G is a cyclic group then any group algebra F[G] has f.r.t. (Hint: Apply exercise
19 to early results of $8.1.) 21. Suppose H is a subgroup of a finite group G, and F is a field. Given an F[H]-
module M define the induced F[G]-module fi to be F[G] M. Let b , , ., . ,b, be a transversal of H in G, i.e., [G:H] = t. If M is an F[G]-module then forming fi by viewing M naturally as F[H]-module (by restricting scalars) show there is a map M -+ fi given by x -+ = bi @ b i 'x. If t is invertible in F this is a split monic, so in this case M is a summand of fi as F[G]-modules. 22. (Higman [54]) Suppose H is a Sylow p-subgroup of a finite group G, and char(F) = p. Every indecomposable F[G]-module is a summand of a module induced from an indecomposable F[H]-module. Likewise, every indecomposable F[H]-module is a summand of some indecomposable F[G]-module (viewed naturally as F[H]-module). Thus F[G] has f.r.t. iff F[H] has f.r.t.. Using exercise 20 conclude F[G] has f.r.t. iff H is cyclic. (Hint: If M is an indecomposable F[G]-module then as F[S]-module M x M, @ ... @ M, for Mi indecomposable; thus fi x fi,0 ... @ fi,so some f i i is a summand of M by exercise 21 and the exchange property. This proves the first assertion, and the second assertion is analogous; now use exercise 23.) 23. If F is a field of characteristic p and G is a noncyclic p-group then F[G] does not have f.r.t. (Hint: Show G has a normal subgroup N with GIN x (Z/PZ)(~)and thus use exercise 22 to reduce to the case G z (Z/PZ)'~);now proceed as in example 2.9.25.
1: ,
52.I0 1. Using the pushout reprove that if every monic from E splits then E is injective.
(Hint: If N
EM
and f:N
-+
E is a map then forming a suitable pushout M' yields
332
Basic Structure Theory
a monic E + M’ which splits; travelling around the square yields a map M + E extending f.)
2. Q is the injective hull of Z in E-Mod.If R
3. 4.
5.
6.
7.
8.
9.
=
(0” i)
for a division ringD
then M,(D) is the injective hull of R in R - & o ~ . (Hint: It is injective by Baer and is an essential extension.) A ring R is (left) hereditary iff E / M is injective for every injective R-module E and every M I E. (Hint: (a) If L < R then any map 5 L - t E / M lifts to f:L + E; now apply Baer. (e) Reverse the argument, using proposition 2.10.27.) Suppose E is an injective R-module and M is an R-S-bimodule which is flat in R - & u ~ . Then Hom,(M,E) is injective in yll$d-S. (Hint: As in proof of proposition 2.10.12.) Suppose R is a finite dimensional algebra over a field F. If M E R-%%.~od then the injective hull of M is in R-F~ww!. (Hint: By exercise 4 we may replace R by F; but all F-modules are injective.) An f.g. left R-module E is injective iff E is a finite direct sum of duals of principal indecomposable right modules. (Hint: Use exercise 1.5.5 repeatedly.) A categorical view of injective. Suppose E is an injective module containing M. Then any map f: M + M can be viewed as f:M + E and thus induces a map g: E + E. Dualizing proposition 2.8.42 show the following assertions are equivalent to E being the injective hull of M: (i) For every monk f: M 4 M every endomorphism (of E) induced by f is monic. (ii) Every endomorphism induced by a given isomorphism is an isomorphism. (iii) Every endomorphism induced by 1, is an isomorphism. (iv) Every endomorphism induced by 1, is monic. A functorial method of embedding a module into an injective. Let E be the injective hull of O L < , R / L . Then any R-module M has a monic into EHom(M-E), cf., remark 4.2.3 below. Conclude that the direct limit of injectives is injective. Suppose E is an injective module. If Ann{r,, ...,r,} = 0 for r,,. .. ,rn in R then x r , E = E. (Hint: Define f:R + R(”) by f r = (rrlr...,rr,,}. For any x in E right multiplication by x can be extended to a map g: R(”) + E, i.e., rx = gfr. Then write g = (g,,. ..,g,,) where each 9,:R 4E and note x = gfl = x r i g i l . The following facts hold for any indecomposable injective module E: (i) Every monic f:E + E is onto (Hint: fE a summand of E). (ii) Every nonzero submodule of E is large. (iii) If f € E n d R E then f or 1 - f is an isomorphism. (Hint: ker(1 - f ) n kerf = 0). (iv) End, E is local. (v) E x E(R/L) for some L c R. (Hint: There is a nonzero map f:R + E, so R/ker f is a nonzero submodule of E.
Criteria for a Ring to be Left Noetherian 10. R is left Noetherian iff the direct sum of injective R-modules must be injective. Hint: (*) Suppose f:L+@E, for L I R. ThenjL is f.g. so is contained in a finite direct sum of E,, which is injective, so f extends to R. (e) Suppose L , s L2 I ... and put L = L i . Define f : L - @ E ( R / L , ) by (fa),= a + L i . Extend f to R
u
Exercises
333
= 0 for all j 2 i, and L/Li = 0 proves L = Li.) 11. If R is left Noetherian then every injective left R-module E is a direct sum of indecomposable injective R-modules. (Hint: For any x in E note R x cannot contain an infinite direct sum, so neither can E(Rx), implying E ( R x ) has an inde-
and put x = f l . Then there is i such that xi
composable summand. Zorn's lemma yields a maximal set of indecomposable summands of E, whose sum is injective and thus a summand of E.) The converse is also true, by exercise 13 below. 12. (Faith-Walker theorem) R is left Noetherian iff there is a cardinal c such that every injective R-module is a direct sum of modules each spanned by a t most c elements. (Hint: (+) By exercise 9 the indecomposable injective R-modules belong to {E(R/L):L c R}, a set, so take c to be the maximal cardinality. (e) One must show any direct sum Oie, Ei of injectives is injective; we may assume each Ei is spanned by < c elements. Let E ' = n Ei injective, let E" be the direct product of 2' copies of E', and write E" = OjEJ E; where each E;' is spanned by < c elements. Each element of El is in at most a finite number of E;I, so El < 1 E ; I summed over a set of cardinality I c. Continuing inductively partition J into { J i : i s I } such that Ei I ~ { E ; I : ~ E JHence ~ } . Ei is a summand of x E ; I so E = O E i is a summand of E" and is thus injective.) 13. If there is a set Y of modules such that each injective R-module is a direct sum of modules isomorphic to modules in Y then R is left Noetherian. (Hint: exercise 12.) Conclude the converse of exercise 11: If every injective left R-module is a direct sum of indecomposables then R is left Noetherian. (Hint: exercise 9(v).) 14. Any decomposition of an injective into indecomposables (if such exists) complements summands. Every indecomposable injective projective R-module M is a summand of R. (Hint: Take 0 # x E M . Then M = E ( R x ) is a summand of some free module R("; x E R(")for some n implies M R("),and n = 1 since M is indecomposable.) 15. (Miller-Turnidge [73]) A Noetherian injective module which is not Artinian. Let M be a right vector space having countable base {xi: i E N} over a division ring D, and let R be the subring { r E End M D :rxk E x i D } . Let el be the primitive idempotent of R sending x 1 + x 1 and all other x i -+ 0. Then Re, is a summand of End M , and is thus injective as well as Noetherian but is not Artinian.
x!=
Quasi-Injective Modules and Self-Injective Rings An R-module M is qwasiinjective (QI) if for every N 5 M each map f :N + M extends to an endomorphism of
M. R is self-injective if R is injective as R-module. The theory of self-injective rings will be examined closely in 53.3; the following exercises provide an optional foretaste. 16. (i) If R itself is QI in R-.&d then R is self-injective. (ii) Every simple R-module is QI. On the other hand, if a E R is regular then for any left ideal L > RaR the module R / L is not injective in R-.&d. (Hint: The map f:Ra + R / L given by
f ( r a ) = r + L does not extend t o R.) In particular, if Z ( R ) is a domain which is not a field then there exist simple QI-modules which are not injective. 16'. Any summand of a QI module is QI. If M is QI then M(")is QI. However, the direct sum of QI modules need not be QI (Take M = Q 8 (Z/pZ). The map J Z 0 Z/pZ -+ M given by f(m,n p Z ) = (0, m p Z ) does not extend.)
+
+
334
Basic Structure Theory
17. (Generalizationof Jacobson's density theorem) Suppose M is QI and D = End, M . Show for any t > 0 and each x # 0 in M(') that every homomorphism Rx + M
extends to a homomorphism M(')+ M . Conclude the double annihilator property, if x,,. . ., x n E M and x 4 x,D then there is r in R with rx # 0 and rx, = 0 for all i 5 n. (Hint: Look at the density theorem's proof.) Moreover, D is a division ring iff the following two conditions hold: (i) Every nonzero submodule of M is large, and (ii) for any x , y in M if Ann x 3 Ann y then x = 0. 18. Suppose Q is QI. If M is a closed submodule and N _< Q with M n N = 0 then the projection z: M Q3 N + M extends to a projection Q + M . (Indeed, extend n to a map s':K + M with K 5 Q maximal possible and extend K' to a map p : Q + Q . Let M' be an essential complement of M in Q . If pQ $ M then M ' n p Q # 0 so calling this M , one has M n M , = 0. Let K , = p-'(M + M I ) . The restriction of p to K , followed by the projection M Q3 M , -+ M extends n' to a map from K , to M, contraction.) 19. Taking N = 0 in exercise 18, conclude any closed submodule of a QI module Q is a summand and is thus QI.By Zorn's lemma any M < Q has a maximal essential extension M in Q, which is thus closed and QI. M is called the quasi-injectioe hull of M in Q, introduced by Johnson and Wong. 20. Suppose Q is QI and E = Hom(Q, Q ) and J = {f E E: kerf is a large submodule of Q } Prove the following facts:
cy=,
(i) For any f in E there is g in E with fgf - f E J. (Hint: Let N be an essential complement of ket f in Q . Then N -+ f N is an isomorphism so its inverse fN -+ N c Q extends to a map g: Q + Q ; N Q3 kerf ker(fgf - f ) so f g f - f~ J.) (ii) J = Jac(E). (Hint: First show J is a left ideal: If f , g E J and h E E then ker( f - g ) 2 kerf n ker g so f - g E J, and ker hf 2 kerf so hf E J . Next show J is quasi-invertible: If f E J then 0 = ker(1 - f )n kerf so ker(1 - f ) = 0; the inverse map (1 - f ) Q Q extends to a left inverse for 1 - f. Thus J c Jac(E). The reverse inclusion follows easily from (i).) (iii) Let ,!? = E / J . For each ii in B there is 6 in ,!? such that aba = ii, by (i); such a ring is called uon Neumann regular and is studied in 52.1 1. (iv) J is idempotent-lifting. (Hint: If f - f2 E J then ker(1 - f ) + kerf 2 ker(f - f2) is large; taking M to be the quasi-injective hull of ker(1 - f )one can extend the projection M @ kerf --+ M to a map e: Q + M . Then e(ker f ) = 0 and ker(e - f ) 2 ker(1 - f ) + kerf implying e - f E J; thus f lifts to e.) (v) If e , , e z are idempotents of E with e l Q n e,Q = 0 then elQ Q3 e z Q is a summand of Q and thus equals e3Q for some idempotent e 3 . - (vi) If e l , ez are idempotents in E such that e l Q n e , Q # 0 then e , E n e2E # 0 in B. (Hint: Take N an essential complement of e l Q n ezQ in Q , and let M be a quasi-injectivehull of e l Q n e,Q in the QI module e l Q . Then there is a projection n,:Q + M with N E ker A,. Viewing n, as an idempotent of E one has elnl = x,; switching the roles of e l and e, find another idempotent n2 with e2x2 = z2such that N E kern, and n z Q is aquasi-injective hull of e l Q n e , Q in e,Q. Then n, - z2E J and 0 # El = E2 E e , E n e2E. (vii) is a right self-injectivering. (Hint: Given a right ideal I of E take a maximal set of idempotents ei such that 5 E Tand the sum x j e j F is direct. Then the ejM are independent. Given $T-P pick f, in E such that = fq. Letting gj be
e
335
Exercises
the restriction of fj to ejM one has a map g = @ g j : C e j Q + Q which extends to a map h: Q + Q. Show left multiplication by h agrees with f on 1)Note: To get a (left)self-injective ring one must pass to the opposite ring of E, cf., definition 1.5.6.) 21. If V is a vector space over a division ring D then End, V = Hom( V, V ) O P is a regular self-injective ring but is not right self-injective if V is infinite dimensional. (Hint: To prove E = End, V is not right self-injective show V is a summand of E which is not injective. This is seen by taking a base {xi:i E I} of V over D and letting x i : V - + xiD be the projection. Then letting A be the right ideal of E spanned by the ni there is a map f: A + V given by n,+ xi but f does not extend to E, proving V is not injective. However, V z x,E and E x x,E @ Ann,x,.)
V-Rings R is a (left) V-ring in case each simple R-module is injective, also, cf., exercises 2.1 1.29. 3.2.10ff.
-
22. The following are equivalent: (i) R is a V-ring. (ii) Rad(M)= O for every R-module M. (iii) Each left ideal is the intersection of maximal left ideals. (Hint: (i) (ii) If 0 # x E M then Rx x R/L for some L < R. Taking a maximal left ideal L' > L we have a nonzero map Rx + R / L which extends to f:M + R/L' since simple modules are injective; but then x $ kerf so x $ Rad(M), proving Rad(M) = 0. (ii) (iii) obvious. (iii) (i) Use Baer's criterion: Suppose N is simple. Given f:L + N where L < R, take a maximal left ideal A > kerf such that LA$ A. Then & A = R. But L n A = kerf since kerf is maximal in L, so definef: R + N by f(x a) = f x where x E L and a E A.) 23. Every V-ring is semiprimitive. Every V-ring which is a domain is simple (by exercise 16), although not necessarily Artinian (cf., exercise 3.2.12). Thus we get a natural class of simple rings, the V-domains, which are one of the key topics of Cozzens-Faith [75B1. 24. (Kirshan [70]) The following are equivalent for a ringR: (i) R is left Noetherian and every simple module is injective. (ii) Every completely reducible module is injective. (iii) Every countable completely reducible module is injective. (Hint: (iii)* (i) To show a chain L,< L, .. . of f.g. left ideals terminates, insert L, I L']I L, I L', IL , . . . such that each Li+,/Lj is a simple R-module and is thus injective and a summand of E(R/LI). Conclude as in the proof of exercise 10 using the injective module @Li+ ,/Lf in place of @E(R/LI).)
-
-
+ +
52.1I 0. A converse of exact: Suppose V is an abelian cftegory. If f: A + A' and g: A'
-+ A" /"Hom(A, C) is exact are morphisms such that 0 + Hom(A", C) 4 Hom(A', C) + for all C in Gf: then A A' 3 A" + 0 is exact. (Hint: First try this for R-Aod. Note A" + cok g is 0 so g is epic. Clearly fA E ker g, and equality holds because the map A' ---* cok f factors through 9.) 1. A functor F is half-exact if 0 + A + B + C + 0 exact implies F A + FB + FC is exact. If F: R - A d + V is half exact and I is a nilpotent ideal of R such that FM = 0 for every module M annihilated by 1 then F = 0. (Hint: Consider 0 + I" ' M + I'M + I'M/l'+'M -+ 0 where 1"'M = 0.)
L
336
Basic Structure Theory
2. (Osofsky [64a]) The following are equivalent: (i) R is semisimple Artinian. (ii) Every R-module is injective. (iii) Every Eg. R-module is injective. (iv) Every cyclic R-module is injective. (Hint: The only hard part is (iv) 3 (i). First note by proposition 2.1 1.20 that R is regular; it suffices to prove R has a finite complete set of orthogonal idempotents, which is implied by the next exercise, gleaned from Osofsky [68].) 3. Say an infinite set {e,:i E I} of orthogonal idempotents satisfies the Osofsky condition if for each J c I there is an element r, in R with eirJ = ei for all i in J, and rJei = 0 for all i in I - J .
(i) If R is a regular ring with R = E(R) then any infinite set of orthogonal idempotents satisfies the Osofsky condition. (Hint: Let n, be the projection from R onto E ( C j , Rej) and let r, = n 1. For i J let a = rJei and take b with aba = a. Note eja = 0. Then Rub ,Rej = 0 so 0 = n,(ab) = abr, implying a = abr,ei = 0.) (ii) If {e,: i E I} satisfies the Osofsky condition and n:R -+ Rei is defined by nr = (re,) and L = kern Re, < R then R = R/L is not injective as Rmodule. (Extensive hint: Note L={rER:almost all rei=O}. Let 6 be a given infinite partition of I with each S in Yoinfinite. By Zorn enlarge Yoto Y c 9 ( I ) maximal with respect to each S in Y infinite and S n T finite for all T in 3 Define f:CseYRrs + R by the rule frs = 0 if S E 9’ - Yoand frs = Fs if S E 9,. This is a well-defined map which does not extend to R. Indeed, iff were given by right multiplication by ii then r,a = rs b rueiufor suitable b in ker 71, so { i E S: eiaei = e,} is infinite for each S in Yo;form So by taking one i from each of these sets. Let r = rso. Obviously So $ Yoso 0 = fr = f i and thus raei = 0 for almost all i. But So n S is infinite for some S in Y (by maximality) and for all i in So n S we have eiraei = eiaei = e, # 0, contradiction.)
,
nkjE
+
fl
+ cis,
s,
+ +
Flat Modules 4. The direct limit of flat modules is flat. (Hint: Apply proposition 1.8.10 to the
definition.) Note the following corollaries: (i) Every central localization of a flat module is flat. (ii) A module is flat iff every f.g. submodule is flat. rixi = 0 (for xi in F) there exist y , ,...,y, in F 5. F is flat iff for every relation for some m and rij in R with ririj = 0 for each j and rijyj = xi for each i. (In other words, each relation has a reason.) (Hint: (e) by the flat test. (a)Let 1 = riR and let V be a free right R-module with base u l , ..., u,. Definingj: V + 1 byfui=riandletting K = k e r f o n e h a s O + K @ F + V @ F + I 6 F + 0 and ui 6 xi E ker(f 6 1) = K 6 F. Write ui 6 x i = kj 6yj for kj = C uirijE K, and match components.) 5’. (Villamayor) Suppose F is free and 0 4 K + F -+ M + 0 is exact. The following are equivalent: (i) M is flat. (ii) For every x in K there is n: F + K with nx = x. (iii) For every xl,. .. ,x, in K there is n: F K with nx, = x i for 1 5 i n. (Hint: riui and let I = riR. Then (i) * (ii) Choose a basis { u , } of F and let x = x E K n l F = I K s o writingx=Cr,y,definen bynui=yifor I s i s n a n d nui = 0 for all other i. (ii)*(i) If x ~ K n l F then x = n x ~ l K(ii)*(iii) . By induc-
xy=
cy=
XI=,
1
1
-+
1
337
Exercises
tion on n. Take n,: F -+ K with n,x, = x, and apply induction to {x’,,. ..,xh} where xi = xi - nnx,.) 6. If P is a projective module and N << P with P I N flat then N = 0. (Hint: You may assume P is free; by exercise 5’ if x E N then Rx is a small summand of P so x = 0.) Conclude that any flat module with a projective cover is itself projective. 7. R is a left perfect ring iff every flat R-module is projective. (Hint: (3) by exercise 6. (-e)It suffices to show there is no infinite descending chain r,R > r,r2R > r,r2r3R > ... Let F bea free module with base x,,x,, ... and let N = CRy, 5 F where yi = xi - rixi+ By exercise 2.8.16 the y , are independent; using the idea at the end of exercise 2.9.10, it suffices to prove N is a summand of F. This is true by hypothesis if F I N is flat. But F I X : = , Ry, is free for each k, so apply proposition 2.1 1.17.) 8. An f.g. module P is projective iff P is flat and finitely presented. (Hint: (*) by $2.8. (-=) One should show that if M -+ N is epic then Hom(P, M ) -+ Hom(P, N ) is epic or, equivalently, that the map of character modules Hom(P, N)” -+ Hom(P, M)“ is monic. But N ” -+ M ” is monic so N ” 8 P -+ M ” @I P is monic. It remains to show M # 8 P x Hom(P, M)” canonically. Let F , F2 P 0 be a presentation of P with F , , F, f.g. free. Then M” @I 4 x Horn(&,M)” by exercise 2.8.7‘,so proposition 2.1 1.15 yields the result.) In particular, every f.g. flat module over a left Noetherian ring is projective. 9. The following are equivalent: (i) M is f.g. (ii) $: N i )€3 M 63 M ) given by $((xi)@I y ) = (xi 8 y ) is onto. (iii) The canonical map R’ 63M M’ as in (ii) is onto, for any set I. (Hint: (iii) (i) Take the element of M M whose component corresponding to x E M is x, and write it as the image of C(rJi €3 xi. Then M =~RX,). 9‘.M is finitelypresented iff the canonical map R’ @ M M’ given by (r,) 8 x (rix) is an isomorphism for each I. Hint: Consider 0 K F -+ M -+ 0 and apply exercise 7 to K in the diagram
,.
-+
-+
(n
-+
-+
fl(& -+
-+
-+
R’OK - R ~ @ F
I
-+RIOM
I
I
-
-+
-+
0
Coherent Rings and Modules A module M is coherent if every f.g. submodule is
finitely presented. A ring R is (left) coherent if R is coherent as R-module. Examples of coherent rings include (i) Noetherian rings; (ii) semihereditary rings. 10. (Chases’ theorem) The following are equivalent: (i) R is coherent. (ii) Every pro-
-
duct of flat right R-modules is flat. (iii) F’ is a flat right module for every free right module F and every set I. (iv) Every finitely presented R-module is coherent. (Hint: (i) (iii) Apply theorem 2.1 1.13. If xiri = 0 where xi E F’ and ri E R then take 0 -+ K -+ R(‘)-+ Rri -+ 0 with K finitely presented. (iii) * (ii) Suppose M i :i E I are flat and take F free with 0 -+ K , -+ F K 0 exact. Then 0 -+ K , -+ F -+ K -+ 0 where the middle term is flat by assumption. Conclude with the flat test. (ii) 3 (iv) If N < M is f.g. then the canonical map (R‘) 8 N -+ N’ is monk because this is true for M using exercise 9’. (iv)* (i) obvious.)
n
c:=,
xf=,
-+
-+
n
fl
338
Basic Structure Theory
s
11. An exact sequence 0 + K + M + N + 0 is pure if it remains exact upon tensoring by any right module H. (Compare this to the definition of flat.) In this case f K is
called a pure submodule of M. Note that N is redundant since the crux of the issue is whether 1, @ 5 H @ K + H @ M is monic. Every split short exact sequence is pure, and every summand of M is a pure submodule. A submodule K of M is pure iff IK = K n IM for every right ideal I of R. An alternate necessary and sufficient condition for K to be a pure submodule of M: Given a, in K for 1 < j < t and given r,, in R, if all a, E r,,M then all a, E r,,K. For M flat we have K < M pure iff M/K is a flat module.
c:=,
I:=,
Localization and Module Properties
z,
12. An example where S-’Hom,(M,N) 5 Horn,-~,(S-’M,S-’N): Take R = S = Z - {0}, M = Q, N = Z. (compare with exercise 2.8.7) This motivates the following chain of ideas leading to exercise 15 below. 13. Suppose R is a C-algebra, and K is a commutative C-algebra. Any R-module is to
be viewed as C-module, and tensors are to be taken over C. If M is a finitely presented R-module and N is a K @ R-module then there is an isomorphism HomR(M,N )+ HomK,,(K @ M,N )sending f to ?where f ( a @ x ) = afx. (Hint: This is true for M = R and thus M free; pass down a finite presentation using the “five lemma.”) 14. Assumptions as in exercise 13 suppose, furthermore, K is a flat C-module. Then K @ Hom,(M, N )x Hom(M, K @ N )given by a @I f + f, where f,x = a @ f x . (Hint: as in exercise 13.) 15. Assumptions as in exercise 14 then K @ HomR(M,N) x Horn,@,(K@ M, K @ N). (Hint: K @ HOmR(M, I N )X HOmR(M, K @ N )Z HOmK@R(K@ M, K @ N )by exercises 13, 14). Two applications: (i) C is a field and K is a field extension of C. (ii) K = S-’C for a suitable submonoid S of Z(R). 151. Suppose E is an f.g. injective right T-module, M E R - A d - T , and N is a finitely presented R-module. Then Hom,(M, E) @, N x Hom,(Hom,(N, M), E). (Hint: Use the same map as in exercise 2.8.7’ and, as in the previous exercises, we may assume N is free.)
Regular Rings 16. (Kaplansky) Let F be a field and let R be the set of sequences of 2 x 2 matrices which are eventually upper triangular, i.e., R = {(ai)E M,(F): a, is upper triangular for all i > i,, for suitable i,}. R is a semiprime ring, and is not regular (for x = (e12)has no y with xyx = x). On the other. hand, any prime homomorphic image of R is isomorphic to F or to M , ( F ) and thus is regular. (Hint: Write e, for the element of R whose only nonzero component is the identity matrix in the i position. Suppose P is a given prime ideal of R. If ei 4 P and r E R then (r - rei) Re, = O so r - re, E P, implying RIP is a homomorphic image of Re, x M , ( F ) and thus RIP x Re,. On the other hand, if each ei E P then RIP is commutative regular.) 17. Suppose R is regular. Show (i) eRe is regular for every idempotent e of R. (ii) M,(R) is regular. (Hint: (i) easy. (ii) Use flat modules; alternatively one can prove it by direct computation, for it is enough to assume n = 2.)
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Exercises
339
18. R is regular iff every f.g. submodule M of a projective module P is a summand (Hint: (-=) Take P = R. (a) We may assume P is free, so assume P = R("). Then Hom(R'"),M) is a left ideal of M,,(R) and thus is a summand; thus M x Hom,(R,M) is projective in R - A d . This argument previews Morita equivalence). 19. R is regular iff every finitely presented module is projective. (Hint: exercise 8.) 20. The following are equivalent: (i) R is regular. (ii) R/L is a flat module for every left ideal L of R. (iii) R/Rr is flat for every r in R. (iv) L n 1 = IL for every left ideal L and right ideal 1. (Hint: (i) =. (ii) = (iii) is clear and (ii) (iv) by a criterion for flatness. Likewise, (iv) * (i) since Rr n rR = rRr.) 21. Suppose R is regular and E E R - A d is injective as R/Ann, E-module. Then E is injective as R-module. (Hint: Let A = Ann, E. Suppose cp: L + E is a map. If r E L nA taking r = rr'r yields cpr = rr'cpr = 0; the map @ (L A)/A -+ E extends to R I A and thus cp extends. 22. Suppose R is regular, and every primitive image is simple Artinian. Then each ideal A # 0 of R contains a nonzero central idempotent. In particular, if R is prime then R itself is simple Artinian. (Hint: Say {eij:1 < i , j < n} is a set of partial n x n matric units if eijevv= djuei,; we do not require eii = 1. Given such a set of partial n x n matric units one can show that if e, Re, has zero-divisors or if eii is not central then there is a new set of (n 1) x (n 1) partial matric units {e;: 1 I i, j I n 1) such that e', E el ,Re, cf., Jacobson CUB, p.2381 for the explicit computation which relies on exercise 17. Starting with any idempotent 0 # e E A, inductively build a set of partial n x n matric units {d$: 1 < i, j I n} such that ey:"EeyiReyi E A. In view of exercise 2.1.17, this procedure cannot continue indefinitely. Thus for some n we have eyi is central, and $4 Re:! is a reduced ring D. Since each e!:' = e ~ ~E A) this e ~proves ~ the first assertion. To see the second assertion note a prime ring has no central idempotent, so R x M,,(D);D is a regular domain and thus is a division ring.)
+
+
,
+ ,,
, , ,
1
+
,
Strongly Regular Rings 23. Every idempotent of a reduced ring is central. (Hint: (er - ere)' = 0 so er
= ere: likewise ere = re.) 24. Call R strongly regular if r E Rr2 for each r in R. The following are equivalent: (i) R is strongly regular. (ii) R is reduced and regular. (iii) Every principal left ideal is generated by a central idempotent and is thus two-sided. (Hint: (i)*(ii) If r = xr2 then (r - rxr)' = 0 so r = rxr. (ii) -(iii) by exercise 23. (iii)*(i) R is regular so write r = rxr; putting e = rx note er = r'e for some r', so r = er = r'e' = ere = re.) 25. A reduced ring R is regular iff every prime homomorphic image is a division ring. (Hint (=.) by exercise 24. (-=) If R is not strongly regular then R is not left n-regular so some prime image is not left n-regular by lemma 2.7.39.) 26. Every strongly regular ring R is a subdirect product of division rings. (Hint: Use structure of reduced rings.)
Fully Idempotent Rings R is jully lefi idempotent if L2 = L for every left ideal L . This is a useful common generalization of regular rings and V-rings (cf., exercise 29.).
340
Basic Structure Theory
27. Every fully left idempotent ring is semiprime. 28. The following are equivalent: (i) R is fully left idempotent. (ii) r E RrRr for every r in R. (iii) RIA is a flat module for each A a R. (iv) L n A = AL for every L < R and A Q R. (Hint: Cyclic proof, analogous to the corresponding results for regular
rings.) 29. Examples of fully left idempotent rings: (i) simple rings; (ii) regular rings; (iii) V-
rings. In fact, V-rings satisfy the stronger property that if L , I L , are left ideals then either L , L,L, or L,L, = L,. (Hint: Otherwise, take a maximal left ideal M 3 LIL, such that L , $ M. The canonical map L , + L,/(M n L,) cannot extend to R.) 30. If R is fully left idempotent then Z(R) is regular. (Hint: If z E Z(R) and z = c r l i z r 2 i z = c r l i r z i z 2thene=CrlirZizisidempotentand [r,e] ~ R z n A n n z = O for all r in R.) In particular, this holds if R is regular or if R is a V-ring. Further results interrelating fully idempotent rings with V-rings and regular rings can be found in the survey article of Fisher [74]; some of the noncommutative results are proved by means of left n-regular rings and lemma 2.7.39.
Rings Spanned by their Units 31. An element r of R is unit regular if r = rur for some unit u of R. In this case show that iff E R and R is regular then r is a sum of two units. (Hint: Ann(2ru - 1) = 0
so 2ru - 1 is invertible, and r
= (24-l
+ (2ru - 1)(2u)-'.)
32. If R is regular and each primitive factor ring is Artinian then each element of R is
unit regular. (Hint: Take r in R; if r is not unit regular then by Zorn's lemma there is A a R maximal with respect to the image of r not being unit regular in RIA. Clearly, RIA is not simple Artinian so by exercise 22 has a central idempotent e; reach a contradiction since the image of r is unit regular in (R/A)e and in (RIA)(1 - e).)
92.12 1. Suppose S is a submonoid of Z(R) and I a R with I n S = 0.Then
S-'R z S-'R/S-'I where-denotes the image in R/I. In particular, S-'R is simple if I is maximal such that I n S = 0. 2. If S is a submonoid of Z(R) then S-'R is a flat R-module. (Hint: If f: N + M is monic then S-'N + S - ' M is monic.) Conclude that there is an exact functor F: R-&d -+ S - l R - d ~ dgiven by FM = S ' M . 3. The R-module @{Rp: P E Spec(Z(R))} is faithfully flat. If M p is flat for every maximal ideal P of Z(R) then M is flat. 4. Suppose f:M N is a map of R-modules. Then f is monic (resp. epic) if the induced map fi M p + Np is monic (resp. epic) for each maximal ideal P of Z(R). Jacobson rings (also see exercises 8.4.13 and 8.4.10) 5. (Pearson-Stephenson) Example of a skew polynomial extension of a commuta-
tive local (but not Noetherian) Jacobson ring which is prime but not semiprimi-
Exercises
341
tive (and thus not Jacobson). Let C = F[A]/A where F is a field, A = {Ai:i E Z } is an infinite set of commuting indeterminates, and A is the ideal generated by all monomials Aicl,"'Li(t) where t 2 2, i(1) < ... i(t), and i(t) - i(1) < t 2 . Then ii++ li+,defines an automorphism a of F[A] and so gives an automorphism of C. Then the maximal ideal M of C consisting of the images of nonconstant polynomials is nil (because Lt E A ) ; hence M is the unique prime ideal of C, and C is local Jacobson. Let T = C[p;a] where p is a new (skew) indeterminate. T is prime but not semiprimitive. (Hint: To show T is prime suppose as in exercise 1.6.1 I that 0 # A 4 C and B 4 (C, a) with AB = 0. Then aumb= 0 for all a in A and b in B. But checking leading terms under the lexicographic order shows aa"b # 0 for m large enough unless a = 0 or b = 0. On the other hand, the image zo of A, in C generates a nil left ideal in T because cipi)zo)' = 0.) Incidentally, an Ore extension of a left Noetherian Jacobson ring is Jacobson as well as left Noetherian, cf., exercise 8.4.10. 6. Any finite centralizing extension of a Jacobson ring is Jacobson. (Hint: One may assume both rings are prime and use corollary 2.5.30.)
((xi=,
Invertible Projectives 7. The following are equivalent for M in C - A o d : (i) M is invertible projective. (ii) M is f.g. projective and End, M z C. (iii) M is finitely presented and M p x C, for every maximal ideal P of C. (iv) There is an isomorphism Hom,(M, C) Q M -+ C given by f Q x + f x . (v) N Q M x C for some N. (Hint: (i) (ii) * (iii) * (iv) Localize and use exercise 4 where appropriate; (iv) * (v) obvious; (v) * (i) count ranks by localizing.) 8. Using exercise 7 conclude {isomorphism classes of invertible projective modules over a commutative ring C} form an abelian group under tensor products, which is called the Picard group, written Pic(C). If H is a commutative C-algebra then there is a group homomorphism Pic(C) + Pic(H) given by M M Q, H. Pic(C) is trivial iff every invertible projective is free. 9. If C is a central subring of a prime ring R then C n Rc = Cc for any c in C. Consequently, LO holds for height 1 primes of C. -+
Passman's Approach to Normalizing Extensions The next exercises sketch the approach of Passman [81] to the question of comparing prime ideals of finite normalizing extensions. The approach is quicker than that of Heinecke-Robson, but fails to capture that elusive INC, which is proved by Heinecke-Robson. The basic construction is of a suitable large primitive ring. Write C{{X}} for the ring of formal power series in a set of noncommuting indeterminates X, and write * for the free product. 10. Suppose R is a prime C-algebra. Let X be a set of noncommuting indeterminates
of cardinality = max(N,,(RI); let T = R{{X}}, let Y be a set of noncommuting indeterminates of cardinality [TI, and let d = T { Y}. Then d is primitive. (Hint: As in exercise 2.1.14. Namely, index X by the elements of R, i.e., as {X,:rER}; and noting Id1 = IT1 write Y = { Y W : w € d }Letting . a=xrERXrr
Basic Structure Theory
342
co+w6J+
show L= d ( l away,) is a proper left ideal comaximal with every nonzero ideal. To show L # d one supposes 1 E L and reaches a contradiction by taking leading terms in Y and then lowest terms in X.) 11. Given A a R write 2 for the set of elements of I? whose coefficients are all in a, i.e., (A * C { { X } } )* C{ Y}. Then i n R = A and (R/A)- = R/A; A,)” = 2,; if A is a prime (resp. semiprime) ideal of R then A’ is a primitive (resp. semiprimitive) ideal of d. 12. If B is annihilator ideal of d then ( B n R j E B, and if P E Spec@) is an annihilator ideal then P n R E Spec(R). (Hint: If r E B n R then r annihilates every coefficient of every element of Ann ’B, implying ( B n R j E Ann(Ann ’B) = B.)
(n
n
Now that Passman’s “primitivity machine” has been set up we need some results on normalizing extensions and simple modules. In exercises 13-18 assume R‘ = Ra, where each a,is R-normalizing.
14.
15.
16.
I;=,
da, and the a, are I?-normalizing. If A a R’ with A n R = 0 then AnW=oindl. If L is a maximal left ideal of R then R’IR’L has composition length 5 n as Rmodule. (Hint: (Ra, R’L)/R’L is an image of the module (R + R‘L)/R’L= R/(R n R‘L)which is simple or 0.) (Bit-David and Robson) If A a R and is large as R - R bimodule (i.e., A intersects all ideals nontrivially) then there is A‘ a R‘ with 0 # A ’ n R E A. (Hint: Take A‘ an R - R bisubmodule of R’ such that A’ @ R is a large bisubmodule of R’. Then M = A‘ @ A is large in R’. For each i , j one sees {r’ E R’: a,r’aj E M} is a large bisubmodule of R’, so their intersection I is a large bisubmodule. But I = {r‘ER‘: R‘r‘R’E M} so I E M a n d 0 # I n R E A.) Cutting Down. If P‘E Spec(R’)then P ’ n R is an intersection of 5 n prime ideals. (Hint: One may assume R’ is prime. Then I? has primitive ideals Pl,... ,Pk for some k I n with pi = 0 by corollary 2.5.30; each pi = Annj+,n 3 so pin R E Spec(R) by exercise 12.) In particular if R’ is prime then 0 is a finite intersection of prime ideals of R. LO generalized. If P E Spec(R) then there exists P‘ in Spec(R’) such that P’ n R c P and P is minimal in Y ( P ’ n R). This generalizes LO since if P’ n R were in Spec(R) we would have P = P‘n R. (Hint: As proposition 2.12.43 one may assume R’ is prime and A’ n R $ P for all 0 # A‘ a R’. By exercise 15 there is OZBQR with PnB=O. “Cutting down” shows there are Pl, ..., Pk in Spec(R) with pi = 0; we may assume each pi is minimal prime. But some pi 8 B so P B = 0 E Pi implies P pi so P is minimal.) There are at most n distinct P ; in Spec(R’) for which any P E Spec(R) is minimal in Y ( P ;n R). (Hint: Using the primitive machine one may assume P is primitive. Write P = Core&) for L maximal and let M = R’/PR’ which by exercise 14 has composition length 5 n. Each 4 can be displayed as an annihilator of a suitable composition factor; however, the reader should be warned this argument relies on lemma 2.5 of Passman [81] which takes two pages to prove.) Passman [81] contains related results and examples, and it would be interesting to see if INC could be proved by these methods, to shorten Heinecke-Robson [81]. Prime rad(R) = R n Prime rad(R‘).(Hint: either directly or exercise 2.5.16.)
13. d‘ =
+
n
17.
18.
19.
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343
82.23 1. Every (*)-primitivering is (*)-prime. 2. (R, *) is primitive iff R has a maximal left ideal which contains no nonzero (a)ideal of R. 3. (Density theorem for primitive rings with involution and nonzero socle) Suppose ( R , * ) is primitive and soc(R) # 0. There is a minimal left ideal L with a sesquilinear (*)-compatible form (,) L x L-, D having the following property for any finite D-subspace V of L: Suppose x E L - V and x’ E L with ( V + xD, x ’ ) = 0. For any left ideal A of R with A L # 0 there is a in A satisfying aV= a*V = a*x = 0 and ax = x’. Furthermore, if L*L # 0 then (,) can be taken to be nondegenerate; if L*L = 0 then (,) can be taken to be trivial (i.e., x‘ can be taken arbitrarily). Hint: By exercise 2.1.1 there is a , in A with a , x = x and a, V = 0 Thus it suffices to find a, in L with a 2 x = x’ and a:(V x D ) = 0 since then take a = a,a,. If L*L = 0 this is easy; otherwise, apply theorem 2.13.21, noting there is x , in L with (x,,x) = 1.) 4. Notation as in exercise 3, given x , , . .. ,x, in L and a finite D-subspace V of L not containing x , , one can find d , = 1, d , , . .. ,dnin D such that for any x in L with ( V z x i d , x ) = 0 there is r in R such that rxi = xdi, r*xi = 0 for 1 Ii 2 n, and rV = r*V = 0. (Hint: exercise 3 and idea of exercise 2.1.1.)
+
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Lie and Jordan Ideals of Associative Rings There is a lovely theory of Herstein
[69B] concerning the Lie and Jordan structure of simple rings. It is useful to view ideals as rings without 1. The key results are exercises 5, 9, 12, and 18. We say R is n-torsion free if nr # 0 for all n # 0. Assume throughout that R is semiprime and 2-torsion free, and 2 = Z ( R ) . 5. If [a, [a, R]] = 0 then a E 2. (Hint: Write d = d, for the inner derivation [a, 1. Then d2R = 0 so 0 = d(d(r,r,))= d((drl)r2 r l d r 2 )= 2dr,dr, for all ri in R; implying 0 = drld(rr,) = (drl)rdrl for all r in R, so d r , = 0.) Note this lemma works for rings without 1. Conclude that if [a,[a,l]] = 0 for some 1a R then a E CR(l). 6. A Jordan ideal of a Jordan ringJ is an additive subgroup A satisfying ara E A and rar E A for all a in A and r in J. Conclude rlar2 r2ar1E A and thus r , a a r , E A for all ri in J. Conversely, if f E J and ra ar E A for all r in J then A is a Jordan ideal. 7 . Suppose A is a Lie or Jordan ideal of an associative ring R and Ra E A . Then RaR E A . (Hint: r1ar2= [ r l a , r 2 ] r2r,a;likewise for Jordan.) 8. If A is a Jordan ideal of R satisfying a 2 = 0 for all a in A then A = 0. (Hint: ab + ba = 0 for all a, b in A ; taking b = ar + ra yields 2aRa = 0.) 9. Any nonzero Jordan ideal A of R contains a nonzero (associative) ideal of R. (Hint: [ r , a 2 ] = [ r , a ] a + a [ r , a ] E A so 2ra2 E A for all a in A and r in R; thus 2Rn2 E A so conclude with exercises 7, 8.) 10. Suppose A is both a subring and a Lie ideal of R. Then R [ A , A I R E A . (Hint: By exercise 7 one needs show R [ A , A] c A. But r [ a , b] = [ra, b] [b, r]a E A.) 11. If A is as in exercise 10 then either 0 # R [ A , A I R a R or A E Z . (Hint: If [ A , A] = 0 then exercise 5 is applicable.)
+
+
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+
+
+
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Basic Structure Theory
12. Suppose R is noncommutative. Then the subring of R generated by [R, R ] contains a nonzero ideal of R. (Hint: It satisfies the hypotheses of exercise 10 and is noncentral by exercise 5.) In particular, if R is simple then [R,R ] generates R as
a ring.
13. If A is a noncentral Lie ideal then [I,R] E A for some 0 # I a R. (Hint: Let T ( A )= { r E R: [r,R] E A } . Then T ( A )is a Lie ideal containing A and is a sub-
+
ring since [r,r,,x] = [ r l , r 2 x ] [r2,xr,].Apply exercise 10.)
14. [ [ R , R ] , [ R , R ] ]2 [I,R] for some 0 # I Q R. (Hint: [ [ R , R ] , [ R , R ] ] is a Lie ideal, which one may assume is central. Exercises 5 and 12 show 2 2 [R,R ] and so Z 2 I for some 0 # I a R, so [I,R] = 0.) 15. If A is a Lie ideal of [R,R ] not containing [I, I ] for any nonzero ideal 1 of R,
then [[[A, A ] ,[A, A ] ] ,[ [ A ,A ] ,[A,A ] ] ]= 0. (Hint: Let T ( A )= { r E R: [r, R ] E A } , a subring of R. Let B = [ T ( A ) , T ( A ) ] .The calculation of exercise 10 shows R[B,B] E T(A ). Likewise, [ B , B ] R E T( A ) , so R [ B , B ] R E R[B,B] [ T ( A ) , R ]E T ( A ) + A. However, repeated applications of the Jacobi identity shows [ T ( A )+ A, T ( A ) + A ] E T(A),so by hypothesis R[B, B ] R = 0. Substitute, noting [A, A ] E T(A).) 16. If A is a Lie ideal of [R, R ] and [ A , A ] = 0 then A E Z. (Hint: Fix a in A and let d denote the inner derivation by a. Thus d A = 0 so d Z [ R R , ] E d A = 0. Taking b in A yields 0 = d2[r,sb]= d2([r,s]b+ s[r,b])= 0 + d2(s[r,b])= (d2s)[r,b]+ 2(ds)d[r,b] 0. Taking r in [R,R ] thus yields (d2s)[[R, R ] ,A ] = 0. Let A l = [[R,R ] ,A]. One is done by exercise 5 unless A, # 0. Let L = Ann, A,. Then [L, [R,R ] ] E L so LCR, R ] c L. But LR E L + [L,R ] so L2R E L2 L [ L ,R ] E L and L’RA, = 0; thus L = 0 since one may assume R is prime. Hence d2s = 0 for all s in R implying a E 2.) 17. If A is a Lie ideal of [R, R ] with [A, A ] E Z then A c Z. (Hint: One may assume R is prime. By exercise 16 assume c = [a, b] # 0 for suitable a, b in A, and let d denote the inner derivation by a. Replacing r by ar, if necessary, one may pick r with rl = d2r 4 Z. Then rl = [a,dr] E A and ar, = [a,d(ar)]E A. Now [b,d2(ar)]= [b,ald2r+a[b,d2r]= -crl +c,a for some cI 20, since r , # Z but [ b , d 2 ( a r ) ] ~ Z . like wise,^; = -c’d2(ar) c‘, for suitable c’,cf in C. Equatingd2(ar)= ar, yields a nondegenerate quadratic equation in a; taking the Lie product with b enables one to conclude a E Z , a contradiction. 18. Any noncentral Lie ideal A of [ R , R ] contains [ I , I ] for some 0 # I a R. (Hint: Apply exercise 17 repeatedly to exercise 15.) In particular, if R is simple then [R,R ] / ( Zn [R,R ] )is a simple Lie ring.
+
+
+
+
Herstein [69B] also considers the characteristic 2 case (for simple rings).
Lie and Jordan Structuresof Rings with Involution We continue with Herstein’s theory for a semiprime ring with involution (R,*). As above, we also make the simplifying assumption that R is 2-torsion free. Write S = (R,*)’ and K = ( R ,*)and given A c R write A for the subring of R generated by A. Several of the hypotheses involve S $ Z or K $ 2. As we shall see in exercise 6.1.36 these restrictions are very mild, for if (R,*) is simple with S E Z or K E Z then [ R : Z ] 5 4. 19.
s is both a Lie ideal and a subring of R. (Hint: [s,r] = (sr + r*s) - (r + r*)s.)
Exercises
345
20. If S $ Z then contains a nonzero ideal of ( R ,*) by exercise 11; in particular, 9 = R if (R,*) is simple. 21. K Z is a Lie ideal of R . (Hint: abr - rab = a(6r + r*b) - (ar* + ra)b for a, 6 in K . ) 22. If K Z $ Z then K z (and thus K )contains a nonzero ideal of (R, *). 23. If A is a Jordan ideal of S then A 2 S n I for some 0 # I 4 ( R , *). (Hint: If a E A then a2r + r*a2 = a(ar + r*a) + (ar + r*a)a - a(r + r*)a E A . Squaring yields 2r*a4r E A so we can take I = 4Ra4R unless a4 = 0 for all a in A . Hence 0 = r(a2r + r*az)4a2= (ra')' for all r in R, implying a' = 0 by Levitzki's theorem (proposition 2.6.26(i)). The argument of exercise 8 shows aSa = 0, so axaxa = 0 for all x in K , implying ( r c ~=) ~0 for all r since 2r E S + K . Thus a = 0.) 24. If ( R ,*) is simple then S is simple. The analogue of exercise 24 for K is that [ K , K]/(Z n [ K , K]) is a simple Lie ring, if (R, *) is simple and [ R : Z ] 2 16. This is proved in Herstein [69B, p. 33-46] and in Benkart [76].
Jordan Homomorphisms Suppose that W is another ring. Under the above notation can any Jordan homomorphism cp: S + W (viewing W as a Jordan ring) be extended to a ring homomorphism R -+ W? Martindale has worked extensively on this question and has solved it modulo low dimensions for (R,*)simple, in case R has a nontrivial symmetric idempotent; the remaining case is thought to be implicit in Zelmanov's work on Jordan division rings. We present one of Martindale's theorems. 25. Suppose (R, *) is simple with a nontrivial symmetric idempotent e, where eSe $ Z and (1 - e)S(1 - e ) $ 2.(As stated earlier, this holds if [R:Z] 2 16.) Any Jordan homomorphism cp: S -+ W can be extended to a unique ring homomorphism @ : R + W. (Hint: Let el = e and e2 = 1 - e; put R , = eiRej and f;. = q e i , a n idempotent of W. From now on assume i # j . For rij E R, note r$ E Rji so one must have @rij = 4(ei(rij+ r$)ej)= h@(rij + r $ ) h , so @ is determined and defined on R , by this formula. Note (RijRji+ R , + Rji + Rjj)4 ( R ,*) so RijRji= R i i . Consequently, @ is determined by its action on R, and is thus unique. However, using this rule to define @ on Rii leads to difficulties in proving @ is well-defined, so instead define @rii= ~uqs~'...qsj"' where rii = c:=lsy...sj"' by means of exercise 20, with each s y ) E S. If s E eJe, then matching components of cp(s(rij+ r t ) + (rij + r ; ) s ) in the Peirce decomposition of W shows @(srij)= cps@rij; iterating yields @(sl.. . s,rij)= q s , . .. cpsI@rij for s I ,...,s, in eiSei. O n the other hand, matching components of f;. shows q(rijqji q ; r t ) = @rij@qji @qX@r$ for all qji in Rji;consequently, if
+
.Y = ~ u i - ~ ~ 'EqS' .then ~ ) 2qs = q ( s
+
+ s*) = I@i$)@qr + @ q ~ ' * @ r ~E ~@Rij@Rji. '*
In particular, 2Si = 2cpeii E @[email protected] follows that @ is well-defined, for if rii = s?). ..s:") = 0 then 2
1c~s?'. . qsj") = 2 1c ~ s ? ' .. .~ p s j uE) ~c c ~ s ? ) ... ~ s ~ " ) @ R ~ ~ @ R ~ ~ ,
= @((
c sy).. .S ~ " ) ) R , ~ ) @0.R ~ =
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Basic Structure Theory
A case-by-case analysis of these formulas also show @ is a homomorphism.) Note these methods relied heavily on the existence of a nontrivial idempotent.
s
The structure of and f In the next few exercises some more results relating the structure of a ring with involution (R, *) and the subring f generated by S are given.
+
26. rSr* G f for all r in r in R. (Hint: rs, -**s,,r*= [ ~ , s ~ . . . s , , - ~ ] [ s , , , ~ * ] s1 ...
~ , , - ~ ( r s , , r *~) -~ ~ ~ ~ s , , - ~ r+r (* ~s ,s,~ ~ ~ * s , - ~ ~byexercise * ) s , , E 19andinduction.) S 27. The following properties pass from (R, *) to S: Prime, semiprime, primitive, semiprimitive, semisimple Artinian. (Each of the proof is a few well-chosen lines, cf., Herstein [76B, p. 231ffl for these and similar results.)
3
Rings of Fractions and Embedding Theorems
The main theme of this chapter is embedding a ring into a nicer ring which can be studied more easily. One familiar example from undergraduate study is embedding an integral domain into its field of fractions; another example is embedding a commutative Noetherian domain into a local Noetherian domain. In fact, these are both examples of “localization,” a term which has evolved considerably during the last 30 years. Our first objective will be to find a larger ring in which a given set S of elements becomes invertible; surely this implies S is regular, but possibly other conditions may be required on S as well. When S is central we have “central localization,” which was encountered already in 51.8 and $2.11. A slightly weaker condition is the “Ore condition,” which ensures that localization can be carried out similarly to the commutative procedure, and Ore localization suffices for most applications in ring theory. Ore localization, also called “classical localization,” is studied in depth in $3.1. The most successful application of Ore localization is Goldie’s theorems, proved in $3.2, which characterized those prime and semiprime rings having semisimple Artinian rings of fractions. In particular, every prime Noetherian ring satisfies Goldie’s conditions, and the reader may wish to pass directly from Goldie’s theorem to 53.5, which contains the basic theory of Noetherian rings.
347
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Rings of Fractions and Embedding Theorems
Goldie’s theorem leads naturally to several generalizations. On the one hand, it is of interest to know which rings can be embedded into Artinian rings, since many properties of Artinian rings pass to their subrings. The remainder of 13.2 is spent on questions of this sort, including Small’s theorem. On the other hand, one may wish to find non-Ore localization procedures which generalize Goldie’s theorem and apply to wider classes of rings. One such theory is Johnson’s theory of nonsingular rings, which is presented in some detail in 13.3, including Goodearl’s theorems; however, this topic is not pursued in later chapters. Johnson’s theory has been generalized further to an abstract localization theory, sketched in 13.4; one important application is Martindale’s ring of quotients and, in particular, his central closure, which is an indispensible tool in parts of the structure theory.
83.1 Classical Rings of Fractions In this section we consider general conditions under which a given submonoid S of elements of R can be made invertible, cf., definition 1.10.1. Often we shall require S to be regular, i.e., every element of S is regular in R. Recall s E R is regular if rs # 0 and sr # 0 for all nonzero r in R. Hopefully there will be no confusion with our other usage of “regular”-“regular ring” means von Neumann regular, cf., definition 2.1 1.18. Definition 3.2.2: Suppose S is a (multiplicative) submonoid of R. Q is a (left) ring of fractions of R with respect to S if the following two conditions are satisfied: (1) There is a ring homomorphism v : R + Q such that vs is invertible for all s in S, with ker v = { r E R : sr = 0 for some s in S } , and (2) every element of Q has the form s - l r where s E S and r E R. (Technically, this should be written (vs)-’vr, but we shall write s - l r abusing the notation slightly.) Important Special Case. If S = {all regular elements of R } and if cp: R + Q is an injection then Q is called the classical ring of fractions of R, and R is called an order in Q. Classical rings of fractions need not exist, and our first objective will be to find necessary and sufficient conditions for constructing rings of fractions. Definition 3.2.2: A (left) denominator set (also called Ore set) is a submonoid S of R satisfying the following two conditions: (i) For any s1 and S and rl in R there exist s2 in S and r2 in R such that s2rl = r2s1.
53.1 Classical Rings of Fractiolls
349
(ii) If rs = 0 for r in R, s in S, then there is s' in S with s'r = 0. Proposition 3.1.3: If R has a ring of fractions with respect to S then S is a denominator set.
Proofi We check the conditions of definition 3.1.2. (i) Given s l , r l we have ( l - ' r 1 ) ( s i 1 l )= s-'r for suitable s,r; then l - l s r , = l-'rsl implies srl - rsl E kerv so for some s' in S we have 0 = s'(srl - rs,) = s'srl - s'rs,; let s2 = s's and r2 = s'r. (ii) If rs = 0 then 0 = (l-'rs)(s-'l) = l - l r , implying s'r = 0 for some s' ins. Q.E.D.
Before considering the converse, let us discuss two important special cases. First, any submonoid S of Z ( R ) is obviously a denominator set, and the corresponding ring of fractions S - R exists and was already constructed three different ways (construction 1.10.2, construction 1.10.3, and exercise 1.10.1). On the other hand, suppose S is regular. Then condition (ii) is automatic, so we need only check (i). In case S = {all regular elements of R} we call (i) the Ore condition, and R is called (left) Ore; for example, any PLID is Ore by remark 1.6.19. The proof that Ore rings have classical rings of fraction follows the lines of the general proof presented below, but with the simplification that one does not worry about zero divisors. We are ready now to proceed to the construction of a ring of fractions with respect to an arbitrary denominator set. As usual there are two approaches-the "brute force" construction which follows one's naive intuition but involves verifications at every step of the way, and a slick construction using direct limits. The slick construction yields a much more encompassing result (cf., theorem 3.4.258) but leaves us without a "feel" for computing in the ring of fractions. Consequently, we shall wade through the brute force construction, and the reader who is so inclined can view it as a corollary of theorem 3.4.25 below.
'
Theorem 3.1.4: Suppose S is a denominator set.
-
-
(i) S x R has an equivalence deJined as follows: ( s l , r l ) ( s 2 , r 2 )if there exist r,r' in R such that rrl = r'r2 and rsl = its2 E S. Write S-'R for the set of equivalence classes, and write s-'r for the equivalence class of (s,r). (ii) If s E S and as E S then (as)-'(ar) = s-lr in S-'R. (iii) To check a function f on S-'R is well-deJned it sufices to prove f has the same value on sT'rl and (rs1)-'(rrl)where rsl E S.
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(iv) S-'R has a natural ring structure (given in the proof), which is a ring of fractions for R with respect of S .
Proofi
-
(i) Clearly is reflexive and symmetric; to prove transitivity we need the following technical lemma, which also cleans up the other sticky points in proving the theorem:
-
Lemma: Suppose (sl,rl) (s,, r,) and as, = a's, E S for a, a' in R. Then there is b in R with bar, = ba'r, and bas, = ba's, E S . Proof of lemma: There are I , I' in R satisfying r r , = r'r, and rs, = r's, E S, by hypothesis. Take r", s" such that s"(rsl) = r"(asl).
Then ( ~ " r r"a)sl = 0 so there is s; in S satisfying s;(s"r
- r"a) = 0,
i.e.,
s;s'lr = s;rtta.
Now s;r"(a'sz)=s;r"asl =s;s"rsl =s;s"r's2 so (s;r"af-s;s''ri)sz =O. Hence for some s; we have s;s;r"a' = s;s;s"r'.
Let b = s;s;t" and s = s;s;sff. Then ba' = sr' and
ba = s;(s;Y"u)
= s;(s;s"Y)= SY.
Hence bar, = srr, = sr'r, = ba'r,. Also, by hypothesis, bas, = ba's, and is Q.E.D. for lemma. in S since bas, = s;s;r"(as,) = s;s;s"(ts,) E S.
-
-
-
For transitivity of suppose (s,, r , ) (s,, r,) and (s,, 1,) (s,, 1,). For i = 1 , 2 we take ai, a: in R with airi = a;ri+l and aisi = ais,,, E S. Since S is a denominator set we have s in S and a in R with sa, = aa; then (aal)s, = (aa;)s, = (saz)s, and (saz)sz = s(azsz) E S. Applying the lemma to aa, and sa2 we get b in R with b(aal)rl = b(sa2)r2and b(aal)sl = b(saz)s2E S. Then
(baa,)r, = bsa,r, = (bsa;)r,
-
and
(baa,)sl = bsa,s, = (bsa;)s, E S.
Thus is transitive, and so is an equivalence. (ii) follows at once from definition of -.
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(iii) is immediate for if rr, = r'r, and rsl = r's2 E S then, by hypothesis, f(s;'r1) = f((rs1)-'rrl)
= f((r's,)-lr'r,) = f(s;lr,).
(iv) We shall define, respectively, the product and sum of sT'rl and s i ' r , . Although not needed formally the case s1 = r , = 1 is very instructive, for if we take r in R, s in S with sr, = rs, then "intuitively" one would want (l-'r,)(s;'l) = rls;' = s - l r . This trick for switching s2 past r , motivates the formal definitions: (s;'r,)(s;'r,) = (as1)-'rr2 where r,a E R are chosen such that as, E S and ar, = rs, ( s i ' r , ) + (sz'r,) = (asl)-'(url + rr,) where r,a are chosen such that as, E S and as, = rs,. We shall show multiplication is well-defined in three steps. S t e p 1. Independent of choice of r,a. By definition 3.1.2(i) there are ro in R and a, in S for which aOrl=r0s2, and we shall prove (usJ1rr2 =(aosl)~1ror2 for each choice of r, a. Taking r' in R and s' in S satisfying r'a, = s'a we have s'(rs2)= s'(ar,) = (s'a)rl = r'aorl = r'r0s2,
so (s'r - r'r0)s2 = 0. Hence s(s'r - r'r,) = 0 for some s in S, so ss'r = sr'r,. Moreover, sr'aOsl= ss'asl E S so by (ii) we see indeed ( a , ~ ~ ) - ~ r ,=r (~r'a~s,)-~sr'r~r~ ~ = (ss'as1)-'ss'rr2 = (usl)-'rr2.
-
Step 11. Well-defined in first argument. By (iii) it suffices to show ((bs,)-'br,) (s;'r,) = (s;'r,)(s;'r,) whenever bs, E S. But taking r,a such that a(br,) = rs, and a(bs,)E S we have
((b~,)-~br,)(s;~r,)= (abs,)-'rr,
= (s;'r1)(si1r2)
by definition since (ab)s, E S and (ab)r, = rs,. S t e p 111. Well-defined in second argument. As in Step 11, if bs, E S then taking r, a such that as, E S and ar, = rbs, we have
(~;~r,)((bs,)-~br,)= (us,)-'rbr, = (si'r,)(s;'r,) viewing ar, = ( r b ) ~ , . This proves multiplication is well-defined; addition is well-defined in the same way, and the ring verifications are now routine, cf., exercise 2. Finally note r , E kerv iff I - l r , = l-'O, which occurs when there are r,r' in R such
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that rrl = r'O = 0 and r = r' E S, proving S-'R is indeed a ring of fractions of R with respect to S. Q.E.D.
Properties of Fractions Remark 3.1.5: Suppose S is a regular denominator set. Since the canonical homomorphism v : R -+ S-'R has kernel 0, we can identify R c S-'R in which all elements of S become invertible. Now we shall show the ring of fractions of R with respect to S is unique, by characterizing S ' R as a universal in terms of remark 3.1.5. Given A c R write s-'A for {s-'a:s E S and a E A}.
Theorem 3.1.6: If S is a denominator set then the ring of fractions S-'R (together with v : R -,S ' R ) is the universal of dejnition 1.10.1. In particular, if f:R -, T is a homomorphism with fs invertible for all s in S then f extends uniquely to a homomorphism f: S-'R -, T ; moreover, f is given by
Proofi Any homomorphism f extending f must satisfy (l), so we shall show (1) defines a homomorphism. First note for s1 E S and rsl E S that f(rs,)-'frfs, = 1 so (fsl)-' = f ( r s , ) - ' f r . Now f is well-defined by theorem 3.1.4(ii) since f(rSl)-w~l= ) f(rsl)-'frfrl
= (fsl)-'frl.
To show -is a homomorphism take r, a as in the definition of sum and product in the proof of theorem 3.1.4, where a is chosen in S (which we can do by the dominator set condition).
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To prove the last assertion first note S-' kerf G kerf, so it suffices to show if 0 = f ( s - ' r ) = (fs)-'fr then s-'r E S-'ker f. But this is obvious Q.E.D. since f r = 0. Corollary 3.1.7: Suppose Sl E S, are denominator sets of R . Then there is a canonical homomorphism Si' R + S ; R under which we may view S, R as (ST'S,)-'(S;'R). Moreover, if 1-'s is invertible in S;'R for every s in S, then STIR x S;'R canonically. In particular, if every regular element of R i s invertible then R is its own classical ring of fractions.
'
'
Since every regular element of a left Artinian ring is invertible, we see that every left Artinian ring is its own classical ring of fractions. If we want to study rings in terms of the classical ring of fractions, the natural procedure is to start with a given kind of left Artinian ring and to characterize orders in this kind of ring. For example, those rings which are orders in fields are the commutative domains. Likewise, we have the following basic result due to Ore: Proposition 3.1.8:
R is an order in a division ring iff R is an Ore domain.
Proofi (a) by proposition 3.1.3, noting that every nonzero element of R is invertible in a larger ring and hence is regular in R; (e) is theorem 3.1.4, Q.E.D. taking S = R - (0) (since (s-'r)-' is then F ' s ) . Thus any PLID is an order in a division ring but need not be a right order in a division ring, in view of example 1.6.26. Nevertheless, proposition 3.1.8 is one of the cornerstones of ring theory, along with its generalization by Goldie to orders in semisimple Artinian rings, to be discussed in 53.2. We conclude this section with some useful observations concerning any denominator set S of R. Remark 3.1.9: For any s in S and ri in R we have + r2), by an easy calculation.
firl + f 1 r 2 =
s-'(rl
Lemma 3.1.10: (Common denominator) Given q l , . . .,q. in S-'R one has r , , ..., r n i n R a n d s i n S w i t h q i = s - ' r i for 1 S i l n . Proof: Inductively write qi = s-lri for 1 I i I n - 1 and write qn = si'r,,. Then picking r' in R and s' in S with s'sn = r's we have (s'sn)-'(s'rn) = qn and (s'sn)-'(r'ri) = (r's)-'r'ri = s-'ri = qi for 1 I i n - 1. Q.E.D.
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Proposition 3.1.11: If L is a left ideal of R and q l , . . . , q , E (S-'R)L then there are a', ...,an in L and s in S such that qi = s-'ai for 1 I i I n. (In particular, (S-'R)L = S-'L.)
Eyzl
Proofi Write qi = q,b, for qij E S-'R and b, E L. Then we can find a common s such that qii = s-'rii for all i, j (suitable rii in R). Hence q, = C:", s-'rijbij = s-' ziij b, so we are done by taking "ai= rijbij. Q.E.D.
Corollary 3.1.12:
xy!l
L n S # 0 @ (S-'R)L = S-'R.
Proofi (a) If s E L n S then 1 = s-'s E S-'L. (e) If (S-'R)L = S-'R then 1-'1 = s-'a for some s in S and a in L , implying s'(s - a) = 0 for some s' in S, and sa = s's E L n S. Q.E.D.
For future reference we list several properties that pass from R to S-'R when S is a denominator set. Proposition 3.1.13:
If R is a left Noetherian ring then so is S ' R .
Proofi Any left ideal of S-'R has the form S-'L where L < R. If L= Ra, then S-'L = ~ : , l ( S - ' R ) a i is f.g., proving S-'L is left Noetherian. Q.E.D.
I:=
This can also be seen as part of a more general lattice correspondence (exercise 5). Proposition 3.1.14: nilpotent.
If I d R is nilpotent and S-'I
a S-'R
then S-'I is
Proofi We shall show for any sequence s;'al,s;'a2, ... in S-'I that there are s: in S and a: in I such that for each n we have s;'a;..s~'al = (sk)-'ak*..a;; it follows at once that I" = 0 implies (S-'Z)" = 0. We find the si and a: by induction. Suppose we have found them for all i In - 1; then s- ,1, ~ , a , , ~ , ~ ~ ~=s(; s' a~,- l ) - l a ~ - lLet ~ ~ x~ a=~(s;la,,)(s;-l)-ll . E S - ~ Iby hypothesis. Then x = (sk)-'a; for suitable s; in S and a; in I so s,'s,''-s;'al
= s,'an(s;~')-'a;~'* * * a= ; xl-'a;-, . - . a ; = (s;)-'a;a;-'**.U;
as desired.
Q.E.D.
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Proposition 3.1.15: If S is regular and R is prime (resp. semiprime) then S ' R is prime (resp. semiprime).
Proof;. We prove the result for prime; for semiprime we would take rl = r2 and s1 =s2. View R G S ' R . If s;'r,Rs;'r,=O then O=s1s~'rl(rs2)s~'r2= r1rr2 for all r in R ; thus rl = 0 or r2 = 0. Q.E.D.
Proposition 3.1.16: Suppose R is both lefi and right Ore, having left (resp. right) ring of fractions Q1(resp. Q 2 ) .Then there is an isomorphism cp: Q1+ Q 2 given by cp(s-'r) = (ls-')(rl-'). Proofi cp exists and is 1 : 1 by theorem 3.1.6 applied to the canonical injection R + Q2.On the other hand, cp-' can be constructed symmetrically, implying cp is an isomorphism. Q.E.D.
Localization of Modules It shall be useful to extend several of these results to modules.
Definition 3.1.17: Suppose S is a left denominator set for R, and M E R - A u d . Then define S-'M = S-'R QR M . Remark 3.1.18: Any element of S ' M has the form s-' Q x for s in S and x in M.(Indeed, by lemma 3.1.10 we can rewrite any element x s f l r i Q xi of S-'M in theformxs-'r:Qxi = Cs-'Q rixi = s-' Q E r j x i . ) There is an alternate way of describing S-'M, more in line with our earlier results. Namely, take the set of equivalence classes of S x M under the equivalence ( s l , x l ) ( s 2 , x z )iff there are r,r' in R with rxl = r'x2 and rsl = r's2 E S. Writing the equivalence class of (s,x) as s-'x we can define scalar multiplication over S-'R by defining (s;'rl)(s; ' x ) = (ssl)-'rx where r E R, s E S are chosen such that srl = rs2. Addition of the s-'x defined as in theorem 3.1.4, we can duplicate the proof of theorem 3.1.4 to prove that these operations provide a module structure which we temporarily call Ms.
-
Proposition 3.1.19: S-'M z Ms under the isomorphism given by s-' 0 x
+
s- ' x .
Proof;. Analogous to that of proposition 1.10.18. There is a balanced map S-'R x M + S-'M given by (s-'r,x) + s-'(rx), yielding the desired map S ' R Q M --f M,. It suffices to prove its inverse, given by s-'x+s-' O x ,
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is well-defined, i.e., if s;’xl = s;’x2 then we want s;’ Q x1 = s;’ @ x 2 . Indeed, take r, r’ in R with rxl = r‘x2 and rsl = i t s 2 E S. Then si’ 8 x1 = (rsl)-lr
x1 = (Is1)-’ 8 rxl = (r’s2)-l 8 r’x2
= (rts2)-lr’8 x2 = s;’ @ x2
as desired.
Q.E.D.
An alternate proof is given in exercise 4. Theorem 3.1.20: Define the localization functor F: R - A o d to be S-’R QR-.Then F is an exact functor.
Prooji
+S-
‘R-Aod
We know F is a right exact functor, so it remains to show that if
f: M + N is monic then 1 @ f is monic. Suppose s-l Q x E ker(1 Q f ), i.e., s- l Q f x = 0. In view of proposition 3.1.19, this means s-yx = l-lO, i.e., there are r, r’ in R with rfx = 0 and rs = r’ E S . But f (rx) = rfx = 0 implies rx = 0, yielding s W 1 Qx = 0 by the next remark. Q.E.D. By proposition 3.1.19 we see s-l Q x = 0 iff s-lx = 0, i.e., there exist r, r’ in R with rx = 0 and rs = r’ E S. In particular, S-’M = 0 iff each 1 Q x = 0, iff for each x in M there is s in S such that sx = 0.
Remark 3.1.21:
53.2 Goldie’s Theorems and Orders in Artinian Quotient Rings The main object in studying rings of fractions is to relate rings to “nicer” rings. Since any left Artinian ring is already its own classical ring of fractions, we are led to consider which rings are orders in simple and semisimple Artinian rings. Goldie [58, 601 classified these rings; in particular, every semiprime Noetherian ring is an order in a simple Artinian ring (also, cf., LesieurCroisot [59]). Goldie’s theorems thereby provided a framework unifying Noetherian rings with commutative rings, and the impact on ring theory has been immense. As Goldie’s conditions involve left annihilators and direct sums of left ideals, these aspects of the structure theory have taken on an interest in their own right, and we shall begin by familiarizing ourselves with them.
Annihilators Definition 3.2.1: L c R is a lefi annihilator if L = Ann(S) for some subset S of R, i.e., L = { r E R : rs = 0).Similarly, we define Ann’s = { r E R : Sr = 0 } and call such a set a right annihilator of R . The terminology [(S) and a ( S ) are often used in the literature for Ann(S) and Ann’(S).
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Remark 3.2.2: (i) Every left annihilator is a left ideal of R, and every right annihilator is a right ideal of R. (ii) If S , E S, then Ann S , 2 Ann S,. (iii) S E Ann(Ann’S) and S 5 Ann’(Ann S). (iv) If A = AnnS then A = Ann(Ann’A). (Indeed, ( G )is by (iii); on the other hand, S E Ann’(Ann S) = Ann‘A so A 2 Ann(Ann’ A ) by (ii).) (iv’) If A = Ann’S then A = Ann’(Ann A). (Analogous to (iv)). (v) (7 Ann Si = A n n ( l S i ) (vi) Ann SiE Ann( Si)(but equality does not necessarily hold, as evidenced by Z, in which Ann( mE) = Ann 0 = Z, but Ann(nZ) =
1
nm+,
1 0 = 0.)
1
1
(vii) Ann Si = Ann( Si)if the Si are right annihilators. (Indeed, let Ai = Ann Si; then Si= Ann’ Ai by (iv’), so (v) yields Ann( Si) = Ann(Ann’( Ai))= Ai = Ann Si.) (viii) If R is a subring of T and S c R then Ann, S = R n Ann, S (ix) If Annx = Annx2 then Rx n Ann x = 0 (for if rx E Annx then r x 2 = 0 so rx = 0.)
1
1
1
These easy facts have far-reaching consequences and are used so often that we shall refer to them merely as (i) through (ix) Proposition 3.2.3: 9 = {left annihilators} is a complete sublattice of {left ideals of R} and is isomorphic to the dual of {right annihilators} under the correspondence A + Ann‘A.
Proof: The first assertion follows at once from (v) and (vii). The second Q.E.D. assertion follows from (ii) and (iv). Corollary 3.2.4: If R is a subring of T then there is a lattice injection { leji annihilators in R} -,{ leji annihilators in T } given by A -,Ann, Ann; A. Q.E.D.
Incidentally, the second assertion of proposition 3.2.3 is purely latticetheoretic and can be given in a much more general context.
ACC (Ann) Our interest lies in the case when {left annihilators} satisfies the ascending chain condition, which we shall call ACC(Ann). This condition is quite tractable.
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Remark 3.2.5: If T satisfies ACC(Ann) then every subring of T satisfies ACC(Ann), by corollary 3.2.4. In particular, every subring of a left Noetherian ring satisfies ACC(Ann). Remark 3.2.6: If R is a semiprime ring with ACC(Ann) then R has no nonzero nil left ideals by proposition 2.6.24.
Lemma 3.2.7: If R is semiprime with ACC(Ann) and 0 # r E R then the left ideal Ann r is not large. Proofi Let J = {r E R:Annr is large}, clearly a right ideal of R since Ann ra 2 Ann r for all a in R . We need to prove J = 0; by remark 3.2.6 it suffices to prove J is nil. Well for any z in J we have Ann z E Ann z2 E * implying Annz" = Annz"" = .--for some n (depending on z). Putting x = z, we then have Ann x = Ann x 2 so R x n Ann x = 0 by (ix); but x E J so R x = 0, proving z" = 0. Q.E.D.
Proposition 3.2.8: Ann' r = 0. Proofi
If R is semiprime with ACC(Ann) and Rr is large then
If a E Ann' r then Ann a 2 Rr is large so a = 0 by the lemma. Q.E.D.
Digression 3.2.9: Proposition 3.2.3 shows that ACC(Ann) is equivalent to the DCC(Ann'). This useful observation can be used to prove for example that any primitive ring R satisfying ACC(Ann) and having nonzero socle is simple Artinian. (Indeed, take an orthogonal set of rank 1 idempotents e1,e2,... and observe that Ann'e, > Ann'{e,,e,} > .-.shows that the rank is bounded, implying by the density theorem that R is simple Artinian, cf., remark 2.1.26.)
Goldie Rings Definition 3.2.10: R satisfies ACC @ if R does not have an infinite independent set of left ideals. R is (left) Goldie if R satisfies ACC(Ann) and More generally, we say a module M satisfies ACC @ if M does not ACC have an infinite independent set of submodules.
0.
Any semisimple Artinian ring R has finite length in R-Mod and thus satisfies ACC @ as well as ACC(Ann) and so is left Goldie. The condition
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ACC @ is rather strong, permitting us to relate regular elements to large left ideals; this turns out to be one of the key aspects of Goldie's theorem. Proposition 3.2.11: Suppose Ann r = 0. ( i )I f L n Rr = 0 then {Lri:i E N} is independent left ideals. ( i i ) I f R satisfies ACC @ then Rr is a large left ideal of R.
a set of
Proof: (i) Otherwise, take Cy=,airi = 0 with a, E L and a, # 0; then E L n Rr = 0, contradiction. (ii) By (i) we have L n Rr # 0 for every 0 # L < R. Q.E.D.
I::,"a,+,ri E Ann rm= 0 so a, = -1;:'' a,+,ri Proposition 3.2.12:
If R is semiprime Goldie and Ann r = 0 then r is regular.
Proof: Rr is large, so proposition 3.2.8 implies Ann'r Q.E.D. given Ann r = 0 we conclude r is regular.
= 0;
since we are
Proposition 3.2.13: Every large left ideal L of a semiprime Goldie ring R contains a regular element.
Proofi We have just seen it suffices to find r in L with Ann r = 0. We start with r l non-nilpotent such that Annr, is maximal possible. Then Annr: = Ann r , so Rr, nAnn rl = 0 by (ix). Continuing inductively, suppose we have chosen non-nilpotent r I , . . ., r , in L such that Rr,, . . .,Rr,, and A, are independent left ideals, where A, = L n Ann r , n . n Ann r,. If A, # 0 then A, is not nil so we can pick non-nilpotent r k + , in A, with Annr,,, maximal =L n possible. Then R r k + , n A n n r k + =, o SO Rr,, ..., Rrk+1, and Ann r n ... nAnn r k + are clearly independent. We have thereby established an inductive procedure, which must terminate because R satisfies ACC(@), Ann ri = 0 since L is large. Let i.e., for some k we have A, = 0 so r= r,. If 0 = ar = ar, then each ar, = 0 (since the Rr, are independent) Q.E.D. so a E Ann ri = 0, proving Ann r = 0.
,
n
,
1
n:=
To prove the next result we need the following special case of proposition 3.3.0 below (which can also be verified directly): Given a left ideal L of R and r E R define Lr-' = {a E R : ar E L } ; if L is large then Lr-' is large.
Theorem 3.2.14: (Goldie's second theorem) The following assertions are equivalent for a ring R :
( i ) R is semiprime Goldie.
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(ii) A left ideal of R is large iff it contains a regular element. (iii) R is Ore and its (classical) ring of fractions is semisimple Artinian. Proof: (i)=s(ii) by 3.2.11 and 3.2.13. (ii) =. (iii) First we show R is Ore. Given a, r E R with a regular we need to find a', r' with a' regular and a'r = r'a. Ra is large, so Rar-' is also large (in R ) and thus has a regular element a'; then a'r E Ra as desired. Let Q be the ring of fractions of R. We shall prove Q is semisimple Artinian by showing Q has no large left ideals other than Q itself (cf., theorem 2.4.9). Well suppose L < Q is large. Then L n R is large in R. (Indeed, if 0 # r E R then L n Qr # 0 so there is in Q with 0 # a;'rlr E L and then rlr E L n R.) Hence L n R has a regular element which thus is invertible in Q, proving L = Q. (iii)*(i) R has ACC(Ann) by remark 3.2.5. Next we claim if L is a large left ideal of R then L has a regular element. Indeed, let Q = S-'R be the classical ring of fractions. Then Q L is large in Q (since for any s-lr E Q one has 0 # L n Rr E QL n Qs-lr). Hence Q L = Q, so L has a regular element by Corollary 3.1.12. To show R is semiprime suppose N 4 R with N 2 = 0 and let L < R be an essential complement of N, i.e., L n N = 0 and L + N is large (cf., proposition 2.4.4). By the previous paragraph there is a regular element a in L + N, and Nu G N L E L n N = 0; since a is regular we have N = 0 proving R is semiprime. It remains to prove R satisfies A C C O . This follows at once from the following lemma: Lemma 3.2.15: Suppose R is an order in Q. If { L , :i E I} is an independent set of left ideals of R then {QL,:i E I } is an independent set of left ideals of Q.
c',=
Proofi If q, = 0 for q, in QLiUthen writing q, = s-lr, for 1 Ii It with ru E LJby proposition 3.1.11) we have x r , = 0 so each r, = 0. Q.E.D. The most important direction of Goldie's second theorem is (i) 3 (iii) since it enables us to study R through its ring of fractions. Condition (ii) is also very useful since it contains the technical information needed in many proofs about Goldie rings. Theorem 3.2.16: (Goldie's Jirst theorem) R is prime Goldie iff R is an order in a simple Artinian ring Q.
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361
Proof;. ( a )Q is semisimple Artinian by theorem 3.2.14. If Q has simple components Q1 # Q2 then (Q1n R)(Q2n R ) = 0, contrary to R prime; hence Q is simple Artinian. (e) R is Goldie, so we need to show R is prime. Suppose 0 # A 4R. Then 0 # QAQ 4Q so 1 = 4ilai4i, for suitable qij in Q and a, in A. Write 4 i , = s-’ril for 1 I i I t with ril, s E R and s regular. Then s = ri1ai4i2E AQ, implying (Ann, A ) s = 0. But s is regular so Ann, A = 0, proving R is prime. Q.E.D.
I:=,
Goldie Rank and Uniform Dimension Of course, once Goldie’s theorem was established the Goldie conditions became objects of study in their own right, so we digress a little to introduce the relevant notions. R has uniform (or Goldie) dimension n if there is no set of n + 1 independent left ideals of R, with n minimal possible. R satisjies ACC,(Ann) if every chain of left annihilators has length I n. R has Goldie rank n if R has uniform dimension n and satisfies ACCJAnn).
Remark 3.2.17: Suppose R C T. If T satisfies ACC,(Ann) then R satisfies ACC,(Ann) by corollary 3.2.4. Proposition 3.2.18: has Goldie rank n.
If R is semiprime Goldie then there is n in N such that R
Proof;. The classical ring of fractions Q of R has a composition series of some length n, so Q satisfies ACC,(Ann) and has uniform dimension 5 n; these properties pass to R by remark 3.2.17 and lemma 3.2.15. Q.E.D
Remark 3.2.18: Suppose R is prime Goldie, and let Q = M,(D) be its simple Artinian ring of fractions. Obviously n is the Goldie rank of Q, which also is the Goldie rank of R. This easy formulation of the Goldie rank of a prime ring is used repeatedly in Chapter 8. ACC(Ann) does not imply ACC,(Ann) for suitable n, cf., exercises 8,9. However, see proposition 3.2.19. The uniform dimension of a ring has a very useful generalization to modules, whose basic properties we shall record here for later use. The uniform (or Goldie) dimension of a module M is the largest n (if it exists) for
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which there is a set of n independent submodules; a module of uniform dimension 1 is called uniform. In other words, M is uniform iff no submodule of M can be written as a direct sum of two proper nonzero submodules. It follows at once that every submodule of a uniform module is large and is uniform as a module itself. Let us use uniform modules to characterize the uniform dimension.
Proposition 3.2.19: Suppose M satifles ACC dimension n. More precisely,
0. Then M has some uniform
(i) Every submodule N of M contains Q unform submodule. (ii) There are independent uniform submodules U , , .. .,U,, of M such that Ui is large in M. (iii) The number n of (ii) is uniquely determined and is the uniform dimension of M.
c:=, Proofi
(i) We are done unless N is not uniform, so N 2 No 0 N ‘ for suitable independent submodules No and N’; continuing with N’ we have an inductive procedure to build an infnite set of independent submodules, contrary to hypothesis unless at some stage we reach a uniform. (ii) By (i) we have a uniform U, IM . Inductively, given U , , ..., U, independent uniform submodules of M let Mt be an essential complement of U1@ 6 U,. If Mt # 0 it contains a uniform Q+,; this gives us an inductive procedure for finding an infinite set of independent uniforms, contrary to hypothesis, so some M,, = 0; i.e., U16 * * * @ V,, is large in M. (iii) Assume, on the contrary, M has independent submodules V,, ...,V ,+ . Choosing the 6 such that 6 E ( U , , . ..,U,,} for 1 I i Iu with u maximal possible,wehaveuIn.Let V = x ; = , t $ I f V n v j = O t h e n { V , , ..., V,,,qJ would be a set of independent submodules, contrary to choice of u. Thus V n q # 0 for each j , implying V is large in @ by proposition 2.4Sf(ii), and thus V is large in M , contrary to V,, n V = 0. Q.E.D.
,
,
Note. (Cohn) This proof could be streamlined by noting that (uniform modules} have a weak dependence relation (cf., remark 0.3.3) given by V E,,~,, Y if V n C( U E Y }= 0.
Digression 3.2.19: An alternate proof of proposition 3.2.19 actually yields stronger results. The idea is to translate these notions to the injective hull E(M), viewed as the maximal essential extension of M , viz. theorem 2.10.20.
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U is uniform iff E( U ) is indecomposable. Furthermore, N IM implies E ( N ) is a summand of E ( M ) , so the uniform dimension of M is precisely the number of summands in a decomposition of E ( M ) , which is unique by theorem 2.10.33. This approach makes available other invariants, most notably Ann E( Ui). We return to the uniform dimension in exercises 3.3.18ff and in 53.5.
Annihilator Ideals in Semiprime Rings Historically, of course, Goldie’s first theorem was proved first; there is an interesting argument (due to Herstein) that it implies Goldie’s second theorem, based on annihilators; the key points also have other applications.
Lemma 3.2.20: Suppose R is semiprime and A 4R. The following are then equivalent for B 4R : (i) (ii) (iii) (iv)
B = Ann A. B is maximal (as ideal) such that A n B = 0. B is an essential complement of A (as left ideal). B = Ann’ A.
(i) => (ii) (A n B ) 2 c BA = 0, so A n B = 0. If A n B’ = 0 for B’ 4R then B’A E A n B’ = Oso B’ s Ann A = B. (ii) (i) Clearly, Ann A 4 R since A 4R , so A n Ann A = 0 as above. But BA E A n B = 0 so B E Ann A, implying B = Ann A by maximality of B. (ii) o (iv). analogous to (i) (ii) above. (ii), (iv) (iii) Suppose ( A + B) n Rr = 0. Then ARr E A n Rr = 0 implying Rr c Ann’ A = B so Rr G B n Rr = 0. This proves A + B is large, and we are given A n B = 0. (iii) (ii) by definition of essential complement. Q.E.D. Proof:
-
One consequence of this lemma is that Ann and Ann’ mean the same when applied to ideals of a semiprime ring.
Proposition 3.2.21: and A 4R :
The following assertions are equivalent for R semiprime
(i) A is large as a left ideal. (ii) A is large as a right ideal. (iii) A n I # 0 for every 0 # I 4R .
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(iv) Ann A = 0 (v) Ann’A = 0.
Proofi (iii) * (iv) * (i) is a special case of the lemma, and (i) + (iii) is trivial. Q.E.D. (iii) +(v) 9 (ii) (iii) follows by left-right symmetry of (iii). In view of these two results we are led to consider annihilators of twosided ideals and formally define an annihilator ideal to be an ideal which is a left annihilator. We aim for the surprising result that in a semiprime ring ACC(annihi1ator ideals) is equivalent to having a finite number of prime ideals with intersection 0.
Remark 3.2.22: If A is an annihilator ideal of a semiprime ring R then A = Ann(Ann A) so A is the left annihilator of an ideal. (Immediate by lemma 3.2.20.) Proposition 3.2.23: lf M E R-Aod and i f the lattice {Ann N: 0 # N < M } has a maximal member P then P is a prime ideal of R. Proof: Suppose A 1 P, B z P are ideals with AB G P. Then ABN = 0 for some 0 # N < M so A(BN) = 0. But BN 4 N so A = P by hypothesis unless BN = 0, in which case B = P. Q.E.D. Remark 3.2.23’: In a semiprime ring, each annihilator ideal is a semiprime ideal. (Indeed if B 2 c Ann A then (B n A)3 = 0 so B n A = 0 implying B E Ann A) Theorem 3.2.24: (i) If R is semiprime with ACC(annihi1ator ideals) then R has a jinite set PI,.. .,P,, of prime ideals with PI n -..n P,, = 0. (ln fact, PI,.. .,P,, are precisely the maximal annihilator ideals.) (ii) Conuersely, if ni=I P = 0 for prime ideals PI,.. .,P,, then R is semiprime with ACC(annihilator ideals). In fact, every annihilator ideal then is an intersection of some of the 5, so, in particular, every chain of annihilator ideals has length In.
Proofi (i) Let 9= {maximal annihilator ideals}. Then all members of 9 are prime ideals by proposition 3.2.23. We shall see 9 is finite with intersection 0. Take distinct P , , P 2 , . . . in 9’ noting 5 $ 4 for all i # j , and let Bk =
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Ann(n:_, pi). Then Ann Pk+l$Z Bk (for, otherwise, (Ann Pk+,)P1. . . P k = O C Pkt implying Ann Pk+ G Pk+ and (Ann Pkt1 ) 2 =O; thus Ann Pk+ =0, contrary to remark 3.2.22). Hence B k c Bk+ for each k, so by ACC(annihi1ator ideals) we cannot have an infinite number of &, proving 9 has a finite number of members P , , . . .,P,. Let A = 4. If A # 0 then B, = Ann A implies B, c pi for some i ; rearranging the indices we may assume B, E P,. Then (En((-)::: pi))’ E BnA = 0 implying B, E En- contrary to what we saw above. Hence A = 0, as desired. (ii) Let I = { 1,. ..,n } . We shall show every annihilator ideal A has the form pi for suitable I‘c I, from which all the assertions follow at once. Let B = A n n A a n d I ’ = { i E I : A ~ ~and } I”={iEI:BE~}.SinceAB= Oc qfor each i we have I =I‘ul“.Then A = A n n B ~ A n n ( n { p i : i ~ 1 ” } ) 2 Q.E.D. { &: i E l ‘ }2 A, so equality holds at every stage.
,
,,
,
,
,
,
n,,,,
n
Remark 3.2.25: The last sentence of the proof shows that Ann pi =
n
j+i
5.
Corollary 3.2.26: If R is semiprime with ACC(annihi1ator ideals) then the maximal annihilator ideals are precisely the minimal prime ideals of R, and these are jinite in number. Proo) By theorem 3.2.24 and proposition 2.12.8 each minimal prime is a maximal annihilator ideal. But each maximal annihilator ideal P contains a Q.E.D. minimal prime which then must be P itself.
Theorem 3.2.27: Suppose R has a jinite set PI,.. . ,P, of minimal prime ideals with intersection 0, and each R / & is prime Goldie with ring of fractions Q,. Then R is Goldie with ring of fractions isomorphic to Q,.
ni=,
Proof: Take 0 # ri E nj+,5for each j. Put R, = R / 4 . Letting-denote the canonical image in R , we have 7, # 0; thus RriR is large in R , and by Goldie’s theorem contains a regular element 6, with a, E nj+,5. Note a = El=,a, is regular. R , given by r --t (r + Pl, . . .,r + P,), so Consider the injection R + that a, corresponds to an element whose only nonzero component is in R,. If 6 E R is regular then b = b + & is regular in R , (for if & = 0 then bra, = 0 so ra, = 0, implying F = 0; likewise, if rb = 0 then F = 0). Thus regular elements of R are invertible in Q = Qi,where each Q , is the simple Artinian ring of fractions of R,. Using the previous injections to view R E R, E Q,= Q, we claim Q is the ring of fractions of R.
fll=
n n
n;=,
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Indeed, we must show that any q E Q has the form b-'r-- where b E-R with b 1qi where qi E Qi and write qi = bi ri with bi regular .regular. Write q = aibi and r = If= airi. Clearly, b is regular and q = b-'r in Take b = Q.E.D. in Q, by componentwise verification.
q.
,
,
We are led to an analogue of the Chinese Remainder Theorem after a useful preliminary about idempotents.
Lemma 3.2.28: Suppose R is a semiprime ring and A is an ideal which is a summand of R as left module. Then A = Re for a central idempotent e, and R x Re x R( 1 - e) as rings. Proofi A = Re for some idempotent e (since A is a summand of R ) and Ann' A = Ann A Q R by lemma 3.2.20. But Ann' A = (1 - e)R so (1 - e)Re = 0 andalsoeR(1-e)EA(l-e)=Osince A 4 R.Thus R = e R e + ( l - e ) R ( l - e ) by the Pierce decomposition, yielding R x R , x R 2 where R , = eRe and R , = (1 - e)r(l - e). Since e, 1 - e are the respective unit elements of R , , R , we Q.E.D. see they are central in R.
Theorem 3.2.29: A semiprime ring R is isomorphic to a Jinite direct product of prime images iff R satisfies ACC(Annihi1ator ideals) and every maximal annihilator ideal has the form Re for e idempotent (and central by the lemma).
pi = 0 so the pi are the maximal Proofi (a) If R x nl=,RIP, then annihilator ideals by theorem 3.2.25, and pi = Re, for e, = (1,. . .,1,0,1,. . .,1) where 0 appears in the i position. (e) Letting Pl,. . .,P, be the maximal annihilator ideals of R we have pi = 0. By the Chinese Remainder Theorem it remains to show pi + 6 = R for each i # j. By hypothesis, we can write pi = Re, and 6 = Rej. Then 1 - ei E Ann P, c pi (cf., remark 3.2.25) so 1 = ei + (1 - e,) E 8 + p j as desired. Q.E.D.
,
Corollary 3.2.30: A semiprime Goldie ring R is a direct product of prime Goldie rings iff every maximal annihilator ideal can be written in the form Re for a suitable idempotent e.
Proposition 3.2.31: Suppose R satisjes ACC(annihi1ator ideals)and A, B Q R with A $ B.i If A - B E a finite union of annihilator ideals then A is contained in some annihilator ideal, so, in particular, Ann' A # 0.
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Pro08 A - B c finite union of maximal annihilators; each of which is prime, so apply proposition 2.12.7. Q.E.D.
Digression: Orders in Semilocal Rings Motivated by Goldie's theorem one might wish to characterize orders in left Artinian rings or, more specifically, Noetherian orders in Artinian rings. This problem was solved independently by two prominent ring theorists, Robson [67] and Small [66], each then at the start of his career. We shall introduce the relevant concepts in the text but leave most of the proofs for the exercises. The underlying idea is to pass down the nilradical; more generally, we deal with arbitrary I a R and let S be a submonoid of R . Definition 3.2.32: S is a denominator set for I if the following two conditions are satisfied:
(i) For any s1 E S and rl E I there exist s2 in S and r2 in I with s2rl = r2s1. (ii) If rs = 0 for r in I , s in S, then there is s' in S with s'r = 0. Remark 3.2.33:
Suppose S is a denominator set of R. Then S is a denominator set for I iff S-'I a S-IR. (Indeed (ii) holds a fortiori, and (i) is equivalent to (l-'r)(s-'l) E S-'I for all r in I . ) In case S = {regular elements} this property is called I-quorite in the literature. We aim to generalize remark 2.12.12 and thus consider the following three properties, where-denotes the image in R = R / I : (1) S is a denominator set of R . (2) S is a denominator set for 1.
(3)
5 is a denominator set of R .
Weakening definition 3.2.32(i) by means of exercise 2, one can easily prove Proposition 3.2.34: (=exercise 3) If two of the properties (l), (2), (3) hold then the third also holds, and S-lE FZ S-'R/S-'I.
Now suppose (1) holds and each element of S is regular. Let Q = S-'R, and view R E Q canonically. Given A a Q we put I = R n A a R ; letting Q = Q / A we have an injection R -P Q. Moreover, is invertible in Q and thus regular in R, implying Q = S-'R. Thus (1) and (3) hold, so (2) also holds. In particular, when S = {all regular elements of R } we have shown S is regular in R for all regular s in R , leading us to consider the converse.
s
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Definition 3.2.35: R satisfies the regularity condition (over I ) if F regular in R implies r is regular in R. When 1 is not specified we take I = Nil(R).
Theorem 3.2.36: (=exercise 4) R is contained naturally in a semilocal ring of fractions Q = S-’R, for suitable I 4 R and S = { r E R: Fis regular in R/I}, iff the following conditions are satisfied: (i) R is semiprime Goldie. (ii) Euery s in S is regular. (iii) S is a denominator set for I. Moreover, in this case S-’I = Jac(Q). Two important consequences of this result: Robson’s theorem: R is an order in a semilocal ring Q iff there is I a R such that ( i ) R satisfies the regularity condition over I ; (ii) R is semiprime Goldie, and (iii) {regular elements of R } is a denominator set for I.
Small’s theorem: A left Noetherian ring is an order in a left Artinian ring iff it satisfies the regularity condition. Robson’s theorem follows at once from theorem 3.2.36 taking S = {regular elements of R}. Small’s theorem requires a further argument, and is proved in exercise 7; a more direct proof is given in exercise 3.5.3 using more machinery. The recent literature contains several other characterizations of when Noetherian rings are orders in Artinian rings.
Embedding of Rings Among other things Goldie’s theorem says any semiprime left Noetherian ring can be embedded in a simple Artinian ring. This already has interesting consequences, as we shall soon see, and we are led to consider procedures of embedding rings into “nicer” rings. In particular, we shall present a general procedure of embedding a prime ring into a primitive ring and shall consider the problem of embedding rings into matrix rings over commutative rings.
Universal Sentences and Ernbeddings The theme pervading this discussion is that any universal sentence holding in a ring R‘ also holds in each subring R. Thus to verify such a property on R we might do best by finding an appropriate overring. One example is the
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sentence Vxy(xy = 1 => y x = 1) which defines weakly 1-finite (cf., definition 1.3.30);likewise, weakly n-finite could be defined by an appropriate universal sentence, for each n.
Theorem 3.2.37: Every subringR of a left Noetherian ring is weakly finite. Proof: We may assume R itself is left Noetherian, and we want to prove R is weakly n-finite for any given n, i.e., that M,,(R) is weakly 1-finite. M,(Nil(R)) is nilpotent since Nil(R) is nilpotent. Since invertibility passes up the Jacobson radical we may replace R by R/Nil(R) which is semiprime and thus is an order in a semisimple Artinian ring R’. But then M,,(R) E M,(R’) which is semisimple Artinian and thus 1-finite, as desired. Q.E.D. Using Small’s theorem we could improve this result, but the statement would become rather technical. A more subtle use of universal sentences asserts the nilpotence of a given subset.
Theorem 3.2.38: If R is a subring of a left Noetherian ring then there is some k such that Sk = 0 for every nil, weakly closed subset S of R (cf., proposition 2.6.30). Proof;. The nilradical N of R is nilpotent, so we may pass to RIN, which is an order in a semisimple Artinian ring, so we are done by theorem 2.6.31. Q.E.D. There are more general nil => nilpotent theorems, cf., exercises 32ff, but this result shows the elegance of a very powerful method. For the rest of the section we shall collect some embedding theorems but shall also pay attention as to the manner of embedding which, although irrelevant in the above applications, can be of supreme importance in other applications.
Embeddings into Primitive and Semiprimitive Rings A certain embedding procedure was outlined preceding theorem 2.5.23, and we elaborate it here for further reference, because of its importance in the structure theory: Definition 3.2.39: A primitive ringR is closed if it has a faithful simple module M such that F = End, M is a field.
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Definition 3.2.40: A central embedding is an injection R + R' with respect to which R' is a central extension of R. Construction 3.2.41: A central embedding of a primitive ring R into a closed primitive ring. (This will rely heavily on centralizers.) Let M be a faithful R-module, and D = End, M. Taking any maximal subfield F of D let T = End MF. Then R E End MD E T naturally. On the other hand, we can inject F + T under the action a + pa where pa: M + M is given by pax = xa. Since F E C,(R) we have the subring R F of T; we claim R F is a closed primitive ring, dense in End MF. Indeed, R F acts on M by (ra)x = (rx)a for r in R, a in F, and x in M, whereby we see M is an RF-module which is simple (since any proper submodule would be an R-submodule and thus 0), and, moreover, RF c End MF shows M is faithful. But F E End,, M E {d E D:x(ad - da) = 0 for all x in M and a in F} = CD(F)= F by proposition 0.0.6, so F = End,,F, proving the claim. Thus R F is the desired closed primitive ring containing R, and we would like to add one more observation. Remark 3.2.42: Z(RF) = F. (Indeed Z(RF) E C,(R) E D so Z ( R F ) is a commutative subring of D containing F, implying Z(RF) = F.) Embedding sequence 3.2.43: Embedding a semiprime ring R into a direct product of endomorphism rings of vector spaces over fields. Step I. R can be embedded in a ring R, which is a homomorphic image of a direct product of copies of R, with Nil(Rl) = 0. (Proof: theorem 2.6.27, since N(R) = 0.) Step 11. R2 = R1[A] is semiprimitive by Amitsur's theorem. Step 111. R, is a subdirect product of primitive images R3i. Step IV. Each R3i can be centrally embedded in a closed primitive ring R,, . Step V. Each R,i is a dense subring of the endomorphism ring of a vector space Mi over a field F,, by construction 3.2.41. The nice feature of this embedding is that at each step I through IV we only took direct products, homomorphic images, and central embeddings, so sentences preserved under these operations would pass from R to its overring. This will be important when we study polynomial identities. In case R is prime there is a further embedding which we can apply.
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Theorem 3.2.44: (Amitsur [67]) If R is a prime ring contained in a semiprimitive ring R' then R can be embedded into a primitive ring R" that satisjes all sentences holding in each primitive homomorphic image of R'. Proof.- Write R' as a subdirect product of {Ri:i E I} and introduce the following filter on I: Writing ri for the canonical image of r in Ri under the composition R + R' + Ri, let I, = { i E I : ri # 0). Then I, # 125 for each r # 0. Moreover, for any nonzero a, b in R we have arb # 0 for some r in R, so I, n I, 3 I,,*. Thus { I r : r # 0} are the base of a filter F0. Let 9 be any ultrafilter containing 9-; let R" be the ultraproduct (Hi,,Ri)/& Then R" satisfies any elementary sentence satisfied by all the Ri. Moreover, since being a faithful simple module is a property which can be described elemenM , ) / 9 is a faithful simple R"-module. Thus R" tarily, we see M" = is primitive, as desired. Q.E.D.
(Hi,,
Remark 3.2.45: Amitsur' proof, based on a proof by'M. Rabin of an earlier theorem of A. Robinson, actually yields more: If % is a class of rings defined by elementary sentences then any prime ring which is a subring of a direct product of rings from %, can be embedded in a ring from W.
Corollary 3.2.46: The first three steps of 3.2.43 combined with theorem 3.2.44 provides an embedding of a prime ring R into a primitive ring R". Of course, there is a more straightforward embedding-let S = Z(R) - {0} for any prime ringR. Then R' = S-'R is an algebra over the field F = S-'Z(R) so R c R' E End R; via the regular representation. However, one has lost all hold on the elementary sentences of R in this procedure.
L Supplement: Embeddings into Matrix Rings It is useful to know when a ring can be embedded into matrices over a commutative ring. We start with an interesting counterexample of Bergman. We say R is quasi-local if R/Jac(R) is simple Artinian. Remark 3.2.47: If e is an idempotent of a quasi-local ring R then ze # 0 for all z # 0 in Z(R). (Indeed, Ann z 4 R so Ann z E Jac(R) which has no trivial idempotents.) Example 3.2.48: (Bergman) A finite ring R which cannot be embedded in M,(C) for any n in N and any commutative ring C. Let G = Z / p Z 0 Z / p 2 Z ,
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a group of order p3, and let R = Hom,(G, G,). Let nibe the projection of G to the i-th component for i = 1,2; then n, and 71, are idempotent with n, + n2 = 1, so R = n,Rnl 0 n,Rn, 0 n2Rnl 0 n2Rn2 by the Peirce decomposition. It is easy to see each of these components is a cyclic abelian group with respective generators n1,n, ,n2,, and n2,where 71, :Z/p2Z + Z/pZ is given by n12(1+ p2Z) = 1 + pE, and n2,:Z/pZ Z/p2Z is given by n2,(1 + pZ) = p + p2Z; these generators have respective order p,p,p, and p2, so IRI = p5. Suppose we had R E M,,(C). Viewing C as scalar matrices, localize at some maximal ideal P E Annc(pn2). Then l-'pn, # 0 in the quasi-local ringM,(C,). In particular, 1-'p # 0 and l-'nl # 0 since pn, = n21n1n12. But (l-'p)(l-'n,) = l-lpn, = 0, contrary to remark 3.2.47.
,
,
Bergman-Vovski have proved the surprising result that all nilpotent algebras (without 1) over a field are embeddible, cf., exercise 38. We return to the question of embeddibility in matrices in theorem 6.1.27 and $6.3, since it belongs to the realm of PI-theory.
L Supplement: Embedding into Division Rings Let us now turn to the famous problem of embeddings into division rings and, more generally, into left Artinian rings. The question of whether every domain can be embedded in a division ring arose in van der Waerden [49B]. Remark 3.2.45 shows any domain which is a subring of a direct product of division rings is in fact embeddible into a division ring. However, Mal'cev [37] came up with the following counterexample: Example 3.2.49: A domain which cannot be embedded in a division ring. Let R = Q{X,,..., X,, Y,,..., & } / A where the ideal A is generated by XIYl X2Y3,X3Y,X4Y3, and X3Y2 X4Y4.Let x i , y i denote the respective images of X i , 4 in R. To check R is a domain, suppose f ( x ,y)g(x,y) = 0 in R. We may assume f , g are homogeneous in total degree and that each monomial of f ends in suitable x i and each monomial of g begins in suitable y j ; then checking the various cases shows that not all the products of monomials cancel, contradiction. On the other hand, the relations from A imply
+
+
+
If R were embeddible in a division ring D then multiplying both sides on the
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left by
for suitable x;, y; in D, and this is impossible. It turns out that the multiplicative monoid R - (0) cannot be embedded in a group, so Mal’cev refined the question as follows: If R - (0) is embeddible in a group then is R embeddible in a division ring? Counterexamples were found in 1966 independently by Bokut [69], Bowtell [67], and Klein [67]. Bokut’s example was a monoid algebra, and he based his proof on a theory he had developed concerning embeddibility of monoid algebras. Klein’s approach was more ring-theoretic. We say a ring R has bounded index n if r n = 0 for every nilpotent element r of R . Klein studied the condition N,,, which says that M,,(R) has bounded index n. Any domain satisfies N , , but A’,, may fail for n > 1. Klein [69] showed that if R satisfies N,, for all n then R - (0) is embeddible in a group. (Also such R must be weakly finite). However, Bergman [74a] found such R which is not embeddible in a division ring. This leads one to search for more intricate conditions for a ring to be embedded in a division ring. Cohn started examining subtler conditions on matrix rings over R . We call a matrix A E M,,(R) full if one cannot write A = PQ for P an n x (n - 1) matrix and Q an (n - 1) x n matrix. Given two matrices A , B which agree in all but the last row we define the determinantal sum to be that matrix whose first n - 1 rows are the same as those of A and B, and whose last row is the sum of those of A and B. Likewise, one can define the determinantal sum of any two matrices which differ in only one row (or column). Noting that the identity matrix cannot be a repeated determinantal sum of nonfull matrices in M,,(D), and thus in M,,(R) if R is a subring of D, Cohn utilized this condition to characterize which domains are embeddible in division rings. Cohn [77B] contains the theory in full detail, although Malcolmson [78] found a modification which shortens and clarifies the construction. The major application of Cohn’s results is for “firs” and “semifirs.” A ring R is a It$ fir, or free ideal ring, if every left ideal is free of unique rank. Every fir is a weakly finite hereditary ring, and, conversely, many of the
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examples of hereditary rings are firs, such as any PLID or the free ring. More generally, R is a semifir if every f.g. left ideal is free of unique rank. Cohn found a matric-theoretic description of semifirs. Given (ri) in R(”) and writing for the transpose we have (ri)(si)[= risi.If (ri)(si)’= 0 we say this is a relation in R‘”); such a relation is trivial if ri = 0 or si = 0 for all i. A relation (ri)(si)‘= 0 is triuializable if there is P in GL(n,R) such that ((ri)P)(P-,(si)‘) 0 is a trivial relation.
,
Remark 3.2.50: R is a semifir iff all relations in R(”) are trivializable, for is clear because any relation can be expressed in terms of every n E N. ((a) the canonical base and must be trivial there. (-=)Write a left ideal L = C:=, Rai with t minimal. Any nonzero relation among the ai could be trivialized, thereby enabling us to shrink the number of generators of L, which is impossible; thus a, ,.. .,a, are independent and so is a base. The uniqueness of the rank of L is standard, cf., theorem 0.3.2. Every semifir is a weakly finite domain, cf., exercise 39, and in fact satisfies Cohn’s criteria to be embeddible in a division ring. Cohn [82] has embarked on a major study of semifirs.
L Supplement: Embedding into Artinian Rings The next question one might ask is, “Which rings are embeddible into simple Artinian rings?” Necessary conditions certainly include “weakly finite,” nilpotence of nil subrings, by theorems 3.2.37 and 3.2.38, and bounded length on chains of left (or right) annihilators. Schofield [SSB] found a lovely characterization of these rings in terms of “rank functions” of maps and modules, which we describe here very briefly. Suppose $: R + T is a ring homomorphism from R to a simple Artinian ring T = M,,(D).Viewing T as T - R bimodule by means of $, we can define a rank function p : R - Y i m o d + Z by p M = t where T @OR M has length t in T-.Mod, i.e., T @JR M x N(‘)where N is the simple T-module. Then p satisfies the following properties: (i) pR = n. (ii) p ( M , (33 M 2 ) = p M l + p M 2 . (iii) If M“ --t M + M’ + 0 is exact then p M ‘ I p M I p M ” functor T 6- is right exact).
+ p M ‘ (since the
(Actually Schofield “normalizes” p by dividing through by pR.) Note that if
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T is a division ring then n = 1 so pR = 1, and Cohn’s theory can be modified to show any rank function p satisfying pR = 1 arises from a homomorphism from R to a division ring. More generally, Schofied shows (for R an algebra over a field F ) that any rank function p with pR = n gives rise to a homomorphism R + M,(D) for a suitable division ring D. He achieves this by means of the free product R M J F ) which by proposition 1.1.3is isomorphic to M,(R‘) for some R‘. Indeed, the rank function p lifts to a rank function on M,(R’)-Aud and thus on R-Afud by theorem 1.1.17, whose proof, in fact, shows pR’ = n/n = 1. Hence there is a homomorphism from R‘ to a division ring D,yielding the composition R
-+
R
M,(D) x M,(R’) + M,,(D’).
Examples:
(i) If R is prime left Noetherian then its ring of fractions Q(R) is simple Artinian, and the ensuing rank function p is called the reduced rank, one of the key tools of Noetherian theory in 53.5. (ii) If R is left Noetherian then any P in Spec(R) gives rise to the composition R -+ R/P + Q(R/P); the ensuing rank function is essentially Stafford‘s important y^ function defined before remark 3.5.64. (iii) If R is left Artinian then the composition length of an f.g. module defines a rank function; it follows that any left Artinian algebra over a field is embeddible into a simple Artinian algebra. (For R semiprime this is an easy exercise.) Note that the ring of example 3.2.48 cannot be embedded in a simple Artinian ring because there is an element annihilated by p2 but not by p . Thus the condition of being an algebra over a field is necessary in Schofield’s theorem. Westreich [87] has generalized the Cohn-Malcolmson construction. On the other hand, Small constructed a left Noetherian ring failing to have minimal left annihilators and thus not embeddible in a left (or right) Artinian ring, cf., exercise 40. Such a construction is impossible for left and right Noetherian rings, but recently Dean-Stafford [86] have shown a certain image of the universal enveloping algebra of the Lie algebra SL(2, C) is embeddible into an Artinian ring, and Dean conjectures that every enveloping algebra of a finite dimensional semisimple Lie algebra has a nonembeddible image. As a consolation we shall obtain a positive result for commutative Noetherian rings. Let us start with some generalities. We say A 4 R is irreducible if A , n A , 3 A whenever A , , A , =I A ; R is an irreducible ring if 0 is an irreducible ideal.
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Remark 3.2.51: Suppose R satisfies ACC(idea1s). Then every ideal A is an intersection of a finite number of irreducible ideals. (For a maximal counterexample A would not be irreducible, so A = At n A 2 , where each Ai =I A and thus is a finite intersection of irreducibles, contradiction.) In particular, taking A = 0 we see R is a finite subdirect product of irreducible rings.
Proposition 3.2.52: If C is a commutative, irreducible Noetherian ring then C is an order in an Artinian ring. Proof: Claim. Every element of C is regular or nilpotent. If this is so then every element of the ring of fractions Q is invertible or nilpotent, so Q/Nil(Q) is a division ring; but Q is then Noetherian semiprimary and thus Artinian, as desired. To prove the claim suppose c E C is not nilpotent and take i E N such that Ann,ci is maximal. Replacing c by c i we may assume Ann c = Ann c2. Given c’c = 0 we shall prove c’ = 0. Otherwise, by hypothesis there is some a E Cc n Cc’. Write a = bc. Then 0 = ac = bc2 so 0 = bc = a, as claimed. Q.E.D.
Theorem 3.2.53: Every commutative Noetherian ring C can be embedded in a commutative Artinian ring. Proofi C is a subdirect product of irreducible C,, . . .,C, by remark 3.2.51; x C, E Q1 x ... x Q, where each Qi is the Artinian ring hence C + CI x of fractions of Ci. Q.E.D. Any attempt to noncommute this approach runs into the following obstacle: Example 3.2.54: (Small) Let M = R = F [ 2 ] , viewing M as R-module in the usual way, but as right R-module by the action x2 = 0, i.e., xp(2) = xp(0). Let R‘ = R 8 M ,viewed as ring under the action ( r l , x l ) ( r 2 , x 2= ) (rlr2,r1x2+ xlr2).Then R‘ is irreducible and Noetherian (since R’ x R ( 2 ) ) but , is not an order in an Artinian ring, by Small’s theorem. Nevertheless, R is embeddible. In summary, we have the following sufficient conditions for embeddibility of a left Noetherian ring R into an Artinian ring: (i) R semiprime, by Goldie’s theorem (ii) Small’s regularity condition
$3.3 Localization of Nonsingular Rings and Their Modules
377
(iii) R commutative (iv) Gordon [75] found ways of generalizing theorem 3.2.53 to certain classes of noncommutative rings.
However, these examples all rely ultimately on Ore localization. A new result is described in digression 3.5.7.
Existentially Closed Rings Instead of inquiring which conditions are needed to embed a ring into a specific kind of ring, one can turn the question around and try to determine the “best” sort of ring into which one can embed a given ring. Generalizing the definition of “algebraic closure,” logicians say a structure 9’of a theory 5 is existentially closed if every existential sentence holding in some structure containing Y must hold in 9’ itself. For example, every algebraically closed field is existentially closed; it turns out every division ring can be embedded into an existentially closed division ring, and every ring can be embedded into an existentially closed ring, cf., the excellent survey BelyaevTaitslin [79, $61. Existentially closed rings are simple (ibid, lemma 6.2) so, in particular, every ring can be embedded in a simple ring; this was originally proved by Bokut.
83.3 Localization of Nonsingular Rings and Their Modules In this section we shall reexamine classical localization in a more idealtheoretic manner, which will permit us to bypass some of the computations while at the same time preparing the ground for a new construction, Johnson’s ring of quotients of a nonsingular ring. This ring has several very nice features, being self-injective,(von Neumann) regular, and quickly yields a proof of Goldie’s theorem (exercise 14). Also the ideals are totally ordered by Goodearl’s theorem (exercise 41; also see exercises 51, 52, 53). This is rather impressive, since most of the rings we shall study are nonsingular. Furthermore, the obstruction to being nonsingular is an ideal, called the singular ideal. Nonetheless, the Johnson-Goodearl theory has been difficult to apply, because a primitive, self-injective,regular ring whose ideals are totally ordered still is a far cry from semisimple Artinian! For example, it need not be weakly finite, cf., example 1.3.33. The philosophy of this section is to use large left ideals in place of regular elements (cf., theorem 3.2.14 (i) o (ii)),and it will be useful to deal more generally with large submodules. Accordingly, we shall be extending some of the
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results of 52.4. Given R-modules N IM and given x E M,define Nx-' = { r E R:rx E N } . The tools of this section lead naturally to techniques of "hard module theory-closed submodules, self-injective rings, quasi-Frobenius rings; these are treated in the exercises. Proposition 3.3.0: Suppose N is a large submodule of M. (i) N x - ' is a large left ideal of R for each x in M . ( i i ) lf f: M' + M is any map then f -'N is a large submodule of M'. Proofi Clearly (ii) implies (i), taking f R + M to be right multiplication by x. To prove (ii), suppose N' < M' with N' n f -'N = 0. Then f N ' n N I f ( N ' n f - ' N ) = 0 so flv' = 0; hence N ' If - ' N so N' = N ' n f - ' N = 0. Q.E.D.
Johnson's Ring of Quotients of a Nonsingular Ring Construction 3.3.1: Let 9= {large left ideals}, viewed as a sublattice of the dual lattice of Y(R), i.e., L1 5 L , iff L , 2 L 2 . Given M in R-Mud define Mi = Horn&,, M) for each L, in 9; for Li 2 L j define Mi+ Mj by sending a map f: Li+M to its restriction f: Lj+M. Let M denote l h ( M i ;cpi) together with the canonical maps p i : Mi + M .
cpi:
Remark 3.3.2: In view of theorem 1.8.7 we can describe M explicitly as an R-module, as follows. Let Y = {pairs (L,f ) where L E 9 and f: L + R is an R-module map}. Define the equivalence on Y by saying (L,,fl ) (L,, f,) if fl and f2 agree on some L in 9contained in L1n L,, i.e., their restrictions to L are the same. Then M is the set of equivalence classes Y / - , made into an R-module under the following operations, writing [ L , f ] for the equivalence class of (L, f ) :
-
CL1,fll
+ CL,,f,l
= CLl
"L29fl
r [ L l , f l ] = [Llr-',rfl]
-
+f21. where the map rfl is given by (rf1) a = fl (ar).
(This is reminiscent of remark 1.5.18'.) The verifications of these assertions are an easy consequence of the following observation: Remark 3.3.3: 9 = {large left ideals of R} is a filter on Y(R) satisfying the following extra properties (by proposition 3.3.0):
(i) If L E and~r e R then Lr-' E X (ii) More generally, if L1,L, E B and f: L1 + R is a map then f - ' L , E X
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Before using remark 3.3.2 we should like to relate M to M given in the next result. Remark 3.3.4: Modifying the proof of proposition 1.5.19 we can define a canonical map :p,r M + M by the rule rp,x = [R, px] where px:R + M is right multiplication by x. Then ker rpu = {x E M: Ann, x E S}.(Indeed, [ R , p , ] = 0 iff px is 0 on a large left ideal L i.e., Lx = 0, iff Annx is large.) This leads us to the following fundamental definition. Definition 3.3.5: The singular submodule Sing(M) = {xEM:Ann, x is large). (Thus Sing(M) = ker q,). M is singular if Sing(M) = M; M is nonsingular if Sing(M) = 0. This definition should be compared with a related notion. We say a module M is torsion if Ann, x contains a regular element of R for each x in M; M is torsionfree if Ann, x does not contain a regular element of R for each x in M. For R semiprime Goldie we see by Goldie's theorem that Ann, x contains a regular element iff xESing(M); thus we see for such R that torsion =singular and torsion-free = nonsingular. A deeper explanation can be found via torsion theory, which is described in 83.4. Remark 3.3.6: f(Sing(M)) I Sing(N) for any map f: M + N (for if Lx = 0 then Lfx = f(Lx) = 0). In particular, if M I N then Sing(M) ISing(N). Let us now consider the important case where M Theorem 3.3.7:
= R.
k is a ring, under the multiplication CL 1 fl I CL, fz 1 = 3
7
cf
IL2 f i
fl I .
The canonical map rp:, R + d is then a ring homomorphism whose kernel is Sing(R), i.e., we have a natural ring injection R/Sing(R) + k. fro08 Note that f i ' L z E S by remark 3.3.3(ii). Since 9 is closed under finite intersections we can verify the ring properties of k on a large left ideal contained in all the domains of the fi. Since 40, was already seen to be a module map it remains to verify rp,(ab) = rp,Uq,b for all a, b in R, i.e., that [R, pa] [R, pb] = [&,,I. By definition
We call k the Johnson ring of quotients.
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Corollary 3.3.8: Sing(R) 4 R. (This can also be proved directly using remark 3.3.3.) In view of this theorem we want to obtain many examples of nonsingular rings.
Lemma 3.3.9: lf R satisfies ACC(Ann) then Sing(R) is nil. Proof.- Given r E Sing(R) take n E N with Ann(r") maximal. Let a = I" E Sing(R). Then Ann a = Ann a2 so remark 3.2.2 (ix) shows Ra n Ann a = 0, implying Ra = 0, i.e., r" = 0. Q.E.D. In fact, under this hypothesis Sing(R) is nilpotent, cf., exercise 6. However, a nilpotent ideal need not be in Sing(R), as we shall see. Example 3.3.10:
(i) Any semiprime ring with ACC(Ann) is nonsingular by lemma 3.3.9 coupled with proposition 2.6.24. This enables us to deduce Goldie's theorem from these results (cf., exercises 12-14). (ii) A commutative ringC is semiprime iff Sing(C) = 0. (Proof (a) If O Z C E Cthen (CcnAnnc)'G(Annc)Cc=O so C c n A n n c = O . Annc is not large. Conversely, if cz = 0 and I is an essential complement of Cc then Ic E I n Cc = Oso Annc 2 I + Ccislarge.) Example 3.3.11:
The semiprimary ring R =
(," i)
is nonsingular, for
any field F. (Proof Lemma 3.3.9 shows Sing(R) c Jac(R) = Fe,,, but Anne,, = Fe,, + Fe,, which is not large, so Sing(R) = 0.)
The Injective Hull of a Nonsingular Module We shall now describe M and hull E(M).
fi much more concisely, using the injective
Remark 3.3.12: If N IM is large then M/N is a singular module. Conversely, if M is nonsingular and M/N is singular then N is large in M. (Proof (*) For any x in M proposition 3.3.0 shows Nx-' is a large left ideal; but ( N x - ' ) x IN so x + N E Sing(M/N). (e) For any x in M we have Lx I N for a large left ideal L , so 0 # Lx G N n Rx.)
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Remark 3.3.13:
If M is nonsingular and h: K -+ M has ker h large in K then h = 0. (Indeed, K/kerh is a singular module isomorphic to h K , so hK I Sing(M) = 0.
Lemma 3.3.14: Suppose M is nonsingular. ( i ) The injectioe hull E(M) is also nonsingular. ( i i ) I f h: K --+ E(M) with ker h large in K then h = 0. Proofi (i) M n SingE(M) E Sing(M) = 0 so SingE(M) = 0, since M is large in E(M). (ii) Apply remark 3.3.13 to E(M). Q.E.D.
Proposition 3.3.15: I f M is a nonsingular module then Mx E(M) canonically. Proofi Let us define the isomorphism cp: E(M) -+ M.Given x in E(M) we have L = Rx-' large in R by proposition 3.3.0, so define cpx = [ L , px] where px is right multiplication by x. Surely cp is a map and is epic since any J L-+M (with L large in R) can be viewed as a map f:L-+E(M) which, by remark 2.10.3 is given by right multiplication by some x in E(M). Finally cp is monic by lemma 3.3.14. Q.E.D.
Theorem 3.3.16:
k z End,
E ( R ) as rings, for any nonsingular ring R.
Proofi First note as modules that there is an isomorphism II/: End, E ( R ) -+ Hom,(R, E(R))which sends a map f:E(R)-+E(R)to its restriction f:R-+E(R); indeed Ic/ is epic since E(R) is injective, and Ic/ is monic by lemma 3.3.14(ii). Thus we have a composite module isomorphism End, E ( R )
-+
Horn,@, E(R)) x E ( R )
-+
8,
by proposition 3.3.15. But this composite preserves the ring multiplication, as Q.E.D. one sees instantly by comparing definition 1.5.6 to theorem 3.3.7.
Carrying Structure of
R
to R
Having constructed k twice for R nonsingular, we shall study how to transfer structure from R to d and then shall examine some of the special properties of R. First note by proposition 3.3.15 that k is an essential ring extension of R, in the sense that k is a ring containing R which is essential as R-module.
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Proposition 3.3.17: Suppose Q is any essential ring extension of R. If R is prime (resp. semiprime) then Q is prime (resp. semiprime). Proofi Suppose A, B 4 Q with AB = 0. Then (A n R)(B n R) = 0 so A n R or B n R = 0, implying A = 0 or B = 0. Q.E.D. Further information requires a more careful analysis.
Proposition 3.3.18: Suppose M, N E k - A d and Sing(N) = 0. ( i ) Any Rmodule map f:M + N is also an k-module map. (ii) I f N is a summand of M in R-Mud then N is a summand of M in k-Aud. Proofi (i) Fix x in M and define h: k + N by hq = f(qx) - qfx. Then h is an Rmodule map with hR = 0, so h = 0 by remark 3.3.13. (ii) The projection a: M+N is an I?-module map, by (i), so M xN@ker TC in R - ~ o d . Q.E.D.
Proposition 3.3.19: Suppose Sing(R)=O. (i)A left ideal L of is large in R; (ii)i? is a nonsingular ring.
d is large iff LnR
Proofi (i) (=.) by proposition 3.3.0 (where f:R + d is the inclusion map). (e) k is an essential extension of R and thus of L n R in R - A d , so k is an essential extension of L in R-Mud and a fortiori in k-Aud. (ii) Immediate from (i). Q.E.D.
Proposition 3.3.20: If M is a nonsingular R-module then M is a nonsingular injective k-module under the action CL,,flCL,,hl= Cf-IL2,hfl forf:L,+R
and
h:L2+MwhereLl,L2~9.
Proofi Note f -'L2 E 9by remark 3.3.3(ii). The argument of theorem 3.3.7 shows M is an k-module. To see Sing(M) = 0 suppose h: L + M has Anns[L,h] large in k. Then L'= Ann,[L,h] is large in R by proposition 3.3.19(i). For any a in L'n L we have some large left ideal L" La-' satisfying O= h(L"a)=L"ha; thus haESing(M)=O. This proves h(L'nL)=O so [ L ,h] = 0 by definition. It remains to show M is injective as k-module. Otherwise, M has a proper essential extension fi. But M is a summand of fi as R-modules (since
-=
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53.3 Localization of Nonsingular Rings and Their Modules
fi z E(M)) and thus as k-modules by proposition 3.3.18(ii), contradiction.
Q.E.D.
Let us push this result a bit further. Theorem 3.3.21: Suppose Sing(R) = 0. There is an exact functor Fj: R - M u d +k - A ~ d given by FjM = fi, where N = M/Sing(M) is nonsingular. Thus FJ sends R-modules to nonsingular injective k-modules. Proofi Write Mo = Sing(M). First we want to show N = M/Mo is nonsingular. Write Sing(N) = M‘/M,. For any x in M‘ we take an essential complement L of Ann,x in R. Right multiplication by x provides an isomorphism L + Lx, so Sing(Lx) w Sing&) = 0. Thus Lx n M, = 0. But then
Lx w Lx/(Lx n M,) w (Lx + M,)/M, c Sing(N) is singular so Lx = 0, implying L = 0. Hence Annx is large so x E M,, proving Sing(N) = M,/M, = 0. Now fi is nonsingular and injective in R-Mod by proposition 3.3.20, so we have I$ acting on R-modules. Recalling fi z E(N) we shall now define FJ acting on a map f:M + M‘. Indeed, the composition
f M’ M+
+
M’/Sing(M‘) = N’
has kernel containing Sing(M) since f(Sing(M)) E Sing(M’), so we have an induced map f: N + N‘, Now fcomposed with the injection N’ + E ( N ’ ) gives a map N + E(N’) which (since E ( N ‘ )is injective) extends to a map 5 E ( N ) + E(N’);f is unique in view of lemma 3.3.14(ii).(Indeed, iffl and f2 both extend f thenfl - f2 has kernel containing N and is thus 0). We denote f as F’f. Given f:M + M’ and f’:M’ + M” the argument of the last paragraph shows FJ(f‘f) = Fjf‘Fjf (since they agree on N), proving FJ is a functor. To prove FJ is exact consider the short exact sequence 0 + M I M’ + f --f
M “ + 0, i.e., f is monic and f ’ is epic and f M = kerf’. Putting g = FJf and g’ = FJf‘ we want to show g is monic, g’ is epic, and g(E(N))= ker g’ where
(as above) N = M/Sing(M) and N’ = M‘/Sing(M’) and N” = M”/Sing(M”). Write f: N + N’ and 7:N’ + N“ for the maps induced from f and f’, respectively, and for ease of notation view M I M‘ via f. First note Sing(M) = M n Sing(M‘) so f is also monic and 0 = kerf = N n ker g , implying ker g = 0 and g is monic. In particular, g ( E ( N ) )is the injective hull of gN and so is a maximal essential extension. On the other hand, g’g = FJ(f‘f) = FJ(0)= 0 so g ( E ( N ) )c kerg’; to prove equality it suffices
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to show kerg’ is essential over gN = fN. Since kerg’ is essential over N ‘ nkerg’ = kerf’ we want kerf’ to be essential over fN. But kerf’ = f’-’(Sing M”)/Sing M ’so (ker f)/fN is a homomorphic image of f’-’(Sing M”)/fM% f’-’(Sing M”)/kerf’ x Sing M”, a singular module. Since kerf’ 5 N ’ is nonsingular we conclude by remark 3.3.12 that fN is large in ker J;’ proving g(E(N)) = ker g’. It remains to show g’ is epic. Note g’N‘ = f’Nf = N“ (since f ’is epic), which is large in E(N”), so it suffices to show g’(E(N’)) is injective. But the sequence 0 + ker g‘ + E ( N ‘ )--* g‘(E(N’))+ 0 splits since we saw above ker 9’ = g(E(N)) x E ( N ) is injective. Hence g’(E(N’)) is a summand of E ( N ’ ) and is injective, as desired. Q.E.D. The real reason that k is useful is that its structure is so nice, as we begin to see in the next result.
Definition 3.3.22: A ring R is self-injective if R is injective as R-module. Theorem 3.3.23: If R is nonsingular then injectioe ring.
k is a nonsingular, regular, self-
Proofi k is nonsingular and self-injective by proposition 3.3.20 (taking M = R).To prove k is regular suppose [ L ,f ] E k, i.e., f: L + R is a map with L a large left ideal. Take an essential complement N of kerf in L; then f: N + j N is an isomorphism. Letting N‘ be an essential complement of fN in R we define g: fN + N’ 4 R by g(fx + x’) = x. For any x in N and y in kerf we have f g f ( x + Y ) = f g f x = f ( g f 4 = f x = f ( x + Y), implying (fgf - f)(N
+ ket f ) = 0. Thus
CL,flCfN proving R is regular.
+ ”,slCL,fl
= CLfl
Q.E.D.
83.4 Noncommutative Localization As beautiful as Goodearl’s results of the last section are, we are left with the problem that rings need not be nonsingular. There is a way of circumventing this problem by using “dense” submodules instead of “large submodules,” cf., definition 3.4.1. However, a construction of Martindale dealing with
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385
two-sided ideals instead of left ideals has proved more useful for prime rings and has several important consequences for the prime spectrum (cf., theorem 3.4.138). All of these constructions can be gathered under a single umbrella construction, due to Gabriel [62], which we describe briefly at the end. Gabriel’s theory has developed into an entire subject, discussed in depth in Golan [86B].
Digression: The Maximal Ring of Quotients Our first two constructions do away with the singular ideal. The first construction, due to Findlay-Lambek and Utumi, works for any ring and is called the maximal ring of quotients. Recall that in the construction of R we had [L,f] = 0 iff kerf is large. In this case f induces a map f:L/kerf + R, so in order to avoid this contingency we introduce the following definition. Definition 3.4.1: N is a dense R-submodule of M if Hom,(M‘/N, M) = 0 for all N IM’ IM. (In other words, if f M’ + M is a map and N I kerf then f = 0.)
Remark 3.4.2: Every dense submodule N of M is large. (Indeed if K nN = 0 then the projection a: K + N + K has kernel N, so a = 0 implying K = 0.) On the other hand, a large submodule need not be dense (exercise 1). However, a large submodule of a nonsingular module is dense, by remark 3.3.13. Example 3.4.3: If M = Q then every Z-submodule N of M is dense, because M/N is torsion and M is torsion-free. For this reason, we say M is a rational extension of N when N is dense in M. Proposition 3.4.4: FD= {dense submodules of M} is a jlter; moreover, if L < R is dense and r E R then Lr- is also a dense left ideal.
Proofi If N < M is dense and N IN’ I M then N’ is dense in M, for if N ’ I M ’ I M a n d f : M ’ + M w i t h fN’=Othen f N = O s o f =O.Toshow FDis a filter it remains to show N n N‘ is dense if N and N’ are dense (in M). Suppose N n N’ I M’ I M and f:M’ + M with f ( N n N‘) = 0. Then we have an induced map T ( N n M’)/(N n N’) + M which we combine with the isomorphism ((N n M’) + N‘)/N’ x N n M‘/N n N’ to produce a map ((NnM’)+N’)/N’+M which must be 0 by hypothesis. Hence f(NnM‘)=O. But now we get an induced map (N + M’)/N x M‘/(N n M‘)+ M which is 0 so fM’ = 0, i.e., f = 0.
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To prove the second assertion suppose Lr-' IL' IR and f:L' + R is a map with f(Lr-') = 0. Then we have a map g:(L'r + L) + R given by g(ar b) = fa for a in L' and b in L; g is well-defined since if ar + b = 0 t h e n a r = - b E L s o a E L r - ' a n d f a = O . But g L = O s o g = O ( s i n c e L is dense); hence f = 0. (This entire argument is generalized in exercise 3.) Q.E.D.
+
Armed with this result we form the following ring of quotients.
-
Let 9 = {dense left ideals}, and let Y = {pairs (L,f) where L E 9 and f: L + R is an R-module map}. Define the equivalence on Y by (L,,f,) (L,,f,) if f, and f, agree on L , n L,, and form Qmax(M) to be the set of equivalence classes Y/-, made into an R-module under the following operations (where [L,f ] denotes the equivalence class of ( L ,f)): Construction 3.4.5:
-
CL1,fll
+ [L2,f21 = [Ll nL,,f, + fzl. where rfl is given by (rf& = fl(ar).
r[Ll,fll = [L1r-',rfl]
(Note that this is analogous to remark 3.3.2, since here fi and f, agree on a dense submodule L of L1 n L z iff (fl - f2)L= 0, ifffl - f, = 0 on L1 n L,. Hence this construction could be viewed as a direct limit as in construction 3.3.1.)
Theorem 3.4.6:
Q,,,(R)
is a ring, under multiplication
CLl, fll CL,
9
f z l = Cf;%
9
fifl I.
The canonical map qR:R + Q,,,(R) gioen by qRr = [R, pr] (where pr is right multiplication by r) is a ring injection.
Proofi Analogous to theorem 3.3.7; clearly r E ker p R iff [R,pr] = [O,pr] iff r = lr = Or = 0. Q.E.D. Key results about dense submodules are given in exercise 1-9. Note Q,, generalizes Johnson's ring of quotients, in view of remark 3.4.2.
The Martindale-Amitsur Ring of Quotients As pleasing as the maximal ring of quotients is, there is the severe drawback that in practice it is hard to identify the dense left ideals of an arbitrary ring. A more successful alternative (for applications) was discovered by Martindale
53.4 Noncommutative Localization
387
[69] for prime rings and later generalized by Amitsur [72a] for semiprime rings. The key observation here is that a (two-sided) ideal A of a semiprime ring is large as left ideal iff Ann A = 0 (by proposition 3.2.21). In particular, every ideal of a prime ring is large, for its annihilator is 0; consequently, we can inject every prime ring into a very useful ring of quotients, described as follows:
Construction and theorem 3.4.7: Suppose R is a semiprime ring. Let 9 = {large 2-sided ideals of R } , and let Y ={ ( A ,f )where A E and~ f:A + R is an R-module map). DeJine the equivalence on Y: by ( A , , f,) (A,, f,) iff fl and f, agree on some L in 9contained in L 1 n L, . Denoting the equivalence class of(A,f)by[A,f] wecanforrntheringQ,(R) = {equivalenceclassesofY/-}; made into a ring under the operations
-
CA1,flI
+ CA29f21 = “41
nA 2 , f l
[(A,
C ~ l , f l I C ~ 2 ~= f21
-
”
+f2l
and
A2)21f2f11.
There is an injection cp: R -, Q,(R) given by cpr = [ R , p,], where p, is right multiplication by r. Q,(R) is a direct limit of abelian groups (under addition) and so is an abelian group under the designated addition. Multiplication is defined because Ann(A, n A,)’ = 0 by iteration, implying (A, n A,), E and f,(A, nA,)2 c (A, nA,)fl(A1 nA,) c A, nA, (on which f2 is defined); clearly multiplication is well-defined. The map cp: R + Qo(R)is a ring homomorphism, as in theorem 3.3.7, and r E ker cp iff Ar = 0 for some A E 9, iff r = 0, by proposition 3.2.21. Q.E.D. Proof:
This construction is fast and efficient, and provides a theory of localization. However, we continue the present discussion with an observation in a different direction. Define the extended centroid of R to be the subring {[A, f ] E Q,(R): 9: A + R is an R - R bimodule map}. Remark 3.4.8: The extended centroid of R is a regular ring containing Z(R) and which is contained in Z(Q,(R)). (Indeed, if f:A + R is a bimodule map then for all [A1,fl] in Q o ( R )and all a,a’ in A n A, we have
ff,(aa’)= f(af1a’) = (faIf1a’ = f1((af)af)= f l f ( W , proving [A, f ] E Z(Q,(R)). Clearly, pz: R -, R is a bimodule map for any z in Z . The extended centroid is regular by the same type of argument as in
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388
theorem 3.3.23: Given a bimodule map f:A + R let B = Ann(fA) 4 R and define g : B + fA + R by g ( b f a ) = a. Then
+
CAlfICB + f A , g l C A , f l =
by a simple calculation.
CAfI
Q.E.D.
DeBnition 3.4.9: The central closure d of a semiprime ring R is the subring R 2 of Qo(R),where 2 = Z(Q,(R)).R is centrally closed if R = R g in Q,(R). The normal closure of R is RN where N = (4 E Q,(R):q normalizes R } .
In this text we shall only deal with the central closure, but the normal closure is used in an analogous way to prove results about normalizing extensions. Note d is an essential central extension of R, since Qo(R) is essential over R. We shall see that centrally closed rings behave much like central simple algebras. Before examining how this is so, let us note that if R is simple then 2 = Z ( R )and d = R. Indeed, iff: A + R is a nonzero bimodule homomorphism then A = R and f l E Z ( R ) since rfl = f r = flr; hence [A,f]can be identified with f l . Proposition 3.4.10: The central closure d of R is itself centrally closed, and its extended centroid is 2 which is also Z ( d ) . Proofi Let 2' be the extended centroid of R. We have a natural injection
cp: 2 + 2' defined by the regular representation; equivalently c p [ A , f ] =
[ d A d , f ] where J d A R + d is given by fCi?ilaifi2= C ? i , ( f a i F i 2 .If we could show cp were onto then we could identify Z with Z' so R = d 2 is centrally closed and 2 E Z ( d )E Z' = 2,which would prove the proposition. Given any large A^ Q d and an d - d bimodule map 5 A^ + d we let A = A^ n f - ' ( R ) and f:A + R be the restriction of f to A. Since A^ and f - ' R are large in k? we see A is large in d and thus in R, so [ A , f ] is admissible, Q.E.D. f ] , proving cp is onto. and clearly q [ A ,f] =
[a,
When R is prime its central closure d is prime by proposition 3.3.17, and
Z ( R ) is a field; furthermore, we can generalize certain important properties of the tensor product from theorem 1.7.27 and corollary 1.7.28.
Theorem 3.4.11: Suppose R is prime and centrally closed. If R' = R W where W is a Z(R)-subalgebra of CR,(R)without R-torsion then every nonzero ideal of R' contains an ideal 1,1, # 0 where I , 4 R and I , Q W. Consequently R' x R @ Z ( R , W canonically if R' is prime.
389
83.4 Noncommutative Localization
Proof: First we claim any 0 # A -4 R W contains an element rw # 0. Indeed, take 0 # a = C:= riwi E A with t minimal; then the wi are clearly Z ( R ) independent. We claim t = 1. Otherwise, define f:R r l R -+ R r 2 R by
1
f Cajrlbj =cajrzbj ( i
for aj,bj in R.
If some xjajrlbj = 0 then writing si as xajribj in R we have a' = siwi = x j a j a b j E A so a' must be 0 by minimality of t ; thus each si = 0 and, in particular, f is well-defined. Now [ R r , R , f ] = z for some z in 2 and
+ z w 2 ) + 1 riwi, I
a = rl(wl
i=3
contrary to the minimality of t. Hence t = 1, as claimed. Now 0 # RrR W w W = R WrwR W C A, proving the first assertion. Also note the claim applied to R Q W shows any nonzero ideal of R Q W contains some r Q w # 0. Now we prove the last assertion by showing the canonical surjection q: R Q W + R' given by q ( r Q w ) = rw is monic; otherwise, taking 0 # r Q w in ker q we get 0 = Wq(r Q w)R = WrwR = rR'w so r = 0 or w = 0, contrary Q.E.D. to r 0 w # 0. Corollary 3.4.12: Suppose R is prime and centrally closed, and R' = RF with F = Z ( R ' )a jield. Then R' is prime and centrally closed. Proof;. If A'B' = 0 for nonzero ideals A', B' of R' then taking 0 # A, B a R with A F c A' and B F c B' we have AB = 0 in R, impossible; thus R' is prime. To prove R' is centrally closed suppose f:A' --* R' is a bimodule map and A' a R'. By theorem 3.4.11 we have some element 0 # a E A ' n R. Let f a = c r i a i where {ai:i E I } is a Z(R)-base of F. Defining A: RaR -,R via La = ri we see [ A ' , f ] = l [ R a R , f i ] a iE Z ( R ) F C F, as desired. (The f;. are Q.E.D. well-defined since R' z R Q F.) We now have the tools at hand to use the central closure effectively. Theorem 3.4.13: (Bergman)Any Jinite centralizing extension R' of R satisfies INC. In fact, if P' E Spec(R') and P' c B' 4 R' then B' n R 3 P' n R. Proof;. Passing to R'IP' and R/(P' n R ) we may assume R and R' are prime with P' = 0. Write R' = Rri with ri E C J R ) . Let fi,fi' denote the respective central closures of R and R'. We claim that fi G fi' canonically. Since R E 8' it suffices to view any [A,f]in the extended centroid of R as
c;=,
Rings of Fractions and Embedding Theorems
390
an element of the extended centroid of R’. The obvious way is to define f’: Ar; + R’ by f’(Cair:)= x ( f a i ) r : .This is well-defined, for if air: = 0 then for any a in A we have
x
C ( f u i ) r : a= C f a , a r ; = C a i ( f a ) r := ( C a i r : ) f a= 0, implying C ( f a i ) r :E Ann,. AR‘ = 0. Furthermore, f ‘is an R‘ - R‘ bimodule map for if r’ E R‘ then writing rir‘ = xjrijri for suitable rij in R we have f ’ ( C a i r i ) r ‘= C(fai)r:r’= C(fai)rijrJ= x f ( a i r i j ) r ;= f’(Cair;r’) i. j
and r’j’(xairi) = f ’ ( x r ’ a i r ; )by an analogous argument. Let F = Z(R’). Then kF is centrally closed by corollary 3.4.12. But every ideal of RF intersects fi nontrivially by theorem 3.4.1 1 and thus intersects R nontrivially; replacing R, R‘ and B’, respectively, by kF, ff‘ and RIB’, we may assume R and R’ are centrally closed with the same center F. Shrink { r ; , ...,I : } to an F-independent set which we can call { r ; , . . .,la}, and let W = Fri. W is closed under multiplication. Indeed, if rir; = Curijurlthen for each r in R we get 0 = [r,rirJ = ~ u [ r , r i j u implying ]r~ [r,riju] = 0 since R‘ x R 8 C,,(R); hence each rijuE Z(R) = F as desired. Consequently W is a prime finite dimensional F-algebra and so is simple Artinian. Now R‘ x R BFW by theorem 3.4.11, so B’ n R # 0 by theorem 1.7.27. Q.E.D. Let us summarize these results.
Theorem 3.4.13‘: (i) Any finite centralizing extension R’ of R satisfies LO, GU, and INC. (ii) lf K is any algebraic field extension of F then R & K as an extension of R satisJies LO, GU, and INC. Proofi (i) Restatement of theorem 2.12.48 and theorem 3.4.13. (ii) LO and G U have already been shown in corollary 2.12.50 without any assumptions on K. To check INC suppose PI 3 P2in Spec(R 8 K) lie over P. Then taking x E P, - P2 we see x E R BFK O for some finitely generated extension K O of F; hence [KO:F] rx) and thus R’ = R BFK O is a finite centralizing extension of R. But R‘ n P, 3 R‘ n P2 in Spec(R’) by proposition 2.12.39. Q.E.D.
-=
53.4 Noncommutative Localization
391
Corollary 3.4.14: If R' is a finite centralizing extension of R and P' E Spec(R') then P' is a primitive ideal of R' if P' n R is a primitive ideal of R. Proof: Passing to RIP' we may assume P' = 0. Suppose L is a maximal left ideal having core 0. We claim R'L is a proper left ideal of R'. Indeed, if R'L = R' then using the notation of the proof of theorem 2.12.48 we have B = BR'L E TL so B E R n TL = L by sublemma 2.5.32', contrary to core(L) = 0. Having proved the claim we see R' is primitive by proposition 2.1.1 1 unless A' + R'L < R'L for some A' 4 R'. But then ( A ' n R) + L < R so A ' n R c L, contrary to A ' n R # 0 by the theorem. Q.E.D.
Corollary 3.4.15: If R' is a jinite centralizing extension of R and P' is a G-ideal of R' then P' n R is a G-ideal of R. Proof: B' n R
n
Let B' = {prime ideals of R' containing P'} 2 P'. Then P' n R c c { prime ideals of R containing P' n R} since LO holds. Q.E.D.
n
We close the discussion with a fairly explicit description for primitive rings. Example 3.4.16: Suppose a primitive ring R is displayed as a dense subring of End MD where M is a faithful simple R-module and D = End, M . Given an R - R bimodule map f: A + R with A a R we define f: M + M by ?(ax) = ( f a ) xfor a in A and x in M . This is well-defined since if a l x l = a 2 x 2 then 0 = ( f A ) ( a , x , - a 2 4 = A ( ( f a , ) x , - (fa2)x2) so ( f a d x 1 = ( f a 2 N 2 . But A M = M so f is defined on all of M , and clearly ?E D. Furthermore, f E Z ( D ) by inspection, so we have an injection cp: 2 + Z ( D ) given by q [ A ,f ] = f. On the other hand, multiplication by any element z of Z(D) yields a map pz: M + M so we can view Z(D) E End M,, and then we have R c RZ(D) E End MD. Thus by corollary 3.4.12 we see RZ(D) is centrally closed. When soc(R) # 0 we must have 2 = Z ( D ) since plsoc(R) = soc(R); thus in this case we get R = RZ(D). I do not know a specific example where 2 # Z ( D ) but suspect one could be constructed by taking a simple ring (e.g., a skew polynomial ring) whose center is "too small." A
C Supplement: Idempotent Filters and Serre Categories Having several examples of quotients at our disposal we should like now to find general principles which tie them together. There are several places where one could start; the filter of left ideals used to build the quotient ring
392
Rings of Fractions and Embedding Theorems
seems as good a place as any. In an attempt to give some of the flavor of the subject we shall skip some details. Definition 3.4.17: A filter 9 of left ideals of R is topologizing if Lr-' E 9 for every L in 9 and every r in R. A topologizing filter is idempotent if L' E 9whenever L'a-l E 9for all a in some L in 9% The reason for the word "topologizing" is that every such filter defines a topological ring structure on R, via a neighborhood system at 0. Let us describe our previous examples in this context. Example 3.4.18: (i) Let S be a denominator set of R. Then 9 = {L < R: L n S # @} is an idempotent filter. (Indeed if s E L n S and s' E L' n S then there is some s2 in S and r2 in R with s2s = rs' E L n L' n S,proving 9 is a filter. Likewise, if s E L n S and r E R then taking s2 in S and r2 in R with s2r = r2s we have s2 E Lr-', proving % is topologizing. Finally suppose L E 9 and L'a-l E 9 for all a in L. Taking a E L n S and s E L'a-' n S gives sa E L' n S, proving 9is idempotent. (ii) If Sing(R) = 0 then 9'= {large left ideals} is an idempotent filter. Indeed, % is a topologizing filter by remark 3.3.3. To prove 9is idempotent suppose L'r-' is large for all r in R. We want L' is large. But if L n Rr = 0 then (Lr-l)r E L n Rr = 0 implying r E Sing(R) = 0. (iii) {Dense left ideals of R } is an idempotent filter, as can be seen readily from exercise 4. Details are left to the reader. (iv) For Martindale-Amitsur's ring of quotients left 9 = {left ideals of R containing a large two-sided ideal}. 4F is clearly a filter and is topologizing since if 14 R then Ir E I so I c Lr-' for every r in R and every left ideal L 3 I.
We can use topologizing filters to mimic the results of 53.3. Definition 3.4.19: A radical is a left exact functor Rad: R - A u d + R - A m ? satisfying the following properties for all modules M, N: (i) Rad M I M. (ii) If f E Hom(M, N) then Radf is the restriction of f to Rad M. (Thus f(Rad M) c Rad N). (iii) Rad(M/Rad M) = 0.
Rad is a torsion (or hereditary) radical if N n Rad M = Rad N for all N I M.
43.4 Noncommutative Localization
393
For example, Rad(M) considered in g2.5 is a radical (not torsion), but the point of view here is different based on the following results. Given a filter S define FF(M)= {x E M: Ann,x E 9.} Proposition 3.4.20:
lf 9 is an idempotent Jilter then F9 is a torsion radical.
Pro08 (i) and (ii) of definition 3.4.19 are clear. To see (iii) let N = FF(M) and suppose x + N E F ~ ( M / N )Then . Lx c N for some L in X Let B = Ann x. For any a in L we have ax E N so Ba-’ E implying B E S and x E N. Hence FFis a radical, which is clearly torsion. Q.E.D. Example 3.4.21: Let us consider the torsion radicals corresponding, respectively, to the first three (idempotent) filters of example 3.4.18. (i) FF(M)= {x E M: sx = 0 for some s in S}, called the torsion submodule of M (with respect to S). This justifies the use of the name “torsion.” (ii) .FF(M) = {x E M: Annx is large} = Sing(M). = {x E M: Annx is dense} = 0. (iii) FF(M)
Example 3.4.22: When % is as in example 3.4.18(iv) then FF(M) = {x E M: l x = 0 for some large ideal 1 of R}. Even though this need not be a torsion radical FF(R) = 0 for R semiprime.
-
Given a topologizing filter 9 define Q s ( M ) = lim{Hom(L, M / F s ( M ) ) :L E S}for every R-module M, under the system analogous to construction 3.3.1; i.e., putting M = M / F 9 ( M ) let M i= Hom(Li,M),and for Li 2 L j define cp;: M i+ Mj by sending a map f : Li + M to its restriction f : L~ + M. Construction 3.4.23:
-
Q,(M) can be explicitly described as an R-module as follows: Writing (L,f) for a map f : L -,M define the equivalence on Y = {Horn& M):L E S}, by ( L , , f , ) ( L 2 , f 2 iff ) there is L G L , n L, in 9 with L I ker(f - 9); then Q s ( M )= y/-. Thus M + Q , ( M ) defines a functor from R-Aod to R-Aud, called the 1ocaZization functor with respect to O% An alternate description in exercise 13 yields many of its important functorial properties.
-
Remark 3.4.24: (as in remark 3.3.2) If 9is a topologizing filter then Q F ( M ) is an R-module under the operations
where f r : Lr-’ + M is defined by ( f r ) a = f(ar).
Rings of Fractions and Embedding Tbeorems
394
There is a map M + Q,(M) sending x to right multiplication by x; the kernel is F,( M).
Theorem 3.4.25:
Q,(R) is a ring, under multiplication.
and the map R -,Q,(R) of remark 3.4.24 is a ring homomorphism.
Proof:
As in theorem 3.3.7.
Q.E.D.
This result unifies Theorems 3.1.4, 3.3.7, 3.4.6, and 3.4.7, in view of examples 3.4.21 and 3.4.22. Note that it gives us a functor from R-dod to Q,(R)-dod. We shall view these results more categorically in exercises l8R perhaps the ideal medium for localization is Grothendieck categories, c.f., exercise 4.2.13ff.
53.5 Left Noetherian Rings In this section we generalize the classical theory of commutative Noetherian rings to a noncommutative setting. The flavor is entirely different from the results surrounding the Faith-Walker theorem (exercises 9-13 of $2.10)which described left Noetherian rings in terms of direct sums of injectives. Instead we shall analyze left Noetherian rings in terms of the Noether-JacobsonLevitzki structure theory, also relying heavily on Goldie’s theorems. Assume until dejinition 3.5.24 that R is a leji Noetherian ring. By Noetherian we mean “left and right Noetherian.” We have already obtained the following basic facts: 1. Any homomorphic image of R is left Noetherian as R-module (by proposition 0.2.19)and thus as a ring itself; any localization of R with respect to a denominator set is left Noetherian (proposition 3.1.13). 2. Every f.g. R-module is a Noetherian module (corollary 0.2.21); consequently, every submodule of an f.g. module is f.g. 3. Nil(R) is nilpotent (theorem 2.6.23); more generally, any nil, weakly closed subset is nilpotent (theorem 3.2.38). 4. Suppose R is semiprime. Then R is Goldie and thus an order in a semisimple Artinian ring (Goldie’s theorem). Furthermore, there is a finite number of prime ideals with intersection 0, and these are the minimal prime ideals; equivalently,these are the maximal annihilator ideals (theorem 3.2.24 and proposition 3.2.26).
395
93.5 Left Noetherian Rings
5. In general, R has a finite set of prime ideals whose product is 0, taken with suitable repetitions. (Indeed, this is true for R/Nil(R) by #4, so apply # 3.)
We shall also make heavy use of Noetherian induction: When proving an assertion about a Noetherian module M we may assume the assertion holds in every proper homomorphic image of M ;indeed, if M were a counterexample we could take N c M maximal for which M / N is a counterexample, and then replace M by MIN. Likewise, when proving an assertion about R we may assume the assertion holds in RIA for every nonzero ideal A of R. Digression: There is a converse of fact 2, due to Formanek-Jategaonkar [74] and generalizing earlier work of Eakin and Nagata: If R is a commutative subring of a left Noetherian ring which is f.g. as R-module then R is left Noetherian. This cannot be generalized to arbitrary noncommutative rings,
as seen by taking R to be the subring
' '> ' M,(Q).
(0 for another result of theirs in a similar vein.
of
See exercise 2.5.15
Much of the material of the first third of this section is drawn from Chatters-Hajarnavis [80B].
Constructing Left Noe ther ian Rings First we shall see that the class of left Noetherian rings is large enough to be interesting. Remark 3.5.1: If the ring T 2 R is f.g. as R-module then T also is a left Noetherian ring (since T is a Noetherian R-module and thus a fortiori a Noetherian T-module). In particular, M,(R) is left Noetherian.
The other elementary way of constructing commutative Noetherian rings is by means of the Hilbert basis theorem, which says the polynomial ring C [ l ] is Noetherian if C is Noetherian. Of course, the free algebra F{X,,X,} over a field F is not left Noetherian, but there is the following result: Proposition 3.5.2: (Hilbert Basis Theorem generalized) Suppose W is generated as a ring by R and an element a such that R + aR = R + Ra. Then W is also lefi Noetherian. (In particular, this holds if a is R-normalizing.) We aim to show every left ideal L of W is f.g. Define Li = {r E R: there is ~ ~ = O a in u rLu with ri = r and all ru E R}, and write $(r) for such
froo$
3%
Rings of Fractions and Embedding Theorems
an element ca'r,. Despite having written the a" on the left, we see each Li is a left ideal since R + aR = R Ra. Lo E L , E L , . . . , so for some m we have Li = Li+ for all i 2 m. Let us write Li = I ? Rsij and ! let! L' = WG(sij). L' I L by definition of the G(r). We claim L' = L. Indeed we shall prove by induction on k that given any w = k a%,, in L (with r,, in R) we have w E L'. First suppose k Im. Writing rk = c r k j s k and j noting akr = rLjak+ x v < k a " Rfor suitable r i j in R (since aR C Ra + R) we see w - fr&G(skj) E L has degree < k and so is in L' by induction on k. Thus w E L' as desired. Hence we may assume k > m. Now rk E L, = L,. As in the previous paragraph i k appears as the leading coefficient of an element w' of L' of degree m; thus rk is the "leading coefficient" of ak-"'w', and w - ak-'"w' has degree c k. Q.E.D. By induction w - ak-'"w' E L', implying w E L'.
+
cy=oc;",
xu=
Corollary 3.5.3. The Ore extension R[I; I J , ~also ] is left Noetherian, provided a is an automorphism. (The assertion may fail i f a is not onto.)
The Reduced Rank In studying noncommutative Noetherian rings one would rather not face the awesome hurdle of noncommutative localization, even in prime Noetherian rings. An attractive alternative introduced by Goldie [64] is the reduced rank, which came to the fore later when used in the proof of several important theorems. We start by recalling from 83.1 that if M is an R-module and S is a left denominator set for R then we can form the S-'R-module S-'M = S-'R & M. Suppose M is an f.g. module over a left Noetherian ring R. Define the reduced rank p(M) as follows:
Definition 3.5.4.
Case ( i ) . If R is semiprime then p(M) is the (composition) length of S-'M as S-'R-module, where S = {regular elements of R}. Case (ii). N = Nil(R) is nilpotent of index t > 1. If NM = 0 then viewing M naturally as R/N-module, take p(M) to be its reduced rank as R/Nmodule. In general, by induction on i such that N'M = 0 define p(M) as p(NM)+p(M/NM). Explicitly p(M)=C:= p(Mi) where Mi= N'-'M/N'M. Remark 3.5.5: If xl,. , . ,x, span M a s R-module then 1 Q xl,. .., 1 @I x, span S-'M as S-'R-module (notation as in (i)); since S-'R is semisimple Artinian by Goldie's theorem we see that case (i) makes sense.
$3.5 Left Noetherian Rings
397
Proposition 3.5.6: p is additive, i.e., if K I M then p ( M ) = p ( K ) + p ( M / K ) . Proo) First assume R is semiprime, i.e., case (i) holds. By theorem 3.1.20 the sequence 0 + S-'K --t S-'M -P S - ' ( M / K ) + 0 is exact, so length(S-'M) = length(S-'K) + length(S-'(M/K)) as S-'R-module, as desired. Next we consider case (ii), i.e., N = Nil@) is nilpotent. The argument is rather computational. By induction on the smallest j such that N j M = 0, we note we are done by case (i) if j = 1 since M is then an R/N-module. We shall appeal repeatedly to Noetherian induction, and assume the assertion holds for every proper homomorphic image of M. By definition and p(K) = p ( N K ) + p ( K / N K ) .
+
p(M)=p(NM) p(M/NM)
By induction p(NM)=p(NK)
+ p ( N M / N K )= p ( K ) - p ( K / N K ) + p ( N M / N K )
so
Thus we may assume NK = 0. But then case (i) applies to K and p(K) = p ( K n N M ) + p ( K / K n NM). By induction
+ p ( N M / K nN M ) = p ( K ) - p ( K / K nN M ) + p ( N M / K nN M ) .
p ( N M ) = p ( K nN M )
Thus p(M)= ( p ( K ) -
nN M )
= p ( K ) - p ( K / K nN M )
+ p(NM/K
N W ) + P(M/NM)
+ p ( M / K nN M )
=p(K)+ p(M/K)
by Noetherian induction unless K n NM = 0. But now K z (K I M I N M , so p ( M / N M ) = p ( K ) + p ( M / K + NM) and p ( M ) = ( P W ) + p ( M / K + N M ) ) + p(NW =p(K)
+NM)/NM
+ p ( M / K + N M ) + p ( K + N M / K ) = p(K) + p ( M / K )
by Noetherian induction.
Q.E.D.
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This proof could be formulated directly without Noetherian induction, cf., Chatters-Hajarnavis [8OB]. Nevertheless, the proof presented here has a certain inevitability. It is useful to know when p ( M ) = 0. Proposition 3.5.7: p ( M ) = 0 iff for each x in M there is an r in Ann, x such that r + N is regular in RIN. (Notation as in definition 3.5.4 (ii).)
Proof: Induction on the smallest j such that N i M = 0. If j = 1 then we may assume N = 0 and case (i) holds. But then p ( M ) = 0 iff S-’ 63 M = 0, iff (by remark 3.1.21) S n Ann, x # 0 for each x in M, as desired. For j > 1 we see using case (ii) that N * - ’ M / N ’ M has reduced rank 0 for each i < j ; by case (i) for each x in N i - ’ M there is si in S such that s,x E N‘M. But then for any x in M we have suitable sl,...,s, in S with sj***slx = 0, as desired. Q.E.D. Digression 3.5.7: Noetherian orders. Although our immediate interest in the reduced rank is in proving the principal ideal theorem, other important applications are given in exercises 1,2, and 3. These include a direct proof of Small’s characterization of left Noetherian orders in left Artinian rings. One of the main tools in studying such rings is the Artinian radical A(R) defined below in 3.5.19. Another method is studying those prime ideals consisting of zero divisors; P. F. Smith showed these are precisely the minimal prime ideals iff R is an order in an Artinian ring. Small-Stafford [82] have fitted Smith’s result into a general theory of regularity modulo prime ideals. Stafford [82a] has shown that a Noetherian ring R is its own classical ring of fractions iff Ann A(R)n Ann’ A(R) c Jac(R); in particular R must be semilocal.
Quite recently interest has returned to the theory of Noetherian rings which are embeddible into Artinian rings. We discussed this topic in $3.2; Small has observed that the reduced rank p yields a “Sylvester rank function” in the sense of Schofield [SSB] and thus a homomorphism from an arbitrary left Noetherian ringR to a simple Artinian ring, whose kernel No is { r E R: p(R/Rr) = p(R)}. In view of proposition 3.5.6, No = { r E R: p(Rr) = O}; since No is f.g. as R-module we get p ( N o ) = 0, so, in particular, No E Nil(R). If R is also right Noetherian then No is f.g. as right R-module and thus annihilates an element regular modulo Nil(R). In certain instances No = 0 (e.g., when R is “Krull homogeneous,” cf., after proposition 3.5.46 below).
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399
The Principal Ideal Theorem One of the key tools in the study of Spec(C) for a commutative Noetherian domain C is the Principal Ideal Theorem, which states that any prime ideal minimal over a nonzero element c is a minimal prime ideal of C . In other words, if c E P and 0 # P, c P then c E P,. For a while the following example of Procesi discouraged attempts to generalize this result to noncommutative Noetherian rings: Example 3.5.8:
Let R
=
M2(Z[1]) and r =
(i :),
a regular element.
Then P = RrR = 2R + 1R is a prime ideal since R I P z M2(Z/2Z), but P is not a minimal prime since 1 R E Spec(R). Nevertheless, Jategaonkar [74a] proved a noncommutative principal ideal theorem, which we present below. Jategaonkar's proof was a masterpiece in the application of abstract torsion theory to a concrete problem about rings. On the positive side this was a triumph for torsion theory, and recently Jategaonkar has used his theory to extend this theorem still further. However, hardly anyone understood the proof, as evidenced by the long interval between Jategaonkar's results, and most ring theorists breathed a sigh of relief when Chatters-Goldie-Hajarnavis-Lenagan[79] recast the proof in terms of the reduced rank.
Lemma 3.5.9: Suppose a E R is a normalizing element and P E Spec(R) with a 4 P. If ra E P then r E P. Proof: rRa = raR
c P so r E P.
Q.E.D.
Theorem 3.5.10: (Principal ideal theorem) Suppose R is prime (left Noetherian) and a # 0 is a normalizing element of R. If P i s a prime ideal of R minimal
over a then P is a minimal (nonzero) prime ideal of R. Proof: Suppose on the contrary, there is PI in Spec(R) with 0 c Pl c P. Since R is prime P, contains a regular element s. Also note a is regular by lemma 3.5.9. Let L, = a-kRs = { r E R:akr E Rs}, a left ideal since a is normalizing. Then Lo I L , I L , I ... so for some m we have Lk = L,, for all k 2 m. Replacing a by amwe may assume m = 1. Thus L , = L 2 . Consequently, if a2r E Rs then ar E Rs.
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400
Let M = (Ra + Rs)/Ra2, viewed naturally as an R/Ra’-module, and let M , = (Ra’ + Rs)/Ra2 < M . We claim p ( M / M , ) = 0, i.e., p ( M ) = p ( M l ) . To prove this claim we use proposition 3.5.6. Let M’ = Ra/Ra2, M i = (Ra’ Rsa)/Ra’, and MY = (Ra’ Ras)/Ra2. Right multiplication by a yields an epic M + M i with kernel M’, so p ( M ) = p ( M ’ ) p ( M ; ) . On the other hand, there is a ring automorphism $: R --f R given by $r = ara-’. Since $(Ra2) = aRa = Ra’ we see that $ induces a ring automorphism of RIRa’, which sends M i (as a subset of R/Ra2) to MY. Thus p ( M ; ) = p(M;‘). We can prove the claim by finding an epic f Ra + M , / M Y having kernel Ra’, for then f would yield an isomorphism M ‘ + M J M ; ’ implying p ( M , ) = p ( M ’ ) p(MY) = p ( M ’ ) p(M;) = p ( M ) , as desired. Our candidate for f is given by f ( r a ) = rs + (Ra’ Ras). Clearly f is onto and RaZ E kerf, so it remains to show that kerf s Ra’. If ra E kerf then rs E RaZ + Ras = a2R Ras; writing rs = a2rl r2as we get a2r1E Rs so ar, E Rs by the first paragraph. Writing ar, = r’s we have rs = ar’s + r,as so r = ar’ + r2a E Ra and ra E Ra’; hence kerf E Ra’, proving the claim. Write - for the canonical image in R = (R/Ra’)/Nil(R/Ra’), a semiprime Goldie ring. By proposition 3.5.7 applied to M / M 1 there is some element r of R such that Fis regular in R and ra E Ra’ Rs. Write ra = r1a2 + r2s for ri in R. Then (r - rla)a = rzs E P, implying r - r,a E Pl c P. Hence r E P, so F is a minimal prime ideal of R containing the regular element 6 contrary to F being an annihilator ideal (fact 4). This proves the theorem. Q.E.D.
+
+
+
+
+
+
+
+
+
Corollary 3.5.11: has height I 1. Proof:
Any prime ideal P minimal over a normalizing element in R
Otherwise P 3 Pl
3
P2; pass to RIP,.
Q.E.D.
This result can be improved still further, with Ra replaced by any invertible ideal, cf., Chatters-Hajarnavis [80B, p. 45-50].
Heights of Prime Ideals For commutative Noetherian rings one proves directly from the principal ideal theorem that every prime ideal has finite height. Although no such sweeping result is available for the noncommutative case, we do have a good partial result.
Lemma 3.5.12:
Suppose P E Spec(R) contains a normalizing element a of R.
$3.5 Left Noetherian Rings
401
Then any chain of prime ideals P = Po 3 Pl 3 chain P = Po 3 P', 3 ... 3 Pi-, 3 P, with a E
3
P, can be modijied to a
Proof: Choose P = Po 3 P', 3 *.. 3 Pi- such that a E P; with k maximal possible. We are done if k = t - 1, so assume k I t - 2. Let Pi = P,. In the prime ring R = R / P ; + we have P ; = i';, 3 0. Thus i';+ is not minimal over ii, by theorem 3.5.10, so there is F i + l c P; with i i i'z+l. ~ Replacing Pk+ by Pi+ yields a contradiction. Q.E.D. Theorem 3.5.13: ( R is 1eJt Noetherian) Suppose P E Spec(R) and i' contains a nonzero normalizing element of R for every prime homomorphic image R of R in which # 0. Then P has jinite height.
Proof: By Noetherian induction we may assume for any nonzero ideal A c P that P/A has finite height in RIA. Passing to R/Nil(R) we may assume R is semiprime, for this does not change the prime spectrum. Since R has only a finite number of minimal prime ideals we may assume furthermore R is prime. Take 0 # a E P with a normalizing in R. Then i' = P/Ra has some finite height t in = R/Ra. But this implies height(P) I t + 1 in R, for, otherwise, given a chain P = Po 3 * * * 1 P , + 2 , we could assume a E P,+ by in R, a contradiction. Q.E.D. lemma 3.5.12, yielding i' 3 3 ... 3
e+,
It will turn out that this condition can indeed be verified in several important cases (such as PI-rings).
Digression: Decomposition Theorems of Noetherian Rings Recall that the ring R is called Noetherian if R is both left and right Noetherian. Many additional theorems can be proved under this slightly stronger hypothesis. Furthermore, there is an alternate approach obtained by decomposing a Noetherian ring (when possible) as a direct product of a semiprime ring and an Artinian ring. We illustrate this approach with a pretty result of Robson, starting with two lemmas of independent interest:
Lemma 3.5.14: Suppose R is a lefi order in a semisimple Artinian ringQ. Suppose, furthermore, that some simple component Q' of Q has a module M # 0 which is f.g. when viewed as an R-module (i.e., jirst extend the action of M trivially to Q and then restrict it to R). Then Q' is an f.g. R-module.
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402
Proof: M is a sum of copies of a minimal left ideal L of Q', implying L also is f.g. as R-module. But Q' is a finite sum of copies of L and so is itself an f.g. R-module. Q.E.D.
Lemma 3.5.15: Suppose R is a left order in an arbitrary ring Q and Q' is an
ideal of Q which is f . g . as an R-module. Then Q' c R under either of the following conditions: ( i ) R is also a right order in Q; ( i i ) Q' = Q.
Proof: (i) Write Q' = CfiniteRris-'. Then Q' = Q's = x R r i E R. (ii) For each regular s in R we have Rs-' IRs-' I in Q so Rs-' = Rs-('+ ') for Q.E.D. some i, implying s-' E R. 0
Let N = Nil(R) and W ( N )= { r E R : r
.
.
+ N is regular in R / N } .
Theorem 3.5.16: (Robson's decomposition theorem) A Noetherian ring R is isomorphic to the direct product of a semiprime ring and an Artinian ring iff sN = N s = N for all s E '$(N). Proof.- (a) We may assume R is semiprime or R is left Artinian. In the former case the assertion is vacuous; in the latter case the assertion is obvious since any regular element s + N in R/N is invertible, implying s is invertible in R. (e) Let-denote the canonical image in R = R / N , a left and right Goldie ring. Also write 4 for N'/N'+', naturally viewed as an R - R bimodule. Applying proposition 3.1.16 to Goldie's theorem we see R is both a left and right order in a semisimple Artinian ring Q. Let f , : N + N be given by right multiplication by s. By hypothesis f, is epic and is thus an isomoris a right Q-module under the action phism by lemma 2.9.6. Hence (x + N i + l ) (rs- - l ) - f,-'(xr) N"' for X E N'; this extends the given right R-module action. Write Q = Qk where each Q, is a simple Artinian ring; we view the Qk as minimal (two-sided) ideals of Q. Let Mik= s i Q k ,a right R-submodule of 4 which is thus f.g.. If Mik # 0 for some i then by lemma 3.5.14 Qk is an f.g. right R-module, so Qk E R by lemma 3.5.15. Likewise if Qk& # 0 then Qk G 8. Thus letting 2-= {k:QkNi= 0 = &Qk} we have Q x Q' x Q" where Q' = n{Q,:k 4 X } and Q" = k E X } .Furthermore Q' E R. Writing R" = Q" n R we see R Q' R" and 8"Ni = NiR" - 0 for all i.
+
4
nF=
fl{Qk:
Write R" = R B where B is idempotent in R; by idempotent-lifting we may assume e is idempotent. Then eN' E N"' and N'e c N"' for all i, from
$3.5 Left Noetherinn Rings
403
which we see eN = Ne = 0 since N is nilpotent. But eR(l - e) c N so we conclude eR(1 - e) = e2R(1 - e) G eN = 0; likewise ( 1 - e)Re = 0, so the Peirce decomposition shows R = eRe + ( 1 - e)R(1 - e). Now eRe x eRe/O = eRe/eNe !z ZRZ= R F % R” is semiprime. ( 1 - e)R(1 - e) is a semiprimary Noetherian ring which is thus (left and right) Artinian by theorem 2.7.2. Q.E.D. Digression 3.5.26‘: Robson’s criterion is satisfied for P P rings, described in exercise 3.2.20. Indeed, if r E W ( N ) then Ann r c N is generated by an idempotent which must thus be 0, and, likewise, Ann‘r = 0. Thus Robson’s theorem reduces the study of hereditary rings to the semiprime case, which has been studied intensively in the literature. Indeed Goldie [72] observed that hereditary Noetherian prime rings were a framework of studying Asano orders, and Robson [72] showed how HNP rings could be obtained from “maximal orders.” However, the H N P theory is limited by a collection of “bad” examples in Stafford-Warfield [SS].
Webber [70] proved some pretty results about simple hereditary Noetherian rings, which have led to some surprises. First, he showed by an easy argument that given any three left ideals L,, L,, and L, one can find a fourth left ideal L, for which L , @ L, x L, @ L,. O n the other hand, the Weyl algebra R = A,(F) is a simple Noetherian hereditary domain and satisfies R 0 L =: R 0 R for every left ideal; so taking L, = R above one gets L, @ L, x R @ R for any nonzero left ideals L , , L , of R = A,(F). This enabled S.P. Smith [81] to show that M,(R) x M , ( T ) does not imply R z T when R is the Weyl algebra, thereby answering negatively a question of Jategaonkar. We shall discuss other results of this type in 54.1.
Artinian Properties of Noetherian Rings: Prelude to Jacobson 3 Conjecture Let us now introduce of Artinian techniques, i.e., we consider ideals which are Artinian as left modules.
Lemma 3.5.17: Suppose A
404
Rings of Fractions and Embedding Theorems
Artinian ring Q. Write A = aiR and let us also identify A naturally as an f.g. R-module. Then ra = 0 iff Fa = 0. Let S = {r E R: Fis regular in R } and A' = {a E A: S n Anna # 0).Then A' q R (seen either by noting A' = Sing(A) or by applying the Ore property directly). Hence A' = 0 or A' = A. But A' # A for, otherwise, there are si in S with siai = 0 for 1 5 i I t; using the Ore condition we could find s in S with sai = 0 for all i, implying sA = 0 contrary to S # 0 in R = R/P. Thus A' = 0. Now for any s in S we have A 2 sA 2 Since A is right Artinian we have siA = s'+'A for some i, implying A = sA by the preceding paragraph. Thus A naturally becomes a Q-module, implying Q = R by lemmas 3.5.14 Q.E.D. and 3.5.15. Thus R is simple Artinian. * . a .
Proposition 3.5.18: Suppose A 4 R is Artinian as a right R-module. Then R/Ann, A is a left Artinian ring, so, in particular, A is Artinian as a left R-module. Proofi Take A , = A and inductively given Ai 4 R take A i + 4 R maximal with respect to Ai+ c A i . (This is possible since Ai is a left submodule of R and is thus Noetherian.) Then A, 3 Al 3 I . . so this chain must terminate when viewed as right modules. In other words, some A, = 0. Then Ai/Ai+' is a minimal ideal of Ri/Ai+', so by lemma 3.5.17 there is a maximal ideal pi in R such that &Ai E A i + l , and, furthermore, R / q is simple Artinian. Let B = 4, Then R/B is semisimple Artinian by the Chinese Remainder Theorem. But E'A E 8- * * . P,A = 0 so B' E Ann, A. Consequently, R/Ann, A is a semiprimary left Noetherian ring which is thus left Artinian. Now A is an f.g. module over R/Ann,A and is thus Artinian, and so is Artinian as R-module. Q.E.D. Digression 3.5.19: (Also see exercise 12). These considerations lead one to define the Artinian radical A(R) to be the sum of all the left ideals of R which are Artinian (as R-modules). If L < R is Artinian then so is Lr for any r in R, so A ( R ) 4 R. Since R is left Noetherian we see A(R) is f.g. and is thus Artinian itself. In other words, A(R) is Artinian and contains all Artinian left ideals. In this context proposition 3.5.18 shows this definition is left-right symmetric for R Noetherian, i.e., A(R) is Artinian as a right R-module and contains all Artinian right ideals. The Artinian radical is one of the central features of Chatters-Hajarnavis [SOB]. (Also see Stafford [82a] in this regard.) Remark 3.5.20: If R is Noetherian and R is a left Artinian image of R then R is Artinian. (Indeed, R is semiprimary and thus right Artinian by theorem 2.7.2.)
43.5 Left Noetherian Rings
405
The interplay between Noetherian and Artinian is quite enlightening.
Lemma 3.5.21: Artinian.
If A,B 4 R and RIA and RIB are Artinian then RIAB is
Proof: BIAB is an f.g. module over RIA and so is Artinian, and thus is Artinian as RIAB-module. Hence RIAB > B/AB > 0 refines to a composition series of RIAB as module over itself. Q.E.D. We are ready to introduce the key Ginn-Moss prime ideal.
Lemma 3.5.22: (Ginn and Moss) Suppose R is Noetherian and M E R-Aoa! such that RIAnnR M is not Artinian. Then there is P E Spec(R) such that P = Ann N for some N I M and RIP is not Artinian. Proof: Take P Q R maximal with respect to RIP not Artinian but P = Ann N for suitable N I M. We are done unless P is not prime, i.e., AB c P for ideals A,B 2 P. Then ABN = 0; replacing A by AnnR(BN) we have RIA Artinian by assumption on P. Now B=B/AB is an f.g. RIA-module and is thus Artinian. Hence B is Artinian as module over R = R/AB by proposition 3.5.18. Let 7 = Ann6 B,i.e., BI = AB. The right-handed version of proposition 3.5.18 shows R I I is Artinian. But BIN = 0 so putting I’ = Ann(1N) 2 B 3 P we see RII‘ is Artinian. By lemma 3.5.21, R/l‘l is Artinian. Q.ED. But 1’1 E Ann N = P, so RIP is Artinian, as desired.
Jacobson’s Conjecture
nieN
Let J = Jac(R). The celebrated Krull Intersection Theorem states J’M = 0 for every f.g. module M when R is commutative Noetherian, and one of the major projects in ring theory has been to verify this for various classes of noncommutative rings. Jacobson’s conjecture is that J ’ = 0 for an arbitrary Noetherian ring R. Jacobson’s conjecture implies the Krull intersection theorem in the commutative theory, since R 0 M can be viewed as a Noetherian ring by taking multiplication (rl,xl)(r2,x2) = r1r2 + rlx2 + r 2 x l . An attempt to “noncommutativize” this trick by means of upper triangular matrices actually leads to a famous counterexample.
nisN
Example 3.5.23: (Herstein) Let P = 2B, C = H,, and R =
view of example 1.1.9 R is left Noetherian. On the other hand, J =
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seen easily because R/Qe,, z Q x C implies J / Q e l , = PpezZ.Hence e l , = (2-ke1,)(2e,,)kE J k + ' for each k, implying 0 # e l , E J'.
nicN
The status of Jacobson's conjecture for (left and right) Noetherian rings is still open. However, there is an additional condition which guarantees the conjecture. Definition 3.5.24: R is left bounded if every large left ideal contains a twosided ideal. R is almost bounded if a large submodule of a faithful f.g. module is necessarily faithful. R is lejit fully bounded (resp. almost fully bounded) if every prime homomorphic image of R is left bounded (resp. almost bounded).
Proposition 3.5.25: Every left bounded prime ring R is almost bounded. Every left fully bounded ring is almost fully bounded. Proof: We verify the first assertion, which formally implies the second. Suppose M = Rxi is faithful, and N 5 M is large. By proposition 3.3.0 each Nxf' is a large left ideal which thus contains a two-sided ideal A i . Let A = A l - . . A , #O. Then AM 5 XARx, E N, so (AnnN)A c AnnM = 0, Q.E.D. implying Ann N = 0 since R is prime. Every commutative ring is left and right fully bounded, and we shall see in 86.1 that this hypothesis passes to rings with polynomial identity. Jategaonkar [74] verified Jacobson's conjecture for left and right fully bounded Noetherian rings. This result was subsequently improved by Schelter [75] and by Cauchon [76], who removed "right fully bounded from the hypothesis. Later Jategaonkar [81] proved the result for almost fully bounded rings. This is a substantial advance because virtually all of the Noetherian rings of interest today are almost fully bounded, whereas many are not left fully bounded. On the other hand, the almost fully bounded case is much more complicated than the left fully bounded case. We present the easier result now, postponing Jategaonkar's stronger theorem until the end of the section. Note that a left bounded prime Goldie ring satisfies the property that if s E R is regular then Rs contains a nonzero ideal. Remark 3.5.26: If R is primitive and left bounded and satisfies ACC(@) then R is simple Artinian. (Indeed let L be a maximal left ideal of R having core 0, Then L is not large, so has a nonzero essential complement L'. Clearly L' is a minimal nonzero left ideal of R, implying soc(R) # 0. By hypothesis
407
53.5 Left Noetherinn Rings
soc(R) is a sum of a finite number of minimal left ideals since soc(R) is completely reducible; hence R is semisimple Artinian by remark 2.1.26.)
Lemma 3.5.27: Suppose R is almost fully bounded Noetherian with primitive images Artinian. (ln particular, this would hold if R is fully bounded, by proposition 3.5.25 and remark 3.5.26.) Then every f.g. module M with a unique minimal nonzero submodule M,, is Artinian. Proofi Replacing R by RIAnn M , we may assume that M is a faithful R-module. Suppose R is not Artinian. By lemma 3.5.22 we have P E Spec(R) such that R = RIP is not Artinian and P = Ann, M' for some 0 # M ' < M. Since M' # 0 we see M , IM , so M , is a large submodule of M'. But M ' is a faithful module over RIP so, by hypothesis, M , is faithful as well as simple as R-module, implying R is primitive and thus Artinian, contradiction. Q.E.D.
nieN
Theorem 3.5.28: (Jategaonkar-Schelter-Cauchon) J'M = 0 for every f.g. module M , whenever R is an almost fully bounded Noetherian ring whose primitive images are Artinian; in particular, it holds for fully bounded Noetherian rings. Proofi It suffices to show that for every x in M there is some m E N with x # J"M. By Zorn's lemma there is N < M maximal with respect to x # N ; thus it suffices to prove J"M IN. This certainly holds by "Nakayama's lemma" if M / N has finite composition length. On the other hand, every nonzero submodule of M I N contains Rx + N , so passing from M to M I N we are in position to apply lemma 3.5.27. Q.E.D. Digression: The key idea in this proof was clearly the reduction to lemma 3.5.27, so we would like to know whether anything was lost in this transition. If R is semilocal and Jacobson's conjecture holds then indeed the J ' M = 0. Then conclusion of lemma 3.5.27 is true; indeed, suppose M, $ J"M for some m, so J"M = 0. Thus M is an f.g. module over RIJ", a semiprimary Noetherian ring which is thus Artinian, implying M is Artinian.
n
On the other hand, the conclusion of lemma 3.5.27 can fail even when Jacobson's conjecture holds, should primitive images of R not be Artinian. Here is a counterexample, over any field F of characteristic 0.
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Example 3.5.29: (Musson) Let R , be the polynomial ring F [ p ] and R be the Ore extension R,[A; 1 , 6 ] where 6p = p, i.e., Ip - pA = p. R is Noetherian by corollary 3.5.3, and p R = R p by inspection. Let M = R / L where L = R ( p - l)(A - l), and M, = R x , where x , = (A - 1 ) + L E M. We shall show every nonzero submodule of M contains M , , but M is not Artinian. In fact letting-denote the image in R = R / L we shall show that M , and RpiT are the only nonzero submodules of M. We start with some easy computations. since p(A - 1 ) - (A - 1 ) = ( p - l ) ( I - 1) E L . (i) p x , = x , obtained by iterating Ap = pA
(ii) [A,pi] = ipi (iii)
1= T + x , .
(iv)
X, =
.-
+ p.
.-
.-
pixo = P I ~-, piT = (npiT - ipiT) - p l l = (A - i - 1 ) p ' l .
In particular, M, c Rp'f for each n. Since Rp"' the following two assertions:
'1 5 RpnT it suffices to prove
( 1 ) For any x # 0 in M we have R x = M, or R x = RpnT for suitable n. (2) p n i # Rpn+'T for each n. Indeed, (1) would imply M, and the RpnT are the only submodules of M and (2) would show that M , is the only Artinian submodule, although it is large in all the submodules. To prove (l), note pA = p I - 1 = p x , modulo L; hence x has the form x' + x : = n a i p i Tfor suitable x'in M,. First assume each mi = 0; then Rx' E M , , so we want to show x , E Rx'. But x' = c ~ = , f l i A i x 0for suitable fli in F, by (i). Furthermore, for i 2 1 we have inductively (1 - P L ) ~ A ~ x= , i ! x , (verfication left for the reader); on the other hand, (1 - p ) x , = x o - p x , = 0 so ( 1 - pyAjxo = 0 for j c i. Consequently, (1 - p ) " ' ~= ' rn!fl,,,x, so x , E Rx' as desired. Thus we may assume some ai # 0; choosing n, t appropriately, we have a, # 0 so each piT E R p " i and thus x E R p n i . To prove ( 1 ) it remains to prove p"T E R x ; in fact, we see the proofs of both ( 1 ) and (2) reduce to proving the following claim (where for (2) we take x = 0):
+
If x
= x'
+
+ c a i p i i with x' E M, and a, = 1 then pnT E R x . f
i=n
We prove this by induction on t. Multiply both sides by A - t - 1 ; noting by (iv) that (A - i - l)piT E M , we have
(A - t - l ) x E M ,
+ 1ai(i - t)pif i=n
$3.5 Left Noetherian Rings
409
and by induction we have p"f E R ( 1 -
t -
l ) x G Rx, as needed.
Musson's example is not a counterexample to Jacobson's conjecture, since it is primitive (by exercise 2.1.4)! On the other hand, I do not know of any counterexample to the conclusion of lemma 3.5.27 when R is Noetherian with all primitive images Artinian.
D Supplement: Noetherian Completion of a Ring In this supplement we do not assume R is Noetherian. Theorem 3.5.28 gave Ji=O where J=Jac(R). Then we can form the J-adic completion k. (This notation has no connection to 83.3.) We shall examine conditions under which d also is left Noetherian; this would be useful since the module theory of complete Noetherian rings is richer than that of Noetherian rings in general. We present the results in the more general setting of rings R with filtration over ( E , +). Not only does this include the A-adic completion (or, more generally, the case where R(0) = R, cf., example 1.8.14(ii)ff.) but also enveloping algebras (as we shall see in 88.3.)
us a criterion for
niEN
Definition 3.5.30: Suppose R is a ring with filtration over (Z, +). Define the associated graded ring G ( R ) to be the additive group O i e n R ( i ) / R ( i l), made into a ring by defining multiplication
+
(ri
where
+ R(i + l))(r;+ R ( j + 1)) = rir; + R(i + j + 1) ri E R ( i )
and
r> E R ( j ) .
If M E R - A o the ~ associated graded module is O i , , R ( i ) M / R ( i + 1)M, made into a module over the associated graded ring by means of scalar multiplication (ri
where
+ R(i + l))(xj+ R ( j + l ) M ) = rixj + R(i + j + l)M ri E R ( i )
and
xj E R ( j ) M .
An important special case is the A-adic filtration; then we write G,(R) for A i / A i + ' , and G J M ) for the associated the associated graded ring OiEN graded module. Of course, there is a functor R - A u d + G , ( R ) - A d sending M to G,(M). Proposition 3.5.31: Suppose R is a ring with Jiltration over (Z, +). (i) If R is complete and M E R - A u d such that R ( i ) M = 0 and G ( M ) is f.g. as G(R)-module,then M is f.g. ( i i ) If R(0) = R then G ( R ) % G(k).
nisN
Rings of Fractions and Embedding Theorems
410
Proofi
I:=:=,
(i) Write M = G(R)x,, where we may take the xu homogeneous, i.e., xu = y, + R(n, + l ) M for suitable nu in N and y, in R(n,)M. For any given 0 # y E M there is a smallest n for which y # R(n + 1)M; looking in G ( M ) we have y' = y - I r n u y uE R(n + l ) M for suitable r,, in R(n); likewise, y' - C u r n + l . u y u E R(n + 2)M, for suitable r , , , , , in R(n + 1). Continuing ad infiniturn we see for each u that ~ m ; l n r Emdu = R and y ,(Cmrn r,,,,,)y, E R(m)M = 0, proving y,, + . .,y, generate M. (ii) From example 1.8.14 we have R n d(n)= R(n), so R(n)n d ( n + 1 ) = R(n + 1); also d(n)= R(n) + d(n 1). Now
x:=
n,
+
k ( n ) / f i ( n+ 1) = (R(n)+ k(n = R(n)/R(n3-
proving (ii).
+ l))/k(n + 1) x R(n)/(R(n)n k(n + 1)) l),
Q.E.D.
Corollary 3.5.32: left Noetherian.
Suppose R is a ring with Jiltration over (Z, +), and G ( R )is
(i) If R ( l ) = 0 then R is left Noetherian. (ii) If R(0) = R then d is left Noetherian.
Proofi (i) R is complete (cf., example 1.8.14(i)), so apply the proposition with M = R. (i) If L c k then G(L) is a left ideal of G ( d ) z G ( R )so is f.g., implying L is f.g. in i. Q.E.D. Our immediate interest is the special case of (ii)for the A-Adic completion: If G,(R) is left Noetherian then d also is left Noetherian. Thus we are led to ask when G,(R) is left Noetherian. This is an easy application of the Hilbert Basis Theorem for R commutative, and with a little more effort we can obtain a slightly more general result.
Proposition 3.5.33: Suppose A = Noetherian then G,(R) and
,Rzi with each zi in Z(R).If RIA is left
d are left Noetherian.
Proofi Write S = {zl,. ..,zt}. Then A = RS and, inductively, A' = RS' for each i. But R , = (R/A)[A,,...,At] is left Noetherian by the Hilbert Basis
53.5 Left Noetherian Rings
411
Theorem, and there is a surjection R , + G,(R) sending I t to the element z ; + A"' in A ' / A ' + ' , so G,(R) is also left Noetherian. Q.E.D. We shall now prove a rather sweeping result of McConnell [79] based on work of Roseblade. We say r E R is central modulo an ideal A if r + A E Z ( A ) . An ideal A of R is polycentral if there is a chain = 0 of ideals of R such that Ai = IRa, for suitable elements {ai,: 1 Iu It ( i ) } each central modulo Ai+1. If we only have the weaker condition that the a,, + A i + l normalize R I A , , , we say A is a polynormal ideal. Definition 3.5.34: A = A, 3 A , 2
+
3
These definitions embrace several examples from group algebras and enveloping algebras to be pursued in 58.4. To study polycentral ideals we construct the ring r?, defined to be the subring of R [ 1 ] generated by Ai2" for all 0 Ii It and all 1 Iv I 2'. Then r? is N-graded (according to the powers of I ) since the generators of r? are homogeneous elements of RCI]. Write I?,+ = R and for each i It let r?, be the subring of r? generated by r?', and all { A i l " :1 Iu I2i); thus r? = r?,.
,
Lemma 3.5.35: d, i s generated by R i + , and a set S = {aijI":1 Ij It ( i ) , 1 I v I2 ' ) (notation as in dejnition 3.5.34);furthermore, [s,s'] E r?'+ and [s, Ri + ,I E Ri + for all s, s' in S.
-
Proof.- The first assertion is immediate. To prove the next assertion take s = a i j l uand s' = aij.l'" in S ; then [s,s'] E A i + , I u + " 'c R , , , since u + v' I 2' + 2' = 2 i + l. It remains to show [ S , R ' + ~E] R i + , . Write a = aij and
-
-
I
T= Ri+l. Note [s, R ] = [a, RIA" E A i + I" E T. But T is spanned by R and products of length m of generating elements for any m > 0, i.e., terms of the form x = b , .b,l" where each b, E A,. for suitable u' 2 i 1 and rn I n I 2"'. Thus it suffices to show for each x of this form that [ s , x ] E T. Now
+
rn
it suffices to show each b l . . . bk - l [ a , b k ] b k + l . . . b m I " + nT. E Since b, E + Rakrjfor a k ' j central modulo A,, + we can break this further into two cases: b, E A,#+ and b, E Rak,jfor some j .
SO
& = A,. +
,,
1
,
Case I. If b, E A,"+ then [a, b,] E + so replacing k' by k' n v I1 2 " 'as desired, and we are done.
+
+ 1 we have
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412
+
Case ZZ. If bk = rak,jfor suitable r in R then [a, bk] = [a,rakgj]= [a,r]akrj r[a,ak,j] EAi+lAk, Ak,+1. By case I we can handle the part from A k , + l rso it remains to show b i ~ ~ ~ b k - , A i + l A k . b k + l ~ ~E~ bT., ,Take , l " +any " elements b; from A i + , and b;+ from A k * Now . bl bk- bib;+ b,+ . . . b,IV+"E T is clear if we replace m by m + 1 5 u + n <- 2"' 2".This disposes of case 11, proving the lemma. Q.E.D.
+
-
,
+ I:=,
Theorem 3.5.36: If R has a polycentral ideal A such that R I A is left Noetherian then.the A-adic completion ff is lefi Noetherian. Proofi We modify the proof of proposition 3.5.33. Let A = A , 3 . * .3 A,+ I = 0 and I? and I?i be as in definition 3.5.34q and write k = BiIi where B, = R, B, = A , and each Bi Q R. Since I? is N-graded we have each BiBj c B i + j ;thus A' G Bi. Also note if i > 2k' then Bi c A k since the highest power of I available in the generators of I? is 2'. Let R" be the graded ring associated to the filtration R > B, > B, > of R. There is a canonical graded ring surjection O:k R" sending Bini to Bi/Bi+ obtained by cancelling I . We claim Ok is left Noetherian. Indeed, @I?'+, =:R = R / A is left Noetherian by hypothesis, so it suffices to prove that if ORi+ is left Noetherian then ORi is also left Noetherian. Write T = OR,,, and note from lemma 3.5.35 that ORi is generated by T and elements sI,... ,s, such that [ s i , T ] c T and [si,sj] E T for each i, j . Write q for the subring of @ki generated by T and s,,. .. ,sk. We shall show by induction that if &- is left Noetherian then is left Noetherian; clearly this would conclude the proof of the claim. A typical element of q- is a product of terms of the form rl - * . I - , , , where each ri E T u {s,, . ..,sk- }. Then [sk,r , * r,,,] = rl rj- , [ s k ,rj]rj+ * . * r m and each [ s k , r j ] E T, so [s,,r,-*.r,] E qP1. Hence [ S k , q - I ] E q - l ,so & is left Noetherian by proposition 3.5.2, establishing the claim. Combining the assertions we see R" is left Noetherian. We want to conclude k is left Noetherian,just as in corollary 3.5.32.But this is seen by copying out the proof of proposition 3.5.33 using R" instead of G,(R), in view of the fact A' E Ei c A k for all i > 2k' established earlier. Q.E.D.
zisN
--f
,
,
,
Remark
,
cj
z
3.5.37:
(i) The same proof yields a stronger result, cf., exercise 14. (ii) In the proof we gave an argument which shows that if R is left Noethe-
43.5 Left Noetherian Rings
rian then concept.
413
k is left Noetherian. This is useful in connection with the following
We say an ideal A a R has the strong Artin-Rees property if for each f.g. module M and submodule N there is some n > 0 such that N n A'M = A'-"(N n A " M ) for all i > n. The strong Artin-Rees property is focal in the study of commutative Noetherian rings. When A is generated by central elements then the usual commutative proof carries over, cf., Jacobson [80B, p.4391. However, the behavior of noncommutative rings is so poor with respect to arbitrary ideals that (taking i = n + 1) one usually examines the weaker property N n A ' M c A N for suitable i . By exercise 20 this property is equivalent to its special case L n A'
c AL
for all L < R ,
which we shall call the AR-property for the ideal A . Every polycentral ideal satisfies the AR-property, by exercise 19. The strong AR-property was used by Artin and Rees in their proof of the Krull intersection theorem, and the AR-property is relevant to the noncommutative case because of exercise 21. See Smith [8 11 for an excellent survey of the AR-property. "Almost bounded" implies the AR-property because of the next result. Proposition 3.538: To verify the AR-property for an ideal A it sufices t o show i s a large submodule of an f.g. module fi and A N = 0 then A"M = 0 that if for some n > 0.
Proof: Suppose M is f.g. and N < M. Take M' < M maximal with respect to M' n N = A N . Let M = M / M ' . Identify N I A N with ( N + M ' ) / M ' , a large submodule of M, since if 0 # M " / M ' < M we have M " n N $ A N . But A ( N / A N ) = 0 so, by hypothesis, A"M = 0 for some n; thus N n A"M E N n M' E A N as desired.
Q.E.D.
This discussion calls for a Small insertion to show that the AR-property can fail for very well-behaved rings. Example 3.5.38': (Small) A Noetherian algebra R f.g. over its Noetherian center, whose Jacobson radical fails to satisfy the AR-property. Let C be any local Noetherian integral domain, and let R = (,[I'
E),
viewing
c as
Rings of Fractions and Embedding Theorems
414
i(o))
C[I]-module via the action I c = 0. The elements of Z ( R ) have the form
( : ‘ I
for p ( l ) E ~ [ I I , so Z ( R ) x ~ [ I I Let . A=
(: J”)
where
J = Jac(C). Clearly, A is quasi-invertibleand R I A z C[I] so A = Jac(R). But
taking L = (oJ
C 4 o) c
we see AL = 0, whereas L n A’ 2 Ji-’e12 # 0.
Krull Dimension for Noncommutative Left Noetherian Rings The “localization” theory for Noetherian rings has been a rich source both of research and of headaches. The main difficulty is to determine when a prime ideal P is left localizable, i.e., when {elements of R modulo P} is a left denominator set. There has been considerable progress in the last five years, to be discussed at the end of this section and in $6.3, but localization remains a very delicate procedure. Fortunately, the reduced rank can serve as a substitute as we have already seen. Coupling the reduced rank with the noncommutative “Krull dimension,” to be defined presently, one can often bypass localization altogether. We shall look at the rudiments of Krull dimension and then study two applications, due to Stafford-Coutinho and to Jategaonkar, which illustrate how to obtain noncommutative versions of important theorems. One of the fundamental ties from commutative ring theory to geometry is Krull’s definition of the dimension of a local Noetherian ring as the height of its maximal ideal, i.e., the length of a maximal chain of prime ideals. There are several candidates generalizing this definition to noncommutative rings. Of course the first candidate is the obvious one-the maximal length of a chain of prime ideals (if this exists); this is the commutative ring theorist’s definition of Krull dimension, which we shall call the little Krull dimension, and is useful for certain classes of noncommutative rings (such as PI-rings, cf. $6.3). Various module-theoretic definitions have proved more successful for noncommutative Noetherian rings. Unfortunately there is a problem in terminology-since these dimensions were considered by Gabriel and generalize Krull dimension, the appellation “Gabriel-Krull” could be applied to each of them; worse yet, there is another dimension introduced by Gelfand-Kirillov (to be considered in Chapter 6) which is now called GK dimension! Ideally the dimension considered here would honor RentschlerGabriel [67] who invented it, and Gordon-Robson [73] who developed the comprehensive theory. However, the title of Gordon and Robson’s treatise, and the name almost universally used, is “Krull dimension.”
83.5 Left Noetherian Rings
415
Definition 3.5.39: (We are not assuming R is left Noetherian.) The Krull dimension of M, abbreviated as K-dim M, is defined as follows (via transfinite induction): K-dim(O)= - 1, and, inductively, K-dim M is the smallest ordinal a (possibly infinite) such that in each descending chain M = M, 2 M, 2 M, 2 one must have K-dim(Mi-,/Mi) < a
for almost all 1.
Let us write out this definition explicitly for low values of a. Example 3.5.40: K-dim M = 0 iff M is Artinian. K-dim M = 1 iff there is no infinite descending chain M = M, > M, > M, > ... such that each factor Mi- ,/Mi is not Artinian. Krull dimension is possibly the principal dimension used today in general noncommutative ring theory. However, the reader should be aware that its one-sidedness makes it a bit awkward in handling rings which are not fully bounded. As with most other dimensions, we define the Krull dimension of a ringR to be its K-dim as R-module. Modules need not have Krull dimension. In fact, if F is a field then M = F(“) is an F-module; let M’ = M(N),and let Mi be the submodule of M‘ consisting of all vectors whose first i components are 0. Then clearly M‘ = M, > MI > is an infinite descending chain with each factor isomorphic to M, so we would have K-dim M‘ > K-dim M. But M‘ z M; consequently, M cannot have Krull dimension. Similarly, any infinite direct product of copies of a field is a ring which does not have Krull dimension. Nevertheless, Krull dimension exists in many important cases, as we shall soon see. Remark 3.5.40’: If K-dim M > a then one can find an infinite chain M = M, > M, > ... such that K-dim(Mi-,/Mi) 2 a for each i. Proposition 3.5.41: K-dim M = sup{K-dim N, K-dim M/N} for any N IM, provided either side exists.
Proofi The inequality “2”is clear since any descending chain from N or from M/N can be transferred to M. To prove “I” we modify the proof of proposition 0.2.19. Suppose M > M, > M, > ... is an infinite chain. We need to show that if K-dim N Ia and K-dim M/N Ia then K-dim M I a, i.e., almost all factors Mi/Mi+ have K-dim < a. Letting Mi denote (Mi + N)/N we have M 2 MI 2 M 22 so
,
a * . ,
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416
,
by hypothesis almost all of the factors &$/Mi+ have K-dim < a. Likewise, almost all (Mi n N)/(Mi+ n N) have K-dim < a. But Mi % Mi/(Mi n N) so the chain Mi > Mi n N > Mi+, n N shows by induction on a that K-dim(Mi/(Mi+ n N)) < a for almost all i. Hence K-dim Mi/Mi+ < a for these i, as desired. Q.E.D.
,
,
,
It follows that K-dim R = max{K-dim M: M E R-Firnod} if either side exists, so K-dim R is categorically defined, cf,, $4.1. For later use let us refine part of the argument of proposition 3.5.41.
Lemma 3.5.41': If M, IMI IM and N I M then K-dim M1/M2 = max{K-dim(M, n N)/(M, n N), K-dim(M, N)/(M, N)}.
+
+
Proof: Let a = K-dim MJM,. There is a monic (M, n N)/(M, n N) + MJM, given by x+(M,nN) H x + M,; likewise, there is an epic M1/M,+ (M, + N)/(M, + N) given by x + M, H (x N) + (M, + N). Thus we have the inequality ( 2 )by proposition 3.5.41. The reverse inequality ( I ) follows by applying proposition 3.5.41 to the chain Mi > Mi n N > Mi+ n N. Q.E.D.
+
,
Remark 3.5.41": In analogy to proposition 3.5.41, we shall call a dimension on modules exact if dim M = max{dim K, dim MIK} for all N IM. In such a case one readily concludes dim M I max{dim K, dim N} for every exact sequence K + M + N. (Indeed, any map g: M + N gives rise to the exact sequence 0 + kerg + M + gM +O, so dim M = max{dim(kerg),dimgM}. But gM IN, and kerg is the image of K in M.) Proposition 3.5.42:
(i) I f K-dim M/N < a for all 0 # N < M then K-dim M exists and I a. (ii) If M is a Noetherian module then K-dim M exists. Proof: (i) If M = M, > M, > M, > * . . is an infinite chain then each K-dim Mi- ,/Mi I K-dim M/Mi < a, so K-dim M Ia. (ii) By Noetherian induction we may assume that K-dim(M/N) exists for each N < M. Let a = sup{K-dim(M/N): N I M}. By (i) K-dim M I a+. Q.E.D. Thus the Krull dimension forges yet another link from Artinian to Noetherian and has proved useful in low dimensions by providing classes of Noetherian rings in which a richer theory can be developed, as indi-
53.5 Left Noetherian Rings
417
cated in exercise 36-38. The following lemma provides a useful induction procedure.
Lemma 3.5.43: Suppose s E R is left regular. If K-dim R = a then Kdim(R/Rs) < a. In particular, if a = 1 then R/Rs is an Artinian module. Pro08 R > Rs > Rs2 > ... is a descending series each of whose factors Rsi/Rsitl !z R/Rs, so by definition a > K-dim(R/Rs). Q.E.D. Remark 3.5.44: (i) K-dim RIA I K-dim R for all A a R. (ii) Suppose R is left Noetherian. K-dim R = K-dim RIP for some P in Spec(R). Proof: (i) by the change-of-rings functor from RIA-&& to R-Mod. (ii) If P, . . . Pk = 0 for prime ideals Pl ,. . .,Pk of R then R > Pk 2 Pk-lPk 2
’*’
20
is a chain each of whose factors is an f.g. R/P,-module for some i and thus has K-dim I K-dimRI8, so we are done by proposition 3.5.41. Q.E.D. Proposition 3.5.45: If P E Spec(R) and P c A then K-dim(R/A) < K-dim(R/P). Pro08 Passing to RIP we may assume P = 0. Then A has a regular element s by Goldie’s theorem, and K-dim(R/P) I K-dim(R/Ra) < K-dim R by lemma 3.5.43. Q.E.D. Corollary 3.5.45’: Suppose K-dim R = n < 00. There are only a finite number of prime ideals P with K-dim RIP = n, and any such P is a minimal prime. Proofi By corollary 3.2.26 it is enough to prove P is minimal. But if P’ c P in Spec(R) then K-dim RIP < K-dim RIP’ I n, contrary to hypothesis. Q.E.D. Proposition 3.5.46: If R is prime ldt Noetherian then K-dim L = K-dim R for every nonzero ldt ideal L of R.
418
Rings of Fractions and Embedding Theorems
Proofi LR = Cfinite Lri for suitable ri in R. But LR 4R and thus has a regular element s. Since R x Rs we get K-dim R = K-dim Rs I K-dim LR Imax{K-dim Lr,} I K-dim L.
Q.E.D
The requirement that all nonzero submodules have the same K-dim occurs in the recent literature; such modules are called Krull homogeneous. There also is a very important connection between K-dim and torsion for modules over prime rings.
Proposition 3.5.47: Suppose R is a prime ring and M is an f.g. module. M is torsion iff K-dim R > K-dim M.
,
Proofi Let o! = K-dim M. (a) Write M = Rxi. Clearly, some Rxi has K-dim a. But Rxi x R/Ann xi so we are done by lemma 3.5.43 since Ann xi has a regular element. ( G ) Suppose, on the contrary, that M is not torsion. By theorem 3.2.13 there is x in M for which Annx is not large and thereby misses some left ideal L. Then L x Lx,so, in view of proposition 3.5.46, we have K-dimR = K-dim L = K-dim Lx I a, contradiction. Q.E.D.
Other Krull Dimensions Before delving into the study of K-dim, let us introduce other noncommutative generalizations of the Krull dimension and see how they compare to K-dim. Definition 3.5.48: (i) The little Krull dimension k dim(R) is the length of a maximal chain of prime ideals of R; we permit chains of any ordinal length. (ii) The Gabriel-Krause dimension of R, called “classical” Krull by Gordon-Robson and denoted cl K-dim(R), is defined inductively as follows: Take Spec-, = 0 and Spec, = {P E Spec(R): Each prime ideal properly containing P is in Specs(R) for some p < a}. Thus Spec,(R) = {maximal ideals of R}. Then cl K-dim(R) is the smallest ordinal o! such that Spec,(R) = Spec(R). (Thus P E Spec,(R) iff cl K-dim(R/P) I a,) Remark 3.5.48’: When clK-dim(R) is finite it is clearly the length of a maximal chain of prime ideals and thus equals k dim R.
53.5 Left Nwtherian Rings
419
To compare k dim(R) with K-dim R we need a couple of preliminary results about prime ideals, which in exercises 31-35 will be seen to hold for all rings having K-dim.
Proposition 3.5.49: K-dim(R). Proof:
If R is left Noetherian then kdim R 5 cl K-dim R I
Let a = K-dim(R). First we prove cl K-dim(R) Ia by showing
P E Spec,(R) for every prime ideal P. Indeed, this is clear if P is maximal. On the other hand, if P c P' in Spec(R) then a 2 K-dim(R/P) > K-dim(R/P') 2 cl K-dim(R/P') by induction on K-dim, implying P' E Spec,JR) for some /3 < a, so P E Spec,(R) by definition. It remains to show k dim(R) I cl K-dim(R). We shall show by induction on /? that if Spec(R) has a descending chain of length p then cl K-dim R 2 /3. If p = y+ then the chain ends in some P 3 P'. But clK-dim(R/P) 2 y by induction, so cl K-dim R 2 y + = jl, as desired. Thus we may assume jl is a limit ordinal. But for every y < 1 we have a chain of length y so by induction cl K-dim(R) 2 y for each y < /3, i.e., cl K-dim(R) 2 jl as desired. Q.E.D.
Of course if R is commutative Noetherian then the finite height of prime ideals shows kdim(R) Io.Yet Gordon-Robson [73] give an example of a commutative Noetherian ring with arbitrarily large infinite K-dim. Thus the next proposition shows that clK-dim can also take on any ordinal value. Lemma 3.5.50: Suppose R is a left bounded, prime left Noetherian ring of K-dim a. For any /3 < a there is P in Spec(R) with K-dim(R/P) = p. Proof: There is an infinite chain R = Lo 2 L, 2 L, 2 ... with each Lk < R such that K-dim(L,2 p for each p. By proposition 3.2.19 there is some k such that R does not have k independent left ideals. It follows at once there is i I k such that Li is large in I,,-,. Let L' be an essential complement of Li_ in R. Then Li L' also is large in R and thus contains an ideal I # 0. Then
+
/? I K-dim(Li-,/Li)
= K-dim((L,-,
I K-dim(R/(L, + L')) I K-dim(R/I) < a.
+ L')/(Li + L'))
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By induction on a there is some prime ideal PI1 of R/l such that K-dim((R/Z)/(P/l))= K-dim(R/P). Q.E.D.
fl =
Proposition 3.5.51: Suppose R is fully bounded left Noetherian. Then cl K-dim(R) = K-dim R.
Proof.- In view of proposition 3.5.49 we need only show “2”.Let a = K dim(R). Choosing Po E Spec(R) with K-dim(R/Po) = a it suffices to prove clK-dim(R/Po) 2 a, so we may pass to RIP, and assume R is prime. We induct on a. For any fl < a there is P E Spec(R) with K-dim(R/P) = p; by Q.E.D. induction jIcl K-dim(R/P). Thus cl K-dim R 2 sup(j Ia} = a. Corollary 3.5.52: If R is fully bounded left Noetherian and k dim(R) is finite then k dim(R) = cl K-dim(R) = K-dim R. Example 3.5.53: Suppose R is the Weyl algebra A,(F) over a field E Then cl K-dim(R) = 0 since R is simple, but K-dim R 2 1 since R is not Artinian. Thus we need the hypothesis of proposition 3.5.51.
In fact K-dim R = 1, cf., exercise 25, although K-dim F[A, p] = 2, cf., exercise 23. Thus K-dim is not preserved under passage to graded algebras. McConnell[84a] computes K-dim A,( T )for a wide range of rings T including uncountable fields.
CriticaI Submodules Dejnittion 3.5.54:
A module M # 0 is critical if K-dim M I N < K-dim M
for all 0 # N < M. Proposition 3.5.55: submodule.
If M # 0 has Krull dimension then M has a critical
Proofi Let a = K-dim M. Clearly, it is enough to show some submodule M’ has a critical submodule, so by induction on a one may assume K-dim M’ = a for all M’ < M. Taking M, = M we build M = M, > MI > inductively as follows. If Mi is noncritical take 0 # M,+, < Mi such that K-dim(Mi/Mi+ = a. By definition of K-dim M, this process must terminate Q.E.D. at some M,, so M, is critical. Remark 3.5.56: Suppose M is critical of K-dima. Every submodule of M is also critical of K-dima. (Indeed, if 0 # M‘ 5 M then K-dim M‘ = a
43.5 Left Noetherian Rings
421
by proposition 3.5.41; now for all 0 # N 5 M’ one has K-dim(M’/N) I K-dim M/N < a.) We say an ideal A of R is a partial annihilator of M if A = Ann N for some 0 # N IM. An assassinator of M is a maximal partial annihilator of M. Any module over a left Noetherian ring clearly has an assassinator, which is a prime ideal by proposition 3.2.23. Remark 3.5.57: Suppose M is an f.g. module over a left Noetherian ring R. Then M has a critical cyclic submodule N whose annihilator P is an assassinator of M, so, in particular, P E Spec(R). (Indeed, let P be an assassinator of M. Then P = Ann N for some N I M. Replacing N by a critical submodule we may assume N is critical; now in view of remark 3.5.56 we may replace N by any nonzero cyclic submodule.)
M Supplement: Basic Dimension One of the games in Noetherian ring theory is to find the “right” chain of submodules. In attempting to “noncommutativize” theorems we often come up against faithful torsion modules, so we are led to the following considerations. Assume throughout that M # 0 is an f.g. module over a left Noetherian ring R . Applying Noetherian induction to proposition 3.5.55 we see M has a series M = M, > M, > > M, = 0 all of whose factors are critical. We shall be interested only in the faithful critical factors. Definition 3.5.58:
A critical-unfaithful series W for M is a chain
such that each Mi/Mi+, is critical or unfaithful. (Coutinho calls this a basic series.) The basic number n(%) of % is max(K-dim(Mi/Mi+,):Mi/Mi+l is faithful (and thus critical)}; by convention we say n(%) = - 1 if each Mi/Mi+l is unfaithful. Now we define the basic dimension B-dim,M = min{n(%‘):%‘ is a critical-unfaithful series for M}. When R is unambiguous we write B-dim M for B-dim, M. The basic dimension was invented by Coutinho [86] to obtain Stafford’s noncommutative version of the Forster-Swan theorem for arbitrary left Noetherian rings, see below. Coutinho’s exposition was developed with the aid of McConnell and Robson (for use in their forthcoming book) and is the
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basis for the treatment here. 1 deeply appreciate their quick and generous communication of these results, in time to include here. A fortiori B-dim M I K-dim M since n(W) Imax{K-dim(Mi/Mi+ 0 5 i < t ) = K-dim M. On the other hand, we have
Proposition 3.5.59: (Exactness of B-dim). B-dim M = max{B-dim N, Bdim MJN} for all N IM. Proof: The inequality “5’’ is seen by juxtaposing series for MJN and N; to prove “ > ” suppose M = M, 2 . - * 2 M, = 0 is a critical-unfaithful series of M. Let Mi=(Mi + N)/N a n d & = M i n N.Then M/N = M, 2 .-.2 M,= O and N = 2 2 fit= 0. If Mi/Mi+l is unfaithful then so are M i / M i + and fii/Gi+1, so (refining these series further if necessary to critical-unfaithful series) lemma 3.5.41’ shows B-dim M/N I B-dim M and B-dim N IB-dim M. Q.E.D.
,
a,,
The following fact enables us to handle the critical part. Remark 3.5.60: Suppose N 5 N’ 5 M with N’JN critical, and L is a left ideal of R with LN’ $ N. Then K-dim(” + LM)/(N + LM) < K-dim N’IN. (Indeed, (N‘ LM)/(N + LM) x N’/(N’n(N LM)) and N < N + LN’ I N ‘ n ( N LM). Thus (N’ + LM)/(N + LM) is a proper image of N‘/N so we are done by criticality.)
+
+
+
Much of the theory will focus on prime rings. Remark 3.5.6Z:
Suppose R is prime.
,
(i) B-dim M = - 1 iff M is unfaithful (for if A,Mi- c Mi then A, ... A, c Ann M). (ii) If M is not torsion then B-dim M = K-dim M = K-dim R. (Indeed, I clearly holds at each stage. On the other hand, there is x E M such that Ann, M misses a left ideal L # 0; Take critical L‘ 5 L. Then L’ x L‘x so
B-dim M 2 B-dim L’x = B-dim L‘ = K-dim L‘ = K-dim R by proposition 3.5.46, as desired). Thus we “know” B-dim except when M is faithful but torsion, and we need a method to handle this case.
Lemma 3.5.62: Suppose R is prime and 0 # A 4 R; let n = B-dim M. Suppose x1,x2 are elements of M such that B-dim(M/(Rx, + Rx,)) < n, and
$3.5 Left Noetherian Rings
Ann,(x, a in A.
423
+ ax,) # 0 for all a in A. Then B-dim M/R(xl + ax2)< n for some
Proof: Let Ax, = M, > > M, = 0 be a critical-unfaithful series of Ax,. We shall prove the following assertion by induction on i: For each i there is a, in A for which B-dim M/(R(xl + aixz) + Mi) < n We shall conclude taking i = t. For i = 0 we have B-dim M/(Rxl + Rx,) < n, by hypothesis, and (Rx, + Rx2)/(Rx, + M,) is unfaithful (annihilated by A), so B-dim M/(Rxl + M,) < n as desired. Suppose by induction that we have found a,-,. Let x’ = x1 + ai-,x,; thus B-dim(M/(Rx’ + Mi- ,)) < n. If Mi- ,/Mi is unfaithful then so is (Rx’ + Mi - ,)/(Rx’ + Mi) and we can take a, = a, - 1. Thus we are done unless M,-l/Mi is faithful and critical. Let L = Annx’ # 0 by remark 3.5.61(ii) Then AL # 0 so ALM,-, $ Mi; hence we have a in AL and y in for which ay 4 Mi. Let xi = x’ + y. Since ay = a(x’ + y) E Rx, we see by an analogous argument to remark 3.5.60 that
K-dim((Rx, + Mi- ,)/(Rx, + Mi)) < K-dim Mi- ,/Mi I n.
,
Moreover, Rx, + Mi- = Rx‘ + Mi- 1, so B-dim M/(Rxi + MiPl)< n by induction hypothesis. Therefore, B-dim M/(Rxi + Mi) < n. But y E M i - E Ax, so y = a’x, for suitable a‘ in A, and thus xi = (x, + ai-,xz) + a’x, E x1 + Ax,, as desired. Q.E.D.
,
Proposition 3.5.62’: Suppose R is prime, A 4R, and M is a faithful torRx,. Then there is x in x1 + Ax, such that sion module with M = B-dim M/Rx < B-dim M.
Xi=,
Xi=,
X:::
Proof: The case t = 2 is the lemma. For t > 2 there is x’ in x1 + Ax, such that B-dim M/(Rx’ + Rx,) < B-dim M. (Indeed, this is obvious for MIRx, unfaithful since then B-dim M/(Rx, + Rx,) = - 1 < B-dim M; for M/Rx, faithful we apply induction to MIRx,.) Applying the lemma to Q.E.D. Rx‘ + Rx, yields the result.
To use the above results we shall pass to M/PM as RIP-module, for each P in Spec(R); accordingly define B-dim(M, P) = B-dim,,, MIPM. If 9’ c Spec(R) define B-dim(M, 9’)= max{B-dim(M, P):P E 9’). Thus B-dim(M, 9’)IB-dim M IK-dim M IK-dim R. Also recall that MIPM z R/PQ M; since RIP Q - is right exact, we conclude via remark 3.5.41“ and proposition 3.5.59 that if N I M then B-dim(M/N, P) I B-dim(M, P) Imax{B-dim(N, P), B-dim(M/N, P)}.
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Primitive ideals come into play because of the following easy fact.
Remark 3.5.62": B-dim(M, P) > - 1 for some primitive ideal P of R. (Indeed, by proposition 0.2.15 M contains a maximal submodule N; letting P be the primitive ideal Ann, MIN we see B-dim,,, MIPM 2 B-dim,,, MIN > - 1 by remark 3.5.61(i).) Proposition 3.5.63: Suppose K-dim R < 00, and Y E Spec(R) satisJies the following two properties: (1) Each primitive ideal of R is in 9'. (2) If Q E Spec(R) is an intersection of prime ideals from Y then Q E 54 Then {Q E 9:B-dim(M, Q) = B-dim(M, Y ) }is Jinite.
Proofi Since B-dim(M, 9) # to prove
-1
by remark 3.5.62" it suffices a fortiori
Lemma 3.5.63': Suppose Y G Spec(R) satisjes property (2) above. Given an K-dim RIP > a } and Fa = {Q E 9: K-dim RIQ = u ordinal a let .4p,= { P E9: and B-dim(M, Q) 2 max { 0, B-dim(M, Then Fa is Jinite.
x)}.
Proof of lemma: Suppose, on the contrary, Fais infinite, and let P be an ideal maximal among those intersections of infinitely many distinct prime ideals of 5 ;write P = ni,,Qi with Qi E 5 ,and I infinite. Then P also is prime; indeed if AB E P then letting I' = {i:A E Qi} and I" = {i:B E Qi} we have I' u I" = I so we may assume I' is infinite, and thus by assumption P = ni,,,Qi 3 A. We aim for a contradiction from the assumption I is infinite. Passing to RIP and MIPM we may assume P = 0, i.e., R is prime. Hence each Qi # 0 so K-dim R > K-dim(M, Qi). On the other hand, each B-dim(M, Qi) 2 0 by assumption so Ann, M/QiM = Qi by remark 3.5.61(i) implying Ann, M E Qi = 0; thus M is faithful, i.e., B-dim M 2 0. Let n = B-dim M. Note K-dim R > K-dim R/Qi = a. Thus 0 E Ya,so n IB-dim(M, 9,) s B-dim(M, Qi). Let A4 = Mo > M, > ... > M, = 0 be a critical-unfaithful series for M. Put A j = Ann Mj-JMj and Nij = Mj + QiM for 0 I j I t. Then Ni, = M 2 Nil 2 2 Nit = Q,M, so max{B-dim(Ni,j-l/Nij,Qi):1 Ij I t } = Bdim(M, Qi) 2 n. We search for the j for which B-dim(Ni,j - l/Nij,Qi) 2 n. If Aj = 0 then Mj- JMj is critical and remark 3.5.60 shows K-dim(N,,jN,j) < K-dim Mj- 1Mj In.
niEN
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+
If A j $ Qithen Qic Aj Qic Ann(N,,j so B-dim(Ni,j - l/Nij,Qi) = -1
M Supplement: The Noncommutative Forster-Swan Theorem Let gR(M)denote the minimal number of elements which span a given f.g. R-module M; we write g(M) if R is unambiguous. How can one estimate g ( M ) ? There is no obvious answer; indeed, it is very difficult even to determine whether a module is cyclic unless we are given its generator. O n the other hand, a good estimate of g(M) would help us greatly in proofs by induction on the number of generators, and the theory to be described presently has important ramifications in K-theory, to be discussed in 55.1. If R is commutative then writing gp for gR‘where P E Spec(R) and R’ = R,, we certainly have g(M) 2 gp(Mp)= gp(Mp/PpMp)by Nakayama’s lemma; the latter number is merely the dimension of Mp/PpMp as vector space over the field Rp/Pp.Thus Nakayama’s lemma provides a lower bound for g(M). An upper bound is given by the Forster-Swan Theorem: Suppose M is an f.g. module over a commutative ring C whose classical Krull dimension is finite. Then y(M) I max{gp(Mp) + K-dim C/P: P E Spec(C)} I
max{g,(M,): P E Max Spec(C)} + K-dim C.
(Recall that K-dim C equals the classical Krull dimension, by proposition 3.5.5 1.) Warfield generalized this to fully bounded Noetherian rings, and Stafford [81] managed to extend Warfield’s result to arbitrary Noetherian rings. Stafford’s proof is very complicated, and a shorter proof of a slightly diluted result appears in Stafford [82]. Recently Coutinho revamped Stafford’s proof by means of the basic dimension and also further generalized the theorem to all left Noetherian rings. Our exposition follows Stafford [82], as modified by Coutinho’s methods. The first task is to find a suitable substitute for gp(Mp) since localization may not be available. As usual the reduced rank is useful, and we define !?(M,
= PR/P(M/pM)/PR/P(R/p),
sometimes called the normalized reduced rank of MIPM). Note we only need
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the reduced rank over a prime ring. It is easy to see in the commutative case that i(M, P) = gp(Mp), but in the noncommutative case $(M, P) need not even be an integer. (Take R = M,(F), P = 0, and M a minimal left ideal.) Thus we define g(M,P) is the smallest integer 2 i(M, P). Note that g(M, P) is relatively easy to compute, being the minimum number of generators of Q @R/p(M/PM) as Q-module, where Q = Q(R/P) is the simple Artinian ring of fractions of RIP. (This is seen through the composition length. Q has a unique simple module M, up to isomorphism, and so Q BRa(M/PM) has the form Mg’ where t = pRlp(M/PM).Letting m = P ~ , ~ ( R / Pthe ) , composition length of Q, we see that Mt’ x Q[’/”’] @ Q’ where Q’ is a module image of Q and is thus cyclic. Hence the number of generators needed for Mg) is the smallest integer 2 t/m, i.e., g(M/P), as desired.) On the other hand, we can tie g(M, P) to other notions we have studied. Remark 3.5.64: g(M, P) = 0 iff M/PM is torsion as (RIP)-module. (Indeed, passing to M/PM and RIP we may assume P =O. Now g(M,O) = 0 iff $(M, 0) = 0 iff p(M) = 0.) The main theorem which we aim for is
Theorem 3.5.65: (Warfield-Staflord-Coutinho) Suppose R is leji Noetherian and M is an f.g. R-module. Then g(M) I max{g(M, P)
+ K-dim RIP: P E Spec(R)}.
In all of the following results we assume R is leji Noetherian, and M is an f.g. R-module. The reader might find it instructive first to work out exercise 44,a special case. We start with the following useful substitute for proposition 3.5.62‘.
Lemma 3.5.66: Suppose R is prime, and 0 # A Q R. For any R-module and any xl, xz in M there is a in A such that g(M/R(x, + ax,),O) = max{g(M/(Rx, + Rxz),O),g(M,O)- I}. Proofi ( 2 ) is clear, so we prove ( I )Of . course, we can transfer the assertion to the reduced rank and want to show p(M/R(x, + axz)) I max{p(M/Rx, Rxz)),pM - pR} for suitable a; we choose x’ = xl + axz such that p(Rx’) is maximal possible. If Annx’ = 0 then pM - pR = pM - p(Rx‘) = p(M/Rx’) so we are done.
+
$3.5 Left Noetherian Rings
427
Thus we may assume Ann x’ # 0. Now p(Rxi+ Rx2) = ~ ( R x ’+ ) p((Rx1 + Ax,)/Rx’)
+ p((Rx,+ Rx,)/(Rx, + AX,)).
The last term is 0 since Rx2/Ax2 is torsion (for any nonzero ideal of a prime ring is large.) Hence it remains to show (Rx, + Ax,)/Rx’ is torsion. But Rx, + Ax, = Rx‘ Ax, so we are done unless Ax, has a uniform torsionfree submodule N # 0 with N n Rx‘ = 0. Then (Annx’)N # 0 so r’a‘x, # 0 for suitable r’ in Ann x’ and a’xl in N. Now r‘a‘x, E N n R(x‘ + u’x,) so (Rx’ + N)/R(x’ + a’x,) = (R(x‘+ a’x,) + N)/R(x’ + a’x,) is torsion since N is uniform. But this says p(R(x’ + a’x,)) = p(Rx’) + p(N) > ~(Rx’),contrary to the choice of x’. Q.E.D.
+
c:=
Lemma 3.5.66‘: Suppose M = Rxi, P E Spec(R), and A Q R with A $-L P. Ifg(M, P) > 0 then there is x in x1 + Axi such that
I:=,
g(M/Rx, P) = g(M, P) - 1. Proofi Passing to RIP and M/PM one may assume P = 0. Now iterate the previous lemma. Q.E.D.
Applications of these results are given at the end of 55.1. We come to a critical definition. Definirion 3.5.67: b(M, P) = B-dim(M, P) + g(M, P) unless g(M, P) = 0, in which case b(M, P) = B-dim(M, P) + 1. For Y c Spec(R) we define b(M, 9’)= max{b(M,P): P E .Y} Let us make several computations. Remark 3.5.68: (i) When g(M, P) 2 1 then M/PM is not RIP-torsion by remark 3.5.64; thus B-dim(M, P) = K-dim RIP by remark 3.5.61(ii) so
b(M, P) = K-dim RIP + g(M, P). In particular, b(M, P) > 0 in this case. (ii) When B-dim(M, P)= - 1 then M/PM is RIP-unfaithful so g(M, P)=O and thus b(M, P) = 0. (iii) For all P we have b(M, P) I K-dim RIP + g(M, P). (Indeed, by (i) we may assume g(M, P) = 0; then M/PM is torsion so K-dim(M/PM) < K-dim RIP by proposition 3.5.47, implying b(M, P) = B-dim(M, P) + 1 I K-dim(M/PM) + 1 I K-dimR/P.)
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Lemma 3.5.69: Suppose Pl, . . .,P, E Spec(R) and M = ci= Rxi with 0 < b(M,<) < CQ for 1 I i I m. Then there is x in x1 + x : = 2 R ~such i that b(M/Rx,e) I b(M,Pj) - 1
for 1 I j I m.
Let b, = b(M,P,) - 1. We order the 4 such that 4 $ pk if j < k, and we proceed by induction on m, assuming there is x’ in x1 + Rxi P,.i such that b(M/Rx‘,4) Ib, for 1 I j < m. Let A = PI n...nP,-l $ If g(M, P,) # 0 then by lemma 3.5.66’ (taking P = P,,,)there is x = x’ + a,x, such that g(M/Rx, P,) I g(M, P,) - 1. Hence b(M/Rx, P,) I b,. But for all j c m we have b(M/Rx, 4) = b(M/Rx‘, 4) I bj, as desired. Thus we may assume g(M,P,) = 0, so M/P,M is RIP,-torsion by remark 3.5.64. Moreover, B-dim(M/P,) + 1 = b(M, P,) > 0 by hypothesis. Hence M/P,M is RIP,-faithful by remark 3.5.61(i), so we conclude as before by means of proposition 3.5.62’. Q.E.D. Proofi
To limit the number of “extremal” P we shall need to appeal to proposition 3.5.63. Accordingly we define J-Spec(R) = { P E Spec(R): P is semiprimitive}, the smallest subset of Spec(R) satisfying properties (1) and (2) of proposition 3.5.63. (Of course, if R satisfies the weak Nullstellensatz then J-Spec(R) = Spec(R).) Proposition 3.5.70: Assume K-dimR c CQ. Let Y = J-Spec(R). There are only finitely many P E Y for which b(M, P) = b(M, 9). Let % = { P E 9’K-dim : RIP > n} and % = { P E 9: K-dim RIP = n and b(M, P) 2 max{ 1, b(M,x)}}. Note % = 0 unless n 5 K-dim R; thus we are done unless some % is infinite. Among all possible f.g. modules which are counterexamples we choose M such that n is as large as possible with % infinite, and within this constraint such that b(M,%) is minimal possible. In particular < + 1 , < + 2 , . . . are each finite; thus Un,,,,<, is finite, and we list it as {Pl,. .. ,P,}. Note if P E % and 1 I b(M, P) = b(M,%) then P E %, for some n’ > n and thus P is one of the 4. By lemma 3.5.69 we have some x in M such that b(M/Rx,4) I b(M,4) - 1 for each 4. Let M’ = M/Rx. Thus b(M’,%) 5 b(M,%) - 1, so we reach a contradiction (replacing M by M’) if we can show b(M’, P) 2 max{ 1, b(M, P) - l } for an infinite number of P in %. Let S L = { P E %:g(M, P) 2 2}. Clearly g(M’, P) 2 g(M, P) - 1 2 1 for P in S k so remark 3.5.68 shows b(M’, P) 2 K-dim RIP + g(M, P) - 1 2 max{ 1, b(M, P) - I}; thus it suffices to show S;is infinite. Proofi
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In other words, we need to show there are only finitely many P in with g(M,P)I 1. But such P have b ( M , P ) = B-dim(M,P) 1 so B-dim(M,P) 2 max{O,B-dim(M,%)}, and we are done by lemma 3.5.63’. Q.E.D.
+
We are now ready to prove a strong version of the Warfield-StaffordCoutinho result. Definition 3.5.71: The stable number of generators s(M) is the smallest Rxi for t > s then number s such that if M =
c:=
.z + r-I
M =
i= 1
R(xi
rix,)
for suitable ri in R.
Compare with the stable range of R, cf., exercise 2.9.15. Obviously g ( M )I s ( M ) ,so theorem 3.5.65 is implied by Theorem 3.5.72: (StafSord [82, theorem 2.63, Coutinho [86, theorem 3.23) Suppose R is a left Noetherian ring of jinite K-dim, and M is an f.g. R-module. Then s ( M ) I b(M,J-Spec(R)) I max{g(M, P)
+ K-dim(R/P):P E J-Spec(R)}.
Proofi Let Y = J-Spec(R). The second inequality is by remark 3.5.68(iii). We prove the first inequality by induction on b = b ( M , Y ) . Note b > 0 by remark 3.562‘‘.Suppose M = Rxi for t 2 b + 1. By proposition 3.5.70 there are only finitely many Pl,. . . , P , in Y for which b ( M , P ) = b ( M , Y ) , and by lemma 3.5.69 we have x = x 1 r)xi for suitable ri in R such that b ( M / R x ,4) I b(M,6)- 1 for 1 I j I m. Write for the image in M = M/Rx. b - 1 so by induction s ( M ) I b ( M , Y ) = b - 1. But M = Then b ( M , Y ) I R i i is generated by t - 1 elements so we have M = R(Zi + riX,) for suitable r 2 , .. . ,rr in R. Hence M = Rx + R(xi + rixr) so writing x, = r”x + C:: r)‘(xi+ rixr)we have .,- 1x, = r’’(x, + r:xi) ry(xi + rixI)
I:=, +I:=
I:=,
c:::
2
+1
t
i=2
= r”xl
+
i=2
(2 r ri + ry)( x i + rixr) 1’
i=2
1:::
c:::
I
I r”r;rix,) + r”rixr). x, E If:: R(xi + rix,), implying -
1-1
i=2
Letting r , = rl r:ri we conclude M = R(xi + rix,) as desired. Q.E.D.
1;:
Taking M = R in theorem 3.5.72 yields s(R) I K-dimR + 1, a noncommutative “stable range” theorem. But there is a direct proof of this fact.
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Theorem 3.5.72': Stable range theorem (Staford [81a]). Suppose R is left Noetherian with K-dim(R) < n. I f R = XI=, Ra, then there are ro,. . . , r n - , in R for which R = R(ai ria,).
xyii
+
Proof: We prove the following more general statement: Claim: If R = L + Ra + are ri in R for which R = L
~~~~
Ra, and K-dim((L + Ra)/L)c n then there
+ xI1,' R(ai + ria).
Clearly this yields the theorem when we let L = 0 and a = a,. To prove the claim we assume by Noetherian induction that L is a maximal counterexample; also we proceed by induction on t = K-dim((L + Ra)/L). If a E L the assertion is trivial, so assume a 4 L. Let P be an assassinator of (L + Ra)/L. By remark 3.5.57 P = AnnL'/L for L'/L cyclic critical. Then L'3Lso R = L'
n-1
+ C R(ai + ria) i=O
for suitable r: in R.
+
Writing a: = ai + ria and L' = L Ra' for suitable a' in Ra we have R = L + Ra' + Rai, yielding the following reduction: We may assume (L + Ra)/L is critical whose annihilator P is prime. Let Ki = La;'. Let us reduce further to the case P = K j for each j. First assume P $ Kj. Then L + Paj =I L so by Noetherian induction we have R =L
+ Paj +
-x- + n- 1
R(ai
rya)
for suitable r:' in R.
i=O
But P U G L implies L+Paj=L+P(aj+r;'a), and thus R=L+CR(ai+r:'a) as desired. Thus we may assume P c Kj for each j . Suppose K j $ P. Then there is d in Kj - P,so dra 4 L for suitable r in R. Putting Lo = L
+ R(aj + ra) 2 L + Rd(uj + ra) 2 L + Rdra =I L
+
we note (Lo + Ra)/Lo x Ra/(Lon Ra) is a homomorphic image of Ra/((L Rdra) n Ra) x ( L Pa)/(L Rdra). Since ( L Ra)/L was assumed critical we have
+
t > K-dim&
+
+
+ Ra)/(L + Rdra) 2 K-dim&, + Ra)/Lo.
On the other hand, Lo+Ra+
i+j
Rai=L+R(aj+ra)+Ra+
C Ra,?
i+i
L+Raj+Ra+
ifj
Ra=R,
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431
so by induction on t we have ri in R for which R = Lo + Ci+jR(ai+ r p ) , yielding the desired result at once. Thus we are done unless P = K j for all 0 I j In. Now right multiplication by a, yields an isomorphism R I P z ( L + Ra,)/L. But P = PR = P(L + Ra + x u i ) G L so writing dim for the uniform dimension as RIP-module, we have dim RIP = dim(L + Ra,)/L = dim(L + Ra,)/P - dim LIP Idim(L + Ra,)/P 5 dim RIP, implying equality holds thoughout. Hence dim LIP = 0 implying L = P ; also ( L + Ra,)/P is large in RIP, implying ( L + Ra) n ( L + Ra,) 3 P = L. Now take Lo = L + Ra,. Then K-dim(L,
+ Ra)/L, = K-dim((L + Ra)/L, n ( L + Ra)) < t
+ xi,,
and Lo + Ra Ra, = R, so by induction we have r , , . . .,rn- for which R = Lo + 1 ;:; R(ai + ria),and so we conclude by taking ro = 0. Q.E.D. Digression: The Stafford-Coutinho theory was not presented in full generality, in order to make the proofs more palatable. We list several useful generalizations, proofs mutandis mutatum. (i) Noting an R-module is f.g. iff it is an image of some R‘”),we fix instead any projective R-module F and define gF(M)= inf{n E N: there is an epic F(”) -+ M } . In particular g R ( M )= g ( M ) . In fact, Stafford and Coutinho used this version in their papers, and it is needed at times in the study of projective modules. (ii) The proofs do not use the full strength of R Noetherian, but rather only that R satisfies ACC(semiprime ideals), and all prime images of R are left Goldie; also each image of MIPM (as RIP-module) is required to have finite K-dim and possess a critical-unfaithful series, for all P in J-Spec(R). (iii) ( S . Dahari[87]) Having freed ourselves of the Noetherian condition on R, we can replace K-dim and “critical”, respectively, by Gabriel dimension and “a-simple’’ (both introduced at the end of this section). The hypotheses of 11, modified in this way, are satisfied by f.g. modules over affine PI-algebras, cf., 56.3, and Dahari extends the Stafford-Coutinho theory to this situation. As stated earlier the normalized reduced rank g(M, P) is a good way to bypass localization. This raises the question of whether g(M,-) has the desirable property of being locally constant (on Spec(R)) for a given f.g. module M . When M is not required to be projective this property can fail even in the
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commutative case, as seen by taking R to be the polynomial ring F [ 1 1 , 1 2 ] and M = RI, + RA,; then i ( M , P) = 2 for P = M and @(M,P) = 0 for all other maximal ideals. Nevertheless, there is an interesting result of WarfieldGoodearl-Stafford-Coutinho:
Theorem 3.5.72": "Patch continuity theorem". If R is lefi Noetherian and M is an J g . R-module then the function Spec(R) + Q given by P I+ G(M,P ) satisjes the following "continuity" property: Given Po in Spec(R) and E > 0 we can find an ideal A 3 Po such that I@(M,P ) - i ( M , Po)l c E for every P in Spec(R) satisfying Po E P and P 8 A. Pro08 Using basic dimension the proof becomes very easy. Passing to R/ Po we may assume Po = 0 and so R is prime. Let a = $(M,O); we need to find 0 # A 4 R such that Ii(M, P) - a1 c E whenever P A. It suffices to find O#A,,A,a Rsuchthat~(M,P)ca+~forallPg2A1and~(M,P)>a-~ for all P 8 A,, for then we take A = A , A , . First we find A,. Of course we could take A, = Ann M if M were unfaithful, so we may assume M is faithful. Write a = rn/n for rn,n in N and take u in N such that un > 2 e - l . Then G(M(""),O)= urn so letting Q be the classical ring of fractions of R we see Q Q R M("")= Q(""');we take a base {s-l 8 xi: 1 I i Iurn} where each xi E M'""), and s E R is regular. Letting N = Rxi we see M'""'/N is torsion, so by proposition 3.5.62' there exists x in M("")for which B-dim(M("")/(Rx + N))c B-dim M. Applying induction on B-dim and noting i(M("")/(Rx + N),O) = 0, we have 0 # A, 4 R such that
XT:,
+
@(M("")/(Rx N ) , P ) c 1 But Rx hence
+N
is generated by urn
i(M(""),P) Ii(M'""'/(Rx
for all P 8 A,.
+ 1 elements so ~ ( R +x N , P ) 5 urn + 1;
+ N ) , P) + ~ ( R +x N , P) I1 + urn + 1
implying i ( M , P) I (urn + 2)/un < a + E for all P g2 A,. To find A , we use a trick. Take an exact sequence 0 -+ K + R(') -+ M + 0. By the previous argument there is A, such that G(K,P) < $(K,O) + E for all P @ A,. But then
i ( M , P ) 2 t - i ( K , P ) > t - g^(K,O)- E = a - E for all P 8 A , .
Q.E.D.
The term "Patch continuity" is used because i ( M , - ) is continuous when
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$3.5 Left Noetherian Rings
Spec(R) is endowed with the appropriate “patch” topology, as discussed in Goodearl [86].
J Supplement :Ja tegaonkar Theory As promised earlier, we return to Jategaonkar’s proof of Jacobson’s conjecture for almost fully bounded rings. The proof introduces various general techniques which are useful in studying other aspects of Noetherian rings, so it is well worthwhile for those who just have successfully struggled through the deep results of Stafford-Coutinho to wade in still further. Some of Jategaonkar’s results have been discovered independently by the British school, notably Lenagan and Brown. Although Jategaonkar [81] rarely makes explicit use of K-dim, some of his claims are easily verified using it. We shall use bimodules throughout, but annihilators are taken only from one side as always. Jategaonkar [82] has introduced a bimodule Krull dimension which extends these results further and recently has published expositions of his localization theory. After proving Jategaonkar’s theorem we shall discuss how the ideas apply to localization. J’M = 0. It Let us start by sketching a general approach for proving turns out that bimodules are easier to handle, so motivated by the proof of theorem 3.5.28 we make the following definitions:
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Definition 3.5.73: A bimodule M is subdirectly irreducible if M has a unique minimal nonzero sub-bimodule. A ring R is subdirectly irreducible if R has a unique minimal nonzero ideal. Remark 3.5.74: Any bimodule (resp. ring) M is a subdirect product of subdirectly irreducible bimodules (resp. rings). (Indeed, for each 0 # x in M take a sub-bimodule M(x) of M, maximal with respect to x # M(x). Then M(x) = 0, so M is a subdirect product of the M/M(x), each of which is subdirectly irreducible since any nonzero sub-bimodule of M / M ( x ) contains the image of RxR.) This can be phrased more generally in universal algebra, also, cf., exercise 2.2.1.
nxeM
Remark 3.5.75: Suppose M is a subdirectly irreducible bimodule, and J = Jac(R). If n i e N J i M= 0 then J’M = 0 for some i. (Indeed, otherwise, M, c J ‘ M for each i, where M, is the unique minimal nonzero sub-bimodule, so M, c n J ’ M ) .
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Remark 3.5.76: If M is a simple R-bimodule (in the sense that 0 is the only proper sub-bimodule) which is f.g. as left R-module then P M < M for some primitive ideal P, by proposition 2.5.25, implying P M = 0.
Armed with these preliminaries !et us outline a plan for proving Jacobson’s conjecture for f.g. bimodules (or modules, for that matter). Applying Noetherian induction to remark 3.5.74 we may assume M is subdirectly irreducible with minimal nonzero submodule M , . By Noetherian induction J ’ ( M / M , ) = 0, and P M , = 0 for some primitive ideal P by remark 3.5.76. If somehow M / M , were subdirectly irreducible then some J’M z M , by remark 3.5.75, so J’+’M E PJ’M = 0 and we would be done. On the other hand, if R were almost bounded then we would have Ann, M # 0 so applying Noetherian induction to the ringR/Ann, M we would be done. Unfortunately, almost fully bounded rings have not been proved to be almost bounded, so we have to worry about primeness of R. This leads us to wonder about finding a chain of sub-bimodules each of whose annihilators is a prime ideal, and Jategaonkar [Sl] can be viewed as an attempt to squeeze the most information from such a chain. In analogy to the Stafford-Coutinho theory, Jategaonkar must cope with the discrepancy between torsion and unfaithful. However, the issue can be sidestepped here, by means of bimodules, as we see now.
n
Proposition 3.5.77: Suppose M is an R - T bimodule which is f g . as right module over T, and R is prime Goldie. M is torsion’ i Ann, M # 0. (e) Ann, M Q R and so contains (a) Write M = xfinitexiT.Each Ann,xi
Proof:
thus is large in R, so Ann, M =
n
a regular element. contains a regular element and Ann xi is large. Q.E.D.
Also relevant to the discussion is proposition 3.5.47 which says (for R prime) an R-module M is torsion iff K-dim M < K-dim R. We shall use these two results throughout. We say a bimodule M is Noetherian if it is Noetherian both as a left module and as a right module. We assume throughout that M is a Noetherian R-T bimodule, and write N IM to denote N is a sub-bimodule. Many ideas from module theory can be transferred to bimodules. For example, if N is an R-submodule of M then N T IM and Ann, N = Ann, NT. Thus the assassinator P of M is also the “bimodule” assassinator, i.e., P = Ann N for suitable N < M. In analogy to the module definition, we say M is a left critical bimodule if K-dim, M / N < K-dim, M for all 0 # N < M .
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Definition 3.5.78: A Noetherian bimodule M is a (left) cell if M is left critical with ASS, M = Ann, M. Right cells are defined analogously. A bi-cell is a right and left cell. Let us record some straightforward properties of cells. Remark 3.5.79: Suppose R is left Noetherian, T is right Noetherian and M is a cell. Let P = Ann, M. (i) Any sub-bimodule N # 0 of M is a cell, P = Ann, N, and K-dim N = K-dim M by remark 3.5.56. (ii) M is torsion-free as RIP-module, by proposition 3.5.77 applied to the torsion R-submodule (which is clearly a bimodule); consequently, K-dim M = K-dim(R/P) by proposition 3.5.47. (iii) M/N is a torsion RIP-module for every 0 # N I M , by proposition 3.5.47.
In analogy with remark 3.5.57 we see every Noetherian bimodule contains a cell, which in turn contains a right cell. In view of remark 3.5.79(i) we conclude that every Noetherian bimodule contains a bi-cell, yielding in turn Proposition 3.5.80: Any Noetherian bimodule has a chain M = Mo 2 M1 2 2 M, for which each Mi-,/Mi is a bi-cell. Proof:
Take a cell N in M and apply Noetherian induction to M/N. Q.E.D.
There is a slight ripple in the uniform dimension theory since a cell need not be uniform as left module. Let us say N I M is large if N n N' # 0 for every 0 < N' I M, and M is biuniform if every nonzero sub-bimodule is large, i.e., M is uniform as R 0 Top-module. Proposition 3.5.81: Every biuniform bimodule M has a unique maximal cell. Proofi It suffices to show if N , and N, are cells in M then N, Let a = K-dim N,. Since N, n N, # 0 we have a = K-dim N, and K-dim(N,
+ N2)/N1 = K-dim N,/(N,
+ N, is a cell.
n N,)
< a.
+ N2 we have K-dim(N, + N2)/N 5 K-dim(N, + N2)/(N, nN) I max{K-dim(Nl + N2)/N1, K-dim N,/(Nl nN)} < a. But now for any 0 # N < N ,
Q.E.D.
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Theorem 3.5.82: ( i ) Every cell is biuniform. ( i i ) Any Noetherian M has a large bisubmodule @:= Mi where each Mi i s a cell. Furthermore, i f Nj is large in M with each 5 a cell then u = t and there i s a permutation '~tin Sym(t) such that Ann, Mi = Ann, Nni and Ann; Mi = Ann; Nnifor 1 5 i 5 t.
,
Proofi (i) If M is a cell and N 1 0 N, = N I M with N , # 0 then remark 3.5.79(i) shows K-dim M = K-dim N I K-dim M I N , , implying N2 = 0 since M is critical. (ii) Take a sub-bimodule maximal of the form M' = @:= Mi; M' is large since otherwise an essential complement would contain a cell, contrary to is also large in M. Taking injective hulls assumption. Now suppose in R@ T'P-Aod we have @ E ( M i ) x E(M) % @E(A$), so by theorem 2.10.33 u = t, and rearranging the 4.we may assume E(&) x E(Mi) for each i. Letting f:E ( M i )-+ E(N,)be the given bimodule isomorphism we have Mi nf - I N , # 0 by proposition 3.3.0 and theorem 2.10.20, so letting = Ann, Mi we have
@r=
8 = Ann,(Mi
,
4
nf -'N,) = Ann,( fMi n N,) = Ann, Ni
by remark 3.5.79(i). Likewise, Ann; Mi = Ann;&. for I I i 5 n.
Q.E.D.
,
If M,, ...,M, are cells for which @f= Mi is large in M we call { M,, ...,M,} a complete set of cells. We shall now bring in the hypothesis of almost fully bounded, which enters in the following way: Remark 3.5.83: Suppose R is almost fully bounded and N is a large submodule of M such that P = Ann, M E Spec(R). Then P = Ann, N , seen by viewing N, M as RIP-modules.
Theorem 3.5.84: If R is almost fully bounded then every biuniform R - T bimodule M has a unique (left) assassinator P, and P = Ann M, for any cell M, in M. Proofi Let M, be a cell in M, and P = Ann, M,. Then P is an assassinator, so it remains to show P is unique. Suppose, on the contrary, P' = Ann N is an assassinator. Then P' = Ann N T , but Mo n N T # 0 since M is biuniform so P = Ann Mo = Ann(M, n N T) 2 Ann N T = P'.
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437
+
Let No = Mo NT. Then P’ G Ann N o E Ann N T = P’ so Ann No = P‘. Take an essential complement M b of M, in N as left module, and let A = Ann,( N o / Mb). Then M, n A N , = 0, so A No = 0, implying A = PI. But M, is a large submodule of N,/Mb so Ann M, = P‘ by remark 3.5.83,proving P = P‘. Q.E.D.
Corollary 3.5.85: If R is almost fully bounded then every (left)assassinator of a bimodule M is the annihilator of a cell. Proofi If P is an assassinator of M then replacing M by a suitable subbimodule one may assume P = Ann M; further replacing M by a biuniform sub-bimodule we may apply the theorem. Q.E.D.
The final ingredient is a relation on Spec(R) which has become increasingly important in the study of left Noetherian rings. Definirion 3.5.86: P E Spec(R) and Q E Spec(T) are bonded primes if there is an RIP - T/Q bimodule which is f.g. and torsion free on both sides. More precisely, an f.g. R-T bimodule M bonds P and Q (via f.g. sub-bimodules M, < M,) if PM, + M,Q I M,, and M,/M, is torsion-free both as left RIP-module and right T/Q-module. (“Bonded” is not to be confused with “bounded”!) Remark 3.5.86‘: In view of proposition 3.5.77 (applied to the torsion submodule on both sides), we see that if there is some ideal A with PQ E A c P n Q, such that ( P n Q ) / A has left assassinator P and right assassinator Q, then R bonds P and Q via the RIP - R/Q bimodules A
Lemma 3.5.87: Suppose R is almost fully bounded, and M, is a maximal cell in a biunform R - T bimodule with P = Ann M,. Furthermore, assume MI,. . . ,M, I M such that { M i = MJM,: 1 Ii It> is a complete set of cells in MIM, such that Ann Mi = Ann N whenever Mo c N IMi.Then P and Ann Mi are bonded primes, for each i.
e=
In the following proof “large” and “essential complement” are as left modules. Proofi We may assume P # 4. Let Mi be an essential complement of M, in Mi. If MI # 0 then P = Ass MI by theorem 3.5.84;but Mi is large in M i so P = 6 by theorem 3.5.84,contradiction.
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Thus we may assume M o is large in M i . Let us collect more information. Write a for K-dim RIP and ai for K-dim RIP,. By proposition 3.5.77 and 3.5.47 we have a = K-dim M , and ai = K-dim Mi, implying K-dim Mi = max{ai,a}. On the other hand, by hypothesis, Mi is not left critical (since M , was chosen maximal such), so there is 0 < N I Mi for which K-dim Mi = K-dim M J N . Replacing N by N n M , we may assume N IM o , so max{ai,a} = K-dimMi/N = max{ai,K-dimM,/N} I max{ai,a - l}, from which we conclude ai 2 a. Thus K-dim Mi = a,. Let A = Ann Mi E pi n P. We claim A # pi A P. Indeed, if A = 4 n P then piPMi E A M , = 0, so Ass PM, 2 4. But P is the unique assassinator of M, by theorem 3.5.84, so P 2 pi and thus A = pi n P = pi; hence pi = P by remark 3.5.83, contrary to assumption. Having proved A c pi n P we shall finally show P and pi are bonded primes. Indeed, Ppi M c P M , = 0 yields P ( 4 n P ) +(pinP)pi c Ppi c A. In view of remark 3.5.86‘ it remains to show (pinP ) / A has left assassinator P and right assassinator pi. Indeed letting B = AnnA’/A for some A c A’ E (pinP ) we have BA’M, c AM, = 0. But A’Mi # 0 so B E ass Mi E P, proving B = P. Similarly, suppose B’ = Ann’ A’ for some A c A’ E (pi n P). Putting N = Mo + B’M, we see Ann N z A‘ 2 A = Ann M i , so by hypothQ.E.D. esis, N = M,, i.e., B’M, E M,; thus B’ = pi as desired. We are now ready to attack Jacobson’s conjecture for almost fully bounded rings.
Lemma 3.5.88: Suppose R is a prime Noetherian ring and the R-T bimodule M is a cell with Ann, M = 0. R is a subdirectly irreducible ring iff M is a subdirectly irreducible bimodule. Proofi (3) Let A be the unique minimal ideal of R . Applying proposisition 3.5.77 to remark 3.5.79(iii) we.see A ( M / N ) = 0 for all 0 # N IM, implying A M is contained in every nonzero sub-bimodule. (e) Replacing M by its unique minimal sub-bimodule, we may assume M is simple as bimodule. Let m and n be the respective uniform dimensions of R and M . Let M ’ be the (M,,(R),M,,(T))-bimodule of n x m matrices with entries in M. Since the uniform dimensions of M ‘ and M,,(R) are the same, we can embed M’ in M J Q ) where Q is the ring of fractions of R . Thus M’ = M,,(R)xi and writing x i in M,,(R)s-’ for suitable regular s we have
xfinite
439
53.5 Left Nwtherinn Rings
M' z M's E Mn(R).But M' is large as left ideal in Mn(R)since they have the same uniform dimension. We shall conclude by showing every nonzero ideal of Mn(R) contains M', for this will show Mn(R) and thus R is subdirectly irreducible. Suppose A 4Mn(R) with A n M' c M'. Then AM' c M' as bimodules; Q.E.D. looking at each component shows AM' = 0, so A = 0 as desired. Remark 3.5.88': As a special case, we note for any bonded primes P, Q that RIP is subdirectly irreducible iff R/Q is subdirectly irreducible (for we take a bi-cell in a torsion-free RIP - R/Q bimodule).
Theorem 3.589: Jacobson's conjecture holds for any Noetherian ring which is almost fully bounded: in fact, in this case Jac(R)'M = 0 for every R - R bimodule M which is 5 9 . as R-module, and, moreover, if M is subdirectly irreducible Jac(R)"M = 0 for some n 2 1 .
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Pro08 Let J = Jac(R). Applying Noetherian induction to remark 3.5.74 we may assume M is subdirectly irreducible and need to show some J"M = 0. In particular, M is biuniform and so has a maximal cell M, by proposition 3.5.81. Pick MI,. . . ,M, I M such that {Mi= Mi/M,: 1 Ii It } is a complete set of cells in M/Mi. Then pick M f with M, < M f IMi such that Ann Mj is maximal possible. Mj 5 Mi is also a cell; replacing Mi by MI we may assume Ann Mi = Ann N whenever M, < N I Mi. Let P = Ann M, and & = Ann M i . Let M i denote the unique minimal nonzero sub-bimodule of M. Then Mg IM, so Ann M i = P by theorem 3.5.84. But J E P by remark 3.5.76 so JM, = 0. It remains to show J " M = 0 for suitable n, where M = M/M,. Mjof M,maximal with respect to Nin Mi Take a sub-bimodule Ni 2 = 0. Then Mi is isomorphic to a large sub-bimodule of M / N i . If we can find numbers n(i) such that J"'")(M/N,)= 0 then taking n = max{n(i): 1 Ii It } we get J"M G Ni = 0. By Noetherian induction we are done if M/Ni is subdirectly irreducible, which is the case if its large subbimodule Mi is subdirectly irreducible. To prove this we shall use lemma 3.5.88 repeatedly. Since M , is subdirectly irreducible, we have RIP subdirectly irreducible. But P is bonded by & by lemma 3.5.87, so R/& is subdirectly irreducible by remark 3.5.88'. Hence Mi is subdirectly irreducible, as desired. Q.E.D.
cj+i
n
Having established Jacobson's conjecture for almost fully bounded Noetherian rings, we would like some examples.
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Lemma 3.5.90: Suppose M is an j g . module over a left Noetherian ring R, and N is a large bisubmodule. If a E R is R-normalizing and aN = 0 then a"M = 0 for some n. Proofi A "Fitting's Lemma"-type argument. Let Mi = {x E M : a ' x = O}. Then M,,= M,,, for some n. Hence a"M n N = 0, for if a"x E N then Q.E.D. x E M,,, = M,,. But N is large so a"M = 0.
,
Proposition 3.5.91: If every ideal of a left Noetherian ringR is polynormal ( c j, dejinition 3.5.34) then R is almost fully bounded (and thus satisjes Jacobson's conjecture). Proofi The hypothesis passes to homomorphic images so we may assume R is prime and need show that R is almost bounded. Suppose M is faithful f.g. and N is a large submodule of M. If Ann N # 0 then by hypothesis aN = 0 for some R-normalizing element a. By the lemma 0 = a"M = a"RM Q.E.D. for some n, so 0 = a"R = (Ra)", contrary to R prime.
Examples of almost bounded Noetherian rings include H N P rings, polycyclic-by-finite group algebras, and enveloping algebras of solvable Lie algebras, cf., Chapter 8. In these two latter cases, however, Jac(R) = 0 whenever F is a field. Jategaonkar [Sl] actually proved a more general result, which involves concepts of independent interest. We say a bimodule M is primary if it has a unique assassinator, and M is semiprimary if it is a subdirect product of primary bimodules. He proves that if every image of M is semiprimary then J'M = 0. Since almost fully bounded rings satisfy this property, theorem 3.5.89 is a special case. He also proves Jac(R) is nilpotent if R is Noetherian with every assassinator semiprimitive.
nioN
Digression 3.5.92: (Localization at prime ideals) There is much of Jategaonkar's theory which we have not touched. In particular, it applies to localization, as described in. detail in Jategaonkar [86B]. Let us indicate some of the basic underlying ideas, following the very well-written expository article of Brown [85]. Recall %'(A)denotes the set of elements which are regular modulo an ideal A of R. One of the principal questions concerning localization is, what is the largest left denominator subset of %'(A)? This is clearly R - A when R is commutative and A E Spec(R). However, when R is noncommutative one
can have rather poor examples, such as R =
"), F
A=(:
;),
and
53.5 Left Noetherian Rings
F
A‘ = (o
F o).
441
Then %?(A)= %(A’)= {units of R}, so we gain nothing here
by localizing. Let us examine this example a bit more. Putting N = Fe12we have Ann N = A and Ann“ = A’. Hence A and A’ are bonded! Thus bimodules arise naturally in the study of localization. The theory of torsionfree modules can be utilized to refine these ideas much further. In dealing with arbitrary Noetherian rings one must strengthen the definitions to obtain nice results. Prime ideals P and Q are linked if there is an ideal A with PQ C A c P n Q, such that ( P n Q ) / A is torsion free as RIP-module and has assassinator 0 as right R/Q-module. The reader can check that this is really what was needed in the proof of lemma 3.5.87. Linked primes, discussed earlier in special cases by Jategaonkar and B. Mueller, constitute one of the recent breakthroughs in Noetherian ring theory; the theory is treated in depth in Jategaonkar [86B]. A closely related approach to Jategaonkar’s theory of bimodules is via afJiated sub-bimodules. A sub-bimodule M, of M is lefi affiliatedif Ann M, = Ass M, and Ann’ Ann M, = M,. Right affiliated is defined analogously. If M is f.g. as left module over a Noetherian ring R then clearly M has a (left and right) affiliated sub-bimodule MI, and applying Noetherian induction to M/M, we get a chain 0 = M, < M, < . . . < M, = M where each Mi/Miis affiliated to M / M i - I ; in this case we call the primes{Ann(Mi/Mi-,) and Ann’(Mi/Mi- 1 2 i 5 t } an afJiliated set of primes of M. Using Small’s theorem, Stafford [82a, proposition 1.33 quickly proves that R is an order in an Artinian ring iff there is an affiliated set of primes of R, all of which are minimal primes; he goes on to prove many interesting results about rings of fractions of Noetherian rings. Affiliated primes are also the key to SmallStafford [82]. A one-sided K-dim approach to the theory was taken by Brown [82], who obtained similar results but then had to deal with the following question: Question 3.5.93: (Krull symmetry) Is the left and right K-dim the same for any Noetherian ring?
This question seems to have survived the vicious attack by Stafford [85a] on conjectures in Noetherian ring theory.
N Supplement: Gabriel Dimension We can define a dimension which is better behaved categorically than K-dim.
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Definition 3.5.94: Define the Gabriel dimension G-dim M and a-simple modules by induction on a, as follows: G-dim0 = 0. M is a-simple if G-dim M/N < a for all 0 # N I M. G-dim M I a if every image of M has a nonzero P-simple submodule for some B I a. For example, the 1-simple modules are just the simple modules, so G-dim M = 1 iff every image of M contains a simple submodule. For M Noetherian this clearly implies M has a composition series, but also any completely reducible module has G-dim 1. Thus a module may have G-dim 1 but fail to have K-dim! On the other hand, when K-dim M exists then the two dimension nearly coincide; a discrepancy of 1 arises since G-dim starts with 0 whereas K-dim starts with - 1.
Proposition 3.5.95: If K-dim M = a then G-dim M exists and G-dim M I K-dim M 1, equality holding when M is Noetherian.
+
Proofi Induction on a. If a = - 1 then M = 0 so G-dim M = 0. In general, let Mo be some image of M.It suffices to prove Mo has a 8-simple submodule for some /.? I; a + 1. Let M1 be a critical submodule of M,. We are done unless M1 does not have G-dim I a + 1. But for every 0 # N < M, we have K-dim MJN < a so G-dim M , / N Ia by induction, implying M , is (a 1)simple, as desired. In case M is Noetherian let a’ = G-dim M. We need to show a I a’ - 1. Take a 8-simple submodule M , for 8 I a’. By Noetherian induction we may assume K-dim(M/M,) = G-dim(M/M,) - 1 I a’ - 1, so it suffices to prove K-dimMo I a’ - 1. But for any 0 # N < M, we see a’ > G-dim(M,/N) 2 K-dim(M,/N) + 1 by induction on G-dim, so K-dim M, I a‘ - 1 by proposition 3.5.42(i). Q.E.D.
+
Thus the Gabriel dimension yields nothing new for Noetherian modules. For M non-Noetherian one can prove a I G-dim M I a + 1 whenever K-dim M = a exists, since then M has finite uniform dimension (exercise 28). Thus Gabriel dimension should be viewed as a possible way of extending the Krull dimension theory to much more general settings. Definition 3.5.96: The Gabriel dimension, or G-dim, of a ring R is its Gabriel dimension as R-module.
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The G-dim of a ring exists in interesting instances where K-dim does not, cf., exercises 29 and §6.3. One might expect this to stimulate work in Gabriel dimension, but thus far the flowering of the Krull dimension and Gelfand-Kirillov dimension (to be defined in §6.2) have stunted its growth. As the reader may have guessed, the definition in Gabriel [62] is categorical, dealing with Grothendieck categories. Most of the basic properties of K-dim can be extended to G-dim, cf., Gordon-Robson [73], Gordon [74] and Dahari [87]. In particular, G-dim is exact in the sense of remark 3.5.41", and G-dim M IG-dim R for every R-module M.
Exercises 1. Prove directly that any Ore domain has a classical ring of quotients, which is a
division ring. 2. Conclude the details of the proof of theorem 3.1.4:Well-definedness of addition
3. 4.
5.
6.
7.
(as with multiplication), associativity, and distributivity. (Hint: Use the technical lemma for associativity.) Prove the analogue of theorem 1.10.1 1 for arbitrary denominator sets S. Display the S-'M and Ms as universals with respect to the same functor and conclude thereby that they are isomorphic. Also display this construction as a direct limit. Let Q = S-'R for a denominator set S . Given M E R - h d and N' IS-'M in Q d ~ define d N = { x E N : s - ' x E N'}.Then N' = S - ' N ; this procedure pulls chains from 9'(&'M) to Y ( , M ) . In particular, if M is a Noetherian (resp. Artinian) R-module then S - 'M is a Noetherian (resp. Artinian) Q-module. If A 4 R and Q = S-'R then for any q l . . .. ,q, in QAQ we can find a,, .. .,a, in AQ and s in S such that qi = s-'ai. This gives a way of comparing ideals of R with ideals of Q. Suppose S is a left denominator set of R. Show the following:
(i) If R is right perfect then S-'R is right perfect. (Hint: Given ri in R there is si in S with Rsiri as small as possible. If S-'Rr, > S-'Rr, > ... then Rs,r, > Rs,r, > ...). (ii) If R is semiprimary then S-'R is semiprimary.
Matric Units and Rings of Fractions 8. If M,,,(R) z M,,(R') where R,R' are left Ore domains then m = n. (Hint: M,,,(R) G M,,(D) for a division ring& so counting orthogonal idempotents shows m < n.) In particular, each Ore domain has IBN. 9. Suppose R is an order in a ring Q having a set of n x n matric units. Then Q has a set E of n x n matric units such that Es c R and s'E G R for suitable regular elements s,s' of R. (Hint: Suppose {eij:1 Ii, j In} is a set of matric units of Q. Writing eij = s-'rij we see seij E R for all i , j . Now let E = {seijs-': 1 Ii, j In}.)
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10. If R has an involution (*) then any denominator set S satisfying S* = S is
also a right donominator set and S-'R has the involution given by (s-'r) =
(1- r *)((s *)- I).
Torsion-Free Modules We say an R-module M is torsion-free if ax # 0 for all x # 0 in M and all regular a in
of an Ore ring R.
R. In what follows take Q to be the ring of fractions
11. A torsion-free R-module M is divisible iff M is a Q-module; in fact, taking y to be that element of M satisfying ay = rx show y = (a-'r)x for a E R regular, r E R,
and x E M.
12. R as above. Any torsion-free R-module M is contained in a Q-module M';if M is f.g. we can take M' f.g. as Q-module. (Hint: Let E = E ( M ) and E , = {x E E: Ann x
c
contains a regular element}. Then E / E , is torsion free and divisible, so a Q-module by exercise 1 1. If M = Rxi then take M' = 1 Qxi in E / E , .) 13. If R is left and right Ore then any f.g. torsion-free R-module M can be embedded in an f.g. free R-module. (Hint: By exercise 12 we can find a monic f:M -+ Qa; each generator of M is in R(')s-', so we have a monk M -+ R(')s-' z R(').) This result is quite useful in Noetherian theory.
93.2 1. If { q : i E N} are distinct height 1 prime ideals in a prime ringR satisfying ACC(idea1s) then pi = 0. (Hint: If 0 # r E pi then R/RrR has only a finite number of minimal prime ideals.)
n
Orders in Semilocal Rings
I? + R/I. S is a denominator set for R if satisfies the following two conditions:
2. Let
is a denominator set of
I? and S
(i) For any s1 in S and a, in I there are sz in S and rz in R with s2al = rzsl (ii) As in definition 3.2.32(ii).
3. Prove proposition 3.2.34. (Hint: (I), (2) imply S-II 4S-'R, and the canonical map F-+ 1-'r S-'I yields (3). If (2), (3) holds then (1) holds by exercise 2. Finally, if (I), (3) holds then S-II is the kernel of the map S-'R --* S-'R obtained from r H i-%) 4. Prove theorem 3.2.36. (Hint: (e) exercise 3 shows S is a denominator set of R, and thus S-'R/S-'I is the ring of fractions of R. Conclude by showing S-'I E Jac(S-'R) since each s-'a in S - ' l is quasi-invertible. (a) Put A = Jac(Q) and I = R n A. R is an order in Q which is semisimple Artinian.) 5. Suppose I a R and S is a denominator set of R with S-'I Q S ' R . If I is nil (resp. 7'-nilpotent) then S-'I is nil (resp. T-nilpotent). (Hint: As in proposition 3.1.14). 6. Suppose R is an order in a semilocal ring Q, and let N = Nil(R). Then Q is right perfect iff N is T-nilpotent; Q is semiprimary iff N is nilpotent. In each case N
+
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is the ideal I of theorem 3.2.36, and, in fact, N = Jac(Q) n R. (Hint: (*) Take I = Jac(Q) n R. Then I is nil so I c N, but R/I is semiprime so I = N. (e) by exercise 5 and proposition 3.1.15.) 7. (Small) (i) A left Noetherian ring R is an order in a left Artinian ring iff it satisfies the regularity condition. (ii) More generally, suppose R is any ring and N = Nil(R). Put T = N n Ann' N'. R is an order in a left Artinian ring iff the following conditions are satisfied: R is left Goldie, R/7; is left Goldie for each i, and R satisfies the regularity condition. (Extensive hint: (-=) for (i): Suppose S = {r E R: Fis regular in R = R/N} can be shown to be a left denominator set. Then proposition 3.2.34 applied to theorem 3.2.36 produces Q = S-'R semilocal with Jac(Q) = S-'N nilpotent; thus Q is semiprimary as well as left Noetherian and is thus left Artinian. To prove S is a left denominator set take t such that N' = 0 and induct on t , weakening the hypothesis on S to assume merely that Ann s = 0 for all s in S . Let N' = N n Ann' N a R, and R' = R/N'. Then S' = {s + N':s E S } satisfies the induction hypothesis and N1-' G N' so S' is a denominator set of R'. By exercise 2 it suffices to show for any s1 in S and a , in N there are s, in S and r, in R with szal = r2s1. Take so in S and ro in R with a', = sOal- rosl EN'. Let L = {r E R:ra', E Rs,}. It suffices to show L n S # 0,for sa', = rs, implies ssOal= (r + sro)s,. But L is large in R; indeed Rs, is large in R so if r E R with F$ there is an element 0 # rlra', E Rs, n Rra', yielding E n Hence L has a regular element of R. (-=) for (ii): N is nilpotent. As in (i) S is a denominator set for N and Q=S-'R is semiprimary. Let Mi = QT, ,/QT, an f.g. Q-submodule of Q/QT since Q/QT as a ring satisfies ACC(@). Each Mi has a composition series so Q is left Artinian by the proof of theorem 2.7.2. (=.) The regularity condition follows from Robson's theorem. To continue for (ii) put S = {regular elements of R}, so Q = S ' R . Remark 3.2.17 shows R is left Goldie so N is nilpotent, and also NS-' E S-'N. Hence QN 4 Q and is nilpotent by proposition 3.1.14, so R n QN = N. It remains to show Q T Q Q and R n QT = 7; for then each R/T is an order in the left Artinian ringQ/QT and is Goldie. To show QT a Q one must show as-' E Q T for each a in T . But as-' = s;'al for some a , in N, and QN'a, = (QN'Q)al = QN'QsY'a, = QN'Qas-' = (QN')as-' = 0, proving a, E T . T o show R n Q T = T suppose r = s-'ai E R n S I T . Then r E R n QN = N and N'r = N'sC'a, c s;'N'ai = 0, proving r E T.)Note: see also Warfield [79]. 8. (Kerr [84]) A commutative ring with ACC(Ann) having annihilator chains of arbitrarily long length. Let 1 = {Aij: 1 I i 5 j < cn} be a set of commuting indeterminates and let R = F [ 1 ] / l where I is the ideal generated by all 1,1,,1,, and all l i j A i S jfor i # i'. Then R is naturally graded as R = Ro 63 R, 0 R , where the polynomials in Ri are homogeneous of degree i. For m I n let A,, = Ann{lin:rnI i 5 n} = R, + FAun. Then A,, < A," < ... < A,, is a chain of annihilators of length n. However, R satisfies ACC(Ann), seen as follows: First note every zero divisor lies in R, + R, by matching components. If A is any annihilator ideal with Ann A =I R, then take r = r , r2 E A, where ri E R,. Then Ann r, = Ann r 3 R, so R, n Ann rl # 0. Let A, (resp. 1"")be the indeterminate of highest lexicographic order appearing in r l (resp. in any element y of R, n Ann(r,)). Then lmnlUv = 0 so u = n, and thus A G R, + xi= F l y , .
1 : ;
+
x:==, ,
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Any chain of annihilators descending from A has finite length by a dimension count; hence R satisfies DCC(Ann) and taking annihilators yields ACC(Ann). 9. Of course exercise 8 is not semiprime, since otherwise it would also be Goldie, in which case there is a bound on the lengths of the chains of annihilators. Nevertheless, it can be made into a prime noncommutakioe ring satisfying ACC(Ann) but without bound on the length of chains of left annihilators, by means of exercise 1.9.8. Kerr [84] presents a different example.
Noetherian V-Rings 10. If R is a left Goldie V-ring then every proper ideal is a left annihilator (since by exercise 2.10.16 no ideal can contain a regular element). Conclude that R is a finite
direct product of simple V-rings. Thus any left Noetherian V-ring is a finite direct product of simple left Noetherian V-rings. 11. Suppose F is a field and is an automorphism of F having infinite order. Then S = {Ai:i E N} is a left and right denominator set for the skew polynomial ringR = F[A;a], and S-'R is a simple domain which is a PLID and PRID but is not a division ring. (Hint: Every ideal of R contains a power of 1 by proposition 1.6.25). 12. (Cozzens domains) In exercise 11 suppose F is algebraically closed of characteristic p and take cr to be the Frobenius automorphism. Then S-'R is a V-ring. (Hint: Every simple R-module has the form R/R(I - a) where a E F; it suffices to show these are divisible R-modules for a # 0 since then they are injective. Given x, r in R one needs to find y in R such that x - ry E R(1- a). Using remark 1.6.20 assume x E F. Trying to solve for y in F yields a polynomial equation and matching components yields an equation in the o'y = y p ' ; this has a solution in F. Note the method degenerates iff a = 0.) By symmetry S-'R is also a right V-ring. Also using the same methods one can show S-'R has a unique simple module up to isomorphism.
Principal Left Ideal Rings (cf., Jategaonkar [7OB] We shall say R is PLI if
each left ideal is principal; more generally, R is semi-PLI if each f.g. left ideal is principal. Recall PLIDs were studied in $1.6, also, cf., corollary 2.8.16. 13. M,,(R) is PLI (resp. semi-PLI) iff every left (resp. f.g. left) R-submodule of R'") is spanned by 2 n elements. 14. "Goldie's third theorem" The following are equivalent: (i) R is a semiprime left
ny=
Goldie semi-PLI ring; (ii) R zz R, where each R, is prime left Goldie semiPLI, and the R, are uniquely determined up to isomorphism and permutation; (iii) R x nl= M J T J where each is left Ore and every tg. Tpbmodule of is spanned by I n, elements. (Extensive hint: (i)*(ii) R is an order in Q = Q i ; write Q,= Qc, for ci a central idempotent in Q,.Write R' = Rc, 2 R; we need R' = R. There is regular s with sci E R for each i ; then R's is a f.g. left ideal of R so R's = Ru for some u in R which must be regular. Then s = ru for r in R, implying R'r = R. Hence r-l E R' so r-' = r-2r E R'r = R implying R' = R. (ii) +(iii) Assume n = 1, i.e., R is prime Goldie with ring of fractions Q = M,(D). It suffices to find a set of n x n matric units lying in R, since then
c') n
n
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exercise 13 would enable us to conclude the proof. Let E be a set of n x n matric units as in exercise 3.1.9, i.e., Es E R and s'E E R. Then REs = Rs, for some s, in R which must be regular, and s, = qs for some q in RE. Hence RE = Rq. Note Er E R iff qr E R iff r E q-'R. But E ( E q - ' ) = Eq-' E R so Eq-' E q-'R and qEq-l E q(q-'R) E R is the desired set of matric units. 15. Suppose A,B 4 R (for R arbitrary). If perchance A = Ra and B = R b then AB =Rab. (Hint: ( E ) aR E AR = A.) 16. (Johnson's first theorem) Call R fully Goldie if each homomorphic image is left Goldie. R is fully Goldie semi-PLI iff R x R, where each Ri is fully Goldie semi-PLI with Ri/Nil(Ri)prime. (Hint: Write R = R/Nil(R) = HE,, i.e., each Ri is isomorphic to an ideal of 8. Write R, = Iiwhere A, a R. Then X A , = R so CAY = R for every u in N, and one needs to show the A" are independent for suitable u. By induction it suffices to prove: If 8 = R/Nil(R) is prime and A + B = R with A T = 0 then A" n B" = 0 for some u. To see this first show by the Chinese Remainder Theorem that AB = Nil(R) = BA. Thus A"B" = (AB)". But (AB)" = 0 for some n so A" E Ann@"). Conclude with remark 3.2.2(ix).)
H;,,
Rings for which Principal Annihilators are Summands (PP Rings) 17. Every maximal left annihilator of a ring has the form Annr for suitable r # 0. 18. If e E R is idempotent then Anne = R(l - e). (Hint: Peirce decomposition.) 19. Suppose e l , e, are idempotents such that Ann,(e,(l - e2))= Re, for some idem-
potent e3. Then there is an idempotent e satisfying Re = Re, n Re,. (Hint: Put L = Re,. Then R(l - e l ) E L so e = e3e1 and e3(l - e l ) are orthogonal idempotents of L whose sum is e3, implying L(l - e l ) = R(l - el). Clearly, Re E L n Ann(1-e,)=LnRe,ERe,nRe, for if re,EL then r e , ( l - e , ) = r e , ( e , ( l - e , ) ) = O implying re, E Ann(1 - e,) = Re,. But Re, n Re, E Re since if rlel = r2e2 then r,e,(e,(l - e,)) = re,(l - e,) = 0 so rlel E L and rlel = (r1e1)e3,implying r l e l = r,e: = ((r,el)e3)elE Re.) 20. We say R is a PP-ring if every principal left ideal is a summand of R. (This is a common generalization of semihereditary and of regular.) R is a PP-ring iff every annihilator of the form Ann(r) is a summand of R, i.e., has the form Re for e idempotent. (Hint: The sequence 0 + Ann r + R + Rr + 0 splits.) Prove any PP-ring has the following properties: (i) If R is left Noetherian then R is an order in a left Artinian ring. (Hint: Verify the regularity condition. Suppose r E R with F regular. Write Annr = Re for e idempotent. Then F = 0 so e = 0. Hence Rr is large. If rr' = 0 then Ann r' is a summand of R containing Rr, so Ann r' = R implying r' = 0.) (ii) If R has no infinite set of orthogonal idempotents then every left annihilator Ann B is a summand of R. (Hint: Construct a descending chain of annihilators as follows: Take any b, in Band put A, = Ann b,; given A, pick bi+ in B, if possible, such that A, $ Ann bi+ and put A , , , = A , n Ann bi+ Using exercise 19 as the inductive step show A, = Re, for e, idempotent, so Re, > Re, > ... which by hypothesis must terminate.) (iii) Under hypothesis of (ii) show any right annihilator has the form eR since it annihilates a left annihilator. Conclude R satisfies ACC(Ann) and DCC(Ann).
,
,.
,
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448
(iv) If R is semiprime Goldie then R is isomorphic to a finite direct product of prime Goldie rings (by theorem 3.2.29). (v) If R is semiprime left Noetherian then R is isomorphic to a finite direct product of prime left Noetherian rings. 21. (Bergman [71]) The center of a P P ring is PP. In fact, if R is PP and Rz = Re with z in Z(R) and e idempotent then e E Z(R). (Hint: Completely elementary. Pick r in R and let r' = [e,r ] . Write Rr' = Re' for e' idempotent. Then r'z = [z, r] =0 so 0 = e'z = ze' and thus ee' = 0. But zr' = 0 implies er' = 0 so er = ere. Thus r'e' = O so O = (e')' = e', proving r' = 0.) In the same article Bergman shows the center of a hereditary ring is a Krull domain. The sharpness of this result is shown in exercise 2.8.6 22. Any uniform left ideal L of a PP ring is minimal. (Hint: If a E L then Ra = Re for e idempotent; conclude L( 1 - e) = 0.) 23. Suppose M is a torsion-free module over a semiprime Goldie ring R . M is injective iff M is divisible. (Hint: (-=) By Baer we need show only that every map f: L M extends to a map g: R M. Furthermore, one may assume L is large and thus contains a regular element a. Define gr = ra-'fa.) -+
-+
Fractional and Invertible Ideals In what follows suppose R is a subring of a
ring Q. A Q-fractional left ideal is an R-submodule of Q; define L-' = { q E Q: Lq E R}. L is called invertible if 1 E L-'L, i.e., x q i a , = 1 for a, in L and qi in L - l . Use this notation below.
24. If r E R is invertible in Q then Rr is an invertible left ideal. 25. Any invertible Q-fractional left ideal is f.g. and projective as R-module. (Hint: L = x(Lqi)ai.Define h:L R by right multiplication by 4,. Then @,,A)is a dual base.) 26. Conversely to exercise 25, if a Q-fractional left ideal L is projective in R-&d and contains an element r of R invertible in Q then L is invertible. (Hint: Letting @,,A)be a dual base, define qi = r-'&ai E Q. Then r = x(rq,)a,, so 1 = C q i a i . ) 27. A semiprime Goldie ring R is hereditary iff every large left ideal is invertible with respect to the classical ring of quotients Q of R. (Hint: (=-) Every left ideal is the summand of a large left ideal.) -+
Rings without 1 and the Faith-Utumi Theorem 28. If a division ring D is an essential ring extension of R (i.e. as R-module) then R is an Ore domain with D as its ring of fractions. 29. Define rings of fractions for rings without 1, with respect to denominator sets. Note the ring of fractions has multiplicative unit s-ls. 30. Suppose Q is an essential ring extension of a semiprime ring R and e is an idempotent of Q such that D = eQe is a division ring with R n eQe # 0. Then D is an essential ring extension of R neQe (possibly as ring without 1). 31. "Faith-Utumi theorem". Suppose R is an order in Q = M,(D). Then R contains M,(T) where T is a suitable order (without 1) of D, with respect to a suitable set of matric units. (Hint: By exercise 3.1.9 pick a set E of n x n matric units of Q
Exercises
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with respect to which Es c R and s'E G R. Let L = {r E R:rE G R}, a large left ideal of R and L" = {r E R: Er E R}. Let To = L'L,a prime subring (without 1) of R. For any r # 0 in R we have L' n L'r # 0 so To n Tor # 0, implying R (and thus Q) is an essential ring extension of To. T = TOne,,Qe,, contains el ,ss'e, I # 0 and is a left order in D by exercise 28 and 30.) Faith [73B, p. 414-4161 uses exercise 31 to determine the structure of semiprime principal left ideal rings and of simple Goldie rings-if such a ringR is an order in M,( W) then R x End, M where T is an order in W and M is a torsion free T-module.
Nil Subsets (also, cf., exercise 3.3.18) 32. (Shock's theorem) Suppose R satisfies ACC(Ann) and N is a nil multiplicative subset of R. If N is not nilpotent then there are r1,r2,... in N with {Rri:i E N } independent, also satisfying A, < A, < .. . where Ai = Ann'( Rri). (Hint: Put I; = Ann N'; some T, is maximal among the IT;.. Take ro = xo E N - T,, with Ann xo maximal possible. Inductively, given ri # T, put Ni = {x E N: rix T,} # 0 since riN $ T,,, and picking xi+I in Ni with Annxi+, maximal, define r i + , = rixi+I = xo. . . xi+,. In the remainder, assume throughout i < j . To prove the properties define rii= 1 and rij=xi+ .*'xi.Then rjxi=(ri- ,xirij)xi for i Ij . Since xirijE N is nilpotent, Annxirijxi > Annxi so xirUxi4 Ni- and so ri- ,xirijxiE T,, i.e., ',xi E T,; thus rjri- l , i + n E rjxiN" = 0. In particular, r i , i + n +El A i + , - A i proving A i < A i + l .It remains to show that the Rri are independent, i.e., if ~ ~ = i a = u 0r ufor aU in R then airi = 0. But 0 = (C:=ia"r")ri,i+n+l= a,riri.i+##+, + 0 = airi+n+l so airi-1 EAnn(xiri.i+n+,)= Annxi by choice of xi, implying 0 = airi- ,xi = airi as desired.) 33. As corollaries to Shock's theorem, prove that nil multiplicative subsets of the following types of rings are necessarily nilpotent: (i) left Goldie rings (Lanski); (ii) rings satisfying both ACC(Ann) and ACC(Ann') (Levitzki; independently Herstein-Small). There does not yet seem to be an example known of a ring satisfying ACC(Ann) and having a nil multiplicative subset which is not nilpotent.
xjr
+
Embeddings 34. Duplicate embedding sequence 3.2.43 for rings with involution. 35. If (R,*) is prime and is contained in (R',*) semiprimitive then (R,*) can be
embedded in a primitive (R", *) satisfying every elementary sentence satisfied by all (*)-primitive images of (R',*). {Hint: Introduce the filter of theorem 3.2.44 noting that for each a, b # 0 in R we have arb # 0 or a*rb # 0 for suitable r, and 1, = I...)
Embeddings of Nilpotent Rings The object of exercises 36-38 is to show any
algebra N without 1 over a field satisfying N" = 0 can be written as a subring of upper triangular matrices of M J C ) for C commutative. This theorem is due independently to Bergman et al. [83] and Anan'in [79]-L'vov [80]. We follow the first approach.
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36. Suppose F is a field and N is an F-algebra without 1. If N" = 0 then we can decompose N = ly: where each ly: s N ' ; indeed, take N n - l = N " - ' , and Nn-' to be a subspace spanned by the preimage in N"-' of a base of N"-'/N"', and continue thus. N acts by right multiplication on F @ N and thereby breaks up into elements of Hom(Ni- 1, A$-1) for 1 < i < j < n. Thus N can be embedded into the ring of upper triangular matrices over F @ N. 37. Suppose we have a system of F-vector spaces Fj for 1 I i < j < n, with bilinear maps Kj x V;, + Kk satisfying the associativity condition ( l ) i j u j k ) l ) k u = v i ] ( ~ j k ~ k v , , ) for i < j < k < u (where vije 7,). Then @ 1 5 i < j S n vi,. can be embedded in a commutative ring C whose multiplication agrees with the above bilinear maps. (Hint: Given a sequence s = i , < jl < i2 < j , < it < j , write V, for Kljl @ Fzj2@ ... @ F,j,, and let C = @ V, summed over all such sequences. To define multiplication it suffices to define the product c = (uilj, @ @ vi,j,)uij where uij E K j with i < j . This is divided into several cases. If some j , = i and j = c =uiljl@
take u = ui.j.(uijui,+lju+l)and
... @ v i u -
Ij u _1
@ 0 @ uiu+I j u + 1 @
@ uitj,.
If somej, < iand j = iU+,, take u = vijui,+lju+l and c = u t.1 j.1
@ * a .
@ vi,,ju @ D @ v i u + z j u + z@ * * * @ V i t j t ,
If some j, = i and j < i,+ c = uil j l @
1,
take u = uisjuuij and
... @ vi,-
ju- I
@ 0 @ vi,,
+
I ju+ I
@ . .. @ ~
ij , .,
In all other cases take c = 0. The proof C actually is a ring involves some subtlety; perhaps the easiest demonstration is by means of the tensor ring.) 38. Combining exercises 37 and 36 show any nilpotent algebra can be embedded in the upper triangular part of Mn(C)for suitable commutative C. Note C is N-graded with Ci = 0 for all i 2 n. If C is to be taken t o be a (commutative) algebra without 1 then we have C" = 0.
Embeddings into Artinian Rings 39. Every semifir is a weakly finite domain. (Hint: If ab = 0 then (au)(u-'b) = 0 is a trivial relation for some invertible u, so a = 0 or b = 0. To prove "weakly finite" suppose A , B E M J R ) with AB = 1. Right multiplication by A gives a map f: R(') + R(")which splits, so coker f = 0 by unique rank.) 40. (Small) A left Noetherian ring which is embeddible neither in a left nor right Artinian ring. Let W be the Weyl algebra A , ( C ) and taking any nonzero left
(rS:).
ideal L let R =
(7 z)
where M = W/L. If S c M then Ann
It follows that R fails DCC(Ann) and thus is not embeddible in
a left Artinian ring. (Hint: Otherwise, take a finite set S c M with 0 # Lo =Ann, S minimal possible. Then Lo = Ann(S u {x}) for any x in M so LoM = 0, implying
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0 # Lo U W, contrary to W simple.) Note R is not right Noetherian and cannot be embedded in a right Noetherian ring. This exercise shows embeddibility does not “lift” modulo the nilradical.
0. ( N x - ’ ) x = N n Rx. Consequently, if Sing(R) = 0 and Lr-’ is a large left ideal then L is large in L Rr; conclude if R r , and Rr, are large in a nonsingular ring R then Rr,r, is large. This is part of the direct proof of Goldie’s theorem by Zelmanowitz [69]. A more structural proof is given in exercises 12-14. 1. Any reduced ring R is nonsingular (since Ann, r 4 R). 2. Modifying triangular matrices construct a ring R where Sing(R/Sing(R)) # 0. 3. A module M is nonsingular iff Hom(K, M) = 0 for every singular module K. 4. The direct product of nonsingular modules is nonsingular. (Hint: Use exercise 3) The ultraproduct of nonsingular rings is a nonsingular ring. 5. If 0 -+ MI -+ M -+ M, -+ 0 is exact with MI, M, nonsingular then M is nonsingular. (Hint: exercise 3). 6. If R satisfies ACC(Ann) then 1 = Sing(R) is nilpotent. (Hint: Assume ‘I # 0 for all k. Take a, in I such that a,l’ # 0 and Anna, is maximal such. Show Ann(a,a) > Ann@,) for all a in 1, implying = 0; thus a,/,’ = 0 so Ann(l“ I ) > Ann(lk) for all k, contradiction.) 7. Define Sing,(M) to be that submodule N where N/Sing(M) = Sing(M/Sing M). Show M/Sing,(M) is a nonsingular module, and R/Sing,(R) is a nonsingular ring. 8. Every PP ring (c.f., exercise 3.2.20) is nonsingular. (Hint: 0 -+ Ann r -+ R -P Rr + 0 splits.) 8‘.Every regular ring is nonsingular. Every semihereditary ring is nonsingular. (Hint: Exercise 8.)
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(t’” :’”)
10. (Osofsky
is nonsingular, but the right annihilator of e l , is large.
[HI) A ringR whose injective hull E (in R-Aud) cannot be given a
ring structure. Let T = Z/42 and R =
(LT ):
c
M,(T), and Li = 2Re,, for
i = 1,2, and let Ei be the injective hull of Li(so E, c E). Since Re,, is an essential extension of L, we may assume eii E E, for i = 1,2; likewise e,, E El. There is an isomorphism f:L, + L , given by f(2e,,) = 2e,,, which extends to an isomorphism g : E , + E , . But L , < R e , , so El <E(Re,,) and is thus a summand. Letting 1 ~E : ( R e , ,) + El be the projection and h = g-%, let x = he,, E E , . Then e l i x = x so 2x = 0 (since, otherwise, there is r in R with 0 # 2rx = 2 r e , , x ~ L , , impossible). If E were a ring then 0 = (ge,,)2x = 2e,,x = h(2e,,) = 2e,,, contradiction. 11 If Q is an essential ring extension of R then Sing(R) E Sing(Q). Consequently, if Q is regular then R is nonsingular, by results of $3.3.
Closed Submodules and Goldie’s Theorem 12. Show the following chain conditions on M are equivalent: (i) ACC(@);
(ii) ACC(c1osed submodules); (iii) DCC(c1osed submodules). Hint: (i) * (ii) by means of essential complements. (ii) (iii) as in exercise 0.0.13. (iii) * (i) Given
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{q:i E N} independent submodules build a descending chain K, > K, > ... of closed submodules inductively by taking Ki+ 2 cEj+ 4 an essential complement of A) in Kj.) 13. Every left annihilator in a nonsingular ring is closed. (Hint: Suppose A = Ann, B is large in A'. Then A'/A is singular so for any a in A' there is large L < R with LaB = 0.) 14. A semiprime ring satisfying ACC(@) is nonsingular iff it is left Goldie. (Hint: (*) exercises 12 and 13.) In this case identify i? with the classical ring of fractions of R. (Hint: s E R is regular iff Anns is large.)
,
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Closed Submodules 15. Suppose N is a submodule of M. If Sing(M/N) = 0 then N is closed; the converse holds when M is nonsingular. (Hint: (*) If N is large in N' I M then N ' / N = 0 by remark 3.3.12. (e) Suppose x N E Sing(M/N). Then (Rx N ) / N is singular, so N is large in Rx N implying x E N.) 16. A prime ring has no proper closed ideals other than 0. The only proper closed ideals of Z x Z are (Z,O) and (0, Z). 17. If A < R is closed as left ideal and R is semiprime self-injective, then A = Re for some central idempotent e of R. (Hint: A is injective and thus a summand of R; apply lemma 3.2.28.)
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Finite Dimensional Modules In the following exercises write dim(M) for the uniform dimension of M.
18. If dim(M)= n then the injective hull E(M) is a direct product of n simple modules.
(Hint: Start with the case n = 1 and apply injective hulls to proposition 3.2.19.) It follows that if M satisfies ACC(@) then E ( M ) has a composition series, so End,E(M) is semiprimary. Using this in conjunction with theorem 2.6.31 and exercise 2.10.20 prove the following theorem of Fisher: Any nil subring of R/Sing(R) is nilpotent. 19. If N is a closed submodule of M then dim(M) = dim(N) + dim(M/N). On the other hand, if N is a large submodule of M then dim(N) = dim(M).
Self-Injective Rings which are Not Necessarily Regular
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20. Suppose R is right self-injective. Then (i) Ann(A n B) = Ann A Ann B for all right ideals A, B of R; (ii) every f.g. left ideal is a left annihilator. (Hint: (i) Given rinAnn(AnB)definef:A B + R byf(a 6) = a (1 r)b.Thenfisa welldefined right module map and thus is given by left multiplication by some x in R. Then (1 r - x)b = (x - l)a for all a, b in R, implying 1 r - x E Ann B and x - 1 E Ann A so r E Ann A Ann B, as desired. (ii) Write L = Rx, and L = Ann' L. Clearly L E Ann L . On the other hand, Ann L = Ann Ann'x, by (i). If y E Ann(Ann'x,) then the map f:xiR + R defined by f(x,r) = yr is given by left multiplication by some w in R, so y = f x , = wxi E L.)
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Exercises
21. If R satisfies ACC (f.g. left ideals) then R is left Noetherian. In particular, this holds if R satisfies ACC(Ann) and every f.g. left ideal is a left annihilator. 22. Suppose R is left self-injective and right perfect. Then (i) every simple module is isomorphic to a left ideal; (ii) Ann(Ann'L) = L for every left ideal L. (Hints: (i) Take a basic set of orthogonal primitive idempotents e,, .. .,el of R. Any simple R-module is also an R/Jac(R)-module by Nakayama's lemma and thus isomorphic to a minimal left ideal of R/Jac(R), so there are at most t of them. But each Rei contains a minimal left ideal Li and Re, = E(Li),so L , , . ..,L, are nonisomorphic and are thus all the simple modules. (ii) Let L' = L + Rx for x in Ann(Ann'L). L is contained in a maximal proper submodule M of L';an injection8 L'/M -+ R (by (i)) yields a map 8 L' -+ R which then can be given by right multiplication by some r in R. Then Lr = 0 so r E Ann'L, implying 0 = L'r = fL',contradiction.)
Quasi-Frobenius Rings A ring is quasi-Frobenius (QF) if it is left and right Artinian and left and right self-injective. The structure of these rings is so rich that they have become an object of close study, especially with respect to their modules, as we shall see later on. Our main object here is to weaken the definition. 23. If R satisfies ACC(Ann) and is left or right self-injective then R is QF. (Hint: Case I. R is right self-injective. Then R is left Noetherian by exercises 20, 21 so R/Jac(R) is left Noetherian and regular by exercise 2.10.20; hence R is semiprimary and thus left Artinian. 1 = Ann' Ann 1 for each right ideal 1 of R by the right-hand version of exercise 22, so R is right Noetherian as well as semiprimary; and thus is right Artinian. To show R is right self-injective it suffices to show for each right ideal 1 that every map f 1 -+ R is given by left multiplication by an element r of R. Do this by induction on t where l=xf=a,R. Namely, let l ' = ~ :aiR ~ and write fa = r'a for a E 1'. Write r" = fa,; then f ( a , r ) = r"r. But r' - r" E Ann(/' n a,R) = Ann(Ann' Ann InAnn' Ann a,) = Ann 1 + Ann a,, so write r' - r" = s' s" where s'l = 0 and s"a, = 0. Put r = r' - s' = r" + s". Case !I. R is left self-injective. Using exercise 20 show R satisfies DCC(principa1 right ideals) and is thus left perfect. In fact, R is semiprimary for if we take n with Ann J" maximal then J" = 0, seen by applying lemma 2.7.37 to a minimal submodule of R/Ann J".Hence every left ideal of R is a left annihilator by exercise 22, so R is left Noetherian and thus left Artinian. Done by Case I.) 24. Suppose R is QF. Then Ann(Ann'L) = L for every left ideal L, so Ann' yields a 1: 1 correspondence between {minimal left ideals} and {maximal right ideals}. Using injectivity show if M is a simple R-module then Hom(M,R) is a simple right R-module under the natural action. Using exercise 2.7.9' show soc(R) = Ann Jac(R). From this conclude that the left and right socles coincide. There is a bijection between principal indecomposable modules and minimal left ideals, by Re -+ soc(R)e, whose inverse correspondence is given by means of the injective hull (c.f., hint for exercise 22). 25. If R is QF then an R-module is projective iff it is injective. (Hint: (e) Any indecomposable injective has the form E(L) where L is a simple module and thus isomorphic to a minimal left ideal; hence E ( L ) is a summand of R.) 26. Every module over a QF-ring is a submodule of a free module. (Hint: Apply exercise 25 to the injective hull.)
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27. If R is a PLID and a # 0 is a normalizing element of R then R/Ra is QF. (Hint:
R/Ra is left Noetherian and selfinjective.)
Frobenius Algebras A finite dimensional algebra R is Frobenius if R x R* = Hom,(R, F) as R-modules (c.f., remark 1.5.18).
28. Any Frobenius algebra is QF. (Hint: Apply exercise 2.10.4.) 29. The following are equivalent: (i) R is Frobenius; (ii) R has a nondegenerate bilinear form (,) which is associative in the sense (ab,c) = (a,bc) for all a,b,c in R; (iii) There is an F-map f R + F whose kernel contains no nonzero left ideal. In this case [R:F]=[L:F]+[Ann'L:F] for every left ideal L. (Hint: (i)+(ii) Given the isomorphism cp: R + R* define (,) by (a, b) = (cpb)a. (ii) =S (iii) Define f R + F by fa = (a, 1). (iii) *(i) Define the monk rp: R --* R* by (cpr)a = f(ar). Since [R*:F]=[R:F] conclude cp is also epic. The last assertion follows from Ann'L= { r E R:(L,r) = O}.) 30. An algebra is symmetric if it has a symmetric associative bilinear form, c.t, exercise 29. Every symmetric algebra has IBN. Every f.d. symmetric algebra is Frobenius. Using trace maps (c.f., definition 1.3.28)one can easily show any group algebra is symmetric, as well as any f.d. semisimple algebra over a field.
Prime Rings having a Maximal Left Annihilator The following exercises due to Koh study rings which have a maximal left annihilator. The next exercise shows this class is much broader than rings with ACC(Ann). 31. Any primitive ring R with nonzero socle has a maximal left annihilator; however,
if R satisfies ACC(Ann) then R is simple Artinian. (Hint: Take annihilators of finite ranked elements). In exercises 32 through 35 assume R is a prime ring with maximal left annihilator A; thus A = Ann r for suitable r in R. 32. Every nil left or right ideal L is 0. (Hint: One may assume L is a left ideal. If 0 # x E LtakeO # rl E rRx. If s E rl Rands" = Oands"-' # Othen Arms"-' = Ann r so sr = 0, i.e., r l Rr = 0, contradiction.) 33. R is nonsingular. (Hint: A n Sing(R) is nil and thus 0.) 34. Let D be the idealizer of A (c.f., exercise 1.5.1). Then D/A is a domain. Let A' = Ann' A. Then A' is naturally a D/A-module and, in fact, is torsion free. The regular representation of R then gives a ring homomorphism cp: R + End,,, A', which is an isomorphism if R is simple. This proves a theorem of Koh: If R is simple then R is isomorphic to the endomorphism ring of a torsion-free module over a domain. (Compare with exercise 2.1.9.) 35. If R is self-injective then R is primitive with nonzero socle. (Hint: Sing(R) = 0 so R is regular. Take an essential complement L of A. L is uniform so L is minimal by exercise 3.2.22.)
Regular Self-Injective Rings and Goodearl's First Theorem 36. Suppose Q is an essential ring extension of R. The following are equivalent: (i) R is nonsingular and Q = k;(ii) Q is regular and self-injective; (iii) Q is self-injective
Exercises
455
and R is nonsingular. (Hint: (ii) =(iii) Q is nonsingular as an R-module. (iii) =s (i) The inclusion R R lifts to a map Q + R, which is a ring injection since the m_ap g: Q -+ R given by gy = jxfy - f(xy) vanishes on R and is thus 0. But then R is an essential ring extension of Q. -+
In ex_ercises37-41 we study a regular self-injective ring Q. By exercise 36 Sing(Q) = 0, and Q = Q.
37. Every f.g. nonsingular Q-module is injective and projective. (Hint: Take f: Q(")+ M epic. Theorem 3.3.21 yields FJf:FJQ(")+ F,M; but FJQ(")= Q(") so FJ f = f . Hence M = fQ(") = M is injective. But kerf = ker F,f = E(ker f )is a summand of Q(")so f splits.) 38. We say a module N is subisomorphic to a module M, written N 5 M , if there is a monic J N + M. For any nonsingular injective Q-modules M, N there is a suitable idempotent e of Q for which e M 5 eN and (1 - e)N 5 (1 - e)M. (Hint: Let 9'= {triples ( M ' ,N',f'): M' I M , N' I N, and f':M' -+ N' is an isomorphism}. By the maximal principle there is a maximal such triple (M', N', f'),in the sense that f' cannot be extended further to a map f":M" -+ N" where M' I M" and N' 2 N". In particular, M' = E ( M ' ) and N' = E ( N ' ) . Hence M = M ' @ J and N = N' @ K for suitable nonsingular injectives J , K. Let A = Anno J . Then Q/A is nonsingular so exercises 15, 17 show A = Qe for some central idempotent e. Then eM = eM' x eN' I eN. Likewise, to show (1 - e)N 5 (1 - e)M one needs to prove (1 - e ) K = 0. So suppose 0 # y E (1 - e)K. Then Q y is projective so Qy x Qe' for some idempotent e'. But e E Ann(Qy) so ee' = 0 and e' E Q( 1 - e). Hence there is 0 # x = e'b E e'J for suitable b. Qx is projective so right multiplication by b yields a split epic Qe' + Qx. Thus Qx is isomorphic to a summand of Qy, and there is a monic g: Qx + Qy; ( M ' + Qx, N' + gQx,f' + g) extends ( M ' ,N',f'), contradiction.) 39. Suppose L is a left ideal of Q, and r E Q. Then r E LQ iff Qr 5 L(")for some n. (Hint: (=.) Write r = aiqifor ai E L and qi E Q. Define the map f:L(")+ Q by Then exercise 37 yields a monic Qr + j-'(Qr) 5 L("). f(bi) = biqi. Then r E fL("). (-=) Qr is a summand of L(")by exercise 37. Let pi:L -+ L(")be the injection to the i-th component, and let II: L(")+ Qr be the projection. Then npi is given by right multiplication by suitable qi in Q so Qr = Lq, 5 LQ.) 40. (Goodearl's first theorem) If P E Spec(Q) then the lattice of ideals of Q/P is totally ordered. (Hint: Given ideals A,B 3 P with B $ A one needs A G B. Take b in B - A and arbitrary a in A. Then Qa, Qb are injective so there is a central idempotent e of with Qea 5 Qeb and Q(l - e)b 5 Q(l - e)a. If e E P then Qb = Q( 1 - e)b + Qeb 5 A(2)so b E AQ = A by exercise 39. Thus e # P. Then 1 - e E P so a E B by analogy. Thus A E B.) 41. If R is prime nonsingular then the lattice of ideals of d is totally ordered.
X
XI=,
1
Directly Finite Modules 42. Say M is directly infinite if M is isomorphic to a proper submodule; otherwise M
is directly ,finite. Show any summand of a directly finite module is directly finite. Examples include modules with composition series.
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43. Suppose M is injective and M(') 5 M. Then E(M("))x M by proposition 2.10.29; in particular, M(') x M if a is finite. Also note la1 5 I MI. 44. Define p ( M ) to be the smallest a for which M(') ,$ M. If M # 0 then p ( M ) 2 2. If p ( M ) > 2 then p ( M ) is infinite since M 2 M(') 2 M ( 4 ) . . . .
Goodearl's Second Theorem In the following assume Q is prime regular selfinjective, and El, E , are nonsingular injective Q-modules # 0.
45. 5 is a total order on isomorphism classes of nonsingular injective Q-modules. (Hint: El 5 E , or E , 5 El by exercise 38; conclude with proposition 2.10.29.) 46. If El 5 E, then p(El) 5 p(E,). (Hint: On the contrary, assume a = p ( E 2 )< p ( E , ) . Then E y ) 5 El and a is infinite. Take a maximal independent set {Mi:i E I} of submodules of E , each satisfying Mi x El. Let E , = ,?(@Mi) and let E b be a complement of E, in E,. Then E , $ Eb so Eb 5 E,. @MY'-+ @Mi = E, extends to a monic + E, so p(E,) > CL. But Ef' 5 5 E , 5 E , = E , @ EL 5 E f ' so E , x E,, contradiction.) 47. The following are equivalent: (i) El is directly infinite; (ii) El x El 8 El; (iii) p ( E l ) is uncountable. (Hint: (i) =s (iii) Let f:El -+ El be nonepic monic. Then fEl is a summand of E,, taking a complement E' let E = E(@fiE'). E x E(E""') SO E") 5 E. But E 5 El by lemma 2.10.28 so p(Ei is uncountable.) 48. If E, $ E y ) for all n in M then E, x E ( E y ) ) for some infinite a 2 p(El). (Hint: El 5 El. Let E,,Eb be as in exercise 46. Then El $ EL so Eb 5 El, so E , ,< E , @ E , x E(EY))@ El x E ( E y ) ) for some a, clearly infinite. Conclude E, x E(E?)).) 49. p(El @ E,) = max{p(E,),p(E,)} if El or E , are directly infinite. (Hint: Assume El 5 E,. Then p ( E , ) is infinite so El @ E , 5 Ei2) 5 E , 5 El 8 E , ; thus El @
Et'
Et'
E2 x E,.) 50. If El and E , are directly finite then E = El 8 E, is directly finite. (Hint: Otherwise E x E(" by exercise 47. Using proposition 2.10.32 take Mi 8 4 = Ei with M, @ M 2 x E and Nl @ N , x E. If M, 5 Nl then E 5 El; if 5 M, then
E 5 EZ. 51. (Goodearl's second theorem) Put H(a) = {r E Q:p(Qr) 5 a}. Then H(a) d Q and
every nonzero ideal of Q has this form. Thus {ideals of Q} is a well-ordered chain. (Hint: First note every principal left ideal is injective by exercise 37. To show H(a) d Q note for r l , r 2 in H(a) that p(Q(rl - rz)) I p(Qrl @ Qr,) I a, p(Qrrl) < p(Qrl) I u , and the epic Qrl -+ Qrlr splits so p(Qrlr) I p(Qrl). On the other hand, given A Q Q take the smallest a for which A I H(a). Claim A = H(a). Otherwise, there is r in H(a) - A. Pick a in A and L = Qa. Then Qr ,$ L(") for all n by exercise 39, so Qr x E(L'"')for some j 2 p ( L ) .Then a 2 p(Qr) > j 2 p(L).Hence there is a' in A - H(B); Qa' $ L so L 5 Qa' so Qr 5 E((QU')(~)) 5 Qa'. Thus r E Qa'Q s A, contradiction.) 52. If R is prime nonsingular then {ideals of 8) is a well-ordered chain; in particular, 8 is primitive. In fact {nonzero ideals of 8 } xsome interval of the chain of infinite cardinals, in view of the next exercise. 53. Notation as in exercise 51, there are suitable cardinals, al,cc2 such that {ideals of Q} = {H(a):al I a 5 cc,}, and H(a) c H(a') if a < a'. (Hint: It suffices to prove the following claim: If E is nonsingular injective and a is a cardinal 2 p ( E ) then
Exercises
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p ( D ) = fi where D = E(E(')) and fl is the smallest cardinal > a (for then D is isomorphic to a summand of Q). To prove the claim write E(') = @ { E i : i E I} where a = 1 '1 and each Ei % E. Then p ( D ) > a so p ( D ) 2 fl. To prove equality assume, on the contrary, p ( D ) > fi. Write E'fl) = @{EJ: j E J } where each EJ o E; by assumption there is a monic f: E'fl)-+ D. Given a finite subset I' of I put J(I') = { j E J : fEJ n C i s , . B i# 0 ) and note J ( I ' ) = J. Hence some IJ(I')l = B. Let M = Ei, so p(M) = a, and 4= E ; n f - ' M . Let j , = { j E J ( I ' ) : p ( 4 ) is minimal}, and J, = { j E J(I'): %o I N Y ) } .Then J(I') = J,, so some IJ,,I = 8. Now
u
fi < p(E(Nif'))I p ( @ { % : j
xisf,
u E
J,}) I p(M'"')
= p(M) = a,
contradiction.)
53.4 1. 2 / 4 2 has the large 2-submodule 22/42 which is not dense. 2. If N 2 M is dense and y E M then L = N y - ' satisfies Ann:, L = 0. (Hint: R/L % ( N R y ) / N so Hom(R/L, M) = 0.) 3. If N I M is dense and f : M ' - + M is a map with M' M then f - ' N is a dense submodule of M'. (Hint: M'/f -'N % ( N + f M ' ) / N . ) 4. N I M is dense iff for every nonzero x , y in M there is T in R with r x # 0 and r y E N . (Hint:(*) by exercise 2: (e) Suppose there is nonzero f:M' -+ M with N I kerf. Take x = fy # 0, and get a contradiction.) In other words, N is dense iff N y - ' x # 0 for all x , y in M. 5. Suppose N I N' I M. N I M is dense iff N 5 N ' and N' I M are dense. (Hint: (-=) Suppose N I M ' I M and J M' -+ M with f N = 0. Restrict f to a map g : M ' n N ' n f - ' N ' N ' , which is 0 by hypothesis. Hence f ( M ' n N ' ) = 0 by exercise 3 and conclude as in proposition 3.4.4.) 6. Given M E R - M d define E,(M) = r){kerJf E End,(E(M)) and f M = O}. This is a rational extension of M and if N is any rational extension of M then the inclusion M E,(M) extends to a monic N -+ Er(M), implying Er(M) has no proper rational extensions. 7. Identify E,(M) as the module constructed in theorem 3.4.6. 8. A (left) quotient ring of R is a ringQ which is a rational extension of R (as R-module). Show the maximal quotient ring Q is maximal in the sense that for any quotient ring Q' of R the injection R -+ Q extends to a ring injection Q' Q. 9. The maximal ring of quotients Q of R is self-injective iff Rx-' is dense in R for each x in E(R). In this case Q % Hom,(E(R), E(R)). 10. If 8 is a topologizing filter of R and R E Qp(R) then Q,(R) is an essential ring extension of R and thus cannot be nonsingular if Sing(R) # 0. This applies, in particular, to the maximal ring of quotients and the Martindale-Amitsur construction, which cannot be regular unless R is nonsingular. 11. (An alternate description of the localization functor) Given an idempotent filter B of left ideals of R and any R-module M define Q(,,(M) = %{Hom(L, M): L E 8). (Thus Q(*,(M) = Q,(M) iff 9JM) = 0.) This provides a left exact functor F : R - A d -+ R-Mod given by F M = Q(F)(M). There is a canonical map p: M Q(F,(M), whose kernel is FF(M). Show FF(M) = M iff Q,,,(M) = 0.
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Moreover, N = Q,,,(M)/(pM is Ys-torsion. Let K = Y s ( M ) . Applying the functor F to the exact sequence 0 + M / K -+ Q ( 4 ) ( M )+ N + 0 yields
so Q,(M) = Q ( s , ( M / K ) x lim{Hom(L', lim Hom(L, M ) ) :L, L E 9}. Using the ideas of theorem 3.3.21 give sufficient conditions for the localization functor to be exact. 12. (Martindale) If R is centrally closed prime and a, b E R such that arb = bra for all r in R then a = bz for some z in Z(R). (Hint: Define f:RaR + R such that fa = b.) This modest fact has proved instrumental in many results of structure theory and lies at the foundations of the theory of generalized identities.)
Torsion Theories A torsion theory for a category V is a pair ( Z 9 )of subclasses of Ob(V) such that Hom(T,F) = 0 for all T in 9 and F in 9, where I and 9 are maximal such. 9 is called the torsion class; f is called the torsion-free class. 13. There are natural 1:l correspondences between the following three concepts in
R - A d : idempotent filters, torsion radicals, and torsion theories. (Hint: We saw in the text that any idempotent filter yields a torsion radical. If Rad is a torsion radical then define a torsion theory by taking M E 9 iff Rad M = M ; define the filter of M E 9iff Rad(M) = 0. Finally, given a torsion theory (Z9) {left ideals L such that R/L E 9}. 14. Suppose (Fo,S0) satisfies Hom(90,fo) = 0 for all T in 9 and F in 9. Then (90,fo) can be expanded to a torsion theory. (Hint: Take 9 = {F:Hom(T, F) = 0 for all T in Yo}and then 9 = { T: Hom(T, F) = 0 for all F in f}.) In particular, taking So= 0 any class can be expanded to a torsion class of a suitable torsion theory. 15. If (g9} is a torsion theory of an abelian category V then ( 9 , Iis ) a torsion theory of the dual category. Thus we can dualize theorems in torsion theories of abelian categories. 16. A class V of R-modules is closed under extensions if M E V whenever 0 + K .+ M + N + 0 is exact with K , N E %. Show that 9 c R - A d is a torsion class with respect to a suitable torsion theory iff 9 is closed under extensions, homomorphic images, and direct sums. Dually 9 is a torsion-free class i f f 9 is closed under extensions, submodules and direct products.
(z9)
53.5 In exercises 1 through 7 assume R is left Noetherian, and S = { r E R: Ann, r = 0 ) and N = Nil(R). We say r E %(N)if r + N is regular in R/N.
Reduced Rank 1. (Goldie) If r E R and s E S then there are r',s' in R with s'r = r's, such that s' E V(N). (Hint: R x Rs so p(R/Rs) = 0. Thus there is s' regular mod N such
that s'(r + Rs) = 0 i.e., s'r E Rs.)
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2. (Djalabi) If s E S then s E V(N). (Hint: Rs has an element in V(N) by exercise 1, implying s is regular mod N, seen by passing to the ring of fractions of R/N.) 3. A better proof of Small's theorem (exercise 3.2.7(i)). Suppose R satisfies Small regularity condition, i.e., every element in V(N) is regular in R. Then S is a left denominator set by exercises 1 and 2. To show Q = S-'R is left Artinian note that if L , < L, < Q then S n Ann,(L,/L,) # 0 so p((L, n R)/L, n R)) > 0 and p(L, n R ) < p(L, n R); this implies length (Q) p(R). 4. A Noetherian ringR is a left order in a left Artinian ring Q, iff R is a right order in Q and Q is a right Artinian ring. (Hint: Small and proposition 3.1.16.) 5. Show for R =
(," :)
and F = Z/2Z that R is Noetherian but the left and
right reduced ranks of N are unequal. 6. Show in exercise 5 that Ns = N holds for all s in V(N). Generalize Robson's
theorem to show that if Ns = N for all s in V(N) then R has an idempotent e such that eR(l - e) = 0 with eRe Artinian and (1 - e)R(1 - e) semiprime, and conversely. 7. If R is as in exercise 6 and R/N is prime then R is prime or Artinian.
More Counterexamples of Jategaonkar 8. (Jategaonkar) Suppose T is a PLID contained in a division ringD and suppose there is an injection u:D T. The skew power series ring R = T [ [ l ;u]] then has the following properties: (i) R is a PLID; (ii) r E R is a unit iff its constant term is a unit in T. (iii) For any a in T we have 1" = (a"a-')l"a E Ra for each n. (iv) Every left ideal L of R has the form Ral" where a E T. (v) Ral" Q T iff au"b E Ta for all b in T; (vi) Suppose Jac( T) = Tb # 0. Letting J = Jac(R) one then has J = Rb and 1E J'. In particular, Jacobson's conjecture fails. (vii) If T is local then R is local. (Hints: (i) as in proposition 1.6.24, using terms of lowest degree; (ii) as in proposition 1.2.24; (iii) clear; (iv) Write L = Rrl" where r has constant term a # 0, and apply (iii); (v) clear; (vi) Suppose L = Ral" is a maximal left ideal. If n 2 1 then n = 1 and Ra = R; but R 1 c Rb by (iii), contradiction. Hence n = 0, so L = Ra. It follows that Rb E J. Hence 1"E J for each n, by (iii), so r E J iff its constant term is in J . Consequently, J = Rb. 1ERb"G J " for each n;(vii) clear from (vi). 9. (A local PLID in which Jacobson's conjecture fails) Suppose A1, A2 are countable sets of commuting indeterminates over a field F,. Let F = Fo(Al) and let T be any localization of F[A,] in D = F(A2).Then there is an injection u:D F since D = Fo(Al uA,) x F. Thus we take R = T [ [ l ; u ] ]and apply exercise 8. Note R / J x F is a field. 10. In exercise 9 the commutative primary decomposition theorem fails for J l where J = Jac(R). (Indeed, write J = Rb and note b" 4 J 1 and 1 4 J1, but J l is not the intersection of ideals properly containing it.) Likewise, the Artin-Rees property fails for M = R, N = J I , for J"M nN = N 3 J"-'+ '1 = J"-'(J'M nN) for t < n. (This uses the fact J' = Rb', cf., exercise 3.2.15. 11. Exercise 9 can be improved using example 2.1.36 (also due to Jategaonkar) Continuing with the notation there, let S = {f E R,: when written in standard form f has some summand in D = R,}, e.g., 1 + 1'1, + 1,1, E S . Show every element in R - {0} has the form s4 where s E S and 4 is a monk monomial in -+
n,?=
-+
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Rings of Fractions and Embedding Theorems
standard form. Furthermore S is a left denominator set, and S-'R is a local ring. If we define inductively !fl= J : for fl = y+ a successor ordinal and J, = ny
nz,
+
Fully Bounded Rings and the H-Condition (Cauchon [76a]) 13. (Cauchon's theorem on fully bounded rings) A ring R satisjes the H-condition if
for every f.g. module M there is a finite set V c M such that Ann M = Ann V. Prove the following assertion holds for any left Noetherian ringR: (i) Every homomorphic image of R is left bounded; (ii) R is fully bounded; (iii) R satisfies the H-condition. (Hint: (i) =. (ii) obvious. (iii) * (i) Note the H-condition passes to homomorphic images of R, so it suffices to prove R is left bounded. Suppose L < R is large. Write M = R/L and take finite V c M with Ann V = Ann M. But Ann V = n { L u P ;D E V} is large. (ii) (iii) The hard direction. By Noetherian induction there are submodules M,, ...,M,, of M such that M, n * . . n M,, = 0 and each M/Mi is uniform. It suffices to show each M/Mi has a finite subset with Ann(&) = Ann(M/Mi) since then taking V to be a union of representatives of I( in M yields Ann M = Ann V. Thus one may assume M is uniform and faithful (replacing R by R/Ann M). Let P U R be maximal with respect to being Ann M' for suitable M' M. Then P E Spec@), and Ann;, P is large as a left ideal in R. Let M' = {x E M:Px = 0},so P = Ann M'. Then M' is a faithful module over the prime ring RIP, implying Sing(M) = 0. (Otherwise, if Sing(M) = Cf= Rx, then Ann,,,xi contains a nonzero ideal A i for 1 i 5 t, so 0 # Ai E Ann M,
n
,
461
Exercises
contradiction.) But then p ( M " ) = 1 for each M" I M ' since p ( M ' / M " ) = 0. Now build V inductively as follows. Take u , arbitrarily in M . Suppose u l , . ..,uk have already been chosen, and let Bk = Ann,{ u , , . .. ,u k } . We are done unless B 3 0 = Ann M so B; = Bk n Ann; P # 0. Thus there is u k + in M such that B ; V k + # 0. PB;Uk+, = 0 so B ; U k + , 5 M ' and P ( B ; U k + , ) = 1. But right multiplication by u k + gives an epic B; + B ; U k + having kernel B;+ so p(B;/B;+ ,) = 1. Thus the process must stop after p ( B ; ) steps.) Every fully bounded left Noetherian ring is left bounded, by Cauchon's theorem.
,
, ,
,
,
Completions 14. (McConnell [79]) If A 4 R and there is a polycentral ideal B of R such that
B + A* = A and R/A is left Noetherian then the A-adic completion of R is left Noetherian. (Hint: The proof of theorem 3.5.36 must be modified considerably since one no longer has A E B , where I? = BJ'. Let BI = B , + A', B;' = B ; = A, and, inductively, BI' = BI + BLBY-,,. Now the BI' provide a filtration of R, and, as in the original proof, the graded ring R" is left Noetherian.) 15. (Hinohara [60]) Suppose J = Jac(R) is f.g. as left R-module, and J' = 0. If R is complete and semilocal then every tg. submodule N of any f.g. projective R-module P is closed in the J-adic topology of P,i.e., N = N + J'P, and, furthermore, J'M = 0 for every finitely presented R-module M. (Hint: It is enough to prove the first assertion, since the second follows from considering 0 + N + F + M + 0 exact where N is f.g. and F is f.g. free. The main step is showing N = N + where N denotes the closure of N ; indeed, then J'N = J'N + J"Nfor each i (in view of exercise 0.2.3) and since R is complete one passes to the limit and gets N = N . To prove N = N Jxsuppose x E N.Then for any u one can find xu in N such that x - xu E J"P, so x - xu E (Rx + N) nJ"P, and one wants (Rx + N ) n J"P c for some u. Let N' = (Rx + N + J N )/JN, - which is an f.g. R/J-module and thus Artinian. Hence N ' n ( J " P J N ) / J N = N' n( J u t 'P + m)/JNfor suitable u, from which one concludes N' n (J"P + J N ) / J N = 0.) 16. (McConnell [79]) Suppose A 4 R and A = Rai such that a,R = Ra, and [ai,R] G RaUfor each i t . If R/A is Artinian then k is Noetherian. (Hint: Assume 0; J' = 0. Let R, = R/Ra,. By exercise 15 the kernel B of the map k + k, is ka,. By symmetry also B = a,k. But 6, is Noetherian by induction on t , and Grflkl is generated by k, and the element a, + B Z , so Gr,R is Noetherian.) 17. Suppose a E R is regular and normalizing. Then a induces an automorphism o on R by ar = (ar)a. There are examples where R has a prime ideal P with P # UP. Explicitly, let R, be any Noetherian ring with a prime ideal Po not fixed by an automorphism a; let R = R,[I; 01, a = 1,and P = R 1 + Po. 17'. (McConnell [69]) Let R, = C[pl, p2] where pi are commuting indeterminates, let u be given by up, = p, and c p 2 = p, + p 2 , and let Po = Rop2.Forming R, P as in exercise 17, one has P" = 0 but the P-adic completion of R is not left Noetherian.
1;::
1
nieN niEN
nieN
m,
+
+
If= ,
1;:: ,
nneN
Rings of Fractions and Embedding Theorems
462
The AR-Property 18. (Jordan) Notation as in exercise 17, P fails to satisfy the AR-property, (Hint: (aP)'-'a = aPi-' is contained in Ra n P i but not in Pa.) 19. Any polycentral ideal A of a left Noetherian ringR satisfies the AR-property. (Hint: If N < M are f.g, R-modules then form R" and the corresponding graded module $, and n BiM)Ai where R" = @BiAi. This is generated by a finite number of components; check the next one.) 20. The following are equivalent for A a R: (i) AR property: (ii) N n A'M E AN for all f.g. M; (iii) the condition in proposition 3.5.38. (Hint: (i) *(iii) Write M = Rx,. Then R n Rxi is large in Rxi so one may assume fi = Rx is cyclic; let L = Rx-1.) 21. The following are equivalent for a semilocal left Noetherian ringR with J = Jac(R): (i) n l s N J i M= 0. (ii) Every f.g. essential extension of an Artinian module is Artinian. (iii) Every f.g. essential extension of a simple module is Artinian. (iv) J has the AR-property. (Hint: (i) o (iii) in the text, and the same argument shows (i) => (ii). (ii) (iv) by exercise 2O(iii) since is an R/J-module and thus Artinian; (iv) (iii) also by exercise 2O(iii).)
OiEN(N
Krull Dimension 22. Suppose L is a commuting indeterminate over R. Viewing M[A] = R[A] BRM as R[I]-module prove for any Noetherian R-module M that K-dimRIAl(MII]) = K-dim,(M) 1. (Hint: Since M is Noetherian there is a finite chain M > M , > M , > ... > M, = 0 each of whose factors is critical. Thus we may assume M
+
is critical. The chain M[I] > AM[I] > A2M[I].. * has factors isomorphic to M, proving "2."Thus it suffices to prove "I," i.e., K-dim~~,lM[d]/R[A]p I a for any p in MIA], where a = K-dim(M). Let xpI" be the leading term of p. Given N < M[L] let Ni = {x E M: x is the leading coefficient in M of a polynomial in M[x] of degree i } . The map N + No + NIA + N2AZ+ shows K-dim(M[A]/R[A]p) < K-dim(M[A]/R[A]x,A"). Next consider M[A] > M[L]A > > M[I]I" 1 R[I]xpL". Each factor has K-dima except possibly the last, which is M[L]1"/R[1]xpAn o (M/Rx,)[I]. But K-dim(M/Rx,) < a so K-dim,,,,((~/~x,)[~])I a by induction; piece the chains togethers.) 23. Suppose R is a left Noetherian ring of K-dim a. Then K-dim(R[A]) as a ring is a+, by exercise 22. 24. If L , > L , > ... is a chain of left ideals of F[1,,1,] such that n i e n L i# 0 then there is n such that L,/L,+, has finite composition length for all i > n. (Hint: K-dim F[A,, L, 5 1 by exercise 22.) Using the Nullstellensatz conclude each of these Li/Li+ is finite dimensional over F. 25. K-dim d , ( F ) = 1 for any field F of characteristic 0. (Hint: Let R be the Weyl algebra d l ( F ) , and show R / L is Artinian for every 0 # L < R. Indeed, otherwise, there is an infinite descending chain L1 > L, > ... containing L; passing to the associated graded algebra F[L,p] apply exercise 24 to conclude Li/Li+ is finite dimensional over F for large enough i. But then R has a simple module finite dimensional over F, which is faithful since R is simple, implying R is finite dimensional over F, absurd.)
A2]/n
,
463
Exercises
26. If S is a left denominator set for R then K-dim S - 'R I K-dim R. Conclude that K-dim K OFK = tr deg K / F for any field extension K of F. 27. Recall in example 2.1.36 that there is a Jaregaonkar sequence {R,:BIa } of PLIDs for any ordinal a; R, = R,[I,; a,] for y = fl', where 0,: R, + R, is an injection. Show by induction that K-dim(R,) = fl, proving there are (noncommutative) PLIDs of arbitrarily large K-dim. An example of a commutative Noetherian domain of arbitrarily large K-dim is given in Gordon-Robson [73, Chapter 91.
Krull Dimension of Non-Noetherian Rings 28. Any module with K-dim satisfies ACC(@) and thus has finite uniform dimension. (Indeed, otherwise, take a counterexample M with a = K-dim M minimal, and take M , = NiI M . Taking suitable infinite subsums yields a chain M , 2 M , 2 M , 2 ... where each M i / M i + ,fails ACC(@) and thus by hypothesis has K-dim 2 61, a contradiction.)
oiz1
FCnl fails to have K-dim, even though R is Lg. as ( 0 FI4) right Z(R)-module and Z(R) is Noetherian. (Hint: R fails ACC(@).) 30. If R is not left Noetherian then R[n] does not have K-dim. (Hint: If Lo < L , < L , < . * * t h e an s R [ 1 ] - m o d u l e ( ~ L i . , l i ) / ~ L i Ixi @Li+,/Lilacks K-dim by exercise 28.) Generalize for normalizing elements. 31. If L 5 Sing(R) is critical then L2 = 0. (Hint: Otherwise La # 0 for some a E L Right multiplication by a gives a map f,:L + La. But K-dim(La) = K-dim(L) by remark 3.5.56, so 0 = kerf, = L n Ann a, contradiction.) 32. Every semiprime ring having K-dim is Goldie. (Hint: Sing(R) = 0 by exercise 31 and ACC @ holds by exercise 28.) 33. Suppose R has K-dim. Proposition 3.5.45holds, so R has ACC(prime ideals). Also every semiprime ideal is a finite intersection of prime ideals, and kdim(R) I cl K dim(R) I K-dim(R). One can also prove NiI(R) is nilpotent (cf., GordonRobson [73, Chapter 51). This seems harder. 34. If Y c Spec(R)satisfies ACC then {finite intersections of members of Y }satisfies ACC. (Hint: Otherwise there is an infinite chain A , c A , c ... where each A iis a finite intersection of elements of 9; let $ = {members of Y minimal over Ai}. Make a graph whose vertices correspond to the prime ideals in Yi, and whose edges correspond to pairs (P,P') where P E $, P' E y+ for suitable i, and P c P'. Show by induction that for any n there are only finitely many vertices having a path of length n to a prime in 9,. Hence there are infinite paths, by the Konig Graph theorem.) 35. Any ring with K-dim satisfies ACC(semiprime ideals). 29. (Gordon-Small) R
=
u
Noetherian Rings of Kdim I (Lenagan) 36. Suppose R is Noetherian and K-dim(R) = 1. If the Artinian radical A(R) = 0 then R satisfies Small's regularity condition. (Hint: Induction on t such that N' = 0, where N = Nil(R). Let T = N n Ann N . Clearly A ( R / N ) = 0. Also
Rings of Fractions and Embedding Theorems
464
A(R/Ann N) = 0 since if LIAnn N is Artinian then La = 0 for each a in N. Thus A(R/T) = 0. By induction if s E W(N) then s + T is regular. We want Anns = Ann's = 0. For a E Anns c T define J R/(N + Rs) + Ra by p = ra. But R/(N + Rs) x (R/N)/(N + RsIN) is Artinian so Ra is Artinian and thus 0, as desired.) 37. Hypotheses as in exercise 36, if L < R and R/L is Artinian then L contains a regular element of R. (Hint: By exercises 12, 36 one may assume R is semiprime. Then any essential complement of L is Artinian and thus 0; hence L is large.) 38. Jacobson's conjecture holds for any Noetherian ring of K-dim 1. (Hint: Let J = Jac(R) and J' = J'. For any regular s in R we have RIRs of finite length so J' G Rs. Let A = A(R). One is done by exercise 36 and 12 unless A # 0. Passing to RIA shows J' E A; also RIAnn, A is Artinian so Ann A contains an element s regular modulo A. Take k by Fitting's lemma such that 0 = Rs'nAnnsk 2 J'.
n,:
Assassinatorsof Modules The following exercises link injective modules to Spec(R). 39. If M is uniform then the assassinator is unique, so we can write Ass(M); in this case Ass(M) = {r E R:rN = 0 for some 0 # N 5 M}. 40. Suppose R is left Noetherian. There is an onto map @:{(indecomposable injective R-modules) + Spec(R), given by E + As@). (Hint: If P E Spec(R) then take a uniform left ideal L of RIP and let E = E(L), noting As@) = Ass(L) = P.) To see when the correspondence of exercise 40 is 1:l we embark on a closer examination using Krull dimension. 41. Suppose R has K-dimfl. For any module M there is some a which is the
smallest K-dim of a nonzero submodule. (Note that every cyclic submodule has K-dim I fl.) Define cr dim(M) to be a; show M has a critical cyclic module of K-dim a. 42. If E is injective and P = Ass(E) then E has a critical cyclic module N with Ann N = P and K-dim(N) = cr dim@); thus N is an RIP-module, so cr dim(E) I K-dim(R1P).On the other hand, equality holds if E = E(L) for L as in exercise 40. 43. The following are equivalent for a let Noetherian ring R: (i) R is fully bounded. (ii) cr dim(E) = K-dim(R/Ass E) for every indecomposable injective E. (iii) The correspondence @ of exercise 40 is 1:l. (Hint: (i)*(ii) Take a critical cyclic module M with P = Ann M = AssE and cr dim(E) = K-dim(M) = a. Writing M x RIL one has core(L/P) = 0 in RIP. By hypothesis LIP is not large and so has an essential complement L' in RIP. Then L' RlL x M so by proposition 3.5.46 K-dim(R1P) = K-dim(L') 5 K:dim(M) = a. (ii)=+(iii) Let P = Ass(E) and CL = cr dim(E) = K-dim(R) where R = RIP. Take a critical cyclic submodule N of E with P = Ann N and K-dim(N) = a. Let L l , ...,L, be critical left ideals of R such that T = @Li is large. Then K-dim(R/T) < a so Hom(R/T,N) = 0. Thus TN # 0 so some Lix # 0 for suitable x in N; comparing K-dim shows the map Li -+ Lix is monic, i.e., N contains a copy of Li, so E x E(Li).(iii)*(ii) by exercise 42. (ii)*(i) Assume R is prime. Suppose L < R is large. Then RIL has a critical submodule Ll/L
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Exercises
whose annihilator is Ass(R/L); inductively we have L = Lo < L , < L, < ... where Ann(Li+,/Li)= Ass(Li+,/Li)# 0 by hypotheses. Then some L, = R and 0 # Ann(L,/L)...Ann(L,/L ,_,) E L.)
Stafford-Coutinho Theory Assume R is left Noetherian. 44. Prove directly that if R is simple (non-Artinian) and M is f.g. Artinian then M is cyclic; in fact, if M = ,Rxi then for any r # 0 there are r: in R such that
M = R(x, + ~ ~ = , r r ~ x(Hint: i ) . By induction reduce to the case t = 2. Take regular s in R with sx, = 0. If Rx, is simple then there is r' in R satisfying Rsrr'x, = Rx, = Rs(x, + rr'x,) and thus also x, E R(x, + rr'x,). Thus one is done unless Rx, is not simple. Take a simple submodule Rax, of M. By induction on composition length M = M/Rax, can be written as R(E, + rr'E,) for suitable r' in R. Then M = R(x, rr'x,) Rax, so as above M = R(x, + rr'x, + rr"ax,) for suitable r" in R.) 45. If R is simple and k-dim R = 1 then for any regular s in R there is r' in R for which R = Rsr' + Rs. (Hint: Apply exercise 44 to M = R/Rs.) 46. Suppose M is f.g. and K-dim M = a. (i) If K-dim R/C 5 a and K-dim M / e M < CL for 1 I j I m then K-dim M/( e ) M < a. (ii) There are only finitely many P in Spec(R) for which K-dim R I P = K-dim M/PM = CL (Hint: (i) Each factor of M > PmMn( 5)M > 5 ) M has K-dim < a, the second factor being an f.g. module over R/(P,,,+ PI n...n P , - , ) . (ii) Otherwise, take an infinite set Pl,P2,... of such P, and let A,,, = PI n . . . n P , . Then M > AIM > A,M > ... has factors all of K-dim 2 a, contradiction.) 47. An example of a Noetherian but not left bounded ring over which $(M,-) is not locally constant. Let R be the Weyl algebra .d,(Z) and M = R/RI where I p = p I + 1. Let Pp = RIP + pR for each prime integer p. Clearly i? = R/pR x d , ( Z / p Z ) is prime, and each of its ideals is an extension of the center (Z/pZ)[IP,pp]. Hence p, = R I P E Spec(i?) so Pp E Spec(R), and i ( M , P,) = l/p. If A 4 R then infinitely many P' satisfy A $Z Pp since n p P p= 0.)
+
ny=-,'
+
n (ny=,
Weak Ideal Invariance A left Noetherian ring R is weakly ideal invariant if whenever K-dim M < a = K-dim RIP for P E Spec(R) and M an f.g. module then Kdim P mRM < a. (The original definition stipulated this condition for all ideals, not merely the prime ideals, but is equivalent to this definition by Brown-LenaganStafford [81].) Noetherian rings often are weakly ideal invariant, but Stafford [8Sa] has found a counterexample. 48. To prove R is weakly ideal invariant one need merely verify that if P E Spec(R) and L < R with K-dim R/L < a = K-dim RIP then K-dim P / P L < a. (Hint: First let M = Rx, and L = Ann,x,, and show P / P L x P @ M via the map a + P L --t a 6 x,; so we have weak ideal invariance for M cyclic. Conclude by induction on the number of generators of M, noting P 6- is right exact.) 48'. If the condition of exercise 48 does not hold then taking L with core(L) maximal we have core(L) E Spec(R). (Hint: Let M = R/L. Then Core(L) = Ann M. By Noetherian induction assume M/N is not a counterexample for all 0 # N < M. Taking critical cyclic N I M with Ann N = Ass N show Ann M =Ann N E Spec(R).
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Rings of Fractions and Embedding Theorems
49. If R is Noetherian and every prime ideal has both the left and right AR-property
then R is weakly ideal invariant. (Hint: Suppose K-dim R/L < a = K-dim RIP but K-dim P / P L 2 a. Then K-dim P / P L = a. Take L with Po = core(l) maximal; P E Spec(R). Take L maximal such, and take n such that P" n L C _ P L . Then P" n L = P" n P L so K-dim((P" + P L ) / P L ) < a, implying K-dim P / P ( P " L) = a and so L 2 P - ' . Thus P"-' E Po so P E Po. Pass to RIPZ and assume P z = 0. Take B a R maximal such that K-dimB < a. Then B E P. Passing to RIB one may assume K-dim A = K-dim R for all 0 # A 4 R. Now let P, = {r E P:rs = 0 for some s regular modulo P}. Then P, = 0, for, otherwise, taking a right assassinator Q of P, and Pz = AnnQ one can use the right AR-property to show K-dim Pz 1, implying K-dim P / P s < K-dim P. Thus K-dim P / P L < a, contradiction. 50. (Jategaonkar) R is called AR-separated if for every P, c Pz in Spec(R) there is Pi c I E P2 for which I / P , satisfies the AR property in R/P,. Any left Noetherian AR-separated ring is almost fully bounded. (Hint: Assume R is prime and N is a large submodule of an f.g. faithful module M. If Ann N # 0 take an assassinator P = Ann N, for N, I N.There is 0 # I , c P and n( 1) > 0 for which N, n I;(')M E I,N1 = 0. Thus one is done unless N, is not large in M ; take an essential complement M' of N, and conclude by induction on uniform dimension.)
'+
Gabriel Dimension 51. Suppose G-dimM = a. Let Nl(M) = x{fl-simple submodu1es:fl I a} and inductively let N , + ( M ) be given by N,+(M)/N,(M)= N,(M/N,(M)), and N,(M) = U{N,.(M):y' < y} for limit ordinals. Then N,(M) = M. 52. Applying induction on y to exercise 51, show every submodule of M contains
a fl-simple submodule for suitable fl < a. Conclude N , ( M ) is large in M. 53. Every nonzero submodule of an a-simple module is a-simple. 54. G-dim is exact, by exercise 52. 55. G-dimxis, Mi = supis, G-dim Mi. (Hint: By exercise 54 assume the sum is direct; then take finite direct sum and finally show any image of Ois,Mi contains a submodule of Oit,,Mi for suitable finite I' E I. 56. If G-dim R = a then G-dim M < a for any R-module M.
4
Categorical Aspects of Module Theory
Although space limitations have cut into our categorical treatment of R-Mud, there are times when the subject demands a strictly categorical viewpoint. Prior instances were seen in 41.4 in the discussion of abelian categories (which is needed for categorical duality), in 52.9 (where we touched briefly on representation theory), and in 83.4 in the more abstract treatment of localization and torsion theories. In this short chapter we collect some of the more celebrated category-oriented theorems about modules. In the main text we present Morita's theorem, which describes when rings have equivalent categories of modules. Then we turn to adjoint functors and see that many previous constructions can be viewed as adjoint functors; thus some properties of the adjoint pair M BT- and Hom,(M,-) can be transferred directly to direct limits, inverse limits, and localization functors.
$4.1 The Morita Theorems Recall two categories V and 9 are equioalent if there are functors F: V + 9 and G : 9 + V such that GF N lo and FG N 1, where N denotes a natural isomorphism of functors; in this case F and G are each called a category equioalence. In theorem 1.1.17 we saw there is a category equivalence R - A u d + M,,(R)-Mud sending M to M'").Category equivalences are important because they preserve categorically-defined notions.
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468
Remark 4.1.1: The following notions are categorically defined and thus preserved under categorical equivalence: monic, epic, kernel, cokernel, exact sequence, direct sum, summand, direct product, projective, injective, projective cover (by proposition 2.8.42), and injective hull (by duality). More generally, direct limits and inverse limits are also categorically defined. For example a: P + M is a projective cover in R-Aod iff a(”):P“ + M(”) is a projective cover in M,,(R)-Mod. This idea whets one’s appetite for proving that a given notion is categorically defined. This cannot always be done: “Free” is not categorically defined since R is a free R-module but R(*) is not a free M,(R)-module. Sometimes it requires some subtlety to prove a notion is categorically defined, as in the case of tg. modules. Definition 4.1.2: An object A of a category %is Jinite if A cannot be written as a direct limit of proper subobjects, i.e., if we cannot write A = %(Ai; cpi) where, in the notation of definition 1.8.3, each is monic but not epic.
This definition is categorical, and is useful because of the following result: Proposition 4.1.3: R-Ad.
An R-module M is f i g . iff M is Jinite as an object in
Proofi (=) (The contrapositive) If M is not an f.g. module then M is a direct limit of proper submodules by example 1.8.9, so is not finite in R-Mud. (e) Suppose M = Rx, could be written as a direct limit of proper subobjects Mj. Then we would have suitable j , with x i E Mj,, 1 < i < t, and so some Mi in the system would contain x,,. ..,x, and thus M, contrary to Mj being a proper submodule. Q.E.D.
xi=,
Since R is left Noetherian iff every submodule of an f.g. module is f.g., we conclude instantly that M,(R) is left Noetherian iff R is left Noetherian. Similarly, R is semiperfect iff MJR) is semiperfect (because semiperfect is described categorically by every f.g. module having a projective cover.) More generally,we could translate many ring theoretic properties of R to any ring R‘ for which R’-Aod and R-A’od are equivalent; in this case we say R and R‘ are Morita equivalent in honor of Morita [58] who proved Theorem 4.1.4: (Morita’s theorem) Rings R and R‘ are Morita equivalent iff R’ x End, P for a suitable f i g . projective module P which is a generator in R-Aua!.
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Categorical Notions The remainder of the section will be spent in digesting and proving Morita's theorem. To do this most expeditiously we shall recall some categorical notions along the way. A is a generator of R-A?od if for every module M there is an epic A") + M for a suitable set I. Remark 4.1.5: R is a generator of R-A?ud. If f: N + A is epic and A is a generator then N is a generator. The following tool is useful in determining whether A is a generator. In what follows M* denotes Hom(M,R), which is in Mod-R by defining the product f r by ( f r ) x = ( f x ) r for all x in M, f in M*, and r E R. Definition 4.1.6: The trace ideal T ( A )= { ~ C i n i , e f aE:A f * , a E A}. T ( A )is indeed an ideal of R since r ( f a ) = f ( r a ) and (fa)r = (fr)a.
Lemma 4.1.7: The following are equivalent for A in R - A u d : ( i ) A is a generator of R-Aod; (ii)T ( A )= R; (iii)R is a homomorphic image of A(")for some n. Proof: (i) * (ii) Suppose h: A") + R is epic. Thus 1 = h(ai)for some (ai)in A('). Writing pi:A + A(') for the canonical injection of A into the i-th component of A('), we can define 1;: = h p i :A + R and see 1 = E T(A). (ii) *(iii) Writing 1 = Cy=,&ai we see the map A(")+ R given by (xl,. . . ,x,) + C f i x i is onto. (iii)+(i) A(") is a generator by remark 4.1.5, clearly*jmplying A is a Q.E.D. generator. Since we are interested in proving theorem 4.1.4 we introduce the following terminology. Definition 4.1.8: generator.
A progenerator is an f.g. projective R-module which is a
Remark 4.1.9: Suppose M is an f.g. projective module over R. If M is faithfully projective then M is a progenerator; in particular, this is the case if R is commutative and is faithful as R-module. (Indeed M = T(M)M by the dual basis theorem, so T(M) = R if M is faithfully projective. To prove the second assertion suppose, on the contrary, A U R and AM M. Taking a maximal
-=
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ideal P of R containing A, note Mp is an f.g. Rp-module and M p = A p M p so Mp = 0 by “Nakayama‘s lemma” since R P is local. This implies there is s in R - P annihilating each of the generators of M so sM = 0, contrary to M faithful.) See exercise 2 for an f.g. projective module which is not a progenerator We shall now illustrate (-=) of Morita’s theorem in two useful special cases.
Digression: Two Examples Example 4.1.10: Suppose P = R(“), a free R-module and obviously a progenerator. Let R’ = EndRP x M,,(R). Then R and R’ are Morita equivalent by theorem 1.1.17. Note P E R - A d - R ‘ . On the other hand, let P‘ = Hom(P, R ) E R ’ - A o d - R . In fact, P’ is isomorphic to R(“’ viewed as an R’-module via matrix multiplication, so (P’)‘”)x R‘, implying P‘ is a progenerator of R’. Then R x End,, p‘. Example 4.1.11: Suppose R is a semiperfect ring. Then R has a complete set of orthogonal primitive idempotents e l , , , , , e , ,from which we take a basic set e l , .. .,e, (rearranging the indices if necessary). Recall this means for each j 5 n that Rej x Re, for precisely one value of i between 1 and t. Let e= e i . Then, obviously, Re is a progenerator, and End, Re z eRe by proposition 2.1.21. But this is the basic ring of R, which is Morita equivalent to R by Morita’s theorem. Let us try to see this fact directly. Indeed, define the functor F: R - M o d + eRe-Mod by putting F M = eM and, given f: M + N defining F$eM + eN by (Ff )ex = f ( e x ) = efx. It is not difficult to prove F is a (category) equivalence by constructing the functor G in the opposite direction, but it is useful to have an intrinsic criterion for F to be an equivalence.
c:=
Recall a functor F: V 9 is full (resp. faithful) if the map Hom(A, B) + Hom(FA, FB) given by f + Ff is epic (resp. monk) for all A, B in Ob V. Then we have Proposition 4.1.12: (cf.,Jacobson [SOB, proposition 1.33) A functor F:%+ 9 is a category equivalence iff F is faithful and full such that for each object D in 9 there is an object A in % with FA isomorphic to D (in 9).
Example 4.1.13: Now we can prove directly and easily that the functor F: R - A o d + eRe-&d of example 4.1.11 is an equivalence. Notation as in
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example 4.1.11, recall from proposition 2.7.25 that given Re, x Rej we have eij in eiRej and eji in ejRe, such that eijeji = ei and ejieij = ej. For convenience we take e,, = ei. For any R-module map f:M + N we have f(ejx) = f(ejieieijx) = ejif(eieijx). In particular, if f(eM) = 0 then f(ejM) = 0 for all j , so f = 0, proving F is faithful. On the other hand, the action of f on each ejM is determined by the action of f on the eiM so F is full. It remains to show that any eRe-module N is naturally isomorphic to FM for a suitable R-module M. Take N, = eiN for i I t, and N, to be a copy of N, as an abelian group with a group isomorphism 'pi: N, + 4 whenever 1 < j I n and Re, x Rej. We take as an abelian group; since R = ejRek as an abelian M = @= group it suffices to define (ejrek)x, for all r in R and xu in Nu, where 1 5 j , k , u I n. This is clear: for k # u we take ejrekx, = 0; if k = u find i, u It with Re, x Rej and Re,, x Re, and define ejreux, = cpj(eijreu,,cp~'x). We leave it to the reader to check this extends N to an R-module.
Morita Contexts and Morita's Theorems Let us turn now to the proof of Morita's theorem, which relies on a collection of data obtained from M E R-&ud. Definition 4.1.14: A Morita Context (also called a set of pre-equivalence data) is a six-tuple (R, R', M, M', z, 7') where R, R' are rings, M E R-Mud-R', M' E R'-&ud-R and z: M Q R *M' --t R and z': M' Q R M + R are bimodule maps under which the following diagrams commute:
M Q M ' O MIT@
RQM
1
l@T'
M OR'
,I
M
M ' Q M Q MI-
IT,@
R'O M'
1
l@T
M'QR
,I
M'
It is clear from the context whether the tensors are over R or R', and so they are left undecorated. Similarly, the unlabelled arrows are the canonical isomorphisms. Note that every requirement comes in pairs, to maintain the duality between the R, M, z and R', M', 7'. In what follows we designate arbitrary elements from their sets as follows: r E R, r' E R', x , y E M, x',y' E M'. We write (x, x') for 7(x 0 x'), and [x', x] for z'(x' Q x). Then the two commu-
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tative diagrams translate to the formulas
(2) [x’, xly’ = x y x , y’). Remark 4.1.15: There is a more concise way of describing a Morita context. Namely,
,:( ):
is a ring under the usual matrix operations, cf.,
example 1.1.10.
Example 4.1.16: The Morita context which concerns us is for M an Rmodule and R’ = End, M. Then M E R-Aud R‘. Let M’ = M * = Horn,( M, R), viewed as an R’ - R bimodule as in remark 1.5.18’. The evaluation map cp: M x M’ -,R given by cp(x,f) = f x is balanced over R‘ since cp(xr’,f) = f ( x r ’ )= (r’f)x= cp(x,r ’ f ) ; thus we get 7:M @ MI-* R given by z(x 0 f ) = f x . We wrote f instead of the promised x’ because of its familiarity to the reader, but now we shall continue with x’. Likewise, we shall write the action of R‘ on M from the right (instead of the more familiar left) to fit into the above setting. Define cp’: M’ x M -,R‘ by taking the action of cp’(x’,x) on M to be given by ycp’(x’, x ) = 7 ( y @ x’)x. Then ycp’(x’r,x) = r ( y 8 x’r)x = t ( y 8 x’)rx = ycp’(x’,rx) so cp‘ is balanced and induces M’ 8 M R‘ which is seen at once to be an R‘ - R bimodule map. TI:
--f
Note that by definition of t’ we have ( x ,x’)y which is (1). To see (2) we note for all y that
= t ( x @ x’)y = x[x’, y ] ,
y[x’, xly’ = ( y ,x’)xy’ = yx’(x, y’) as desired. To add insight to this computation let us return to example 4.1.10, where M = R‘”), written as 1 x n matrices, R‘ = Hom, M x Mn(R), and M‘ = Hom(M,R) which we can identify with n x 1 matrices over R. Now we can view 7,t’ as the usual matrix product, i.e., z(x 8 x’) = xx’ and t(x’ 8 x ) = x’x. Note that t ,7’are epic, since Mn(R)is spanned by t ( e i 0 el) where { e l , . . .,en), {ei, ..., ei} are the standard bases of M,M‘, respectively. This is the final ingredient for the decisive theorem on Morita contexts.
Theorem 4.1.17: (“Morita I ” ) Suppose (R,R’, M, M’, T,t’)is a Morita context with t,t’epic. Then (i) M is a progenerator in R-J#ud and in Mud-R’.
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(i‘) M’ is a progenerator in Mod-R and R‘-Aud. (ii) z, are isomorphisms. (iii) M’ x M* = HOmR(M,R) under the R‘ - R bimodule isomorphism x‘ + ( ,x’). There are three analogous isomorphisms M w Hom(M‘, R)R (under x + (x, )), M x HOmR.(M’,R’) (under x + [ ,XI), and M’ w Hom(M, ROR, (under x’ + [x’, 3). Note we write the subscript on the right when we have maps of right modules. (iv) R’ x End, M under the regular representation. (Likewise, R’ x End Mh, and R % End,. M‘ x End MR,.) (v) R-Aud and R‘-Mud are equivalent under the functors M @ , . and M‘ (Likewise, Mud-R and .Mud-R’ are equivalent.) (vi) Hom(M,-) and MI@- are naturally equivalent functors from R-Mud to R’-Mud.
0,.
Because of symmetry we need only prove the first assertion in each part. Define ( ,x’) in M* as the map x + (x,x’). Since z is onto we have c;=,(x,,x:) = 1 for suitable xi. Likewise, write c j ” = l [ y i , y j ]= 1. The proof largely consists of substituting these expressions for 1 whenever appropriate, to wit: Proof:
(i) M is a generator, by lemma 4.1.7. To see M is f.g. projective we note {(yj,A):1 I j I m } is a dual basis, where J = ( ,yJ); indeed, for all x in M,
(ii) We need to show if zk E M and z; E M’ with a = then a = 0. Indeed 0 = za = l ( z k z, ; ) so
zk Q z; E ker z
n
a=Czk@z; ~ ( x i , x ~ ) = ~ z k @ [ z ; , x i ] ~ ~ = ~ ~ k [ ~ ; , x i ] ~ x i= 1
i.k
i.k
(iii) x’ + ( ,x’) is clearly a bimodule homomorphism, and is monic since ( ,x’) = 0 implies x’ = [ y j , yj]x’ = C y ; ( y j , x’) = 0. On the other hand, for any f E M* we have
cj”=
y;fyj). I implying f = ( , (iv) Let p denote the regular representation, i.e., if r’ E R‘ then (pr’):M + M is right multiplication by r‘. Then p is a ring homomorphism from R’ to HomR(M,M)OP= End, M. Also kerp = 0 for if pr’ = 0 then Mr‘ = 0 and
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r' = c [ y > , y j ] r ' = c [ y ; , y j r ' ] = 0.But p is onto since for any f in End, M we have (in analogy to (iii))
1
proving f = P( y>fyj). (v) Suppose N E R-Mud. Then N x R 8, N x ( M B,, M I ) @ , N x M BRt (M' 8, N) in view of (ii). An analogous argument using z' shows for N' in R'-Aud that N' x M' BR(MOR, N ' ) and so the composite of the functors M' 8,and M OR,- in either direction is naturally equivalent to the identity. (We leave it to the reader to draw the appropriate diagram.) (vi) This follows from the chain of isomorphisms HOm,(M, N ) x Q.E.D. Hom,.(M' 8 M , M' 03 N ) x HOmR,(R', M' N ) x M' 63 N .
Remark 4.2.27': The proof of theorem 4.1.17(i) shows more precisely that M is f.g. projective if z' is onto; symmetrically M' is an f.g. projective R'-module if z is onto. Actually two types of symmetry are at work here-one is between R and R', and the other is the left-right symmetry between R - A n d and Aud-R; this is why we wound up with four assertions for the price of one, with an impressive effect in the following result relating the ring-theoretic structures of R and R' (also, cf., exercise 6). Proposition 4.1.18: Notation and hypothesis of theorem 4.1.17. Then the lattice of ideals of R is isomorphic to the lattice of ideals of R'. In fact, the correspondence L --t M ' L is a lattice isomorphism from 9 ( R ) to y k , ( M ' ) which restricts to a lattice isomorphism from {ideals of R} to {R' - R' bisubmodules of M ' } . Proofi It is enough to prove the last assertion, since applying the two kinds of symmetry observed above would also yield a lattice isomorphism {R' - R bisubmodules of M ' } + {ideals of R'}, as desired. L --t M ' L is obviously order-preserving (under set inclusion) so it remains to show there is an order-preserving inverse correspondence, 0:Y R t ( M ' + ) Y(R), given by N ' + ( M , N ' ) = { ~ ( z , . , a ; ) : z , ~ M , z ; ~ NIndeed, '}. L = RL=(M,M')L= (M,M'L)=(D(M'L) and N ' = [ M ' , M ] N ' = M ' ( M , N ' ) = M ' O N ' . Q.E.D. Morita contexts can be tied in with progenerators in the following nice way.
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Proposition 4.1.19: In the Morita context of example 4.1.16, t is epic if M is a generator; t’ is epic if M is f . g . projective. Consequently, Morita 1 applies if M is a progenerator. Proof: t is epic since T ( M ) = R by lemma 4.1.7. If M is f.g. projective then the dual basis lemma gives us x i in M and xi in M * for 1 Ii It satisfying x = x , ( x ; , x )= C [ x i , x J x for all x in P; thus C [ X , , X : = ] 1, proving t‘also is epic. Q.E.D.
c:=
Proof of Morita ’s Theorem and Applications Proof of Theorem 4.2.4: (i) (ii). Let F : W - d u d + R-Mud be a category equivalence. Then P = FW is a progenerator of R, and W x Endw W x End, FW = End, P. (ii) 3 (i) Construct the Morita context (R, W,P, P*, z, z’) of example 4.1.16, noting T,7’are onto by proposition 4.1.19. Then R and W are Morita equivalent by Morita I. Q.E.D. Remark 4.1.20: Morita’s theorem gives fast proofs of the following facts that previously required considerable effort:
(i) Wedderburn-Artin Theorem. If R is simple Artinian then R is a direct sum of minimal left ideals, each of which are isomorphic to each other by proposition 2.1.15. In other words, any minimal left ideal L of R is a progenerator; thus R x End L,, where D = End, L is a division ring by Schur’s lemma. (ii) If M is an f.g. module over a semisimple Artinian ring R then End, M is semisimple Artinian (since M is projective.) (iii) If P is an f.g. projective module over a semiperfect ring R then End, P is semiperfect. Morita’s theorem is tied more closely to matrices in exercises 7,8. Note that this theory gives very explicit information, and often one builds a Morita context in order to pass information concretely from one ring to a Morita equivalent ring. We conclude by introducing some ideas which “spin off” from Morita I. Remark 4.2.22: The proof of theorem 4.1.4, coupled with Morita I, shows that any categorical equivalence F: W - d o d + R-Modis naturally isomorphic to the functor P Ow- where P = FW. This give rise to question as to
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which functors can in fact be given by tensors; an answer is given by Watt’s theorem (exercise 4.2.8).
Digression: Simple Noetherian Rings Of course Morita theory can be applied by showing a given ring is Morita equivalent to a ring with more manageable structure and then transfering Morita-invariant information. Along these lines, example 4.1.13 shows any semiperfect ring is Morita equivalent to its basic subring. Hart and Robson proved a Goldie ring is Morita equivalent to an Ore domain iff it has a uniform left ideal which is projective. This thread was carried further by Cozzens-Faith [75B, theorem 2.171, which characterizes a simple ring with a uniform left ideal as the endomorphism ring of an f.g. projective right module over an Ore domain having at most one proper nonzero ideal. Their book contains other interesting results along these lines, and it was hoped that there might be a way of classifying simple Noetherian rings as matrix rings over suitable domains. This project hit a snag when Zaleskii-Neroslavskii found a simple Noetherian ring without idempotents, and then Stafford [78] showed their example is not Morita equivalent to a domain. The verifications are still rather complicated, but we shall describe some of the constructions.
Example 4.1.22: In each of these examples R is a simple Noetherian hereditary ring which is not a matrix ring over a domain. Details for (i) and (ii) are in Goodearl [78]. (i) Let Ro = dl(R),the Weyl algebra over the reals, cf., example 1.6.32, and let a be the automorphism sending the generators I and p, respectively, to -1 and - p . Then G = { l,a} is a group of automorphisms of R,, and the skew group ring R = R , * G has uniform dimension 2. (Indeed, R is not a domain since (1 a)(l - a) = 0, but is free of rank 2 over R,.) Obviously R is Noetherian since R , is Noetherian, and it is easy to see R is also simple and hereditary. If R were a matrix ring M,(D) for D a domain one must have n = 2, so there would be an element whose square is - 1, and this is shown to be impossible. Recently Goodearl [84] has found a much more direct verification of this example using results from K-theory, along the lines of (iii) below. Furthermore, replacing R by a field containing a primitive n-th root 5 of 1 and taking a instead on Ro such that a I = CI and ap = cp, he obtains an example of a simple hereditary Noetherian ring of uniform dimension n which is not a matrix ring over a domain, for any n 2 2.
+
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(ii) Let F be a field of characteristic 2, and C = F(A)[p,p-'] where p,1 are commuting indeterminates over F. Then C has an automorphism a of infinite order given by r p = 1 - l ~let ; R , = C * (7). Now let a be the automorphism of Ro over F(1) given by ap = /.-'and or = r-'. R , is Noetherian by the Hilbert basis theorem (modified for skew polynomial rings) and then localization; hence arguing as in (i) we get R simple Noetherian not a domain. It turns out that R has no nontrivial idempotents, and so by results of Stafford [78] has Krull dimension 1 and is not Morita-equivalent to a domain. (iii) Stafford-Warfield [83] have found a method of constructing simple hereditrary Noetherian domains R whose modules have unusual but precisely described behavior. Their example 3.3 possesses a nonfree projective P such that P @ R(") for n > 1 is never of the form Q'").Consequently, End(P 0 R'")) cannot be a matrix ring over a domain. (This trick was first used by Hart.) Example 4.1.22(ii) has infinite global dimension (cf., Chapter 5), so there is still hope that every simple Noetherian ring with finite global dimension is Morita equivalent to a domain. Having wandered into the existence question, we might as well become entangled with the uniqueness question as well. Namely, if M J D ) z M,(D') for Ore domains D and D' need D z D'? S. P. Smith [81] has found a counterexample: D is the Weyl algebra dland D' = End,L where L = D1' + D(1p + 1) and n = 2. The endomorphism ring of any projective left ideal L of dlhas come under close scrutiny, largely because L'" x d\') and thus M,(End, L ) x M 2 ( d 1by ) Webber [70]. Stafford [86] has vastly generalized Smith's result by proving End L x dl whenever L is noncyclic and has many other interesting results along these lines.
84.2 Adjoints The purpose of this section is to see what lies behind the adjoint isomorphism (proposition 2.10.9). In fact, certain categorical principles are just beneath the suface and can be used in sundry applications. Definition 4.2.1: Suppose F : W --+ 9 and G: 9 + W are functors. (F, G) is an adjoint pair if for each C in ObW and D in O b 9 there is a bijection qC,,: Hom,(FC, D)+ Hom,(C, G D ) which is natural in the sense that qu is a natural isomorphism from Hom(FC,-) to Hom(C,-)G and q-,, is a I-
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natural isomorphism from Horn(-, D)F to Horn(-, an adjunction, and F is called a lejit adjoint of G.
GD). (F, G, q) is called
Example 4.2.2: Take W = S-Aud and 9 = R-Mud and A E R-dud-S. By proposition 2.10.9 the functors F = A BS- and G = HOmR(A,-) are an adjoint pair.
These two functors have entered into many categorically-oriented theorems about modules, so it is natural to ask what one can conclude merely from the fact they are an adjoint pair. The results are rather startling. As a preliminary we give a slight generalization of lemma 2.1 1.12. Given a ringR define its test module E to be the injective hull of O L X R R / L . E is injective. Moreover, for any R-module N and any given --* E with fx # 0. (Indeed, since E is injective we may replace N by its submodule Rx x RIL where L = Ann, x; hence take f to be the composition of canonical monics Rx + 0RIL + E.) Remark 4.2.3:
x in N there is a map f: N
A module satisfying the properties given in remark 4.2.3 is called an injective cogenerator (cf., exercise 4.1.4).
Lemma 4.2.4: Let E be the test module for R, and let # denote the contravariant functor Horn(-, E ) from R-.+ud-S to S - A d . Then a sequence K+ fM b '+ N in R-Aud-S is exact iff N * 8, M # -f," K' is exact. Proof;. Analogous to lemma 2.11.12, so details are left to the reader. Q.E.D.
Proposition 4.2.5: Suppose F: S-Aud +R-Aud and G:R-Aud +S - A d are an adjoint pair. Then F is right exact and G is left exact. Proof;. Suppose 0 + K + M + N is exact in R - A d . By naturality we have a commutative diagram 0 + HOmR(FK,E ) + HomR(FM,E ) + HOmR(FN,E )
3330 + Homs(K, G E ) + Homs(M, G E ) + Hom,(N, G E )
where E is the test module for R, and the vertical lines are isomorphisms. The bottom row is exact since Horn(-,A) is left exact for any module A; hence the top row is exact, so FN + F M + FK + 0 is exact by 4.2.4.
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We shall see that proposition 4.2.5 is actually a special case of theorem 4.2.9 (cf., exercise 5). A lovely theorem of Watts (given in exercise 8) shows that any left adjoint F : R-Aud -+ T-&od is given by a tensor functor; dually any right adjoint G: R-Aud -+ d b is given by a Hom functor. Nevertheless, there are many other interesting examples of adjoints, as we now see.
Adjoint Pairs and Universals Proposition 4.2.6: Suppose G:%? -P 9 is a functor such that for every object X in %? there is a universal from X to G. Calling this universal (Ux,ux) we can define a functor F: 9 -+ % as follows: FX = W,, and for any morphism f:X + Y in 9 take Ff to be that (unique) morphism such that G ( F f ) completes the diagram
X
ux
>,GU,
i.e., (GFf )ux = u yf . Then (F,G ) is an adjoint pair.
Proof;. We use the uniqueness of the universal property. Clearly F1, = IF,. Also iff: X -+ Y and g: Y + Z are morphisms then G(FgFf )ux = GFgGFfu, = (GFg)u,f = uzgf proving FgFf = F ( g f ). Thus F is indeed a functor. To prove (F,G) is an adjoint pair we must find the natural bijection v,,~: Hom(FX, A ) --* Hom(X, GA) for all X in Ob%?and A in O b 9 . Given f:FX + A we take q X . Af to be the composition (Gf )ux:X -+ GU, --* GA. This is clearly natural, and it remains to find the inverse correspondence. Indeed given g: X + GA take the unique f: U, + A such that Gf completes the diagram
Then g
=
Gfux = qX,.,f , as desired.
Q.E.D.
Actually every adjoint pair arises in this manner, cf., exercise 2. This point of view is particularly satisfying since it enables us to view the construction
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of universals functorially; we already did this in several special cases, most notably the localization functor, which we should like now to view in this setting. Corollary 4.2.7: Under the hypothesis of proposition 4.2.6, the functor F taking each object to its universal is right exact. Example 4.2.8: The direct limit is a universal when it exists and is thus a right exact functor from W' to W.
One special case of a direct limit is the localization functor of theorem 3.3.21, for which most of the proof of its exactness could be eliminated using corollary 4.2.7. (Also, cf., exercise 3.1.4.) This approach also makes exercise 3.4.11 much easier. The localization functor is characterized in exercises 13ff. Let us return to adjoints in general. Theorem 4.2.9: limits.
If F: %?+ 9 is a leji adjoint then F commutes with direct
Proofi First note that if ( A i ;cpi) is a system in W' then (FA,; F q j ) is a system in 9';furthermore, if L = %1 Ai with p i :A , + L then we claim F L and Fp, satisfy the requirements of definition 1.8.3. Indeed let (F, G, q) denote the adjunction, and write qi for qA,, for given D in O b D. Given any family of morphisms g,: F A , + D satisfying gi = gjFcp; for all i I j we put f i = q i g i : A , + GD. Naturality of q provides the following commutative diagram: H o m ( F A j ,D )
%
,H o m ( A j ,G D ) .c.
.c. H o m ( F A i ,D)
H o m ( A , , GD)
Thus &cpi = (cp{)"& = (cp{)#qjgj= qi(Fq{)#gj= qigi = f i for all i Ij , implying there is a unique map f: L + D satisfying fpi = fi for all i. Take g = q& f . Then the commutativity of H o m ( F L ,D )
+
qL'D
Horn& G D ) J.
Hom(FA,,D) A H o m ( A , , GD)
Exercises
481
shows q i ( g F p i )= qi(Fpi)#g = p#qL,Dg = p? f = f p i = 1;: = q i g i ; since qi is 1:1, we conclude g F p i = gi. To show g is unique note that if g'Fpi = gi for all i then the last square yields (qL,Dg')pi = 1;: for all i, implying qL,Dg' = f and so g' = g . Q.E.D. This result generalizes proposition 1.8.10 but actually is much wider in scope, since there are so many examples of adjoint pairs. Recalling that the direct sum of modules is a special case of direct limit (where I is given the trivial preorder, we have Example 4.2.10: Suppose for each j that { Mij: i E I } is a direct system of Mij % lim Mij. In particular, the localization modules. Then lim functor of theorem 3.3.21 commutes with direct limits. Dual to theorem 4.2.9 we have
Oj
Theorem 4.2.11:
Oj
If G is a right adjoint then G commutes with inverse limits.
fro08 Apply theorem 4.2.9 to the dual categories. Corollary 4.2.12: Hom(M,-) lar, preserves direct products). Corollary 4.2.13:
commutes with inverse limits (and, in particu-
Completions commute with inverse limits.
Thus we see for any adjoint pair (F,G), F commutes with direct limits and G commutes with inverse limits. This circle of ideas is completed in the exercises.
Exercises $4.1
XI=,
1. Any left ideal L of a simple ringR is a generator. (Hint: If 1 = airi for a, E L and ri E R then the map L(")+ R given by ( x , , .. . ,x,) + xiri is epic.) 2. Example of a faithful cyclic projective module which is not a generator: Let R = {lower triangular matrices over a field K}, and P = Re, Note T(P)= P # R.
1
,.
3. The following are equivalent: (i) A is a generator of R - A o ~(ii) . The functor Hom(A,-) is faithful. (iii) For any module M there is an epic A"' + M where I = Hom(A,M).
Categorical Aspects of Module Theory
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4. A cogenerator of a category V is an object A such that Hom( , A ) is a faithful functor. Show A is a cogenerator of R-&d iff for any module M there is a monic M + A' for a suitable set 1. (Hint: 1 = Hom(M, A).) Further results on cogenerators are in Anderson-Fuller [74B]. 5. Q/Z is an injective cogenerator for Z - J h d . In general, R-Aod has an injective cogenerator (see remark 4.2.3). 6. If R, R' are Morita equivalent then Z(R) x Z(R'). (Hint: As in theorem 4.1.176~) view R z End M,. and R' s End, M in End, M, where they are centralizers and note their centers are both End, M n End MR,.) 7. Suppose e E R is idempotent. Re is a progenerator of R-.Mod iff R = ReR. (Hint: (=s) 1 E T(Re) implies 1 = x f i ( r i e )= Criesi where si = fie E R. (e) Go backwards, where J is right multiplication by si.) 8. R and R' are Morita equivalent iff for some n there is an idempotent e of W = M,,(R) such that WeW = Wand ewe x R'. (Hint: (*) Let P be a progenerator with R' x End, P. P is a summand of R'") for suitable n. Then End, R(")x W and P = R(")eso End, P x ewe. Identifying R with el We, show P x el We and T ( P )= e,, WeWe,,; since el, E T(P)conclude WeW = W. (-=) Using exercise 7 note We is a progenerator of W, so W is Morita equivalent to ewe x R'.) 9. Consider the monic @: Hom(P, P,) -,Hom(P, 4) sending (gi) to the map x -,(gix). A projective module P is f.g. iff @ is an isomorphism when each 4 = P. (Hint: (e) Use a dual base {(xi,fi):iE I} to show there is g: P -,0 6 given by gx = ((fix)xi).) 10. (Eilenberg) If P is a progenerator for R then P(') x R"). (Hint: by remark 2.8.4') 11. (Camillo [84]) Rings R and T are Morita equivalent iff End, R") z End, T'") as rings. (Hint: (*) Write R,,, for End, R'". Using exercise lOshow R(,) x End, P(,) x (End, P)") for a progenerator P. (-=) the hard direction. Let cp: R(N)4 qN)be the isomorphism; letting el R'" + R'" denote the projection onto the first component show (pel ,)T") is a progenerator, using exercise 9.)
,
Oi,,
,
OiEN
Balanced Modules and the Wedderburn- Artin Theorem The following exercises due to Faith [67] show how Rieffel's short proof of the Wedderburn-Artin theorem (exercise 2.1.8) fits into the Morita theory. In what follows let M E R-Aud, R' = End, M, and R"=End MR,. We say M is balanced if the regular representation yields an isomorphism from R to R". In particular, any balanced module is faithful. 12. If there is an epic I[: M + R then M is balanced. (Hint: R is projective so x splits. Write M = R fl3 N.It suffices to prove any r" in R" is given by left multiplication
by r"(1,O). Given x in M there is r' in R' such that (1,O)r' = x. Viewing 7~ E R' shows r"(1,O) = (r"(1,O))nE R so r"x = (r"(1,O))r' = r"(l,O)((l,O)r') = r"(1,O)x.) 13. If M'") is balanced then M is balanced. (Hint: Look at the diagonal.) 14. M is a generator of R - d o d iff M is a balanced R-module and an f.g. projective R'-module. (Hint: (*) There is an epic M(") -+ R, so M(") x R @ N and M is balanced. Moreover (R')'")x Hom(M("),M) x M @ Hom(N, M). (+) If R"") x M @ N in Aod-R' then M'") x Hom(R', M)'") x R" @ Hom(N, M) so M is a generator for R" x R. 15. If R is simple and L is a left ideal then L is an f.g. projective right module over
Exercises
483
T = End, L , and R 2: End L,. T is a simple ring iff L is f.g. projective as an R-module, in which case R and Tare Morita equivalent. (Hint: L is a generator, by exercise 1). Use this result to refine Koh’s theorem (exercise 3.3.34.) concerning simple rings having a maximal left annihilator.
$4.2 1. Using exercise 2.1 1.0 generalize proposition 4.2.5to arbitrary abelian categories.
2. (Converse to proposition 4.2.6)Suppose (F, G, q ) is an adjunction, where F: 9 -+ V and G: V -+ 9. For every X in Ob 9 there is a universal (U, u ) from X to G , where U = F X and u = qx,FxlFx. (Hint: Yoneda’s lemma) Moreover, F is then the functor obtained from G using proposition 4.2.6. 3. Using exercise 2 show that any two left adjoints of a functor G are naturally iso-
morphic. State the dual result. 4. Write the direct limit explicitly as part of an adjoint pair. 5. If F commutes with direct limits then F is right exact, by exercise 1.8.2. Dually, if G commutes with inverse limits then G is left exact. Deduce proposition 4.2.5 as a
consequence of theorem 4.2.9.
6. The pushout commutes with sums; the pullback commutes with products. 7. State precisely the following facts: Direct limits commute; inverse limits commute.
Watt’s Theorems 8. Suppose F : A d - R + A d - T is a right exact functor which preserves direct sums. Then F is naturally equivalent to - @,I M where M = FR, viewed as an R - T bimodule under the action rx = (Fp,)x where p,: R + R is left multiplication by r. (Hint: More generally, for any right R-module N there is a balanced map N x M + F N given by (a,x) + (Fp,,)x for a in N ; this induces a map qN: N @I M + FN, and q is a natural transformation from - @IR M to F. Note qN is an isomorphism for N = R, and thus for N free. But any R-module N has an exact sequence F, + F2-+ N + 0 for F, ,F2 free; conclude using the five lemma as in exercise 2.11.13.) 9. The localization functor for a nonsingular ring R is given by taking the tensor with the ring of quotients of R. 10. The following are equivalent for any functor F : R - A u d + T-Aud:(i) F is the left adjoint of some functor G : T - A d + R - A d ; (ii) F commutes with direct limits; (iii) F is right exact and commutes with direct sums; (iv) F is naturally equivalent to F R @IR-. 11. Suppose G: R - A d + d b is a functor which commutes with inverse limits. Then G 2: Hom(M,-) for some R-module M. (Hint: Let E be an injective cogenerator for R - A d and let E’ = E’ where 1 is an index set of cardinality GE. Note the identity map l G Eis in GE’ 2: (GE)’, and the x-th coordinate is x. Projecting onto each coordinate show the map n: Hom(E’, E ) + G E given by nf = (Gf)lGE is epic. Let M = { N 5 E’: 1 G E E G N } , noting that if N E‘ then G sends the inclusion map N + E’ to a monic G N + GE’, so we can view G N I GE’. In fact, there is a
Categorical Aspects of Module Theory
484
natural transformation q: Hom(M,-) + G,where qA:Hom(M, A ) -S G A is given by f -+ ( G f ) l G E . On the other hand, M = &{N: IGE E G N } by definition, so GM =@G N = G N contains lGE. The inclusion map M + E' yields an epic Hom(E', E) -+ Hom(M, E) since E is injective, and the kernel is ker n. Thus qE is an isomorphism, implying qA is an isomorphism for any product A of copies of E. But since E is a cogenerator, any module A is a submodule of a product of copies of E, so for suitable sets l', I" one has 0 -+ A -+ E" -+ El" exact; conclude using the five lemma.) 12. The following assertions are equivalent for any functor G:R-&d + d d : (i) G has a left adjoint F: d d -+ R - d ~ d (ii) . G commutes with inverse limits. (iii) G is left exact and commutes with direct products. (iv) G is naturally isomorphic to Hom,(M, -) for some R-module M.
n
Grothendieck Categories and the Popescu-Gabriel Theorem, cf., C supplements. An abelian category V is Grothendieck if V has coproducts and a generator, and the lattice Y ( V )is upper continuous (cf., exercise 0.0.4). 13. R 4 0 d is a Grothendieck category, as is R-dod-R.
A category W is Giraud if there is a faithful full functor G: V -+ R-Aod which has a retraction which preserves kernels. We can view G as the inclusion functor and say V is a Giraud subcategory of R-&d.
(x9)
14. If is a torsion theory for R - d o d then 9 is a Giraud subcategory of R-Aod, taking FM = M/Rad(M) where Rad is the corresponding radical. 15. (Converse of Popescu-Gabriel) Any Giraud subcategory of R - d u d is a Grothendieck category with generator F R (where F is the retraction). 16. Define injective object and essential extension for an abelian category, and define the injective hull E ( A ) to be an injective object which is an essential extension of A. Prove in every Grothendieck category V that any object has an injective hull. (Hint: Let C be a generator for W. Then Hom(C,-) identifies W with a subcategory of d6,which has injective hulls, so Zorn's lemma shows every object of V has a maximal essential extension, which thus is an injective hull in V.) 17. (Popescu-Gabriel) Suppose W is a Grothendieck category with generator C. Put R = Hom(C, C)op.Then Vcan be identified with a Giraud subcategory of R-&od via the full and faithful functor G = Hom(C,-). (Hint: First note Hom(C, A ) can be viewed naturally as R-module by means of composition of morphisms. G is thus a functor, and is faithful since C is a generator. To prove G is full one needs to show every map cp: Hom(C, A ) + Hom(C, A') is of the form f# for suitable f: A -+ A', that is cpg = fg for each g:C -+ A . Write Hom(C, A ) = { g , : i E I } . The g i : C + A induce a morphism h' = @cpgi: C(" -+ A'. But h'(ker h) = 0 since C is a generator, so h' = fh for suitable J A -+ A', an cp = . Now define F: R - d o d + by taking R(') R(J)4M -+ 0 and letting FM = cokerf' where f': C'') -+ C J and ) Gf' = f.Since G is full and faithful, one defines F to be the inverse of G on morphisms, and now it is clear that F is a left adjoint of G . Hence F is right exact. Suppose F can also be seen to be left exact. Let V' denote the full subcategory GV of R - d u d . Then G induces a category equivalence
e p
Appendix A
485
G’:V + V‘, and G‘F: R-Aod + W‘ is exact and a left adjoint of the inclusion functor V’ R-Aod. The proof F is left exact follows Takeuchi [71]. First show that if cp: M -+ GA is monic then Fcp: FM -+ FGA is monic; this is seen by showing the map f of the first paragraph is monic. The remainder requires some ideas from Chapter 5. Take a free resolution ... -+ P, -+ Pn- -+ * . * + M 0 of M. If K, is the n-th syzygy theneach F K , + F P , i s m o n i c , s 0 ~ ~ ~ - + F P , - + F P , ~ ~ - , ~ ~ ~ ~ aFpM r o--+ 0 i s jective resolution of FM. Thus the first derived functor L,F of F is 0. But now F is exact by corollary 5.2.24,also, cf., exercise 5.2.4. 18. There is a 1: 1 correspondence between idempotent filters 9of left ideals of R and Giraud subcategories of R-Aod. (Hint: Given f let A ’ = { M E R - A d : the canonical map Hom(R, M )-+ Hom(L, M )given by restriction is 1:l and onto for all L in f }called , the f-closed modules. Then A ’ is a Giraud subcategory of R - A d . Conversely, let V be a Giraud subcategory of R - k o d with retraction F: R - A d -+ V and F = { M E R-Aod: FM = O}.) 19. Any Grothendieck category with torsion radical F gives rise to the “localization” E , ( M / Y M ) . This is the “modern” point of view; Fis also called a torsion theory. -+
-+
Appendix A. The Proof of Magnus’ Theorem and the Magnus-Witt Theorem (Theorem 1.3.38) Proof of Mugnus’ Theorem; Let B be the ideal of Z{X} generated by X f : i E I, and identify X i with its image in R = Z { X } / B . Then we can view the elements of R as “polynomials” and talk of the (total) “degree” of the terms in the X i . Note, however, that X z = 0, so (1 + Xi)-’ = 1 - Xi. Letting H be the multiplicative subgroup of R generated by 1 + X i and 1 - Xifor each i, we thus obtain an onto group homomorphism cp: G + H given by cpX, = 1 + Xiand q X ; ’ = 1 - X i . Then cp(Xy) = (1 + X,)”’ = 1 + rnXi for g ~in reall rn in Z. Writing an element a in H in the form g ~ * . . - as mark 1.3.36weseea = (1 + rn,g,)...(l + rnrgt)= r n , ~ ~ ~ r n , g ,+ ~ ~termsof ~g, lower degree, so, in particular, cpa # 1 for a # 1, proving cp is an isomorphism. Moreover, the constant term of cpa is 1. Letting a(u) be the sum of all terms of cpa having degree u we see a = 1 + a(u).(For example, if a = X,X , then cpu = 1 + X , + X , + X , X 2 so a(1) = X, + X2and 4 2 ) = X,X,.) We claim if a E G, then a(u) = 0 for all 1 Iu I n. This will show for aE G,,that a(u) = 0 for all u 2 1, so a = 1, proving the theorem. To verify the claim we proceed inductively on n, noting it is vacuous for n = 0. Write a = g-’b-’gb for suitable g in G and b in G,-,. Then b-’ E G,-, so b(u) = 0 = b-’(u) for 1 5 u In - 1, by induction. Since we are interested only in terms of degree I n, we let A = {polynomials whose nonzero terms all have degree > n } ; noting A a R we pass t o a = R / A , for a(u) = a(#) for
xu?
Categorical Aspects of Module Theory
486
+ +
+
1 I u I n. But cpb = 1 b(n) and cpb-’ = 1 b-’(n), so 1 = cpbcpb-’ = (1 b(n))(l b-’(n)) = 1 b(n) b-’(n), proving b-’(n) = -b(n). Hence
+
+
+
= cpg-lcpg = 1
Q.E.D.
proving a(u) = 0 for 1 I u I n as desired.
Proof of the Mugnus-Witt Theorem. The key new concept of this proof is a certain ordered class of commutators, called basic (group) cornmututors, plus a “collection procedure” of P. Hall which gives much more information about G; a fundamental connection between ring commutators (defined by [a, b] = ab - ba) and group commutators (defined by (9,h) = g-lh-lgh) is revealed in the proof. To develop the ideal fully we bring in some group theory and then introduce the ring theory connection to complete the proof. Basic group commutators are defined in terms of weight as follows:
(9 Each X i is called a “basic group commutator”
of weight 1 (although it is not a commutator). (ii) The basic group commutators of weight n are ordered arbitrarily with respect to each other but are declared to be greater than all basic commutators of weight < n. (iii) Inductively, given all basic commutators of weight I n we define the basic group commutators of weight n 1 to be of the form (cy,c,) where cu > c, are basic commutators whose weights have sum n 1, and such that if cu = (ci,c,) then c, 2 c,.
+
+
This definition is motivated by the attempt to write a string of elements in ascending order. For example, suppose we wanted to rewrite the element a = XzX3Xl. First we would try to move X1to the left; we “switch” X, with X3 by using the formula X3X,= XlX3(X3,Xl) and continue as follows: a = xZx1~3(x3~x1) = xlxZx3(xZ~xl)((xZ~
=x
lxZ(~Z~~l)~3(~3~~1)
x l ) ~ x3)(x3~
xl)
= X ~ X ~ ~ ~ ( X Z , xl)((XZ~xl)~ X~)(X~Y x 3 ~ ~ ~ ~ x Z ~ x 1 ~ ~ x 3 ~ ~ ~ x 3 Y x 1 ~ ~ ~
Appendix A
487
Thus a is finally written in ascending order in terms of the basic (group) Xl1, ((X,, X1), X 3 ) ,and (((X,, Xl1, X 3 ) , commutators Xl ,X,,X, (X,Xl 1, (X,, (X,,Xl)). The process of arranging the basic commutators in ascending order is called the collection procedure but need not terminate because every switch generates a new basic commutator. However, it does terminate modulo G,, for any n, as we shall see now by examining the collection procedure more carefully. Starting with a word w = Xi, *.axik viewed as a product of k basic group commutators of weight 1 (namely Xi, ... Xi,), we define the collection steps, given inductively as follows. Suppose after the previous collection step we have written w = c1 " ' c , where each c, is a basic commutator. The collected part is that part c1 ...ck (for k maximal) such that c1 < ... < ck; any remaining c, is called uncollected. Take the smallest uncollected basic commutator in its leftmost occurrence c,. Then c,c, (for, otherwise, c,I c, so c U p lis collected by hypothesis, implying c, also is collected, a contradiction.) The collection step is to rewrite c,- lc, = c,c,- l(cu- c,). If c,- = (ci,cj), then we must have collected cj in order to form cup1, so (c,- 1, c,) is a basic commutator. The collection step thus creates a new basic commutator while pushing c, one place to the left. Thus in order to collect c, we may have to create many basic commutators of the form (g,c,). If c, E G, then (g,c,) E G,+ 1, so after a finite number of steps we can collect all basic commutators not in G,,, proving the following result: For given n, any word in the Xi can be written in the form c1 * - . c,a where cl I ... I c, are basic group commutators and a E G,,. A more intricate argument enables one to prove that any element in the free group can be written in the form c;l. * . c: a where c1 I * * Ic, are basic commutators, a E G,,, and each ci = & 1. To see this we need a collection procedure for inverses of basic commutators, i.e., to rewrite expressions c,ci', c;'cj, and c i ' c i ' where ci > cj are basic group commutators. This is a tedious procedure, which we describe for tic;'; the other cases are similar, cf., Hall [59B,p. 1671 for greater detail. Define uo = ciand, inductively, urn+ for (u,, cj).Then for each u, we have the equality
=-
1 = (u,, c;
l)(U,,
Cj)((U,, Cj), c
Solving and iterating yields
i
= (urn,c;l)u,+ l(U, + 1, CJ
).
Categorical Aspects of Module Theory
488
Clearly, urnE G,, for m 2 n so we get (Ci,C7')
= (uo,c;')
= u,(u,,c;')u~' = u2u4(u4,c;')u;'u;'
= *..
= U 2 U q U 6 ' * * U 5- 1 0 3 -1 01 -1 (modG,,)
and qc;' = c~'ciu2u4u6~~~u;'u;'u;'(mod G,,),the desired step switching ci and c; since each urn is a basic commutator. Our ultimate goal is to show that the images of basic group commutators form a base for the (free) abelian group G,,/G,,+l. We have seen they generate the group, so it remains to show they have no dependence relation. Unfortunately, it is difficult to describe multiplicative dependence relations, so we shall simplify matters by introducing a ring. Let A,, be the ideal of R = Z{X} generated by {polynomials whose nonzero terms all have total degree > n } , and let R , = R / A i . We proceed as in the proof of Magnus' theorem, this time focusing exclusively on low order terms. Since (1 + Xi)-' = C:=,( - l)iXiwe have a group homomorphism cp from G to the given by substituting multiplicative subgroup of R generated by the 1 + Xi, 1 + X i for X i . For any g in G write g(u) for the sum of those terms of cpg of degree u. Given a basic group commutator ci, we define the corresponding basic ring commutator di to be obtained by substituting [ , 3 for each appearance of ( , ). Viewing the Xi now in R, we can evaluate every basic ring commutator in R . Note that the degree of di obviously equals the weight of c i . As in the proof of Magnus' theorem we see readily that cpc, = 1 + ci(u).
'
Lemma A: Suppose ci is a basic commutator of weight m. Then ci(u)= 0 for 1 I u < m, and ci(m) = d,. Proofi We work in R , (since what happens in degree > m is irrelevant). For m = 1 the result is instant (since X i = 1 + Xi) so we proceed inductively on m, writing ci= (cj,ck)where, by induction hypothesis, cpcj = 1 + dj + and CpCk = 1 + dk + "'. Write U = CpCj - 1, a' = V(C,;') - 1, b = cpck - 1, b' = &;)' - 1. Then 1 . 1
+ a')(l + a) = 1 + a' + a + a'a, so a' + a + a'a = 0 and cpc;' = 1 - dj + ... also ab = didk = a'b' Likewise, b' + b + b'b = 0 and cpc;' = 1 - dk + 1 = cpCj'cpCj = (1
m e - ;
Appendix A
489
because all other terms have degree 2 m. Hence
+ a')(l + b'))((l+ a)(l + b)) = (I + a' + b' + djdk)(l + U + b + = ( 1 + Q' + b')(l + u + b) + 2djdk = 1 + a' + b' + u + a'a + b'a + b + arb + b'b + 2djdk = 1 + (a' + u + U'U) + (b' + b + b'b) + b'a + a'b + 2djdk =1- djdk + 2djdk = 1 + [dj,d k ] , as desired. Q.E.D. for lemma A.
C ~ C ;= (( 1
djdk)
dkdj
Lemma A implies immediately that the lowest-order nonconstant term of c:'..c: is mudu (if nonzero), so the Magnus-Witt theorem will be proved if we can show the basic ring commutators of degree n are linearly independent over 2'. This is a much easier problem, since noncommutative multiplication has been replaced by commutative addition and is obtained by proving the following stronger statement about Q { X } :
,
Lemma B: Let V , be the Q-subspace of Q { X , , . . .,xk}spanned by all monomials of degree n, and let B = { f E V,: f is a product of basic ring commutators d i , ... dit where i , < i , < ... < i, ( t arbitrary)}. Then B is a base of V,. Since any dependence among the basic ring commutators of degree n would violate lemma B, we shall be done once this lemma is proved.
Proof of lemma B: Clearly, [ V . : Q ] = k" since this is the number of monomials in Q { X,,...,xk}of degree n. Thus we must show B spans V , and IBI = k". These assertions follow from a ring-theoretic collection procedure akin to the group-theoretic one described earlier. Given a product d l d , of basic ring commutators we say that the collected part is the substring d l * * a d k with k maximal such that d , < *..< dk. There is a collection procedure for monomials which consists of a series of collection steps to produce sums of collected products of basic ring commutators: Namely, take the smallest uncollected d in its leftmost position d, and replace d,- ,d by [du- ,,d] + dduthen
,;
d l . . . d , = d l ~ ~ ~ d u - 2 [ d u - l , d ] d" *, + d ,1+ d , ...d,-2ddu-ldu+l...dr.
By induction on t one sees easily that a finite number of applications produces a sum of elements from B, proving B spans V,. Note that all basic ring
Categorical Aspects of Module Theory
490
commutators created by collecting d are of the form [r, d ] , so all the collection d ] for d , > d. steps involved in collecting d have the form [[. .. [[d,, d ] , d ] To prove IBI = k" we count the basic ring commutators according to their order. Put Bj = { f V~, : f is product of basic ring commutators which has been obtained by collecting all basic ring commutators up to but not including the j-th ring commutator} Thus B, is the set of monomials (since nothing has been collected) and B, = B for suitably large j. We shall conclude by proving lBjl = lBj+,l for each j, for then IBI = lBll = k". We define a correspondence cp: Bj+ + Bj as follows: letting d be the j-th basic ring commutator we take any element of B j + , and erase all brackets containing d. As observed above, this means replacing each basic ring commutator [[...[[d,-,,d],d]...],d]by d,-,dd-..d. Thus elements of J / B j + , have the form a ] ,
,
i.e., in the uncollected part any string of repeated occurrences d is preceded immediately by some d , > d. But we have observed this is the typical form of an element of Bj, i.e., in which every basic ring commutator preceding d has been collected, so J/ is onto and is obviously 1:l. Therefore lBjl = lBj+,1 as Q.E.D. desired, proving (BI = k".
Exercise 1. Every element of the free group G can be written uniquely mod G,, as a product of basic commutators in ascending order. (Hint: Otherwise 1 is a nontrivial product of basic commutators.) Use this to order G by saying g1 < g2 if modulo some C,, the product of basic commutators for g1 is of lower order lexicographically than that for g2. (Hint: if g1 < g2 and g E G then g,g < g 2 g because gig = ggi(gi,g).)
Appendix B. Normed Algebras and Banach Algebras One of the sources for the structure theory of rings is the theory of Banach algebras, which arises in turn from LP-spaces.Consequently, we shall take a brief diversion into mathematical analysis to lay the algebraic foundation for Banach algebras which will subsequently be used as an occasional source for examples as the structure theory unfolds.The reader should consult Rudin [66B, p. 131 for the analytical background. Definition Bl:
Suppose S is a measurable set with positive measure p, and
Appendix B
491
let I I denote the usual absolute value of @. For any positive integer p define
The most usual cases for S are 88, the unit interval [0,1], or the unit disc. We have defined Lp(S)as a subset of CSand obviously it is an additive subgroup, but in general L’(S) is not a subring of @‘ (since the function x-% integrable on [0,1] but x-l = (x-l/’)’ is not integrable on [0,1]). Of greater interest (for S = R and p the Lebesgue measure divided by is the convolution f l * fz of f l and fz , defined by
6)
a,
(fl
*f z b= j
-a,
f l ( X - Y)fZ(Y)dP(Y)
for each x in R. Using * as our multiplication, we see by means of Rudin [66B, theorem 7.143 that L’(R) is closed under multiplication and is easily seen to be a ring. In fact Cop. cit. p. 1481 Lp(R)is an L’(R)-module for each p, where convolution is the scalar multiplication.
Banach Algebras There is a standard way to algebraize the above notions. Let R + denote the nonnegative real numbers. Definition B2: An absolute value on a field F is a function satisfying the following properties for all ai in F:
I I: F + R +
(0 1% + a21 5 bll + b 2 1 .
(4 Ia1azI = l ~ l l l ~ 2 1 .
(iii) la1 = 0 iff CI = 0.
Note that 111’ = 111 so, by (iii), 111 = 1. In what follows below, the usual absolute value on R and C.
I I denotes
Definition B3: Let F be a subfield of @. A E F-dc!g is normed if there is a function 11 11: A + R + satisfying the following properties for all a, in A:
0) lbl + azll 5 Iblll + IIQZII. (ii) Ila1a211 Ilalllllazll* for all a in F. (iii) llaall = Ialllall (iv) If llall = 0 then a = 0. (v) 11111 = 1.
Categorical Aspects of Module Theory
492
Using the norm one can define a metric d(a,,a,) = (la, - a,Il, which in turn induces a topology whose open sets have the base of "open balls" B(a;6)= {a' E A : /la' - all < S} for all a in A, 6 in Iw'. Addition and multiplication are continuous, since for Ila; - alll < 6' and [la; - a,ll < 6, we have [[(a; a;) - (al az)ll 6, + 6 , and Ila',a; - alazll = Ila;(a; - a,) + (a', - al)azll < (6, + lla111)d2 6111azll (which is small if 6,,6, are suitably small). Using this metric one can define a sequence (al, a2,. . .) to be Cauchy if for every 0 # 6 E Iw' there is some n such that [lai- ajll < 6 for all i > j > n; the sequence converges to a E A if for every 6 E Iw' there is some n such that llai - all < 6 for all i > n. Clearly, every convergent sequence is Cauchy; a Banach algebra is a normed algebra in which every Cauchy sequence converges to an element of the algebra (i.e., it is complete under the induced metric). Note that 11 11 induces an absolute value on the completion of the base field F, so we may assume F = Iw or F = C by Ostrowski's theorem (cf., exercise 3). The invertible elements of a Banach algebra behave in the following nice way.
+
+
-=
+
Proposition B4: Suppose A i s a Banach algebra. ( i ) If llall< 1 then(1 -a)isinuertible;infact,(l - a ) - ' = 1 + a + a Z + . . . . ( i i ) The set of invertible elements is open. ( i i i ) The function a --f a-' is continuous on the set of open elements.
Proof:
(i) The sequence (1,l + a, 1 + a + a', ...) is Cauchy and thus converges to some element a'. But, for large enough n, (1 - a)a' is close to (1 - a) ( ~ : ~ ~=a1 "- )a", which is arbitrarily close to 1, proving (1 - a)a' = 1; likewise a'(1 - a) = 1. (ii) If a is invertible and b is close to a, then a-'b is close to a-'a = 1 , so by (i) is invertible. Then b-' = ( u - ' ~ ) - ' u - ' . (iii) By the same computation, if b is close to a then b-' is close to a-'. Q.E.D. The foremost example of Banach algebras in analysis is L' = I!,'@)/% where %={f EL'(R): f is 0 on a set whose complement has measure O}; L1 has the norm 11f 11 = S?J(f(x)I dp(x). L' can be identified with a subalgebra of { h E C': h is continuous and limt+mht = 0) by means of the Fourier transform, cf., Rudin [66B, Chapter 91. Our interest lies in the category
Appendix B
493
of Banach algebras. Accordingly, if A,, A, are Banach algebras (over the same field), we say a Banach homomorphism is an algebra homomorphism f: A, -,A, such that whenever a sequence (a,, a,,. . .) converges to a E A, we have (fa,, fa,, . ..) converging to fa in A,. Proposition B5: ( i ) If f:Al --* A , is a Banach homomorphism then kerf is closed (in the topology of A,). (ii) Suppose A is a Banach algebra and 14 A. Then its closure 7 4 A; moreover, A / i is a Banach algebra and the canonical homomorphism A -,AJris a Banach homomorphism.
Proofi (i) If a is in the closure of kerf then there is a sequence (a,,a,,. ..) of elements of kerf converging to a; applying f to each term yields (0,0, . . .) converging to fa, so fa = 0. (ii) The fact iis an ideal follows at once from the continuity ofaddition and multiplication. To prove Tis proper, note that 1 cannot contain any invertible elements (since 1 is proper), so 1 4 T by proposition B4. We define the norm on AITby IJU-I-
= inf{ (la
+ XII: x E I }
and need to verify conditions (i) through (v) of definition B3. Conditions (i) through (iii) and (v)are straightforward. To see (iv) suppose ()a fil = 0. Then there are x, in 1 such that {Ila + x,ll:nE N } 4 0 , implying {xl,x z,...} is Cauchy with some limit x in 1 Then Ila + xII = 0 so a = --x E rand a T= 0. Thus A/iis normed; the Banach properties follow easily from the fact A is Q.E.D. Banach and are left to the reader.
+
+
Exercises 1. Suppose C is a commutative ring with a primitive n-th root (' of 1, and for z fixed in C define a function g by ga = n - ' ~ ~ = , ( a(k z)-1 when this exists. Then ga = (a - z(a-'z)"-')-' when this expression exists. (Hint: Use the "logarithmic derivative" of the equality 1"- z" = II(d - rkz),divide by n,and substitute a for A,) 2. Suppose A is a C-algebra which is normed as an R-algebra. For any a in A there is c in C such that a - cis not invertible. (Hint: Otherwise, we can define a continuous function f: C -+ A by f c = (a - c)-'; for c large, fc = c-'(ac-' - l)-' is small so { IIfcll:c E C) has a maximal element p. Put S = {c E C : f c = p}, a closed
nonempty bounded set. Replacing a by Q - c for some c in S we may assume ~ ~ = p.a S cannot ~ ' be open ~ ~so there exists JzI < p-lsuch that Il(a - z)-'1( < p - E for some positive E ; then Il(a - z')-'ll < p - E for all z' sufficiently near z. In
494
3. 4. 5.
6.
7.
Categorical Aspects of Module Theory
particular, for n large enough there is some positive ratio v between 0 and 1 such that Il(a - t;kz)-lll < p - E for all 0 5 k 5 vn (where t; is a primitive n-th root of l), so n-lxll(a - c k z ) - l 11 < n-'(vn(p - E ) + (n - vn)p) = p - Y E c p. For n large enough exercise 1 shows ((gal(+ lla-lll = p whereas we see 11gall p - ve.) (Ostrowski's theorem) The only field =I R with an absolute value extending the absolute value on R is C itself. (Hint: If necessary adjoin i = and define the norm given by Ila + bill = la1 + lbl.) The only Banach algebras which are fields are R and C. If A is a normed algebra then the set of Cauchy sequences over A becomes a subalgebra A' of A N (i.e., operations are componentwise) and has an ideal 1 = {Cauchy sequences converging to O}; A'/l has a norm given by ll(al,a2,...) + III=limllaill and, in fact, A'/l is a Banach algebra containing an isomorphic copy of A. View this as a completion (example 1.8.12). Let Iz be a commuting indeterminate over C;C[A] is a normed algebra under the normx c i l i = Zlc,l, so conclude there exist integral domains which are Banach C-algebras. On the other hand, the only Banach C-algebra which is a division ring is C itself. (Hint: For any such division ring A and a E A we form the subfield @(a) of A which is a commutative Banach algebra and thus C itself. This is an algebraic proof of a theorem from analysis due to Gelfand-Mazur.) The only division rings which are Banach algebras are R,C, and the real quaternions: (Hint: Any subfield is isomorphic to C,so appeal to Frobenius' theorem, cf., Herstein CUB, p. 3271.)
-=
J-1
The Basic Ring-Theoretic Notions and Their Characterizations A ring R is lefi Artinian if R satisfies the DCC (or, equivalently, the minimum condition) on left ideals. Other characterizations:
1. Every f.g. R-module is Artinian (Corollary 0.2.21); 2. R has a composition series as left R-module (Remark 2.3.2 and Characterization 4.); 3. Every f.g. R-module has a composition series (Remark 2.3.2); 4. Left Noetherian and semiprimary (Theorem 2.7.2); 5. Left Noetherian and perfect (Exercise 2.7.7); 6. R is perfect and Jac(R)/Jac(R)’ is f.g. as R-module (Exercise 2.7.8); 7. K-dim R = 0 (Example 3.5.40). A ring R is simple Artinian if R is a simple ring which is left Artinian. Other characterizations:
1. Matrix ring over a division ring (Remark 2.3.8) (Note: This criterion is left-right symmetric, so the right-handed version of all subsequent characterizations are applicable; 2. Prime ring and left Artinian (Theorem 2.3.9.); 3. Morita equivalent to division ring (Morita’s Theorem Proposition 4.1.18); 4. Simple ring with minimal left ideal (Corollary 2.1.25’); 5. Ring of fractions of prime (left) Goldie ring (Goldie’s Theorem);
+
495
4%
The Basic Ring-Theoretic Notions and Their Characterizations
6. Primitive with ACC (Ann) and socle # 0 (Digression 3.2.9); 7. Prime regular with all primitive images Artinian (Exercise 2.1 1.22). A ring R is semisimple (Artinian) if it satisfies properties (l), (2), or (3) below, all equivalent by Theorem 2.3.10. (The name is justified by property (4)).This is the most important kind of ring. Other characterizations: 1. Finite direct product of simple Artinian rings Note: This criterion is left-right symmetric, so the right-handed versions of all subsequent characterizations are applicable); 2. Semiprime and left Artinian; 3. Completely reducible as R-module, i.e. R = soc(R) (Theorem 2.3.10); 4. Left Artinian, and the intersection of the maximal ideals is 0 (by theorem 2.3.9 + Chinese Remainder Theorem); 5. Every R-module is completely reducible (Theorem 2.4.9); 6. Complemented as R-module (Theorem 2.4.8); 7. Every R-module is complemented (Theorem 2.4.9); 8. R has no proper large left ideal (Theorem 2.4.8); 9. Ring of fractions of semiprime (left) Goldie ring (Goldie’s Theorem); 10. R is semiprime, each nonzero left ideal contains a minimal left ideal, and soc(R) is a sum of a finite number of minimal left ideals (Corollary 2.3.1 1); 11. Semiprime with a complete set of orthogonal primitive idempotents (Corollary 2.3.12); 12. Every short exact sequence of R-modules splits (Theorem 2.4.9; Remark 2.1 1.7); 13. Every R-module is injective (Remark 2.1 1.7); 14. Every cyclic R-module is injective (Exercise 2.1 1.2); 15. Every left ideal is injective as R-module (by Theorem 2.4.9 + Proposition 2.10.14); 16. Every R-module is projective (Remark 2.1 1.7); 17. All cyclic modules are projective (since every left ideal is then a summand); 18. Left Noetherian and (von Neumann) regular ring (Corollary 2.1 1.21); 19. Regular ring with a complete set of orthogonal primitive idempotents (Proposition 2.1 1.20 + Theorem 2.4.8); 20. Regular ring without an infinite set of orthogonal idempotents (Exercise 2.11.3); 21. Every additive functor from R-Aud to W-AULLis exact (Proposition 2.1 1.8); 22. Every left exact functor from R-M& to Z - M O is ~ exact (proof of Proposition 2.1 1.8); 23. Global dimension is 0 (Example 5.1.17).
The Basic Ring-Theoretic Notiom and Their Characterizations
497
A ring R is iejt Noetherian if R satisfies the ACC (or equivalently the maximum condition) on left ideals:
1. 2. 3. 4.
Every f.g. R-module is Noetherian (Proposition 0.2.21); ACC on f.g. left ideals (Exercise 3.3.21); The direct sum of injective modules must be injective (Exercise 2.10.10); Every injective is a direct sum of indecomposable injectives (Exercises 2.10.1 1,2.10.13); 5. There is a cardinal c such that every injective R-module is a direct sum of modules each spanned by at most c elements. (Exercise 2.10.12); 6. The polynomial ring R[L] has K-dim (Exercise 3.5.30).
A ring R is primitive if R has a faithful simple module. Other characterizations: 1. R has a maximal left ideal whose core is 0 (proposition 2.1.9); 2. R has a left ideal with core 0, which is comaximal with all ideals (Proposition 2.1.1 1); 3. R is prime and has a faithful module of finite length (Exercise 2.3.3).
The Jacobson radical Jac(R) is the intersection of the primitive ideals of R. Other characterizations: 1. Intersection of the right primitive ideals, by note below. 2. The quasi-invertible ideal which contains all quasi-invertible left ideals (Proposition 2.5.4); 3. The right-handed version of property 2 (by discussion after Proposition 2.5.4) Note: This criterion is left-right symmetric, so the right-handed versions of all subsequent characterizations are applicable; 4. The intersection of the maximal left ideals (Proposition 2.5.2); 5. The intersection of the left annihilators of all simple modules (Proposition 0.2.14 + Proposition 2.5.2); 6. The intersection of the cores of maximal left ideals (Exercise 2.5.3); 7. The intersection of the cores of maximal inner ideals (Exercise 2.5.4); 8. The sum of the small left ideals of R (Exercise 2.8.1 1). Note the Jacobson radical itself is a small left ideal, by property 4; 9. { r E R: r + a is invertible for all invertible a in R ) (Exercise 2.5.2).
This Page Intentionally Left Blank
Major Ring- and Module-Theoretic Results Proved in Volume I (Theorems and Counterexamples; also cf. “Char acterizations”) E means “exercise”. Result preceded by (*) appears in stronger form elsewhere in the text.
Theorems Chapter 0 0.3.2 Existence and “uniqueness” of a base.
Chapter 1 1.1.17 I . 1.25 1.2.23 1.2.24’
R - A o d and M,(R)-Aod are equivalent categories. Matrix units can be lifted via idempotent-lifting ideals. R ( ( S ) )is a domain, for any domain R and ordered monoid S . D((G))is a division ring, for any ordered group G .
499
Major Ring- and Module-Theoretic Results Proved in Volume I
500
Magnus’ Theorem, Magnus- Witt Theorem on the lower central series of the free group (proofs in Appendix A). 1.3.39 The free group G is ordered. (Hence the free algebra is contained in the division ring F((G)).) 1.4.15 (Los’s Theorem) Elementary sentences pass to ultraproducts. Algebras have coproducts. 1.4.30 1.5.13 End M is a homomorphic image of a subring of a matrix ring, for any f.g. module M. An Ore extension of a division ring is a PLID (also cf., 1.6.15, 1.6.24). 1.6.21 1.7.27 The tensor product of a simple algebra and a central simple algebra is simple. (Corollaries include results about centralizers in tensor products.) Direct limits of directed systems exist. 1.8.7 1.9.24 Construction of the free product of rings over a common subring W. The free product of domains (which are algebras over a common 1.9.26 subfield) is a domain. *1.10.11 The closed ideals of R correspond to the ideals of S-’R. 1.10.20 Properties of the central localization functor. E1.4.25 Y ( A ) is a modular lattice. 1.3.38
Chapter 2 *2.1.6 *2.1.8 2.1.25 2.2.1 2.3.3 2.3.9 2.3.10 2.3.13 *2.3.27 2.4.7 2.4.8
(Density theorem) Every primitive ring is dense in a ring of endomorphisms (also cf., E2.4.1). (Wedderburn- Artin) A primitive Artinian ring is matrices over a division ring (also cf., 2.1.25‘,4.1.20(i), E2.1.8, E2.1.9). soc(R) = {elements of finite rank} for R primitive. (Chinese remainder theorem) A finite subdirect product is direct. (Jordan-Holder and Schreier) If a module has a composition series then any chain of submodules can be refined to an equivalent composition series. The prime ideals of a left Artinian ring are maximal and finite in n um ber . Semisimple Artinian = finite direct product of simple Artinian rings. Structure of semisimple Artinian rings in terms of idempotents. R @ RoPis split for R central simple. The socle is the intersection of the large submodules. Completely reducible module = complemented = no proper large submodule.
Major Ring- and Module-Theoretic Results Proved in Volume I
2.4.9 2.5.16 2.5.19 2.5.22 2.5.23 “2.5.24 j2.5.35 2.5.36 “2.537
2.5.40 j2.5.42 2.5.5 1 “2.i53 2.6.7 2.6.1 1 2.6.17 2.6.19 “2.8.23 2.6.27 2.6.3 1 2.6.35 2.7.2 2.7.7 2.7.30 2.7.33 2.7.36 2.8.14 2.8.26 2.8.40 *2.9.2 *2.9.7 2.9.10
501
R semisimple Artinian = R is a complemented R-module. The Jacobson radical of a left Artinian ring is nilpotent. The Jacobson radical of an algebraic algebra is nil. Jac(R) is nil whenever R is an algebra over a suitably large field. (Amitsur) If Nil(R) = 0 then R[1] is semiprimitive. (“Nakayama’s lemma”) Jac(R)M # M for any Eg. module M # 0. Jac(R) = R n Jac(T) for any tensor extension T of R (also cf., E2.5.13). Jac(R) 0 H = Jac(H) for H separable over F. (Wedderburn’s principal theorem) Every finite dimensional algebra over a perfect field is a direct sum (as vector spaces) of the radical and a semisimple algebra (also cf., E2.5. 11). If R is Z-Graded then Jac(R) is a graded ideal. (Gives reproof of 2.5.23.) If Nil(R) = 0 then R(I1) is semiprimitive. (G(Jac(RG)G Jac(R) for any finite group of automorphisms G . (Bergman-Isaacs) If RG = 0 then lGlR is nilpotent. Nil(R) is the intersection of a certain class of prime ideals. If R is reduced then RIP is a domain for every minimal prime ideal P of R. semiprime = no nonzero nilpotent (left) ideals. (Levitzki) The prime radical can be built using transfinite induction on nilpotent ideals. Every nil left or right ideal of a left Noetherian ring is nilpotent. Embedding a semiprime ring into a ring with upper nilradical 0. Every “weakly closed” nil subset of a semiprimary ring is nilpotent. Conditions equivalent to Koethe’s conjecture. Artinian = Noetherian + semiprimary. Semiprimary = bounded length for chains of principal left ideals. Every semiperfect ring has a basic ring which modulo the radical is a direct product of division rings (also cf. E2.8.23). Right perfect = semilocal + Jac(R) T-nilpotent. Right perfect rings characterized by chain conditions (also cf. E2.7.5, E2.7.6, E2.8.27, E2.8.28). Over a hereditary ring, submodules of projectives are projective. (Schanuel’s lemma) Short exact sequences ending in M and with a projective in the middle are “almost” unique. semiperfect = complete set of orthogonal LE-idempotents. Every module of length t has a decomposition of length It. Fitting’s lemma. An indecomposable with finite length is LE. End M semiprimary if M has finite length.
Major Ring- and Module-Theoretic Results Proved in Volume I
502
*2.9.15 2.9.17 2.9.18 2.9.29 2.9.40
(“Exchange property”) The summands in two LE-decompositions can be interchanged. (Wedderburn-Krull-Schmidt-Remak-Azumaya = Krull-Schmidt) All decompositions of a module having an LE-decomposition are equivalent. A complete set of idempotents of a semiperfect ring is unique, up to inner automorphisrn. (Harada-Sai) Any sequence of nonisomorphisms of indecomposables having bounded composition length is finite. First Brauer Thrall conjecture verified (bounded representation type implies finite representation type).
(Also sesqui-BT is proved modulo 2.9.42; other major theorems from the representation theory of Artin algebras are described but not proved) 2.10.3‘ (Baer’s criterion) Injectivity can be checked on left ideals. 2.10.10 There are “enough” injectives. 2.10.20 Injective hull = maximal essential extension. 2.10.33 “Uniqueness” of decomposition of injective. 2.11.13 F is a flat module iff FY is injective. 2.11.14 F flat iff I @ F x IF for any f.g. right ideal I (also cf. E2.11.5’). 2.12.22 The rank of f.g. projective modules is locally constant. 2.12.28 Nullstellensatz for algebras over “large” fields. 2.12.36 “Generic flatness” implies the Nullstellensatz. 2.12.48 LO and G U hold for finite centralizing extensions. 2.12.50 LO and G U hold when tensoring up by a field. 2.12.52 Analogue of 2.5.35, 2.5.36 for prime ideals. 2.13.21 Structure of semiprime (*)-rings with nonzero socle, in terms of a sesquilinear form. 2.13.29 Characterizations of involutions on matrix rings. E2.1.15 A free product of algebras over a field is usually primitive. Density theorem for completely reducible modules. E2.4.1 E2.5.10 (Koethe-Noether-Jacobson) Every algebraic division algebra contains a separable element. E2.9.4 Exchange property for a module having an LE-decomposition into countable submodules. (This implies every projective over a local ring is free, generalizing example 2.8.9.)
Chapter 3 *3.1.4
S-’R is a ring of fractions for R with respect to S, for any denominator set S.
Major Ring- and Module-Theoretic Results Proved in Volume I
3.1.6 *3.1.20 3.2.14
503
The ring of fractions is a universal and is thus unique. The Ore localization functor is exact. (Goldie’s second theorem) Semiprime (left) Goldie = order in semisimple Artinian ring. 3.2.16 (Goldie’s first theorem) Prime Goldie = order in simple Artinian ring. 3.2.27 More explicit structure of semiprime Goldie rings. Direct product of prime rings = semiprime + ACC (annihilator 3.2.29 ideals) + every maximal annihilator is a summand of R. 3.2.36 Characterizes orders in semilocal rings. Also implies Small’s theorem: A left Noetherian ring with the regularity condition is an order in a left Artinian ring. 3.2.37, 8 Every subring of a left Noetherian ring is weakly finite, and every nil weakly closed subset is nilpotent. 3.2.44 A prime semiprimitive ring can be embedded into a primitive ring which satisfies many of the same properties. Every commutative Noetherian ring is embeddible into a commu3.2.53 tative Artinian ring. *3.3.7 The Johnson ring of quotients k is a ring. R is isomomorphic to End E ( R ) , for R nonsingular. 3.3.16 If R is nonsingular then the Johnson localization functor is exact. *3.3.21 R is nonsingular then d is regular nonsingular and self-injective. 3.3.23 The Martindale-Amitsur ring of quotients is a ring. *3.4.7 If R is centrally closed prime then extensions are tensor extensions. 3.4.1 1 INC for finite centralizing extensions. 3.4.13 Tensoring by algebraic field extension satisfies LO, GU, INC. 3.4.13’ Q,(R) is a ring. 3.4.25 (Hilbert basis theorem) An Ore extension of a Noetherian ring 3.5.2 (with respect to an automorphism) is Noetherian. (Principal ideal theorem) Every prime ideal minimal over a nor3.5.10 malizing element is minimal. Certain prime ideals have finite height. 3.5.13 Noetherian ring = direct product of Artinian and semiprime, 3.5.16 provided sN = Ns = N for all s regular mod N. Jacobson’s conjecture holds for almost fully bounded Noetherian *3.5.28 rings whose primitive images are Artinian. The completion with respect to a polycentral ideal is Noetherian. 3.5.36 3.5.49, 5 1 cl K-dim IK-dim; equality for fully bounded left Noetherian. 3.5.72 The stable number of generators of a module M is bounded by local information (i.e. on R I P and M I P M ) . The reduced rank function satisfies “patch continuity” on Spec(R). 3.5.72“
Major Ring- and Module-Theoretk Results Proved in Volume I
504
3.5.82 3.5.89 E3.2.31 E3.2.38 E3.3.14 E3.3.40 E3.3.53
Any Noetherian bimodule has a large bimodule which is a direct sum of cells, and this decomposition is “unique”. Jacobson’s conjecture holds for almost fully bounded Noetherian rings. (Faith-Utumi) Any order in M,(D) contains a subring (without 1) of the form M,(T) where T is an order in D. Any nilpotent algebra over a field is embeddible into an algebra of upper triangular matrices. Goldie = nonsingular, for any semiprime ring satisfying ACC( (Goodearl) The ideals of Q / P are totally ordered, for any prime ideal P of a regular self-injective ring Q. (Goodearl) The lattice of ideals of a prime regular self-injective ring is isomorphic to an interval of ordinals.
0).
Chapter 4 4.1.4
(Morita’s theorem) R‘ is Morita equivalent to R if R’ x End P for a progenerator P. 4.1.17 More explicit description of Morita equivalence. A left adjoint preserves direct limits. 4.2.9 4.2.11 A right adjoint preserves inverse limits. E4.2.8 (Watt) Any right exact functor preserving direct sums is a tensor functor. E4.2.11 (Watt) Any left exact functor preserving direct products is a Hom functor. E4.2.17 (Popescu-Gabriel) Any Grothendieck category with generator Cis a category of modules over the ring Hom(C, C).
Counterexamples Chapter 0 0.1.4 A non-onto epic in W i q .
Chapter I 1.1.29 (Zelinski) An uncountable set of idempotents cannot be lifted. 1.3.33 A ring lacking IBN.
Major Ring- and Module-Theoretic Results Proved in Volume 1
505
1.6.26 PLID which is not PRID. 1.7.25 A tensor product of fields which has nilpotent elements. E1.4.5 Nonisomorphic rings each with an injection into the other (based on 1.4.27).
Chapter 2 2.1.33f *2.1.36
Simple domains having socle 0. Jategaonkar’s counterexample- Right but not left primitive, also PLID and other nice properties. A left but not right Artinian ring. 2.3.14 A local ring which is a union of a chain of semiprimitive rings. 2.5.31 A semiprimary ring which is neither left nor right Artinian. 2.7.1 1 2.7.22 A commutative semilocal ring which is not semiperfect. 2.7.38 A right but not left perfect ring. Projective but not free (also cf. E2.8.1). 2.8.8 2.8.35 A module without a projective cover. Krull-Schmidt fails over arbitrary Noetherian rings. 2.9.22 2.9.23 A commutative Noetherian ring possessing a f.g. module failing the following properties: uniqueness of Krull-Schmidt length, direct sum cancellation, and power cancellation. 2.9.25 A group algebra not having finite representation type (also cf. E2.9.22, E2.9.23). E2.5.21 A simple radical rng. Levitski’s transfinite procedure for defining the prime radical may E2.6.3 require arbitrarily many steps. E2.6.15 A prime rng R whose fixed subrng RG is 0. E2.6.16 A simple ring R whose fixed subring RG is not semiprime. E2.7.22 A semiprimary ring with a cyclic module having an onto map which is not 1:l. A left but not right hereditary ring. E2.8.5 A left hereditary Noetherian ring whose center is any given Krull E2.8.6 domain. E2.9.14 A local Noetherian ring over which Krull-Schmidt fails. E2.9.17 Direct sum cancellation fails in Z(’). E2.10.15 A Noetherian injective module which is not Artinian. E2.10.21 A regular left but not right self-injective ring. E2.11.16 A nonregular semiprime ring whose prime images all are regular. E2.12.5 A skew polynomial extension of a commutative local Jacobson ring, but which is not Jacobson.
506
Major Ring- and Module-TheoreticResults Proved in Volume I
Chapter 3 A finite ring which cannot be embedded into a matrix ring. A domain not embeddable into a division ring. An irreducible Noetherian ring which is not an order in an Artinian ring. A nonsingular, but not semiprime, Artinian ring. 3.3.1 1 A noncommutative Noetherian ring failing the principal ideal 3.5.8 theorem. A left Noetherian ring failing Jacobson’s conjecture. 3.5.23 A Noetherian ring having a non-Artinian module which is an 3.5.29 essential extension of a simple module. A Noetherian ring whose radical fails the AR-property. 3.5.38’ cl K-dim < K dim for the Weyl algebra. 3.5.53 Nonlocalizable prime ideals of an Artinian ring. 3.5.92 A commutative ring with ACC (Ann) but having arbitrarily long E3.2.8 chains of annihilators. E3.2.12 A simple Noetherian PLID which is a V-ring with a unique simple module, but which is not Artinian. A left Noetherian ring which cannot be embedded in a left E3.2.40 Artinian ring. (A more powerful &.ample has recently been / discovered by Dean- Stafford.) A left but not right nonsingular ring. E3.3.9 E3.3.10 A ring whose injective hull does not have a ring structure. E3.4.1 A large submodule which is not dense. E3.5.5 A Noetherian ring whose nilradical has unequal left and right reduced ranks. E3.5.9,10 A local PLID for which Jacobson’s conjecture fails, and in which primary decomposition fails. E3.5.17’ A Noetherian ring whose P-adic completion is non-Noetherian. A prime ideal failing the AR-property. E3.5.18 E3.5.27 A PLID with arbitrary large K-dim. E3.5.29 A right Noetherian ring lacking (left) K-dim.
3.2.48 3.2.49 3.2.54
Chapter 4 4.1.22 A simple Noetherian hereditary ring not Morita equivalent to a domain (given without proof). E4.1.2 A faithful cyclic projective module which is not a progenerator.
References
Bibliography of Books This list contains most of the advanced ring theory books, plus books from other subjects that bear heavily on the material of Volume I. ALBERT, A. A. [61] Structure of Algebras (second edition). AMS Colloq. Pub. 24. American Mathematical Society, Providence ANDERSON, F., and FULLER, K. [74] Rings and Categories of Modules. Springer-Verlag, Berlin ARTIN,E., NFSBIT,C. J., and THRALL, R. [44] Rings with Minimum Condition. University of Michigan Press, Ann Arbor BARBILIAN, D. [56] Teoria Aritmetica a fdealilor (in inele necornutatiue). Ed. Acad. Rep. Pop. Romine, Bucaresti BARWISE,J. (ED.) [78] Handbook of Logic. North-Holland, Amsterdam BJORK,J. [79] Rings of Differential Operators. North-Holland, Amsterdam BOKUT', L. A. [81] Associative Rings 1,2 (Russian). NGU, Novosibirsk B~URBAKI, N. [72] Commutative Aloebra (transl. from French). Elements de Mathematique. Hemann, Paris CARTAN, H., and EILENBERG, S. [561 Homological Algebra. Princeton University Press, Princeton CHATTERS, A. W., and HAJARNAVIS, C. [80] Rings with Chain Conditions. Pitman, London
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COHN,P. M. [74] Algebra I. Wiley, London [77] Skew Field Constructions. Cambridge University Press, Cambridge [81] Universal Algebra (second edition). Reidel, Dordrecht [85] Free Rings and Their Relations (second edition). Academic Press, New York [88] Algebra 1 1 , I I I (second edition, in press). Wiley, London COZZENS, C., and FAITH,C. [75] Simple Noetherian Rings. Cambridge University Press, Cambridge CURTIS, C., and REINER,1. [62] Representation Theory of Finite Groups and Associative Algebras. Interscience, New York [8l] Methods of Representation Theory, Vol. 1. Wiley, New York (Vol. 2, 1987) DEURING, M. [66] Algebren. Springer-Verlag, Berlin DICKSON, L. E. [23] Algebras and Their Arithmetics. University of Chicago, Chicago DIVINSKY, N. J. [65] Rings and Radicals. Toronto University, Toronto FAITH,C. [67] Lectures on Injective Modules and Quotient Rings. Lecture Notes in Mathematics 49. Springer-Verlag. Berlin [73] Algebra: Rings, Modules, and Categories. Springer-Verlag, Berlin [76] Algebra 11: Ring Theory. Springer-Verlag, Berlin [82] Injective Modules and Injective Quotient Rings. Marcel Dekker, New York FAITH, C., and PAGE,S. [84] FPF Ring Theory. London Mathematical Society Lecture Notes 88. Cambridge University Press, Cambridge FREYD, P. [64] Abelian Categories: An Introduction to the Theory of Functors. Academic Press, New York GILLMAN, L., and JERISON, M. [60] Rings of Continuous Functions. Academic Press, New York GOLAN, J. [75] Localization of Noncommutative Rings. Marcel Dekker, New York [86] Torsion Theories. Longman Scientific & Technical, Harlow, England GWDEARL, K. [76] Ring Theory: Nonsingular Rings and Modules. Marcel Dekker, New York [79] von Neumann Regular Rings. Pitman, New York HALL,M. [59] The Theory of Groups. MacMillan, New York HERSTEIN, I. N. [64] Topics in Algebra (second edition). Xerox, Lexington, MA [68] Noncommutative Rings. Carus Mathematical Monographs 15. American Mathematical Society, Providence [69] Topics in Ring Theory. University of Chicago Press, Chicago [76] Rings with Involution. University of Chicago Press, Chicago JACOBSON, N. [43] Theory of Rings. AMS Surveys 1. American Mathematical Society, Providence [62] Lie Algebras. Wiley, New York [64] Structure of Rings (second edition). AMS Colloq. Pub. 37. American Mathematical Society, Providence [SO] Basic Algebra 11. Freeman, San Francisco [8 11 Structure of Jordan Algebras. Lecture Notes. University of Fayetteville. [85] Basic Algebra I (second edition). Freeman, San Francisco
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JATEGAONKAR, A. V. [70] Left Principal Ideal Rings. Lecture Notes in Mathematics 123. Springer-Verlag, Berlin [86] Localization in Noetherian Rings. London Mathematical Society Lecture Notes 98. Cambridge University Press, Cambridge 1. KAPLANSKY, [68] Rings of Operators. Benjamin, New York KASCH,F. [82] Modules and Rings (trans]. from German). London Mathematical Society Monographs 17. Academic Press, New York KELLEY, J. [55] General Topology. van Nostrand, New York KRUSE,R. L., and PRICE,D. [69] Nilpotent Rings. Gordon & Breach, New York LAMBEK, J. [66] Lectures on Rings and Modules. Blaisdell, Massachussetts LANC,S. [65] Algebra (second edition, 1984). Addison-Wesley, Reading MCCONNEL, J., and ROBSON,J. C. [88] Noetherian rings. Wiley, London MCDONALD, B. [74] Finite Rings with Identity. Marcel Dekker, New York MATLIS,E. [72] Torsion-Free Modules. University of Chicago Press, Chicago MITCHELL, B. [65] Theory of Categories. Pure and Applied Mathematics 17. Academic Press, New York MONTGOMERY, S. [80] Fixed Rings of Finite Automorphism Groups of Associatiue Rings. Lecture Notes in Mathematics 8 18. Springer-Verlag, Berlin F. NATASESCU, C., and V A N OYSTAEYEN, [82] Graded Ring Theory. Mathematical Library 28. North-Holland, Amsterdam PASSMAN, D. [77] The Algebraic Structure of Group Rings. Wiley, New York PIERCE,R. S. [82] Associatiue Algebras. Springer-Verlag, Berlin REINER, I. [75] Maximal Orders. Academic Press, London RENAULT, J. [75] AlgGbre Noncommutatiue. Gauthier-Villars, Paris RINCEL [84] Tame algebras and integral quadratic forms. Lecture Notes in Mathematics, 1099. Springer-Verlag. Berlin RUDIN,W. [66] Real and Complex Analysis. McGraw-Hill, New York SCHOFIELD, A. H. [85] Representations of Rings ouer Skew Fields. London Mathematical Society Lecture Notes 92. Cambridge University Press, Cambridge SHARPE, D. W., and VAMOS,P. [72] Injectiue Modules. Cambridge University Press, Cambridge SMALL,L. (ED.) [8 11 Reviews in Ring Theory. American Mathematical Society, Providence [86] Reviews in Ring Theory 11. American Mathematical Society, Providence STENSTROM, B. [75] Rings of Quotients: An Introduction to Methods of Ring Theory. Springer-Verlag, Berlin
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Collections of Papers Collections which are in print, by year of publication. 1972
R. Gordon (Ed.). Ring Theory. Academic Press, New York Proceedings of Conference on Orders, Group Rings, and Related Topics: Ohio St. Conference. Lecture Notes in Mathematics 353. Springer-Verlag, Berlin Proceedings of Ring Theory Conference, Tulane University: Lectures on the Applications of Sheaves to Ring Theory. Lecture Notes in Mathematics 248. Springer-Verlag, Berlin Proceedings of Ring Theory Conference, Tulane University: Lectures on Rings and Modules. Lecture Notes in Mathematics 246. Spring-Verlag, Berlin 1973
A. Kertesz (Ed.). Rings, Modules and Radicals. Colloquium Mathematica Societatis Janos Bolyai 6. North-Holland, Amsterdam 1974
V. Dlab and P. Gabriel (Eds.). Representations of Algebras: Ottawa Conference. Lecture Notes in Mathematics 488. Springer-Verlag, Berlin B. R. McDonald, A. Magid, and K. Smith (Eds.).Ring Theory: Oklahoma University Conference. Lecture Notes in Mathematics 7. Marcel Dekker, New York 1975
J. H.Cozens and F. L. Sandomierski (Eds.). Noncommutative Ring Theory: Kent State Conference. Lecture Notes in Mathematics 545. Springer-Verlag, Berlin B.R. McDonald and R. Morris (Eds.). Ring Theory 11: Second Oklahoma University Conference. Lecture Notes in Mathematics 26. Marcel Dekker, New York D. Zelinsky (Ed.). Proceedings of Brauer Group Conference: Evanston. Lecture Notes in Mathematics 549. Springer-Verlag, Berlin 1977 S. K. Jain (Ed.). Ring Theory: Ohio State University Conference, 1976. Lecture Notes in Mathematics 25. Marcel Dekker, New York
M. P. Malliavin (Ed.). Stminaire GAlgtbre Paul Dubreil, Paris (1975-76). Lecture Notes in Mathematics 586. Springer-Verlag, Berlin
1978
R. Gordon (Ed.). Representation Theory of Algebras: Temple University Conference, 1976.Lecture Notes in Mathematics 37. Marcel Dekker, New York M. P. Malliavin (Ed.). Stminaire GAlgtbre Paul Dubreil, Paris (1976-77).Lecture Notes in Mathematics 641. Springer-Verlag, Berlin F.Van Oystaeyen (Ed.). Ring Theory: Antwerp Conference, 1977.Lecture Notes in Mathematics 40.Marcel Dekker, New York
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D. Handleman and J. Lawrence(Eds.). Ring Theory: Waterloo Conference, 1978. Lecture Notes in Mathematics 734. Springer-Verlag, Berlin M. P. Malliavin (Ed.). SCminaire d’AlgPbre Paul Dubreil, Paris (1977-78). Lecture Notes in Mathematics 740. Springer-Verlag, Berlin F. Van Oystaeyen (Ed.). Ring Theory: Antwerp Conference, 1978. Lecture Notes in Mathematics 51. Marcel Dekker, New York 1980
V. Dlab (Ed.). Representation Theory 1: Carleton University Conference, 1979. Lecture Notes in Mathematics 831. Springer-Verlag, Berlin V. Dlab (Ed.). Representation Theory I1 : Carleton University Conference, 1979. Lecture Notes in Mathematics 832. Springer-Verlag, Berlin B. R. McDonald (Ed.). Ring Theory and Algebra 111: Third Oklahoma University Conference. Lecture Notes in Mathematics 55. Marcel Dekker, New York M. P. Malliavin (Ed.). Shinaire d’AlgPbre Paul Dubreil et Marie-Paule Malliauin, Paris (1978- 79). Lecture Notes in Mathematics 795. Springer-Verlag. Berlin F. Van Oystaeyen (Ed.). Ring Theory: Antwerp Conference, 1980. Lecture Notes in Mathematics 825. Springer-Verlag, Berlin 1981
M. Kervaire and M. Ojanguren (Eds.). Groupe de Brauer: Seminaire, Les Plans-Sur-Bex. Lecture Notes in Mathematics 844. Springer-Verlag, Berlin M. P. Malliavin (Ed.). Sdminaire d’dlgdbre Paul Dubreil et Marie-Paule Malliauin, Paris (1979-80). Lecture Notes in Mathematics 867. Springer-Verlag, Berlin 1982
S. A. Amitsur, G. Seligman, and D. Saltman (Eds.). N . Jacobson: An Algebraists’ Homage: Yale University 1980. Contemporary Mathematics 13. American Mathematical Society, Providence P. J. Fleury (Ed.). Advances in Noncommutative Ring Theory: Plattsburg Conference, 1981. Lecture Notes in Mathematics 951. Springer-Verlag, Berlin M. P. Malliavin (Ed.). Siminaire d’AlgPbre Paul Dubreil et Marie-Paule Malliauin, Paris (1981-82). Lecture Notes in Mathematics 924. Springer-Verlag. Berlin Representations of Algebras (Pueblo).Lecture Notes in Mathematics 944. Springer-Verlag, Berlin F. Van Oystaeyen and A. Verschoren (Eds.). Brauer Groups in Ring Theory and Algebraic Geometry. Lecture Notes in Mathematics 917. Springer-Verlag, Berlin 1983
M. P. Malliavin (Ed.). Seminaire d’dlgkbre Paul Dubreil et Marie-Paule Malliauin, Paris (1982-83). Lecture Notes in Mathematics 1029. Springer-Verlag, Berlin 1984
B. Srinivasan and J. Sally (eds.). Emmy Noether in Bryn Mawr. Springer-Verlag, Berlin F. Van Oystaeyen (Ed.). Methods in Ring Theory. NATO Advanced Science Institutes, Reidel, Dordrecht 1985
M. P. Malliavin (Ed.) Siminaire d’Algebre Paul Dubreil et Marie-Paule Malliauin, Paris (1983-84). Lecture Notes in Mathematics 1146. Springer-Verlag, Berlin L. Marki and R. Wiegandt (Eds.). Radical Theory, Colloquium Mathematica Societatis Janos Bolyai 38. North-Holland, Amsterdam
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S. Montgomery (Ed.). Group Actions on Rinp. Contemporary Mathematics 43. American Mathematical Society, Providence I. Reiner and K. Roggenkamp. (Eds.). Orders and their Applications. Lecture Notes in Mathematics 1142. Springer-Verlag, Berlin 1986 V. Dlab, P. Gabriel, and G. Michler (Eds.). Representation Theory I: Carleton University Conference, 1984. Lecture Notes in Mathematics 1177. Springer-Verlag, Berlin F. Van Oystaeyen (Ed.). Ring Theory: Antwerp Conference, 1985. Lecture Notes in Mathematics 1197. Springer-Verlag, Berlin
Bibliography of Articles General note: This referencelist is only for Volume I; the combined list of references is given at the end of Volume 11. However, these lists are not intended as a comprehensive bibliography of ring theory, for which the reader could turn to Small’s compilations of reviews. Rather, 1have selected those articles most relevant to the areas covered in this volume. Articles wholly included in books by the same author have sometimes been deleted. MR denotes the location of the review in Mathematical Reviews. S. A. AMITSUR, [54] A general theory of radicals. 11. Radicals in rings and bicategories. 111. Applications. Amer. J. Math.76, 100-136. MR 15 #499b,c. [56] Algebras over infinite fields. Proc. Amer. Math. Soc. 7,35-48. MR 17 # 822b. [57] Derivations in simple rings. Proc. London Math. Soc. (3), 57 87-1 12. MR 19 # 525. [67] Prime rings having polynomial identities with arbitrary coefficients. Proc. London Math. SOC.17,470-486. MR 36 #209. [71] Nil radicals. Historical notes and some new results. Rings, Modules, and Radicals. Proc. Collog. Keszthely, pp. 47-65. North-Holland, Amsterdam. MR 50 Z 374. [71a] Rings of quotients and Morita contexts. J. Algebra 17,273-298. MR 54 82704. [71b] Embeddings in matrix rings. Pacific J . Math 36,21-29. MR 43 #2017. [72] On rings of quotients. Symp. Math. Inst. Naz. Alt. Matematica, 149-164. Academic Press, London. MR 48 # 11180. [73] On universal embeddings in matrix rings. J. Math. Soc. Japan 25, 322-328. MR 47 # 3444. AMITSUR, S. A,, and SMALL,L. W. [78] Polynomials over division rings. Israel J. Math. 31,353-358. MR 8Of # 116022. ANAN’IN,A. Z. [79] Embedding of algebras in algebras of triangular matrices. Mar. Sbornik 108, 168-186. MR 81c # 16031. ANAN’IN, A. Z., and ZJABKO, E. M. [74] On a question due to faith. Algebra i Logika 13, 125-131. MR 50 # 13168. ANDRUNAKIEVIC, V. A., and RJABUHIN, Ju. M. [68] Rings without nilpotent elements and completely prime rings. Soviet Math. Dokl. 9, 565-568. MR 37 #6320. T133. ARMENDARIZ, E. P., FISHER,I. W., and SNIDER,R. L. [78] On injective and surjective endomorphisms of finitely generated modules. Comm. Alg. 6 , 659-672. MR 57 #9754. AUSLANDER, M. [74] Representation theory of Artin algebras I, I1 Comm. Algebra 1,177-310. MR 50 #2240.
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[78] The diamond lemma for ring theory, Adu. Math. 29,178-218. MR 81b # 16001. [86] Radicals, tensor products, and algebraicity. Preprint. G. M., and ISAACS, I. M. BERGMAN, [73] Rings with fixed-point-free group actions, Proc. London Math. SOC.27 (3), 69-87. MR 49 # 10743. BERGMAN,G. M., and OGOUS,A. [72] Nakayama's lemma for half-exact functors. Proc. Amer. Math. SOC.31, 67-74. MR 46 # 1777. BERGMAN, G. M., VOVSKI, S. M., BRITTEN D. J., and LEMIRE F.W. [83] Embedding rings in completed graded rings. I. Triangular embeddings. 11.Algebras over a field. 111. Algebras over general k. IV. Commutative algebras. J . Algebra 84,14-106. MR 85i # 16001. I. M., and PONOMAREV, V. A. BERNSTEIN, I. N., GELFAND, [73] Coxeter functors and Gabriel's theorem. Russian Math. Suroeys 28, 17-32. MR 52 # 13876. BIORK,J.-E. [71] Conditions which imply that subrings of Artinian rings are Artinian. J . Reine Angew. Math. 241,123-138. MR 43 #6249. [71a] Conditions which imply that subrings of semiprimary rings are semiprimary. J . AIgebra 19,384-395. MR 44 # 1686. S. BLATTNER, R., COHEN,M., and MONTGOMERY, [86] Crossed products and inner actions of Hopf algebras. Trans. h e r . Math. SOC.298, 671-711. BLOCK, R. E. [81] The irreducible representations of the Lie algebra sI(2) and of the Weyl algebra. Ado. Math. 39,69-110. MR 83c # 17010. BOKUT, L. A. [69] The problem of Mal'cev. Siberian Math. J . 10,706-739. MR 41 #267b. [69a] Groups of fractions of multiplicative semigroups of certain rings, I, 11, 111. Siberian Math. J . 10.172-203,541-600. MR 41 #267a. BONGARTZ,K. [85] Indecomposables are standard. Comment. Math. Helu. 60, 400-410. MR 87d # 16039. BOWTELL,A. J. [67] On a question of Mal'cev. J . AIgebra 7, 126-139. MR 37 #6310. BROWN,B., and McCoy, N. H. [SO] The maximal regular ideal of a ring. Proc. Amer. Math. SOC.1, 165-171. MR 11 #638. BROWN,K. A. [81] Module extensions over Noetherian rings. J. Algebra 69,247-260. MR 83g # 16027. [82] The Nullstellensatz for certain group rings. J . London Math. SOC.(2)26,425-434. MR 84c # 16013. [85] Ore sets in Noetherian rings. Lecture Notes in Mathematics 1146, pp. 355-366. SpringerVerlag, Berlin. T. H. BROWN,K. A., and LENAGAN, [82] A note on Jacobson's conjecture for right Noetherian rings. GIasgow Math. J. 23,7-8. MR 83b # 16010. T. H., and STAFFORD,J. T. BROWN,K. A., LENACAN, [81] K-theory and stable structure of some Noetherian group rings. Proc. London Math. SOC. (3) 42,193-230. MR 82g # 16017. BROWN,K. A,, and WARFIELD R. B., JR. [84] Krull and Gabriel dimensions of fully bounded Noetherian rings. Proc. Amer. Math. Soc. 92,169-174. MR 86d # 16019.
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CAMILLO, V. P., and FULLER, K. R. [74] On the Loewy length of rings. Pacific J. Math. 53,347-354. MR 52 # 10798. CAUCHON, G. [76] Anneaux semipremiers Noetheriens a identites polyn6miales. Bull. Soc. Math. France 104, 99-111. MR 53 # 10859. [76a] Les T-anneaux, la condition (H) de Gabriel, et ses consequences. Comm. Algebra 4, 11-50. MR 49 #2836. [79] ldeaux bilateres et centre des anneaux de polyn6mes de Ore sur les anneaux quasisimples. Lecture Notes in Mathematics 740, pp. 397-407. Springer-Verlag, Berlin. MR 82e # 16001.
[86] Centralisateurs dans les Corps de Weyl. Comm. Alg. 14, 1403-1428. MR 87m # 16039. CAUCHON, G., and LESIEUR,L. [78] Localisation classique en un ideal premier d’un anneau Noetherien a Gauche. Comm. Alg. 6,109-1 108. MR 58 # 10985. CHASE,S . U. [60] Direct products of modules. Trans. Amer. Math. Soc. 97,457-473. MR 22 # 11017. [61] A generalization of the ring of triangular matrices. Nagoya Math. J. 18, 13-25. MR 23A #919. CHASE,S. U., and FAITH,C [65] Quotient rings and direct products of full linear rings, Math. Z. 88, 250-264. MR 31 #2281. CHATTERS, A. W. [76] Two results on P. P. rings. Comm. Algebra 4,881-891. MR 54 #7522. CHATTERS, A. W., GOLDIE,A. W., HAJARNAVIS, C. J., and LENAGAN, T. H. [79] Reduced rank in Goldie rings. J. Algebra 61,582-589. MR 81i # 16041. COHEN,M. [78] Centralizers of algebraic elements. Comm. Algebra 6, 1505-1519. MR 80a # 16028. [85] Hopf algebras acting on semiprime algebras. Contemp. Math. 43, 49-61. MR 87a # 16016. COHEN,M., and MONTGOMERY, S. [84] Group-graded rings, smash products, and group actions. Trans. h e r . Math. Soc. 282, 237-258. MR 85i # 16002. COHN,P. M. [56] The complement of a f.g. direct summand of an Abelian group. Proc. Amer. Math. Soc. 7, 520-521. MR 17 # 1182. [59] On the free product of associative rings. Math. Z. 71, 380-398. MR 21 #5648. [60] On the free product of associative rings. 11. The case of (skew)fields. Math. Z. 73,433-456. MR 22 #4747. [61] On the embedding of rings in skew fields. Proc. London Math SOC.(3) 11,511-530. MR 25 # 100. [66] Some remarks on the invariant basis property. Topology 5,215-228. MR 33 #5676. [68] On the free product of associative rings 111, J. Algebra 8,376-383; correction ibid. 10,23 (1968).MR 36 #5170. [71] The embedding of firs into skew fields. Proc. London Math. Soc. (3) 23, 193-213. MR 45 # 6866. [74] Progress in free associative algebras. Israel J. Math. 19, 109-151. MR 52 #460.
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Subject Index
Abelian category, 79f, 140ff absolute value, 49 1 ACC, see ascending chain condition additive functor, 78 adjoining I , 87f adjoint, 4778 functor, 478tT isomorphism, 264 pair, 4778 adjunction, 478,483 algebra, 5f algebraic, 185 algebraic element, 25, I85 almost split sequence, 2538 amalgamated sum, see free product Amitsur’s theorem, 187, 194ff, 307,318 annihilator, 19f, 82f. 363tT, see also left ideal antisymmetric, 297 AR-property, see Artin-Rees Artin algebra, 256,258 Artinian module, 22f, 165.403tT, 460 radical, 404,460 ring, I67ff, 402,405 left, 22, 1678, 211,215,249, 277,321, 402,4038
semisImple, 169lT, 171f, 179,209,215, 310,336,475 simple, 1688, 173f, 215, 310, 339 Artin-Rees property, 413,462 separated, 466 ascending chain condition, 22,205 on @, 358ff, 362,453,463 on Ann, 205,3578, 380,445,447,453 on annihilator ideals, 364f on cyclic submodules, 223,321 on ideals, 205, 376 on principal left ideals, 223 on semiprime ideals, 43 I, 463 assassinator, 421,435,464 atom, 27 automorphism, 88f fixed subring under, 88, 198f, 319f Baer’s criterion, 262,268 balanced map, lOlf balanced module, 482f Banach algebra, 4928 base, 54 basic idempotent, 221 basic subring, 221
531
532 Bergman's example (nonembeddibility), 371 Bergman-Isaacs theorem, 198f biendomorphism ring, 84 bimodule, 15f, 103, llof, 434ff free, 111 biuniform, 435 bisubmodule, see sub-bimodule Block decomposition, 322 bonds, 437 Boolean algebra, 28 bounded (left), 406,460 almost, 406 almost fully, 406,407,437ff fully, 406f, 420,460 bounded index, 206,373 Brauer-Thrall conjectures, 251,257 sesqui-BT, 258 C-Alg, 25 cancellation, see direct sum Cartan-Brauer-Hua theorem, 3 16 category, 13ff Abelian, 79f, 14M equivalence, 467,468 Grothendieck, 4841 preadditive, 78 cell (left), 435 bi-cell, 435 complete set of, 436 maximal, 435 center, 5 central closure, 388f centralizer, 5, 107, 110,388 centralizes, 6 centralizing extension, 106ff finite, 2931,3891. see also normalizing extension central localization, see localization, central; module centrally closed, 388f, 458 central simple algebra, see simple algebra character module, 2731 Chinese Remainder Theorem, 162 Clifford algebra, 147f coefficient, 42 cogenerator, 482 coherent ring, 337 cokernel, 79 comaximal, 153,162 commutative diagram, 60 commutativity theorems, 31%
Subject Index
commutator (in ring), 89f complement, 9, 176 complements summands, 329f completely reducible, 176, 178f completion, 116ff, 409,481 A-adic, 117f,409f, 412,461 composition length, 167 composition series, 165,310,313 concrete category, 56 coproduct, 74ff, 80 core, 152 countable module, 328 cover (in lattice), 26 cover of module, 325, see also projective (module), cover C o m n s domain, 446 Crawley-JQnsson-Warfield theorem, 328 critical module, 420f, 464 cutting down, 342 cyclic algebra, 97 cyclic module, 19,227,234,268,278 DCC, see descending chain condition decomposition (of modules), 237ff, 244ff, 269 denominator set, 348 for ideal, 367 dense (in Spec(R)),282 dense subring, 151,158 density theorem, 151,308,311,334 for (*)-rings, 343 dependence, 23ff dependent modules, 54 derivation, 89f, 91f, 97ff, 143,207 inner, 89 u-derivation, 92 descending chain condition, 22 on cyclic submodules, 223 on f.g. submodules, 223, 320 on left ideals, see Artinian ring, left on principal left ideals, 171,221, see also perfect ring diagram-chasing, 275 differential polynomial ring, 97f dimensions of modules and rings basic, 421ff exact, 416,422,466 Gabriel, 442f, 466 K-dimension, see Krull uniform, 361ff, 452,463 directed set, 6, 123
Subject Index
direct limit, 112ff, 146,480,483 directly finite, 455 direct product, 66ff, 139f direct sum, 67ff, 74,269 (internal), 55 cancellation in, 248,330 of rings, 75 divisible module, 263f, 444,448 division ring, 2, 56f, 66. 139, 156, 158, 182, 183,221,312,372 domain, 2,46,50,93, 119, 127, 158,202, 353,3728,443 dual base, 86 basis lemma, 230 category, 14 functor, 142 lattice, 7,26f numbers, 147 poset, 6f Eilenberg’s trick, 226 embedding, 3688,449f central, 370 of nilpotent rngs, 449f endomorphism ring, 81ff, 142,15Off, 154, 156,216,236,2381 epic, 15,226, 236 equivalent categories, 38 essential complement, 176, 362 essential over, 266 essential ring extension, 381f, 448 exact seqcence, 67ff pure, 338 short, 68 exchange property, 241,270 extended centroid, 387 extension, see centralizing extension; normalizing extension exterior algebra, 123 f.g., see finitely generated f.r.t., see finite representation type faithful, 82, 83, 150 faithfully flat, 278f, 340 faithfully projective, 279 Faith-Utumi theorem, 448 filter, 9,691f cofinite, 9, 72 idernpotent, 392f, 485 of left ideal, 378, 386, 392f, 485 principal, 9
533 filtration, 45f, 409 A-adic, 117 valuated, 45 finitely generated, 18f, 73, 119,277,278, 337,401,444,468 minimally, 425 projective, 468ff finitely presented, 232ff, 338 finite module, see finitely generated finite object in category, 468 finite representation type, 249ff, 259f, 331 fir, see free ideal ring Fitting’s lemma, 239,252, 323 five lemma, 275 fixed subring, see automorphism flat, 230ff, 265,213ff, 277,337,338,340
Forster-Swan-Warfield-Stafford-Coutinho theorem, 426 fractions, 348, see also quotients free algebra, 57ff, 66,75 commutative ring, 59 group, 64ff,485ff, 490 ideal ring, 373 module, 53ff, 55f,226f, 229,325,321 object, 56 product, 77, 124ff, 147,308f ring, 57ff R-ring, 124 freely generated module, 42 Frobenius algebra, 453f, 454 full subcategory, 16 fully idempotent ring, 339 functor, 16f additive, 270 exact, 271ff, 335,356,383 faithful, 470 full, 470 half exact, 335 left exact, 271 right exact, 271,480 fundamental theorem of Abelian groups, 2 Gabriel’s theorem, 259 G-domain, 286 generator, 469,475,481,484 generic flatness, 288ff G-ideal, 286 going up, 292ff, 294,390 Goldie ring, 358ff, 361,452,463
534 Goldie’s theorems, 3568, 45 If first, 360 second, 359 Goodearl’s theorems, 4548 graded, 1188 map, 119 module, 119 associated, 409 ring, 119,146,194fT.319 associated, 409 submodule, 119 group, If acting on ring, 143f. 198,319f commutator, 53 ordered, see ordered group symmetric, 2 torsion-free, 2, 52 group ring, 43 GU, see going up Harada-Sai lemma, 252 H-condition, 460 hereditary (ring), 228f, 324,333,447 Noetherian prime, 403,478 Herstein’s example, 405 Hilbert Basis Theorem, 395 Hilbert’s Nullstellensatz, 184 Hom(A, -), 16 homomorphism, 2 IBN, see invariant base number ideal 3, see also prime; primitive; semiprime G-ideal, 286, 391 inner, 31 1 maximal, 10, 134, 168 polycentral, 41 1,461,462 polynormal, 41 1 proper, 3 ring with involution, 299 ideal-compatible, 291 idealizer, 142 idempotent, 28,35tT, 137, 165, 170, 183, 215f,2198,223,276,281,310,366, 447,482 central, 366 complete set of primitive, 218f, 221,235 primitive, 39, 170 idempotent-lifting, 398, 117,217, 218, 234 INC, 292,293,389f incidence algebra, 138
Subject Index indecomposable, 2378, 2498 injective, 270,332,464 independent modules, 54f indeterminate, 44 inductive poset, 10 injective (module), 2618,268,2698, 332f, 338,448,456,464,483 cogenerator, 478,482 hull, 2671,333,363,380f invariant base number, 618,454 inverse limit, 115, 146,481,483f invertible element, 181, 182, 185,492 projective module, 286,341 involution, 296 canonical symplectic, 298,303 of first kind, 300,305 of second kind, 300 transpose, see transpose irreducible module, see simple module irreducible ring, 37% isomorphic categories, 17 isomorphism, 23 isomorphism theorem, 4,324,338 Jac(R), see Jacobson radical Jacobi identity, 90 Jacobson radical, 1798, 210,213f, 225,307, 3118,314 Jacobson ring, 2861,289,290,340f Jacobson’s conjecture, 405,439,464 Jategaonkar’s example, 95, 159f, 309,459, 463 Jategaonkar’s ring, 160 Jordan homomorphism, 345 ideal, 343 structure of ring, 344 subring, 297 Jordan-Holder theorem, 165,3 10 J-Spec(R), 428 kernel, 3f of morphism, 78 Kerr’s counterexample, 445f Koethe-Noether-Jacobson theorem, 3 12 Koethe’s conjecture, 209f, 318 Kronecker delta, 29 Krull dimension, 4158, 4621 classical, 418f little, 418f
Subject Index
Krull dimension of non-Noetherian rings, 463 Krull dimension I, 463 Krull homogeneous, 418 Krull-Schmidt, 237ff theorem, 241f Krull symmetry, 441 large, see left ideal; submodule lattice, 7ff Boolean, 28 compactly generated, 26 complemented, 9,27 distributive, 28 isomorphism, 7 modular, 9,27 upper continuous, 26 Laurent extension, 101 Laurent series, 48ff skew, 101 LE-decomposition, 330,328 left annihilator, 19, 356ff maximal, 447,454 left ideal, 3 fractional, 448 independent, 358ff invertible, 448 large, 358f, 3782 382 maximal, 10, 134, 152, 180 minimal, 153, 155f, 157f, 169,301 modular, 196ff principal, 212ff, 221ff, 276 left multiplication, 8 I Leibniz’s formula, 98 LE-module, 240ff, 253 length (of module), 167 Levy’s counterexample, 245ff Levy’s theorem, 138 Lichtman’s example, 309f Lie algebra, 90f homomorphism, 90 ideal, 91, 99, 343 product, 90 structure of ring, 344 lifting property, 225 LO, see lying over localization, 129,338, 377ff. 414 abstract, 458f, 483,485 central, 129IT, 148,283f of ring with involution, 300
535 Ore, 348ff, 443ff of modules, 355 at prime ideals, 44Of local ring, 182f,218,227, 235,238,239,243, 245,324,330 Loewy chain, 322 lower central series, 53 lowest order term, 46 lying over, 292ff, 342, 390 Magnus’ theorem, 65,485 Magnus-Witt theorem, 65,485 Mal’cev’s example, 372 map, 3 matrix ring, 29ff, 137ff, 152, 183, 211, 305f, 403,443,448,470,476f infinite, 137, 327 upper triangular, 33f, 228 matrix unit, 30 maximal commutative subring, 6, 10f left ideal, see left ideal ring of quotients, 386 maximum condition, see ascending chain condition minimum condition, see descending chain condition modular, see lattice module, 3, see also individual modules central localization of, 134tT induced, 331 Ore localization of, 355 monic, I5,28,266 monoid, 1 ordered, 45f, 49f monoid ring, 43 Morita, 467tT.472, 476 context, 471 equivalent, 468,476f, 482 ring, 35, 146 Multiplicative Basis Theorem, 260 Musson’s example, 408 N(R), 206 Nakayama’s lemma, 188,224,233,278,313 Nazarova-Roiter functor, 250 nil,41, 184ff, 187, 196, 200ff, 209f, 314, 318, 444 subset, 207ff, 449 Nil(R), see nilradical, upper
536 nilpotent, 184f, l99,202f, 354,444 generalized, 3 12 of index n, 202 locally, 318f nilradical, 199ff lower, see prime, radical upper, 2 W , 203,205ff, 307 Noetherian, 22 induction, 395 module, 22f, 165,238,244,313,395 ring, 401f left, 22,205,211,215, 244r, 277, 332r, 354,369,394ff, 445,453,458$476 nonassociative, 90 nonsingular, 379& 451,454ff normalizing element, 189,399f normalizing extension, 189ff, 313,341lT normed algebra, 491f Nullstellensatz, 286ff Hilbert, 184 maximal, 2871 strong, 287 weak, 184ff, 286,287,290 order(left), 348,367,402 in Artinian ring, 368,376,398,445,447, 459 Noetherian, 398 in semilocal ring, 3679 444 ordered group, 50,5 If, 138 ordered monoid, 455 49f ordinal, 1If Ore condition, 349 extension, 93,34Of, 396 property, 94 ring, 94,349,353,360 Ostrowski's theorem, 494 partial annihilator, see assassinator partial order, 6 Passman's primitivity machine, 341ff Patch continuity theorem, 432 Peirce decomposition, 36 perfect field, 192 perfect ring, 221ff, 320f. 326,329,337,443, 453 Picard group, 341 PLID, see principal left ideal, domain PO, 6 polynomial ring, 43ff,96f, 187, 195,228
Subject Index
Popescu-Gabriel theorem, 484 poset, 6f positive cone, 52 power series, 48 P P ring, 403,4471 preorder, 6 prime ideal, 163,164,168,203,28off, 417f affiliated set of, 441 bonded, 437 height of, 296,401 height 1,444 minimal, 202,282,296,365, 365,399f, 417 radical, 203f, 225,3 18,342 ring, 143f, 153ff, 355,360,388f. 417f, 419, 422ff, 456 with involution, 296ff, 307 spectrum, 28M, 287,432 prime avoidance, 282 primitive ideal, 179f. 188,391,424 idempotent, see idempotent ring, ISM, 154,215,309,310,369,371, 39 1 with involution, 306,343 Principal ideal theorem, 3991 principal left ideal domain, 93ff, 143, 160,228,229,2635 454, 459,463 ring, 4 48 product (in category), 73f. 80, see also direct product progenerator, 469ff, 482 projective (module), 225ff, 244,2682 2852 323f, 337,481f cover, 233ff. 236,326 generator, 469tT, 482 invertible, 286, 341 projective limit, see inverse limit pullback, 145,260,483 pushout, 143,254,260,331,483 quadratic form, 147 quasi-Frobenius ring, 453f quasi-injective module, 3341 quasi-invertible, 180ff, 195, 197 quasi-local ring, 219,371 quaternions, 32, 148,494 quiver, 259
Subject Index quotients (ring of), 348 abstract, 393 central, 134 Johnson, 379 Martindale-Amitsur, 386 maximal, 385,457 radical, 199, see also Artinian; Jacobson radical; prime (abstract), 392 Baer, see prime, radical Levitzki, 318 of module, 189,325 rank of free module, 61 function on f.g. modules, 374 Goldie, 361 of projective module, 285f reduced, 3965 425,4581 reduced product, 70, 140, 146 reduced rank, 3961, 425,4581 reduced ring, 201f, 317 reduced trace and norm, 175 regular element, 128, 129, 359,398, 417 regular representation, 82 regular ring, 2768, 334, 3381 384, 454 n-regular ring, 2241 323 strongly, 339 regular subset, 129,348,352 resolvent set, 185 retraction, 16 right ideal, 3,278 see also left ideal ring 2, 14f,see also indioidual rings change of, 8 commutative, 2f continuous functions, 140 with involution, 2968 Rng, 87f, 143, 1968, 308,314 radical, 314 Robson’s decomposition theorem, 402 Robson’s theorem, 368,444 Sasiada’s example, 3 14 scalar multiplication, 3 Schanuel’s lemma, 232 Schreier refinement theorem, 165ff Schur’s lemma, 150 self-injective, 384, 4548 semifir, 374,450 semigroup, 1 semihereditary ring, 324
537 semilocal ring, 211,215f, 221,225 semiperfect ring, 2178, 221,2428,321f, 323, 326,329 semiprimary ring, 21 1,213,217,239,323,
444
semiprime ideal, 163,28 I, 463, see also semiprime ring ring, 164f. 169f, 172, I86,203f,205,3588, 3638,402,463 with involution, 302f, 307 semiprimitive ring, 179f, 187, 31 1 semisimple module, 176, 178f semisimple ring, see Artinian, semisimple separable, 25, 192 algebra, 194,312 simple algebra, 97, 109f central, 173f, 310 module, 19f, 37f. 150, 154,335 ring, 19f,97,403,465,476,482 with involutim, 299,304 singular submodule, 379ff, 451 skeletally small, 142 skew field, see division ring group ring, 101 Laurent series, 101 polynomial ring, 96f power series ring, 100 symmetric, 297 small category, 14 Small’s examples, 376,413,450 Small’s theorem, 368,376, 445,459 socle, 156, 169, 308, 31 1 of module, 27,176 spanned, 18 Spec(R), see prime, spectrum split algebra, 171f epic, 69,226 exact sequence, 68,47 monk, 69,266 splitting field, 1718 S-prime, 133f stable number of generators, 4291 range, 330,4298 sub-bimodule, 433ff large, 435 subdirectly irreducible, 310,433,4381
538 subdirect product, 164,201,202 submodule, 3 closed, 269,451f critical, 421 dense, 385,457 large, 176,178f, 378 maximal, 21,164,188,189,325 small, 233,326,327 subring, 2 summand, 68,191,226,234,243,265,382 support, 42 symmetric, 297 symmetric algebra,(first definition), 123 (second definition), 454 symmetric word monoid, 30 system, 112f, 113 T-condition, 255 tensor algebra, 122f, 147 tensor product, lOlff, ll4,121ff, 144f, 192, 196,23Of, 272f, 295,337,388,390, 483 over a field, l08ff tensor ring, 121ff test module, 478 T-nilpotent, 2226 325,327,444 torsion, 2 module, 379,418,434 theory, 458 torsion-free, 2 module, 379,435,444,448 trace ideal, 257,469 trace map, 62f, 139
Subject Index
transcendence base, 25 transcendence degree, 25 transfinite induction, 1If transpose, 32,297,303 ultrafilter, 9, 10 72 ultraproduct, 71ff, 139 uniform module, 362, see also dimensions of modules and rings, uniform unit, 89 unit regular, 340 universal, 6Of, 480 vanishing, see T-nilpotent von Neumann regular, see regular ring V-ring, 335,446 Watt’s theorems, 483f weakly closed, 208 weakly finite, 63f, 139, 369,450 weakly ideal invariant, 465
Wedderburn-Artin(-Hopkins-Levitzki) theorem, 152,308,475,482
Wedderburn-Krull-Schmidt-RemakAzumaya theorem, 241f Wedderburn’s principal theorem, 192f, 3 12 well-ordered, 1If Weyl algebra, 98f, 144, 159,308,403,420, 462,465,477 word, 43 Zariski topology, 1748 Zorn’s lemma, lOf, 27
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