De Gruyter Graduate
Celine Carstensen Benjamin Fine Gerhard Rosenberger
Abstract Algebra Applications to Galois Theo...
66 downloads
1744 Views
2MB Size
Report
This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form
De Gruyter Graduate
Celine Carstensen Benjamin Fine Gerhard Rosenberger
Abstract Algebra Applications to Galois Theory, Algebraic Geometry and Cryptography
De Gruyter
Mathematics Subject Classification 2010: Primary: 12-01, 13-01, 16-01, 20-01; Secondary: 01-01, 08-01, 11-01, 14-01, 94-01.
This book is Volume 11 of the Sigma Series in Pure Mathematics, Heldermann Verlag.
ISBN 978-3-11-025008-4 e-ISBN 978-3-11-025009-1 Library of Congress Cataloging-in-Publication Data Carstensen, Celine. Abstract algebra : applications to Galois theory, algebraic geometry, and cryptography / by Celine Carstensen, Benjamin Fine, and Gerhard Rosenberger. p. cm. ⫺ (Sigma series in pure mathematics ; 11) Includes bibliographical references and index. ISBN 978-3-11-025008-4 (alk. paper) 1. Algebra, Abstract. 2. Galois theory. 3. Geometry, Algebraic. 4. Crytography. I. Fine, Benjamin, 1948⫺ II. Rosenberger, Gerhard. III. Title. QA162.C375 2011 5151.02⫺dc22 2010038153
Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.d-nb.de. ” 2011 Walter de Gruyter GmbH & Co. KG, Berlin/New York Typesetting: Da-TeX Gerd Blumenstein, Leipzig, www.da-tex.de Printing and binding: AZ Druck und Datentechnik GmbH, Kempten ⬁ Printed on acid-free paper Printed in Germany www.degruyter.com
Preface
Traditionally, mathematics has been separated into three main areas; algebra, analysis and geometry. Of course there is a great deal of overlap between these areas. For example, topology, which is geometric in nature, owes its origins and problems as much to analysis as to geometry. Further the basic techniques in studying topology are predominantly algebraic. In general, algebraic methods and symbolism pervade all of mathematics and it is essential for anyone learning any advanced mathematics to be familiar with the concepts and methods in abstract algebra. This is an introductory text on abstract algebra. It grew out of courses given to advanced undergraduates and beginning graduate students in the United States and to mathematics students and teachers in Germany. We assume that the students are familiar with Calculus and with some linear algebra, primarily matrix algebra and the basic concepts of vector spaces, bases and dimensions. All other necessary material is introduced and explained in the book. We assume however that the students have some, but not a great deal, of mathematical sophistication. Our experience is that the material in this can be completed in a full years course. We presented the material sequentially so that polynomials and field extensions preceded an in depth look at group theory. We feel that a student who goes through the material in these notes will attain a solid background in abstract algebra and be able to move on to more advanced topics. The centerpiece of these notes is the development of Galois theory and its important applications, especially the insolvability of the quintic. After introducing the basic algebraic structures, groups, rings and fields, we begin the theory of polynomials and polynomial equations over fields. We then develop the main ideas of field extensions and adjoining elements to fields. After this we present the necessary material from group theory needed to complete both the insolvability of the quintic and solvability by radicals in general. Hence the middle part of the book, Chapters 9 through 14 are concerned with group theory including permutation groups, solvable groups, abelian groups and group actions. Chapter 14 is somewhat off to the side of the main theme of the book. Here we give a brief introduction to free groups, group presentations and combinatorial group theory. With the group theory material in hand we return to Galois theory and study general normal and separable extensions and the fundamental theorem of Galois theory. Using this we present several major applications of the theory including solvability by radicals and the insolvability of the quintic, the fundamental theorem of algebra, the construction of regular n-gons and the famous impossibilities; squaring the circling, doubling the cube and trisecting an angle. We
vi
Preface
finish in a slightly different direction giving an introduction to algebraic and group based cryptography. October 2010
Celine Carstensen Benjamin Fine Gerhard Rosenberger
Contents
Preface 1 Groups, Rings and Fields 1.1 Abstract Algebra . . . . . . . . . . . . 1.2 Rings . . . . . . . . . . . . . . . . . . 1.3 Integral Domains and Fields . . . . . . 1.4 Subrings and Ideals . . . . . . . . . . . 1.5 Factor Rings and Ring Homomorphisms 1.6 Fields of Fractions . . . . . . . . . . . 1.7 Characteristic and Prime Rings . . . . . 1.8 Groups . . . . . . . . . . . . . . . . . . 1.9 Exercises . . . . . . . . . . . . . . . .
v
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
1 1 2 4 6 9 13 14 17 19
2 Maximal and Prime Ideals 2.1 Maximal and Prime Ideals . . . . . . . . . . 2.2 Prime Ideals and Integral Domains . . . . . . 2.3 Maximal Ideals and Fields . . . . . . . . . . 2.4 The Existence of Maximal Ideals . . . . . . . 2.5 Principal Ideals and Principal Ideal Domains . 2.6 Exercises . . . . . . . . . . . . . . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
21 21 22 24 25 27 28
. . . . . . .
29 29 35 38 41 45 51 51
. . . . .
53 53 55 57 58 65
. . . . . . . . .
. . . . . . . . .
3 Prime Elements and Unique Factorization Domains 3.1 The Fundamental Theorem of Arithmetic . . . . 3.2 Prime Elements, Units and Irreducibles . . . . . 3.3 Unique Factorization Domains . . . . . . . . . . 3.4 Principal Ideal Domains and Unique Factorization 3.5 Euclidean Domains . . . . . . . . . . . . . . . . 3.6 Overview of Integral Domains . . . . . . . . . . 3.7 Exercises . . . . . . . . . . . . . . . . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
4 Polynomials and Polynomial Rings 4.1 Polynomials and Polynomial Rings . . . . . . . . . . . 4.2 Polynomial Rings over Fields . . . . . . . . . . . . . . 4.3 Polynomial Rings over Integral Domains . . . . . . . . 4.4 Polynomial Rings over Unique Factorization Domains 4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . .
. . . . .
. . . . . . .
. . . . .
. . . . . . .
. . . . .
. . . . . . .
. . . . .
. . . . . . .
. . . . .
. . . . . . .
. . . . .
. . . . . . .
. . . . .
viii 5
Contents
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
66 66 69 70 74 75 78
6 Field Extensions and Compass and Straightedge Constructions 6.1 Geometric Constructions . . . . . . . . . . . . . . . . . . . 6.2 Constructible Numbers and Field Extensions . . . . . . . . . 6.3 Four Classical Construction Problems . . . . . . . . . . . . 6.3.1 Squaring the Circle . . . . . . . . . . . . . . . . . . 6.3.2 The Doubling of the Cube . . . . . . . . . . . . . . 6.3.3 The Trisection of an Angle . . . . . . . . . . . . . . 6.3.4 Construction of a Regular n-Gon . . . . . . . . . . . 6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
80 80 80 83 83 83 83 84 89
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
91 91 94 100 100 101 105 109 111
8 Splitting Fields and Normal Extensions 8.1 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
113 113 115 118
9 Groups, Subgroups and Examples 9.1 Groups, Subgroups and Isomorphisms 9.2 Examples of Groups . . . . . . . . . 9.3 Permutation Groups . . . . . . . . . . 9.4 Cosets and Lagrange’s Theorem . . . 9.5 Generators and Cyclic Groups . . . . 9.6 Exercises . . . . . . . . . . . . . . .
119 119 121 125 128 133 139
7
Field Extensions 5.1 Extension Fields and Finite Extensions . . . . 5.2 Finite and Algebraic Extensions . . . . . . . 5.3 Minimal Polynomials and Simple Extensions 5.4 Algebraic Closures . . . . . . . . . . . . . . 5.5 Algebraic and Transcendental Numbers . . . 5.6 Exercises . . . . . . . . . . . . . . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
Kronecker’s Theorem and Algebraic Closures 7.1 Kronecker’s Theorem . . . . . . . . . . . . . . . . . . 7.2 Algebraic Closures and Algebraically Closed Fields . . 7.3 The Fundamental Theorem of Algebra . . . . . . . . . 7.3.1 Splitting Fields . . . . . . . . . . . . . . . . . 7.3.2 Permutations and Symmetric Polynomials . . . 7.4 The Fundamental Theorem of Algebra . . . . . . . . . 7.5 The Fundamental Theorem of Symmetric Polynomials 7.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . . . .
. . . . . .
. . . . . .
. . . . . . . .
. . . . . .
. . . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
ix
Contents
10 Normal Subgroups, Factor Groups and Direct Products 10.1 Normal Subgroups and Factor Groups . . . . . . . . 10.2 The Group Isomorphism Theorems . . . . . . . . . . 10.3 Direct Products of Groups . . . . . . . . . . . . . . 10.4 Finite Abelian Groups . . . . . . . . . . . . . . . . . 10.5 Some Properties of Finite Groups . . . . . . . . . . . 10.6 Exercises . . . . . . . . . . . . . . . . . . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
141 141 146 149 151 156 160
11 Symmetric and Alternating Groups 11.1 Symmetric Groups and Cycle Decomposition 11.2 Parity and the Alternating Groups . . . . . . 11.3 Conjugation in Sn . . . . . . . . . . . . . . . 11.4 The Simplicity of An . . . . . . . . . . . . . 11.5 Exercises . . . . . . . . . . . . . . . . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
161 161 164 167 168 170
. . . . .
171 171 172 175 177 179
. . . . .
. . . . .
. . . . .
. . . . .
12 Solvable Groups 12.1 Solvability and Solvable Groups . . . . . . . . . . . 12.2 Solvable Groups . . . . . . . . . . . . . . . . . . . . 12.3 The Derived Series . . . . . . . . . . . . . . . . . . 12.4 Composition Series and the Jordan–Hölder Theorem 12.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . 13 Groups Actions and the Sylow Theorems 13.1 Group Actions . . . . . . . . . . . . . . . . 13.2 Conjugacy Classes and the Class Equation . 13.3 The Sylow Theorems . . . . . . . . . . . . 13.4 Some Applications of the Sylow Theorems 13.5 Exercises . . . . . . . . . . . . . . . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
180 180 181 183 187 191
14 Free Groups and Group Presentations 14.1 Group Presentations and Combinatorial Group Theory 14.2 Free Groups . . . . . . . . . . . . . . . . . . . . . . . 14.3 Group Presentations . . . . . . . . . . . . . . . . . . . 14.3.1 The Modular Group . . . . . . . . . . . . . . 14.4 Presentations of Subgroups . . . . . . . . . . . . . . . 14.5 Geometric Interpretation . . . . . . . . . . . . . . . . 14.6 Presentations of Factor Groups . . . . . . . . . . . . . 14.7 Group Presentations and Decision Problems . . . . . . 14.8 Group Amalgams: Free Products and Direct Products . 14.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
192 192 193 198 200 207 209 212 213 214 216
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
x
Contents
15 Finite Galois Extensions 15.1 Galois Theory and the Solvability of Polynomial Equations 15.2 Automorphism Groups of Field Extensions . . . . . . . . 15.3 Finite Galois Extensions . . . . . . . . . . . . . . . . . . 15.4 The Fundamental Theorem of Galois Theory . . . . . . . 15.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
217 217 218 220 221 231
16 Separable Field Extensions 16.1 Separability of Fields and Polynomials 16.2 Perfect Fields . . . . . . . . . . . . . 16.3 Finite Fields . . . . . . . . . . . . . . 16.4 Separable Extensions . . . . . . . . . 16.5 Separability and Galois Extensions . . 16.6 The Primitive Element Theorem . . . 16.7 Exercises . . . . . . . . . . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
233 233 234 236 238 241 245 247
17 Applications of Galois Theory 17.1 Applications of Galois Theory . . . . 17.2 Field Extensions by Radicals . . . . . 17.3 Cyclotomic Extensions . . . . . . . . 17.4 Solvability and Galois Extensions . . 17.5 The Insolvability of the Quintic . . . . 17.6 Constructibility of Regular n-Gons . . 17.7 The Fundamental Theorem of Algebra 17.8 Exercises . . . . . . . . . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
248 248 248 252 253 254 259 261 263
18 The Theory of Modules 18.1 Modules Over Rings . . . . . . . . . . . . . . . . . . . . 18.2 Annihilators and Torsion . . . . . . . . . . . . . . . . . . 18.3 Direct Products and Direct Sums of Modules . . . . . . . 18.4 Free Modules . . . . . . . . . . . . . . . . . . . . . . . . 18.5 Modules over Principal Ideal Domains . . . . . . . . . . . 18.6 The Fundamental Theorem for Finitely Generated Modules 18.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
265 265 270 271 273 276 279 283
. . . .
285 285 286 288 294
19 Finitely Generated Abelian Groups 19.1 Finite Abelian Groups . . . . . . . . . . . . . . . . . 19.2 The Fundamental Theorem: p-Primary Components 19.3 The Fundamental Theorem: Elementary Divisors . . 19.4 Exercises . . . . . . . . . . . . . . . . . . . . . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
xi
Contents
20 Integral and Transcendental Extensions 20.1 The Ring of Algebraic Integers . . . 20.2 Integral ring extensions . . . . . . . 20.3 Transcendental field extensions . . . 20.4 The transcendence of e and . . . . 20.5 Exercises . . . . . . . . . . . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
295 295 298 302 307 310
21 The Hilbert Basis Theorem and the Nullstellensatz 21.1 Algebraic Geometry . . . . . . . . . . . . . . . . . . . 21.2 Algebraic Varieties and Radicals . . . . . . . . . . . . 21.3 The Hilbert Basis Theorem . . . . . . . . . . . . . . . 21.4 The Hilbert Nullstellensatz . . . . . . . . . . . . . . . 21.5 Applications and Consequences of Hilbert’s Theorems 21.6 Dimensions . . . . . . . . . . . . . . . . . . . . . . . 21.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
312 312 312 314 315 317 320 325
22 Algebraic Cryptography 22.1 Basic Cryptography . . . . . . . . . . . . . . . . . 22.2 Encryption and Number Theory . . . . . . . . . . 22.3 Public Key Cryptography . . . . . . . . . . . . . . 22.3.1 The Diffie–Hellman Protocol . . . . . . . . 22.3.2 The RSA Algorithm . . . . . . . . . . . . 22.3.3 The El-Gamal Protocol . . . . . . . . . . . 22.3.4 Elliptic Curves and Elliptic Curve Methods 22.4 Noncommutative Group based Cryptography . . . 22.4.1 Free Group Cryptosystems . . . . . . . . . 22.5 Ko–Lee and Anshel–Anshel–Goldfeld Methods . . 22.5.1 The Ko–Lee Protocol . . . . . . . . . . . . 22.5.2 The Anshel–Anshel–Goldfeld Protocol . . 22.6 Platform Groups and Braid Group Cryptography . 22.7 Exercises . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
326 326 331 335 336 337 339 341 342 345 349 350 350 351 356
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
Bibliography
359
Index
363
Chapter 1
Groups, Rings and Fields
1.1
Abstract Algebra
Abstract algebra or modern algebra can be best described as the theory of algebraic structures. Briefly, an algebraic structure is a set S together with one or more binary operations on it satisfying axioms governing the operations. There are many algebraic structures but the most commonly studied structures are groups, rings, fields and vector spaces. Also widely used are modules and algebras. In this first chapter we will look at some basic preliminaries concerning groups, rings and fields. We will only briefly touch on groups here, a more extensive treatment will be done later in the book. Mathematics traditionally has been subdivided into three main areas – analysis, algebra and geometry. These areas overlap in many places so that it is often difficult to determine whether a topic is one in geometry say or in analysis. Algebra and algebraic methods permeate all these disciplines and most of mathematics has been algebraicized – that is uses the methods and language of algebra. Groups, rings and fields play a major role in the modern study of analysis, topology, geometry and even applied mathematics. We will see these connections in examples throughout the book. Abstract algebra has its origins in two main areas and questions that arose in these areas – the theory of numbers and the theory of equations. The theory of numbers deals with the properties of the basic number systems – integers, rationals and reals while the theory of equations, as the name indicates, deals with solving equations, in particular polynomial equations. Both are subjects that date back to classical times. A whole section of Euclid’s elements is dedicated to number theory. The foundations for the modern study of number theory were laid by Fermat in the 1600s and then by Gauss in the 1800s. In an attempt to prove Fermat’s big theorem Gauss introduced the complex integers a C bi where a and b are integers and showed that this set has unique factorization. These ideas were extended by Dedekind and Kronecker who developed a wide ranging theory of algebraic number fields and algebraic integers. A large portion of the terminology used in abstract algebra, rings, ideals, factorization comes from the study of algebraic number fields. This has evolved into the modern discipline of algebraic number theory. The second origin of modern abstract algebra was the problem of trying to determine a formula for finding the solutions in terms of radicals of a fifth degree polynomial. It was proved first by Ruffini in 1800 and then by Abel that it is impossible to find a formula in terms of radicals for such a solution. Galois in 1820 extended
2
Chapter 1 Groups, Rings and Fields
this and showed that such a formula is impossible for any degree five or greater. In proving this he laid the groundwork for much of the development of modern abstract algebra especially field theory and finite group theory. Earlier, in 1800, Gauss proved the fundamental theorem of algebra which says that any nonconstant complex polynomial equation must have a solution. One of the goals of this book is to present a comprehensive treatment of Galois theory and a proof of the results mentioned above. The locus of real points .x; y/ which satisfy a polynomial equation f .x; y/ D 0 is called an algebraic plane curve. Algebraic geometry deals with the study of algebraic plane curves and extensions to loci in a higher number of variables. Algebraic geometry is intricately tied to abstract algebra and especially commutative algebra. We will touch on this in the book also. Finally linear algebra, although a part of abstract algebra, arose in a somewhat different context. Historically it grew out of the study of solution sets of systems of linear equations and the study of the geometry of real n-dimensional spaces. It began to be developed formally in the early 1800s with work of Jordan and Gauss and then later in the century by Cayley, Hamilton and Sylvester.
1.2
Rings
The primary motivating examples for algebraic structures are the basic number systems; the integers Z, the rational numbers Q, the real numbers R and the complex numbers C. Each of these has two basic operations, addition and multiplication and form what is called a ring. We formally define this. Definition 1.2.1. A ring is a set R with two binary operations defined on it, addition, denoted by C, and multiplication, denoted by , or just by juxtaposition, satisfying the following six axioms: (1) Addition is commutative: a C b D b C a for each pair a; b in R. (2) Addition is associative: a C .b C c/ D .a C b/ C c for a; b; c 2 R. (3) There exists an additive identity, denoted by 0, such that a C 0 D a for each a 2 R. (4) For each a 2 R there exists an additive inverse, denoted by a, such that a C .a/ D 0. (5) Multiplication is associative: a.bc/ D .ab/c for a; b; c 2 R. (6) Multiplication is left and right distributive over addition: a.b C c/ D ab C ac and .b C c/a D ba C ca for a; b; c 2 R.
3
Section 1.2 Rings
If in addition (7) Multiplication is commutative: ab D ba for each pair a; b in R. then R is a commutative ring. Further if (8) There exists a multiplicative identity denoted by 1 such that a 1 D a and 1 a D a for each a in R. then R is a ring with identity. If R satisfies (1) through (8) then R is a commutative ring with an identity. A set G with one operation, C, on it satisfying axioms (1) through (4) is called an abelian group. We will discuss these further later in the chapter. The numbers systems Z; Q; R; C are all commutative rings with identity. A ring R with only one element is called trivial. A ring R with identity is trivial if and only if 0 D 1. A finite ring is a ring R with only finitely many elements in it. Otherwise R is an infinite ring. Z; Q; R; C are all infinite rings. Examples of finite rings are given by the integers modulo n, Zn , with n > 1. The ring Zn consists of the elements 0; 1; 2; : : : ; n1 with addition and multiplication done modulo n. That is, for example 4 3 D 12 D 2 modulo 5. Hence in Z5 we have 4 3 D 2. The rings Zn are all finite commutative rings with identity. To give examples of rings without an identity consider the set nZ D ¹nz W z 2 Zº consisting of all multiples of the fixed integer n. It is an easy verification (see exercises) that this forms a ring under the same addition and multiplication as in Z but that there is no identity for multiplication. Hence for each n 2 Z with n > 1 we get an infinite commutative ring without an identity. To obtain examples of noncommutative rings we consider matrices. Let M2 .Z/ be the set of 2 2 matrices with integral entries. Addition of matrices is done componentwise, that is a2 b2 a1 C a2 b1 C b2 a1 b1 C D c1 d1 c2 d2 c1 C c2 d1 C d2 while multiplication is matrix multiplication
a 1 b1 c1 d1
a2 b2 a a C b1 c2 a1 b2 C b1 d2 D 1 2 : c2 d 2 c1 a2 C d1 c2 c1 b2 C d1 d2
Then again it is an easy verification (see exercises) that M2 .Z/ forms a ring. Further since matrix multiplication is noncommutative this forms a noncommutative ring. However the identity matrix does form a multiplicative identity for it. M2 .nZ/ with n > 1 provides an example of an infinite noncommutative ring without an identity. Finally M2 .Zn / for n > 1 will give an example of a finite noncommutative ring.
4
1.3
Chapter 1 Groups, Rings and Fields
Integral Domains and Fields
Our basic number systems have the property that if ab D 0 then either a D 0 or b D 0. However this is not necessarily true in the modular rings. For example 2 3 D 0 in Z6 . Definition 1.3.1. A zero divisor in a ring R is an element a 2 R with a ¤ 0 such that there exists an element b ¤ 0 with ab D 0. A commutative ring with an identity 1 ¤ 0 and with no zero divisors is called an integral domain. Notice that having no zero divisors is equivalent to the fact that if ab D 0 in R then either a D 0 or b D 0. Hence Z; Q; R; C are all integral domains but from the example above Z6 is not. In general we have the following. Theorem 1.3.2. Zn is an integral domain if and only if n is a prime. Proof. First of all notice that under multiplication modulo n an element m is 0 if and only if n divides m. We will make this precise shortly. Recall further Euclid’s lemma which says that if a prime p divides a product ab then p divides a or p divides b. Now suppose that n is a prime and ab D 0 in Zn . Then n divides ab. From Euclid’s lemma it follows that n divides a or n divides b. In the first case a D 0 in Zn while in the second b D 0 in Zn . It follows that there are no zero divisors in Zn and since Zn is a commutative ring with an identity it is an integral domain. Conversely suppose Zn is an integral domain. Suppose that n is not prime. Then n D ab with 1 < a < n, 1 < b < n. It follows that ab D 0 in Zn with neither a nor b being zero. Therefore they are zero divisors which is a contradiction. Hence n must be prime. In Q every nonzero element has a multiplicative inverse. This is not true in Z where only the elements 1; 1 have multiplicative inverses within Z. Definition 1.3.3. A unit in a ring R with identity is an element a which has a multiplicative inverse, that is an element b such that ab D ba D 1. If a is a unit in R we denote its inverse by a1 . Hence every nonzero element of Q and of R and of C is a unit but in Z the only units are ˙1. In M2 .R/ the units are precisely those matrices that have nonzero determinant while in M2 .Z/ the units are those integral matrices that have determinant ˙1. Definition 1.3.4. A field F is a commutative ring with an identity 1 ¤ 0 where every nonzero element is a unit. The rationals Q, the reals R and the complexes C are all fields. If we relax the commutativity requirement and just require that in the ring R with identity each nonzero element is a unit then we get a skew field or division ring.
5
Section 1.3 Integral Domains and Fields
Lemma 1.3.5. If F is a field then F is an integral domain. Proof. Since a field F is already a commutative ring with an identity we must only show that there are no zero divisors in F . Suppose that ab D 0 with a ¤ 0. Since F is a field and a is nonzero it has an inverse a1 . Hence a1 .ab/ D a1 0 D 0 H) .a1 a/b D 0 H) b D 0: Therefore F has no zero divisors and must be an integral domain. Recall that Zn was an integral domain only when n was a prime. This turns out to also be necessary and sufficient for Zn to be a field. Theorem 1.3.6. Zn is a field if and only if n is a prime. Proof. First suppose that Zn is a field. Then from Lemma 1.3.5 it is an integral domain, so from Theorem 1.3.2 n must be a prime. Conversely suppose that n is a prime. We must show that Zn is a field. Since we already know that Zn is an integral domain we must only show that each nonzero element of Zn is a unit. Here we need some elementary facts from number theory. If a; b are integers we use the notation ajb to indicate that a divides b. Recall that given nonzero integers a; b their greatest common divisor or GCD d > 0 is a positive integer which is a common divisor, that is d ja and d jb, and if d1 is any other common divisor then d1 jd . We denote the greatest common divisor of a; b by either gcd.a; b/ or .a; b/. It can be proved that given nonzero integers a; b their GCD exists, is unique and can be characterized as the least positive linear combination of a and b. If the GCD of a and b is 1 then we say that a and b are relatively prime or coprime. This is equivalent to being able to express 1 as a linear combination of a and b. Now let a 2 Zn with n prime and a ¤ 0. Since a ¤ 0 we have that n does not divide a. Since n is prime it follows that a and n must be relatively prime, .a; n/ D 1. From the number theoretic remarks above we then have that there exist x; y with ax C ny D 1: However in Zn the element ny D 0 and so in Zn we have ax D 1:
6
Chapter 1 Groups, Rings and Fields
Therefore a has a multiplicative inverse in Zn and is hence a unit. Since a was an arbitrary nonzero element we conclude that Zn is a field. The theorem above is actually a special case of a more general result from which Theorem 1.3.6 could also be obtained. Theorem 1.3.7. Each finite integral domain is a field. Proof. Let F be a finite integral domain. We must show that F is a field. It is clearly sufficient to show that each nonzero element of F is a unit. Let ¹0; 1; r1 ; : : : ; rn º be the elements of F . Let ri be a fixed nonzero element and multiply each element of F by ri on the left. Now if ri rj D ri rk then ri .rj rk / D 0: Since ri ¤ 0 it follows that rj rk D 0 or rj D rk . Therefore all the products ri rj are distinct. Hence R D ¹0; 1; r1 ; : : : ; rn º D ri R D ¹0; ri ; ri r1 ; : : : ; ri rn º: Hence the identity element 1 must be in the right-hand list, that is there is an rj such that ri rj D 1. Therefore ri has a multiplicative inverse and is hence a unit. Therefore F is a field.
1.4
Subrings and Ideals
A very important concept in algebra is that of a substructure that is a subset having the same structure as the superset. Definition 1.4.1. A subring of a ring R is a nonempty subset S that is also a ring under the same operations as R. If R is a field and S also a field then its a subfield. If S R then S satisfies the same basic axioms, associativity and commutativity of addition for example. Therefore S will be a subring if it is nonempty and closed under the operations, that is closed under addition, multiplication and taking additive inverses. Lemma 1.4.2. A subset S of a ring R is a subring if and only if S is nonempty and whenever a; b 2 S we have a C b 2 S , a b 2 S and ab 2 S.
Section 1.4 Subrings and Ideals
7
Example 1.4.3. Show that if n > 1 the set nZ is a subring of Z. Here clearly nZ is nonempty. Suppose a D nz1 ; b D nz2 are two element of nZ. Then a C b D nz1 C nz2 D n.z1 C z2 / 2 nZ a b D nz1 nz2 D n.z1 z2 / 2 nZ ab D nz1 nz2 D n.nz1 z2 / 2 nZ: Therefore nZ is a subring. Example 1.4.4. Show that the set of real numbers of the form p S D ¹u C v 2 W u; v 2 Qº is a subring of p p p R. Here 1 C 2 2 S , so S is nonempty. Suppose a D u1 C v1 2, b D u2 C v2 2 are two element of S. Then p p p a C b D .u1 C v1 2/ C .u2 C v2 2/ D u1 C u2 C .v1 C v2 / 2 2 S p p p a b D .u1 C v1 2/ .u2 C v2 2/ D u1 u2 C .v1 v2 / 2 2 S p p p a b D .u1 C v1 2/ .u2 C v2 2/ D .u1 u2 C 2v1 v2 / C .u1 v2 C v1 u2 / 2 2 S: Therefore S is a subring. We will see this example later as an algebraic number field. In the following we are especially interested in special types of subrings called ideals. Definition 1.4.5. Let R be a ring and I R. Then I is a (two-sided) ideal if the following properties holds: (1) I is nonempty. (2) If a; b 2 I then a ˙ b 2 I . (3) If a 2 I and r is any element of R then ra 2 I and ar 2 I . We denote the fact that I forms an ideal in R by I G R. Notice that if a; b 2 I , then from (3) we have ab 2 I and ba 2 I . Hence I forms a subring, that is each ideal is also a subring. ¹0º and the whole ring R are trivial ideals of R. If we assume that in (3) only ra 2 I then I is called a left ideal. Analogously we define a right ideal.
8
Chapter 1 Groups, Rings and Fields
Lemma 1.4.6. Let R be a commutative ring and a 2 R. Then the set hai D aR D ¹ar W r 2 Rº is an ideal of R. This ideal is called the principal ideal generated by a. Proof. We must verify the three properties of the definition. Since a 2 R we have that aR is nonempty. If u D ar1 ; v D ar2 are two elements of aR then u ˙ v D ar1 ˙ ar2 D a.r1 ˙ r2 / 2 aR so (2) is satisfied. Finally let u D ar1 2 aR and r 2 R. Then ru D rar1 D a.rr1 / 2 aR
and
ur D ar1 r D a.r1 r/ 2 aR:
Recall that a 2 hai if R has an identity. Notice that if n 2 Z then the principal ideal generated by n is precisely the ring nZ, that we have already examined. Hence for each n > 1 the subring nZ is actually an ideal. We can show more. Theorem 1.4.7. Any subring of Z is of the form nZ for some n. Hence each subring of Z is actually a principal ideal. Proof. Let S be a subring of Z. If S D ¹0º then S D 0Z so we may assume that S has nonzero elements. Since S is a subring if it has nonzero elements it must have positive elements (since it has the additive inverse of any element in it). Let S C be the set of positive elements in S . From the remarks above this is a nonempty set and so there must be a least positive element n. We claim that S D nZ. Let m be a positive element in S . By the division algorithm m D q n C r; where either r D 0 or 0 < r < n. Suppose that r ¤ 0. Then r D m q n: Now m 2 S and n 2 S . Since S is a subring it is closed under addition so that q n 2 S . But S is a subring so m q n 2 S. It follows that r 2 S. But this is a contradiction since n was the least positive element in S. Therefore r D 0 and m D q n. Hence each positive element in S is a multiple of n. Now let m be a negative element of S . Then m 2 S and m is positive. Hence m D q n and thus m D .q/n. Therefore every element of S is a multiple of n and so S D nZ. It follows that every subring of Z is of this form and therefore every subring of Z is an ideal.
Section 1.5 Factor Rings and Ring Homomorphisms
9
We mention that this is true in Z but not always true. For example Z is a subring of Q but not an ideal. An extension of the proof of Lemma 1.4.2 gives the following. We leave the proof as an exercise. Lemma 1.4.8. Let R be a commutative ring and a1 ; : : : ; an 2 R be a finite set of elements in R. Then the set ha1 ; : : : ; an i D ¹r1 a1 C r2 a2 C C rn an W ri 2 Rº is an ideal of R. This ideal is called the ideal generated by a1 ; : : : ; an . Recall that a1 ; : : : ; an are in ha1 ; : : : ; an i if R has an identity. Theorem 1.4.9. Let R be a commutative ring with an identity 1 ¤ 0. Then R is a field if and only if the only ideals in R are ¹0º and R. Proof. Suppose that R is a field and I C R is an ideal. We must show that either I D ¹0º or I D R. Suppose that I ¤ ¹0º then we must show that I D R. Since I ¤ ¹0º there exists an element a 2 I with a ¤ 0. Since R is a field this element a has an inverse a1 . Since I is an ideal it follows that a1 a D 1 2 I . Let r 2 R then, since 1 2 I , we have r 1 D r 2 I . Hence R I and hence R D I . Conversely suppose that R is a commutative ring with an identity whose only ideals are ¹0º and R. We must show that R is a field or equivalently that every nonzero element of R has a multiplicative inverse. Let a 2 R with a ¤ 0. Since R is a commutative ring and a ¤ 0, the principal ideal aR is a nontrivial ideal in R. Hence aR D R. Therefore the multiplicative identity 1 2 aR. It follows that there exists an r 2 R with ar D 1. Hence a has a multiplicative inverse and R must be a field.
1.5
Factor Rings and Ring Homomorphisms
Given an ideal I in a ring R we can build a new ring called the factor ring or quotient ring of R modulo I . The special condition on the subring I that rI I and I r I for all r 2 R, that makes it an ideal, is specifically to allow this construction to be a ring. Definition 1.5.1. Let I be an ideal in a ring R. Then a coset of I is a subset of R of the form r C I D ¹r C i W i 2 I º with r a fixed element of R.
10
Chapter 1 Groups, Rings and Fields
Lemma 1.5.2. Let I be an ideal in a ring R. Then the cosets of I partition R, that is any two cosets are either coincide or disjoint. We leave the proof to the exercises. Now on the set of all cosets of an ideal we will build a new ring. Theorem 1.5.3. Let I be an ideal in a ring R. Let R=I be the set of all cosets of I in R, that is R=I D ¹r C I W r 2 Rº: We define addition and multiplication on R=I in the following manner: .r1 C I / C .r2 C I / D .r1 C r2 / C I .r1 C I / .r2 C I / D .r1 r2 / C I: Then R=I forms a ring called the factor ring of R modulo I . The zero element of R=I is 0 C I and the additive inverse of r C I is r C I . Further if R is commutative then R=I is commutative and if R has an identity then R=I has an identity 1 C I . Proof. The proofs that R=I satisfies the ring axioms under the definitions above is straightforward. For example .r1 C I / C .r2 C I / D .r1 C r2 / C I D .r2 C r1 / C I D .r2 C I / C .r1 C I / and so addition is commutative. What must be shown is that both addition and multiplication are well-defined. That is, if r1 C I D r10 C I and r2 C I D r20 C I then and
.r1 C I / C .r2 C I / D .r10 C I / C .r20 C I / .r1 C I / .r2 C I / D .r10 C I / .r20 C I /:
Now if r1 C I D r10 C I then r1 2 r10 C I and so r1 D r10 C i1 for some i1 2 I . Similarly if r2 C I D r20 C I then r2 2 r20 C I and so r2 D r20 C i2 for some i2 2 I . Then .r1 C I / C .r2 C I / D .r10 C i1 C I / C .r20 C i2 C I / D .r10 C I / C .r20 C I / since i1 C I D I and i2 C I D I . Similarly .r1 C I / .r2 C I / D .r10 C i1 C I / .r20 C i2 C I / D r10 r20 C r10 i2 C r20 i1 C r10 I C r20 I C I I D .r10 r20 / C I since all the other products are in the ideal I .
Section 1.5 Factor Rings and Ring Homomorphisms
11
This shows that addition and multiplication are well-defined. It also shows why the ideal property is necessary. As an example let R be the integers Z. As we have seen each subring is an ideal and of the form nZ for some natural number n. The factor ring Z=nZ is called the residue class ring modulo n denoted Zn . Notice that we can take as cosets 0 C nZ; 1 C nZ; : : : ; .n 1/ C nZ: Addition and multiplication of cosets is then just addition and multiplication modulo n, as we can see, that this is just a formalization of the ring Zn , that we have already looked at. Recall that Zn is an integral domain if and only if n is prime and Zn is a field for precisely the same n. If n D 0 then Z=nZ is the same as Z. We now show that ideals and factor rings are closely related to certain mappings between rings. Definition 1.5.4. Let R and S be rings. Then a mapping f W R ! S is a ring homomorphism if f .r1 C r2 / D f .r1 / C f .r2 / for any r1 ; r2 2 R f .r1 r2 / D f .r1 / f .r2 /
for any r1 ; r2 2 R:
In addition, (1) f is an epimorphism if it is surjective. (2) f is an monomorphism if it is injective. (3) f is an isomorphism if it is bijective, that is both surjective and injective. In this case R and S are said to be isomorphic rings which we denote by R Š S. (4) f is an endomorphism if R D S , that is a ring homomorphism from a ring to itself. (5) f is an automorphism if R D S and f is an isomorphism. Lemma 1.5.5. Let R and S be rings and let f W R ! S be a ring homomorphism. Then (1) f .0/ D 0 where the first 0 is the zero element of R and the second is the zero element of S. (2) f .r/ D f .r/ for any r 2 R. Proof. We obtain f .0/ D 0 from the equation f .0/ D f .0 C 0/ D f .0/ C f .0/. Hence 0 D f .0/ D f .r r/ D f .r C .r// D f .r/ C f .r/, that is f .r/ D f .r/.
12
Chapter 1 Groups, Rings and Fields
Definition 1.5.6. Let R and S be rings and let f W R ! S be a ring homomorphism. Then the kernel of f is ker.f / D ¹r 2 R W f .r/ D 0º: The image of f , denoted im.f /, is the range of f within S. That is im.f / D ¹s 2 S W there exists r 2 R with f .r/ D sº: Theorem 1.5.7 (ring isomorphism theorem). Let R and S be rings and let f WR!S be a ring homomorphism. Then (1) ker.f / is an ideal in R, im.f / is a subring of S and R= ker.f / Š im.f /: (2) Conversely suppose that I is an ideal in a ring R. Then the map f W R ! R=I given by f .r/ D r C I for r 2 R is a ring homomorphism whose kernel is I and whose image is R=I . The theorem says that the concepts of ideal of a ring and kernel of a ring homomorphism coincide, that is each ideal is the kernel of a homomorphism and the kernel of each ring homomorphism is an ideal. Proof. Let f W R ! S be a ring homomorphism and let I D ker.f /. We show first that I is an ideal. If r1 ; r2 2 I then f .r1 / D f .r2 / D 0. It follows from the homomorphism property that f .r1 ˙ r2 / D f .r1 / ˙ f .r2 / D 0 C 0 D 0 f .r1 r2 / D f .r1 / f .r2 / D 0 0 D 0: Therefore I is a subring. Now let i 2 I and r 2 R. Then f .r i / D f .r/ f .i / D f .r/ 0 D 0
and
f .i r/ D f .i/ f .r/ D 0 f .r/ D 0
and hence I is an ideal. Consider the factor ring R=I . Let f W R=I ! im.f / by f .r C I / D f .r/. We show that f is an isomorphism. First we show that it is well-defined. Suppose that r1 C I D r2 C I then r1 r2 2 I D ker.f /. It follows that f .r1 r2 / D 0 so f .r1 / D f .r2 /. Hence f .r1 C I / D f .r2 C I / and the map f is well-defined.
13
Section 1.6 Fields of Fractions
Now f ..r1 C I / C .r2 C I // D f ..r1 C r2 / C I / D f .r1 C r2 / D f .r1 / C f .r2 / D f .r1 C I / C f .r2 C I / and f ..r1 C I / .r2 C I // D f ..r1 r2 / C I / D f .r1 r2 / D f .r1 / f .r2 / D f .r1 C I / f .r2 C I /: Hence f is a homomorphism. We must now show that it is injective and surjective. Suppose that f .r1 CI / D f .r2 CI /. Then f .r1 / D f .r2 / so that f .r1 r2 / D 0. Hence r1 r2 2 ker.f / D I . Therefore r1 2 r2 C I and thus r1 C I D r2 C I and the map f is injective. Finally let s 2 im.f /. Then there exists and r 2 R such that f .r/ D s. Then f .r C I / D s and the map f is surjective and hence an isomorphism. This proves the first part of the theorem. To prove the second part let I be an ideal in R and R=I the factor ring. Consider the map f W R ! R=I given by f .r/ D r C I . From the definition of addition and multiplication in the factor ring R=I it is clear that this is a homomorphism. Consider the kernel of f . If r 2 ker.f / then f .r/ D r C I D 0 D 0 C I . This implies that r 2 I and hence the kernel of this map is exactly the ideal I completing the theorem. Theorem 1.5.7 is called the ring isomorphism theorem or the first ring isomorphism theorem. We mention that there is an analogous theorem for each algebraic structure. In particular for groups and vector spaces. We will mention the result for groups in Section 1.8.
1.6
Fields of Fractions
The integers are an integral domain and the rationals Q are a field that contains the integers. First we show that Q is the smallest field containing Z. Theorem 1.6.1. The rationals Q are the smallest field containing the integers Z. That is if Z F Q with F a subfield of Q then F D Q. Proof. Since Z F we have m; n 2 F for any two integers m; n. Since F is a subfield, it is closed under taking division, that is taking multiplicative inverses and hence the fraction m n 2 F . Since each element of Q is such a fraction it follows that Q F . Since F Q it follows that F D Q.
14
Chapter 1 Groups, Rings and Fields
Notice that to construct the rationals from the integers we form all the fractions m2 1 with n ¤ 0 and where m n1 D n2 if m1 n2 D n1 m2 . We then do the standard operations on fractions. If we start with any integral domain D we can mimic this construction to build a field of fractions from D that is the smallest field containing D. m n
Theorem 1.6.2. Let D be an integral domain. Then there is a field F containing D, called the field of fractions for D, such that each element of F is a fraction from D, that is an element of the form d1 d21 with d1 ; d2 2 D. Further F is unique up to isomorphism and is the smallest field containing D. Proof. The proof is just the mimicking of the construction of the rationals from the integers. Let F ? D ¹.d1 ; d2 / W d1 ; d2 ¤ 0; d1 ; d2 2 Dº: Define on F ? the equivalence relation .d1 ; d2 / D .d10 ; d20 / if d1 d20 D d2 d10 : Let F be the set of equivalence classes and define addition and multiplication in the usual manner as for fractions where the result is the equivalence class. .d1 ; d2 / C .d3 ; d4 / D .d1 d4 C d2 d3 ; d2 d4 / .d1 ; d2 / .d3 ; d4 / D .d1 d3 ; d2 d4 /: It is now straightforward to verify the ring axioms for F . The inverse of .d1 ; 1/ is .1; d1 / for d1 ¤ 0 in D. As with Z we identify the elements of F as fractions dd12 . The proof that F is the smallest field containing D is the same as for Q from Z. As examples we have that Q is the field of fractions for Z. A familiar but less common example is the following. Let RŒx be the set of polynomials over the real numbers R. It can be shown that RŒx forms an integral domain. The field of fractions consists of all formal functions f .x/ where f .x/; g.x/ are real polynomials with g.x/ ¤ 0. The corresponding field g.x/ of fractions is called the field of rational functions over R and is denoted R.x/.
1.7
Characteristic and Prime Rings
We saw in the last section that Q is the smallest field containing the integers. Since any subfield of Q must contain the identity, it follows that any nontrivial subfield of Q must contain the integers and hence be all of Q. Therefore Q has no nontrivial subfields. We say that Q is a prime field.
Section 1.7 Characteristic and Prime Rings
15
Definition 1.7.1. A field F is a prime field if F contains no nontrivial subfields. Lemma 1.7.2. Let K be any field. Then K contains a prime field F as a subfield. Proof. Let K1 ; K2 be subfields of K. If k1 ; k2 2 K1 \K2 then k1 ˙k2 2 K1 since K1 is a subfield and k1 ˙ k2 2 K2 since K2 is a subfield. Therefore k1 ˙ k2 2 K1 \ K2 . Similarly k1 k21 2 K1 \ K2 . It follows that K1 \ K2 is again a subfield. Now let F be the intersection of all subfields of K. From the argument above F is a subfield and the only nontrivial subfield of F is itself. Hence F is a prime field. Definition 1.7.3. Let R be a commutative ring with an identity 1 ¤ 0. The smallest positive integer n such that n 1 D 1 C 1 C C 1 D 0 is called the characteristic of R. If there is no such n, then R has characteristic 0. We denote the characteristic by char.R/. Notice first that the characteristic of Z; Q; R are all zero. Further the characteristic of Zn is n. Theorem 1.7.4. Let R be an integral domain. Then the characteristic of R is either 0 or a prime. In particular the characteristic of a field is zero or a prime. Proof. Suppose that R is an integral domain and char.R/ D n ¤ 0. Suppose that n D mk with 1 < m < n, 1 < k < n. Then n 1 D 0 D .m 1/.k 1/. Since R is an integral domain we have no zero divisors and hence m 1 D 0 or k 1 D 0. However this is a contradiction since n is the least positive integer such that n 1 D 0. Therefore n must be a prime. We have seen that every field contains a prime field. We extend this. Definition 1.7.5. A commutative ring R with an identity 1 ¤ 0 is a prime ring if the only subring containing the identity is the whole ring. Clearly both the integers Z and the modular integers Zn are prime rings. In fact up to isomorphism they are the only prime rings. Theorem 1.7.6. Let R be a prime ring. If char.R/ D 0 then R Š Z, while if char.R/ D n > 0 then R Š Zn . Proof. Suppose that char.R/ D 0. Let S D ¹r D m 1 W r 2 R; m 2 Zº. Then S is a subring of R containing the identity (see the exercises) and hence S D R. However the map m 1 ! m gives an isomorphism from S to Z. It follows that R is isomorphic to Z. If char.R/ D n > 0 the proof is identical. Since n 1 D 0 the subring S of R defined above is all of R and isomorphic to Zn .
16
Chapter 1 Groups, Rings and Fields
Theorem 1.7.6 can be extended to fields with Q taking the place of Z and Zp , with p a prime, taking the place of Zn . Theorem 1.7.7. Let K be a prime field. If K has characteristic 0 then K Š Q while if K has characteristic p then K Š Zp . Proof. The proof is identical to that of Theorem 1.7.6; however we consider the smallest subfield K1 of K containing S. We mention that there can be infinite fields of characteristic p. Consider for example the field of fractions of the polynomial ring Zp Œx. This is the field of rational functions with coefficients in Zp . We give a theorem on fields of characteristic p that will be important much later when we look at Galois theory. Theorem 1.7.8. Let K be a field of characteristic p. Then the mapping W K ! K given by .k/ D k p is an injective endomorphism of K. In particular .a C b/p D ap C b p for any a; b 2 K. This mapping is called the Frobenius homomorphism of K. Further if K is finite, is an automorphism. Proof. We first show that is a homomorphism. Now .ab/ D .ab/p D ap b p D .a/.b/: We need a little more work for addition. ! ! p p1 X X p p i pi p p ab ai b pi C b p .a C b/ D .a C b/ D Da C i i i D0
i D1
by the binomial expansion which holds in any commutative ring. However ! p p.p1 / .p i C 1/ D i i .i 1/ 1 p and it is clear that pj i for 1 i p 1. Hence in K we have pi 1 D 0 and so we have .a C b/ D .a C b/p D ap C b p D .a/ C .b/: Therefore is a homomorphism. Further is always injective. To see this suppose that .x/ D .y/. Then .x y/ D 0 H) .x y/p D 0: But K is a field so there are no zero divisors so we must have x y D 0 or x D y. If K is finite and is injective it must also be surjective and hence an automorphism of K.
17
Section 1.8 Groups
1.8
Groups
We close this first chapter by introducing some basic definitions and results from group theory, that mirror the results, that were presented for rings and fields. We will look at group theory in more detail later in the book. Proofs will be given at that point. Definition 1.8.1. A group G is a set with one binary operation (which we will denote by multiplication) such that (1) The operation is associative. (2) There exists an identity for this operation. (3) Each g 2 G has an inverse for this operation. If, in addition, the operation is commutative, the group G is called an abelian group. The order of G is the number of elements in G, denoted by jGj. If jGj < 1; G is a finite group otherwise G is an infinite group. Groups most often arise from invertible mappings of a set onto itself. Such mappings are called permutations. Theorem 1.8.2. The group of all permutations on a set A forms a group called the symmetric group on A which we denote by SA . If A has more than 2 elements then SA is nonabelian. Definition 1.8.3. Let G1 and G2 be groups. Then a mapping f W G1 ! G2 is a (group) homomorphism if f .g1 g2 / D f .g1 /f .g2 /
for any g1 ; g2 2 G1 :
As with rings we have further (1) f is an epimorphism if it is surjective. (2) f is an monomorphism if it is injective. (3) f is an isomorphism if it is bijective, that is both surjective and injective. In this case G1 and G2 are said to be isomorphic groups, which we denote by G1 Š G2 . (4) f is an endomorphism if G1 D G2 , that is a homomorphism from a group to itself. (5) f is an automorphism if G1 D G2 and f is an isomorphism.
18
Chapter 1 Groups, Rings and Fields
Lemma 1.8.4. Let G1 and G2 be groups and let f W G1 ! G2 be a homomorphism. Then (a) f .1/ D 1 where the first 1 is the identity element of G1 and the second is the identity element of G2 . (b) f .g 1 / D .f .g//1 for any g 2 G1 . If A is a set, jAj denotes the size of A. Theorem 1.8.5. If A1 and A2 are sets with jA1 j D jA2 j then SA1 Š SA2 . If jAj D n with n finite we call SA the symmetric group on n elements which we denote by Sn . Further we have jSn j D nŠ. Subgroups are defined in an analogous manner to subrings. Special types of subgroups called normal subgroups take the place in group theory that ideals play in ring theory. Definition 1.8.6. A subset H of a group G is a subgroup if H ¤ ; and H forms a group under the same operation as G. Equivalently, H is a subgroup if H ¤ ; and H is closed under the operation and inverses. Definition 1.8.7. If H is a subgroup of a group G, then a left coset of H is a subset of G of the form gH D ¹gh W h 2 H º. A right coset of H is a subset of G of the form Hg D ¹hg W h 2 H º. As with rings the cosets of a subgroup partition a group. We call the number of right cosets of a subgroup H in a group G then index of H in G, denoted jG W H j. One can prove that the number of right cosets is equal to the number of left cosets. For finite groups we have the following beautiful result called Lagrange’s theorem. Theorem 1.8.8 (Lagrange’s theorem). Let G be a finite group and H a subgroup. Then the order of H divides the order of G. In particular jGj D jH jjG W H j: Normal subgroups take the place of ideals in group theory. Definition 1.8.9. A subgroup H of a group G is a normal subgroup, denoted H C G, if every left coset of H is also a right coset, that is gH D Hg for each g 2 G. Note that this does not say that g and H commute elementwise, just that the subsets gH and Hg are the same. Equivalently H is normal if g1 Hg D H for any g 2 G. Normal subgroups allow us to construct factor groups just as ideals allowed us to construct factor rings.
19
Section 1.9 Exercises
Theorem 1.8.10. Let H be a normal subgroup of a group G. Let G=H be the set of all cosets of H in G, that is G=H D ¹gH W g 2 Gº: We define multiplication on G=H in the following manner .g1 H /.g2 H / D g1 g2 H: Then G=H forms a group called the factor group or quotient group of G modulo H . The identity element of G=H is 1H and the inverse of gH is g 1 H . Further if G is abelian then G=H is also abelian. Finally as with rings normal subgroups, factor groups are closely tied to homomorphisms. Definition 1.8.11. Let G1 and G2 be groups and let f W G1 ! G2 be a homomorphism. Then the kernel of f , denoted ker.f /, is ker.f / D ¹g 2 G1 W f .g/ D 1º: The image of f , denoted im.f /, is the range of f within G2 . That is im.f / D ¹h 2 G2 W there exists g 2 G1 with f .g/ D hº: Theorem 1.8.12 (group isomorphism theorem). Let G1 and G2 be groups and let f W G1 ! G2 be a homomorphism. Then (1) ker.f / is a normal subgroup in G1 . im.f / is a subgroup of G2 and G1 = ker.f / Š im.f /: (2) Conversely suppose that H is a normal subgroup of a group G. Then the map f W G ! G=H given by f .g/ D gH for g 2 G is a homomorphism whose kernel is H and whose image is G=H .
1.9
Exercises
1. Let W K ! R be a homomorphism from a field K to a ring R. Show: Either .a/ D 0 for all a 2 K or is a monomorphism. 2. Let R be a ring and M ¤ ; an arbitrary set. Show that the following are equivalent: (i) The ring of all mappings from M to R is a field. (ii) M contains only one element and R is a field.
20
Chapter 1 Groups, Rings and Fields
3. Let be a set of prime numbers. Define ² ³ a Q D W all prime divisors of b are in : b (i) Show that Q is a subring of Q. (ii) Let R be a subring of Q and let 1 2 R. b
a b
2 R with coprime integers a; b. Show that
(iii) Determine all subrings R of Q. (Hint: Consider the set of all prime divisors of denominators of reduced elements of R.) 4. Prove Lemma 1.5.2. 5. Let R be a commutative ring with an identity 1 2 R. Let A, B and C be ideals in R. A C B WD ¹a C b W a 2 A; b 2 Bº and AB WD .¹ab W a 2 A; b 2 Bº/. Show: (i) A C B G R, A C B D .A [ B/ (ii) AB D ¹a1 b1 C C an bn W n 2 N; ai 2 A; bi 2 Bº, AB A \ B (iii) A.B C C/ D AB C AC , .A C B/C D AB C BC , .AB/C D A.BC / (iv) A D R , A \ R ¤ ; (v) a; b 2 R ) hai C hbi D ¹xa C yb W x; y 2 Rº (vi) a; b 2 R ) haihbi D habi. Here hai D Ra D ¹xa W x 2 Rº. 6. Solve the following congruence: 3x 5 mod 7: Is this congruence also solvable mod 17? 7. Show that the set of 2 2 matrices over a ring R forms a ring. 8. Prove Lemma 1.4.8. 9. Prove that if R is a ring with identity and S D ¹r D m 1 W r 2 R; m 2 Zº then S is a subring of R containing the identity.
Chapter 2
Maximal and Prime Ideals
2.1
Maximal and Prime Ideals
In the first chapter we defined ideals I in a ring R and then the factor ring R=I of R modulo the ideal I . We saw further that if R is commutative then R=I is also commutative and if R has an identity then so does R=I . This raises further questions concerning the structure of factor rings. In particular we can ask under what conditions does R=I form an integral domain and under what conditions does R=I form a field. These questions lead us to define certain special properties of ideals, called prime ideals and maximal ideals. For motivation let us look back at the integers Z. Recall that each proper ideal in Z has the form nZ for some n > 1 and the resulting factor ring Z=nZ is isomorphic to Zn . We proved the following result. Theorem 2.1.1. Zn D Z=nZ is an integral domain if and only if n D p a prime. Further Zn is a field again if and only if n D p is a prime. Hence for the integers Z a factor ring is a field if and only if it is an integral domain. We will see later that this is not true in general. However what is clear is that the special ideals nZ leading to integral domains and fields are precisely when n is a prime. We look at the ideals pZ with p a prime in two different ways and then use these in subsequent sections to give the general definitions. We first need a famous result, Euclid’s lemma, from number theory. For integers a; b the notation ajb means that a divides b. Lemma 2.1.2 (Euclid). If p is a prime and pjab then pja or pjb. Proof. Recall that the greatest common divisor or GCD of two integers a; b is an integer d > 0 such that d is a common divisor of both a and b and if d1 is another common divisor of a and b then d1 jd . We express the GCD of a; b by d D .a; b/. It is known that for any two integers a; b their GCD exists and is unique and further is the least positive linear combination of a and b, that is the least positive integer of the form ax C by for integers x; y. The integers a; b are relatively prime if their GCD is 1, .a; b/ D 1. In this case 1 is a linear combination of a and b. Now suppose pjab where p is a prime. If p does not divide a then since the only positive divisors of p are 1 and p it follows that .a; p/ D 1. Hence 1 is expressible
22
Chapter 2 Maximal and Prime Ideals
as a linear combination of a and p. That is ax C py D 1 for some integers x; y. Multiply through by b, so that abx C pby D b: Now pjab so pjabx and pjpby. Therefore pjabx C pby, that is, pjb. We now recast this lemma in two different ways in terms of the ideal pZ. Notice that pZ consists precisely of all the multiples of p. Hence pjab is equivalent to ab 2 pZ. Lemma 2.1.3. If p is a prime and ab 2 pZ then a 2 pZ or b 2 pZ. This conclusion will be taken as the definition of a prime ideal in the next section. Lemma 2.1.4. If p is a prime and pZ nZ then n D 1 or n D p. That is, every ideal in Z containing pZ with p a prime is either all of Z or pZ. Proof. Suppose that pZ nZ. Then p 2 nZ so p is a multiple of n. Since p is a prime it follows easily that either n D 1 or n D p. In Section 2.3 the conclusion of this lemma will be taken as the definition of a maximal ideal.
2.2
Prime Ideals and Integral Domains
Motivated by Lemma 2.1.3 we make the following general definition for commutative rings R with identity. Definition 2.2.1. Let R be a commutative ring. An ideal P in R with P ¤ R is a prime ideal if whenever ab 2 P with a; b 2 R then either a 2 P or b 2 P . This property of an ideal is precisely what is necessary and sufficient to make the factor ring R=I an integral domain. Theorem 2.2.2. Let R be a commutative ring with an identity 1 ¤ 0 and let P be a nontrivial ideal in R. Then P is a prime ideal if and only if the factor ring R=P is an integral domain. Proof. Let R be a commutative ring with an identity 1 ¤ 0 and let P be a prime ideal. We show that R=P is an integral domain. From the results in the last chapter we have that R=P is again a commutative ring with an identity. Therefore we must show that there are no zero divisors in R=P . Suppose that .a C I /.b C I / D 0 in R=P . The zero element in R=P is 0 C P and hence .a C P /.b C P / D 0 D 0 C P H) ab C P D 0 C P H) ab 2 P:
Section 2.2 Prime Ideals and Integral Domains
23
However P is a prime ideal so we must then have a 2 P or b 2 P . If a 2 P then a C P D P D 0 C P so a C P D 0 in R=P . The identical argument works if b 2 P . Therefore there are no zero divisors in R=P and hence R=P is an integral domain. Conversely suppose that R=P is an integral domain. We must show that P is a prime ideal. Suppose that ab 2 P . Then .a C P /.b C P / D ab C P D 0 C P . Hence in R=P we have .a C P /.b C P / D 0: However R=P is an integral domain so it has no zero divisors. It follows that either a C P D 0 and hence a 2 P or b C P D 0 and b 2 P . Therefore either a 2 P or b 2 P so P is a prime ideal. In a commutative ring R we can define a multiplication of ideals. We then obtain an exact analog of Euclid’s lemma. Since R is commutative each ideal is 2-sided. Definition 2.2.3. Let R be a commutative ring with an identity 1 ¤ 0 and let A and B be ideals in R. Define AB D ¹a1 b1 C C an bn W ai 2 A; bi 2 B; n 2 Nº: That is AB is the set of finite sums of products ab with a 2 A and b 2 B. Lemma 2.2.4. Let R be a commutative ring with an identity 1 ¤ 0 and let A and B be ideals in R. Then AB is an ideal. Proof. We must verify that AB is a subring and that it is closed under multiplication from R. Le r1 ; r2 2 AB. Then r1 D a1 b1 C C an bn and Then
0 0 bm r2 D a10 b10 C C am
for some ai 2 A; bi 2 B for some ai0 2 A; bi0 2 B:
0 0 r1 ˙ r2 D a1 b1 C C an bn ˙ a10 b10 ˙ ˙ am bm
which is clearly in AB. Further 0 0 bm : r1 r2 D a1 b1 a10 b10 C C an bn am
Consider for example the first term a1 b1 a10 b10 . Since R is commutative this is equal to .a1 a10 /.b1 b10 /: Now a1 a10 2 A since A is a subring and b1 b10 2 B since B is a subring. Hence this term is in AB. Similarly for each of the other terms. Therefore r1 r2 2 AB and hence AB is a subring.
24
Chapter 2 Maximal and Prime Ideals
Now let r 2 R and consider rr1 . This is then rr1 D ra1 b1 C C ran bn : Now rai 2 A for each i since A is an ideal. Hence each summand is in AB and then rr1 2 AB. Therefore AB is an ideal. Lemma 2.2.5. Let R be a commutative ring with an identity 1 ¤ 0 and let A and B be ideals in R. If P is a prime ideal in R then AB P implies that A P or B P. Proof. Suppose that AB P with P a prime ideal and suppose that B is not contained in P . We show that A P . Since AB P each product ai bj 2 P . Choose a b 2 B with b … P and let a be an arbitrary element of A. Then ab 2 P . Since P is a prime ideal this implies either a 2 P or b 2 P . But by assumption b … P so a 2 P . Since a was arbitrary we have A P .
2.3
Maximal Ideals and Fields
Now, motivated by Lemma 2.1.4 we define a maximal ideal. Definition 2.3.1. Let R be a ring and I an ideal in R. Then I is a maximal ideal if I ¤ R and if J is an ideal in R with I J then I D J or J D R. If R is a commutative ring with an identity this property of an ideal I is precisely what is necessary and sufficient so that R=I is a field. Theorem 2.3.2. Let R be a commutative ring with an identity 1 ¤ 0 and let I be an ideal in R. Then I is a maximal ideal if and only if the factor ring R=I is a field. Proof. Suppose that R is a commutative ring with an identity 1 ¤ 0 and let I be an ideal in R. Suppose first that I is a maximal ideal and we show that the factor ring R=I is a field. Since R is a commutative ring with an identity the factor ring R=I is also a commutative ring with an identity. We must show then that each nonzero element of R=I has a multiplicative inverse. Suppose then that r D r CI 2 R=I is a nonzero element of R=I . It follows that r … I . Consider the set hr; I i D ¹rx C i W x 2 R; i 2 I º. This is also an ideal (see exercises) called the ideal generated by r and I , denoted hr; I i. Clearly I hr; I i and since r … I and r D r 1 C 0 2 hr; I i it follows that hr; I i ¤ I . Since I is a maximal ideal it follows that hr; I i D R the whole ring. Hence the identity element 1 2 hr; I i and so there exist elements x 2 R and i 2 I such that 1 D rx C i . But then 1 2 .r C I /.x C I / and so 1 C I D .r C I /.x C I /.
Section 2.4 The Existence of Maximal Ideals
25
Since 1 C I is the multiplicative identity of R=I is follows that x C I is the multiplicative inverse of r C I in R=I . Since r C I was an arbitrary nonzero element of R=I it follows that R=I is a field. Now suppose that R=I is a field for an ideal I . We show that I must be maximal. Suppose then that I1 is an ideal with I I1 and I ¤ I1 . We must show that I1 is all of R. Since I ¤ I1 there exists an r 2 I1 with r … I . Therefore the element r C I is nonzero in the factor ring R=I and since R=I is a field it must have a multiplicative inverse x C I . Hence .r C I /.x C I / D rx C I D 1 C I and therefore there is an i 2 I with 1 D rx C i. Since r 2 I1 and I1 is an ideal we get that rx 2 I1 . Further since I I1 it follows that rx C i 2 I1 and so 1 2 I1 . If r1 is an arbitrary element of R then r1 1 D r1 2 I1 . Hence R I1 and so R D I1 . Therefore I is a maximal ideal. Recall that a field is already an integral domain. Combining this with the ideas of prime and maximal ideals we obtain: Theorem 2.3.3. Let R be a commutative ring with an identity 1 ¤ 0. Then each maximal ideal is a prime ideal Proof. Suppose that R is a commutative ring with an identity and I is a maximal ideal in R. Then from Theorem 2.3.2 we have that the factor ring R=I is a field. But a field is an integral domain so R=I is an integral domain. Therefore from Theorem 2.2.2 we have that I must be a prime ideal. The converse is not true in general. That is there are prime ideals that are not maximal. Consider for example R D Z the integers and I D ¹0º. Then I is an ideal and R=I D Z=¹0º Š Z is an integral domain. Hence ¹0º is a prime ideal. However Z is not a field so ¹0º is not maximal. Note however that in the integers Z a proper ideal is maximal if and only if it is a prime ideal.
2.4
The Existence of Maximal Ideals
In this section we prove that in any ring R with an identity there do exist maximal ideals. Further given an ideal I ¤ R then there exists a maximal ideal I0 such that I I0 . To prove this we need three important equivalent results from logic and set theory. First recall that a partial order on a set S is a reflexive, transitive relation on S. That is a a for all a 2 S and if a b; b c then a c. This is a “partial” order since there may exist elements a 2 S where neither a b nor b a. If A is any set then it is clear that containment of subsets is a partial order on the power set P .A/. If is a partial order on a set M , then a chain on M is a subset K M such that a; b 2 K implies that a b or b a. A chain on M is bounded if there exists an
26
Chapter 2 Maximal and Prime Ideals
m 2 M such that k m for all k 2 K. The element m is called an upper bound for K. An element m0 2 M is maximal if whenever m 2 M with m0 m then m D m0 . We now state the three important results from logic. Zorn’s lemma. If each chain of M has an upper bound in M then there is at least one maximal element in M . Axiom of well-ordering. Each set M can be well-ordered, such that each nonempty subset of M contains a least element. i 2 I º be a nonempty collection of nonempty sets. Then Axiom of choice. Let ¹Mi W S there is a mapping f W I ! i 2I Mi with f .i/ 2 Mi for all i 2 I . The following can be proved. Theorem 2.4.1. Zorn’s lemma, the axiom of well-ordering and the axiom of choice are all equivalent. We now show the existence of maximal ideals in commutative rings with identity. Theorem 2.4.2. Let R be a commutative ring with an identity 1 ¤ 0 and let I be an ideal in R with I ¤ R. Then there exists a maximal ideal I0 in R with I I0 . In particular a ring with an identity contains maximal ideals. Proof. Let I be an ideal in the commutative ring R. We must show that there exists a maximal ideal I0 in R with I I0 . Let M D ¹X W X is an ideal with I X ¤ Rº: Then M is partially ordered by containment. We want to show first that each chain in M has a maximal element. If K D ¹Xj W Xj 2 M; j 2 J º is a chain let [ X0 D Xj : j 2J
If a; b 2 X 0 then there exists an i; j 2 J with a 2 Xi ; b 2 Xj . Since K is a chain either Xi Xj or Xj Xi . Without loss of generality suppose that Xi Xj so that a; b 2 Xj . Then a ˙ b 2 Xj X 0 and ab 2 Xj X 0 since Xj is an ideal. Further if r 2 R then ra 2 Xj X 0 since Xj is an ideal. Therefore X 0 is an ideal in R. Since Xj ¤ R it follows that 1 … Xj for all j 2 J . Therefore 1 … X 0 and so 0 ¤ R. It follows that under the partial order of containment X 0 is an upper bound X for K. We now use Zorn’s lemma. From the argument above we have that each chain has a maximal element. Hence for an ideal I the set M above has a maximal element. This maximal element I0 is then a maximal ideal containing I .
Section 2.5 Principal Ideals and Principal Ideal Domains
2.5
27
Principal Ideals and Principal Ideal Domains
Recall again that in the integers Z each ideal I is of the form nZ for some integer n. Hence in Z each ideal can be generated by a single element. Lemma 2.5.1. Let R be a commutative ring and a1 ; : : : ; an be elements of R. Then the set ha1 ; : : : ; an i D ¹r1 a1 C C rn an W ri 2 Rº forms an ideal in R called the ideal generated by a1 ; : : : ; an . Proof. The proof is straightforward. Let a D r1 a1 C C rn an ;
b D s1 a1 C C sn an
with r1 ; : : : ; rn ; s1 ; : : : ; sn elements of R, be two elements of ha1 ; : : : ; an i. Then a ˙ b D .r1 ˙ s1 /a1 C C .rn ˙ sn /an 2 ha1 ; : : : ; an i ab D .r1 s1 a1 /a1 C .r1 s2 a1 /a2 C C .rn sn an /an 2 ha1 ; : : : ; an i so ha1 ; : : : ; an i forms a subring. Further if r 2 R we have ra D .rr1 /a1 C C .rrn /an 2 ha1 ; : : : ; an i and so ha1 ; : : : ; an i is an ideal. Definition 2.5.2. Let R be a commutative ring. An ideal I R is a principal ideal if it has a single generator. That is I D hai D aR
for some a 2 R:
We now restate Theorem 1.4.7 of Chapter 1. Theorem 2.5.3. Every nonzero ideal in Z is a principal ideal. Proof. Every ideal I in Z is of the form nZ. This is the principal ideal generated by n. Definition 2.5.4. A principal ideal domain or PID is an integral domain in which every ideal is principal. Corollary 2.5.5. The integers Z are a principal ideal domain. We mention that the set of polynomials KŒx with coefficients from a field K is also a principal ideal domain. We will return to this in the next chapter. Not every integral domain is a PID. Consider KŒx; y the set of polynomials over K in two variables x; y. Let I consist of all the polynomials with zero constant term.
28
Chapter 2 Maximal and Prime Ideals
Lemma 2.5.6. The set I in KŒx; y as defined above is an ideal but not a principal ideal. Proof. We leave the proof that I forms an ideal to the exercises. To show that it is not a principal ideal suppose I D hp.x; y/i. Now the polynomial q.x/ D x has zero constant term so q.x/ 2 I . Hence p.x; y/ cannot be a constant polynomial. Further if p.x; y/ had any terms with y in them there would be no way to multiply p.x; y/ by a polynomial h.x; y/ and obtain just x. Therefore p.x; y/ can contain no terms with y in them. But the same argument using s.y/ D y shows that p.x; y/ cannot have any terms with x in them. Therefore there can be no such p.x; y/ generating I and so I is not principal and KŒx; y is not a principal ideal domain.
2.6
Exercises
1. Consider the set hr; I i D ¹rx C i W x 2 R; i 2 I º where I is an ideal. Prove that this is also an ideal called the ideal generated by r and I , denoted hr; I i. 2. Let R and S be commutative rings and let W R ! S be a ring epimorphism. Let M be a maximal ideal in R. Show: .M/ is a maximal ideal in S if and only if ker./ M. Is .M/ always a prime ideal of S? 3. Let A1 ; : : : ; A t be ideals of a commutative ring R. Let P be a prime ideal of R. Show: T (i) tiD1 Ai P ) Aj P for at least one index j . T (ii) tiD1 Ai D P ) Aj D P for at least one index j . 4. Which of the following ideals A are prime ideals of R? Which are maximal ideals? (i) A D .x/, R D ZŒx. (ii) A D .x 2 /, R D ZŒx. p p (iii) A D .1 C 5/, R D ZŒ 5. (iv) A D .x; y/, R D QŒx; y. p 5. Let w D 12 .1 C 3/. Show that h2i is a prime ideal and even a maximal ideal of ZŒw, but h2i is neither a prime ideal nor a maximal ideal of ZŒi. 6. Let R D ¹ ab W a; b 2 Z; b oddº. Show that R is a subring of Q and that there is only one maximal ideal M in R. 7. Let R be a ring with an identity. Let x; y 2 R and x ¤ 0 not be a zero divisor. Further let hxi be a prime ideal with hxi hyi ¤ R. Show that hxi D hyi. 8. Consider KŒx; y the set of polynomials over K in two variables x; y. Let I consist of all the polynomials with zero constant term. Prove that the set I is an ideal.
Chapter 3
Prime Elements and Unique Factorization Domains
3.1
The Fundamental Theorem of Arithmetic
The integers Z have served as much of our motivation for properties of integral domains. In the last chapter we saw that Z is a principal ideal domain and furthermore that prime ideals ¤ ¹0º are maximal. From the viewpoint of the multiplicative structure of Z and the viewpoint of classical number theory the most important property of Z is the fundamental theorem of arithmetic. This states that any integer n ¤ 0 is uniquely expressible as a product of primes where uniqueness is up to ordering and the introduction of ˙1, that is units. In this chapter we show that this property is not unique to the integers and there are many other integral domains where this also holds. These are called unique factorization domains and we will present several examples. First we review the fundamental theorem of arithmetic, its proof and several other ideas from classical number theory. Theorem 3.1.1 (fundamental theorem of arithmetic). Given any integer n ¤ 0 there is a factorization n D cp1 p2 pk where c D ˙1 and p1 ; : : : ; pn are primes. Further this factorization is unique up to the ordering of the factors. There are two main ingredients that go into the proof; induction and Euclid’s lemma. We presented this in the last chapter. In turn however Euclid’s lemma depends upon the existence of greatest common divisors and their linear expressibility. Therefore to begin we present several basic ideas from number theory. The starting point for the theory of numbers is divisibility. Definition 3.1.2. If a; b are integers we say that a divides b, or that a is a factor or divisor of b, if there exists an integer q such that b D aq. We denote this by ajb. b is then a multiple of a. If b > 1 is an integer whose only factors are ˙1; ˙b then b is a prime, otherwise b > 1 is composite. The following properties of divisibility are straightforward consequences of the definition.
30
Chapter 3 Prime Elements and Unique Factorization Domains
Lemma 3.1.3. The following properties hold: (1) ajb ) ajbc for any integer c. (2) ajb and bjc implies ajc. (3) ajb and ajc implies that aj.bx C cy/ for any integers x; y. (4) ajb and bja implies that a D ˙b. (5) If ajb and a > 0; b > 0 then a b. (6) ajb if and only if cajcb for any integer c ¤ 0. (7) aj0 for all a 2 Z and 0ja only for a D 0. (8) aj ˙ 1 only for a D ˙1. (9) a1 jb1 and a2 jb2 implies that a1 a2 jb1 b2 . If b; c; x; y are integers then an integer bx C cy is called a linear combination of b; c. Thus part (3) of Lemma 3.1.3 says that if a is a common divisor of b; c then a divides any linear combination of b and c. Further, note that if b > 1 is a composite then there exists x > 0 and y > 0 such that b D xy and from part (5) we must have 1 < x < b, 1 < y < b. In ordinary arithmetic, given a; b we can always attempt to divide a into b. The next result called the division algorithm says that if a > 0 either a will divide b or the remainder of the division of b by a will be less than a. Theorem 3.1.4 (division algorithm). Given integers a; b with a > 0 then there exist unique integers q and r such that b D qa C r where either r D 0 or 0 < r < a. One may think of q and r as the quotient and remainder respectively when dividing b by a. Proof. Given a; b with a > 0 consider the set S D ¹b qa 0 W q 2 Zº: If b > 0 then b C a 0 and the sum is in S. If b 0 then there exists a q > 0 with qa < b. Then b C qa > 0 and is in S . Therefore in either case S is nonempty. Hence S is a nonempty subset of N [ ¹0º and therefore has a least element r. If r ¤ 0 we must show that 0 < r < a. Suppose r a, then r D a C x with x 0 and x < r since a > 0. Then b qa D r D a C x ) b .q C 1/a D x. This means that x 2 S. Since x < r this contradicts the minimality of r which is a contradiction. Therefore if r ¤ 0 it follows that 0 < r < a. The only thing left is to show the uniqueness of q and r. Suppose b D q1 a C r1 also. By the construction above r1 must also be the minimal element of S . Hence r1 r and r r1 so r D r1 . Now b qa D b q1 a H) .q1 q/a D 0 but since a > 0 it follows that q1 q D 0 so that q D q1 .
Section 3.1 The Fundamental Theorem of Arithmetic
31
The next idea that is necessary is the concept of greatest common divisor. Definition 3.1.5. Given nonzero integers a; b their greatest common divisor or GCD d > 0 is a positive integer which is a common divisor, that is d ja and d jb, and if d1 is any other common divisor then d1 jd . We denote the greatest common divisor of a; b by either gcd.a; b/ or .a; b/. Certainly, if a; b are nonzero integers with a > 0 and ajb then a D gcd.a; b/. The next result says that given any nonzero integers they do have a greatest common divisor and it is unique. Theorem 3.1.6. Given nonzero integers a; b their GCD exists, is unique and can be characterized as the least positive linear combination of a and b. Proof. Given nonzero a; b consider the set S D ¹ax C by > 0 W x; y 2 Zº: Now a2 C b 2 > 0 so S is a nonempty subset of N and hence has a least element d > 0. We show that d is the GCD. First we must show that d is a common divisor. Now d D ax C by and is the least such positive linear combination. By the division algorithm a D qd C r with 0 r < d . Suppose r ¤ 0. Then r D a qd D a q.ax C by/ D .1 qx/a qby > 0. Hence r is a positive linear combination of a and b and therefore is in S. But then r < d contradicting the minimality of d in S. It follows that r D 0 and so a D qd and d ja. An identical argument shows that d jb and so d is a common divisor of a and b. Let d1 be any other common divisor of a and b. Then d1 divides any linear combination of a and b and so d1 jd . Therefore d is the GCD of a and b. Finally we must show that d is unique. Suppose d1 is another GCD of a and b. Then d1 > 0 and d1 is a common divisor of a; b. Then d1 jd since d is a GCD. Identically d jd1 since d1 is a GCD. Therefore d D ˙d1 and then d D d1 since they are both positive. If .a; b/ D 1 then we say that a; b are relatively prime. It follows that a and b are relatively prime if and only if 1 is expressible as a linear combination of a and b. We need the following three results. Lemma 3.1.7. If d D .a; b/ then a D a1 d and b D b1 d with .a1 ; b1 / D 1. Proof. If d D .a; b/ then d ja and d jb. Hence a D a1 d and b D b1 d . We have d D ax C by D a1 dx C b1 dy:
32
Chapter 3 Prime Elements and Unique Factorization Domains
Dividing both sides of the equation by d we obtain 1 D a1 x C b1 y: Therefore .a1 ; b1 / D 1. Lemma 3.1.8. For any integer c we have that .a; b/ D .a; b C ac/. Proof. Suppose .a; b/ D d and .a; b C ac/ D d1 . Now d is the least positive linear combination of a and b. Suppose d D axCby. d1 is a linear combination of a; bCac so that d1 D ar C .b C ac/s D a.cs C r/ C bs: Hence d1 is also a linear combination of a and b and therefore d1 d . On the other hand d1 ja and d1 j.b C ac/ and so d1 jb. Therefore d1 jd so d1 d . Combining these we must have d1 D d . The next result, called the Euclidean algorithm, provides a technique for both finding the GCD of two integers and expressing the GCD as a linear combination. Theorem 3.1.9 (Euclidean algorithm). Given integers b and a > 0 with a − b, form the repeated divisions b D q1 a C r1 ;
0 < r1 < a
a D q2 r1 C r2 ;
0 < r2 < r1
:: : rn2 D qn rn1 C rn ;
0 < rn < rn1
rn1 D qnC1 rn : The last nonzero remainder rn is the GCD of a; b. Further rn can be expressed as a linear combination of a and b by successively eliminating the ri ’s in the intermediate equations. Proof. In taking the successive divisions as outlined in the statement of the theorem each remainder ri gets strictly smaller and still nonnegative. Hence it must finally end with a zero remainder. Therefore there is a last nonzero remainder rn . We must show that this is the GCD. Now from Lemma 3.1.7 the gcd .a; b/ D .a; b q1 a/ D .a; r1 / D .r1 ; aq2 r1 / D .r1 ; r2 /. Continuing in this manner we have then that .a; b/ D .rn1 ; rn / D rn since rn divides rn1 . This shows that rn is the GCD. To express rn as a linear combination of a and b notice first that rn D rn2 qn rn1 :
Section 3.1 The Fundamental Theorem of Arithmetic
33
Substituting this in the immediately preceding division we get rn D rn2 qn .rn3 qn1 rn2 / D .1 C qn qn1 /rn2 qn rn3 : Doing this successively we ultimately express rn as a linear combination of a and b.
Example 3.1.10. Find the GCD of 270 and 2412 and express it as a linear combination of 270 and 2412. We apply the Euclidean algorithm 2412 D .8/.270/ C 252 270 D .1/.252/ C 18 252 D .14/.18/: Therefore the last nonzero remainder is 18 which is the GCD. We now must express 18 as a linear combination of 270 and 2412. From the first equation 252 D 2412 .8/.270/ which gives in the second equation 270 D 2412 .8/.270/ C 18 H) 18 D .1/.2412/ C .9/.270/ which is the desired linear combination. The next result that we need is Euclid’s lemma. We stated and proved this in the last chapter but we restate it here. Lemma 3.1.11 (Euclid’s lemma). If p is a prime and pjab then pja or pjb. We can now prove the fundamental theorem of arithmetic. Induction suffices to show that there always exists such a decomposition into prime factors. Lemma 3.1.12. Any integer n > 1 can be expressed as a product of primes, perhaps with only one factor. Proof. The proof is by induction. n D 2 is prime so its true at the lowest level. Suppose that any integer 2 k < n can be decomposed into prime factors, we must show that n then also has a prime factorization. If n is prime then we are done. Suppose then that n is composite. Hence n D m1 m2 with 1 < m1 < n, 1 < m2 < n. By the inductive hypothesis both m1 and m2 can be expressed as products of primes. Therefore n can, also using the primes from m1 and m2 , completing the proof.
34
Chapter 3 Prime Elements and Unique Factorization Domains
Before we continue to the fundamental theorem we mention that the existence of a prime decomposition, unique or otherwise, can be used to prove that the set of primes is infinite. The proof we give goes back to Euclid and is quite straightforward. Theorem 3.1.13. There are infinitely many primes. Proof. Suppose that there are only finitely many primes p1 ; : : : ; pn . Each of these is positive so we can form the positive integer N D p1 p2 pn C 1: From Lemma 3.1.12 N has a prime decomposition. In particular there is a prime p which divides N . Then pj.p1 p2 pn C 1/: Since the only primes are assumed p1 ; p2 ; : : : ; pn it follows that p D pi for some i D 1; : : : ; n. But then pjp1 p2 pi : : : pn c so p cannot divide p1 pn C 1 which is a contradiction. Therefore p is not one of the given primes showing that the list of primes must be endless. We can now prove the fundamental theorem of arithmetic. Proof. We assume that n 1. If n 1 we use c D n and the proof is the same. The statement certainly holds for n D 1 with k D 0. Now suppose n > 1. From Lemma 3.1.12, n has a prime decomposition n D p1 p2 pm : We must show that this is unique up to the ordering of the factors. Suppose then that n has another such factorization n D q1 q2 qk with the qi all prime. We must show that m D k and that the primes are the same. Now we have n D p1 p2 pm D q1 qk : Assume that k m. From n D p1 p2 pm D q1 qk it follows that p1 jq1 q2 qk . From Lemma 3.1.11 then we must have that p1 jqi for some i. But qi is prime and p1 > 1 so it follows that p1 D qi . Therefore we can eliminate p1 and qi from both sides of the factorization to obtain p2 pm D q1 qi 1 qi C1 qk : Continuing in this manner we can eliminate all the pi from the left side of the factorization to obtain 1 D qmC1 qk : If qmC1 ; : : : ; qk were primes this would be impossible. Therefore m D k and each prime pi was included in the primes q1 ; : : : ; qm . Therefore the factorizations differ only in the order of the factors, proving the theorem.
Section 3.2 Prime Elements, Units and Irreducibles
3.2
35
Prime Elements, Units and Irreducibles
We now let R be an arbitrary integral domain and attempt to mimic the divisibility definitions and properties. Definition 3.2.1. Let R be an integral domain. (1) Suppose that a; b 2 R. Then a is a factor or divisor of b if there exists a c 2 R with b D ac. We denote this, as in the integers, by ajb. If a is a factor of b then b is called a multiple of a. (2) An element a 2 R is a unit if a has a multiplicative inverse within R, that is there exists an element a1 2 R with aa1 D 1. (3) A prime element of R is an element p ¤ 0 such that p is not a unit and if pjab then pja or pjb. (4) An irreducible in R is an element c ¤ 0 such that c is not a unit and if c D ab then a or b must be a unit. (5) a and b in R are associates if there exists a unit e 2 R with a D eb. Notice that in the integers Z the units are just ˙1. The set of prime elements coincides with the set of irreducible elements. In Z this are precisely the set of prime numbers. On the other hand if K is a field every nonzero element is a unit so in K there are no prime elements and no irreducible elements. Recall that the modular rings Zn are fields (and integral domains) when n is a prime. In general if n is not a prime then Zn is a commutative ring with an identity and a unit is still an invertible element. We can characterize the units within Zn . Lemma 3.2.2. a 2 Zn is a unit if and only if .a; n/ D 1. Proof. Suppose .a; n/ D 1. Then there exist x; y 2 Z such that ax C ny D 1. This implies that ax 1 mod n which in turn implies that ax D 1 in Zn and therefore a is a unit. Conversely suppose a is a unit in Zn . Then there is an x 2 Zn with ax D 1. In terms of congruence then ax 1
mod n H) njax 1 H) ax 1 D ny H) ax ny D 1:
Therefore 1 is a linear combination of a and n and so .a; n/ D 1. If R is an integral domain then the set of units within R will form a group. Lemma 3.2.3. If R is a commutative ring with an identity then the set of units in R form an abelian group under ring multiplication. This is called the unit group of R denoted U.R/.
36
Chapter 3 Prime Elements and Unique Factorization Domains
Proof. The commutativity and associativity of U.R/ follow from the ring properties. The identity of U.R/ is the multiplicative identity of R while the ring multiplicative inverse for each unit is the group inverse. We must show that U.R/ is closed under ring multiplication. If a 2 R is a unit we denote its multiplicative inverse by a1 . Now suppose a; b 2 U.R/. Then a1 ; b 1 exist. It follows that .ab/.b 1 a1 / D a.bb 1 /a1 D aa1 D 1: Hence ab has an inverse, namely b 1 a1 (D a1 b 1 in a commutative ring) and hence ab is also a unit. Therefore U.R/ is closed under ring multiplication. In general irreducible elements are not prime. Consider for example the subring of the complex numbers (see exercises) given by p p R D ZŒi 5 D ¹x C i 5y W x; y 2 Zº: This is a subring of the complex numbers C and hence can have no zero divisors. Therefore R is an integralpdomain. For an element x C iy 5 2 R define its norm by p p N.x C iy 5/ D jx C iy 5j D x 2 C 5y 2 : Since x; y 2 Z it is clear that the norm of an element in R is a nonnegative integer. Further if a 2 R with N.a/ D 0 then a D 0. We have the following result concerning the norm. Lemma 3.2.4. Let R and N be as above. Then (1) N.ab/ D N.a/N.b/ for any elements a; b 2 R. (2) The units of R are those a 2 R with N.a/ D 1. In R the only units are ˙1. Proof. The fact that the norm is multiplicative is straightforward and left to the exercises. If a 2 R is a unit then there exists a multiplicative inverse b 2 R with ab D 1. Then N.ab/ D N.a/N.b/ D 1. Since both N.a/ and N.b/ are nonnegative integers we must have N.a/ D N.b/ D 1. p Conversely suppose that N.a/ D 1. If a D x C iy 5 then x 2 C 5y 2 D 1. Since x; y 2 Z we must have y D 0 and x 2 D 1. Then a D x D ˙1. Using this lemma we can show that R possesses irreducible elements that are not prime. p Lemma 3.2.5. Let R be as above. Then 3 D 3 C i 0 5 is an irreducible element in R but 3 is not prime.
Section 3.2 Prime Elements, Units and Irreducibles
37
Proof. Suppose that 3 D ab with a; b 2 R and a; b nonunits. Then N.3/ D 9 D N.a/N.b/ with p neither N.a/ D 1 nor N.b/ D 1. Hence both N.a/ D 3 and N.b/ D 3. Let a D x C iy 5. It follows that x 2 C 5y 2 D 3. Since x; y 2 Z this is impossible. Therefore one of a or b must be a unit and 3 is an irreducible element. p p We show that 3 is not prime in R. Let a D 2 C i 5 and b D 2 i 5. Then ab D 9 and hence 3jab. Suppose 3ja so that a D 3c for some c 2 R. Then 9 D N.a/ D N.3/N.c/ D 9N.c/ H) N.c/ D 1: Therefore c is a unit in R and from Lemma 3.2.4 we get c D ˙1. Hence a D ˙3. This is a contradiction, so 3 does not divide a. An identical argument shows that 3 does not divide b. Therefore 3 is not a prime element in R. We now examine the relationship between prime elements and irreducibles. Theorem 3.2.6. Let R be an integral domain. Then (1) Each prime element of R is irreducible. (2) p 2 R is a prime element if and only if p ¤ 0 and hpi D pR is a prime ideal. (3) p 2 R is irreducible if and only if p ¤ 0 and hpi D pR is maximal in the set of all principal ideals of R which are not equal to R. Proof. (1) Suppose that p 2 R is a prime element and p D ab. We must show that either a or b must be a unit. Now pjab so either pja or pjb. Without loss of generality we may assume that pja, so a D pr for some r 2 R. Hence p D ab D .pr/b D p.rb/. However R is an integral domain so p prb D p.1 rb/ D 0 implies that 1 rb D 0 and hence rb D 1. Therefore b is a unit and hence p is irreducible. (2) Suppose that p is a prime element. Then p ¤ 0. Consider the ideal pR and suppose that ab 2 pR. Then ab is a multiple of p and hence pjab. Since p is prime it follows that pja or pjb. If pja then a 2 pR while if pjb then b 2 pR. Therefore pR is a prime ideal. Conversely suppose that pR is a prime ideal and suppose that p D ab. Then ab 2 pR so a 2 pR or b 2 pR. If a 2 pR then pja and if b 2 pR then pjb and so p is prime. (3) Let p be irreducible then p ¤ 0. Suppose that pR aR where a 2 R. Then p D ra for some r 2 R. Since p is irreducible it follows that either a is a unit or r is a unit. If r is a unit we have pR D raR D aR ¤ R since p is not a unit. If a is a unit then aR D R and pR D rR ¤ R. Therefore pR is maximal in the set of principal ideals not equal to R. Conversely suppose p ¤ 0 and pR is a maximal ideal in the set of principal ideals ¤ R. Let p D ab with a not a unit. We must show that b is a unit. Since aR ¤ R and pR aR from the maximality we must have pR D aR. Hence a D rp for some r 2 R. Then p D ab D rpb and as before we must have rb D 1 and b a unit.
38
Chapter 3 Prime Elements and Unique Factorization Domains
Theorem 3.2.7. Let R be a principle ideal domain. Then: (1) An element p 2 R is irreducible if and only if it is a prime element. (2) A nonzero ideal of R is a maximal ideal if and only if it is a prime ideal. (3) The maximal ideals of R are precisely those ideals pR where p is a prime element. Proof. First note that ¹0º is a prime ideal but not maximal. (1) We already know that prime elements are irreducible. To show the converse suppose that p is irreducible. Since R is a principal ideal domain from Theorem 3.2.6 we have that pR is a maximal ideal, and each maximal ideal is also a prime ideal. Therefore from Theorem 3.2.6 we have that p is a prime element. (2) We already know that each maximal ideal is a prime ideal. To show the converse suppose that I ¤ ¹0º is a prime ideal. Then I D pR where p is a prime element with p ¤ 0. Therefore p is irreducible from part (1) and hence pR is a maximal ideal from Theorem 3.2.6. (3) This follows directly from the proof in part (2) and Theorem 3.2.6.
3.3
Unique Factorization Domains
We now consider integral domains where there is unique factorization into primes. If R is an integral domain and a; b 2 R then we say that a and b are associates if there exists a unit 2 R with a D b. Definition 3.3.1. An integral domain D is a unique factorization domain or UFD if for each d 2 D then either d D 0, d is a unit or d has a factorization into primes which is unique up to ordering and unit factors. This means that if r D p1 pm D q1 qk then m D k and each pi is an associate of some qj . There are several relationships in integral domains that are equivalent to unique factorization. Definition 3.3.2. Let R be an integral domain. (1) R has property (A) if and only if for each nonunit a ¤ 0 there are irreducible elements q1 ; : : : ; qr 2 R satisfying a D q1 qr . (2) R has property (A0 ) if and only if for each nonunit a ¤ 0 there are prime elements p1 ; : : : ; pr 2 R satisfying a D p1 pr .
Section 3.3 Unique Factorization Domains
39
(3) R has property (B) if and only if whenever q1 ; : : : ; qr and q10 ; : : : ; qs0 are irreducible elements of R with q1 qr D q10 qs0 then r D s and there is a permutation 2 Sr such that for each i 2 ¹1; : : : ; rº 0 are associates (uniqueness up to ordering and unit the elements qi and q.i/ factors). (4) R has property (C) if and only if each irreducible element of R is a prime element. Notice that properties (A) and (C) together are equivalent to what we defined as unique factorization. Hence an integral domain satisfying (A) and (C) is a UFD. We show next, that there are other equivalent formulations. Theorem 3.3.3. In an integral domain R the following are equivalent: (1) R is a UFD. (2) R satisfies properties (A) and (B). (3) R satisfies properties (A) and (C). (4) R satisfies property (A0 ). Proof. As remarked before the statement of the theorem by definition (A) and (C) are equivalent to unique factorization. We show here that (2), (3) and (4) are equivalent. First we show that (2) implies (3). Suppose that R satisfies properties (A) and (B). We must show that it also satisfies (C), that is we must show that if q 2 R is irreducible then q is prime. Suppose that q 2 R is irreducible and qjab with a; b 2 R. Then we have ab D cq for some c 2 R. If a is a unit from ab D cq we get that b D a1 cq and qjb. Identically if b is a unit. Therefore we may assume that neither a nor b are units. If c D 0 then since R is an integral domain either a D 0 or b D 0 and qja or qjb. We may assume then that c ¤ 0. If c is a unit then q D c 1 ab and since q is irreducible either c 1 a or b are units. If c 1 a is a unit then a is also a unit so if c is a unit either a or b are units contrary to our assumption. Therefore we may assume that c ¤ 0 and c is not a unit. From property (A) we have a D q1 q r b D q10 qs0 c D q100 q t00
40
Chapter 3 Prime Elements and Unique Factorization Domains
where q1 ; : : : qr ; q10 ; : : : ; qs0 ; q100 ; : : : q t00 are all irreducibles. Hence q1 qr q10 qs0 D q100 q t00 q: From property (B) q is an associate of some qi or qj0 . Hence qjqi or qjqj0 . It follows that qja or qjb and therefore q is a prime element. That (3) implies (4) is direct. We show that (4) implies (2). Suppose that R satisfies property (A0 ). We must show that it satisfies both (A) and (B). We show first that (A) follows from (A0 ) by showing that irreducible elements are prime. Suppose that q is irreducible. Then from (A0 ) we have q D p1 pr with each pi prime. It follows without loss of generality that p2 pr is a unit and p1 is a nonunit and hence pi j1 for i D 2; : : : ; r. Thus q D p1 and q is prime. Therefore (A) holds. We now show that (B) holds. Let q1 qr D q10 qs0 where qi ; qj0 are all irreducibles and hence primes. Then q10 jq1 qr and so q10 jqi for some i . Without loss of generality suppose q10 jq1 . Then q1 D aq10 . Since q1 is irreducible it follows that a is a unit and q1 and q10 are associates. It follows then that aq2 qr D q20 qs0 since R has no zero divisors. Property (B) holds then by induction and the theorem is proved. Note that in our new terminology Z is a UFD. In the next section we will present other examples of UFD’s however not every integral domain is a unique factorization domain. As we defined in the last section let R be the following subring of C. p p R D ZŒi 5 D ¹x C i 5y W x; y 2 Zº: R is an integral domain and we showed using the norm 3 is an irreducible in R. p that p are also irreducibles Analogously we can show that the elements 2 C i p5; 2 i 5p in R and further 3 is not an associate of either 2 C i 5 or 2 i 5. Then p p 9 D 3 3 D .2 C i 5/.2 i 5/
41
Section 3.4 Principal Ideal Domains and Unique Factorization
give two different decompositions for an element in terms of irreducible elements. The fact that R is not a UFD also follows from the fact that 3 is an irreducible element which is not prime. Notice also that Euclid’s proof, that there are infinitely many primes in Z, works in any UFD. Theorem 3.3.4. Let R be a UFD. Then R has infinitely many prime elements. Unique factorization is tied to the famous solution of Fermat’s big theorem. Wiles and Taylor in 1995 proved the following. Theorem 3.3.5. The equation x p C y p D z p has no integral solutions with xyz ¤ 0 for any prime p 3. Kummer tried to prove this theorem by attempting to factor x p D z p y p . Call the 2 i conclusion of Theorem 3.3.4 in an integral domain R property .Fp /. Let D e p . Then p1 Y p p .z j y/: z y D j D0
View this equation in the ring R D ZΠD
² p1 X
³ j
aj W aj 2 Z :
j D0
Kummer proved that if R is a UFD then property .Fp / holds. However, independently, from Uchida and Montgomery (1971) R is a UFD only if p 19 (see [41]).
3.4
Principal Ideal Domains and Unique Factorization
In this section we prove that every principal ideal domain (PID) is a unique factorization domain (UFD). We say that an ascending chain of ideals in R I1 I2 In becomes stationary if there exists an m such that Ir D Im for all r m. Theorem 3.4.1. Let R be an integral domain. If each ascending chain of principal ideals in R becomes stationary, then R satisfies property (A). Proof. Suppose that a ¤ 0 is a not a unit in R. Suppose that a is not a product of irreducible elements. Clearly then a cannot itself be irreducible. Hence a D a1 b1 with a1 ; b1 2 R and a1 ; b1 are not units. If both a1 or b1 can be expressed as a product
42
Chapter 3 Prime Elements and Unique Factorization Domains
of irreducible elements then so can a. Without loss of generality then suppose that a1 is not a product of irreducible elements. Since a1 ja we have the inclusion of ideals aR a1 R. If a1 R D aR then a1 2 aR and a1 D ar D a1 b1 r which implies that b1 is a unit contrary to our assumption. Therefore aR ¤ a1 R and the inclusion is proper. By iteration then we obtain a strictly increasing chain of ideals aR a1 R an R : From our hypothesis on R this must become stationary contradicting the argument above that the inclusion is proper. Therefore a must be a product of irreducibles. Theorem 3.4.2. Each principal ideal domain R is a unique factorization domain. Proof. Suppose that R is a principal ideal domain. R satisfies property (C) by Theorem 3.2.7(1), so to show that it is a unique factorization domain we must show that it also satisfies property (A). From the previous theorem it suffices to show that each ascending chain of principal ideals becomes stationary. Consider such an ascending chain a1 R a2 R an R : Now let I D
1 [
ai R:
i D1
Now I is an ideal in R and hence a principal ideal. Therefore I D aR for some a 2 R. Since I is a union there exists an m such that a 2 am R. Therefore I D aR am R and hence I D am R and ai R am R for all i m. Therefore the chain becomes stationary and from Theorem 3.4.1 R satisfies property (A). Since we showed that the integers Z are a PID we can recover the fundamental theorem of arithmetic from Theorem 3.4.2. We now present another important example of a PID and hence a UFD. In the next chapter we will look in detail at polynomials with coefficients in an integral domain. Below we consider polynomials with coefficients in a field and for the present leave out many of the details. If F is a field and n is a nonnegative integer, then a polynomial of degree n over F is a formal sum of the form P .x/ D a0 C a1 x C C an x n with ai 2 F for i D 0; : : : ; n, an ¤ 0, and x an indeterminate. A polynomial P .x/ over F is either a polynomial of some degree or the expression P .x/ D 0, which is called the zero polynomial and has degree 1. We denote the degree of P .x/ by deg P .x/. A polynomial of zero degree has the form P .x/ D a0 and is
Section 3.4 Principal Ideal Domains and Unique Factorization
43
called a constant polynomial and can be identified with the corresponding element of F . The elements ai 2 F are called the coefficients of P .x/; an is the leading coefficient. If an D 1, P .x/ is called a monic polynomial. Two nonzero polynomials are equal if and only if they have the same degree and exactly the same coefficients. A polynomial of degree 1 is called a linear polynomial while one of degree two is a quadratic polynomial. We denote by F Œx the set of all polynomials over F and we will show that F Œx becomes a principal ideal domain and hence a unique factorization domain. We first define addition, subtraction, and multiplication on F Œx by algebraic manipulation. That is, suppose P .x/ D a0 C a1 x C C an x n ; Q.x/ D b0 C b1 x C C bm x m then P .x/ ˙ Q.x/ D .a0 ˙ b0 / C .a1 ˙ b1 /x C ; that is, the coefficient of x i in P .x/ ˙ Q.x/ is ai ˙ bi , where ai D 0 for i > n and bj D 0 for j > m. Multiplication is given by P .x/Q.x/ D .a0 b0 /C.a1 b0 Ca0 b1 /xC.a0 b2 Ca1 b1 Ca2 b0 /x 2 C C.an bm /x nCm ; that is, the coefficient of x i in P .x/Q.x/ is .a0 bi C a1 bi 1 C C ai b0 /. Example 3.4.3. Let P .x/ D 3x 2 C 4x 6 and Q.x/ D 2x C 7 be in QŒx. Then P .x/ C Q.x/ D 3x 2 C 6x C 1 and P .x/Q.x/ D .3x 2 C 4x 6/.2x C 7/ D 6x 3 C 29x 2 C 16x 42: From the definitions the following degree relationships are clear. The proofs are in the exercises. Lemma 3.4.4. Let 0 ¤ P .x/; 0 ¤ Q.x/ in F Œx. Then: (1) deg P .x/Q.x/ D deg P .x/ C deg Q.x/. (2) deg.P .x/ ˙ Q.x// Max.deg P .x/; deg Q.x// if P .x/ ˙ Q.x/ ¤ 0. We next obtain the following. Theorem 3.4.5. If F is a field, then F Œx forms an integral domain. F can be naturally embedded into F Œx by identifying each element of F with the corresponding constant polynomial. The only units in F Œx are the nonzero elements of F . Proof. Verification of the basic ring properties is solely computational and is left to the exercises. Since deg P .x/Q.x/ D deg P .x/ C deg Q.x/, it follows that if neither P .x/ ¤ 0 nor Q.x/ ¤ 0 then P .x/Q.x/ ¤ 0 and therefore F Œx is an integral domain.
44
Chapter 3 Prime Elements and Unique Factorization Domains
If G.x/ is a unit in F Œx, then there exists an H.x/ 2 F Œx with G.x/H.x/ D 1. From the degrees we have deg G.x/ C deg H.x/ D 0 and since deg G.x/ 0, deg H.x/ 0. This is possible only if deg G.x/ D deg H.x/ D 0. Therefore G.x/ 2 F . Now that we have F Œx as an integral domain we proceed to show that F Œx is a principal ideal domain and hence there is unique factorization into primes. We first repeat the definition of a prime in F Œx. If 0 ¤ f .x/ has no nontrivial, nonunit factors (it cannot be factorized into polynomials of lower degree) then f .x/ is a prime in F Œx or a prime polynomial. A prime polynomial is also called an irreducible polynomial. Clearly, if deg g.x/ D 1 then g.x/ is irreducible. The fact that F Œx is a principal ideal domain follows from the division algorithm for polynomials, which is entirely analogous to the division algorithm for integers. Lemma 3.4.6 (division algorithm in F Œx). If 0 ¤ f .x/, 0 ¤ g.x/ 2 F Œx then there exist unique polynomials q.x/; r.x/ 2 F Œx such that f .x/ D q.x/g.x/ C r.x/ where r.x/ D 0 or deg r.x/ < deg g.x/. (The polynomials q.x/ and r.x/ are called respectively the quotient and remainder.) This theorem is essentially long division of polynomials. A formal proof is based on induction on the degree of g.x/. We omit this but give some examples from QŒx. Example 3.4.7.
(a) Let f .x/ D 3x 4 6x 2 C 8x 6, g.x/ D 2x 2 C 4. Then
3x 4 6x 2 C 8x 6 3 D x 2 6 with remainder 8x C 18: 2 2x C 4 2 Thus here q.x/ D 32 x 2 6, r.x/ D 8x C 18. (b) Let f .x/ D 2x 5 C 2x 4 C 6x 3 C 10x 2 C 4x, g.x/ D x 2 C x. Then 2x 5 C 2x 4 C 6x 3 C 10x 2 C 4x D 2x 3 C 6x C 4: x2 C x Thus here q.x/ D 2x 3 C 6x C 4 and r.x/ D 0. Theorem 3.4.8. Let F be a field. Then the polynomial ring F Œx is a principal ideal domain and hence a unique factorization domain. Proof. The proof is essentially analogous to the proof in the integers. Let I be an ideal in F Œx with I ¤ F Œx. Let f .x/ be a polynomial in I of minimal degree. We claim that I D hf .x/i the principal ideal generated by f .x/. Let g.x/ 2 I . We must show that g.x/ is a multiple of f .x/. By the division algorithm in F Œx we have g.x/ D q.x/f .x/ C r.x/
45
Section 3.5 Euclidean Domains
where r.x/ D 0 or deg.r.x// < deg.f .x//. If r.x/ ¤ 0 then deg.r.x// < deg.f .x//. However r.x/ D g.x/ q.x/f .x/ 2 I since I is an ideal and g.x/; f .x/ 2 I . This is a contradiction since f .x/ was assumed to be a polynomial in I of minimal degree. Therefore r.x/ D 0 and hence g.x/ D q.x/f .x/ is a multiple of f .x/. Therefore each element of I is a multiple of f .x/ and hence I D hf .x/i. Therefore F Œx is a principal ideal domain and from Theorem 3.4.2 a unique factorization domain. We proved that in a principal ideal domain every ascending chain of ideals becomes stationary. In general a ring R (commutative or not) satisfies the ascending chain condition or ACC if every ascending chain of left (or right) ideals in R becomes stationary. A ring satisfying the ACC is called a Noetherian ring.
3.5
Euclidean Domains
In analyzing the proof of unique factorization in both Z and F Œx, it is clear that it depends primarily on the division algorithm. In Z the division algorithm depended on the fact that the positive integers could be ordered and in F Œx on the fact that the degrees of nonzero polynomials are nonnegative integers and hence could be ordered. This basic idea can be generalized in the following way. Definition 3.5.1. An integral domain D is a Euclidean domain if there exists a function N from D ? D D n ¹0º to the nonnegative integers such that (1) N.r1 / N.r1 r2 / for any r1 ; r2 2 D ? . (2) For all r1 ; r2 2 D with r1 ¤ 0 there exist q; r 2 D such that r2 D qr1 C r where either r D 0 or N.r/ < N.r1 /. The function N is called a Euclidean norm on D. Therefore Euclidean domains are precisely those integral domains which allow division algorithms. In the integers Z define N.z/ D jzj. Then N is a Euclidean norm on Z and hence Z is a Euclidean domain. On F Œx define N.p.x// D deg.p.x// if p.x/ ¤ 0. Then N is also a Euclidean norm on F Œx so that F Œx is also a Euclidean domain. In any Euclidean domain we can mimic the proofs of unique factorization in both Z and F Œx to obtain the following: Theorem 3.5.2. Every Euclidean domain is a principal ideal domain and hence a unique factorization domain. Before proving this theorem we must develop some results on the number theory of general Euclidean domains. First some properties of the norm.
46
Chapter 3 Prime Elements and Unique Factorization Domains
Lemma 3.5.3. If R is a Euclidean domain then (a) N.1/ is minimal among ¹N.r/ W r 2 R? º. (b) N.u/ D N.1/ if and only if u is a unit. (c) N.a/ D N.b/ for a; b 2 R? if a; b are associates. (d) N.a/ < N.ab/ unless b is a unit. Proof. (a) From property (1) of Euclidean norms we have N.1/ N.1 r/ D N.r/
for any r 2 R? :
(b) Suppose u is a unit. Then there exists u1 with u u1 D 1. Then N.u/ N.u u1 / D N.1/: From the minimality of N.1/ it follows that N.u/ D N.1/. Conversely suppose N.u/ D N.1/. Apply the division algorithm to get 1 D qu C r: If r ¤ 0 then N.r/ < N.u/ D N.1/ contradicting the minimality of N.1/. Therefore r D 0 and 1 D qu. Then u has a multiplicative inverse and hence is a unit. (c) Suppose a; b 2 R? are associates. Then a D ub with u a unit. Then N.b/ N.ub/ D N.a/: On the other hand b D u1 a so N.a/ N.u1 a/ D N.b/: Since N.a/ N.b/ and N.b/ N.a/ it follows that N.a/ D N.b/. (d) Suppose N.a/ D N.ab/. Apply the division algorithm a D q.ab/ C r where r D 0 or N.r/ < N.ab/. If r ¤ 0 then r D a qab D a.1 qb/ H) N.ab/ D N.a/ N.a.1 qb// D N.r/ contradicting that N.r/ < N.ab/. Hence r D 0 and a D q.ab/ D .qb/a. Then a D .qb/a D 1 a H) qb D 1 since there are no zero divisors in an integral domain. Hence b is a unit. Since N.a/ N.ab/ it follows that if b is not a unit we must have N.a/ < N.ab/.
47
Section 3.5 Euclidean Domains
We can now prove Theorem 3.5.2. Proof. Let D be a Euclidean domain. We show that each ideal I ¤ D in D is principal. Let I ¤ D be an ideal in D. If I D ¹0º then I D h0i and I is principal. Therefore we may assume that there are nonzero elements in I . Hence there are elements x 2 I with strictly positive norm. Let a be an element of I of minimal norm. We claim that I D hai. Let b 2 I . We must show that b is a multiple of a. Now by the division algorithm b D qa C r where either r D 0 or N.r/ < N.a/. As in Z and F Œx we have a contradiction if r ¤ 0. In this case N.r/ < N.a/ but r D b qa 2 I since I is an ideal contradicting the minimality of N.a/. Therefore r D 0 and b D qa and hence I D hai. As a final example of a Euclidean domain we consider the Gaussian integers ZŒi D ¹a C bi W a; b 2 Zº: It was first observed by Gauss that this set permits unique factorization. To show this we need a Euclidean norm on ZŒi. Definition 3.5.4. If z D a C bi 2 ZŒi then its norm N.z/ is defined by N.a C bi / D a2 C b 2 : The basic properties of this norm follow directly from the definition (see exercises). Lemma 3.5.5. If ˛; ˇ 2 ZŒi then: (1) N.˛/ is an integer for all ˛ 2 ZŒi. (2) N.˛/ 0 for all ˛ 2 ZŒi. (3) N.˛/ D 0 if and only if ˛ D 0. (4) N.˛/ 1 for all ˛ ¤ 0. (5) N.˛ˇ/ D N.˛/N.ˇ/, that is the norm is multiplicative. From the multiplicativity of the norm we have the following concerning primes and units in ZŒi. Lemma 3.5.6.
(1) u 2 ZŒi is a unit if and only if N.u/ D 1.
(2) If 2 ZŒi and N./ D p where p is an ordinary prime in Z then is a prime in ZŒi.
48
Chapter 3 Prime Elements and Unique Factorization Domains
Proof. Certainly u is a unit if and only if N.u/ D N.1/. But in ZŒi we have N.1/ D 1 so the first part follows. Suppose next that 2 ZŒi with N./ D p for some p 2 Z. Suppose that D 1 2 . From the multiplicativity of the norm we have N./ D p D N.1 /N.2 /: Since each norm is a positive ordinary integer and p is a prime it follows that either N.1 / D 1 or N.2 / D 1. Hence either 1 or 2 is a unit. Therefore is a prime in ZŒi. Armed with this norm we can show that ZŒi is a Euclidean domain. Theorem 3.5.7. The Gaussian integers ZŒi form a Euclidean domain. Proof. That ZŒi forms a commutative ring with an identity can be verified directly and easily. If ˛ˇ D 0 then N.˛/N.ˇ/ D 0 and since there are no zero divisors in Z we must have N.˛/ D 0 or N.ˇ/ D 0. But then either ˛ D 0 or ˇ D 0 and hence ZŒi is an integral domain. To complete the proof we show that the norm N is a Euclidean norm. From the multiplicativity of the norm we have if ˛; ˇ ¤ 0 N.˛ˇ/ D N.˛/N.ˇ/ N.˛/
since N.ˇ/ 1:
Therefore property (1) of Euclidean norms is satisfied. We must now show that the division algorithm holds. Let ˛ D a C bi and ˇ D c C d i be Gaussian integers. Recall that for a nonzero complex number z D x C iy its inverse is z x iy 1 : D 2 D 2 z jzj x C y2 Therefore as a complex number ˇ c di ˛ D ˛ 2 D .a C bi / 2 ˇ jˇj c C d2 ac C bd ac bd D 2 C 2 i D u C iv: 2 c Cd c C d2 Now since a; b; c; d are integers u; v must be rationals. The set ¹u C iv W u; v 2 Qº is called the set of the Gaussian rationals.
Section 3.5 Euclidean Domains
49
If u; v 2 Z then u C iv 2 ZŒi, ˛ D qˇ with q D u C iv, and we are done. Otherwise choose ordinary integers m; n satisfying ju mj 12 and jv nj 12 and let q D m C i n. Then q 2 ZŒi. Let r D ˛ qˇ. We must show that N.r/ < N.ˇ/. Working with complex absolute value we get ˇ ˇ ˇ ˇ˛ jrj D j˛ qˇj D jˇjˇˇ q ˇˇ: ˇ Now
s ˇ ˇ 2 2 q ˇ ˇ˛ 1 1 ˇ q ˇ D j.u m/ C i.v n/j D .u m/2 C .v n/2 C < 1: ˇ ˇˇ 2 2 Therefore jrj < jˇj H) jrj2 < jˇj2 H) N.r/ < N.ˇ/ completing the proof. Since ZŒi forms a Euclidean domain it follows from our previous results that ZŒi must be a principal ideal domain and hence a unique factorization domain. Corollary 3.5.8. The Gaussian integers are a UFD. Since we will now be dealing with many kinds of integers we will refer to the ordinary integers Z as the rational integers and the ordinary primes p as the rational primes. It is clear that Z can be embedded into ZŒi. However not every rational prime is also prime in ZŒi. The primes in ZŒi are called the Gaussian primes. For example we can show that both 1 C i and 1 i are Gaussian primes, that is primes in ZŒi. However .1 C i/.1 i / D 2 so that the rational prime 2 is not a prime in ZŒi. Using the multiplicativity of the Euclidean norm in ZŒi we can describe all the units and primes in ZŒi. Theorem 3.5.9.
(1) The only units in ZŒi are ˙1; ˙i.
(2) Suppose is a Gaussian prime. Then is either: (a) a positive rational prime p 3 mod 4 or an associate of such a rational prime. (b) 1 C i or an associate of 1 C i. (c) a C bi or a bi where a > 0, b > 0, a is even and N./ D a 2 C b 2 D p with p a rational prime congruent to 1 mod 4 or an associate of a C bi or a bi . Proof. (1) Suppose u D x C iy 2 ZŒi is a unit. Then from Lemma 3.5.6 we have N.u/ D x 2 C y 2 D 1 implying that .x; y/ D .0; ˙1/ or .x; y/ D .˙1; 0/. Hence u D ˙1 or u D ˙i.
50
Chapter 3 Prime Elements and Unique Factorization Domains
(2) Now suppose that is a Gaussian prime. Since N./ D and 2 ZŒi it follows that jN./. N./ is a rational integer so N./ D p1 pk where the pi ’s are rational primes. By Euclid’s lemma jpi for some pi and hence a Gaussian prime must divide at least one rational prime. On the other hand suppose jp and jq where p; q are different primes. Then .p; q/ D 1 and hence there exist x; y 2 Z such that 1 D px C qy. It follows that j1 a contradiction. Therefore a Gaussian prime divides one and only one rational prime. Let p be the rational prime that divides. Then N./jN.p/ D p 2 . Since N./ is a rational integer it follows that N./ D p or N./ D p2 . If D a C bi then a2 C b 2 D p or a2 C b 2 D p 2 . If p D 2 then a2 C b 2 D 2 or a2 C b 2 D 4. It follows that D ˙2; ˙2i or D 1 C i or an associate of 1 C i . Since .1 C i/.1 i / D 2 and neither 1 C i nor 1 i are units it follows that neither 2 nor any of its associates are primes. Then D 1 C i or an associate of 1 C i . To see that 1 C i is prime suppose 1 C i D ˛ˇ. Then N.1 C i / D 2 D N.˛/N.ˇ/. It follows that either N.˛/ D 1 or N.ˇ/ D 1 and either ˛ or ˇ is a unit. If p ¤ 2 then either p 3 mod 4 or p 1 mod 4. Suppose first that p 3 mod 4. Then a2 C b 2 D p would imply from Fermat’s two-square theorem (see [35]) that p 1 mod 4. Therefore from the remarks above a2 C b 2 D p 2 and N./ D N.p/. Since jp we have D ˛p with ˛ 2 ZŒi. From N./ D N.p/ we get that N.˛/ D 1 and ˛ is a unit. Therefore and p are associates. Hence in this case is an associate of a rational prime congruent to 3 mod 4. Finally suppose p 1 mod 4. From the remarks above either N./ D p or N./ D p 2 . If N./ D p 2 then a2 C b 2 D p 2 . Since p 1 mod 4 from Fermat’s two square theorem there exist m; n 2 Z with m2 C n2 D p. Let u D m C i n then the norm N.u/ D p. Since p is a rational prime it follows that u is a Gaussian prime. Similarly its conjugate u is also a Gaussian prime. Now uu D p 2 D N./. Since jN./ it follows that juu and from Euclid’s lemma either ju or ju. If ju they are associates since both are primes. But this is a contradiction since N./ ¤ N.u/. The same is true if ju. It follows that if p 1 mod 4, then N./ ¤ p 2 . Therefore in this case N./ D p D a2 C b 2 . An associate of has both a; b > 0 (see exercises). Further since a2 C b 2 D p one of a or b must be even. If a is odd then b is even then i is an associate of with a even completing the proof. Finally we mention that the methods used in ZŒi cannot be applied to all quadratic p integers. For example we have seen that there is not unique factorization in ZŒ 5.
Section 3.6 Overview of Integral Domains
3.6
51
Overview of Integral Domains
Here we present some additional definitions for special types of integral domains. Definition 3.6.1. (1) A Dedekind domain D is an integral domain such that each nonzero proper ideal A (¹0º ¤ A ¤ R) can be written uniquely as a product of prime ideals A D P1 Pr with each Pi a prime ideal and the factorization is unique up to ordering. (2) A Prüfer ring is an integral domain such that A .B \ C / D AB \ AC for all ideals A; B; C in R. Dedekind domains arise naturally in algebraic number theory. It can be proved that the rings of algebraic integers in any algebraic number field are Dedekind domains (see [35]). If R is a Dedekind domain it is also a Prüfer Ring. If R is a Prüfer ring and a unique factorization domain then R is a principal ideal domain. In the next chapter we will prove a theorem due to Gauss that if R is a UFD then the polynomial ring RŒx is also a UFD. If K is a field we have already seen that KŒx is a UFD. Hence the polynomial ring in several variables KŒx1 ; : : : ; xn is also a UFD. This fact plays an important role in algebraic geometry.
3.7
Exercises
1. Let R be an integral domain and let 2 R n .U.R/ [ ¹0º/. Show: (i) If for each a 2 R with − a there exist ; 2 R with C a D 1 then is a prime element of R. (ii) Give an example for a prime element in an UFD R, which does not satisfy the conditions of (i). 2. Let R be an UFD and let a1 ; : : : ; a t be pairwise coprime elements of R. If a1 a t is an m-th power (m 2 N), then all factors ai are an associate of an m-th power. Is each ai necessarily an m-th power? p p p 3, ZŒ 5 and ZŒ 7 3. Decide if the unit pis finite or infinite. For which pgroup of ZŒ p a 2 Z are .1 5/ and .a C 5/ associates in ZŒ 5? p p 4. Let k 2 Z andp k ¤ x 2 for all x 2 Z. Let ˛ D a C b k and ˇ D c C d k be elements of ZŒ k and N.˛/ D a2 kb 2 , N.ˇ/ D c 2 kd 2 . Show:
52
Chapter 3 Prime Elements and Unique Factorization Domains
(i) The equality of the absolutepvalues of N.˛/ and N.ˇ/ is necessary for the association of ˛ and ˇ in ZŒ k. Is this constraint also sufficient? p (ii) Sufficient for the irreducibility of ˛ in ZŒ k is the irreducibility of N.˛/ in Z. Is this also necessary? 5. In general irreducible elements are not prime. Consider the set of complex number given by p p R D ZŒi 5 D ¹x C i 5y W x; y 2 Zº: Show that they form a subring of C. p 6. For an element x C iy 5 2 R define its norm by p p N.x C iy 5/ D jx C iy 5j D x 2 C 5y 2 : Prove that the norm is multiplicative, that is N.ab/ D N.a/N.b/. 7. Prove Lemma 3.4.4. 8. Prove that the set of polynomials RŒx with coefficients in a ring R forms a ring. 9. Prove the basic properties of the norm of the Gaussian integers. If ˛; ˇ 2 ZŒi then: (i) N.˛/ is an integer for all ˛ 2 ZŒi. (ii) N.˛/ 0 for all ˛ 2 ZŒi. (iii) N.˛/ D 0 if and only if ˛ D 0. (iv) N.˛/ 1 for all ˛ ¤ 0. (v) N.˛ˇ/ D N.˛/N.ˇ/, that is the norm is multiplicative.
Chapter 4
Polynomials and Polynomial Rings
4.1
Polynomials and Polynomial Rings
In the last chapter we saw that if K is a field then the set of polynomials with coefficients in K, which we denoted KŒx, forms a unique factorization domain. In this chapter we take a more detailed look at polynomials over a general ring R. We then prove that if R is a UFD then the polynomial ring RŒx is also a UFD. We first take a formal look at polynomials. Let R be a commutative ring with an identity. Consider the set RQ of functions f from the nonnegative integers N D N [¹0º into R with only a finite number of values nonzero. That is RQ D ¹f W N ! R W f .n/ ¤ 0 for only finitely many nº: On RQ we define the following addition and multiplication .f C g/.m/ D f .m/ C g.m/ X f .i/g.j /: .f g/.m/ D i Cj Dm
If we let x D .0; 1; 0; : : :/ and identify .r; 0; : : :/ with r 2 R then x 0 D .1; 0; : : :/ D 1 and
x i C1 D x x i :
Now if f D .r0 ; r1 ; r2 ; : : :/ then f can be written as f D
1 X i D0
ri x i D
m X
ri x i
i D0
for some m 0 since ri ¤ 0 for only finitely many i. Further this presentation is unique. now call x an indeterminate over R and write each element of RQ as f .x/ D PWe m i Q iD0 ri x with f .x/ D 0 or rm ¤ 0. We also now write RŒx for R. Each element of RŒx is called a polynomial over R. The elements r0 ; : : : ; rm are called the coefficients of f .x/ with rm the leading coefficient. If rm ¤ 0 the natural number m is called the degree of f .x/ which we denote by deg f .x/. We say that f .x/ D 0 has degree 1. The uniqueness of the representation of a polynomial implies that two nonzero
54
Chapter 4 Polynomials and Polynomial Rings
polynomials are equal if and only if they have the same degree and exactly the same coefficients. A polynomial of degree 1 is called a linear polynomial while one of degree two is a quadratic polynomial. The set of polynomials of degree 0 together with 0 form a ring isomorphic to R and hence can be identified with R, the constant polynomials. Thus the ring R embeds in the set of polynomials RŒx. The following results are straightforward concerning degree. Lemma 4.1.1. Let f .x/ ¤ 0; g.x/ ¤ 0 2 RŒx. Then: (a) deg f .x/g.x/ deg f .x/ C deg g.x/. (b) deg.f .x/ ˙ g.x// Max.deg f .x/; deg g.x//. If R is an integral domain then we have equality in (a). Theorem 4.1.2. Let R be a commutative ring with an identity. Then the set of polynomials RŒx forms a ring called the ring of polynomials over R. The ring R identified with 0 and the polynomials of degree 0 naturally embeds into RŒx. RŒx is commutative if and only if R is commutative. Further RŒx is uniquely determined by R and x. P P Proof. Set f .x/ D niD0 ri x i and g.x/ D jmD0 sj x j . The ring properties follow directly by computation. The identification of r 2 R with the polynomial r.x/ D r provides the embedding of R into RŒx. From the definition of multiplication in RŒx if R is commutative then RŒx is commutative. Conversely if RŒx is commutative then from the embedding of R into RŒx it follows that R must be commutative. Note that if R has a multiplicative identity 1 ¤ 0 then this is also the multiplicative identity of RŒx. Finally if S is a ring that contains R and ˛ 2 S then ³ ²X i ri ˛ W ri 2 R and ri ¤ 0 for only a finite number of i RŒ˛ D i 0
is a homomorphic image of RŒx via the map X X ri x i 7! ri ˛ i : i 0
i 0
Hence RŒx is uniquely determined by R and x. If R is an integral domain then irreducible polynomials are defined as irreducibles in the ring RŒx. If R is a field then f .x/ is an irreducible polynomial if there is no factorization f .x/ D g.x/h.x/ where g.x/ and h.x/ are polynomials of lower degree than f .x/. Otherwise f .x/ is called reducible. In elementary mathematics polynomials are considered as functions. We recover that idea via the concept of evaluation.
Section 4.2 Polynomial Rings over Fields
55
Definition 4.1.3. Let f .x/ D r0 C r1 x C C rm x n be a polynomial over a commutative ring R with an identity and let c 2 R. Then the element f .c/ D r0 C r1 c C C rn c n 2 R is called the evaluation of f .x/ at c. Definition 4.1.4. If f .x/ 2 RŒx and f .c/ D 0 for c 2 R, then c is called a zero or a root of f .x/ in R.
4.2
Polynomial Rings over Fields
We now restate some of the result of the last chapter for KŒx where K is a field. We then consider some consequences of these results to zeros of polynomials. Theorem 4.2.1. If F is a field, then F Œx forms an integral domain. F can be naturally embedded into F Œx by identifying each element of F with the corresponding constant polynomial. The only units in F Œx are the nonzero elements of F . Proof. Verification of the basic ring properties is solely computational and is left to the exercises. Since deg P .x/Q.x/ D deg P .x/ C deg Q.x/, it follows that if neither P .x/ ¤ 0 nor Q.x/ ¤ 0 then P .x/Q.x/ ¤ 0 and therefore F Œx is an integral domain. If G.x/ is a unit in F Œx, then there exists an H.x/ 2 F Œx with G.x/H.x/ D 1. From the degrees we have deg G.x/ C deg H.x/ D 0 and since deg G.x/ 0, deg H.x/ 0. This is possible only if deg G.x/ D deg H.x/ D 0. Therefore G.x/ 2 F . Now that we have F Œx as an integral domain we proceed to show that F Œx is a principal ideal domain and hence there is unique factorization into primes. We first repeat the definition of a prime in F Œx. If 0 ¤ f .x/ has no nontrivial, nonunit factors (it cannot be factorized into polynomials of lower degree) then f .x/ is a prime in F Œx or a prime polynomial. A prime polynomial is also called an irreducible polynomial over F . Clearly, if deg g.x/ D 1 then g.x/ is irreducible. The fact that F Œx is a principal ideal domain follows from the division algorithm for polynomials, which is entirely analogous to the division algorithm for integers. Theorem 4.2.2 (division algorithm in F Œx). If 0 ¤ f .x/; 0 ¤ g.x/ 2 F Œx then there exist unique polynomials q.x/; r.x/ 2 F Œx such that f .x/ D q.x/g.x/ C r.x/ where r.x/ D 0 or deg r.x/ < deg g.x/. (The polynomials q.x/ and r.x/ are called respectively the quotient and remainder.) This theorem is essentially long division of polynomials. A formal proof is based on induction on the degree of g.x/. We omit this but give some examples from QŒx.
56
Chapter 4 Polynomials and Polynomial Rings
Example 4.2.3.
(a) Let f .x/ D 3x 4 6x 2 C 8x 6, g.x/ D 2x 2 C 4. Then
3x 4 6x 2 C 8x 6 3 D x 2 6 with remainder 8x C 18: 2 2x C 4 2 Thus here q.x/ D 32 x 2 6 and r.x/ D 8x C 18. (b) Let f .x/ D 2x 5 C 2x 4 C 6x 3 C 10x 2 C 4x, g.x/ D x 2 C x. Then 2x 5 C 2x 4 C 6x 3 C 10x 2 C 4x D 2x 3 C 6x C 4: x2 C x Thus here q.x/ D 2x 3 C 6x C 4 and r.x/ D 0. Theorem 4.2.4. Let F be a field. Then the polynomial ring F Œx is a principal ideal domain and hence a unique factorization domain. We now give some consequences relative to zeros of polynomials in F Œx. Theorem 4.2.5. If f .x/ 2 F Œx and c 2 F with f .c/ D 0 then f .x/ D .x c/h.x/; where deg h.x/ < deg f .x/. Proof. Divide f .x/ by x c. Then by the division algorithm we have f .x/ D .x c/h.x/ C r.x/ where r.x/ D 0 or deg r.x/ < deg.x c/ D 1. Hence if r.x/ ¤ 0 then r.x/ is a polynomial of degree 0, that is a constant polynomial, that is r.x/ D r for r 2 F . Hence we have f .x/ D .x c/h.x/ C r: This implies that 0 D f .x/ D 0h.c/ C r D r and therefore r D 0 and f .x/ D .x c/h.x/. Since deg.x c/ D 1 we must have that deg h.x/ < deg f .x/. If f .x/ D .x c/k h.x/ for some k 1 with h.c/ ¤ 0 then c is called a zero of order k. Theorem 4.2.6. Let f .x/ 2 F Œx with degree 2 or 3. Then f is irreducible if and only if f .x/ doesn’t have a zero in F . Proof. Suppose that f .x/ is irreducible of degree 2 or 3. If f .x/ has a zero c then from Theorem 4.2.4 we have f .x/ D .x c/h.x/ with h.x/ of degree 1 or 2. Therefore f .x/ is reducible a contradiction and hence f .x/ cannot have a zero. Conversely from Theorem 4.2.4 if f .x/ has a zero and if of degree greater than 1 then f .x/ is reducible.
Section 4.3 Polynomial Rings over Integral Domains
4.3
57
Polynomial Rings over Integral Domains
Here we consider RŒx where R is an integral domain. Definition 4.3.1. Let R be an integral domain. Then a1 ; a2 ; : : : ; an 2 R are coprime if the set of all common divisors of a1 ; : : : ; an consists only of units. Notice for example that this concept depends on the ring R. For example 6 and 9 are not coprime over the integers Z since 3j6 and 3j9 and 3 is not a unit. However 6 and 9 are coprime over the rationals Q. Here 3 is a unit. Pn i Definition 4.3.2. Let f .x/ D i D1 ri x 2 RŒx where R is an integral domain. Then f .x/ is a primitive polynomial or just primitive if r0 ; r1 ; : : : ; rn are coprime in R. Theorem 4.3.3. Let R be an integral domain. Then (a) The units of RŒx are the units of R. (b) If p is a prime element of R then p is a prime element of RŒx. Proof. If r 2 R is a unit then since R embeds into RŒx it follows that r is also a unit in RŒx. Conversely suppose that h.x/ 2 RŒx is a unit. Then there is a g.x/ such that h.x/g.x/ D 1. Hence deg f .x/ C deg g.x/ D deg 1 D 0. Since degrees are nonnegative integers it follows that deg f .x/ D deg g.x/ D 0 and hence f .x/ 2 R. Now suppose that p is a prime element of R. Then p ¤ 0 and pR is a prime ideal in R. We must show that pRŒx is a prime ideal in RŒx. Consider the map W RŒx ! .R=pR/Œx given by X X n n i ri x D .ri C pR/x i : i D0
i D0
Then is an epimorphism with kernel pRŒx. Since pR is a prime ideal we know that R=pR is an integral domain. It follows that .R=pR/Œx is also an integral domain. Hence pRŒx must be a prime ideal in RŒx and therefore p is also a prime element of RŒx. Recall that each integral domain R can be embedded into a unique field of fractions K. We can use results on KŒx to deduce some results in RŒx. Lemma 4.3.4. If K is a field then each nonzero f .x/ 2 KŒx is a primitive. Proof. Since K is a field each nonzero element of K is a unit. Therefore the only common divisors of the coefficients of f .x/ are units and hence f .x/ 2 KŒx is primitive.
58
Chapter 4 Polynomials and Polynomial Rings
Theorem 4.3.5. Let R be an integral domain. Then each irreducible f .x/ 2 RŒx of degree > 0 is primitive. Proof. Let f .x/ be an irreducible polynomial in RŒx and let r 2 R be a common divisor of the coefficients of f .x/. Then f .x/ D rg.x/ where g.x/ 2 RŒx. Then deg f .x/ D deg g.x/ > 0 so g.x/ … R. Since the units of RŒx are the units of R it follows that g.x/ is not a unit in RŒx. Since f .x/ is irreducible it follows that r must be a unit in RŒx and hence r is a unit in R. Therefore f .x/ is primitive. Theorem 4.3.6. Let R be an integral domain and K its field of fractions. If f .x/ 2 RŒx is primitive and irreducible in KŒx then f .x/ is irreducible in RŒx. Proof. Suppose that f .x/ 2 RŒx is primitive and irreducible in KŒx and suppose that f .x/ D g.x/h.x/ where g.x/; h.x/ 2 RŒx KŒx. Since f .x/ is irreducible in KŒx either g.x/ or h.x/ must be a unit in KŒx. Without loss of generality suppose that g.x/ is a unit in KŒx. Then g.x/ D g 2 K. But g.x/ 2 RŒx and K \RŒx D R. Hence g 2 R. Then g is a divisor of the coefficients of f .x/ and as f .x/ is primitive g.x/ must be a unit in R and therefore also a unit in RŒx. Therefore f .x/ is irreducible in RŒx.
4.4
Polynomial Rings over Unique Factorization Domains
In this section we prove that if R is a UFD then the polynomial ring RŒx is also a UFD. We first need the following due to Gauss. Theorem 4.4.1 (Gauss’ lemma). Let R be a UFD and f .x/; g.x/ primitive polynomials in RŒx. Then their product f .x/g.x/ is also primitive. Proof. Let R be a UFD and f .x/; g.x/ primitive polynomials in RŒx. Suppose that f .x/g.x/ is not primitive. Then there is a prime element p 2 R that divides each of the coefficients of f .x/g.x/. Then pjf .x/g.x/. Since prime elements of R are also prime elements of RŒx it follows that p is also a prime element of RŒx and hence pjf .x/ or pjg.x/. Therefore either f .x/ or g.x/ is not primitive giving a contradiction. Theorem 4.4.2. Let R be a UFD and K its field of fractions. (a) If g.x/ 2 KŒx is nonzero then there is a nonzero a 2 K such that ag.x/ 2 RŒx is primitive. (b) Let f .x/; g.x/ 2 RŒx with g.x/ primitive and f .x/ D ag.x/ for some a 2 K. Then a 2 R. (c) If f .x/ 2 RŒx is nonzero then there is a b 2 R and a primitive g.x/ 2 RŒx such that f .x/ D bg.x/.
Section 4.4 Polynomial Rings over Unique Factorization Domains
59
Pn ri i Proof. (a) Suppose that g.x/ D i D0 ai x with ai D si , ri ; si 2 R. Set s D s0 s1 sn . Then sg.x/ is a nonzero element of RŒx. Let d be a greatest common divisor of the coefficients of sg.x/. If we set a D ds then ag.x/ is primitive. (b) For a 2 K there are coprime r; s 2 R satisfying a D rs . Suppose that a … R. Then there is a prime element p 2 R dividing s. Since g.x/ is primitive p does not divide all the coefficients of g.x/. However we also have f .x/ D ag.x/ D rs g.x/. Hence sf .x/ D rg.x/ where pjs and p doesn’t divide r. Therefore p divides all the coefficients of g.x/ and hence a 2 R. (c) From part (a) there is a nonzero a 2 K such that af .x/ is primitive in RŒx. Then f .x/ D a1 .af .x//. From part (b) we must have a1 2 R. Set g.x/ D af .x/ and b D a1 . Theorem 4.4.3. Let R be a UFD and K its field of fractions. Let f .x/ 2 RŒx be a polynomial of degree 1. (a) If f .x/ is primitive and f .x/jg.x/ in KŒx then f .x/ divides g.x/ also in RŒx. (b) If f .x/ is irreducible in RŒx then it is also irreducible in KŒx. (c) If f .x/ is primitive and a prime element of KŒx then f .x/ is also a prime element of RŒx. Proof. (a) Suppose that g.x/ D f .x/h.x/ with h.x/ 2 KŒx. From Theorem 4.4.2 part (a) there is a nonzero a 2 K such that h1 .x/ D ah.x/ is primitive in RŒx. Hence g.x/ D a1 .f .x/h1 .x/. From Gauss’ lemma f .x/h1 .x/ is primitive in RŒx and therefore from Theorem 4.4.2 part (b) we have a1 2 R. It follows that f .x/jg.x/ in RŒx. (b) Suppose that g.x/ 2 KŒx is a factor of f .x/. From Theorem 4.4.2 part (a) there is a nonzero a 2 K with g1 .x/ D ag.x/ primitive in RŒx. Since a is a unit in K it follows that g.x/jf .x/ in KŒx
implies
g1 .x/jf .x/ in KŒx
and hence since g1 .x/ is primitive g1 .x/jf .x/ in RŒx: However by assumption f .x/ is irreducible in RŒx. This implies that either g1 .x/ is a unit in R or g1 .x/ is an associate of f .x/. If g1 .x/ is a unit then g1 2 K and g1 D ga and hence g 2 K, that is g D g.x/ is a unit. If g1 .x/ is an associate of f .x/ then f .x/ D bg.x/ where b 2 K since g1 .x/ D ag.x/ with a 2 K. Combining these it follows that f .x/ has only trivial factors in KŒx and since by assumption f .x/ is nonconstant it follows that f .x/ is irreducible in KŒx.
60
Chapter 4 Polynomials and Polynomial Rings
(c) Suppose that f .x/jg.x/h.x/ with g.x/; h.x/ 2 RŒx. Since f .x/ is a prime element in KŒx we have that f .x/jg.x/ or f .x/jh.x/ in KŒx. From part (a) we have f .x/jg.x/ or f .x/jh.x/ in RŒx implying that f .x/ is a prime element in RŒx. We can now state and prove our main result. Theorem 4.4.4 (Gauss). Let R be a UFD. Then the polynomial ring RŒx is also a UFD Proof. By induction on degree we show that each nonunit f .x/ 2 RŒx; f .x/ ¤ 0, is a product of prime elements. Since R is an integral domain so is RŒx, and so the fact that RŒx is a UFD then follows from Theorem 3.3.3. If deg f .x/ D 0 then f .x/ D f is a nonunit in R. Since R is a UFD f is a product of prime elements in R. However from Theorem 4.3.3 each prime factor is then also prime in RŒx. Therefore f .x/ is a product of prime elements. Now suppose n > 0 and that the claim is true for all polynomials f .x/ of degree < n. Let f .x/ be a polynomial of degree n > 0. From Theorem 4.4.2 (c) there is an a 2 R and a primitive h.x/ 2 RŒx satisfying f .x/ D ah.x/. Since R is a UFD the element a is a product of prime elements in R or a is a unit in R. Since the units in RŒx are the units in R and a prime element in R is also a prime element in RŒx it follows that a is a product of prime elements in RŒx or a is a unit in RŒx. Let K be the field of fractions of R. Then KŒx is a UFD. Hence h.x/ is a product of prime elements of KŒx. Let p.x/ 2 KŒx be a prime divisor of h.x/. From Theorem 4.4.2 we can assume by multiplication of field elements that p.x/ 2 RŒx and p.x/ is primitive. From Theorem 4.4.2 (c) it follows that p.x/ is a prime element of RŒx and further from Theorem 4.4.3 (a) that p.x/ is a divisor of h.x/ in RŒx. Therefore f .x/ D ah.x/ D ap.x/g.x/ 2 RŒx where (1) a is a product of prime elements of RŒx or a is a unit in RŒx, (2) deg p.x/ > 0, since p.x/ is a prime element in KŒx, (3) p.x/ is a prime element in RŒx, and (4) deg g.x/ < deg f .x/ since deg p.x/ > 0. By our inductive hypothesis we have then that g.x/ is a product of prime elements in RŒx or g.x/ is a unit in RŒx. Therefore the claim holds for f .x/ and therefore holds for all f .x/ by induction. If RŒx is a polynomial ring over R we can form a polynomial ring in a new indeterminate y over this ring to form .RŒx/Œy. It is straightforward that .RŒx/Œy is isomorphic to .RŒy/Œx. We denote both of these rings by RŒx; y and consider this as the ring of polynomials in two commuting variables x; y with coefficients in R.
Section 4.4 Polynomial Rings over Unique Factorization Domains
61
If R is a UFD then from Theorem 4.4.4 RŒx is also a UFD and hence RŒx; y is also a UFD. Inductively then the ring of polynomials in n commuting variables RŒx1 ; x2 ; : : : ; xn is also a UFD. Corollary 4.4.5. If R is a UFD then the polynomial ring in n commuting variables RŒx1 ; : : : ; xn is also a UFD. We now give a condition for a polynomial in RŒx to have a zero in KŒx where K is the field of fractions of R. Theorem 4.4.6. Let R be a UFD and K its field of fractions. Let f .x/ D x n C rn1 x n1 C C r0 2 RŒx. Suppose that ˇ 2 K is a zero of f .x/. Then ˇ is in R and is a divisor of r0 . Proof. Let ˇ D
r s
where s ¤ 0 and r; s 2 R and r; s are coprime. Now f
r n1 rn r D 0 D n C rn1 n1 C C r0 : s s s
Hence it follows that s must divide r n . Since r and s are coprime s must be a unit and then without loss of generality we may assume that s D 1. Then ˇ 2 R and r.r n1 C C a1 / D a0 and so rja0 . Note that since Z is a UFD, Gauss’ theorem implies that ZŒx is also a UFD. However ZŒx is not a principal ideal domain. For example the set of integral polynomials with even constant term is an ideal but not principal. We leave the verification to the exercises. On the other hand we saw that if K is a field KŒx is a PID. The question arises as to when RŒx actually is a principal ideal domain. It turns out to be precisely when R is a field. Theorem 4.4.7. Let R be a commutative ring with an identity. Then the following are equivalent: (1) R is a field. (2) RŒx is Euclidean. (3) RŒx is a principal ideal domain. Proof. From Section 4.2 we know that (a) implies (b) which in turn implies (c). Therefore we must show that (c) implies (a). Assume then that RŒx is a principal ideal domain. Define the map W RŒx ! R
62
Chapter 4 Polynomials and Polynomial Rings
by .f .x// D f .0/: It is easy to see that is a ring homomorphism with RŒx= ker. / Š R. Therefore ker. / ¤ RŒx. Since RŒx is a principal ideal domain it is an integral domain. It follows that ker. / must be a prime ideal since the quotient ring is an integral domain. However since RŒx is a principal ideal domain prime ideals are maximal ideals and hence ker. / is a maximal ideal. Therefore R Š RŒx= ker. / is a field. We now consider the relationship between irreducibles in RŒx for a general integral domain and irreducibles in KŒx where K is its field of fractions. This is handled by the next result called Eisenstein’s criterion. Theorem 4.4.8 (Eisenstein’s Let R be an integral domain and K its field Pcriterion). n i 2 RŒx of degree n > 0. Let p be a prime a x of fractions. Let f .x/ D i D0 i element of R satisfying (1) pjai for i D 0; : : : ; n 1. (2) p does not divide an . (3) p 2 does not divide a0 . Then: (a) If f .x/ is primitive then f .x/ is irreducible in RŒx. (b) Suppose that R is a UFD. Then f .x/ is also irreducible in KŒx. Proof. (a) Suppose that f .x/ D g.x/h.x/ with g.x/; h.x/ 2 RŒx. Suppose that g.x/ D
k X
bi x i ;
bk ¤ 0 and
h.x/ D
i D1
l X
cj x j ;
cl ¤ 0:
i D1
Then a0 D b0 c0 . Now pja0 but p 2 does not divide a0 . This implies that either p doesn’t divide b0 or p doesn’t divide c0 . Without loss of generality assume that pjb0 and p doesn’t divide c0 . Since an D bk cl and p does not divide an it follows that p does not divide bk . Let bj be the first coefficient of g.x/ which is not divisible by p. Consider aj D bj c0 C C b0 cj where everything after the first term is divisible by p. Since p does not divide both bj and c0 it follows that p does not divide bj c0 and therefore p does not divide aj which implies that j D n. Then from j k n it follows that k D n. Therefore deg g.x/ D deg f .x/ and hence deg h.x/ D 0. Thus h 2 R. Then from f .x/ D hg.x/ with f primitive it follows that h is a unit and therefore f .x/ is irreducible.
Section 4.4 Polynomial Rings over Unique Factorization Domains
63
(b) Suppose that f .x/ D g.x/h.x/ with g.x/; h.x/ 2 RŒx. The fact that f .x/ was primitive was only used in the final part of part (a) so by the same arguments as in part (a) we may assume without loss of generality that h 2 R K. Therefore f .x/ is irreducible in KŒx. We give some examples. Example 4.4.9. Let R D Z and p a prime number. Suppose that n; m are integers such that n 1 and p does not divide m. Then x n ˙ pm is irreducible in ZŒx and 1 QŒx. In particular .pm/ n is irrational. Example 4.4.10. Let R D Z and p a prime number. Consider the polynomial ˆp .x/ D
xp 1 D x p1 D x p2 C C 1: x1
Since all the coefficients of ˆp .x/ are equal to 1, Eisenstein’s criterion is not directly applicable. However the fact that ˆp .x/ is irreducible implies that for any integer a the polynomial ˆp .x C a/ is also irreducible in ZŒx. It follows that p x p C p1 x p1 C C p1 x C 1p 1 .x C 1/p 1 D ˆp .x C 1/ D .x C 1/ 1 x ! ! p p1 p D x p1 C x C C : 1 p1 p D p is not Now pj pi for 1 i p 1 (see exercises) and moreover p1 2 divisible by p . Therefore we can apply the Eisenstein criterion to conclude that ˆp .x/ is irreducible in ZŒx and QŒx. P Theorem 4.4.11. Let R be a UFD and K its field of fractions. Let f .x/ D niD0 ai x i 2 RŒx be a polynomial of degree 1. Let P be a prime ideal in R with an … P . Let R D R=P and let ˛ W RŒx ! RŒx be defined by ˛
X m i D0
ri x
i
D
m X
.ri C P /x i :
i D0
˛ is an epimorphism. Then if ˛.f .x// is irreducible in RŒx then f .x/ is irreducible in KŒx. Proof. By Theorem 4.4.3 there is an a 2 R and a primitive g.x/ 2 RŒx satisfying f .x/ D ag.x/. Since an … P we have that ˛.a/ ¤ 0 and further the highest coefficient of g.x/ is also not an element of P . If ˛.g.x// is reducible then ˛.f .x// is also reducible. Thus ˛.g.x// is irreducible. However from Theorem 4.4.4 g.x/ is
64
Chapter 4 Polynomials and Polynomial Rings
irreducible in KŒx so f .x/ D ag.x/ is also irreducible in KŒx. Therefore to prove the theorem it suffices to consider the case where f .x/ is primitive in RŒx. Now suppose that f .x/ is primitive. We show that f .x/ is irreducible in RŒx. Suppose that f .x/ D g.x/h.x/, g.x/; h.x/ 2 RŒx with f .x/; g.x/ nonunits in RŒx. Since f .x/ is primitive g; h … R and so deg g.x/ < deg f .x/ and deg h.x/ < deg f .x/. Now we have ˛.f .x// D ˛.g.x//˛.h.x//. Since P is a prime ideal R=P is an integral domain and so in RŒx we have deg ˛.g.x// C deg ˛.g.x// D deg ˛.f .x// D deg f .x/ since an … P . Since R is a UFD it has no zero divisors and so deg f .x/ D deg g.x/ C deg h.x/: Now deg ˛.g.x// deg g.x/ deg ˛.h.x// deg h.x/: Therefore deg ˛.g.x// D deg g.x/ and deg ˛.h.x// D deg h.x/. Therefore ˛.f .x// is reducible and we have a contradiction. It is important to note that ˛.f .x// being reducible does not imply that f .x/ is reducible. For example f .x/ D x 2 C 1 is irreducible in ZŒx. However in Z2 Œx we have x 2 C 1 D .x C 1/2 and hence f .x/ is reducible in Z2 Œx. Example 4.4.12. Let f .x/ D x 5 x 2 C 1 2 ZŒx. Choose P D 2Z so that ˛.f .x// D x 5 C x 2 C 1 2 Z2 Œx: Suppose that in Z2 Œx we have ˛.f .x// D g.x/h.x/. Without loss of generality we may assume that g.x/ is of degree 1 or 2. If deg g.x/ D 1 then ˛.f .x// has a zero c in Z2 Œx. The two possibilities for c are c D 0 or c D 1. Then If c D 0 then 0 C 0 C 1 D 1 ¤ 0: If c D 1 then 1 C 1 C 1 D 1 ¤ 0: Hence the degree of g.x/ cannot be 1.
65
Section 4.5 Exercises
Suppose deg g.x/ D 2. The polynomials of degree 2 over Z2 Œx have the form x 2 C x C 1;
x 2 C x;
x 2 C 1;
x2:
The last three, x 2 Cx; x 2 C1; x 2 all have zeros in Z2 Œx so they can’t divide ˛.f .x//. Therefore g.x/ must be x 2 C x C 1. Applying the division algorithm we obtain ˛.f .x// D .x 3 C x 2 /.x 2 C x C 1/ C 1 and therefore x 2 CxC1 does not divide ˛.f .x//. It follows that ˛.f .x// is irreducible and from the previous theorem f .x/ must be irreducible in QŒx.
4.5
Exercises
1. For which a; b 2 Z does the polynomial x 2 C 3x C 1 divide the polynomial x 3 C x 2 C ax C b? 2. Let a C bi 2 C be a zero of f .x/ 2 RŒx. Show that also a i b is a zero of f .x/. 3. Determine all irreducible polynomials over R. Factorize f .x/ 2 RŒx in irreducible polynomials. 4. Let R be an integral domain, A G R an ideal and f 2 RŒx a monic P polynomial. N Define f 2 RŒx=AŒx by the mapping RŒx ! R=AŒx, f D ai x i 7! fN D P i N aNi x , where aN WD a C A. Show: If f 2 R=AŒx is irreducible then also f 2 RŒx is irreducible. 5. Decide if the following polynomials f 2 RŒx are irreducible: (i) f .x/ D x 3 C 2x 2 C 3, R D Z. (ii) f .x/ D x 5 2x C 1, R D Q. (iii) f .x/ D 3x 4 C 7x 2 C 14x C 7, R D Q. (iv) f .x/ D x 7 C .3 i /x 2 C .3 C 4i /x C 4 C 2i , R D ZŒi. (v) f .x/ D x 4 C 3x 3 C 2x 2 C 3x C 4, R D Q. (vi) f .x/ D 8x 3 4x 2 C 2x 1, R D Z. 6. Let R be an integral domain with characteristic 0, let k 1 and ˛ 2 R. In RŒx define the derivatives f .k/ .x/, k D 0; 1; 2; : : : , of a polynomial f .x/ 2 RŒx by f 0 .x/ WD f .x/; 0
f .k/ .x/ WD f .k1/ .x/: Show that ˛ is a zero of order k of the polynomial f .x/ 2 RŒx, if f .k1/ .˛/ D 0, but f .k/ .˛/ ¤ 0. 7. Prove that the set of integral polynomials with even constant term is an ideal but not principal. 8. Prove that pj pi for 1 i p 1.
Chapter 5
Field Extensions
5.1
Extension Fields and Finite Extensions
Much of algebra in general arose from the theory of equations, specifically polynomial equations. As discovered by Galois and Abel the solutions of polynomial equations over fields is intimately tied to the theory of field extensions. This theory eventually blossoms into Galois Theory. In this chapter we discuss the basic material concerning field extensions. Recall that if L is a field and K L is also a field under the same operations as L then K is called a subfield of L. If we view this situation from the viewpoint of K we say that L is an extension field or field extension of K. If K; L are fields with K L we always assume that K is a subfield of L. Definition 5.1.1. If K; L are fields with K L then we say that L is a field extension or extension field of K. We denote this by LjK. Note that this is equivalent to having a field monomorphism i WK!L and then identifying K and i.K/. As examples we have that R is an extension field of Q and C is an extension field of both C and Q. If K is any field then the ring of polynomials KŒx over K is an integral domain. Let K.x/ be the field of fractions of KŒx. This is called the field of rational functions over K. Since K can be considered as part of KŒx it follows that K K.x/ and hence K.x/ is an extension field of K. A crucial concept is that of the degree of a field extension. Recall that a vector space V over a field F consists of an abelian group V together with scalar multiplication from F satisfying: (1) f v 2 V if f 2 F , v 2 V . (2) f .u C v/ D f u C f v for f 2 F , u; v 2 V . (3) .f C g/v D f v C gv for f; g 2 F , v 2 V . (4) .fg/v D f .gv/ for f; g 2 F , v 2 V . (5) 1v D v for v 2 V .
67
Section 5.1 Extension Fields and Finite Extensions
Notice that if K is a subfield of L then multiplication of elements of L by elements of K are still in L. Since L is an abelian group under addition, L can be considered as a vector space over K. Thus any extension field is a vector space over any of its subfields. Using this we define the degree jL W Kj of an extension K L as the dimension dimK .L/ of L as a vector space over K. We call L a finite extension of K if jL W Kj < 1. Definition 5.1.2. If L is an extension field of K then the degree of the extension LjK is defined as the dimension, dimK .L/, of L, as a vector space over K. We denote the degree by jL W Kj. The field extension LjK is a finite extension if the degree jL W Kj is finite. Lemma 5.1.3. jC W Rj D 2 but jR W Qj D 1. Proof. Every complex number can be written uniquely as a C i b where a; b 2 R. Hence the elements 1; i constitute a basis for C over R and therefore the dimension is 2, that is jC W Rj D 2. The fact that jR W Qj D 1 depends on the existence of transcendental numbers. An element r 2 R is algebraic (over Q) if it satisfies some nonzero polynomial with coefficients from Q. That is, P .r/ D 0, where 0 ¤ P .x/ D a0 C a1 x C C an x n
with ai 2 Q:
Any q 2 Q is algebraic since if P .x/ D p x q then P .q/ D 0. However, many p irrationals are also algebraic. For example 2 is algebraic since x 2 2 D 0 has 2 as a root. An element r 2 R is transcendental if it is not algebraic. In general it is very difficult to show that a particular element is transcendental. However there are uncountably many transcendental elements (see exercises). Specific examples are e and . We will give a proof of their transcendence later in this book. Since e is transcendental, for any natural number n the set of vectors ¹1; e; e 2 ; : : : ; e n º must be independent over Q, for otherwise there would be a polynomial that e would satisfy. Therefore, we have infinitely many independent vectors in R over Q which would be impossible if R had finite degree over Q. Lemma 5.1.4. If K is any field then jK.x/ W Kj D 1. Proof. For any n the elements 1; x; x 2 ; : : : ; x n are independent over K. Therefore as in the proof of Lemma 5.1.3 K.x/ must be infinite dimensional over K. If LjK and L1 jK1 are field extensions then they are isomorphic field extensions if there exists a field isomorphism f W L ! L1 such that fjK is an isomorphism from K to K1 . Suppose that K L M are fields. Below we show that the degrees multiply. In this situation where K L M we call L an intermediate field.
68
Chapter 5 Field Extensions
Theorem 5.1.5. Let K; L; M be fields with K L M . Then jM W Kj D jM W LjjL W Kj: Note that jM W Kj D 1 if and only if either jM W Lj D 1 or jL W Kj D 1. Proof. Let ¹xi W i 2 I º be a basis for L as a vector space over K and let ¹yj W j 2 J º be a basis for M as a vector space over L. To prove the result it is sufficient to show that the set B D ¹xi yj W i 2 I; j 2 J º is a basis for M as a vector space over K. To show this we must show that B is a linearly independent set over K and that B spans M . Suppose that X kij xi yj D 0 where kij 2 K: i;j
We can then write this sum as XX j
kij xi yj D 0:
i
P But i kij xi 2 L. Since ¹yj W j 2 JP º is a basis for M over L the yj are independent over L and hence for each j we get, i kij xi D 0. Now since ¹xi W i 2 I º is a basis for LP over K it follows that the xi are linearly independent and since for each j we have i kij xi D 0 it must be that kij D 0 for all i and for all j . Therefore the set B is linearly independent over K. Now suppose that m 2 M . Then since ¹yj W j 2 J º spans M over L we have mD
X
cj yj
with cj 2 L:
j
However ¹xi W i 2 I º spans L over K and so for each cj we have cj D
X
kij xi
with kij 2 K:
i
Combining these two sums we have mD
X
kij xi yj
ij
and hence B spans M over K. Therefore B is a basis for M over K and the result is proved.
69
Section 5.2 Finite and Algebraic Extensions
Corollary 5.1.6. (a) If jL W Kj is a prime number then there exists no proper intermediate field between L and K. (b) If K L and jL W Kj D 1 then L D K. Let LjK be a field extension and suppose that A L. Then certainly there are subrings of L containing both A and K, for example L. We denote by KŒA the intersection of all subrings of L containing both K and A. Since the intersection of subrings is a subring it follows that KŒA is a subring containing both K and A and the smallest such subring. We call KŒA the ring adjunction of A to K. In an analogous manner we let K.A/ be the intersection of all subfields of L containing both K and A. This is then a subfield of L and the smallest subfield of L containing both K and A. The subfield K.A/ is called the field adjunction of A to K. Clearly KŒA K.A/. If A D ¹a1 ; : : : ; an º then we write KŒA D KŒa1 ; : : : ; an and
K.A/ D K.a1 ; : : : ; an /:
Definition 5.1.7. The field extension LjK is finitely generated if there exist a1 ; : : : ; an 2 L such that K D K.a1 ; : : : ; an /. The extension LjK is a simple extension if there is an a 2 L with L D K.a/. In this case a is called a primitive element of LjK. Later we will look at an alternative way to view the adjunction constructions in terms of polynomials.
5.2
Finite and Algebraic Extensions
We now turn to the relationship between field extensions and the solution of polynomial equations. Definition 5.2.1. Let LjK be a field extension. An element a 2 L is algebraic over K if there exists a polynomial p.x/ 2 KŒx with p.a/ D 0. L is an algebraic extension of K if each element of L is algebraic over K. An element a 2 L that is not algebraic over K is called transcendental. L is a transcendental extension if there are transcendental elements, that is they are not algebraic over K. For the remainder of this section we assume that LjK is a field extension. Lemma 5.2.2. Each element of K is algebraic over K. Proof. Let k 2 K. Then k is a root of the polynomial p.x/ D x k 2 KŒx.
70
Chapter 5 Field Extensions
We tie now algebraic extensions to finite extensions. Theorem 5.2.3. If LjK is a finite extension then LjK is an algebraic extension. Proof. Suppose that LjK is a finite extension and a 2 L. We must show that a is algebraic over K. Suppose that jL W Kj D n < 1 then dimK .L/ D n. It follows that any n C 1 elements of L are linearly dependent over K. Now consider the elements 1; a; a 2 ; : : : ; an in L. These are n C 1 distinct elements in L so they are dependent over K. Hence there exist c0 ; : : : ; cn 2 K not all zero such that c0 C c1 a C C cn an D 0: Let p.x/ D c0 C c1 x C C cn x n . Then p.x/ 2 KŒx and p.a/ D 0. Therefore a is algebraic over K. Since a was arbitrary it follows that L is an algebraic extension of K. From the previous theorem it follows that every finite extension is algebraic. The converse is not true, that is there are algebraic extensions that are not finite. We will give examples in Section 5.4. The following lemma gives some examples of algebraic and transcendental extensions. Lemma 5.2.4. CjR is algebraic but RjQ and CjQ are transcendental. If K is any field then K.x/jK is transcendental. Proof. Since 1; i constitute a basis for C over R we have jC W Rj D 2. Hence C is a finite extension of R and therefore from Theorem 5.2.3 an algebraic extension. More directly if ˛ D a C i b 2 C then ˛ is a zero of x 2 2ax C .a2 C b 2 / 2 RŒx. The existence of transcendental numbers (we will discuss these more fully in Section 5.5) shows that both RjQ and CjQ are transcendental extensions. Finally the element x 2 K.x/ is not a zero of any polynomial in KŒx. Therefore x is a transcendental element so the extension K.x/jK is transcendental.
5.3
Minimal Polynomials and Simple Extensions
If LjK is a field extension and a 2 L is algebraic over K then p.a/ D 0 for some polynomial p.x/ 2 KŒx. In this section we consider the smallest such polynomial and tie it to a simple extension of K. Definition 5.3.1. Suppose that LjK is a field extension and a 2 L is algebraic over K. The polynomial ma .x/ 2 KŒx is the minimal polynomial of a over K if
Section 5.3 Minimal Polynomials and Simple Extensions
71
(1) ma .x/ has leading coefficient 1, that is, it is a monic polynomial. (2) ma .a/ D 0. (3) If f .x/ 2 KŒx with f .a/ D 0 then ma .x/jf .x/. Hence ma .x/ is the monic polynomial of minimal degree that has a as a zero. We prove next that every algebraic element has such a minimal polynomial. Theorem 5.3.2. Suppose that LjK is a field extension and a 2 L is algebraic over K. Then (1) The minimal polynomial ma .x/ 2 KŒx exists and is irreducible over K. (2) KŒa Š K.a/ Š KŒx=.ma .x// where .ma .x// is the principal ideal in KŒx generated by ma .x/. (3) jK.a/ W Kj D deg.ma .x//. Therefore K.a/jK is a finite extension. Proof. (1) Suppose that a 2 L is algebraic over K. Let I D ¹f .x/ 2 KŒx W f .a/ D 0º: Since a is algebraic I ¤ ;. It is straightforward to show (see exercises) that I is an ideal in KŒx. Since K is a field we have that KŒx is a principal ideal domain. Hence there exists g.x/ 2 KŒx with I D .g.x//. Let b be the leading coefficient of g.x/. Then ma .x/ D b 1 g.x/ is a monic polynomial. We claim that ma .x/ is the minimal polynomial of a and that ma .x/ is irreducible. First it is clear that I D .g.x// D .ma .x//. If f .x/ 2 KŒx with f .a/ D 0 then f .x/ D h.x/ma .x/ for some h.x/. Therefore ma .x/ divides any polynomial that has a as a zero. It follows that ma .x/ is the minimal polynomial. Suppose that ma .x/ D g1 .x/g2 .x/. Then since ma .a/ D 0 it follows that either g1 .a/ D 0 or g2 .a/ D 0. Suppose g1 .a/ D 0. Then from above ma .x/jg1 .x/ and since g1 .x/jma .x/ we must then have that g2 .x/ is a unit. Therefore ma .x/ is irreducible. (2) Consider the map W KŒx ! KŒa given by X X i ki x D ki a i : i
i
Then is a ring epimorphism (see exercises) and ker. / D ¹f .x/ 2 KŒx W f .a/ D 0º D .ma .x// from the argument in the proof of part (1). It follows that KŒx=.ma .x// Š KŒa:
72
Chapter 5 Field Extensions
Since ma .x/ is irreducible we have KŒx=.ma .x// is a field and therefore KŒa D K.a/. (3) Let n D deg.ma .x//. We claim that the elements 1; a; : : : ; an1 are a basis for KŒa D K.a/ over K. First suppose that n1 X
ci a i D 0
i D1
P i with not all ci D 0 and ci 2 K. Then h.a/ D 0 where h.x/ D n1 i D0 ci x . But this contradicts the fact that ma .x/ has minimal degree over all polynomials in KŒx that have a as a zero. Therefore the set 1; a; : : : ; an1 is linearly independent over K. Now let b 2 KŒa Š KŒx=.ma .x//. Then there is a g.x/ 2 KŒx with b D g.a/. By the division algorithm g.x/ D h.x/ma .x/ C r.x/ where r.x/ D 0 or deg.r.x// < deg.ma .x//. Now r.a/ D g.a/ h.a/ma .a/ D g.a/ D b: If r.x/ D 0 then b D 0. If r.x/ ¤ 0 then since deg.r.x// < n we have r.x/ D c0 C c1 x C C cn1 x n1 with ci 2 K and some ci but not all might be zero. This implies that b D r.a/ D c0 C c1 a C C cn1 an1 and hence b is a linear combination over K of 1; a; : : : ; an1 . Hence 1; a; : : : ; an1 spans KŒa over K and hence forms a basis. Theorem 5.3.3. Suppose that LjK is a field extension and a 2 L is algebraic over K. Suppose that f .x/ 2 KŒx is a monic polynomial with f .a/ D 0. Then f .x/ is the minimal polynomial if and only if f .x/ is irreducible in KŒx. Proof. Suppose that f .x/ is the minimal polynomial of a. Then f .x/ is irreducible from the previous theorem. Conversely suppose that f .x/ is monic, irreducible and f .a/ D 0. From the previous theorem ma .x/jf .x/. Since f .x/ is irreducible we have f .x/ D cma .x/ with c 2 K. However since both f .x/ and ma .x/ are monic we must have c D 1 and f .x/ D ma .x/. We now show that a finite extension of K is actually finitely generated over K and further it is generated by finitely many algebraic elements.
Section 5.3 Minimal Polynomials and Simple Extensions
73
Theorem 5.3.4. Let LjK be a field extension. Then the following are equivalent: (1) LjK is a finite extension. (2) LjK is an algebraic extension and there exist elements a1 ; : : : ; an 2 L such that L D K.a1 ; : : : ; an /. (3) There exist algebraic elements a1 ; : : : ; an 2 L such that L D K.a1 ; : : : ; an /. Proof. (1) ) (2). We have seen in Theorem 5.2.3 that a finite extension is algebraic. Suppose that a1 ; : : : ; an are a basis for L over K. Then clearly L D K.a1 ; : : : ; an /. (2) ) (3). If LjK is an algebraic extension and L D K.a1 ; : : : ; an / then each ai is algebraic over K. (3) ) (1). Suppose that there exist algebraic elements a1 ; : : : ; an 2 L such that L D K.a1 ; : : : ; an /. We show that LjK is a finite extension. We do this by induction on n. If n D 1 then L D K.a/ for some algebraic element a and the result follows from Theorem 5.3.2. Suppose now that n 2. We assume then that an extension K.a1 ; : : : ; an1 / with a1 ; : : : ; an1 algebraic elements is a finite extension. Now suppose that we have L D K.a1 ; : : : ; an / with a1 ; : : : ; an algebraic elements. Then jK.a1 ; : : : ; an / W Kj D jK.a1 ; : : : ; an1 /.an / W K.a1 ; : : : ; an1 /jjK.a1 ; : : : ; an1 / W Kj: The second term jK.a1 ; : : : ; an1 / W Kj is finite from the inductive hypothesis. The first term jK.a1 ; : : : ; an1 /.an / W K.a1 ; : : : ; an1 /j is also finite from Theorem 5.3.2 since it is a simple extension of the field K.a1 ; : : : ; an1 / by the algebraic element an . Therefore jK.a1 ; : : : ; an / W Kj is finite. Theorem 5.3.5. Suppose that K is a field and R is an integral domain with K R. Then R can be viewed as a vector space over K. If dimK .R/ < 1 then R is a field. Proof. Let r0 2 R with r0 ¤ 0. Define the map from R to R given by .r/ D rr0 : It is easy to show (see exercises) that this is a linear transformation from R to R considered as a vector space over K. Suppose that .r/ D 0. Then rr0 D 0 and hence r D 0 since r0 ¤ 0 and R is an integral domain. It follows that is an injective map. Since R is a finite dimensional vector space over K and is an injective linear transformation it follows that must also be surjective. This implies that there exists and r1 with .r1 / D 1. Then r1 r0 D 1 and hence r0 has an inverse within R. Since r0 was an arbitrary nonzero element of R it follows that R is a field.
74
Chapter 5 Field Extensions
Theorem 5.3.6. Suppose that K L M is a chain of field extensions. Then M jK is algebraic if and only if M jL is algebraic and LjK is algebraic. Proof. If M jK is algebraic then certainly M jL and LjK are algebraic. Now suppose that M jL and LjK are algebraic. We show that M jK is algebraic. Let a 2 M . Then since a is algebraic over L there exist b0 ; b1 ; : : : ; bn 2 L with b0 C b1 a C C bn an D 0: Each bi is algebraic over K and hence K.b0 ; : : : ; bn / is finite dimensional over K. Therefore K.b0 ; : : : ; bn /.a/ D K.b0 ; : : : ; bn ; a/ is also finite dimensional over K. Therefore K.b0 ; : : : ; bn ; a/ is a finite extension of K and hence an algebraic extension K. Since a 2 K.b0 ; : : : ; bn ; a/ it follows that a is algebraic over K and therefore M is algebraic over K.
5.4
Algebraic Closures
As before suppose that LjK is a field extension. Since each element of K is algebraic over K there are certainly algebraic elements over K within L. Let AK denote the set of all elements of L that are algebraic over K. We prove that AK is actually a subfield of L. It is called the algebraic closure of K within L. Theorem 5.4.1. Suppose that LjK is a field extension and let AK denote the set of all elements of L that are algebraic over K. Then AK is a subfield of L. AK is called the algebraic closure of K in L. Proof. Since K AK we have that AK ¤ ;. Let a; b 2 AK . Since a; b are both algebraic over K from Theorem 5.3.4 we have that K.a; b/ is a finite extension of K. Therefore K.a; b/ is an algebraic extension of K and hence each element of K.a; b/ is algebraic over K. Now a; b 2 K.a; b/, if b ¤ 0, and K.a; b/ is a field so a ˙ b; ab and a=b are all in K.a; b/ and hence all algebraic over K. Therefore a ˙ b; ab; a=b, if b ¤ 0, are all in AK . It follows that AK is a subfield of L. In Section 5.2 we showed that every finite extension is an algebraic extension. We mentioned that the converse is not necessarily true, that is there are algebraic extensions that are not finite. Here we give an example. Theorem 5.4.2. Let A be the algebraic closure of the rational numbers Q within the complex numbers C. Then A is an algebraic extension of Q but jA W Qj D 1. Proof. From the previous theorem A is an algebraic extension of Q. We show that it cannot be a finite extension. By Eisenstein’s criterion the rational polynomial f .x/ D x p C p is irreducible over Q for any prime p. Let a be a zero in C of f .x/. Then
Section 5.5 Algebraic and Transcendental Numbers
75
a 2 A and jQ.a/ W Qj D p. Therefore jA W Qj p for all primes p. Since there are infinitely many primes this implies that jA W Qj D 1
5.5
Algebraic and Transcendental Numbers
In this section we consider the string of field extensions Q R C. Definition 5.5.1. An algebraic number ˛ is an element of C which is algebraic over Q. Hence an algebraic number is an ˛ 2 C such that f .˛/ D 0 for some f .x/ 2 QŒx. If ˛ 2 C is not algebraic it is transcendental. We will let A denote the totality of algebraic numbers within the complex numbers C, and T the set of transcendentals so that C D A [ T . In the language of the last subsection, A is the algebraic closure of Q within C. As in the general case, if ˛ 2 C is algebraic we will let m˛ .x/ denote the minimal polynomial of ˛ over Q. We now examine the sets A and T more closely. Since A is precisely the algebraic closure of Q in C we have from our general result that A actually forms a subfield of C. Further since the intersection of subfields is again a subfield it follows that A0 D A \ R the real algebraic numbers form a subfield of the reals. Theorem 5.5.2. The set A of algebraic numbers forms a subfield of C. The subset A0 D A \ R of real algebraic numbers forms a subfield of R. Since each rational is algebraic it is clear p that there are algebraic numbers. Further there are irrational algebraic numbers, 2 for example, since it satisfies the irreducible polynomial x 2 2 D 0 over Q. On the other hand we haven’t examined the question of whether transcendental numbers really exist. To show that any particular complex number is transcendental is in general quite difficult. However it is relatively easy to show that there are uncountably infinitely many transcendentals. Theorem 5.5.3. The set A of algebraic numbers is countably infinite. Therefore T the set of transcendental numbers and T 0 D T \ R, the real transcendental numbers, are uncountably infinite. Proof. Let Pn D ¹f .x/ 2 QŒx W deg.f .x// nº: Since if f .x/ 2 Pn , f .x/ D qo C q1 x C C qn x n with qi 2 Q we can identify a polynomial of degree n with an .n C 1/-tuple .q0 ; q1 ; : : : ; qn / of rational numbers. Therefore the set Pn has the same size as the .n C 1/-fold Cartesian product of Q: QnC1 D Q Q Q:
76
Chapter 5 Field Extensions
Since a finite Cartesian product of countable sets is still countable it follows that Pn is a countable set. Now let [ Bn D ¹roots ofp.x/º; p.x/2Pn
that is Bn is the union of all roots in C of all rational polynomials of degree n. Since each such p.x/ has a maximum of n roots and since Pn is countable it follows that Bn is a countable union of finite sets and hence is still countable. Now AD
1 [
Bn
nD1
so A is a countable union of countable sets and is therefore countable. Since both R and C are uncountably infinite the second assertions follow directly from the countability of A. If say T were countable then C D A [ T would also be countable which is a contradiction. From Theorem 5.5.3 we know that there exist infinitely many transcendental numbers. Liouville in 1851 gave the first proof of the existence of transcendentals by exhibiting a few. He gave as one the following example. Theorem 5.5.4. The real number cD
1 X 1 10j Š j D1
is transcendental. P Proof. First of all since 101j Š < 101j , and j1D1 101j is a convergent geometric series, it follows from the comparison test P that the infinite series defining c converges and defines a real number. Further since j1D1 101j D 19 , it follows that c < 19 < 1. Suppose that c is algebraic so that g.c/ D 0 for some rational nonzero polynomial g.x/. Multiplying through by the least common multiple of all the denominators Pn in g.x/j we may suppose that f .c/ D 0 for some integral polynomial f .x/ D j D0 mj x . Then c satisfies n X j D0
for some integers m0 ; : : : ; mn .
mj c j D 0
77
Section 5.5 Algebraic and Transcendental Numbers
If 0 < x < 1 then by the triangle inequality ˇ n ˇ X n ˇX ˇ 0 j 1 ˇ ˇ j mj x jj mj j D B jf .x/j D ˇ ˇ j D1
j D1
where B is a real constant depending only on the coefficients of f .x/. Now let k X 1 ck D 10j Š j D1
be the k-th partial sum for c. Then jc ck j D
1 X j DkC1
1 1 < 2 .kC1/Š : j Š 10 10
Apply the mean value theorem to f .x/ at c and ck to obtain jf .c/ f .ck /j D jc ck jjf 0 . /j for some with ck < < c < 1. Now since 0 < < 1 we have jc ck jjf 0 . /j < 2B
1 10.kC1/Š
:
On the other hand, since f .x/ can have at most n roots, it follows that for all k large enough we would have f .ck / ¤ 0. Since f .c/ D 0 we have ˇ ˇX ˇ n 1 j ˇˇ ˇ jf .c/ f .ck /j D jf .ck /j D ˇ mj ck ˇ > nkŠ 10 j D1 since for each j , mj ckj is a rational number with denominator 10j kŠ . However if k is chosen sufficiently large and n is fixed we have 2B 1 > .kC1/Š nkŠ 10 10 contradicting the equality from the mean value theorem. Therefore c is transcendental. In 1873 Hermite proved that e is transcendental while Lindemann in 1882 showed that is transcendental. Schneider in 1934 showed that ab is transcendental if a ¤ 0, a and b are algebraic and b is irrational. Later in the book we will prove that both e and are transcendental. An interesting open question is the following: Is transcendental over Q.e/? To close this section we show that in general if a 2 L is transcendental over K then K.a/jK is isomorphic to the field of rational functions over K.
78
Chapter 5 Field Extensions
Theorem 5.5.5. Suppose that LjK is a field extension and a 2 L is transcendental over K. Then K.a/jK is isomorphic to K.x/jK. Here the isomorphism W K.x/ ! K.a/ can be chosen such that .x/ D a. Proof. Define the map W K.x/ ! K.a/ by f .a/ f .x/ D g.x/ g.a/ for f .x/; g.x/ 2 KŒx with g.x/ ¤ 0. Then is a homomorphism and .x/ D a. Since ¤ 0 it follows that is an isomorphism.
5.6 1.
Exercises Let a 2 C with a3 2a C 2 D 0 and b D a2 a. Compute the minimal polynomial mb .x/ of b over Q and compute the inverse of b in Q.a/.
Determine the algebraic closure of R in C.x/. p 2n 2 2 R, n D 1; 2; 3; : : : and A WD ¹an W n 2 Nº and E WD Q.A/. 3. Let an WD Show:
2.
(i) jQ.an / W Qj D 2n . (ii) jE W Qj D 1. S (iii) E D 1 nD1 Q.an /. (iv) E is algebraic over Q. 4.
Determine jE W Qj for p p (i) E D Q. 2; 2/. p p p (ii) E D Q. 3; 3 C 3 3/.
p ; 1Ci p /. (iii) E D Q. 1Ci 2 2 p p p p p 5. Show that Q. 2;p 3/pD ¹a C b 2 C c 3 C d 6 W a; b;p c; d p 2 Qº. Determine p 2; 3/ over Q. Further show that Q. 2; 3/ D Q. 2 C the degree of Q. p 3/.
6.
Let K, E be fields and a 2 E be transcendental over K. Show: (i) Each element of K.a/jK is transcendental over K. (ii) an is transcendental over K for each n > 1. 3
a (iii) If L WD K. aC1 / then a is algebraic over L. Determine the minimal polynomial ma .x/ of a over L.
79
Section 5.6 Exercises
7.
Let K be a field and a 2 K.x/jK. Show: (i) x is algebraic over K.a/. (ii) If L is a field with K L K.x/, then jK.x/ W Lj < 1. (iii) a is transcendental over K.
8.
Suppose that a 2 L is algebraic over K. Let I D ¹f .x/ 2 KŒx W f .a/ D 0º: Since a is algebraic I ¤ ;. Prove that I is an ideal in KŒx.
9.
Prove that there are uncountably many transcendental numbers. To do this show that the set A of algebraic numbers is countable. To do this: (i) Show that Qn Œx the set of rational polynomials of degree n is countable (finite Cartesian product of countable sets). (ii) Let Bn D ¹Zeros of polynomials in Qn º. Show that B is countable. S (iii) Show that A D 1 nD1 Bn and conclude that A is countable. (iv) Show that the transcendental numbers are uncountable.
10. Consider the map W KŒx ! KŒa given by X X i ki x D ki a i : i
i
Show that is a ring epimorphism. 11. Suppose that K is a field and R is an integral domain with K R. Then R can be viewed as a vector space over K. Let r0 2 R with r0 ¤ 0. Define the map from R to R given by .r/ D rr0 : Show that this is a linear transformation from R to R considered as a vector space over K.
Chapter 6
Field Extensions and Compass and Straightedge Constructions
6.1
Geometric Constructions
Greek mathematicians in the classical period posed the problem of constructing certain geometric figures in the Euclidean plane using only a straightedge and a compass. These are known as geometric construction problems. Recall from elementary geometry that using a straightedge and compass it is possible to draw a line parallel to a given line segment through a given point, to extend a given line segment, and to erect a perpendicular to a given line at a given point on that line. There were other geometric construction problems that the Greeks could not determine straightedge and compass solutions but on the other hand were never able to prove that such constructions were impossible. In particular there were four famous insolvable (to the Greeks) construction problems. The first is the squaring of the circle. This problem is, given a circle, to construct using straightedge and compass a square having area equal to that of the given circle. The second is the doubling of the cube. This problem is given a cube of given side length, to construct using a straightedge and compass, a side of a cube having double the volume of the original cube. The third problem is the trisection of an angle. This problem is to trisect a given angle using only a straightedge and compass. The final problem is the construction of a regular n-gon. This problems asks which regular n-gons could be constructed using only straightedge and compass. By translating each of these problems into the language of field extensions we can show that each of the first three problems are insolvable in general and we can give the complete solution to the construction of the regular n-gons.
6.2
Constructible Numbers and Field Extensions
We now translate the geometric construction problems into the language of field extensions. As a first step we define a constructible number. Definition 6.2.1. Suppose we are given a line segment of unit length. An ˛ 2 R is constructible if we can construct a line segment of length j˛j in a finite number of steps from the unit segment using a straightedge and compass.
Section 6.2 Constructible Numbers and Field Extensions
81
Our first result is that the set of all constructible numbers forms a subfield of R. Theorem 6.2.2. The set C of all constructible numbers forms a subfield of R. Further, Q C . Proof. Let C be the set of all constructible numbers. Since the given unit length segment is constructible, we have 1 2 C. Therefore, C ¤ ;, and thus to show that it is a field we must show that it is closed under the field operations. Suppose ˛; ˇ are constructible. We must show then that ˛ ˙ ˇ; ˛ˇ, and ˛=ˇ for ˇ ¤ 0 are constructible. If ˛; ˇ > 0, construct a line segment of length j˛j. At one end of this line segment extend it by a segment of length jˇj. This will construct a segment of length ˛ C ˇ. Similarly, if ˛ > ˇ, lay off a segment of length jˇj at the beginning of a segment of length j˛j. The remaining piece will be ˛ ˇ. By considering cases we can do this in the same manner if either ˛ or ˇ or both are negative. These constructions are pictured in Figure 6.1. Therefore, ˛ ˙ ˇ are constructible.
Figure 6.1 In Figure 6.2 we show how to construct ˛ˇ. Let the line segment OA have length j˛j. Consider a line L through O not coincident with OA. Let OB have length jˇj as in the diagram. Let P be on ray OB so that OP has length 1. Draw AP and then find Q on ray OA such that BQ is parallel to AP . From similar triangles we then have jOP j jOBj
D
jOAj jOQj
)
j˛j 1 D : jˇj jOQj
Then jOQj D j˛jjˇj, and so ˛ˇ is constructible.
Figure 6.2
82
Chapter 6 Field Extensions and Compass and Straightedge Constructions
A similar construction, pictured in Figure 6.3, shows that ˛=ˇ for ˇ ¤ 0 is constructible. Find OA; OB; OP as above. Now, connect A to B and let PQ be parallel to AB. From similar triangles again we have jOQj j˛j 1 D H) D jOQj: jˇj j˛j jˇj Hence ˛=ˇ is constructible.
Figure 6.3 Therefore, C is a subfield of R. Since char C D 0, it follows that Q C . Let us now consider analytically how a constructible number is found in the plane. Starting at the origin and using the unit length and the constructions above, we can locate any point in the plane with rational coordinates. That is, we can construct the point P D .q1 ; q2 / with q1 ; q2 2 Q. Using only straightedge and compass, any further point in the plane can be determined in one of the following three ways. 1. The intersection point of two lines each of which passes through two known points each having rational coordinates. 2. The intersection point of a line passing through two known points having rational coordinates and a circle whose center has rational coordinates and whose radius squared is rational. 3. The intersection point of two circles each of whose centers has rational coordinates and each of whose radii is the square root of a rational number. Analytically, the first case involves the solution of a pair of linear equations each with rational coefficients and thus only leads to other rational numbers. In cases two and three we must solve equations of the form x 2 C y 2 C ax C by C c D 0, with a; b; c 2 Q. These will then be quadratic equations over Q, and thus the solutions p will either be in Q or in a quadratic extension Q. ˛/ of Q. Once a real quadratic extension of Q is found, the process can be iterated. Conversely it can be shown that p if ˛ is constructible, so is ˛. We thus can prove the following theorem.
Section 6.3 Four Classical Construction Problems
83
Theorem 6.2.3. If is constructible with … Q, then there exists a finite number of elements ˛1 ; : : : ; ˛r 2 R with ˛r D such that for i D 1; : : : ; r, Q.˛1 ; : : : ; ˛i / is a quadratic extension of Q.˛1 ; : : : ; ˛i 1 /. In particular, jQ. / W Qj D 2n for some n 1. Therefore, the constructible numbers are precisely those real numbers that are contained in repeated quadratic extensions of Q. In the next section we use this idea to show the impossibility of the first three mentioned construction problems.
6.3
Four Classical Construction Problems
We now consider the aforementioned construction problems. Our main technique will be to use Theorem 6.2.3. From this result we have that if is constructible with
… Q, then jQ. / W Qj D 2n for some n 1.
6.3.1 Squaring the Circle Theorem 6.3.1. It is impossible to square the circle. That is, it is impossible in general, given a circle, to construct using straightedge and compass a square having area equal to that of the given circle. Proof. Suppose the given circle has radius 1. It is then constructible and would have p an area of . A corresponding square would then have to have a side of length . To be constructible a number ˛ must have jQ.˛/ W Qj D 2m < 1 and hence ˛ must p be algebraic. However is transcendental, so is also transcendental and therefore not constructible.
6.3.2 The Doubling of the Cube Theorem 6.3.2. It is impossible to double the cube. This means that it is impossible in general, given a cube of given side length, to construct using a straightedge and compass, a side of a cube having double the volume of the original cube. Proof. Let the given side length be 1, so that the original volume is also 1. To double this we would have to construct a side of length 21=3 . However jQ.21=3 / W Qj D 3 since the minimal polynomial over Q is m21=3 .x/ D x 3 2. This is not a power of 2 so 21=3 is not constructible.
6.3.3 The Trisection of an Angle Theorem 6.3.3. It is impossible to trisect an angle. This means that it is impossible in general to trisect a given angle using only a straightedge and compass.
84
Chapter 6 Field Extensions and Compass and Straightedge Constructions
Proof. An angle is constructible if and only if a segment of length jcos j is constructible. Since cos.=3/ D 1=2, therefore =3 is constructible. We show that it cannot be trisected by straightedge and compass. The following trigonometric identity holds cos.3 / D 4 cos3 . / 3 cos. /: Let ˛ D cos.=9/. From the above identity we have 4˛ 3 3˛ 12 D 0. The polynomial 4x 3 3x 12 is irreducible over Q, and hence the minimal polynomial over Q is m˛ .x/ D x 3 34 x 18 . It follows that jQ.˛/ W Qj D 3, and hence ˛ is not constructible. Therefore, the corresponding angle =9 is not constructible. Therefore, =3 is constructible, but it cannot be trisected.
6.3.4 Construction of a Regular n-Gon The final construction problem we consider is the construction of regular n-gons. The algebraic study of the constructibility of regular n-gons was initiated by Gauss in the early part of the nineteenth century. Notice first that a regular n-gon will be constructible for n 3 if and only if 2 the angle 2 n is constructible, which is the case if and only if the length cos n is a 2 constructible number. From our techniques if cos n is a constructible number then m necessarily jQ.cos. 2 n // W Qj D 2 for some m. After we discuss Galois theory we see that this condition is also sufficient. Therefore cos 2 n is a constructible number if 2 m and only if jQ.cos. n // W Qj D 2 for some m. The solution of this problem, that is the determination of when jQ.cos. 2 n / W Qj D m 2 involves two concepts from number theory; the Euler phi-function and Fermat primes. Definition 6.3.4. For any natural number n, the Euler phi-function is defined by .n/ D number of integers less than or equal to n and relatively prime to n: Example 6.3.5. .6/ D 2 since among 1; 2; 3; 4; 5; 6 only 1; 5 are relatively prime to 6. It is fairly straightforward to develop a formula for .n/. A formula is first determined for primes and for prime powers and then pasted back together via the fundamental theorem of arithmetic. Lemma 6.3.6. For any prime p and m > 0, 1 : .p m / D p m p m1 D p m 1 p
85
Section 6.3 Four Classical Construction Problems
Proof. If 1 a p then either a D p or .a; p/ D 1. It follows that the positive integers less than or equal to p m which are not relatively prime to p m are precisely the multiples of p that is p; 2p; 3p; : : : ; p m1 p. All other positive a < p m are relatively prime to p m . Hence the number relatively prime to p m is p m p m1 : Lemma 6.3.7. If .a; b/ D 1 then .ab/ D .a/.b/. Proof. Given a natural number n a reduced residue system modulo n is a set of integers x1 ; : : : ; xk such that each xi is relatively prime to n, xi ¤ xj mod n unless i D j and if .x; n/ D 1 for some integer x then x xi mod n for some i. Clearly .n/ is the size of a reduced residue system modulo n. Let Ra D ¹x1 ; : : : ; x.a/ º be a reduced residue system modulo a, Rb D ¹y1 ; : : : ; y.b/ º be a reduced residue system modulo b, and let S D ¹ayi C bxj W i D 1; : : : ; .b/; j D 1; : : : ; .a/º: We claim that S is a reduced residue system modulo ab. Since S has .a/.b/ elements it will follow that .ab/ D .a/.b/. To show that S is a reduced residue system modulo ab we must show three things: first that each x 2 S is relatively prime to ab; second that the elements of S are distinct; and finally that given any integer n with .n; ab/ D 1 then n s mod ab for some s 2 S . Let x D ayi C bxj . Then since .xj ; a/ D 1 and .a; b/ D 1 it follows that .x; a/ D 1. Analogously .x; b/ D 1. Since x is relatively prime to both a and b we have .x; ab/ D 1. This shows that each element of S is relatively prime to ab. Next suppose that ayi C bxj ayk C bxl
mod ab:
Then abj.ayi C bxj / .ayk C bxl / H) ayi ayk
mod b:
Since .a; b/ D 1 it follows that yi yk mod b. But then yi D yk since Rb is a reduced residue system. Similarly xj D xl . This shows that the elements of S are distinct modulo ab. Finally suppose .n; ab/ D 1. Since .a; b/ D 1 there exist x; y with ax C by D 1. Then anx C bny D n: Since .x; b/ D 1 and .n; b/ D 1 it follows that .nx; b/ D 1. Therefore there is an si with nx D si C t b. In the same manner .ny; a/ D 1 and so there is an rj with
86
Chapter 6 Field Extensions and Compass and Straightedge Constructions
ny D rj C ua. Then a.si C t b/ C b.rj C ua/ D n H) n D asi C brj C .t C u/ab H) n ari C bsj
mod ab
and we are done. We now give the general formula for .n/. e Theorem 6.3.8. Suppose n D p1e1 pkk then
.n/ D .p1e1 p1e1 1 /.p2e2 p2e2 1 / .pkek pkek 1 /: Proof. From the previous lemma we have e
.n/ D .p1e1 /.p2e2 / .pkk / D .p1e1 p1e1 1 /.p2e2 p2e2 1 / .pkek pkek 1 / D p1e1 .1 1=p1 / pkek .1 1=pk / D p1e1 pkek .1 1=p1 / .1 1=pk / Y D n .1 1=pi /: i
Example 6.3.9. Determine .126/. Now 126 D 2 32 7 H) .126/ D .2/.32 /.7/ D .1/.32 3/.6/ D 36: Hence there are 36 units in Z126 . An interesting result with many generalizations in number theory is the following. Theorem 6.3.10. For n > 1 and for d 1 X .d / D n: d jn
Proof. We first prove the theorem for prime powers and then paste together via the fundamental theorem of arithmetic. Suppose that n D p e for p a prime. Then the divisors of n are 1; p; p 2 ; : : : ; p e , so X .d / D .1/ C .p/ C .p 2 / C C .p e / d jn
D 1 C .p 1/ C .p 2 p/ C C .p e p e1 /: Notice that this sum telescopes, that is 1 C .p 1/ D p; p C .p 2 p/ D p 2 and so on. Hence the sum is just p e and the result is proved for n a prime power.
87
Section 6.3 Four Classical Construction Problems
We now do an induction on the number of distinct prime factors of n. The above argument shows that the result is true if n has only one distinct prime factor. Assume that the result is true whenever an integer has less than k distinct prime factors and suppose n D p1e1 pkek has k distinct prime factors. Then n D p e c where p D p1 , e D e1 and c has fewer than k distinct prime factors. By the inductive hypothesis X
.d / D c:
d jc
Since .c; p/ D 1 the divisors of n are all of the form p ˛ d1 where d1 jc and ˛ D 0; 1; : : : ; e. It follows that X X X X .d / D .d1 / C .pd1 / C C .p e d1 /: d jn
d1 jc
d1 jc
d1 jc
Since .d1 ; p ˛ / D 1 for any divisor of c this sum equals X
.d1 / C
d1 jc
D
X
X
.p/.d1 / C C
d1 jc
.d1 / C .p 1/
d1 jc
X
X
.p e /.d1 /
d1 jc
.d1 / C C .p e p e1 /
d1 jc 2
X
.d1 /
d1 jc e
D c C .p 1/c C .p p/c C C .p p
e1
/c:
As in the case of prime powers this sum telescopes giving a final result X
.d / D p e c D n:
d jn
Example 6.3.11. Consider n D 10. The divisors are 1; 2; 5; 10. Then .1/ D 1, .2/ D 1, .5/ D 4, .10/ D 4. Then .1/ C .2/ C .5/ C .10/ D 1 C 1 C 4 C 4 D 10: We will see later in the book that the Euler phi-function plays an important role in the structure theory of abelian groups. We now turn to Fermat primes. Definition 6.3.12. The Fermat numbers are the sequence .Fn / of positive integers defined by n Fn D 22 C 1; n D 0; 1; 2; 3; : : : : If a particular Fn is prime it is called a Fermat prime.
88
Chapter 6 Field Extensions and Compass and Straightedge Constructions
Fermat believed that all the numbers in this sequence were primes. In fact F0 ; F1 ; F2 ; F3 ; F4 are all prime but F5 is composite and divisible by 641 (see exercises). It is still an open question whether or not there are infinitely many Fermat primes. It has been conjectured that there are only finitely many. On the other hand if a number of the form 2n C 1 is a prime for some integer n then it must be a Fermat prime. Theorem 6.3.13. If a 2 and an C 1, n 1, is a prime then a is even and n D 2m for some nonnegative integer m. In particular if p D 2k C 1, k 1, is a prime then k D 2n for some n and p is a Fermat prime. Proof. If a is odd then an C 1 is even and hence not a prime. Suppose then that a is even and n D kl with k odd and k 3. Then akl C 1 D a.k1/l a.k2/l C C 1: al C 1 Therefore al C 1 divides akl C 1 if k 3. Hence if an C 1 is a prime we must have n D 2m . We can now state the solution to the constructibility of regular n-gons. Theorem 6.3.14. A regular n-gon is constructible with a straightedge and compass if and only if n D 2m p1 pk where p1 ; : : : ; pk are distinct Fermat primes. Before proving the theorem notice for example that a regular 20-gon is constructible since 20 D 22 5 and 5 is a Fermat prime. On the other hand a regular 11-gon is not constructible. Proof. Let D e
2 i n
be a primitive n-th root of unity. Since 2 i 2 2 e n D cos C i sin n n
is easy to compute that (see exercises) 2 1 : C D 2 cos n Therefore Q. C we will prove that
1 /
D Q.cos. 2 n //. After we discuss Galois theory in more detail ˇ ˇ ˇ ˇ ˇQ C 1 W Qˇ D .n/ ˇ ˇ 2
where .n/ is the Euler phi-function. Therefore cos. 2 n / is constructible if and only .n/ if 2 and hence .n/ is a power of 2.
89
Section 6.4 Exercises
Suppose that n D 2m p1e1 pkek , all pi odd primes. Then from Theorem 6.3.8 .n/ D 2m1 .p1e1 p1e1 1 /.p2e2 p2e2 1 / .pkek pkek 1 /: If this was a power of 2 each factor must also be a power of 2. Now piei piei 1 D piei 1 .pi 1/: If this is to be a power of 2 we must have ei D 1 and pi 1 D 2ki for some ki . Therefore each prime is distinct to the first power and pi D 2ki C 1 is a Fermat prime proving the theorem.
6.4
Exercises
1. Let be a given angle. In which of the following cases is the angle from the angle by compass and straightedge? (a) D (b) D (c) D
13 , 33 , 7,
constructible
26 . D 11 . D 12 .
D
2. (The golden section) In the plane let AB be a given segment from A to B with length a. The segment AB should be divided such that the proportion of AB to the length of the bigger subsegment is equal to the proportion of the length of the bigger subsegment to the length of the smaller subsegment: b a D ; b ab where b is the length of the bigger subsegment. Such a division is called division x by the golden section. If we write b D ax, 0 < x < 1, then x1 D 1x , that is 2 D 1 x. Show: x (a)
1 x
D
p 1C 5 2
D ˛.
(b) Construct the division of AB by the golden section with compass and straightedge. (c) If we divide the radius r > 0 of a circle by the golden section, then the bigger part of the so divided radius is the side of the regular 10-gon with its 10 vertices on the circle. 3. Given a regular 10-gon such that the 10 vertices are on the circle with radius R > 0. Show that the length of each side is equal to the bigger part of the, by the golden section divided, radius. Describe the procedure of the construction of the regular 10-gon and 5-gon.
90
Chapter 6 Field Extensions and Compass and Straightedge Constructions
4. Construct the regular 17-gon with compass and straightedge. Hint: We have to 2 i 17 . First, construct construct the number 12 .! C ! 1 / D cos 2 17 , where ! D e 2 the positive zero !1 of the polynomial x C x 4; we get 1 p !1 D . 17 1/ D ! C ! 1 C ! 2 C ! 2 C ! 4 C ! 4 C ! 8 C ! 8 : 2 Then, construct the positive zero !2 of the polynomial x 2 !1 x 1; we get q p 1 p 17 1 C 34 2 17 D ! C ! 1 C ! 4 C ! 4 : !2 D 4 From !1 and !2 construct ˇ D 12 .!22 !1 C !2 4/. Then !3 D 2 cos 2 17 is the biggest of the two positive zeros of the polynomial x 2 !2 x C ˇ. 5. The Fibonacci-numbers fn , n 2 N [ ¹0º, are defined by f0 D 0, f1 D 1 and fnC2 D fnC1 C fn for n 2 N [ ¹0º. Show: (a) fn D
˛ n ˇ n ˛ˇ
with ˛ D
p 1C 5 2 ,
ˇD
f
(b) . nC1 / converges and limn!1 fn n2N 0 1 n fn1 fn (c) 1 1 D fn fnC1 , n 2 N.
p 1 5 2 .
fnC1 fn
D
p 1C 5 2
D ˛.
(d) f1 C f2 C C fn D fnC2 1, n 1. (e) fn1 fnC1 fn2 D .1/n , n 2 N. (f) f12 C f22 C C fn2 D fn fnC1 , n 2 N. (g) The Fermat numbers F0 ; F1 ; F2 ; F3 ; F4 . are all prime but F5 is composite and divisible by 641. 6. Let D e
show that
2 i n
be a primitive n-th root of unity. Using 2 i 2 2 e n D cos C i sin n n 2 1 C D 2 cos : n
Chapter 7
Kronecker’s Theorem and Algebraic Closures
7.1
Kronecker’s Theorem
In the last chapter we proved that if LjK is a field extension then there exists an intermediate field K A L such that A is algebraic over K and contains all the elements of L that are algebraic over K. We call A the algebraic closure of K within L. In this chapter we prove that starting with any field K we can construct an extension field K that is algebraic over K and is algebraically closed. By this we mean that there are no algebraic extensions of K or equivalently that there are no irreducible nonlinear polynomials in KŒx. In the final section of this chapter we will give a proof of the famous fundamental theorem of algebra which in the language of this chapter says that the field C of complex numbers is algebraically closed. We will present another proof of this important result later in the book after we discuss Galois theory. First we need the following crucial result of Kronecker which says that given a polynomial f .x/ in KŒx where K is a field we can construct an extension field L of K in which f .x/ has a root ˛. We say that L has been constructed by adjoining ˛ to K. Recall that if f .x/ 2 KŒx is irreducible then f .x/ can have no roots in K. We first need the following concept. Definition 7.1.1. Let LjK and L0 jK be field extensions. Then a K-isomorphism is an isomorphism W L ! L0 that is the identity map on K, that is fixes each element of K. Theorem 7.1.2 (Kronecker’s theorem). Let K be a field and f .x/ 2 KŒx. Then there exists a finite extension K 0 of K where f .x/ has a root. Proof. Suppose that f .x/ 2 KŒx. We know that f .x/ factors into irreducible polynomials. Let p.x/ be an irreducible factor of f .x/. From the material in Chapter 4 we know that since p.x/ is irreducible the principal ideal hp.x/i in KŒx is a maximal ideal. To see this suppose that g.x/ … hp.x/i, so that g.x/ is not a multiple of p.x/. Since p.x/ is irreducible, it follows that .p.x/; g.x// D 1. Thus there exist h.x/; k.x/ 2 KŒx with h.x/p.x/ C k.x/g.x/ D 1:
92
Chapter 7 Kronecker’s Theorem and Algebraic Closures
The element on the left is in the ideal .g.x/; .p.x//, so the identity, 1, is in this ideal. Therefore, the whole ring KŒx is in this ideal. Since g.x/ was arbitrary, this implies that the principal ideal hp.x/i is maximal. Now let K 0 D KŒx=hp.x/i. Since hp.x/i is a maximal ideal it follows that K 0 is a field. We show that K can be embedded in K 0 and that p.x/ has a zero in K 0 . First consider the map ˛ W KŒx ! K 0 by ˛.f .x// D f .x/ C hp.x/i. This is a homomorphism. Since the identity element 1 2 K is not in hp.x/i it follows that ˛ restricted to K is nontrivial. Therefore ˛ restricted to K is a monomorphism since if ker.˛jK / ¤ K then ker.˛jK / D ¹0º. Therefore K can be embedded into ˛.K/ which is contained in K 0 . Therefore K 0 can be considered as an extension field of K. Consider the element a D x C hp.x/i 2 K 0 . Then p.a/ D p.x/ C hp.x/i D 0 C hp.x/i since p.x/ 2 hp.x/i. But 0 C hp.x/i is the zero element 0 of the factor ring KŒx=hp.x/i. Therefore in K 0 we have p.a/ D 0 and hence p.x/ has a zero in K 0 . Since p.x/ divides f .x/ we must have f .a/ D 0 in K 0 also. Therefore we have constructed an extension field of K in which f .x/ has a zero. We now outline a slightly more constructive proof of Kronecker’s theorem. From this construction we say that the field K 0 constructed by adjoining the root ˛ to K. Proof of Kronecker’s theorem. We can assume that f .x/ is irreducible. Suppose that f .x/ D a0 C a1 x C C an x n with an ¤ 0. Define ˛ to satisfy a0 C a1 ˛ C C an ˛ n D 0: Now define K 0 D K.˛/ in the following manner. We let K.˛/ D ¹c0 C c1 ˛ C C cn1 ˛ n1 W ci 2 Kº: Then on K.˛/ define addition and subtraction componentwise and define multiplication by algebraic manipulation, replacing powers of ˛ higher than ˛ n by using ˛n D
a0 a1 ˛ an1 ˛ n1 : an
We claim that K 0 D K.˛/ then forms a field of finite degree over K. The basic ring properties follow easily by computation (see exercises) using the definitions. We must show then that every nonzero element of K.˛/ has a multiplicative inverse. Let g.˛/ 2 K.˛/. Then the corresponding polynomial g.x/ 2 KŒx is a polynomial of degree n 1. Since f .x/ is irreducible of degree n it follows that f .x/ and g.x/ must be relatively prime, that is .f .x/; g.x// D 1. Hence there exist a.x/; b.x/ 2 KŒx with a.x/f .x/ C b.x/g.x/ D 1: Evaluate these polynomials at ˛ to get a.˛/f .˛/ C b.˛/g.˛/ D 1:
93
Section 7.1 Kronecker’s Theorem
Since by definition we have f .˛/ D 0 this becomes b.˛/g.˛/ D 1: Now b.˛/ might have degree higher than n 1 in ˛. However using the relation that f .˛/ D 0 we can rewrite b.˛/ as b.˛/ where b.˛/ now has degree n 1 in ˛ and hence is in K.˛/. Therefore b.˛/g.˛/ D 1 and hence g.˛/ has a multiplicative inverse. It follows that K.˛/ is a field and by definition f .˛/ D 0. The elements 1; ˛; : : : ; ˛ n1 form a basis for K.˛/ over K and hence jK.˛/ W Kj D n: Example 7.1.3. Let f .x/ D x 2 C 1 2 RŒx. This is irreducible over R. We construct the field in which this has a root. Let ˛ be an indeterminate with ˛ 2 C 1 D 0 or ˛ 2 D 1. The extension field R.˛/ then has the form R.˛/ D ¹x C ˛y W x; y 2 R; ˛ 2 D 1º: It is clear that this field is R-isomorphic to the complex numbers C, that is, R.˛/ Š R.i/ Š C. In Chapter 5 we showed that if LjK is a field extension and a 2 L is algebraic over K then there is a smallest algebraic extension K.a/ of K within L. Further K.a/ is determined by the minimal polynomial ma .x/. The difference between this construction and the construction in Kronecker’s theorem is that in the proof of Kronecker’s theorem ˛ is defined to be the root and we constructed the field around it, whereas in the previous construction ˛ was assumed to satisfy the polynomial and K.˛/ was an already existing field that contained ˛. However the next theorem says that these constructions are the same up to K-isomorphism. Theorem 7.1.4. Let p.x/ 2 KŒx be an irreducible polynomial and let K 0 D K.˛/ be the extension field of K constructed in Kronecker’s theorem in which p.x/ has a zero ˛. Let L be an extension field of K and suppose that a 2 L is algebraic with minimal polynomial m˛ .x/ D p.x/. Then K.˛/ is K-isomorphic to K.a/. Proof. If LjK is a field extension and a 2 L is algebraic then the construction of the subfield K.a/ within L is identical to the construction outlined above for K.˛/. If deg.p.x// D n then the elements 1; a; : : : ; a n1 constitute a basis for K.a/ over K and the elements 1; ˛; : : : ; ˛ n1 constitute a basis for K.˛/ over K. The mapping W K.a/ ! K.˛/ defined by .k/ D k if k 2 K and .a/ D ˛ and then extended by linearity is easily shown to be a K-isomorphism (see exercises).
94
Chapter 7 Kronecker’s Theorem and Algebraic Closures
Theorem 7.1.5. Let K be a field. Then the following are equivalent. (1) Each nonconstant polynomial in KŒx has a zero in K. (2) Each nonconstant polynomial in KŒx factors into linear factors over K. That is, for each f .x/ 2 KŒx there exist elements a1 ; : : : ; an ; b 2 K with f .x/ D b.x a1 / .x an /: (3) An element of KŒx is irreducible if and only if it is of degree one. (4) If LjK is an algebraic extension then L D K. Proof. Suppose that each nonconstant polynomial in KŒx has a zero in K. Let f .x/ 2 KŒx with deg.f .x// D n. Suppose that a1 is a zero of f .x/ then f .x/ D .x a1 /h.x/ where the degree of h.x/ is n 1. Now h.x/ has a zero a2 in K so that f .x/ D .x a1 /.x a2 /g.x/ with deg.g.x// D n 2. Continue in this manner and f .x/ factors completely into linear factors. Hence (1) implies (2). Now suppose (2), that is that each nonconstant polynomial in KŒx factors into linear factors over K. Suppose that f .x/ is irreducible. If deg.f .x// > 1 then f .x/ factors into linear factors and hence is not irreducible. Therefore f .x/ must be of degree 1 and (2) implies (3). Now suppose that an element of KŒx is irreducible if and only if it is of degree one and suppose that LjK is an algebraic extension. Let a 2 L. Then a is algebraic over K. Its minimal polynomial ma .x/ is monic and irreducible over K and hence from (3) is linear. Therefore ma .x/ D x a 2 KŒx. It follows that a 2 K and hence K D L. Therefore (3) implies (4). Finally suppose that whenever LjK is an algebraic extension then L D K. Suppose that f .x/ is a nonconstant polynomial in KŒx. From Kronecker’s theorem there exists a field extension L and a 2 L with f .a/ D 0. However L is an algebraic extension so by supposition K D L. Therefore a 2 K and f .x/ has a zero in K. Therefore (4) implies (1) completing the proof. In the next section we will prove that given a field K we can always find an extension field K with the properties of the last theorem.
7.2
Algebraic Closures and Algebraically Closed Fields
A field K is termed algebraically closed if K has no algebraic extensions other than K itself. This is equivalent to any one of the conditions of Theorem 7.1.5.
Section 7.2 Algebraic Closures and Algebraically Closed Fields
95
Definition 7.2.1. A field K is algebraically closed if every nonconstant polynomial f .x/ 2 KŒx has a zero in K. The following theorem is just a restatement of Theorem 7.1.5. Theorem 7.2.2. A field K is algebraically closed if and only it satisfies any one of the following conditions. (1) Each nonconstant polynomial in KŒx has a zero in K. (2) Each nonconstant polynomial in KŒx factors into linear factors over K. That is, for each f .x/ 2 KŒx there exist elements a1 ; : : : ; an ; b 2 K with f .x/ D b.x a1 / .x an /: (3) An element of KŒx is irreducible if and only if it is of degree one. (4) If LjK is an algebraic extension then L D K. The prime example of an algebraically closed field is the field C of complex numbers. The fundamental theorem of algebra says that any nonconstant complex polynomial has a complex root. We now show that the algebraic closure of one field within an algebraically closed field is algebraically closed. First we define a general algebraic closure. Definition 7.2.3. An extension field K of a field K is an algebraic closure of K if K is algebraically closed and KjK is algebraic. Theorem 7.2.4. Let K be a field and LjK an extension of K with L algebraically closed. Let K D AK be the algebraic closure of K within L. Then K is an algebraic closure of K. Proof. Let K D AK be the algebraic closure of K within L. We know that KjK is algebraic therefore we must show that K is algebraically closed. Let f .x/ be a nonconstant polynomial in KŒx. Then f .x/ 2 LŒx. Since L is algebraically closed f .x/ has a zero a in L. Since f .a/ D 0 and f .x/ 2 KŒx it follows that a is algebraic over K. However K is algebraic over K and therefore a is also algebraic over K. Hence a 2 K and f .x/ has a zero in K. Therefore K is algebraically closed. We want to note the distinction between being algebraically closed and being an algebraic closure. Lemma 7.2.5. The complex numbers C is an algebraic closure of R but not an algebraic closure of Q. An algebraic closure of Q is A the field of algebraic numbers within C.
96
Chapter 7 Kronecker’s Theorem and Algebraic Closures
Proof. C is algebraically closed (the fundamental theorem of algebra) and since jC W Rj D 2 it is algebraic over R. Therefore C is an algebraic closure of R. Although C is algebraically closed and contains the rational numbers Q it is not an algebraic closure of Q since it is not algebraic over Q since there exist transcendental elements. On the other hand, A, the field of algebraic numbers within Q, is an algebraic closure of Q from Theorem 7.2.4. We now show that every field has an algebraic closure. To do this we first show that any field can be embedded into an algebraically closed field. Theorem 7.2.6. Let K be a field. Then K can be embedded into an algebraically closed field. Proof. We show first that there is an extension field L of K in which each nonconstant polynomial f .x/ 2 KŒx has a zero in L. Assign to each nonconstant f .x/ 2 KŒx the symbol yf and consider R D KŒyf W f .x/ 2 KŒx the polynomial ring over K in the variables yf . Set I D
²X n
³ fj .yfj /rj W rj 2 R; fj .x/ 2 KŒx :
j D1
It is straightforward that I is an ideal in R. Suppose that I D R. Then 1 2 I . Hence there is a linear combination 1 D g1 f1 .yf1 / C C gn fn .yfn / where gi 2 I D R. In the n polynomials g1 ; : : : ; gn there are only a finite number of variables, say for example yf1 ; : : : ; yfn ; : : : ; yfm : Hence 1D
n X
gi .yf1 ; : : : ; yfm /fi .yfi /. /:
i D1
Successive applications of Kronecker’s theorem lead us to construct an extension field P of K in which each fi has a zero ai . Substituting ai for yfi in (8) above we get that 1 D 0 a contradiction. Therefore I ¤ R. Since I is a ideal not equal to the whole ring R it follows that I is contained in a maximal ideal M of R. Set L D R=M . Since M is maximal K is a field. Now K \ M D ¹0º. If not suppose that a 2 K \ M with a ¤ 0. Then a1 a D 1 2 M
Section 7.2 Algebraic Closures and Algebraically Closed Fields
97
and then M D R. Now define W K ! L by .k/ D k C M . Since K \ M D ¹0º it follows that ker. / D ¹0º so is a monomorphism. This allows us to identify K and .K/ and shows that K embeds into L. Now suppose that f .x/ is a nonconstant polynomial in KŒx. Then f .yf C M / D f .yf / C M: However by the construction f .yf / 2 M so that f .yf C M / D M C M D the zero element of L: Therefore yf C M is a zero of f .x/. Therefore we have constructed a field L in which every nonconstant polynomial in KŒx has a zero in L. We now iterate this procedure to form a chain of fields K K1 .D L/ K2 such that each nonconstant polynomial of Ki Œx has a zero in Ki C1 . S Now let KO D I Ki . It is easy to show (see exercises) that KO is a field. If f .x/ is O a nonconstant polynomial in KŒx then there is some i with f .x/ 2 Ki Œx. Therefore O Hence f .x/ has a zero in KO and KO is algebraically f .x/ has a zero in Ki C1 Œx K. closed. Theorem 7.2.7. Let K be a field. Then K has an algebraic closure. Proof. Let KO be an algebraically closed field containing K which exists from Theorem 7.2.6. Now let K D AKO be the set of elements of KO that are algebraic over K. From Theorem 7.2.4 KO is an algebraic closure of K. The following lemma is straightforward. We leave the proof to the exercises. Lemma 7.2.8. Let K; K 0 be fields and W K ! K 0 a homomorphism. Then Q W KŒx ! K 0 Œx given by X X n n i Q ki x D ..ki //x i i D1
i D0
Q If is is also a homomorphism. By convention we identify and Q and write D . Q an isomorphism then so is .
98
Chapter 7 Kronecker’s Theorem and Algebraic Closures
Lemma 7.2.9. Let K; K 0 be fields and W K ! K 0 an isomorphism. Let f .x/ 2 KŒx be irreducible. Let K K.a/ and K 0 K 0 .a0 / where a is a zero of f .x/ and a0 is a zero of .f .x//. Then there is an isomorphism W K.a/ ! K 0 .a0 / with 0 is uniquely determined. jK D and .a/ D a . Further Proof. This is a generalized version of Theorem 7.1.4. If b 2 K.a/ then from the construction of K.a/ there is a polynomial g.x/ 2 KŒx with b D g.a/. Define a map W K.a/ ! K 0 .a0 / by .b/ D .g.x//.a0 /: We show that is an isomorphism. First is well-defined. Suppose that b D g.a/ D h.a/ with h.x/ 2 KŒx. Then .g h/.a/ D 0. Since f .x/ is irreducible this implies that f .x/ D cma .x/ and since a is a zero of .g h/.x/ then f .x/j.g h/.x/. Then .f .x//j..g.x// .h.x///: Since .f .x//.a0 / D 0 this implies that .g.x//.a0 / D .h.x//.a0 / and hence the map is well-defined. It is easy to show that is a homomorphism. Let b1 D g1 .a/, b2 D g2 .a/. Then b b1 1 D g1 g2 .a/. Hence .b1 b2 / D ..g1 g2 //.a0 / D .g1 /.a0 /.g2 /.a0 / D
.b1 / .b2 /:
In the same manner we have .b1 C b2 / D .b1 / C .b2 /. Now suppose that k 2 K so that k 2 KŒx is a constant polynomial. Then .k/ D ..k//.a0 / D .k/. Therefore restricted to K is precisely . As is not the zero mapping it follows that is a monomorphism. Finally since K.a/ is generated from K and a, and restricted to K is it follows that is uniquely determined by and .a/ D a0 . Hence is unique. Theorem 7.2.10. Let LjK be an algebraic extension. Suppose that L1 is an algebraically closed field and is an isomorphism from K to K1 L1 . Then there exists a monomorphism from L to L1 with jK D .
Section 7.2 Algebraic Closures and Algebraically Closed Fields
99
Before we give the proof we note that the theorem gives the following diagram:
In particular the theorem can be applied to monomorphisms of a field K within an algebraic closure K of K. Specifically suppose that K K where K is an algebraic closure of K and let ˛ W K ! K be a monomorphism with ˛.K/ D K. Then there exists an automorphism ˛ of K with ˛jK D ˛. Proof of Theorem 7.2.10. Consider the set M D ¹.M; / W M is a field with K M L; where there exists a monomorphism W M ! L1 with jK D º: Now the set M is nonempty since .K; / 2 M. Order M by .M1 ; 1 / < .M2 ; 2 / if M1 M2 and .2 /jM1 D 1 . Let K D ¹.Mi ; i / W i 2 I º be a chain in M. Let .M; / be defined by M D
[
Mi
with .a/ D i .a/ for all a 2 Mi :
i 2I
It is clear that M is an upper bound for the chain K. Since each chain has an upper bound it follows from Zorn’s lemma that M has a maximal element .N; /. We show that N D L. Suppose that N ¨ L. Let a 2 L n N . Then a is algebraic over N and further algebraic over K since LjK is algebraic. Let ma .x/ 2 N Œx be the minimal polynomial of a relative to N . Since L1 is algebraically closed .ma .x// has a zero a0 2 L1 . Therefore there is a monomorphism 0 W N.a/ ! L1 with 0 restricted to N the same as . It follows that .N; / < .N.a/; 0 / since a … N . This contradicts the maximality of N . Therefore N D L completing the proof. Combining the previous two theorems we can now prove that any two algebraic closures of a field K are unique up to K-isomorphism, that is up to an isomorphism that is the identity on K.
100
Chapter 7 Kronecker’s Theorem and Algebraic Closures
Theorem 7.2.11. Let L1 and L2 be algebraic closures of the field K. Then there is a K-isomorphism W L ! L1 . Again by K-isomorphism we mean that is the identity on K. Proof. From Theorem 7.2.7 there is a monomorphism W L1 ! L2 with the identity on K. However since L1 is algebraically closed so is .L1 /. Then L2 j .L1 / is an algebraic extension and since L2 is algebraically closed we must have L2 D .L1 /. Therefore is also surjective and hence an isomorphism. The following corollary is immediate. Corollary 7.2.12. Let LjK and L0 jK be field extensions with a 2 L and a0 2 L0 algebraic elements over K. Then K.a/ is K-isomorphic to K.a0 / if and only if jK.a/ W Kj D jK.a0 / W Kj and there is an element a00 2 K.a0 / with ma .x/ D ma00 .x/.
7.3
The Fundamental Theorem of Algebra
In this section we give a proof of the fact that the complex numbers form an algebraically closed field. This is known as the fundamental theorem of algebra. First we need the concept of a splitting field for a polynomial. In the next chapter we will examine this concept more deeply.
7.3.1 Splitting Fields We have just seen that given an irreducible polynomial over a field F we could always find a field extension in which this polynomial has a root. We now push this further to obtain field extensions where a given polynomial has all its roots. Definition 7.3.1. If K is a field and 0 ¤ f .x/ 2 KŒx and K 0 is an extension field of K, then f .x/ splits in K 0 (K 0 may be K), if f .x/ factors into linear factors in K 0 Œx. Equivalently, this means that all the roots of f .x/ are in K 0 . K 0 is a splitting field for f .x/ over K if K 0 is the smallest extension field of K in which f .x/ splits. (A splitting field for f .x/ is the smallest extension field in which f .x/ has all its possible roots.) K 0 is a splitting field over K if it is the splitting field for some finite set of polynomials over K. Theorem 7.3.2. If K is a field and 0 ¤ f .x/ 2 KŒx, then there exists a splitting field for f .x/ over K. Proof. The splitting field is constructed by repeated adjoining of roots. Suppose without loss of generality that f .x/ is irreducible of degree n over F . From Theorem 6.2.2 there exists a field F 0 containing ˛ with f .˛/ D 0. Then f .x/ D .x˛/g.x/ 2 F 0 Œx
Section 7.3 The Fundamental Theorem of Algebra
101
with deg g.x/ D n 1. By an inductive argument g.x/ has a splitting field and therefore so does f .x/. In the next chapter we will give a further characterization of splitting fields.
7.3.2 Permutations and Symmetric Polynomials To obtain a proof of the fundamental theorem of algebra we need to go a bit outside of our main discussions of rings and fields and introduce symmetric polynomials. In order to introduce this concept we first review some basic ideas from elementary group theory which we will look at in detail later in the book. Definition 7.3.3. A group G is a set with one binary operation which we will denote by multiplication, such that (1) The operation is associative, that is, .g1 g2 /g3 D g1 .g2 g3 / for all g1 ; g2 ; g3 2 G. (2) There exists an identity for this operation, that is, an element 1 such that 1g D g for each g 2 G. (3) Each g 2 G has an inverse for this operation, that is, for each g there exists a g 1 with the property that gg1 D 1. If in addition the operation is commutative (g1 g2 D g2 g1 for all g1 ; g2 2 G), the group G is called an abelian group. The order of G is the number of elements in G, denoted jGj. If jGj < 1; G is a finite group. H G is a subgroup if H is also a group under the same operation as G. Equivalently, H is a subgroup if H ¤ ; and H is closed under the operation and inverses. Groups most often arise from invertible mappings of a set onto itself. Such mappings are called permutations. Definition 7.3.4. If T is a set, a permutation on T is a one-to-one mapping of T onto itself. We denote by ST the set of all permutations on T . Theorem 7.3.5. For any set T , ST forms a group under composition called the symmetric group on T . If T; T1 have the same cardinality (size), then ST Š ST1 . If T is a finite set with jT j D n, then ST is a finite group and jST j D nŠ. Proof. If ST is the set of all permutations on the set T , we must show that composition is an operation on ST that is associative and has an identity and inverses. Let f; g 2 ST . Then f; g are one-to-one mappings of T onto itself. Consider f ıg W T ! T . If f ıg.t1 / D f ıg.t2 /, then f .g.t1 // D f .g.t2 // and g.t1 / D g.t2 /, since f is one-to-one. But then t1 D t2 since g is one-to-one. If t 2 T , there exists t1 2 T with f .t1 / D t since f is onto. Then there exists t2 2 T with g.t2 / D t1 since g is onto. Putting these together, f .g.t2 // D t , and
102
Chapter 7 Kronecker’s Theorem and Algebraic Closures
therefore f ı g is onto. Therefore, f ı g is also a permutation and composition gives a valid binary operation on ST . The identity function 1.t / D t for all t 2 T will serve as the identity for ST , while the inverse function for each permutation will be the inverse. Such unique inverse functions exist since each permutation is a bijection. Finally, composition of functions is always associative and therefore ST forms a group. If T; T1 have the same cardinality, then there exists a bijection W T ! T1 . Define a map F W ST ! ST1 in the following manner: if f 2 ST , let F .f / be the permutation on T1 given by F .f /.t1 / D .f . 1 .t1 ///. It is straightforward to verify that F is an isomorphism (see the exercises). Finally, suppose jT j D n < 1. Then T D ¹t1 ; : : : ; tn º. Each f 2 ST can be pictured as t1 : : : t n : f D f .t1 / : : : f .tn / For t1 there are n choices for f .t1 /. For t2 there are only n 1 choices since f is one-to-one. This continues down to only one choice for tn . Using the multiplication principle, the number of choices for f and therefore the size of ST is n.n 1/ 1 D nŠ: For a set with n elements we denote ST by Sn called the symmetric group on n symbols. Example 7.3.6. Write down the six elements of S3 and give the multiplication table for the group. Name the three elements 1; 2; 3. The six elements of S3 are then: 1 2 3 1 2 3 1 2 3 1D ; aD ; bD 1 2 3 2 3 1 3 1 2 1 2 3 1 2 3 1 2 3 cD ; dD ; eD : 2 1 3 3 2 1 1 3 2 The multiplication table for S3 can be written down directly by doing the required composition. For example, 1 2 3 1 2 3 1 2 3 ac D D D d: 2 3 1 2 1 3 3 2 1 To see this, note that a W 1 ! 2; 2 ! 3; 3 ! 1; c W 1 ! 2; 2 ! 1; 3 ! 3 and so ac W 1 ! 3; 2 ! 2; 3 ! 1. It is somewhat easier to construct the multiplication table if we make some observations. First, a2 D b and a3 D 1. Next, c 2 D 1, d D ac, e D a2 c and finally ac D ca2 .
103
Section 7.3 The Fundamental Theorem of Algebra
From these relations the following multiplication table can be constructed:
1 a a2 c ac a2 c
1 1 a a2 c ac a2 c
a a a2 1 a2 c c ac
a2 a2 1 a ac a2 c c
c c ac a2 c 1 a a2
ac ac a2 c c a2 1 a
a2 c a2 c c ac : a a2 1
To see this, consider, for example, .ac/a2 D a.ca2 / D a.ac/ D a2 c. More generally, we can say that S3 has a presentation given by S3 D ha; cI a3 D c 2 D 1; ac D ca2 i: By this we mean that S3 is generated by a; c, or that S3 has generators a; c and the whole group and its multiplication table can be generated by using the relations a3 D c 2 D 1, ac D ca2 . An important result, the form of which we will see later in our work on extension fields, is the following. Lemma 7.3.7. Let T be a set and T1 T a subset. Let H be the subset of ST that fixes each element of T1 , that is, f 2 H if f .t / D t for all t 2 T1 . Then H is a subgroup. Proof. H ¤ ; since 1 2 H . Now suppose h1 ; h2 2 H . Let t1 2 T1 and consider h1 ı h2 .t1 / D h1 .h2 .t1 //. Now h2 .t1 / D t1 since h2 2 H , but then h1 .t1 / D t1 since h1 2 H . Therefore, h1 ı h2 2 H and H is closed under composition. If h1 fixes t1 then h1 1 also fixes t1 so H is also closed under inverses and is therefore a subgroup. We now apply these ideas of permutations to certain polynomials in independent indeterminates over a field. We will look at these in detail later in this book. Definition 7.3.8. Let y1 ; : : : ; yn be (independent) indeterminates over a field K. A polynomial f .y1 ; : : : ; yn / 2 KŒy1 ; : : : ; yn is a symmetric polynomial in y1 ; : : : ; yn if f .y1 ; : : : ; yn / is unchanged by any permutation of ¹y1 ; : : : ; yn º, that is, f .y1 ; : : : ; yn / D f . .y1 /; : : : ; .yn //. If K K 0 are fields and ˛1 ; : : : ; ˛n are in F 0 , then we call a polynomial f .˛1 ; : : : ; ˛n / with coefficients in K symmetric in ˛1 ; : : : ; ˛n if f .˛1 ; : : : ; ˛n / is unchanged by any permutation of ¹˛1 ; : : : ; ˛n º.
104
Chapter 7 Kronecker’s Theorem and Algebraic Closures
Example 7.3.9. Let K be a field and k0 ; k1 2 K. Let h.y1 ; y2 / D k0 .y1 C y2 / C k1 .y1 y2 /. There are two permutations on ¹y1 ; y2 º, namely 1 W y1 ! y1 , y2 ! y2 and
2 W y1 ! y2 , y2 ! y1 . Applying either one of these two to ¹y1 ; y2 º leaves h.y1 ; y2 / invariant. Therefore, h.y1 ; y2 / is a symmetric polynomial. Definition 7.3.10. Let x; y1 ; : : : ; yn be indeterminates over a field K (or elements of an extension field K 0 of K). Form the polynomial p.x; y1 ; : : : ; yn / D .x y1 / .x yn /: The i-th elementary symmetric polynomial si in y1 ; : : : ; yn for i D 1; : : : ; n, is .1/i ai , where ai is the coefficient of x ni in p.x; y1 ; : : : ; yn /. Example 7.3.11. Consider y1 ; y2 ; y3 . Then p.x; y1 ; y2 ; y3 / D .x y1 /.x y2 /.x y3 / D x 3 .y1 C y2 C y3 /x 2 C .y1 y2 C y1 y3 C y2 y3 /x y1 y2 y3 : Therefore, the three elementary symmetric polynomials in y1 ; y2 ; y3 over any field are (1) s1 D y1 C y2 C y3 . (2) s2 D y1 y2 C y1 y3 C y2 y3 . (3) s3 D y1 y2 y3 . In general, the pattern of the last example holds for y1 ; : : : ; yn . That is, s1 D y1 C y2 C C yn s2 D y1 y2 C y1 y3 C C yn1 yn s3 D y1 y2 y3 C y1 y2 y4 C C yn2 yn1 yn :: : sn D y1 yn : The importance of the elementary symmetric polynomials is that any symmetric polynomial can be built up from the elementary symmetric polynomials. We make this precise in the next theorem called the fundamental theorem of symmetric polynomials. We will use this important result several times, and we will give a complete proof in Section 7.5.
Section 7.4 The Fundamental Theorem of Algebra
105
Theorem 7.3.12 (fundamental theorem of symmetric polynomials). If P is a symmetric polynomial in the indeterminates y1 ; : : : ; yn over a field K, that is, P 2 KŒy1 ; : : : ; yn and P is symmetric, then there exists a unique g 2 KŒy1 ; : : : ; yn such that f .y1 ; : : : ; yn / D g.s1 ; : : : ; sn /. That is, any symmetric polynomial in y1 ; : : : ; yn is a polynomial expression in the elementary symmetric polynomials in y1 ; : : : ; yn . From this theorem we obtain the following two lemmas, which will be crucial in our proof of the fundamental theorem of algebra. Lemma 7.3.13. Let p.x/ 2 KŒx and suppose p.x/ has the roots ˛1 ; : : : ; ˛n in the splitting field K 0 . Then the elementary symmetric polynomials in ˛1 ; : : : ; ˛n are in K. Proof. Suppose p.x/ D c0 C c1 x C C cn x n 2 KŒx. In K 0 Œx, p.x/ splits, with roots ˛1 ; : : : ; ˛n , and thus in K 0 Œx, p.x/ D cn .x ˛1 / .x ˛n /: The coefficients are then cn .1/i si .˛1 ; : : : ; ˛n /, where the si .˛1 ; : : : ; ˛n / are the elementary symmetric polynomials in ˛1 ; : : : ; ˛n . However, p.x/ 2 KŒx, so each coefficient is in K. It follows then that for each i, cn .1/i si .˛1 ; : : : ; ˛n / 2 K, and hence si .˛1 ; : : : ; ˛n / 2 K since cn 2 K. Lemma 7.3.14. Let p.x/ 2 KŒx and suppose p.x/ has the roots ˛1 ; : : : ; ˛n in the splitting field K 0 . Suppose further that g.x/ D g.x; ˛1 ; : : : ; ˛n / 2 K 0 Œx. If g.x/ is a symmetric polynomial in ˛1 ; : : : ; ˛n , then g.x/ 2 KŒx. Proof. If g.x/ D g.x; ˛1 ; : : : ; ˛n / is symmetric in ˛1 ; : : : ; ˛n , then from Theorem 7.3.12 it is a symmetric polynomial in the elementary symmetric polynomials in ˛1 ; : : : ; ˛n . From Lemma 7.3.13 these are in the ground field K, so the coefficients of g.x/ are in K. Therefore, g.x/ 2 KŒx.
7.4
The Fundamental Theorem of Algebra
We now present a proof of the fundamental theorem of algebra. Theorem 7.4.1 (fundamental theorem of algebra). Any nonconstant complex polynomial has a complex root. In other words, the complex number field C is algebraically closed. The proof depends on the following sequence lemmas. The crucial one now is the last, which says that any real polynomial must have a complex root. Lemma 7.4.2. Any odd-degree real polynomial must have a real root.
106
Chapter 7 Kronecker’s Theorem and Algebraic Closures
Proof. This is a consequence of the intermediate value theorem from analysis. Suppose P .x/ 2 RŒx with deg P .x/ D n D 2k C 1 and suppose the leading coefficient an > 0 (the proof is almost identical if an < 0). Then P .x/ D an x n C (lower terms) and n is odd. Then, (1) limx!1 P .x/ D limx!1 an x n D 1 since an > 0. (2) limx!1 P .x/ D limx!1 an x n D 1 since an > 0 and n is odd. From (1), P .x/ gets arbitrarily large positively, so there exists an x1 with P .x1 / > 0. Similarly, from (2) there exists an x2 with P .x2 / < 0. A real polynomial is a continuous real-valued function for all x 2 R. Since P .x1 /P .x2 / < 0, it follows from the intermediate value theorem that there exists an x3 , between x1 and x2 , such that P .x3 / D 0. Lemma 7.4.3. Any degree-two complex polynomial must have a complex root. Proof. This is a consequence of the quadratic formula and of the fact that any complex number has a square root. If P .x/ D ax 2 C bx C c, a ¤ 0, then the roots formally are x1 D
b C
p b 2 4ac ; 2a
x2 D
b
p b 2 4ac : 2a
From DeMoivre’s theorem every complex number has a square root, hence x1 ; x2 exist in C. They of course are the same if b 2 4ac D 0. To go further we need the concept of the conjugate of a polynomial and some straightforward consequences of this idea. Definition 7.4.4. If P .x/ D a0 C C an x n is a complex polynomial then its conjugate is the polynomial P .x/ D a0 C C an x n . That is, the conjugate is the polynomial whose coefficients are the complex conjugates of those of P .x/. Lemma 7.4.5. For any P .x/ 2 CŒx we have: (1) P .z/ D P .z/ if z 2 C. (2) P .x/ is a real polynomial if and only if P .x/ D P .x/. (3) If P .x/Q.x/ D H.x/ then H .x/ D .P .x//.Q.x//.
Section 7.4 The Fundamental Theorem of Algebra
107
Proof. (1) Suppose z 2 C and P .z/ D a0 C C an z n . Then P .z/ D a0 C C an z n D a0 C a1 z C C an z n D P .z/: (2) Suppose P .x/ is real then ai D ai for all its coefficients and hence P .x/ D P .x/. Conversely suppose P .x/ D P .x/. Then ai D ai for all its coefficients and hence ai 2 R for each ai and so P .x/ is a real polynomial. (3) The proof is a computation and left to the exercises. Lemma 7.4.6. Suppose G.x/ 2 CŒx. Then H.x/ D G.x/G.x/ 2 RŒx. Proof. H .x/ D G.x/G.x/ D G.x/G.x/ D G.x/G.x/ D G.x/G.x/ D H.x/. Therefore, H.x/ is a real polynomial. Lemma 7.4.7. If every nonconstant real polynomial has a complex root, then every nonconstant complex polynomial has a complex root. Proof. Let P .x/ 2 CŒx and suppose that every nonconstant real polynomial has at least one complex root. Let H.x/ D P .x/P .x/. From Lemma 7.4.6, H.x/ 2 RŒx. By supposition there exists a z0 2 C with H.z0 / D 0. Then P .z0 /P .z0 / D 0, and since C is a field it has no zero divisors. Hence either P .z0 / D 0 or P .z0 / D 0. In the first case z0 is a root of P .x/. In the second case P .z0 / D 0. Then from Lemma 7.4.5 P .z0 / D P .z0 / D P .z0 / D 0. Therefore, z0 is a root of P .x/. Now we come to the crucial lemma. Lemma 7.4.8. Any nonconstant real polynomial has a complex root. Proof. Let f .x/ D a0 C a1 x C C an x n 2 RŒx with n 1, an ¤ 0. The proof is an induction on the degree n of f .x/. Suppose n D 2m q where q is odd. We do the induction on m. If m D 0 then f .x/ has odd degree and the theorem is true from Lemma 7.4.2. Assume then that the theorem is true for all degrees d D 2k q 0 where k < m and q 0 is odd. Now assume that the degree of f .x/ is n D 2m q. Suppose K 0 is the splitting field for f .x/ over R in which the roots are ˛1 ; : : : ; ˛n . We show that at least one of these roots must be in C. (In fact, all are in C but to prove the lemma we need only show at least one.) Let h 2 Z and form the polynomial H.x/ D
Y i<j
.x .˛i C ˛j C h˛i ˛j //:
108
Chapter 7 Kronecker’s Theorem and Algebraic Closures
This is in K 0 Œx. In forming H.x/ we chose pairs of roots ¹˛i ; ˛j º, so the number of such pairs is the number of ways of choosing two elements out of n D 2m q elements. This is given by .2m q/.2m q 1/ D 2m1 q.2m q 1/ D 2m1 q 0 2 with q 0 odd. Therefore, the degree of H.x/ is 2m1 q 0 . H.x/ is a symmetric polynomial in the roots ˛1 ; : : : ; ˛n . Since ˛1 ; : : : ; ˛n are the roots of a real polynomial, from Lemma 7.3.14 any polynomial in the splitting field symmetric in these roots must be a real polynomial. Therefore, H.x/ 2 RŒx with degree 2m1 q 0 . By the inductive hypothesis, then, H.x/ must have a complex root. This implies that there exists a pair ¹˛i ; ˛j º with ˛i C ˛j C h˛i ˛j 2 C: Since h was an arbitrary integer, for any integer h1 there must exist such a pair ¹˛i ; ˛j º with ˛i C ˛j C h1 ˛i ˛j 2 C: Now let h1 vary over the integers. Since there are only finitely many such pairs ¹˛i ; ˛j º, it follows that there must be at least two different integers h1 ; h2 such that z1 D ˛i C ˛j C h1 ˛i ˛j 2 C
and
z2 D ˛i C ˛j C h2 ˛i ˛j 2 C:
Then z1 z2 D .h1 h2 /˛i ˛j 2 C and since h1 ; h2 2 Z C it follows that ˛ ˛i j 2 C. But then h1 ˛i ˛j 2 C, from which it follows that ˛i C ˛j 2 C. Then, p.x/ D .x ˛i /.x ˛j / D x 2 .˛i C ˛j /x C ˛i ˛j 2 CŒx: However, p.x/ is then a degree-two complex polynomial and so from Lemma 7.4.3 its roots are complex. Therefore, ˛i ; ˛j 2 C, and therefore f .x/ has a complex root. It is now easy to give a proof of the fundamental theorem of algebra. From Lemma 7.4.8 every nonconstant real polynomial has a complex root. From Lemma 7.4.7 if every nonconstant real polynomial has a complex root, then every nonconstant complex polynomial has a complex root proving the fundamental theorem. Theorem 7.4.9. If E is a finite dimensional field extension of C, then E D C. Proof. Let a 2 E. Regard the elements 1; a; a2 ; : : : . These elements become linearly dependent over C and we get a nonconstant polynomial over C with root a. By the fundamental theorem of algebra we know that a 2 C. Corollary 7.4.10. If E is a finite dimensional field extension of R, then E D R or E D C.
Section 7.5 The Fundamental Theorem of Symmetric Polynomials
7.5
109
The Fundamental Theorem of Symmetric Polynomials
In the proof of the fundamental theorem of algebra that was given in the previous section we used the fact that any symmetric polynomial in n indeterminates is a polynomial in the elementary symmetric polynomials in these indeterminates. In this section we give a proof of this theorem. Let R be an integral domain with x1 ; : : : ; xn (independent) indeterminates over R and let RŒx1 ; : : : ; xn be the polynomial ring in these indeterminates. Any polynomial f .x1 ; : : : ; xn / 2 RŒx1 ; : : : ; xn is composed of a sum of pieces of the form ax1i1 : : : xnin with a 2 R. We first put an order on these pieces of a polynomial. j j The piece ax1i1 : : : xnin with a ¤ 0 is called higher than the piece bx11 : : : xnn with b ¤ 0 if the first one of the differences i1 j1 ; i2 j2 ; : : : ; in jn that differs from zero is in fact positive. The highest piece of a polynomial f .x1 ; : : : ; xn / is denoted by HG.f /. Lemma 7.5.1. For f .x1 ; : : : ; xn /; g.x1 ; : : : ; xn / 2 RŒx1 ; : : : ; xn we have HG.fg/ D HG.f / HG.g/: Proof. We use an induction on n, the number of indeterminates. It is clearly true for n D 1, and now assume that the statement holds for all polynomials in k indeterminates with k < n and n 2. Order the polynomials via exponents on the first indeterminate x1 so that f .x1 ; : : : ; xn / D x1r r .x2 ; : : : ; xn / C x1r1 r1 .x2 ; : : : ; xn / C C 0 .x2 ; : : : ; xn / g.x1 ; : : : ; xn / D x1s
s .x2 ; : : : ; xn /
C C Then HG.fg/ D x1rCs HG.r
s /.
HG.r
C x1s1
s1 .x2 ; : : : ; xn /
0 .x2 ; : : : ; xn /:
By the inductive hypothesis s/
D HG.r / HG.
s /:
Hence HG.fg/ D x1rCs HG.r / HG.
s/
D .x1r HG.r //.x1s HG.
s //
D HG.f / HG.g/:
110
Chapter 7 Kronecker’s Theorem and Algebraic Closures
The elementary symmetric polynomials in n indeterminates x1 ; : : : ; xn are s1 D x1 C x2 C C xn s2 D x1 x2 C x1 x3 C C xn1 xn s3 D x1 x2 x3 C x1 x2 x4 C C xn2 xn1 xn :: : sn D x1 xn : These were found by forming the polynomial p.x; x1 ; : : : ; xn / D .x x1 / .x xn /. The i -th elementary symmetric polynomial si in x1 ; : : : ; xn is then .1/i ai , where ai is the coefficient of x ni in p.x; x1 ; : : : ; xn /. In general, X sk D xi1 xi2 xik ; i1
where the sum is taken over all the kn different systems of indices i1 ; : : : ; ik with i1 < i2 < < ik . Further, a polynomial s.x1 ; : : : ; xn / is a symmetric polynomial if s.x1 ; : : : ; xn / is unchanged by any permutation of ¹x1 ; : : : ; xn º, that is, s.x1 ; : : : ; xn / D s. .x1 /; : : : ; .xn //. Lemma 7.5.2. In the highest piece ax1k1 xnkn ; a ¤ 0, of a symmetric polynomial s.x1 ; : : : ; xn / we have k1 k2 kn . Proof. Assume that ki < kj for some i < j . As a symmetric polynomial, s.x1 ; : : : ; k
xn / also must then contain the piece ax1k1 xi j xjki xnkn , which is higher than k
ax1k1 xiki xj j xnkn , giving a contradiction. kn1 kn kn Lemma 7.5.3. The product s1k1 k2 s2k2 k3 sn1 sn with k1 k2 kn k1 k2 kn has the highest piece x1 x2 xn .
Proof. From the definition of the elementary symmetric polynomials we have that HG.skt / D .x1 x2 xk /t ;
1 k n; t 1:
From Lemma 7.4.2, kn1 kn kn sn / HG.s1k1 k2 s2k2 k3 sn1 kn1 kn /.x1 xn /kn D x1k1 k2 .x1 x2 /k2 k3 .x1 xn1
D x1k1 x2k2 xnkn :
111
Section 7.6 Exercises
Theorem 7.5.4. Let s.x1 ; : : : ; xn / 2 RŒx1 ; : : : ; xn be a symmetric polynomial. Then s.x1 ; : : : ; xn / can be uniquely expressed as a polynomial f .s1 ; : : : ; sn / in the elementary symmetric polynomials s1 ; : : : ; sn with coefficients from R. Proof. We prove the existence of the polynomial f by induction on the size of the highest pieces. If in the highest piece of a symmetric polynomial all exponents are zero, then it is constant, that is, an element of R and there is nothing to prove. Now we assume that each symmetric polynomial with highest piece smaller than that of s.x1 ; : : : ; xn / can be written as a polynomial in the elementary symmetric polynomials. Let ax1k1 xnkn , a ¤ 0, be the highest piece of s.x1 ; : : : ; xn /. Let kn1 kn kn sn : t .x1 ; : : : ; xn / D s.x1 ; : : : ; xn / as1k1 k2 sn1
Clearly, t .x1 ; : : : ; xn / is another symmetric polynomial, and from Lemma 7.4.5 the highest piece of t .x1 ; : : : ; xn / is smaller than that of s.x1 ; : : : ; xn /. Therefore, kn1 kn kn sn t .x1 ; : : : ; xn / and hence s.x1 ; : : : ; xn / D t .x1 ; : : : ; xn / C as1k1 k2 sn1 can be written as a polynomial in s1 ; : : : ; sn . To prove the uniqueness of this expression assume that s.x1 ; : : : ; xn / D f .s1 ; : : : ; sn / D g.s1 ; : : : ; sn /. Then f .s1 ; : : : ; sn / g.s1 ; : : : ; sn / D h.s1 ; : : : ; sn / D .x1 ; : : : ; xn / is the zero polynomial in x1 ; : : : ; xn . Hence, if we write h.s1 ; : : : ; sn / as a sum of products of powers of the s1 ; : : : ; sn , all coefficients disappear because two different products of powers in the s1 ; : : : ; sn have different highest pieces. This follows from previous set of lemmas. Therefore, f and g are the same, proving the theorem.
7.6 1.
Exercises Suppose that f .x/ D a0 C a1 x C C an x n 2 KŒx, K a field, with an ¤ 0 and f .x/ irreducible. Define ˛ to satisfy a0 C a1 ˛ C C an ˛ n D 0 and define K 0 D K.˛/ in the following manner. We let K.˛/ D ¹c0 C c1 ˛ C C cn1 ˛ n1 W ci 2 Kº: Then on K.˛/ define addition and subtraction componentwise and define multiplication by algebraic manipulation, replacing powers of ˛ higher than ˛ n by using a0 a1 ˛ an1 ˛ n1 : ˛n D an Prove that K 0 D K.˛/ forms a ring.
112 2.
Chapter 7 Kronecker’s Theorem and Algebraic Closures
Let f; g 2 KŒx be irreducible polynomials of degree 2 over the field K. Let ˛1 ; ˛2 respectively ˇ1 ; ˇ2 be zeros of f and g. For 1 i; j 2 let ij D ˛i C ˇj . Show: (a) jK.ij / W Kj 2 ¹1; 2; 3º. (b) For fixed f; g there are at most two different degrees in (a). (c) Decide, which combinations of degrees in (b) (with f; g variable) are possible, and give an example in each case.
3.
Let LjK be a field extension, let 2 L and f .x/ 2 LŒx a polynomial of degree 1. Let all coefficients of f .x/ be algebraic over K. If f ./ D 0, then is algebraic over K.
4.
Let LjK be a field extension and let M be an intermediate field. The extension M jK is algebraic. For 2 L the following are equivalent: (a) is algebraic over M . (b) is algebraic over K.
5.
Let LjK be a field extension and 1 ; 2 2 L. Then the following are equivalent: (a) 1 and 2 are algebraic over K. (b) 1 C 2 and 1 2 are algebraic over K.
6.
Let LjK be a simple field extension. Then there is an extension field L0 of L of the form L0 D K.1 ; 2 / with: (a) 1 and 2 are transcendental over K. (b) The set of all over K algebraic elements of L0 is L.
7.
In the proof of Theorem 7.1.4 show that the mapping W K.a/ ! K.˛/
defined by .k/ D k if k 2 K and .a/ D ˛ and then extended by linearity is a K-isomorphism. S 8. For each i 2 I let Ki be a field. Now let KO D i 2I Ki . Show that KO is a field. 9.
Prove Lemma 7.2.8.
10. If T; T1 are sets with the same cardinality, then there exists a bijection W T ! T1 . Define a map F W ST ! ST1 in the following manner: if f 2 ST , let F .f / be the permutation on T1 given by F .f /.t1 / D .f . 1 .t1 ///. Prove that F is an isomorphism. 11. Prove that if P .X /; Q.x/; H.x/ 2 C then if P .x/Q.x/ D H.x/ then H .x/ D .P .x//.Q.x//.
Chapter 8
Splitting Fields and Normal Extensions
8.1
Splitting Fields
In the last chapter we introduced splitting fields and used this idea to present a proof of the fundamental theorem of algebra. The concept of a splitting field is essential to the Galois theory of equations so in this chapter we look more deeply at this idea. Definition 8.1.1. Let K be a field and f .x/ a nonconstant polynomial in KŒx. An extension field L of K is a splitting field for f .x/ over K if (a) f .x/ splits into linear factors in LŒx. (b) If K M L and M ¤ L then f .x/ does not split into linear factors in M Œx. From part (b) in the definition the following is clear. Lemma 8.1.2. L is a splitting field for f .x/ 2 KŒx if and only if f .x/ splits into linear factors in LŒx and if f .x/ D b.x a1 / .x an / with b 2 K then L D K.a1 ; : : : ; an /. Example 8.1.3. The field C of complex numbers is a splitting field for the polynomial p.x/ D x 2 C 1 in RŒx. In fact since C is algebraically closed it is a splitting field for any real polynomial f .x/ 2 RŒx which has at least one nonreal zero. The field Q.i/ adjoining i to Q is a splitting field for x 2 C 1 over QŒx. The next result was used in the previous chapter. We restate and reprove it here. Theorem 8.1.4. Let K be a field. Then each nonconstant polynomial in KŒx has a splitting field. Proof. Let K be an algebraic closure of K. Then f .x/ splits in KŒx, that is f .x/ D b.x a1 / .x an / with b 2 K and ai 2 K. Let L D K.a1 ; : : : ; an /. Then L is the splitting field for f .x/ over K. We next show that the splitting field over K of a given polynomial is unique up to K-isomorphism.
114
Chapter 8 Splitting Fields and Normal Extensions
Theorem 8.1.5. Let K; K 0 be field and W K ! K 0 an isomorphism. Let f .x/ be a nonconstant polynomial in KŒx and f 0 .x/ D .f .x// its image in K 0 Œx. Suppose that L is a splitting field for f .x/ over K and L0 is a splitting field for f 0 .x/ over K 0 . (a) Suppose that L0 L00 . Then if W L ! L00 is a monomorphism with jK D then is an isomorphism from L onto L0 . Moreover maps the set of zeros of f .x/ in L onto the set of zeros of f 0 .x/ in L0 . The map is uniquely determined by the values of the zeros of f .x/. (b) If g.x/ is an irreducible factor of f .x/ in KŒx and a is a zero of g.x/ in L and a 0 is a zero of g 0 .x/ D .g.x// in L0 then there is an isomorphism from L to L0 with jK D and .a/ D .a0 /. Before giving the proof of this theorem we note that the following important result is a direct consequence of it. Theorem 8.1.6. A splitting field for f .x/ 2 KŒx is unique up to K-isomorphism. Proof of Theorem 8.1.5. Suppose that f .x/ D b.x a1 / .x an / 2 LŒx and that f 0 .x/ D b 0 .x a10 / .x an0 / 2 L0 Œx. Then f 0 .x/ D .f .x// D
.f .x// D . .b//.x
.a1 // .x
.an //:
We have proved that polynomials have unique factorization over fields. Since L0 L00 it follows that the set of zeros . .a1 /; : : : ; .an // is a permutation of the set of zeros .a10 ; : : : ; an0 /. In particular this implies that .ai / 2 L0 , that is im. / D L0 D K 0 .a1 ; : : : ; an0 /: Since the image of is K 0 .a1 ; : : : ; an0 / D K 0 . .ai /; : : : ; .an // it is clear that is uniquely determined by the images .ai /. This proves part (a). For part (b) embed L0 in an algebraic closure L00 . Hence there is a monomorphism 0 W K.a/ ! L00 with j0K D and 0 .a/ D a0 . Hence there is a monomorphism W L ! L00 with 0 W L ! L0 is an isomorphism. jK.a/ D . Then from part (a) it follows that Example 8.1.7. Let f .x/ D x 3 7 2 QŒx. This has no zeros in Q and since it is of degree 3 it follows that it must be irreducible in QŒx. p Let ! D 12 C 23 i 2 C. Then it is easy to show by computation that ! 2 D 12
p
3 2 i
and ! 3 D 1. Therefore the three zeros of f .x/ in C are a1 D 71=3 a2 D ! 71=3 a3 D ! 2 71=3 :
Section 8.2 Normal Extensions
115
Hence L D Q.a1 ; a2 ; a3 / the splitting field of f .x/. Since the minimal polynomial of all three zeros over Q is the same .f .x// it follows that Q.a1 / Š Q.a2 / Š Q.a3 /: Since Q.a1 / R and a2 ; a3 are nonreal it is clear that a2 ; a3 … Q.a1 /. Suppose that Q.a2 / D Q.a3 /. Then ! D a3 a21 2 Q.a2 / and so 71=3 D ! 1 a2 2 Q.a2 /. Hence Q.a1 / Q.a2 / and therefore Q.a1 / D Q.a2 / since they have the same degree over Q. This contradiction shows that Q.a2 / and Q.a3 / are distinct. By computation we have a3 D a11 a22 and hence L D Q.a1 ; a2 ; a3 / D Q.a1 ; a2 / D Q.71=3 ; !/: Now the degree of L over Q is jL W Qj D jQ.71=3 ; !/ W Q.!/jjQ.!/ W Qj: Now jQ.!/ W Qj D 2 since the minimal polynomial of ! over Q is x 2 C x C 1. Since no zero of f .x/ lies in Q.!/ and the degree of f .x/ is 3 it follows that f .x/ is irreducible over Q.!/. Therefore we have that the degree of L over Q.!/ is 3. Hence jL W Qj D .2/.3/ D 6. We now have the following lattice diagram of fields and subfields:
We do not know however if there are any more intermediate fields. There could for example be infinitely many. However as we will see when we do the Galois theory there are no others.
8.2
Normal Extensions
We now consider algebraic field extensions L of K which have the property that if f .x/ 2 KŒx has a zero in L then f .x/ must split in L. In particular we show that if L is a splitting field of finite degree for some g.x/ 2 KŒx then L has this property.
116
Chapter 8 Splitting Fields and Normal Extensions
Definition 8.2.1. A field extension L of a field K is a normal extension if (a) LjK is algebraic. (b) Each irreducible polynomial f .x/ 2 KŒx that has a zero in L splits into linear factors in LŒx. Note that in Example 8.1.7 the extension fields Q.˛i /jQ are not normal extensions. Although f .x/ has a zero in Q.˛i / the polynomial f .x/ does not split into linear factors in Q.˛i /Œx. We now show that LjK is a finite normal extension if and only if L is the splitting field for some f .x/ 2 KŒx. Theorem 8.2.2. Let LjK be a finite extension. Then the following are equivalent. (a) LjK is a normal extension. (b) LjK is a splitting field for some f .x/ 2 KŒx. (c) If L L0 and W L ! L0 is a monomorphism with then is an automorphism of L, that is .L/ D L.
jK
the identity map on K
Proof. Suppose that LjK is a finite normal extension. Since LjK is a finite extension L is algebraic over K and since of finite degree we have L D K.a1 ; : : : ; an / with ai algebraic over K. Let fi .x/ 2 KŒx be the minimal polynomial of ai . Since LjK is a normal extension fi .x/ splits in LŒx. This is true for each i D 1; : : : ; n. Let f .x/ D f1 .x/f2 .x/ fn .x/. Then f .x/ splits into linear factors in LŒx. Since K D K.a1 ; : : : ; an / the polynomial f .x/ cannot have all its zeros in any intermediate extension between K and L. Therefore L is the splitting field for f .x/. Hence (a) implies (b). Now suppose that L L0 and W L ! L0 is a monomorphism with jK the identity map on K. Then the extension field .L/ of K is also a splitting field for f .x/ since jK is the identity on K. Hence maps the zeros of f .x/ in L L0 onto the zeros of f .x/ in .L/ L0 it follows that .L/ D L. Hence (b) implies (c). W L ! L0 is a Finally suppose (c). Hence we assume that if L L0 and monomorphism with jK the identity map on K then is an automorphism of L, that is .L/ D L. As before LjK is algebraic since LjK is finite. Suppose that f .x/ 2 KŒx is irreducible and that a 2 L is a zero of f .x/. There are algebraic elements a1 ; : : : ; an 2 L with L D K.a1 ; : : : ; an / since LjK is finite. For i D 1; : : : ; n let fi .x/ 2 KŒx be the minimal polynomial of ai and let g.x/ D f .x/f1 .x/ fn .x/. Let L0 be the splitting field of g.X/. Clearly L L0 . Let b 2 L0 be a zero of f .x/. From Theorem 8.1.5 there is an automorphism of L0 with .a/ D b and jK the identity on K. Hence by our assumption jL is an automorphism of L. It follows that b 2 L and hence f .x/ splits in LŒx. Therefore (c) implies (a) completing the proof.
117
Section 8.2 Normal Extensions
To give simple examples of normal extensions we have the following lemma. Lemma 8.2.3. If L is an extension of K with jL W Kj D 2 then L is a normal extension of K. Proof. Suppose that jL W Kj D 2. Then LjK is algebraic since its finite. Let f .x/ 2 KŒx be irreducible with leading coefficient 1 and which has a zero in L. Let a be one zero. Then f .x/ must be the minimal polynomial of a. However deg.ma .x// jL W Kj D 2 and hence f .x/ is of degree 1 or 2. Since f .x/ has a zero in L it follows that it must split into linear factors in LŒx and therefore L is a normal extension. Later we will tie this result to group theory when we prove that a subgroup of index 2 must be a normal subgroup. Example 8.2.4. As a first examplepof the lemma consider the polynomial p p f .x/ D the field Q. 2/ is the x 2 2. In R this splits as .x 2/.x C 2/ and hence p splitting field of f .x/ D x 2 2 over Q. Therefore Q. 2/ is a normal extension of Q. Example 8.2.5. As a second example consider the polynomial x 4 2 in QŒx. The zeros in C are 21=4 ;
21=4 i;
21=4 i 2 ;
21=4 i 3 :
Hence L D Q.21=4 ; 21=4 i; 21=4 i 2 ; 21=4 i 3 / is the splitting field of x 4 2 over Q. Now L D Q.21=4 ; 21=4 i; 21=4 i 2 ; 21=4 i 3 / D Q.21=4 ; i /: Therefore we have jL W Qj D jL W Q.21=4 /jjQ.21=4 / W Qj: Since x 4 2 is irreducible over Q we have jQ.21=4 / W Qj D 4. Since i has degree 2 1=4 a normal extension of over any real field we p have jL p W Q.2 /j D 2. Therefore L is1=4 1=4 2 Q.2 / and x 2 2 Q. 2/Œx has the splitting field Q.2 /. Altogether we have LjQ.21=4 /, Q.21=4 /jQ.21=2 /, Q.21=2 /jQ and LjQ are normal extensions. However Q.21=4 /jQ is not normal since 21=4 is a zero of x 4 2 but Q.21=4 / does not contain all the zeros of x 4 2.
118
Chapter 8 Splitting Fields and Normal Extensions
Hence, we have the following figure.
Figure 8.1
8.3
Exercises
1. Determine the splitting field of f .x/ 2 QŒx and its degree over Q in the following cases: (a) f .x/ D x 4 p, where p is a prime. (b) f .x/ D x p 2, where p is a prime. 2. Determine the degree of the splitting field of the polynomial x 4 C 4 over Q. Determine the splitting field of x 6 C 4x 4 C 4x 2 C 3 over Q. 3. For each a 2 Z let fa .x/ D x 3 ax 2 C .a 3/x C 1 2 QŒx be given. (a) fa is irreducible over Q for each a 2 Z. (b) If b 2 R is a zero of fa , then also .1 b/1 and .b 1/b 1 are zeros of fa . (c) Determine the splitting field L of fa .x/ over Q and its degree jL W Qj. 4. Let K be a field and f .x/ 2 KŒx a polynomial of degree n. Let L be a splitting field of f .x/. Show: (a) If a1 ; : : : ; an 2 L are the zeros of f , then jK.a1 ; : : : ; a t / W Kj n .n 1/ .n t C 1/ for each t with 1 t n. (b) L over K is of degree at most nŠ. (c) If f .x/ is irreducible over K then n divides jL W Kj.
Chapter 9
Groups, Subgroups and Examples
9.1
Groups, Subgroups and Isomorphisms
Recall from Chapter 1 that the three most commonly studied algebraic structures are groups, rings and fields. We have now looked rather extensively at rings and fields and in this chapter we consider the basic concepts of group theory. Groups arise in many different areas of mathematics. For example they arise in geometry as groups of congruence motions and in topology as groups of various types of continuous functions. Later in this book they will appear in Galois theory as groups of automorphisms of fields. First we recall the definition of a group given previously in Chapter 1. Definition 9.1.1. A group G is a set with one binary operation which we will denote by multiplication, such that (1) The operation is associative, that is, .g1 g2 /g3 D g1 .g2 g3 / for all g1 ; g2 ; g3 2 G. (2) There exists an identity for this operation, that is, an element 1 such that 1g D g and g1 D g for each g 2 G. (3) Each g 2 G has an inverse for this operation, that is, for each g there exists a g 1 with the property that gg1 D 1 and g 1 g D 1. If in addition the operation is commutative, that is g1 g2 D g2 g1 for all g1 ; g2 2 G, the group G is called an abelian group. The order of G, denoted jGj, is the number of elements in the group G. If jGj < 1, G is a finite group otherwise it is an infinite group. It follows easily from the definition that the identity is unique and that each element has a unique inverse. Lemma 9.1.2. If G is a group then there is a unique identity. Further if g 2 G its inverse is unique. Finally if g1 ; g2 2 G then .g1 g2 /1 D g21 g11 . Proof. Suppose that 1 and e are both identities for G. Then 1e D e since e is an identity and 1e D 1 since 1 is an identity. Therefore 1 D e and there is only one identity. Next suppose that g 2 G and g1 and g2 are inverses for g. Then g1 gg2 D .g1 g/g2 D 1g2 D g2
120
Chapter 9 Groups, Subgroups and Examples
since g1 g D 1. On the other hand g1 gg2 D g1 .gg2 / D g1 1 D g1 since gg2 D 1. It follows that g1 D g2 and g has a unique inverse. Finally consider .g1 g2 /.g21 g11 / D g1 .g2 g21 /g11 D g1 1g11 D g1 g11 D 1: Therefore g21 g11 is an inverse for g1 g2 and since inverses are unique it is the inverse of the product. Groups most often arise as permutations on a set. We will see this, as well as other specific examples of groups, in the next sections. Finite groups can be completely described by their group tables or multiplication tables. These are sometimes called Cayley tables. In general, let G D ¹g1 ; : : : ; gn º be a group, then the multiplication table of G is: g1 g2 :: :
g1 g2 gj
gn
gi gi gj :: : gn : : : The entry in the row of gi 2 G and column of gj 2 G is the product (in that order) gi gj in G. Groups satisfy the cancellation law for multiplication. Lemma 9.1.3. If G is a group and a; b; c 2 G with ab D ac or ba D ca then b D c. Proof. Suppose that ab D ac. Then a has an inverse a1 so we have a1 .ab/ D a1 .ac/: From the associativity of the group operation we then have .a1 a/b D .a1 a/c H) 1 b D 1 c H) b D c: A consequence of Lemma 9.1.3 is that each row and each column in a group table is just a permutation of the group elements. That is each group element appears exactly once in each row and each column. A subset H G is a subgroup of G if H is also a group under the same operation as G. As for rings and fields a subset of a group is a subgroup if it is nonempty and closed under both the group operation and inverses.
Section 9.2 Examples of Groups
121
Lemma 9.1.4. A subset H G is a subgroup if H ¤ ; and H is closed under the operation and inverses. That is, if a; b 2 H then ab 2 H and a1 ; b 1 2 H . We leave the proof of this to the exercises. Let G be a group and g 2 G; we denote by g n , n 2 N, as with numbers, the product of g taken n times. A negative exponent will indicate the inverse of the positive exponent. As usual, let g0 D 1. Clearly group exponentiation will satisfy the standard laws of exponents. Now consider the set H D ¹1 D g 0 ; g; g 1 ; g 2 ; g 2 ; : : :º of all powers of g. We will denote this by hgi. Lemma 9.1.5. If G is a group and g 2 G then hgi forms a subgroup of G called the cyclic subgroup generated by g. hgi is abelian even if G is not. Proof. If g 2 G then g 2 hgi and hence hgi is nonempty. Suppose then that a D g n ; b D g m are elements of hgi. Then ab D gn g m D g nCm 2 hgi so hgi is closed under the group operation. Further a1 D .g n /1 D g n 2 hgi so hgi is closed under inverses. Therefore hgi is a subgroup. Finally ab D g n g m D g nCm D g mCn D g m g n D ba and hence hgi is abelian. Suppose that g 2 G and g m D 1 for some positive integer m. Then let n be the smallest positive integer such that g n D 1. It follows that the set of elements ¹1; g; g 2 ; : : : ; g n1 º are all distinct but for any other power g k we have gk D g t for some k D 0; 1; : : : ; n 1 (see exercises). The cyclic subgroup generated by g then has order n and we say that g has order n which we denote by o.g/ D n. If no such n exists we say that g has infinite order. We will look more deeply at cyclic groups and subgroups in Section 9.5. We introduce one more concept before looking at examples. Definition 9.1.6. If G and H are groups then a mapping f W G ! H is a (group) homomorphism if f .g1 g2 / D f .g1 /f .g2 / for any g1 ; g2 2 G. If f is also a bijection then it is an isomorphism. As with rings and fields we say that two groups G and H are isomorphic, denoted by G Š H , if there exists an isomorphism F W G ! H . This means that abstractly G and H have exactly the same algebraic structure.
9.2
Examples of Groups
As already mentioned groups arise in many diverse areas of mathematics. In this section and the next we present specific examples of groups.
122
Chapter 9 Groups, Subgroups and Examples
First of all any ring or field under addition forms an abelian group. Hence, for example .Z; C/; .Q; C/; .R; C/; .C; C/ where Z; Q; R; C are respectively the integers, the rationals, the reals and the complex numbers, all are infinite abelian groups. If Zn is the modular ring Z=nZ then for any natural number n, .Zn ; C/ forms a finite abelian group. In abelian groups the group operation is often denoted by C and the identity element by 0 (zero). In a field F , the nonzero elements are all invertible and form a group under multiplication. This is called the multiplicative group of the field F and is usually denoted by F . Since multiplication in a field is commutative the multiplicative group of a field is an abelian group. Hence Q ; R ; C are all infinite abelian groups while if p is a prime Zp forms a finite abelian group. Recall that if p is a prime then the modular ring Zp is a field. Within Q ; R ; C there are certain multiplicative subgroups. Since the positive rationals QC and the positive reals RC are closed under multiplication and inverse they form subgroups of Q and R respectively. In C if we consider the set of all complex numbers z with jzj D 1 then these form a multiplicative subgroup. Further within this subgroup if we consider the set of n-th roots of unity z (that is z n D 1) for a fixed n this forms a subgroup, this time of finite order. The multiplicative group of a field is a special case of the unit group of a ring. If R is a ring with identity, recall that a unit is an element of R with a multiplicative inverse. Hence in Z the only units are ˙1 while in any field every nonzero element is a unit. Lemma 9.2.1. If R is a ring with identity then the set of units in R forms a group under multiplication called the unit group of R and is denoted by U.R/. If R is a field then U.R/ D R . Proof. Let R be a ring with identity. Then the identity 1 itself is a unit so 1 2 U.R/ and hence U.R/ is nonempty. If e 2 R is a unit then it has a multiplicative inverse e 1 . Clearly then the multiplicative inverse has an inverse, namely e so e1 2 U.R/ if e is. Hence to show U.R/ is a group we must show that it is closed under product. Let e1 ; e2 2 U.R/. Then there exist e11 ; e21 . It follows that e21 e11 is an inverse for e1 e2 . Hence e1 e2 is also a unit and U.R/ is closed under product. Therefore for any ring R with identity U.R/ forms a multiplicative group. To present examples of nonabelian groups we turn to matrices. If F is a field we let GL.n; F / D ¹n n matrices over F with nonzero determinantº and SL.n; F / D ¹n n matrices over F with determinant oneº:
123
Section 9.2 Examples of Groups
Lemma 9.2.2. If F is a field then for n 2, GL.n; F / forms a nonabelian group under matrix multiplication and SL.n; F / forms a subgroup. GL.n; F / is called the n-dimensional general linear group over F , while SL.n; F / is called the n-dimensional special linear group over F . Proof. Recall that for two n n matrices A; B with n 2 over a field we have det.AB/ D det.A/ det.B/ where det is the determinant. Now for any field the n n identity matrix I has determinant 1 and hence I 2 GL.n; F /. Since determinants multiply the product of two matrices with nonzero determinant has nonzero determinant so GL.n; F / is closed under product. Further over a field F if A is an invertible matrix then det.A1 / D
1 det A
and so if A has nonzero determinant so does its inverse. It follows that GL.n; F / has the inverse of any of its elements. Since matrix multiplication is associative it follows that GL.n; F / form a group. It is nonabelian since in general matrix multiplication is noncommutative. We leave the fact that SL.n; F / forms a subgroup to the exercises. Groups play an important role in geometry. In any metric geometry an isometry is a mapping that preserves distance. To understand a geometry one must understand the group of isometries. We look briefly at the Euclidean geometry of the plane E 2 . An isometry or congruence motion of E 2 is a transformation or bijection T of E 2 that preserves distance, that is d.a; b/ D d.T .a/; T .b// for all points a; b 2 E 2 . Theorem 9.2.3. The set of congruence motions of E 2 forms a group called the Euclidean group. We denote the Euclidean group by E. Proof. The identity map I is clearly an isometry and since composition of mappings is associative we need only show that the product of isometries is an isometry and that the inverse of an isometry is an isometry. Let T; U be isometries. Then d.a; b/ D d.T .a/; T .b// and d.a; b/ D d.U.a/; U.b// for any points a; b. Now consider d.T U.a/; T U.b// D d.T .U.a//; T .U.b/// D d.U.a/; U.b// since T is an isometry. However d.U.a/; U.b// D d.a; b/ since U is an isometry. Combining these we have that T U is also an isometry.
124
Chapter 9 Groups, Subgroups and Examples
Consider T 1 and points a; b. Then d.T 1 .a/; T 1 .b// D d.T T 1 .a/; T T 1 .b// since T is an isometry. But T T 1 D I and hence d.T 1 .a/; T 1 .b// D d.T T 1 .a/; T T 1 .b// D d.a; b/: Therefore T 1 is also an isometry and hence E is a group. One of the major results concerning E is the following. We refer to [25], [26] and [22] for a more thorough treatment. Theorem 9.2.4. If T 2 E then T is either a translation, rotation, reflection or glide reflection. The set of translations and rotations forms a subgroup. Proof. We outline a brief proof. If T is an isometry and T fixes the origin .0; 0/ then T is a linear mapping. It follows that T is a rotation or a reflection. If T does not fix the origin then there is a translation T0 such that T0 T fixes the origin. This gives translations and glide reflections. In the exercises we expand out more of the proof. If D is a geometric figure in E 2 , such as a triangle or square, then a symmetry of D is a congruence motion T W E 2 ! E 2 that leaves D in place. It may move the individual elements of D however. So for example a rotation about the center of a circle is a symmetry of the circle. Lemma 9.2.5. If D is a geometric figure in E 2 then the set of symmetries of D forms a subgroup of E called the symmetry group of D denoted by Sym.D/. Proof. We show that Sym.D/ is a subgroup of E. The identity map I fixes D so I 2 Sym.D/ and so Sym.D/ is nonempty. Let T; U 2 Sym.D/. Then T maps D to D and so does U . It follows directly that so does the composition T U and hence T U 2 Sym.D/. If T maps D to D then certainly the inverse does. Therefore Sym.D/ is a subgroup of E. Example 9.2.6. Let T be an equilateral triangle. Then there are exactly six symmetries of T (see exercises). These are I D the identity r D a rotation of 120ı around the center of T r 2 D a rotation of 240ı around the center of T f D a reflection over the perpendicular bisector of one of the sides f r D the composition of f and r f r 2 D the composition of f and r 2 :
125
Section 9.3 Permutation Groups
Sym.T / is called the dihedral group D3 . In the next section we will see that it is isomorphic to S3 , the symmetric group on 3 symbols.
9.3
Permutation Groups
Groups most often appear as groups of transformations or permutations on a set. In this section we will take a short look at permutation groups and then examine them more deeply in the next chapter. We recall some ideas first introduced in Chapter 7 in relation to the proof of the fundamental theorem of algebra. Definition 9.3.1. If A is a set, a permutation on A is a one-to-one mapping of A onto itself. We denote by SA the set of all permutations on A. Theorem 9.3.2. For any set A, SA forms a group under composition called the symmetric group on A. If jAj > 2 then SA is nonabelian. Further if A; B have the same cardinality, then SA Š SB . Proof. If SA is the set of all permutations on the set A, we must show that composition is an operation on SA that is associative and has an identity and inverses. Let f; g 2 SA . Then f; g are one-to-one mappings of A onto itself. Consider f ı g W A ! A. If f ı g.a1 / D f ı g.a2 /, then f .g.a1 // D f .g.a2 // and g.a1 / D g.a2 /, since f is one-to-one. But then a1 D a2 since g is one-to-one. If a 2 A, there exists a1 2 A with f .a1 / D a since f is onto. Then there exists a2 2 A with g.a2 / D a1 since g is onto. Putting these together, f .g.a2 // D a, and therefore f ı g is onto. Therefore, f ı g is also a permutation and composition gives a valid binary operation on SA . The identity function 1.a/ D a for all a 2 A will serve as the identity for SA , while the inverse function for each permutation will be the inverse. Such unique inverse functions exist since each permutation is a bijection. Finally, composition of functions is always associative and therefore SA forms a group. Suppose that jAj > 2. Then A has at least 3 elements. Call them a1 ; a2 ; a2 . Consider the 2 permutations f and g which fix (leave unchanged) all of A except a1 ; a2 ; a3 and on these three elements f .a1 / D a2 ;
f .a2 / D a3 ;
f .a3 / D a1
g.a1 / D a2 ;
g.a2 / D a1 ;
g.a3 / D a3 :
Then under composition f .g.a1 // D a3 ;
f .g.a2 // D a2 ;
f .g.a3 // D a1
126
Chapter 9 Groups, Subgroups and Examples
while g.f .a1 // D a1 ;
g.f .a2 // D a3 ;
g.f .a3 // D a2 :
Therefore f ı g ¤ g ı f and hence SA is not abelian. If A; B have the same cardinality, then there exists a bijection W A ! B. Define a map F W SA ! SB in the following manner: if f 2 SA , let F .f / be the permutation on B given by F .f /.b/ D .f . 1 .b///. It is straightforward to verify that F is an isomorphism (see the exercises). If A1 A then those permutations on A that map A1 to A1 form a subgroup of SA called the stabilizer of A1 denoted stab.A1 /. We leave the proof to the exercises. Lemma 9.3.3. If A1 A then stab.A1 / D ¹f 2 SA W f W A1 ! A1 º forms a subgroup of SA . A permutation group is any subgroup of SA for some set A. We now look at finite permutation groups. Let A be a finite set, say A D ¹a1 ; a2 ; : : : ; an º. Then each f 2 SA can be pictured as
a1 : : : an f D : f .a1 / : : : f .an / For a1 there are n choices for f .a1 /. For a2 there are only n 1 choices since f is one-to-one. This continues down to only one choice for an . Using the multiplication principle, the number of choices for f and therefore the size of SA is n.n 1/ 1 D nŠ: We have thus proved the following theorem. Theorem 9.3.4. If jAj D n then jSA j D nŠ. For a set with n elements we denote SA by Sn , called the symmetric group on n symbols. Example 9.3.5. Write down the six elements of S3 and give the multiplication table for the group. Name the three elements 1; 2; 3. The six elements of S3 are then 1 2 3 1 2 3 1 2 3 1D ; aD ; bD 1 2 3 2 3 1 3 1 2 1 2 3 1 2 3 1 2 3 cD ; dD ; eD : 2 1 3 3 2 1 1 3 2
127
Section 9.3 Permutation Groups
The multiplication table for S3 can be written down directly by doing the required composition. For example, 1 2 3 1 2 3 1 2 3 ac D D D d: 2 3 1 2 1 3 3 2 1 To see this, note that a W 1 ! 2; 2 ! 3; 3 ! 1; c W 1 ! 2; 2 ! 1; 3 ! 3 and so ac W 1 ! 3; 2 ! 2; 3 ! 1. It is somewhat easier to construct the multiplication table if we make some observations. First, a2 D b and a3 D 1. Next, c 2 D 1, d D ac, e D a2 c and finally ac D ca2 . From these relations the following multiplication table can be constructed:
1 a a2 c ac a2 c
1 1 a a2 c ac a2 c
a a a2 1 a2 c c ac
a2 a2 1 a ac a2 c c
c c ac a2 c 1 a a2
ac ac a2 c c a2 1 a
a2 c a2 c c ac a a2 1
To see this, consider, for example, .ac/a2 D a.ca2 / D a.ac/ D a2 c. More generally, we can say that S3 has a presentation given by S3 D ha; cI a3 D c 2 D 1; ac D ca2 i: By this we mean that S3 is generated by a; c, or that S3 has generators a; c and the whole group and its multiplication table can be generated by using the relations a3 D c 2 D 1, ac D ca2 . A theorem of Cayley actually shows that every group is a permutation group. A group G is a permutation group on the group G itself considered as a set. This result however does not give much information about the group. Theorem 9.3.6 (Cayley’s theorem). Let G be a group. Consider the set of elements of G. Then the group G is a permutation group on the set G, that is G is a subgroup of SG . Proof. We show that to each g 2 G we can associate a permutation of the set G. If g 2 G let g be the map given by g W g1 ! gg1
for each g1 2 G:
It is straightforward to show that each g is a permutation on G.
128
9.4
Chapter 9 Groups, Subgroups and Examples
Cosets and Lagrange’s Theorem
In this section given a group G and a subgroup H we define an equivalence relation on G. The equivalence classes all have the same size and are called the (left) or (right) cosets of H in G. Definition 9.4.1. Let G be a group and H G a subgroup. For a; b 2 G define a b if a1 b 2 H . Lemma 9.4.2. Let G be a group and H G a subgroup. Then the relation defined above is an equivalence relation on G. The equivalence classes all have the form aH for a 2 G and are called the left cosets of H in G. Clearly G is a disjoint union of its left cosets. Proof. Let us show, first of all, that this is an equivalence relation. Now a a since a1 a D e 2 H . Therefore the relation is reflexive. Further a b implies a1 b 2 H , but since H is a subgroup of G we have b 1 a D .a1 b/1 2 H and so b a. Therefore the relation is symmetric. Finally suppose that a b and b c. Then a1 b 2 H and b 1 c 2 H . Since H is a subgroup a1 b b 1 c D a1 c 2 H , and hence a c. Therefore the relation is transitive and hence is an equivalence relation. For a 2 G the equivalence class is Œa D ¹g 2 G W a gº D ¹a 2 G W a1 g 2 H º: But then clearly g 2 aH . It follows that the equivalence class for a 2 G is precisely the set aH D ¹g 2 G W g D ah for some h 2 H º: These classes, aH , are called left cosets of H and since they are equivalence classes they partition G. This means that every element of g is in one and only one left coset. In particular, bH D H D eH if and only if b 2 H . If aH is a left coset then we call the element a a coset representative representative. A complete collection ¹a 2 G W ¹aH º is the set of all distinct left cosets of H º is called a (left) transversal of H in G. One could define another equivalence relation by defining a b if and only if ba1 2 H . Again this can be shown to be an equivalence relation on G, and the equivalence classes here are sets of the form Ha D ¹g 2 G W g D ha for some h 2 H º;
Section 9.4 Cosets and Lagrange’s Theorem
129
called right cosets of H . Also, of course, G is the (disjoint) union of distinct right cosets. It is easy to see that any two left (right) cosets have the same order (number of elements). To demonstrate this consider the mapping aH ! bH via ah 7! bh where h 2 H . It is not hard to show that this mapping is 1-1 and onto (see exercises). Thus we have jaH j D jbH j. (This is also true for right cosets and can be established in a similar manner.) Letting b 2 H in the above discussion, we see jaH j D jH j, for any a 2 G, that is the size of each left or right coset is exactly the same as the subgroup H . One can also see that the collection ¹aH º of all distinct left cosets has the same number of elements as the collection ¹Haº of all distinct right cosets. In other words, the number of left cosets equals the number of right cosets (this number may be infinite). For consider the map f W aH ! Ha1 : This mapping is well-defined: for if aH D bH , then b D ah where h 2 H . Thus f .bH / D H b 1 D H h1 a1 D f .aH /. It is not hard to show that this mapping is 1-1 and onto (see exercises). Hence the number of left cosets equals the number of right cosets. Definition 9.4.3. Let G be a group and H G a subgroup. The number of distinct left cosets, which is the same as the number of distinct right cosets, is called the index of H in G, denoted by ŒG W H . Now let us consider the case where the group G is finite. Each left coset has the same size as the subgroup H and here both are finite. Hence jaH j D jH j for each coset. Further the group G is a disjoint union of the left cosets, that is G D H [ g1 H [ [ gn H: Since this is a disjoint union we have jGj D jH j C jg1 H j C C jgn H j D jH j C jH j C C jH j D jH jŒG W H : This establishes the following extremely important theorem. Theorem 9.4.4 (Lagrange’s theorem). Let G be a group and H G a subgroup. Then jGj D jH jŒG W H : If G is a finite group this implies that both the order of a subgroup and the index of a subgroup are divisors of the order of the group. This theorem plays a crucial role in the structure theory of finite groups since it greatly restricts the size of subgroups. For example in a group of order 10 there can be proper subgroups only of orders 1, 2 and 5.
130
Chapter 9 Groups, Subgroups and Examples
As an immediate corollary, we have the following result. Corollary 9.4.5. The order of any element g 2 G, where G is a finite group, divides the order of the group. In particular if jGj D n and g 2 G then o.g/jn and gn D 1. Proof. Let g 2 G and o.g/ D m. Then m is the size of the cyclic subgroup generated by g and hence divides m from Lagrange’s theorem. Then n D mk and so g n D g mk D .g m /k D 1k D 1: Before leaving this section we consider some results concerning general subsets of a group. Suppose that G is a group and S is an arbitrary nonempty subset of G, S G and S ¤ ;; such a set S is usually called a complex of G. If U and V are two complexes of G, the product U V is defined as follows: U V D ¹g1 g2 2 G W u 2 U; v 2 V º: Now suppose that U; V are subgroups of G. When is the complex U V again a subgroup of G? Theorem 9.4.6. The product U V of two subgroups U , V of a group G is itself a subgroup if and only if U and V commute, that is, if and only if U V D V U . Proof. We note first that when we say U and V commute, we do not demand that this is so elementwise. In other words, it is not required that uv D vu for all u 2 U and all v 2 V ; all that is required is that for any u 2 U and v 2 V uv D v1 u1 , for some elements u1 2 U and v1 2 V . Assume that U V is a subgroup of G. Let u 2 U and v 2 V . Then u 2 U 1 U V and v 2 1 V U V . But since U V is assumed itself to be a subgroup, it follows that vu 2 U V . Hence each product vu 2 U V and so V U U V . In an identical manner U V V U and so U V D V U . Conversely, suppose that U V D V U . Let g1 D u1 v1 2 U V , g2 D u2 v2 2 U V . Then g1 g1 D .u1 v1 /.u2 v2 / D u1 .v1 u2 /v2 D u1 u3 v3 v2 D .u1 u3 /.v3 v2 / 2 U V since v1 u2 D u3 v3 for some u3 2 U and v3 2 V . Further g11 D .u1 v1 /1 D v11 u1 1 D u 4 v4 : It follows that U V is a subgroup.
131
Section 9.4 Cosets and Lagrange’s Theorem
Theorem 9.4.7 (product formula). Let U; V be subgroups of G and let R be a left transversal of the intersection U \ V in U . Then [ rV UV D r2R
where this is a disjoint union. In particular if U; V are finite then jU V j D
jU jjV j : jU \ V j
Proof. Since R U we have that [
rV U V:
r2R
In the other direction let uv 2 U V . Then [ r.U \ V /: U D r2R
It follows that u D rv 0 with r 2 R and v 0 2 U \ V . Hence uv D rv 0 v 2 rV: The union of cosets of V is disjoint so uv 2
[
rV:
r2R
S Therefore U V r2R rV proving the equality. Now suppose that jU j and jV j are finite. Then we have jU V j D jRjjV j D jU W U \ V jjV j D
jU j jU jjV j jV j D : jU \ V j jU \ V j
We now show that index is multiplicative. Later we will see how this fact is related to the multiplicativity of the degree of field extensions. Theorem 9.4.8. Suppose G is a group and U and V are subgroups with U V G. Then if G is the disjoint union [ rV GD r2R
and V is the disjoint union V D
[ s2S
sU
132
Chapter 9 Groups, Subgroups and Examples
then we get a disjoint union for G as GD
[
rsU:
r2R;s2S
In particular if ŒG W V and ŒV W U are finite then ŒG W U D ŒG W V ŒV W U : Proof. Now GD
[
rV D
r2R
[[ r2R
s2S
sU
D
[
rsU:
r2R;s2S
Suppose that r1 s1 U D r2 s2 U . Then r1 s1 U V D r2 s2 U V . But s1 U V D V and s2 U V D V so r1 V D r2 V which implies that r1 D r2 . Then s1 U D s2 U which implies that s1 D s2 . Therefore the union is disjoint. The index formula now follows directly. The next result says that the intersection of subgroups of finite index must again be of finite index. Theorem 9.4.9 (Poincaré). Suppose that U; V are subgroups of finite index in G. Then U \ V is also of finite index. Further ŒG W U \ V ŒG W U ŒG W V : If ŒG W U ; ŒG W V are relatively prime then equality holds. Proof. Let r be the number of left cosets of U in G that are contained in U V . r is finite since the index ŒG W U is finite. From Theorem 9.4.7 we then have jV W U \ V j D r ŒG W U : Then from Theorem 9.4.8 ŒG W U \ V D ŒG W V ŒV W U \ V ŒG W V ŒG W U : Since both ŒG W U and ŒG W V are finite so is ŒG W U \ V . Now ŒG W U jŒG W U \ V and ŒG W V jŒG W U \ V . If ŒG W U and ŒG W V are relatively prime then ŒG W U ŒG W V jŒG W U \ V H) ŒG W U ŒG W V ŒG W U \ V Therefore we must have equality. Corollary 9.4.10. Suppose that ŒG W U and ŒG W V are finite and relatively prime. Then G D U V .
Section 9.5 Generators and Cyclic Groups
133
Proof. From Theorem 9.4.9 we have ŒG W U \ V D ŒG W U ŒG W V : From Theorem 9.4.8 ŒG W U \ V D ŒG W V ŒV W U \ V : Combing these we have ŒV W U \ V D ŒG W U : The number of left cosets of U in G that are contained in V U is equal to the number of all left cosets of U in G. It follows then that we must have G D U V .
9.5
Generators and Cyclic Groups
We saw that if G is any group and g 2 G then the powers of g generate a subgroup of G called the cyclic subgroup generated by g. Here we explore more fully the idea of generating a group or subgroup. We first need the following. Lemma 9.5.1. If U and V are subgroups of a group G then their intersection U \ V is also a subgroup. Proof. Since the identity of G is in both U and V we have that U \ V is nonempty. Suppose that g1 ; g2 2 U \ V . Then g1 ; g2 2 U and hence g11 g2 2 U since U is a subgroup. Analogously g11 g2 2 V . Hence g 1 g2 2 U \ V and therefore U \ V is a subgroup. Now let S be a subset of a group G. The subset S is certainly contained in at least one subgroup of G, namely T G itself. Let ¹U˛ º be the collection of all subgroups of G containing S . Then ˛ U˛ is again a subgroup of G from Lemma 9.5.1. T Further it is the smallest subgroup of G containing S (see the exercises). We call ˛ U˛ the subgroup of G generated by S and denote it by hS i or grp.S/. We call the set S a set of generators for hS i. Definition 9.5.2. A subset M of a group G is a set of generators for G if G D hM i, that is the smallest subgroup of G containing M is all of G. We say that G is generated by M and that M is a set of generators for G. Notice that any group G has at least one set of generators, namely G itself. If G D hM i and M is a finite set then we say that G is finitely generated. Clearly any finite group is finitely generated. Shortly we will give an example of a finitely generated infinite group.
134
Chapter 9 Groups, Subgroups and Examples
Example 9.5.3. The set of all reflections forms a set of generators for the Euclidean group E. Recall that any T 2 E is either a translation, a rotation, a reflection or a glide reflection. It can be shown (see exercises) that any one of these can be expressed as a product of 3 or fewer reflections. We now consider the case where a group G has a single generator. Definition 9.5.4. A group G is cyclic if there exists a g 2 G such that G D hgi. In this case G D ¹g n W n 2 Zº, that is G consists of all the powers of the element g. If there exists an integer m such that g m D 1, then there exists a smallest such positive integer say n. It follows that g k D g l if and only if k l mod n. In this situation the distinct powers of g are precisely ¹1 D g 0 ; g; g 2 ; : : : ; gn1 º: It follows that jGj D n. We then call G a finite cyclic group. If no such power exists then all the powers of G are distinct and G is an infinite cyclic group. We show next that any two cyclic groups of the same order are isomorphic. Theorem 9.5.5. (a) If G D hgi is an infinite cyclic group then G Š .Z; C/ that is the integers under addition. (b) If G D hgi is a finite cyclic group of order n then G Š .Zn ; C/ that is the integers modulo n under addition. It follows that for a given order there is only one cyclic group up to isomorphism. Proof. Let G be an infinite cyclic group with generator g. Map g onto 1 2 .Z; C/. Since g generates G and 1 generates Z under addition this can be extended to a homomorphism. It is straightforward to show that this defines an isomorphism. Now let G be a finite cyclic group of order n with generator g. As above map g to 1 2 Zn and extend to a homomorphism. Again it is straightforward to show that this defines an isomorphism. Now let G and H be two cyclic groups of the same order. If both are infinite then both are isomorphic to .Z; C/ and hence isomorphic to each other. If both are finite of order n then both are isomorphic to .Zn ; C/ and hence isomorphic to each other. Theorem 9.5.6. Let G D hgi be a finite cyclic group of order n. Then every subgroup of G is also cyclic. Further if d jn there exists a unique subgroup of G of order d . Proof. Let G D hgi be a finite cyclic group of order n and suppose that H is a subgroup of G. Notice that if g m 2 H then g m is also in H since H is a subgroup. Hence H must contain positive powers of the generator g. Let t be the smallest positive power of g such that g t 2 H . We claim that H D hg t i the cyclic subgroup
Section 9.5 Generators and Cyclic Groups
135
of G generated by g t . Let h 2 H then h D g m for some positive integer m t . Divide m by t to get m D qt C r
where r D 0 or 0 < r < t:
If r ¤ 0 then r D m qt > 0. Now g m 2 H , g t 2 H so g qt 2 H for any q since H is a subgroup. It follows that g m g qt D gmqt 2 H . This implies that gr 2 H . However this is a contradiction since r < t and t is the least positive power in H . It follows that r D 0 so m D qt . This implies that g m D gqt D .g t /q , that is gm is a multiple of g t . Therefore every element of H is a multiple of g t and therefore g t generates H and hence H is cyclic. Now suppose that d jn so that n D kd . Let H D hgk i, that is the subgroup of G generated by g k . We claim that H has order d and that any other subgroup H1 of G with order d coincides with H . Now .g k /d D g kd D g n D 1 so the order of gk divides d and hence is d . Suppose that .g k /d1 D g kd1 D 1 with d1 < d . Then since the order of g is n we have n D kd jkd1 with d1 < d which is impossible. Therefore the order of g k is d and h D hgk i is a subgroup of G of order d . Now let H1 be a subgroup of G of order d . We must show that H1 D H . Let h 2 H1 so h D g t and hence g td D 1. It follows that njt d and so kd jt d and hence kjt , that is t D qk for some positive integer q. Therefore g t D .gk /q 2 H . Therefore H1 H and since they are of the same size H D H1 . Theorem 9.5.7. Let G D hgi be an infinite cyclic group. Then a subgroup H is of the form H D hg t i for a positive integer t . Further if t1 ; t2 are positive integers with t1 ¤ t2 then hg t1 i and hg t2 i are distinct. Proof. Let G D hgi be an infinite cyclic group and H a subgroup of G. As in the proof of Theorem 9.5.6 H must contain positive powers of the generator g. Let t be the smallest positive power of g such that g t 2 H . We claim that H D hgt i the cyclic subgroup of G generated by g t . Let h 2 H then h D gm for some positive integer m t . Divide m by t to get m D qt C r
where r D 0 or 0 < r < t:
If r ¤ 0 then r D m qt > 0. Now g m 2 H , g t 2 H so g qt 2 H for any q since H is a subgroup. It follows that g m g qt D gmqt 2 H . This implies that gr 2 H . However this is a contradiction since r < t and t is the least positive power in H . It follows that r D 0 so m D qt . This implies that g m D gqt D .g t /q , that is gm is a multiple of g t . Therefore every element of H is a multiple of g t and therefore g t generates H and hence H D hg t i. From the proof above in the subgroup hg t i the integer t is the smallest positive power of g in hg t i. Therefore if t1 ; t2 are positive integers with t1 ¤ t2 then hg t1 i and hg t2 i are distinct.
136
Chapter 9 Groups, Subgroups and Examples
Theorem 9.5.8. Let G D hgi be a cyclic group. Then (a) If G D hgi is finite of order n then g k is also a generator if and only if .k; n/ D 1. That is the generators of G are precisely those powers gk where k is relatively prime to n. (b) If G D hgi is infinite then the only generators are g; g1 . Proof. (a) Let G D hgi be a finite cyclic group of order n and suppose that .k; n/ D 1. Then there exist integers x; y with kx C ny D 1. It follows that g D g kxCny D .gk /x .gn /y D .g k /x since gn D 1. Hence g is a power of g k that implies every element of G is also a power of g k . Therefore g k is also a generator. Conversely suppose that gk is also a generator. Then g is a power of gk so there exists an x such that g D g kx . It follows that kx 1 modulo n and so there exists a y such that kx C ny D 1: This then implies that .k; n/ D 1. (b) If G D hgi is infinite then any power of g other than g 1 generates a proper subgroup. If g is a power of g n for some n so that g D g nx it follows that g nx1 D 1 so that g has finite order contradicting that G is infinite cyclic. Recall that for positive integers n the Euler phi-function is defined as follows. Definition 9.5.9. For any n > 0, let .n/ D number of integers less than or equal to n and relatively prime to n: Example 9.5.10. .6/ D 2 since among 1; 2; 3; 4; 5; 6 only 1; 5 are relatively prime to 6. Corollary 9.5.11. If G D hgi is finite of order n then there are .n/ generators for G where is the Euler phi-function. Proof. From Theorem 9.5.8 the generators of G are precisely the powers gk where .k; n/ D 1. The numbers relatively prime to n are counted by the Euler phi-function. Recall that in an arbitrary group G, if g 2 G, then the order of g, denoted o.g/, is the order of the cyclic subgroup generated by g. Given two elements g; h 2 G in general there is no relationship between o.g/; o.h/ and the order of the product gh. However if they commute there is a very direct relationship.
Section 9.5 Generators and Cyclic Groups
137
Lemma 9.5.12. Let G be an arbitrary group and g; h 2 G both of finite order o.g/; o.h/. If g and h commute, that is gh D hg, then o.gh/ divides lcm.o.g/; o.h//. In particular if G is an abelian group then o.gh/j lcm.o.g/; o.h// for all g; h 2 G of finite order. Further if hgi \ hhi D ¹1º then o.gh/ D lcm.o.g/; o.h//. Proof. Suppose o.g/ D n and o.h/ D m are finite. If g; h commute then for any k we have .gh/k D gk hk . Let t D lcm.n; m/ then t D k1 m; t D k2 n. Hence .gh/t D g t ht D .g m /k1 .hn /k2 D 1: Therefore the order of gh is finite and divides t . Suppose that hgi \ hhi D ¹1º that is the cyclic subgroup generated by g intersects trivially with the cyclic subgroup generated by h. Let k D o.gh/ which we know is finite from the first part of the lemma. Let t D lcm.n; m/. We then have .gh/k D gk hk D 1 which implies that g k D hk . Since the cyclic subgroups have only trivial intersection this implies that g k D 1 and hk D 1. But then njk and mjk and hence t jk. Since kjt it follows that k D t. Recall that if m and n are relatively prime then lcm.m; n/ D mn. Further if the orders of g and h are relatively prime it follows from Lagrange’s theorem that hgi \ hhi D ¹1º. We then get the following. Corollary 9.5.13. If g; h commute and o.g/ and o.h/ are finite and relatively prime then o.gh/ D o.g/o.h/. Definition 9.5.14. If G is a finite abelian group then the exponent of G is the lcm of the orders of all elements of G. That is exp.G/ D lcm¹o.g/ W g 2 Gº: As a consequence of Lemma 9.5.12 we obtain Lemma 9.5.15. Let G be a finite abelian group. Then G contains an element of order exp.G/. e Proof. Suppose that exp.G/ D p1e1 pkk with pi distinct primes. By the definition of exp.G/ there is a gi 2 G with o.gi / D piei ri with pi and ri relatively prime. Let hi D giri . Then from Lemma 9.5.12 we get o.hi / D piei . Now let g D h1 h2 hk . From the corollary to Lemma 9.5.12 we have o.g/ D p1e1 pkek D exp.G/.
If K is a field then the multiplicative subgroup of nonzero elements of K is an abelian group K ? . The above results lead to the fact that a finite subgroup of K ? must actually be cyclic. Theorem 9.5.16. Let K be a field. Then any finite subgroup of K ? is cyclic.
138
Chapter 9 Groups, Subgroups and Examples
Proof. Let A K ? with jAj D n. Suppose that m D exp.A/. Consider the polynomial f .x/ D x m 1 2 KŒx. Since the order of each element in A divides m it follows that am D 1 for all a 2 A and hence each a 2 A is a zero of the polynomial f .x/. Hence f .x/ has at least n zeros. Since a polynomial of degree m over a field can have at most m zeros it follows that n < m. From Lemma 9.5.15 there is an element a 2 A with o.a/ D m. Since jAj D n it follows that mjn and hence m < n. Therefore m D n and hence A D hai showing that A is cyclic. We close this section with two other results concerning cyclic groups. The first proves, using group theory, a very interesting number theoretic result concerning the Euler phi-function. Theorem 9.5.17. For n > 1 and for d 1 X .d / D n: d jn
Proof. Consider a cyclic group G of order n. For each d jn, d 1 there is a unique cyclic subgroup H of order d . H then has .d / generators. Each element in G generates its own cyclic subgroup H1 , say of order d and hence must be included in the .d / generators of H1 . Therefore X .d / D sum of the numbers of generators of the cyclic subgroups of G: d jn
But this must be the whole group and hence this sum is n. We shall make use of the above theorem directly in the following theorem. Theorem 9.5.18. If jGj D n and if for each positive d such that d jn, G has at most one cyclic subgroup of order d , then G is cyclic (and consequently, has exactly one cyclic subgroup of order d ). Proof. For each d jn, d > 0, let .d / D the number of elements of G of order d . Then X .d / D n: d jn
Now suppose that .d / ¤ 0 for a given d jn. Then there exists an a 2 G of order d which generates a cyclic subgroup, hai, of order d of G. We claim that all elements of G of order d are in hai. Indeed, if b 2 G with o.b/ D d and b … hai, then hbi is a second cyclic subgroup of order d , distinct from hai. This contradicts the hypothesis, so the claim is proved. Thus, if .d / ¤ 0, then P .d / D .d /. P In general, we have .d / .d /, for all positive d jn. But n D d jn .d / d jn .d /, by the previous theorem. It follows, clearly, from this that .d / D .d / for all d jn. In
Section 9.6 Exercises
139
particular, .n/ D .n/ 1. Hence, there exists at least one element of G of order n; hence G is cyclic. This completes the proof. Corollary 9.5.19. If in a group G of order n, for each d jn, the equation x d D 1 has at most d solutions in G, then G is cyclic. Proof. The hypothesis clearly implies that G can have at most one cyclic subgroup of order d since all elements of such a subgroup satisfy the equation. So Theorem 9.5.17 applies to give our result. If H is a subgroup of a group G then G operates as a group of permutations on the set ¹aH W a 2 Rº of left cosets of H in G where R is a left transversal of H in G. This we can use to show that a finitely generated group has only finitely many subgroups of a given finite index. Theorem 9.5.20. Let G be a finitely generated group. The number of subgroups of index n < 1 is finite. Proof. Let H be a subgroup of index n. We choose a left transversal ¹c1 ; : : : ; cn º for H in G where c1 D 1 represents H . G permutes the set of cosets ci H by multiplication from the left. This induces a homomorphism H from G to Sn as follows. For each g 2 G let H .g/ be the permutation which maps i to j if gci H D cj H . H .g/ fixes the number 1 if and only if g 2 H because c1 H D H . Now, let H and L be two different subgroups of index n in G. Then there exists g 2 H with g … L and H .g/ ¤ L .g/, and hence H and L are different. Since G is finitely generated there are only finitely many homomorphisms from G to Sn . Therefore the number of subgroups of index n < 1 is finite.
9.6
Exercises
1.
Prove Lemma 9.1.4.
2.
Suppose that g 2 G and g m D 1 for some positive integer m. Let n be the smallest positive integer such that gn D 1. Show the set of elements ¹1; g; g2 ; : : : ; g n1 º are all distinct but for any other power g k we have g k D g t for some k D 0; 1; : : : ; n 1.
3.
Let G be a group and U1 ; U2 be finite subgroups of G. If jU1 j and jU2 j are relatively prime, then U1 \ U2 D ¹eº.
4.
Let A; B be subgroups of a finite group G. If jAj jBj > jGj then A \ B ¤ ¹eº.
140 5.
Chapter 9 Groups, Subgroups and Examples
Let G be the set of all real matrices of the form Show:
a b b a
, where a2 C b 2 ¤ 0.
(a) G is a group. (b) For each n 2 N there is at least one element of order n in G. 6.
Let p be a prime, and let G D SL.2; p/. Show: (a) G has at least 2p 2 elements of order p (their exact number is p 2 1). (b) If p is odd, then x p D 1 for all x 2 G.
7.
Let p be a prime and a 2 Z. Show that ap a mod p.
8.
Let F be a field. Show that the set of n n matrices of determinant 1 over F forms a group.
9.
Here we outline a proof that every planar Euclidean congruence motion is either a rotation, translation, reflection or glide reflection. An isometry in this problem is a planar Euclidean congruence motion. Show: (a) If T is an isometry then it is completely determined by its action on a triangle – equivalent to showing that if T fixes three noncollinear points then it must be the identity. (b) If an isometry T has exactly one fixed point then it must be a rotation with that point as center. (c) If an isometry T has two fixed points then it fixes the line joining them. Then show that if T is not the identity it must be a reflection through this line. (d) If an isometry T has no fixed point but preserves orientation then it must be a translation. (e) If an isometry T has no fixed point but reverses orientation then it must be a glide reflection.
10. Let Pn be a regular n-gon and DN its group of symmetries. Show that jDn j D 2n. (Hint: First show that jDn j 2n and then exhibit 2n distinct symmetries.) 11. If A; B have the same cardinality, then there exists a bijection W A ! B. Define a map F W SA ! SB in the following manner: if f 2 SA , let F .f / be the permutation on B given by F .f /.b/ D .f . 1 .b///. Show that F is an isomorphism. 12. Prove Lemma 9.3.3.
Chapter 10
Normal Subgroups, Factor Groups and Direct Products
10.1
Normal Subgroups and Factor Groups
In rings we saw that there were certain special types of subrings, called ideals, that allowed us to define factor rings. The analogous object for groups is called a normal subgroup which we will define and investigate in this section. Definition 10.1.1. Let G be an arbitrary group and suppose that H1 and H2 are subgroups of G. We say that H2 is conjugate to H1 if there exists an element a 2 G such that H2 D aH1 a1 . H1 ; H2 are the called conjugate subgroups of G. Lemma 10.1.2. Let G be an arbitrary group. Then the relation of conjugacy is an equivalence relation on the set of subgroups of G. Proof. We must show that conjugacy is reflexive, symmetric and transitive. If H is a subgroup of G then 11 H1 D H and hence H is conjugate to itself and therefore the relation is reflexive. Suppose that H1 is conjugate to H2 . Then there exists a g 2 G with g 1 H1 g D H2 . This implies that gH2 g 1 D H1 . However .g 1 /1 D g and hence letting g 1 D g1 we have g11 H2 g1 D H1 : Therefore H2 is conjugate to H1 and conjugacy is symmetric. Finally suppose that H1 is conjugate to H2 and H2 is conjugate to H3 . Then there exist g1 ; g2 2 G with H2 D g11 H1 g1 and H3 D g21 H2 g2 . Then H3 D g21 g11 H1 g1 g2 D .g1 g2 /1 H1 .g1 g2 /: Therefore H3 is conjugate to H2 and conjugacy is transitive. Lemma 10.1.3. Let G be an arbitrary group. Then for g 2 G the map g W a ! g 1 ag is an automorphism on G. Proof. For a fixed g 2 G define the map f W G ! G by f .a/ D g 1 ag for a 2 G. We must show that this is a homomorphism and that it is one-to-one and onto.
142
Chapter 10 Normal Subgroups, Factor Groups and Direct Products
Let a1 ; a2 2 G. Then f .a1 a2 / D g 1 a1 a2 g D .g1 a1 g/.g 1 a2 g/ D f .a1 /f .a2 /: Hence f is a homomorphism. If f .a1 / D f .a2 / then g 1 a1 g D g 1 a2 g. Clearly by the cancellation law we then have a1 D a2 and hence f is one-to-one. Finally let a 2 G and let a1 D gag 1 . Then a D g 1 a1 g and hence f .a1 / D a. It follows that f is onto and therefore f is an automorphism on G. In general a subgroup H of a group G may have many different conjugates. However in certain situations the only conjugate of a subgroup H is H itself. If this is the case we say that H is a normal subgroup. We will see shortly that this is precisely the analog for groups of the concept of an ideal in rings. Definition 10.1.4. Let G be an arbitrary group. A subgroup H is a normal subgroup of G, which we denote by H G G, if g 1 Hg D H for all g 2 G. Since the conjugation map is an isomorphism it follows that if g 1 Hg H then D H . Hence in order to show that a subgroup is normal we need only show g inclusion. 1 Hg
Lemma 10.1.5. Let N be a subgroup of a group G. Then if aNa1 N for all a 2 G, then aNa1 D N . In particular, aNa1 N for all a 2 G implies that N is a normal subgroup. Notice that if g 1 Hg D H then Hg D gH . That is as sets the left coset gH is equal to the right coset Hg. Hence for each h1 2 H there is an h2 2 H with gh1 D h2 g. If H G G this is true for all g 2 G. Further if H is normal then for the product of two cosets g1 H and g2 H we have .g1 H /.g2 H / D g1 .Hg2 /H D g1 g2 .HH / D g1 g2 H: If .g1 H /.g2 H / D .g1 g2 /H for all g1 ; g2 2 G we necessarily have gHg1 D H for all g 2 G. Hence we have proved: Lemma 10.1.6. Let H be a subgroup of a group G. Then the following are equivalent: (1) H is a normal subgroup of G. (2) g 1 Hg D H for all g 2 G. (3) gH D Hg for all g 2 G. (4) .g1 H /.g2 H / D .g1 g2 /H for all g1 ; g2 2 G.
Section 10.1 Normal Subgroups and Factor Groups
143
This is precisely the condition needed to construct factor groups. First we give some examples of normal subgroups. Lemma 10.1.7. Every subgroup of an abelian group is normal. Proof. Let G be abelian and H a subgroup of G. Suppose g 2 G then gh D hg for all h 2 H since G is abelian. It follows that gH D Hg. Since this is true for every g 2 G it follows that H is normal. Lemma 10.1.8. Let H G be a subgroup of index 2, that is ŒG W H D 2. Then H is normal in G. Proof. Suppose that ŒG W H D 2. We must show that gH D Hg for all g 2 G. If g 2 H then clearly H D gH D Hg. Therefore we may assume that g is not in H . Then there are only 2 left cosets and 2 right cosets. That is, G D H [ gH D H [ Hg: Since the union is a disjoint union we must have gH D Hg and hence H is normal.
Lemma 10.1.9. Let K be any field. Then the group SL.n; K/ is a normal subgroup of GL.n; K/ for any positive integer n. Proof. Recall that GL.n; K/ is the group of n n matrices over the field K with nonzero determinant while SL.n; K/ is the subgroup of n n matrices over the field K with determinant equal to 1. Let U 2 SL.n; K/ and T 2 GL.n; K/. Consider T 1 U T . Then det.T 1 U T / D det.T 1 / det.U / det.T / D det.U / det.T 1 T / D det.U / det.I / D det.U / D 1: Hence T 1 U T 2 SL.n; K/ for any U 2 SL.n; K/ and any T 2 GL.n; K/. It follows that T 1 SL.n; K/T SL.n; K/ and therefore SL.n; K/ is normal in GL.n; K/. The intersection of normal subgroups is again normal and the product of normal subgroups is normal. Lemma 10.1.10. Let N1 ; N2 be normal subgroups of the group G. Then (1) N1 \ N2 is a normal subgroup of G. (2) N1 N2 is a normal subgroup of G. (3) If H is any subgroup of G then N1 \H is a normal subgroup of H and N1 H D HN1 .
144
Chapter 10 Normal Subgroups, Factor Groups and Direct Products
Proof. (a) Let n 2 N1 \ N2 and g 2 G. Then g 1 ng 2 N1 since N1 is normal. Similarly g 1 ng 2 N2 since N2 is normal. Hence g1 ng 2 N1 \ N2 . It follows that g 1 .N1 \ N2 /g N1 \ N2 and therefore N1 \ N2 is normal. (b) Let n1 2 N1 ; n2 2 N2 . Since N1 ; N2 are both normal N1 N2 D N2 N1 as sets and the complex N1 N2 forms a subgroup of G. Let g 2 G and n1 n2 2 N1 N2 . Then g 1 .n1 n2 /g D .g 1 n1 g/.g 1 n2 g/ 2 N1 N2 since g 1 n1 g 2 N1 and g 1 n2 g 2 N2 . Therefore N1 N2 is normal in G. (c) Let h 2 H and n 2 N \ H . Then as in part (a) h1 nh 2 N \ H and therefore N \ H is a normal subgroup of H . The same argument as in part (b) shows that N1 H D HN1 . We now construct factor groups or quotient groups of a group modulo a normal subgroup. Definition 10.1.11. Let G be an arbitrary group and H a normal subgroup of G. Let G=H denote the set of distinct left (and hence also right) cosets of H in G. On G=H define the multiplication .g1 H /.g2 H / D g1 g2 H for any elements g1 H; g2 H in G=H . Theorem 10.1.12. Let G be a group and H a normal subgroup of G. Then G=H under the operation defined above forms a group. This group is called the factor group or quotient group of G modulo H . The identity element is the coset 1H D H and the inverse of a coset gH is g 1 H . Proof. We first show that the operation on G=N is well-defined. Suppose that a0 N D aN and b 0 N D bN , then b 0 2 bN and so b 0 D bn1 . Similarly a0 D an2 where n1 ; n2 2 N . Therefore a0 b 0 N D an2 bn1 N D an2 bN since n1 2 N . But b 1 n2 b D n3 2 N , since N is normal, so the right-hand side of the equation can be written as an2 bN D abN: Thus we have shown that if N G G then a0 b 0 N D abN , and the operation on G=N is indeed, well-defined. The associative law is true because coset multiplication as defined above uses the ordinary group operation which is by definition associative. The coset N serves as the identity element of G=N . Notice that aN N D aN 2 D aN
Section 10.1 Normal Subgroups and Factor Groups
145
and N aN D aN 2 D aN: The inverse of aN is a1 N since aNa1 N D aa1 N 2 D N: We emphasize that the elements of G=N are cosets and thus subsets of G. If jGj < 1, then jG=N j D ŒG W N , the member of cosets of N in G. It is also to be emphasized that in order for G=N to be a group N must be a normal subgroup of G. In some cases properties of G are preserved in factor groups. Lemma 10.1.13. If G is abelian then any factor group of G is also abelian. If G is cyclic then any factor group of G is also cyclic. Proof. Suppose that G is abelian and H is a subgroup of G. H is necessarily normal from Lemma 10.1.7 so that we can form the factor group G=H . Let g1 H; g2 H 2 G=H . Since G is abelian we have g1 g2 D g2 g1 . Then in G=H , .g1 H /.g2 H / D .g1 g2 /H D .g2 g1 /H D .g2 H /.g1 H /: Therefore G=H is abelian. We leave the proof of the second part to the exercises. An extremely important concept is when a group contains no proper normal subgroups other than the identity subgroup ¹1º. Definition 10.1.14. A group G ¤ ¹1º is simple provided that N G G implies N D G or N D ¹1º. One of the most outstanding problems in group theory has been to give a complete classification of all finite simple groups. In other words, this is the program to discover all finite simple groups and to prove that there are no more to be found. This was accomplished through the efforts of many mathematicians. The proof of this magnificent result took thousands of pages. We refer the reader to [18] for a complete discussion of this. We give one elementary example. Lemma 10.1.15. Any finite group of prime order is simple and cyclic. Proof. Suppose that G is a finite group and jGj D p where p is a prime. Let g 2 G with g ¤ 1. Then hgi is a nontrivial subgroup of G so its order divides the order of G by Lagrange’s theorem. Since g ¤ 1 and p is a prime we must have jhgij D p. Therefore hgi is all of G, that is G D hgi and hence G is cyclic. The argument above shows that G has no nontrivial proper subgroups and therefore no nontrivial normal subgroups. Therefore G is simple. In the next chapter we will examine certain other finite simple groups.
146
10.2
Chapter 10 Normal Subgroups, Factor Groups and Direct Products
The Group Isomorphism Theorems
In Chapter 1 we saw that there was a close relationship between ring homomorphisms and factor rings. In particular to each ideal, and consequently to each factor ring, there is a ring homomorphism that has that ideal as its kernel. Conversely to each ring homomorphism its kernel is an ideal and the corresponding factor ring is isomorphic to the image of the homomorphism. This was formalized in Theorem 1.5.7 which we called the ring isomorphism theorem. We now look at the group theoretical analog of this result, called the group isomorphism theorem. We will then examine some consequences of this result that will be crucial in the Galois theory of fields. Definition 10.2.1. If G1 and G2 are groups and f W G1 ! G2 is a group homomorphism then the kernel of f , denoted ker.f /, is defined as ker.f / D ¹g 2 G1 W f .g/ D 1º: That is the kernel is the set of the elements of G1 that map onto the identity of G2 . The image of f , denoted im.f /, is the set of elements of G2 mapped onto by f from elements of G1 . That is im.f / D ¹g 2 G2 W f .g1 / D g2 for some g1 2 G1 º: Note that if f is a surjection then im.f / D G2 . As with ring homomorphisms the kernel measures how far a homomorphism is from being an injection, that is, a one-to-one mapping. Lemma 10.2.2. Let G1 and G2 are groups and f W G1 ! G2 a group homomorphism. Then f is injective if and only if ker.f / D ¹1º. Proof. Suppose that f is injective. Since f .1/ D 1 we always have 1 2 ker.f /. Suppose that g 2 ker.f /. Then f .g/ D f .1/. Since f is injective this implies that g D 1 and hence ker.f / D ¹1º. Conversely suppose that ker.f / D ¹1º and f .g1 / D f .g2 /. Then f .g1 /.f .g2 //1 D 1 H) f .g1 g21 / D 1 H) g1 g21 2 ker.f /: Then since ker.f / D ¹1º we have g1 g21 D 1 and hence g1 D g2 . Therefore f is injective. We now state the group isomorphism theorem. This is entirely analogous to the ring isomorphism theorem replacing ideals by normal subgroups. We note that this theorem is sometimes called the first group isomorphism theorem.
Section 10.2 The Group Isomorphism Theorems
147
Theorem 10.2.3 (group isomorphism theorem). (a) Let G1 and G2 be groups and f W G1 ! G2 a group homomorphism. Then ker.f / is a normal subgroup of G1 , im.f / is a subgroup of G2 and G= ker.f / Š im.f /: (b) Conversely suppose that N is a normal subgroup of a group G. Then there exists a group H and a homomorphism f W G ! H such that ker.f / D N and im.f / D H . Proof. (a) Since 1 2 ker.f / the kernel is nonempty. Suppose that g1 ; g2 2 ker.f /. Then f .g1 / D f .g2 / D 1. It follows that f .g1 g21 / D f .g1 /.f .g2 //1 D 1. Hence g1 g21 2 ker.f / and therefore ker.f / is a subgroup of G1 . Further for any g 2 G1 we have f .g1 g1 g/ D .f .g//1 f .g1 /f .g/ D .f .g//1 1 f .g/ D f .g 1 g/ D f .1/ D 1: Hence g 1 g1 g 2 ker.f / and ker.f / is a normal subgroup. It is straightforward to show that im.f / is a subgroup of G2 . Consider the map fO W G= ker.f / ! im.f / defined by fO.g ker.f // D f .g/: We show that this is an isomorphism. Suppose that g1 ker.f / D g2 ker.f / then g1 g21 2 ker.f / so that f .g1 g21 / D 1. This implies that f .g1 / D f .g2 / and hence the map fO is well-defined. Now fO.g1 ker.f /g2 ker.f // D fO.g1 g2 ker.f // D f .g1 g2 / D f .g1 /f .g2 / D fO.g1 ker.f //fO.g2 ker.f // and therefore fO is a homomorphism. Suppose that fO.g1 ker.f // D fO.g2 ker.f // then f .g1 / D f .g2 / and hence g1 ker.f / D g2 ker.f /. It follows that fO is injective. Finally suppose that h 2 im.f /. Then there exists a g 2 G1 with f .g/ D h. Then fO.g ker.f // D h and fO is a surjection onto im.f /. Therefore fO is an isomorphism completing the proof of part (a). (b) Conversely suppose that N is a normal subgroup of G. Define the map f W G ! G=N by f .g/ D gN for g 2 G. By the definition of the product in the quotient group G=N it is clear that f is a homomorphism with im.f / D G=N . If g 2 ker.f / then f .g/ D gN D N since N is the identity in G=N . However this implies that g 2 N and hence it follows that ker.f / D N completing the proof.
148
Chapter 10 Normal Subgroups, Factor Groups and Direct Products
There are two related theorems that are called the second isomorphism theorem and the third isomorphism theorem. Theorem 10.2.4 (second isomorphism theorem). Let N be a normal subgroup of a group G and U a subgroup of G. Then U \ N is normal in U and .UN /=N Š U=.U \ N /: Proof. From Lemma 10.1.10 we know that U \ N is normal in U . Define the map ˛ W UN ! U=U \ N by ˛.un/ D u.U \ N /. If un D u0 n0 then u01 u D n0 n1 2 U \ N . Therefore u0 .U \ N / D u.U \ N / and hence the map ˛ is well-defined. Suppose that un; u0 n0 2 UN Since N is normal in G we have that unu0 n0 2 uu0 N . Hence unu0 n0 D uu0 n00 with n00 2 N . Then ˛.unu0 n0 / D ˛.uu0 n/ D uu0 .U \ N /: However U \ N is normal in U so uu0 .U \ N / D u.U \ N /u0 .U \ N / D ˛.un/˛.u0 n0 /: Therefore ˛ is a homomorphism. We have im.˛/ D U=.U \ N / by definition. Suppose that un 2 ker.˛/. Then ˛.un/ D U \ N N which implies u 2 N . Therefore ker.f / D N . From the group isomorphism theorem we then have UN=N Š U=.U \ N / proving the theorem. Theorem 10.2.5 (third isomorphism theorem). Let N and M be normal subgroups of a group G with N a subgroup of M . Then M=N is a normal subgroup in G=N and .G=N /=.M=N / Š G=M: Proof. Define the map ˇ W G=N ! G=M by ˇ.gN / D gM: It is straightforward that ˇ is well-defined and a homomorphism. If gN 2 ker.ˇ/ then ˇ.gN / D gM D M and hence g 2 M . It follows that ker.ˇ/ D M=N . In particular this shows that M=N is normal in G=N . From the group isomorphism theorem then .G=N /=.M=N / Š G=M:
Section 10.3 Direct Products of Groups
149
For a normal subgroup N in G the homomorphism f W G ! G=N provides a one-to-one correspondence between subgroups of G containing N and the subgroups of G=N . This correspondence will play a fundamental role in the study of subfields of a field. Theorem 10.2.6 (correspondence theorem). Let N be a normal subgroup of a group G and let f be the corresponding homomorphism f W G ! G=N . Then the mapping W H ! f .H / where H is a subgroup of G containing N provides a one-to-one correspondence between all the subgroups of G=N and the subgroups of G containing N . Proof. We first show that the mapping is surjective. Let H1 be a subgroup of G=N and let H D ¹g 2 G W f .g/ 2 H1 º: We show that H is a subgroup of G and that N H . If g1 ; g2 2 H then f .g1 / 2 H1 and f .g2 / 2 H1 . Therefore f .g1 /f .g2 / 2 H1 and hence f .g1 g2 / 2 H1 . Therefore g1 g2 2 H . In an identical fashion g11 2 H . Therefore H is a subgroup of G. If n 2 N then f .n/ D 1 2 H1 and hence n 2 H . Therefore N H showing that the map is surjective. Suppose that .H1 / D .H2 / where H1 and H2 are subgroups of G containing N . This implies that f .H1 / D f .H2 /. Let g1 2 H1 . Then f .g1 / D f .g2 / for some g2 2 H2 . Then g1 g21 2 ker.f / D N H2 . It follows that g1 g21 2 H2 so that g1 2 H2 . Hence H1 H2 . In a similar fashion H2 H1 and therefore H1 D H2 . It follows that is injective.
10.3
Direct Products of Groups
In this section we look at a very important construction, the direct product, which allows us to build new groups out of existing groups. This construction is the analog for groups of the direct sum of rings. As an application of this construction, in the next section we present a theorem which completely describes the structure of finite abelian groups. Let G1 ; G2 be groups and let G be the Cartesian product of G1 and G2 . That is G D G1 G2 D ¹.a; b/ W a 2 G1 ; b 2 G2 º: On G define .a1 ; b1 / .a2 ; b2 / D .a1 a2 ; b1 b2 /: With this operation it is direct to verify the groups axioms for G and hence G becomes a group.
150
Chapter 10 Normal Subgroups, Factor Groups and Direct Products
Theorem 10.3.1. Let G1 ; G2 be groups and G the Cartesian product G1 G2 with the operation defined above. Then G forms a group called the direct product of G1 and G2 . The identity element is .1; 1/ and .g; h/1 D .g1 ; h1 /. This construction can be iterated to any finite number of groups (also to an infinite number but we won’t consider that here) G1 ; : : : ; Gn to form the direct product G1 G2 Gn . Theorem 10.3.2. For groups G1 and G2 we have G1 G2 Š G2 G1 and G1 G2 is abelian if and only if each Gi , i D 1; 2, is abelian. Proof. The map .a; b/ ! .b; a/ where a 2 G1 ; b 2 G2 provides an isomorphism from G1 G2 ! G2 G1 . Suppose that both G1 ; G2 are abelian. Then if a1 ; a2 2 G1 ; b1 ; b2 2 G2 we have .a1 ; b1 /.a2 ; b2 / D .a1 a2 ; b1 b2 / D .a2 a1 ; b2 b1 / D .a2 ; b2 /.a1 ; b1 / and hence G1 G2 is abelian. Conversely suppose G1 G2 is abelian and suppose that a1 ; a2 2 G1 . Then for the identity 1 2 G2 we have .a1 a2 ; 1/ D .a1 ; 1/.a2 ; 1/ D .a2 ; 1/.a1 ; 1/ D .a2 a1 ; 1/: Therefore a1 a2 D a2 a1 and G1 is abelian. Identically G2 is abelian. We show next that in G1 G2 there are normal subgroups H1 ; H2 with H1 Š G1 and H2 Š G2 . Theorem 10.3.3. Let G D G1 G2 . Let H1 D ¹.a; 1/ W a 2 G1 º and H2 D ¹.1; b/ W b 2 G2 º. Then both H1 and H2 are normal subgroups of G with G D H1 H2 and H1 \ H2 D ¹1º. Further H1 Š G1 ; H2 Š G2 and G=H1 Š G2 and G=H2 Š G1 . Proof. Map G1 G2 onto G2 by .a; b/ ! b. It is clear that this map is a homomorphism and that the kernel is H1 D ¹.a; 1/ W a 2 G1 º. This establishes that H1 is a normal subgroup of G and that G=H1 Š G2 . In an identical fashion we get that G=H2 Š G1 . The map .a; 1/ ! a provides the isomorphism from H1 onto G1 . If the factors are finite it is easy to find the order of G1 G2 . The size of the Cartesian product is just the product of the sizes of the factors. Lemma 10.3.4. If jG1 j and jG2 j are finite then jG1 G2 j D jG1 jjG2 j. Now suppose that G is a group with normal subgroups G1 ; G2 such that G D G1 G2 and G1 \ G2 D ¹1º. Then we will show that G is isomorphic to the direct product G1 G2 . In this case we say that G is the internal direct product of its subgroups and that G1 ; G2 are direct factors of G.
Section 10.4 Finite Abelian Groups
151
Theorem 10.3.5. Suppose that G is a group with normal subgroups G1 ; G2 such that G D G1 G2 and G1 \ G2 D ¹1º. Then G is isomorphic to the direct product G1 G2 . Proof. Since G D G1 G2 each element of G has the form ab with a 2 G1 ; b 2 G2 . We first show that each a 2 G1 commutes with each b 2 G2 . Consider the element aba 1 b 1 . Since G1 is normal ba1 b 1 2 G1 which implies that abab 1 2 G1 . Since G2 is normal aba1 2 G2 which implies that aba1 b 1 2 G2 . Therefore aba1 b 1 2 G1 \ G2 D ¹1º and hence aba1 b 1 D 1 so that ab D ba. Now map G onto G1 G2 by f .ab/ ! .a; b/. We claim that this is an isomorphism. It is clearly onto. Now f ..a1 b1 /.a2 b2 // D f .a1 a2 b1 b2 / D .a1 a2 ; b1 b2 / D .a1 ; b1 /.a2 ; b2 / D f ..a1 ; b1 //f .a2 ; b2 // so that f is a homomorphism. The kernel is G1 \ G2 D ¹1º and so f is an isomorphism. Although the end resulting groups are isomorphic we call G1 G2 an external direct product if we started with the groups G1 ; G2 and constructed G1 G2 and call G1 G2 an internal direct product if we started with a group G having normal subgroups as in the theorem.
10.4
Finite Abelian Groups
We now use the results of the last section to present a theorem that completely provides the structure of finite abelian groups. This theorem is a special case of a general result on modules that we will examine in detail later in the book. Theorem 10.4.1 (basis theorem for finite abelian groups). Let G be a finite abelian group. Then G is a direct product of cyclic groups of prime power order. Before giving the proof we give two examples showing how this theorem leads to the classification of finite abelian groups. Since all cyclic groups of order n are isomorphic to .Zn ; C/ we will denote a cyclic group of order n by Zn . Example 10.4.2. Classify all abelian groups of order 60. Let G be an abelian group of order 60. From Theorem 10.4.1 G must be a direct product of cyclic groups of prime power order. Now 60 D 22 3 5 so the only primes involved are 2, 3 and 5. Hence the cyclic group involved in the direct product decomposition of G have order either 2, 4, 3 or 5 (by Lagrange’s theorem they must be divisors of 60). Therefore G
152
Chapter 10 Normal Subgroups, Factor Groups and Direct Products
must be of the form G Š Z4 Z3 Z5 G Š Z2 Z2 Z3 Z5 : Hence up to isomorphism there are only two abelian groups of order 60. Example 10.4.3. Classify all abelian groups of order 180. Now 180 D 22 32 5 so the only primes involved are 2, 3 and 5. Hence the cyclic group involved in the direct product decomposition of G have order either 2, 4, 3, 9 or 5 (by Lagrange’s theorem they must be divisors of 180). Therefore G must be of the form G Š Z4 Z9 Z5 G Š Z2 Z2 Z9 Z5 G Š Z4 Z3 Z3 Z5 G Š Z2 Z2 Z3 Z3 Z5 : Hence up to isomorphism there are four abelian groups of order 180. The proof of Theorem 10.4.1 involves the following lemmas. Lemma 10.4.4. Let G be a finite abelian group and let pjjGj where p is a prime. Then all the elements of G whose orders are a power of p form a normal subgroup of G. This subgroup is called the p-primary component of G, which we will denote by Gp . Proof. Let p be a prime with pjjGj and let a and b be two elements of G of order a power of p. Since G is abelian the order of ab is the lcm of the orders which is again a power of p. Therefore ab 2 Gp . The order of a1 is the same as the order of a so a 1 2 Gp and therefore Gp is a subgroup. Lemma 10.4.5. Let G be a finite abelian group of order n. Suppose that n D e p1e1 pkk with p1 ; : : : ; pk distinct primes. Then G Š Gp1 Gpk where Gpi is the pi -primary component of G. Proof. Each Gpi is normal since G is abelian and since distinct primes are relatively prime the intersection of the Gpi is the identity. Therefore Lemma 10.4.5 will follow by showing that each element of G is a product of elements in the Gp1 .
153
Section 10.4 Finite Abelian Groups
Let g 2 G. Then the order of g is p1f1 pkfk . We write this as pifi m with f
.m; pi / D 1. Then g m has order pi i and hence is in Gpi . Now since p1 ; : : : ; pk are relatively prime there exists m1 ; : : : ; mk with f
f
m1 p1 1 C C mk pk k D 1 and hence
f1
fk
g D .g p1 /m1 .g pk /mk : Therefore g is a product of elements in the Gpi . We next need the concept of a basis. Let G be any finitely generated abelian group (finite or infinite) and let g1 ; : : : ; gn be a set of generators for G. The generators g1 ; : : : ; gn form a basis if G D hg1 i hgn i; that is G is the direct product of the cyclic subgroups generated by the gi . The basis theorem for finite abelian groups says that any finite abelian group has a basis. Suppose that G is a finite abelian group with a basis g1 ; : : : ; gk so that G D hg1 i hgk i. Since G is finite each gi has finite order say mi . It follows than from the fact that G is a direct product that each g 2 G can be expressed as g D g1n1 gknk and further the integers n1 ; : : : ; nk are unique modulo the order of gi . Hence each integer ni can be chosen in the range 0; 1; : : : ; mi 1 and within this range for the element g the integer ni is unique. From the previous lemma each finite abelian group splits into a direct product of its p-primary components for different primes p. Hence to complete the proof of the basis theorem we must show that any finite abelian group of order p m for some prime p has a basis. We call an abelian group of order p m an abelian p-group. Consider an abelian group G of order p m for a prime p. It is somewhat easier to complete the proof if we consider the group using additive notation. That is the operation is considered C, the identity as 0 and powers are given by multiples. Hence if an element g 2 G has order p k then in additive notation p k g D 0. A set of elements g1 ; : : : ; gk is then a basis for G if each g 2 G can be expressed uniquely as g D m1 g1 C C mk gk where the mi are unique modulo the order of gi . We say that the g1 ; : : : ; gk are independent and this is equivalent to the fact that whenever m1 g1 C C mk gk D 0 then mi 0 modulo the order of gi . We now prove that any abelian p-group has a basis. Lemma 10.4.6. Let G be a finite abelian group of prime power order p n for some prime p. Then G is a direct product of cyclic groups.
154
Chapter 10 Normal Subgroups, Factor Groups and Direct Products
Notice that in the group G we have p n g D 0 for all g 2 G as a consequence of Lagrange’s theorem. Further every element has order a power of p. The smallest power of p say p r such that p r g D 0 for all g 2 G is called the exponent of G. Any finite abelian p-group must have some exponent p r . Proof. The proof of this lemma is by induction on the exponent. The lowest possible exponent is p so suppose first that pg D 0 for all g 2 G. Since G is finite it has a finite system of generators. Let S D ¹g1 ; : : : ; gk º be a minimal set of generators for G. We claim that this is a basis. Since this is a set of generators to show its a basis we must show that they are independent. Hence suppose that we have m1 g1 C C mk gk D 0
(1)
for some set of integers mi . Since the order of each gi is p, as explained above we may assume that 0 mi < p for i D 1; : : : ; k. Suppose that one mi ¤ 0. Then .mi ; p/ D 1 and hence there exists an xi with mi xi 1 mod p (see Chapter 4). Multiplying the equation (1) by xi we get modulo p, m1 xi g1 C C gi C C mk xi gk ; and rearranging gi D m1 xi g1 mk xk gk ; But then gi can be expressed in terms of the other gj and therefore the set ¹g1 ; : : : ; gk º is not minimal. It follows that g1 ; : : : ; gk constitute a basis and the lemma is true for the exponent p. Now suppose that any finite abelian group of exponent p n1 has a basis and assume that G has exponent p n . Consider the set G D pG D ¹pg W g 2 Gº. It is straightforward that this forms a subgroup (see exercises). Since p n g D 0 for all g 2 G it follows that p n1 g D 0 for all g 2 G and so the exponent of G p n1 . By the inductive hypothesis G has a basis S D ¹pg1 ; : : : ; pgk º: Consider the set ¹g1 ; : : : ; gk º and adjoin to this set the set of all elements h 2 G satisfying ph D 0. Call this set S1 so that we have S1 D ¹g1 ; : : : ; gk ; h1 ; : : : ; h t º: We claim that S1 is a set of generators for G. Let g 2 G. Then pg 2 G which has the basis pg1 ; : : : ; pgk so that pg D m1 pg1 C C mk pgk : This implies that p.g m1 g1 mk gk / D 0
155
Section 10.4 Finite Abelian Groups
so that g1 m1 g1 mk gk must be one of the hi . Hence g m 1 g 1 m k g k D hi
so that g D m1 g1 C C mk gk C hi
proving the claim. Now S1 is finite so there is a minimal subset of S1 that is still a generating system for G. Call this S0 and suppose that S0 , renumbering if necessary, is S0 D ¹g1 ; : : : ; gr ; h1 ; : : : ; hs º with phi D 0 for i D 1; : : : ; s: The subgroup generated by h1 ; : : : ; hs has exponent p so by inductive hypothesis has a basis. We may assume than that h1 ; : : : ; hs is a basis for this subgroup and hence is independent. We claim now that g1 ; : : : ; gr ; h1 ; : : : ; hs are independent and hence form a basis for G. Suppose that m1 gr C C mr gr C n1 h1 C C ns hs D 0
(2)
for some integers m1 ; : : : ; mr ; h1 ; : : : ; hs . Each mi ; ni must be divisible by p. Suppose for example that some mi is not. Then .mi ; p/ D 1 and then .mi ; p n / D 1. This implies that there exists an xi with mi xi 1 mod p n . Multiplying through by xi and rearranging we then obtain gi D m1 xi g1 ns xi hs : Therefore gi can be expressed in terms of the remaining elements of S0 contradicting the minimality of S0 . An identical argument works if some ni is not divisible by p. Therefore the relation (2) takes the form a1 pg1 C C ar pgr C b1 ph1 C C bs phs D 0:
(3)
Each of the terms phi D 0 so that (3) becomes ap g1 C C ar pgr D 0: The g1 ; : : : ; gr are independent and hence ai p D 0 for each i and hence ai D 0. Now (2) becomes n1 h1 C C ns hs D 0: However h1 ; : : : ; hs are independent so each ni D 0 completing the claim. Therefore the whole group G has a basis proving the lemma by induction. For more details see the proof of the general result on modules over principal ideal domains later in the book. There is also an additional elementary proof for the basis theorem for finitely generated abelian groups.
156
10.5
Chapter 10 Normal Subgroups, Factor Groups and Direct Products
Some Properties of Finite Groups
Classification is an extremely important concept in algebra. A large part of the theory is devoted to classifying all structures of a given type, for example all UFD’s. In most cases this is not possible. Since for a given finite n there are only finitely many group tables it is theoretically possible to classify all groups of order n. However even for small n this becomes impractical. We close the chapter by looking at some further results on finite groups and then using these to classify all the finite groups up to order 10. Before stating the classification we give some further examples of groups that are needed. Example 10.5.1. In Example 9.2.6 we saw that the symmetry group of an equilateral triangle had 6 elements and is generated by elements r and f which satisfy the relations r 3 D f 2 D 1, f 1 rf D r 1 , where r is a rotation of 120ı about the center of the triangle and f is a reflection through an altitude. This was called the dihedral group D3 of order 6. This can be generalized to any regular n-gon. If D is a regular n-gon, then the symmetry group Dn has 2n elements and is called the dihedral group of order 2n. It is generated by elements r and f which satisfy the relations r n D f 2 D 1, f 1 rf D r n1 , where r is a rotation of 2 n about the center of the n-gon and f is a reflection. Hence, D4 , the symmetries of a square, has order 8 and D5 , the symmetries of a regular pentagon, has order 10. Example 10.5.2. Let i; j; k be the generators of the quaternions. Then we have i 2 D j 2 D k 2 D 1;
.1/2 D 1
and
ij k D 1:
These elements then form a group of order 8 called the quaternion group denoted by Q. Since ij k D 1 we have ij D j i , and the generators i and j satisfy the relations i 4 D j 4 D 1, i 2 D j 2 , ij D i 2 j i . We now state the main classification and then prove it in a series of lemmas. Theorem 10.5.3. Let G be a finite group. (a) If jGj D 2 then G Š Z2 . (b) If jGj D 3 then G Š Z3 . (c) If jGj D 4 then G Š Z4 or G Š Z2 Z2 . (d) If jGj D 5 then G Š Z5 . (e) If jGj D 6 then G Š Z6 Š Z2 Z3 or G Š D3 , the dihedral group with 6 elements. (Note D3 Š S3 the symmetric group on 3 symbols.) (f) If jGj D 7 then G Š Z7 .
Section 10.5 Some Properties of Finite Groups
157
(g) If jGj D 8 then G Š Z8 or G Š Z4 Z2 or G Š Z2 Z2 Z2 or G Š D4 , the dihedral group of order 8, or G Š Q the quaternion group. (h) If jGj D 9 then G Š Z9 or G Š Z3 Z3 . (i) If jGj D 10 then G Š Z10 Š Z2 Z5 or G Š D5 , the dihedral group with 10 elements. Recall from Section 10.1 that a finite group of prime order must be cyclic. Hence in the theorem the cases jGj D 2; 3; 5; 7 are handled. We next consider the case where G has order p 2 where p is a prime. Definition 10.5.4. If G is a group then its center denoted Z.G/ is the set of elements in G which commute with everything in G. That is Z.G/ D ¹g 2 G W gh D hg for any h 2 Gº: Lemma 10.5.5. For any group G the following hold: (a) The center Z.G/ is a normal subgroup. (b) G D Z.G/ if and only if G is abelian. (c) If G=Z.G/ is cyclic then G is abelian. Proof. (a) and (b) are direct and we leave them to the exercises. Consider the case where G=Z.G/ is cyclic. Then each coset of Z.G/ has the form g m Z.G/ where g 2 G. Let a; b 2 G. Then since a; b are in cosets of the center we have a D g m u and b D g n v with u; v 2 Z.G/. Then ab D .g m u/.g n v/ D .g m g n /.uv/ D .g n g m /.vu/ D .g n v/.g m u/ D ba since u; v commute with everything. Therefore G is abelian. A p-group is any finite group of prime power order. We need the following. The proof of this is based on what is called the class equation which we will prove in Chapter 13. Lemma 10.5.6. A finite p-group has a nontrivial center of order at least p. Lemma 10.5.7. If jGj D p2 with p a prime then G is abelian and hence G Š Zp2 or G Š Zp Zp . Proof. Suppose that jGj D p 2 . Then from the previous lemma G has a nontrivial center and hence jZ.G/j D p or jZ.G/j D p 2 . If jZ.G/j D p 2 then G D Z.G/ and G is abelian. If jZ.G/j D p then jG=Z.G/j D p. Since p is a prime this implies that G=Z.G/ is cyclic and hence from Lemma 10.5.5 G is abelian.
158
Chapter 10 Normal Subgroups, Factor Groups and Direct Products
Lemma 10.5.7 handles the cases n D 4 and n D 9. Therefore
