Infinite Crossed Products
This is Volume 135 in PURE AND APPLIED MATHEMATICS
H. Bass, A. Borel, J. Moser, and S.-T. Yau, editors Paul A. Smith and Samuel Eilenberg, founding editors A list of titles in this series appears at the end of this volume.
Infinite Crossed Products Donald S . Passman Mathematics Department University of Wisconsin - Madison Madison, Wisconsin
ACADEMIC PRESS, INC. Harcourt Brace Jovanovich, Publishers Boston San Diego New York Berkeley London Sydney Tokyo Toronto
Copyright @ 1989 by Academic Press, Inc. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher.
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U n i t e d K i n g d o m E d i t i o n published by ACADEMIC PRESS INC. (LONDON) LTD. 24-28 Oval Road, London NW1 7DX
Library of Congress Cataloging-in-Publication Data Passman, Donald S., D a t e Infinite crossed products / Donald Passman. p. cm. - (Pure and applied mathematics : v. 135) Bibliography: p. Includes index. ISBN 0-12-546390-1 2. Group rings. 1. Von Neumann algebras-Crossed products. 3. Galois theory. I. Title. 11. Series: Pure and applied mathematics : 135. QA3.P8 vol. 135 [QA3261 510 SAC 19 88-7597 [512'.55] CIP Printed in the United States of America 89 90 91 92 987654321
This book is dedicated to my mother, Fanny Passman, and to the memory of my father, Julius Passman.
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Contents
Preface Chapter 1. Crossed Products and Group-Graded Rings 1. Crossed Products 2. Group-Graded Rings and Duality 3. Induced Modules 4. Maschke’s Theorem
Chapter 2. Delta Methods and Semiprime Rings 5. 6. 7. 8. 9.
Delta Methods Coset Calculus Minimal Forms Sufficient Conditions Polynomial Identities
xi 1 1 10 19 29
39 39 48 56 67 74
Chapter 3. The Symmetric Ring of Quotients
83
10. The Martindale Ring of Quotients
83
vii
viii
Contents
11. Separated Groups 12. X-Inner Automorphisms 13. Free Rings
Chapter 4. Prime Ideals - The Finite Case
94 105 117
131
14. G-Prime Coefficients 15. Prime Coefficients 16. Finite Groups and Incomparability 17. Primeness and Sylow Subgroups 18. Semiprimeness and Sylow Subgroups
131 142 151 163 176
Chapter 5. Prime Ideals - The Noetherian Case
187
19. 20. 21. 22. 23.
Orbitally Sound Groups Polycyclic Group Algebras Polycyclic Crossed Products Jacobson Rings P. I. Algebras
187 197 208 220 232
Chapter 6. Group Actions and Fixed Rings
241
24. 25. 26. 27. 28.
Fixed Points and Traces Integrality Finiteness Conditions Rings With No Nilpotent Elements Prime Ideals and Fixed Rings
Chapter 7. Group Actions and Galois Theory 29. 30. 31. 32.
Traces and Truncation The Galois Correspondence Almost Normal Subgroups Free Rings and Subrings
241 254 264 276 285
297 297 309 319 330
ix
Contents
Chapter 8. Grothendieck Groups and Induced Modules 33. 34. 35. 36.
Grothendieck Groups Graded Rings Group Extensions The Induction Theorem
343 343 356 367 378
Chapter 9. Zero Divisors and Idempotents
391
Zero Divisors and Goldie Rank The Zalesskii-Neroslavskii Example Almost Injective Modules Stably Free Modules
391 405 418 430
37. 38. 39. 40.
References
445
Index
459
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Preface
Crossed products are another meeting place for group theory and ring theory. Historically, they first occurred in the study of finite dimensional division algebras and central simple algebras. More recently, they have become closely related to the study of infinite group algebras, groupgraded rings and the Galois theory of noncommutative rings. This book is mainly concerned with these newer developments. During the past few years, there have been a number of major achievements in this field. These include: 1. Cohen-Montgomery duality, a machine to translate crossed product results into the context of groupgraded rings; 2. Understanding and computing the symmetric Martindale ring of quotients of prime and semiprime rings; 3. Classifying prime and semiprime crossed products and, more generally, the prime ideals in certain important special cases; 4. The Galois theory of prime and semiprime rings, along with skew group ring applications to the subject; 5. Determining the Grothendieck group of a Noetherian crossed product to settle the zero divisor and Goldie rank conjectures.
Indeed, these topics form the core of the book and the reason for its existence.
xi
xii
Preface
Chapter 1 is introductory in nature. It contains many of the basic definitions and proves duality and various versions of Maschke’s theorem. Chapter 2 uses Delta methods, a coset counting technique, to classify the prime and semiprime crossed products. Chapter 3 discusses the left and symmetric Martindale ring of quotients and X-inner automorphisms of rings. Numerous examples are computed. Chapters 4 and 5 study prime ideals in crossed products R*G with either G finite or with G polycyclic-by-finite and R right Noetherian. Chapters 6 and 7 are concerned with group actions on rings. Topics include the existence of fixed points, integrality, prime ideals and the Galois theory of prime rings. Finally, Chapters 8 and 9 consider the Grothendieck groups of Noetherian crossed products. In particular this material includes the induction theorem, the zero divisor and Goldie rank conjectures, the Zalesskii-Neroslavskii example and some specific computations. The book is written in a reasonably self-contained manner. Nevertheless, some facts (usually concerning infinite groups, group algebras or homological algebra) have to be quoted. At such points, the necessary prerequisites are at least precisely stated. The book contains over 200 exercises and a challenge to the reader to overcome certain notational inconsistencies. For example, modules are usually right, but not always. Functions are sometimes written on the left, sometimes on the right and sometimes as exponents and, of course, this effects the way they multiply. In any case, 3 always stands for strict inclusion, while 2 allows for equality. In closing, I would like to thank the following colleagues and friends for their many helpful comments and suggestions: Gerald Cliff, Martin Lorenz, Susan Montgomery, Jim Osterburg and Declan Quinn. Of course, thanks also to the National Science Foundation for its support of my research. Finally, I would like to express my love and appreciation to my family Marjorie, Barbara and Jonathan for their encouragement and support of this project. Donald S. Passman Madison, Wisconsin June 1988
1
Crossed Products and Group-Graded Rings
1. Crossed Products Let K be a field and let H be a multiplicative group. Then the group algebra K [ H ]is an associative K-algebra with basis { z I z E H } and with multiplication defined distributively using the group multiplication in H . If N is a subgroup of H , then certainly K [ N ]C K [ H ] . Furthermore, if N a H , then it is natural to expect that K [ H ]is somehow constructed from the subgroup ring K [ N ]and the quotient group H / N . To this end, let R = K [ N ]and G = H / N . For each z E G let 3 E H be a fixed inverse image. Then the disjoint union H = U, Z N implies that X
X
so G = { 5 I z E G } is an R-basis for K [ H ] . Since N a H , 3-lNZ = N so Z-'K[N]? = K [ N ]and 3 induces a conjugation automorphism u(z)on R. Thus we have a map, although not necessarily a group homomorphism, u: G -+ Aut(R).
1
2
1. Crossed Products and Group-Graded Rings
Furthermore, if
2
E G and r E R then
Note that u is trivial if N is central in H . Next for 2,y E G we have iEN - yN = ZyN so 39 = ?@T(z,y) for some ~ ( 2 , y )E N U(R), the group of units of R. Thus we have a map T:G x G + U(R) which is called the twisting. Note that T is trivial if we can choose a consistent set of coset representatives, that isifH=NMG. What we have shown is that K[H] = R*G = K[N]*(H/N) is a crossed product of H/N over the ring K“].
Definition. Let R be a ring with 1 and let G be a group. Then a crossed product R*G of G over R is an associative ring which contains R and has as an R-basis the set G, a copy of G. Thus each element of R*G is uniquely a finite sum CzEG Zr, with r, E R. Addition is as expected and multiplication is determined by the two rules below. Specifically for 2,y E G we have
zy = z y T ( 2 , y)
(twisting)
where T : G x G -, U = U(R), the group of units of R. Furthermore for z E G and r E R we have
where u:G
+ Aut(R).
Note that, by definition, a crossed product is merely an associative ring which happens to have a particular structure relative to R and G. We will usually assume that R*G is given. However for the rare occasions when we wish to construct such rings, the following is crucial.
Lemma 1.1. The associativity of R*G is equivalent to the assertions that for all 2,y, z E G i. ~ ( x yz, ) ~ ( xy)+) ,
=~
( 2y a ,) ~ ( y2) ,
1. Crossed Products
3
ii. a ( y ) a ( z ) = a(yz)q(y, z ) where q(y, z ) denotes the automorphism of R induced by the unit ~ ( yz,) .
Proof. The associativity of R*G is clearly equivalent to the equality
for all r, s, t E R and z, y, z E G. Simple computation shows that the left-hand expression equals
while the right-hand expression becomes
The result follows by first setting r = s = t = 1. I Equation (i) above asserts that 7 is a 2-cocycle for the action of G on U . Unlike group rings, crossed products do not have a natural basis. Indeed if d: G -, U assigns to each element z E G a unit d,, then G = { 2 = 3d, I z E G } yields an alternate R-basis for R*G which still exhibits the basic crossed product structure. We call this a diagonal change of basis. Now it is easy to see (Exercise 2 ) that the identity element of R*G is of the form 1 = i u for some u E U . Thus, via a diagonal change of basis, we can and will assume that I = 1. The embedding of R into R*G is then given by T H Ir. On the other hand, G is in general not contained in R*G. Rather, each 3 is a unit of the ring and 6 = { 3u 1 x E G , u E U } , the group of trivial units of R*G, satisfies G/U G. Note that B acts on both R*G and R by conjugation and that for z E G, T E R we have 3 - h = T ~ ( ~ ) . Certain special cases of crossed products have their own names. If there is no action or twisting, that is if a(z)= 1 and ~ ( z , y )= 1 for all z,y E G, then R*G = R[G]is an ordinary group ring. If the
4
1. Crossed Products and Group-Graded Rings
action is trivial, then R*G = Rt[G]is a twisted group ring. Finally if the twisting is trivial, then R*G = RG is a skew group ring. We frequently construct the latter rings using the following immediate consequence of Lemma 1.1.
Lemma 1.2. The associativity of RG is equivalent to the map u: G -+ Aut(R) being a group homomorphism. Note that since the twisting is trivial in RG we have %Gy = Zy. Thus we can drop the overbars here and assume that RG 2 G. Historically, crossed products arose in the study of division rings. Let K be a field and let D be a division algebra finite-dimensional over its center K . If F is a maximal subfield of D ,then dimK D = (dimK F ) 2 . Suppose that F / K is normal, although this is not always true. If z E Gal(F/K) = G, then the Skolem-Noether theorem implies that there exists 5 E D \ 0 with %-'f% = f" for all f E F . Furthermore, Zjj and Zy agree in their action on F so Gy-'5-'Zy E C D ( F )= F . Once we show (Exercise 5) that the elements 5 are linearly independent over F , we can then conclude by computing dimensions that D = ~3CzEG %F = F I G . More generally, suppose A is a central simple algebra over K so that A is a simple ring, finite dimensional over its center K . Then A = Mn(D) for some n and division ring D with Z ( D ) = K . Two such algebras A and B are equivalent if they have the same D. The equivalence classes then form a group, the Brauer group, under tensor product 8 ~ Now . given A , one can show that there exists B A with B = F*G. But F*G is determined by the twisting function 7:G x G + F , a 2-cocycle. Thus in this way we obtain the homological characterization of the Brauer group as the 2nd-cohorno20gy group. See [73,Chapter 41 for more details. The thrust of this book is in other directions; namely we are concerned here with the ring theoretic structure of crossed products. For example we will consider when such rings are prime or semiprime and we will discuss the nature of their prime ideals and modules. F'urthermore we will describe crossed product applications to problems in group algebras and to the Galois theory of rings. N
5
1. Crossed Products
We remark that the notation R*G for a crossed product is certainly ambiguous since it does not convey the full (T,.r-structure. Nevertheless it is simpler and hence preferable to something like ( R ,G ,(T, 7 ) . Moreover it rarely if ever causes confusion. For example if a = CxEG S r , E R*G, then the support of Q is given by Supp a = { z E G
I T, # 0 ) .
It follows that if H is a subgroup of G , then
{ a € R*G I S u p p a E H } = R*H is the naturally embedded sub-crossed product. Furthermore the argument at the beginning of this section yields
Lemma 1.3. Let R*G be given and let N
Q
G. Then
R*G = ( R * N ) * ( G / N ) where the latter is some crossed product of the group GIN over the ring R* N . Recall that conjugation by the elements of 8 yields a homomorphism 8 -, Aut(R). Therefore B permutes the ideals of R and, since U obviously fixes all such ideals, we obtain a well-defined permutation action of G g G/U on the set of these ideals. If I is a G-stable ideal of R we write I*G = (R*G)I.
Lemma 1.4. Let R*G be given. i. I f J Q R*G, then J n R is a G-stable ideal of R and J 2 ( J n R)*G. ii. If1 is a G-stable ideal o f R , then I*GaR*G with ( I * G ) n R= I . Moreover ( R * G ) / ( I * G ) ( R / I ) * G where the latter is a suitable crossed product of G over R / I .
Proof: (i) This is clear since both R and J are stable under the action of
8.
6
1. Crossed Products and Group-Graded Rings
(ii) Since I is &stable, it is clear that IB = 61 and hence that I*G = (R*G)I is an ideal of R*G. Furthermore since
we have (I*G) f l R = I . Finally let &R*G --+ (R*G)/(I*G) be the natural homomorphism. Then +(R*G) is an associative ring containing 4(R) S ( R / I )and 4(G). Since the action and twisting equations in R*G map to similar equations in 4(R*G) and since +(G) is clearly a free basis for ~!J(R*G) over 4(R),the result follows. I Suppose N a G . Then R*G = ( R * N ) * ( G / N )so G/N and hence also G permute the ideals of R*N. If I is a G-stable ideal of R*N, we deduce from the above that (R*G)IaR*G and that (R*G)/(R*G)I = [(R*N)/II*(G/N). The closure properties of crossed products exhibited in the previous two lemmas are crucial. The first allows us to study R*G by lifting information from various R*N with N a G; the second allows us to study ideals J of R*G under the simplifying assumption that
JnR=o. Now let H be a subgroup of G. Then there is a natural projection T H : R*G + R*H given by TH
(c
?rx)
xEG
=
C zrx. xEH
Note that both R*G and R*H are R*H-bimodules under left and right multiplication and then clearly
Lemma 1.5. The map T H : R*G
R*H is an R*H-bimodde homo-
morphism.
Two classes of groups play key roles in the study of crossed products. First, of course, there are the finite groups; next there are the polycyclic-by-finite ones. Recall that G is polycyclic-by-finite if G has a subnormal series l = G o a G 1 a ... aG, = G
7
1. Crossed Products
with each Gi+l/Gi either infinite cyclic or finite. The following is a simple extension of the Hilbert Basis Theorem.
Proposition 1.6. I f R is right Noetherian and G is polycyclic-byfinite, then R*G is right Noetherian.
Proof: By induction and Lemma 1.3, it suffices to assume that G is either infinite cyclic or finite. If G = (x)is infinite cyclic, then R*G = (R,Z,3-') is generated by R, Z and 3-l with 3F1R3 = R. It follows from [161, Theorem 10.2.6(iii)] that R*G is right Noetherian. If G is finite, then R*G is a finitely generated R-module and hence a Noetherian right R-module. Thus R*G is also Noetherian as a right R*G-module. I In rare cases we will consider semigroup crossed products. These have the same action and twisting structure as ordinary crossed products, but G is allowed to be a multiplicative semigroup. Thus for example, if G is the infinite cyclic semigroup G = { l , ~ x2,. , . . } we obtain R*G = R [ x ;u]a skew polynomial ring. If G is the free abelian semigroup on n variables, then R*G is just a noncommutative analog of a polynomial ring in n variables. The next result follows from [161, Theorem 10.2.6(ii)] by induction on n.
Lemma 1.7. I f R is right Noetherian and G is the free abelian semigroup on n generators, then R*G is right Noetherian. Obviously Proposition 1.6 and Lemma 1.7 have left analogs. Returning to groups G , we close this section by observing that any crossed product R*G has an untwisted extension. Specifically, there exists an overring S 2 R such that S*G 2 R*G and S*G = SG is a skew group ring. The proof of this result requires an extremely large ring extension of R. If R is given and { ui I i E I } is a collection of symbols, then we let = ( R ,ui,U i l I i E I )
s
8
1. Crossed Products and Group-Graded Rings
be the ring freely generated by R and the various ui and u i ' . Furthermore we insist that l € R is the identity of s. This ring can be constructed by first taking F = (uiI i E I) to be the free group generated by the ui and then forming the free product of the ring R with the integral group ring of F amalgamating the identity elements of the two rings. We will not use any specific structure theorem for S. We will only need the fact that S exists and satisfies an appropriate universal property. To be precise, suppose T is a ring and 7:R + T is an embedding. If ti is a unit of T for each i E I, then 7 extends to a homomorphism 7:S T with ui H ti. The following result generalizes the standard untwisting of 2cocycles in the commutative case. --f
Lemma 1.8. Let R*G be given and set S = (R,u,, u;I I z E G\{l}). Then there exists a crossed product S*G containing R*G such that S*G = SG is a skew group ring. ProuJ We recall the basic crossed product definitions. Thus in R*G we have 3 g = wr(z,y) and rZ = f ~ " (where ~ ) ~ ( zy), E U(R) and ~ ( zE ) Aut(R). Since we assume that I = 1, we have in addition ~ ( z , l=) ~ ( 1 , z=) 1 and o(1) = 1. Furthermore, by Lemma 1.1, associativity in R*G is equivalent to
+Y,
Y)"'"' = +, YZ)T(Y, 4Y>+> = 4YZ)rl(Y,Z) z),+,
2)
(*>
(**>
for all z, y, z E G, where q(y, z ) is the automorphism of R induced by the unit ~ ( yz). , Now let S be as given and set u1 = 1 E S. For each y E G we extend a ( y ) E Aut(R) to an endomorphism of S by T H T " ( Y ) for T E R and u, H ~ ( z , y ) - ~ u , ~ u ;Note '. that each ~ ( z , y ) - ' u , ~ u ; ' is a unit of S and that when z = 1 we have u1 H 1. Thus, by the universal property of S , a ( y ) is indeed an endomorphism of S. We first show that these extended endomorphisms also satisfy (**). This is certainly the case when applied to any T E R so we need only apply these maps to u,. We have (2L5)"(v)u(z' = ( T ( 2 , y)-lzl,yug -1 )''(%I = T(2,Y)-''(')T(Zy,
Z)-'U,y~U~JT(Y, 2)
1. Crossed Products
9
so, by (*), these two are equal. Furthermore note that u(1): u, H u, so a(1)= 1. Hence setting z = y-l in (**) for the extended u's, we get u(y)a(y-l) = q(y, y-'). But the latter is an automorphism of S and hence we conclude that u(y) E Aut(S). We can now form S*G using the same twisting and the extended map u: G 3 Aut(S). Since (*) is inherited by S*G, it follows from Lemma 1.1 that S*G is a crossed product containing R*G. Finally set 5 = Zu, E S*G. Then
Since this is a diagonal change of basis, we conclude that S*G = SG is a skew group ring. I Note that, in the above, if R is commutative we could take S to be the group ring S = R[ux,u;' I z E G \ {l}].Furthermore in the usual case with R a field, we could just take S to be the purely transcendental extension of R generated by the u,.
EXERCISES 1. Show that under a diagonal change of basis, R*G still retains the basic crossed product structure. Determine the new action and twisting functions in terms of the old. Note that T will be changed by a factor which is called a 2-coboundary. 2. Suppose R*G was defined without assuming that it contains R as a subring. In other words, the elements of R*G would then be formal finite sums CsEG Zr, with multiplication given by
10
1. Crossed Products and Group-Graded Rings
Show first that there exists a unit u E U such that e = iu is an idempotent. Then prove that R I G = e(R*G) = (R*G)e. Deduce that e is the identity element of R*G and that the map r H er embeds R into R*G. 3. Let S be a ring containing R and a group G of units. Suppose that g U 1 R g = R for all g E G. Show that this action of G on R gives rise to a homomorphism 0:G -+ Aut(R) and that the embeddings of R and G into S extend to a unique ring homomorphism of RG into S . This yields a universal property of the skew group ring RG. 4. Let B be the group of trivial units of R*G so that B acts on R by conjugation. Show that there is a homomorphism from the skew group ring RG onto R*G obtained by identifying R n Q with U . Obtain a universal property of the crossed product R*G. There are several possibilities. 5. Let D be the division ring discussed after Lemma 1.2. Use a shortest length argument to show that the elements 3 are F-linearly independent. To this end, suppose C,Zfx = 0 and that both y , z E G occur in this expression. Multiply the equation on the left by an appropriate f E F and on the right by f Y and then subtract to obtain a shorter expression. 6. If R*G is commutative and G is free abelian, show that R*G E R[G].If G is any cyclic group, determine when R*G = RG via a diagonal change of basis.
2. Group-Graded Rings and Duality Let S = R*G be a crossed product and for each x E G set S, = ZR. Then S = @CzEGSx and SZSg = Sxg. Thus S is a strongly Ggraded ring.
Definition. Let G be a multiplicative group. An associative ring S is G-graded if S = @ CxEG S, is a direct sum of additive subgroups S, indexed by the elements x E G and if S,S, S ,, for all z,y E G. Clearly R = S1 is a subring of S , its base ring, and each S, is an R-bimodule under left and right multiplication. We say that S is strongly G-graded if SxSy= S,, for all x,y E G.
2. Group-Graded Rings and Duality
11
It is easy to verify (Exercise 1 ) that the identity element 1 E S is contained in R and we will assume this throughout. Furthermore S is strongly graded if and only if 1 E S,S,-l for all II: E G. If X is any subset of G, we write R ( X ) = CzEXS,.Thus R(G) = S and if H is a subgroup of G, then R ( H ) is the naturally contained H-graded subrzng. Furthermore if N Q G, then S is also (GIN)-graded with components SN, = R ( N s ) . Hence the base ring here is R ( N ) and we have S = R(G) = R(N)(G/N), the analog of Lemma 1.3. If g E G, we sometimes write R(g) for R({g})= S,. As we mentioned above, S = R*G is strongly G-graded with base ring R. Indeed it is easy to see (Exercise 2) that a G-graded ring S is a crossed product if and only if each component S, contains a unit. An example of a strongly graded ring which is not a crossed product is given in Exercise 3. Another well-known example comes from the theory of Lie algebras. Let L be a Lie algebra over K and let U ( L )be its universal enveloping abgebra. If A: L -, K is a linear functional, let Sx = { u E U ( L )
I [c,u]= A(C)u for all C E L } .
These are the semi-invariants corresponding to A. Now set S = C x S x . Then one knows that the sum is direct and that SxS, C Sx+w Thus the semicenter S of U ( L ) is graded by the additive group HomK ( L ,K ). We remark that G-graded rings without additional assumptions can frequently have no relationship t o the group G. For example, let R be a ring, let M be an R-bimodule and let G = { 1 , g , . . . } be any nonidentity group. If S is the ring S = R @ M with M 2 = 0, then S becomes G-graded by setting S1 = R, S, = M and all other components zero. Similarly if S is G-graded and T is H-graded, then S @ T is naturally a W-graded ring for any group W containing G and H as disjoint subgroups. Therefore mild assumptions must sometimes be imposed on graded rings to enable the group structure to play its appropriate role. Group-graded rings were introduced in [45] as a formal way to deal with finite group representation problems. Indeed the standard module arguments carried over immediately to that context. In addition, group-graded rings occur naturally in certain Galois theory
12
1. Crossed Products and Group-Graded Rings
situations and, of course, they are related to crossed products. For the most part, our interest in groupgraded rings will center on their relationship to crossed products. We will not go much further afield. It is of course tempting to try to extend all crossed product results to groupgraded rings. One soon discovers however that the old techniques do not carry over. Fortunately there exists a dzlality machine begun in I391 and extended in [180]which translates many of the crossed product results directly to this new context. We will use the more concrete construction of the latter paper and we will deal with both finite and infinite groups at the same time. Because of this, the infinite results we list here are slightly less precise than those of [180]. Note that if s E S , we let s, be its s-component so that s, E S, and s = CxEG s,.
Definition. Let S be a G-graded ring with base ring R = S1 and use (GI to denote the cardinality of the set G. Let MG(S) denote the ring of row and column finite (GIx IGI matrices over S with rows and columns indexed by the elements of G. In particular if Q E MG(S) and s, y E G then we use ~ ( sy), to denote the (2, y)-entry of Q. Let ML(S) denote the set of all matrices in MG(S) having only finitely many nonzero entries. It is clear that ML(S) is an essential right and left ideal of MG( S ) . Now for each g E G we let g E MG(S) be the permutation matrix g = [ C ~ ~ - I , , which ~] has a 1 in the (~,g-~x)-positions and zeros elsewhere. In addition for each s E S we define d E MG(S) to be the matrix satisfying d(s,y) = s,-ly for all X ,y E G.
Lemma 2.1. With the above notation we have i. The map -:G + G = { g I g E G } is a group isomorphism embedding G into MG(S). ii. The map S = { d I s E S } is a ring isomorphism I :
--$
embedding S into MG(S). iii. If g E G, s E S then gd = dg. This is just a simple matrix computation. We now come to a crucial
13
2. Group-Graded Rings and Duality
Definition. Let H be any subgroup of G. We define S { H } E MG(S) bY
Furthermore, D(H) and D-l(H) are the subsets of diag MG(S), the set of diagonal matrices in MG ( S ) ,given by
D(H) = { (u E diag MG(S) I a ( z , z )E R ( H z ) for all z E G } and
D-'(H) = { (u E diag MG(S) I (u(z,z) E R(z-lH) for all z E G } . In particular S{G} = MG(S) and D(G) = D-l(G) = diag MG(S). The applications to group-graded rings come from having alternate descriptions of S { H } . First we have
Proposition 2.2. [180] If H is a subgroup of G, then S { H } is a subring of MG ( S ) . Furthermore D-l(H) * M G ( R ( H ) )* D(H)
S{H}
This is immediate from the formula R ( X ) - R ( Y ) R(XY) for any subsets X , Y C G. Note that the above two equations say that S { H } and MG(R(H)) are generalized conjugates via certain well-understood sets of diagonal matrices. In particular this yields a correspondence between the ideals of S { H } and of R ( H ) which we will consider at the end of this section. For each x E G, let e, E ML(S) C MG(S) denote the idempotent with 1in the (z,z)-position and zeros elsewhere. Since 1 E R(1) we have e, E S { H } and it follows that S { H } b= M&(S)nS{H}is an essential right and left ideal of S { H } . When H = 1, a more precise description of S { H } can be given.
14
1. Crossed Products and Group-Graded Rings
Lemma 2.3. S{l}b= @ C x E G e , S is a free right S-module with basis { ex I x E G}. Furthermore if s = C , s, E S and z, y E G then
- -- e,sey -
e,s,-l,
-
= s,-~,e,.
Proof: Note that S C S{1} and since ex
E S{l}bwe have S{l}b= Let a E e,S{l}, the x-th row of S(1). If t = @C,exS{l}. C,a(z,y) E S , then tX-ly = a ( x , y ) and hence e,t = a. Thus s{1jb = CB C , e,S. Finally let s = C , s, E S so that e,Se, = e,s,-lyey. In particular, if exS = 0 then s = 0. Furthermore since s&, has only one nonzero entry in each row and column we conclude that
e,s,-1, as required.
-
-exsey = s,-lyey I
I
In case G is finite, we recognize the above structure as coming from the theory of Hopf algebras; it is the smash product of the Ggraded ring 3 EZ S by the dual of the group algebra of G. Specifically if S is a G-graded ring with G finite, then the smash product S#G* is an associative ring with 1 having S as a subring. Furthermore there exists a decomposition of 1 into orthogonal idempotents p , , one for each x E G, such that { p , I x E G } is a free right S-basis for S#G* with PxSx-1,
= P X S P , = Sx-lyPy
for all z,y E G and s E S. Since the above assertions uniquely determine the arithmetic in S#G*, we conclude from Lemmas 2.1 and 2.3 that, for G finite, S#G* % S{1} via the map given by - : S --+ 8 and p , H ex. (We remark that the notation S#G* is not standard; we include the * to indicate the presence of the Hopf algebra dual.) The second description of S { H } is contained in the following.
Lemma 2.4. Let S be a G-graded ring.
15
2. Group-Graded Rings and Duality
i. If g,x E G then g-le,g eg-lX and hence G acts as automorphisms on S{ 1) centralizing S . ii. S{G) 2 @ Cg-,,G gS{l) = S{l)G 2 S{G)b where S{l)G is a skew group ring of G over S{ 1). iii. If H is a subgroup of G, then S{ 1)G n S { H } = S{ l)H is the naturally embedded sub-skew group ring. Prooj (i) The formula g-'e,g = eg-lX is a simple matrix computation. Since the 2-th row of S{1) is e,S{l) = ezS{l)b = ezS, by Lemma 2.3, it follows from Lemma 2.l(iii) that G permutes these rows and centralizes S. Thus clearly G normalizes S{ 1). (ii)(iii) We assign a grade to certain elements of MG(S) as follows. We let a E MG(S) have grade g E G if and only if for all 2,y E G the entry a(x,y) satisfies a(x,y) E S, with xzy-l = g . If Tg denotes the set of elements of MG(S) of grade g,then it follows easily that MG(S) 2 T = @CgEGTg 2 ML(S) and that TgTh E Tgh. In particular, T is a G-graded ring. Moreover TI = S{1) and Th = T n S { H )since 2zy-l E H if and only if z E zL1Hy. Note also that g E T has grade g. Thus g-'Tg 2 TI so Tg = gT1 and hence T = @ CgEC Tg= @ CgEG gT1. The result now follows from Lemma 2.l(i) and the above observations.
xhEH
It is now a simple matter to obtain the ~ ~ u Z z theorem. ty
Theorem 2.5. [39]Let S be a G-graded ring with G finite and let S#G* = CxEG p,S be the smash product of G over S . Then G acts on S#G* as automorphisms via (pxs)9= pg-lxs for all x,g E G and s E S . Furthermore, with respect to this action, the skew group ring (S#G*)G satisfies (S#G*)G z Mc(S). Prooj Since G is finite, ML(S) = MG(S) and therefore a number of the inclusions above become equalities. In view of Lemma 2.4(i) and the isomorphisms G ? G and S#G* E S{ 1) we see that G does indeed act as automorphisms on S#G* in the indicated manner. Furthermore by Lemma 2.4(ii)(iii) we have
(S#G*)G
S(1)G = S { G ) 1MG(S)
16
1. Crossed Products and Group-Graded Rings
as required. I
There is another duality theorem for G finite. If R*G is a crossed product, then it is a G-graded ring with (R*G), = 3R and we can form the smash product (R*G)#G*. The theorem of [39] and [199] then asserts that (R*G)#G* 2 MG(R) (see Exercises 7 and 8 of Section 3). In fact if S = R(G) is strongly G-graded, then more generally we have S#G* End(RS). There are also some analagous results when G is infinite. In this case, if S is G-graded, then it is more difficult to define S#G* in the sense of Hopf algebras because of the infinite dimensional dual. Furthermore even when it can be defined (see [25]), it is not in general equal to S{ 1). We close this section by describing the ideal correspondence determined by the generalized conjugation in Proposition 2.2. If S is a G-graded ring, we say that S is component regular if, for each x E G , the component S, has right and left annihilator in S equal to zero. The following result uses the notation of Lemma 2.4. Furthermore if T is any ring we let Z(T)denote its family of two-sided ideals.
Proposition 2.6. [180] Let S be a G-graded ring with base ring R = S1 and let H be a subgroup o f G. Then there exist inclusion preserving maps
$ : Z ( R ( H ) ) Z(S{l}H) --+
and
J:Z(S{l}H) -+ Z ( R ( H ) ) such that i. If A , B a R ( H ) , then A9B4 (AB)4 and A96 = A . ii. If I , J a S(l}H, then I t Je C ( I J ) c . Furthermore if S is component regular or if H = G, then It = 0 implies that I = 0.
Proof: For convenience write D = D ( H ) ,D-' = D-'(H) and let ex,y denote the matrix with 1 in the (z,y)-position and zeros elsewhere. We first observe that D . S { H } b E M k ( R ( H ) ) . D . To this end, let a E S{H}b have only one nonzero entry, namely a ( z , y ) E R(z-'Hy), and let 6 E D. We note that
d' = (6a)(x, y) = 6(~, X)Q(Z,y) E R(Hx)R(x-'Hy)
R(Hy).
2. Group-Graded Rings and Duality
17
Then es,y E M L ( R ( H ) ) , 6' = d'ey,y E D and Scr = e,,,S'
E M L ( R ( H ) )- D
as required. Similarly we have S{H}b . D-' C - D-l . M L ( R ( H ) ) . Now let A a R ( H ) and define A4 = D-' - ML(A) . D. Then by Proposition 2.2 and Lemma 2.4,
Furthermore A4 a S { H } since for example
s
Thus A4 a S{l}H. In addition DD-' C MG(R(H)) yields AdB4 (AB)4 and since e1,lD-l = el,lR(H) = De1,l we have el,lAdel,l = el,lA. In the other direction, let I a S{l}H C S { H } and define the set 1s = {a(l,l)1 a E I } . Since I G S { H } and e l , l R ( H ) G S{H}b C S{1}, it follows that 15 a R ( H ) . Furthermore if A a R ( H ) , then A4E = A since el,lA4el,l = e1,lA. Finally suppose S is component regular and that I # 0. Choose a E I with a ( z ,y) # 0 and note that el,,R(Hz), e,,lR(y-lH) C S { H } b& S{l}H. Thus
and hence R ( H z ) c r ( ~ , y ) R ( y - ~ HC) Is. We conclude that Is # 0 since a ( z ,y) # 0 and since either S is component regular or H = G and 1 E R ( H z )= R(y-'H). 1
Corollary 2.7. Let S be a G-graded ring. i. If S { 1 } H is prime or semiprime, then so is R ( H ) . ii. Suppose that either S is component regular or H = G. If R ( H ) is prime or semiprime, then so is S{ 1 } H .
This is an immediate consequence of Proposition 2.6. It will be used to transfer results on prime or semiprime crossed products
18
1. Crossed Products and Group-Graded Rings
to the context of group-graded rings. Finally we mention a simpler result which holds when H = G.
Proposition 2.8. [31] Let S be a G-graded ring. If P is a prime ideal of S , then P’ = MG(P)n S{l}G is a prime of S{l}G. Furthermore, the map P H P’ is one-to-one.
Proof: By Lemma 2.4(ii), S{l}G 2 ML(S) so it is clear that the map P H P’ is one-to-one. Moreover, for all z, y E G, S{ l}G contains the matrix unit ez,y which has a 1 in the (z,y)-position and zeros elsewhere. Now let A , B a S{l}G with AB E P‘ and write el,lAel,l = e1,lA” and el,lBel,l = el,lB” where A“ and B” are ideals of S. Then e1,lA”B“ = el,lAel,l el,lBel,l
G el,lP’el,l = el,# so A“B” C P. Therefore since P is prime, one of these factors, say A”, is contained in P. It follows that if a E A and z,y E G, then e l , l a ( z , y ) = e1,,aey,l E A so a ( z , y ) E A” s P. Thus a E M/r,(P) n S{l}G = P’ so A C P’ and P‘ is indeed a prime ideal. I EXERCISES 1. Let S be a G-graded ring with 1. Show that 1 E S1 and that S is strongly G-graded if and only if 1 E S,S,-l for all z E G. 2. Let S be G-graded. If u E S, is a unit of S, prove that u-l E S-1. Deduce that S = R*G if and only if each S, contains a unit of S. 3. Let S = M3(K) and let G = { 1 , z } be a group of order 2. Define
S1=
(:
K
K
0) 0
and
S,=
Show that this makes S a strongly G-graded ring, but that S is not a crossed product. For the latter, compute dimK S1 and dimK S,.
19
3. Induced Modules
4. Let G be a finite group and let R be a ring. Show that MG(R) becomes G-graded by assigning a grade of x-ly to the entries in the (2, 3)-position. Prove, in fact, that MG(R) is a crossed product of G over the diagonal matrices. Thus M,(R) is a crossed product over the diagonal matrices by any group G of order n. 5. Let S be a G-graded ring with G finite. Prove directly that S#G* is an associative ring and verify that G acts on S#G* via (PXSP
=P g - l z S .
6. What would happen in the proof of Proposition 2.6 if we defined I€ more naturally by I t = { a(1,l)I a E DID-’ }? 7. Suppose S = R ( G ) is strongly G-graded, fix 2 E G and write 1= a& with ai E S,-I, pi E S,. If T E Z(R), set r” = airpi and prove that rx is uniquely determined by the equation ry = yr” for all y E S,. Conclude that ”: r H T” is an automorphism of Z ( R ) and that G -+ Aut(Z(R)) is a group homomorphism. This action of G on Z ( R ) is called the Miyashita automorphism ([125]). 8. Show that the map P H P’ of Proposition 2.8 is not onto if G is infinite. To this end consider a maximal ideal of S{l}G containing
xi
xi
M m .
3. Induced Modules If S is a crossed product or a group-graded ring, then we are interested in relationships between S and its base ring R. One such concerns the modules of the two rings. We start by considering any ring S and subring R.
Befinition. Let R
S be rings with the same 1. If M = M s is a right S-module, we let MlR denote the restriction of M to R. If VR is a right R-module, we let Vls denote the induced Smodule = VR @ Rs.
v“
R
The S-module structure here is of course given by (w @ s)t = v @ st for all z, E V and s , t E S.
20
1. Crossed Products and Group-Graded Rings
Basic properties of restriction are obvious. For induction we have
Lemma 3.1. Let R C S be rings. i. If VR is finitely generated, then so is Vls. ii. Induction commutes with direct sums. iii. Rls S and hence if VR is free or projective, then so is Vls. iv. If M is an S-module, then there is an S-module epimorphism (MlR)Is + M given by m @ s H ms. v. If S E T and V is an R-module, then (VIS)IT E VIT.
We remark that it is quite possible to have VR# 0 but Vls = 0. For example, let R be the ring of integers, S the rationals and let V be any periodic abelian group. Now if 9: UR ---t VR is an R-homomorphism, then 8 1:Uls Vls is an S-homomorphism and we have
Lemma 3.2. Let 0
U + V + W --$ 0 be a short exact sequence of R-modules. Then with respect to the above tensor maps the sequence Uls -+ Vls + Wls + 0 exact. Furthermore induction is an exact functor if and only if RS is a Aat left R-module. -+
Note that RS flat means precisely that, for all 0
---f
U
-+ V -+
W + 0 as above,
is an exact sequence of abelian groups. Note also that if RS is free or projective, then it is flat. Before we restrict our attention to crossed products, we require an additional
Definition. Let u E Aut(R). If V is an R-module, then there is a conjugate modu2e V" which can be viewed in two different ways. First, if Endz(V) denotes the endomorphism ring of V as an abelian group, then the module structure of V is determined by the ring
21
3. Induced Modules
homomorphism p: R + Endz(V). We can then obtain a different structure by composing p with 0 - l to yield the representation R
-1
d R & Endz(V).
Alternately we can let V" = { w" I w E V } be an isomorphic copy of V as an additive abelian group and we can define the R-module structure on V" via the formula w U f ' = ( w T ) " . It is clear that V and V" have the same lattice of submodules. Hence V is irreducible, completely reducible, indecomposable or Noetherian if and only if V" is. Furthermore if E ess V, then E" ess V". We now consider crossed products.
Lemma 3.3. Let S
= R*G be given and let V be an R-module. Then Vls = @ CzEG V 8 3 is an R-module direct sum with V @ 33 V"(").
Proof. Since S is a free left R-module with basis G we have Vls = V @R S = @ CzEG V 8 2 , a direct sum of additive abelian groups. Furthermore, by definition of the module structure, it follows that for w E V, z E G and T E R we have
(w 8 z)r"(")= v 8 ( 3 4 " ) )= w 8 (7.3)= ( V T ) 8 3. In other words if we write o"@) = w @ 2, then o ' ( z ) ~ a ( z = ) (VT)"(") and V 8 3 is an R-module isomorphic to V"("). I Note that if N a G then S = R*G = (R*N)*(G/N). Thus if V is a n R*N-module, then there is an analogous formula for VlsThe following is a simple extension of the Hilbert Basis Theorem and we just sketch its proof.
Proposition 3.4. Let S = R*G be a crossed product where G is a polycyclic-by-finite group. If V is a Noetherian R-module, then Vls is a Noetherian S-module.
Proof. In view of Lemma 1.3 and the transitivity of induction given in Lemma 3.l(v), it suffices to assume that G is either infinite cyclic or finite.
22
1. Crossed Products and Group-Graded Rings
V“(”)is also a In the latter case, observe that each V 8 3 Noetherian R-module. Hence since Vls = @ C V 8 Z is a finite sum, we see that Vls is Noetherian as an R-module and therefore as an S-module. Now let G = ( g ) be infinite cyclic. Via a diagonal change of basis, we may assume that G = { gi 1 i = . . . , -1,O, 1,.. . } so that W = V ~ ~ = C B ~Set~ ~~ +~ =V CB z~~ ~v .~ ~ j j Let U be an S-submodule of W . The goal is to show that U is finitely generated. To this end, we define for each integer i 2 0
~.
Ui = { w E V I w 8 gi + lower degree terms E U n W +} . It follows that Ui is an R-submodule of V and that Ui C Ui+l since U s = U . In particular, since V is Noetherian, the ascending series Uo U, C_ . . . V must stabilize, say at i = n. Now each of UO,U l , . . . ,U, is a finitely generated R-module and if wi,l,w ~. . .,,wi,k(i) generate Ui, we can choose ui,j E U n W+ with
z
ui,j
= vi,j @ gi
+ lower degree terms .
It is easy to show (Exercise 2), by induction on the degree of the element u E U n W f , that U n W+ ui,jS. Finally if u E U , then ugm E U n W+ for some m so u E (U n W+)g-”. We conclude that U is generated as an S-module by the finite set { ui,j }. I We remark that if R is right Noetherian, then the above result with M = R implies that S = R 8~ S is also Noetherian. This is Proposition 1.6. Our goal now is to discover an analog of this result in the context of group-graded rings. Suppose S is a G-graded ring with R = SI. If V is an R-module, then Vls has a special structure here; it is a graded S-module.
Definition. Let S = R(G) be a G-graded ring. An S-module M is said to be a graded m o d u l e if M = @ C z E G M ,is the direct sum of the additive subgroups M,, indexed by the elements z E G, with MZy for all z,y E G. In particular, each M, is an RM,Sy submodule of M . Furthermore S itself is a graded right S-module.
3. Induced Modules
23
If N = @ CxEG N , is another graded module, then N is a graded M and I?, = N n M , for all x f G . We subrnodde of M if N say that M is graded simple if 0 and M are the unique graded submodules of M . Similarly M is graded Noetherian if the lattice of graded submodules of M satisfies the ascending chain condition. Finally an S-homomorphism 8: M + N is a graded homomorphism if N , for all z E G. It is clear that the kernel of a graded 8(M,) homomorphism is a graded submodule of M .
Lemma 3.5. Let S = R(G) be a G-graded ring. If V is an R-module, then Vls is a graded S-module. Conversely if S is strongly G-graded and if M is a graded S-module, then M = (MI)[’. Hence, in the latter case, there is an isomorphism between the lattice of graded S-submodules of M and the lattice of R-submodules of M I .
EXEC: S,, it follows immediately that W = Vls = and that W is graded with W, = V 8 S,. Now let S be strongly G-graded and let M = @ C M, be any graded S-module. Then
Proof: Since S @
=@
CXEG(V 8 S,)
and thus clearly M, = MIS,. In particular, MI = 0 implies that M = 0. Furthermore we have a graded epimorphism 8: (M1)I’ + M given by ml 8 s e+mls for all ml E MI and s E S. Note that the kernel K of 8 is a graded submodule of (MI)(’ and that K1 = 0. Thus K = 0 and 8 is an isomorphism.
Thus in the above we see that M I is a Noetherian R-module if and only if M is graded Noetherian. The graded analog of Proposition 3.4 is therefore: If M is a graded Noetherian S-module and if G is a polycyclic-by-finite group, then M is a Noetherian S-module. This was essentially proved in [13] for S strongly graded (see Exercise 3) and in [148] for G a finitely generated nilpotent group. The general result requires the duality machine. Let S = R(G) be a G-graded ring. We use the notation of the previous section so that MG(S) is the ring of row and column finite
24
1. Crossed Products and Group-Graded Rings
]GIx )GImatrices over S . Then MG(S) contains the subring S(1) = { QI I O!(z,y) E R(z-ly) = Sx-ly }, the subgroup G = { g = S( 1)G.
1 g E G } and the skew group ring
[ C ~ ~ - I ~ , ~ ]
Definition. Let M be an S-module. We define RowG(M)to be the set of all row finite 1 x IGI row matrices with entries in M. Thus if w E RowG(M)then 21 has only finitely many nonzero entries and its entry in the z-column is denoted by w(z). If M = @ CxEG Mx is also a graded module, then we can define M RowG(M)by &Z = ( w E Rowc(M) 1 ~ ( xE)Mxfor all z E G ) .
Key properties of these sets are as follows.
Lemma 3.6. Let S be a G-graded ring and let M be an S-module. Assume M is graded in parts (ii) and (iii) below. i. RowG(M) is a right S(1)G-module and all submodules are of the form RowG(N)for N an S-submodule of M. ii. M is a right S(1)-module and all submodules are of the form N for N a graded S-submodule of M . iii. RowG(M) is the induced module
RowG(M) = M - ISilP.
Pro05 We first introduce some notation. For each z, y E G let ex,y E MG(S) denote the matrix with a 1 in the (z,y)-position and zeros elsewhere. Note that ex,%E S(1) for all z E G and that S(1)G 2 ML(S) by Lemma 2.4(ii). For each z E G, let n,:Row~(M)-+ M be the projection map into the z-coordinate so that n,(v) = w(z). Furthermore, for convenience, let fx denote the 1 x IGl row matrix with a 1 in the zcoordinate and zeros elsewhere. Thus if w E RowG(M)we can write
25
3. Induced Modules
c,
In particular, RowG(M)= C, M f , and & = l M,f,. (i) It is clear that RowG(M)is a right Mc(S)-module and hence a right S{l}G-module. Let W be a submodule. Since eZ,,S 2 S{l}G it follows that W = C,slr,(W)f, and that each r z ( W )is an S-submodule of M. Furthermore since ez,Y E S{l}G we see that slrz(W) = r Y ( W ) for all z,y. Thus W = z,slrl(W)f, = RoWG (rl(W )). (ii) If v E &f and 01 E S{l), then v ( z ) ~ ( z , y )E M,S,-iY C My and it follows that va E &f. Thus &f is an S{1}-module. Now let W be a submodule of M and set N = C,slr,(W) M. Since ez,yS,-lY C S{1} we see that r,(W)Sz-~Y r Y ( W )for all z,y E G and hence that N = @ C, r,(W) is a graded submodule of M with N, = r,(W). Furthermore since e x , , E S{l} we conclude that
as required. (iii) Finally since f,g = fg-1,
c,
and % = !IM,f, we have
MfY= RowG(M).
= Y
It now follows easily from &flSfl)G
=@CiM@g=@C(M@l)g 9
and the above that
&fls(l)G
9
is isomorphic to RowG(M).
We remark that parts (i) and (ii) above yield lattice isomorphisms between the lattice of submodules of various modules. In particular, one module is Noetherian if and only if the other is. It is now a simple matter to prove
Theorem 3.7. [31]Let S be a G-graded ring with G polycyclic-byfinite and let M be a graded S-module. Then M is Noetherian if and only if it is graded Noetherian.
26
1. Crossed Products and Group-Graded Rings
Proof: Certainly if M is Noetherian then it is graded Noetherian. Conversely assume that M is graded Noetherian and form fi RowG(M). Then, by Lemma 3.6(ii), M is a Noetherian S(1)-module and hence Proposition 3.4 and Lemma 3.6(iii) imply that uIs{'}G E RowG(M) is a Noetherian S(1)G-module since G is polycyclic-byfinite. We conclude from Lemma 3.6(i) that M is a Noetherian Smodule. I
As an immediate consequence of the above and Lemma 3.5 we obtain
Corollary 3.8. [13] Let S be a strongly G-graded ring where G is a polycyclic-by-finite group. If V is a Noetherian S1-module, then the induced module Vls is also Noetherian. Other applications of duality come from the following observation of [39].
Lemma 3.9. Let S be a G-graded ring with G finite. If M is a right S#G*-module, then M becomes a graded S-module by defining M, = Mp, for all x E G. Conversely if M is a graded S-module, then M becomes an S#G*-module by defining m p , = m, for all m E M, x E G.
Proof: Let M be a right S#G*-module and define M, = M p , so
Ex
Ex
that M = @ M, since 1 = p , is an orthogonal decomposition of 1. Furthermore, p,Sy = S y p x y implies that M,Sy G Mxy. Conversely let A4 be a graded S-module and let 71:, M + M, be the natural projection. For any s E S it follows easily that 7rxS,-ly
- 7rxs7ry= S,-ly7ry
and thus M becomes an S#G*-module with p , acting like 7 r x . I Therefore we have a one-to-one inclusion preserving correspondence between graded S-modules and arbitrary S#G*-modules. We return to crossed products R*G and close this section by briefly considering induced modules from sub-crossed products. The following is known as the Muclcey decomposition.
3. Induced Modules
27
Lemma 3.10. Let R*G be
a, crossed product and suppose that H and A are subgroups of G and that V is an R*H-module. Let 23 be a complete set of ( H ,A)-double coset representatives in G. If d E D, then V €3 d is an R*Hd-module and we have
Proof. If X is any subset of G, then we let R*X = { a E R*G I S u p p a C X ) . Since G is the disjoint union G = U d @ HdA, it follows that
d = Note that the above tensor €3 is over the ring R*H and that v V@R*(HdA)is clearly an R*A-submodule of VIR*Gsince R*(HdA) is an (R*H,R*A)-bimodule. It remains to describe each V d . To this end, let T be a right transversal for HdnA in A. It follows (Exercise 6) that dT is a right transversal for H in D = H d A . Thus R*D is a free left R*H-module with basis dT and we have
Now note that V 8 d is an R*Hd-module since d(R*Hd)= (R*H)d. Thus (V @ is an R*(Hd f l A)-module and we can induce it to R*A. By definition of T this yields
where €3' denotes tensor over R*(Hd n A). It is now a simple exercise to show that vd is R*A-isomorphic to this induced module via the map v €3 dt H (v g d) €3' t for all v E V , t E T . I
28
1. Crossed Products and Group-Graded Rings
EXERCISES 1. Verify the equivalence of the two definitions for the module structure of the conjugate module V". Furthermore if u,T E Aut(R) V"'. prove that (V")' 2. In the proof of Proposition 3.4, first verify that each U . is an R-submodule of V. Next suppose that u E U n W + with u = w @gi+ lower degree terms. If i 5 n show that u - Cjui,jrj E U n W+ has degree less than i for suitable rj E R. If i > n show that u - Cjun,jrjgi-n E U f l W + has degree less than i for suitable rj E R. Deduce that U n W + ui,jS. 3. Let S be a strongly G-graded ring with base ring R. Let x E G and write 1 = a& with ai E S,, ,8i E S,-l. Show that S, = aiR and S,-1 = R&. Furthermore, prove that S, is a finitely generated projective right and left R-module. 4. Suppose V is a right S-module where S is G-graded with G finite. Then S C S#G* and we consider the induced module Vls#G*. Show that Vls#G* = @ C, V @I p , and that (w @I p,)sy = 'usy @I p,, for all w E V , z,y E G and sy E S,. In particular if S = K[G] is a group algebra and dimKV = 1 conclude that (VlS#'*) 1s 2 Ss. This example is from [24]. 5. Let G act as permutations on the set R. If K is a field, then the vector space KR, with K-basis R) becomes in a natural manner a K[G]-module; it is called a permutation module. Show that KR is the direct sum of the permutation modules corresponding to the orbits of G on 0. Furthermore suppose G is transitive on R, with H the stabilizer of some point. Prove that KR is induced from a 1dimensional K[H]-module. 6. In Lemma 3.10, let a , b E A. Show that Hda = Hdb if and only if ab-l E H d nA and deduce that dT is a right transversal for H in H d A . Furthermore prove that the map given by w@I& H (w@Id)@'f is an R*A-isomorphism. To this end, let a E A, r E R and define Q E R*(Hd n A) and ,8 E R*H by tar = QGfor some tl E T and da = ,8d. Compute (w @ %)ar and [(w @I d) 8' sir. 7. If S is G-graded, use Lemma 3.9 to show that S is an ( R ,S#G*)-bimodule and that S#G* & End(RS) if S is component
'&
xi
xi
xi
29
4. Maschke’s Theorem
regular. In addition, if S = R*G, prove that End(& MG(R). 8. We continue with the preceding notation and assume further that S is strongly graded. The goal is to show that S#G* = End(RS). Let 0 E End(RS) and choose s E S,, t E S,-Ifor z, y E G. Observe that PE@P, t R A S , S,-R
-
is an R-endomorphism of R. Deduce that s .p,Opy - t = p,r for some r E R. Finally use the facts that C,px = 1 and 1 E SX-lSx to conclude that 0 E S#G*. This is a result of [199].
4. Maschke’s Theorem Maschke’s theorem is probably the first major theorem proved about group algebras of finite groups. It shows the strong effect of [GItorsion, or the lack of it, on the structure of these algebras. As we see below, it converts directly to a result on crossed products. Furthermore there is an essential version of this theorem which is particularly useful for proving the nonexistence of nilpotent ideals. Finally, much of this translates, via duality, into results on groupgraded rings. Recall that an abelian group V is said to have n-torsion for some integer n if there exists 0 # w E V with wn = 0. We begin with the classical version of Maschke’s theorem.
Theorem 4.1. [120] Given R*G with G finite. Let W E V be R*G-modules with no [GI-torsion and let V = W @ U where U is a complementary R-submodule. Then there exists an R*G-submodule
U’ of V with V . IGl C W @ U’.
Proof. We note that G permutes the R-submodules of V. Indeed if U is any R-submodule of V and if 3: E G, then the formula 3rU(”)= rZ implies that U Z is an R-submodule isomorphic to the conjugate module U“(“). Now let V = W @ U be as given. Then for all z E G, V = WZ @ UiE = W @ UZ and we let 7 r x : V + W be the R-homomorphism
30
1. Crossed Products and Group-Graded Rings
determined by this decomposition. Note that if w = w w y = wy uZy so clearly
+
+ u Z , then
7rxy(wy) = wy = 7rx(w)y.
It follows that since
7r
=
Ex
7rx
is an R*G-homomorphism from V to W
c X
=
X
7rx(w)y = .(w)y
X
for a11 w E V, y E G. Set U’ = Ker(.lr) so that U’ is an R*Gsubmodule of V. Now if w E W , then ~ ( w=)w - IGI so W n U’ = 0 since V has no )GI-torsion. Finally if w E V, then w . [GI - ~ ( w )E Ker(7r) and hence [GI E W @ U’as required. I w
e
In particular, if V.(G(= V, then V = WeU’. As a consequence we see that if V IGI = V and if y, is completely reducible, then so is V. This is immediate since V is completely reducible if and only if every submodule is a direct summand. Note that V - \GI = V is automatically satisfied if 1GI-l E R. We now have the following result on J(R*G), the Jacobson radical of R*G.
Theorem 4.2. [201]Let R*G be a crossed product with G finite. Then J(R*G)I~I2 J(R)*G
J(R*G).
Furthermore if 1GI-l E R then J(R*G) = J(R)*G.
PruuJ Let V be an irreducible R*G-module. Then V is a cyclic
R*G-module and hence y~ is finitely generated. It now follows from Nakayama’s lemma that VJ(R) # V. But VJ(R) is easily seen to be an R*G-submodule so VJ(R) = 0. We conclude that J(R) C J(R*G) and hence that J(R)*G J(R*G). In the other direction, let W be an irreducible R-module and R form the induced R*G-module V = WIR*G.By Lemma 3.3, V ~ = @ CsEG W @ 2 is the direct sum of n = [GI irreducible R-modules.
31
4. Maschke’s Theorem
Thus V must have composition length 5 n and V * J(R*G)= = 0. Now let Q = E J(R*G)n. Then for any w E W we have 0 = ( w @ l ) a = z w r g @ gso wrg = 0 and W r , = 0. Since this is true for all such W we have rg E J(R) and therefore J(R*G)= J(R)*G. Finally if [GI-’ E R, then since VjR is completely reducible it follows from Theorem 4.1 that V is completely reducible. Hence in this case VJ(R*G) = 0 for all W and the above argument yields
zgEG~,fi
c
J(R*G) 2 J(R)*G. I To deal with nilpotent ideals, we require the following essential version of Maschke’s theorem.
Proposition 4.3. [112][165]Given R*G with G finite. Let W C V be R*G-modules with no IGI-torsion. Then W ess V if and only if W I Ress Y R .
Proof. If W p ess V ~ Rthen surely W ess V. Conversely assume W ess V. Suppose first that V = W @ U where U is an R-submodule. Then, by Theorem 4.1, there exists an R*G-submodule U‘ of V with V -IGI E W @ U ’ . But W ess V so U‘ = 0 and U . [GI c U n W = 0. Thus U = 0 and W = V. Now for the general case. Choose UR V maximal satisfying uRnw = 0. Then (Wl~Cl3U )ess V j R and we set = nZEG(W@ U)?. Note that (W Cl3 U ) 2 ess V ~ R(see Exercise 1). Thus since the intersection is finite, it follows that E is an R*G-submodule of V with ElR ess y R .Furthermore, W G E C_ W @ U so E = W @ (U n E ) . The result of the preceding paragraph now implies that W = E so W I R ess Y R - 1 This yields the important
Theorem 4.4. [58]Let R be a semiprime ring with no /GI-torsion. Then R*G is semiprime.
Proof: Let N a R*G with N Z = 0. If L = t ~ * c ( N )then , L a R*G and L ess R*G as right ideals. Since R*G has no ]GI-torsion, Proposition 4.3 now implies that LIR ess R*G~Rso ( L n R) ess RR. Since
32
1. Crossed Products and Group-Graded Rings
R is semiprime, we conclude that r R ( L n R ) = 0 and then N rR*G(L n R ) = 0 by the freeness of R*G over R (see Exercise 2). I The proof given above is from [165]but the original techniques of [58] are still needed for certain generalizations which we will consider in Section 18. In the case of group-graded rings, nilpotent ideals can also occur because of certain degeneracies in the structure. Let S be a G-graded ring. If B is a subring of S, possibly without 1, then we say that B is a graded subring if B = CxEG B, with B, = BnS,. We have the following pigeon-hole argument.
Lemma 4.5.[74] [40] Let S = R ( G ) be a G-graded ring with G finite and let B be a graded subring (without 1). Then B is nilpotent if and only if B1 is nilpotent. Indeed if By = 0, then BnlGl = 0.
Proof: If B is nilpotent, then so is B1. Conversely suppose B? = 0 and set m = nlGI. Consider any product
,...,x,
E G and define yo,yl, ...,ym E G by yi = 21x2 * * zi with yo = 1. By the definition of m, at least one value of y must occur n+ 1times here. But observe that if yi = y j with j > i then x i + l . . xj-lzj = 1 so BXi+,- - +Bxj-l Bxj B1. It follows that C & SBYS = 0 and hence, since B is graded, we have Bm = 0. I
with
q,x2
This of course applies with B any graded left or right ideal of S.
Definition. Let S be a G-graded ring. We say that S is graded semiprime if S has no nonzero nilpotent graded ideal. When this occurs, then certainly S has no nonzero nilpotent graded right or left ideal. We say that S is nondegenerate if for all z E G and all 0 # s E S, we have sS,-1 # 0 and S,-ls # 0. We remark that nondegeneracy is a slightly weaker assumption than component regularity. Indeed S = R(G) is component regular
33
4. Maschke's Theorem
if and only if it is nondegenerate with each component S, faithful as a left and right R-module (see Exercise 6). A key relationship between the above two definitions is as follows.
Lemma 4.6. Let S = R(G) be a groupgraded ring with G finite. Then S is graded semiprime if and only if S is nondegenerate and R is semiprime.
ProuJ Suppose S is graded semiprime. If 0 #
s E
S,, then sS is a
nonzero graded right ideal and hence is not nilpotent. By Lemma 4.5, ( s S )= ~ sS,-1 is not zero. Similarly S,-ls # 0 and S is nondegenerate. Moreover if I is a nilpotent ideal of R,then I S is a right ideal of S with ( I S ) 1 = IS1 = I . Thus again by Lemma 4.5, I S is nilpotent so I S = 0 and I = 0. Conversely let S be nondegenerate and let R be semiprime. If I is a graded nilpotent ideal of S , then I1 is a nilpotent ideal of R and hence I1 = 0. Finally IzSz-l C_ I1 = 0 so since S is nondegenerate we have I , = 0 for all z E G. I The goal now is to obtain a graded version of Theorem 4.4.For this we use duality. Recall that the smash product S#G* is given by S#G* = CxpxS with pxs,-ly = pxspy = s,-lypy. Furthermore G acts as automorphisms on S#G* via (p,s)9 = ~ ~ - 1 , s .
Lemma 4.7. Let S be a G-graded ring with G finite and let IaS#G*. Then I fl S is a graded ideal of S . Furthermore I is G-stable if and only if I = ( I nS)(S#G*) = (S#G*)(I n S ) .
ProoJ If
sE
I n S then S, = C p y , - l s p y
E
I ns
Y
for all 2 E G. Thus I n S is graded. If I = ( I n S)(S#G*), then I is surely G-stable. Conversely let I be G-stable and let Q = C a ( z ) p , E I with a(.) E S. Then a(z)px= ap, f I and, since I is G-stable, we have
34
1. Crossed Products and Group-Graded Rings
as required. For the other equality, write
(II
= Cpza‘(z). I
Theorem 4.8. [39]Let S be a G-graded ring with G finite. If S is graded semiprime, then S#G* is semiprime. If in addition S has no [GI-torsion,then S is semiprime.
Proaf: If J is a nilpotent ideal of S#G*, then so is I = CzEG J” and I is G-stable. Since S is graded semiprime, it follows from the preceding lemma that J c I = 0. Thus S#G* is semiprime. If in addition S has no (GI-torsion, then Theorems 2.4 and 4.4imply that Mc(S) is semiprime. We conclude that S is semiprime. I In particular, in view of Lemma 4.6, S is semiprime if it is nondegenerate, has no [GI-torsion and if S1 is semiprime. Thus we have an appropriate graded analog of Theorem 4.4. Now we consider the structure of the Jacobson radical J(S) with S a G-graded ring. This also follows fairly directly from duality and does not require the next two results. Nevertheless, we include them because of their intrinsic interest.
Lemma 4.9. Let S = R ( G ) be a G-graded ring with G finite and let V # 0 be an S-module. Suppose W is an R-submodule of V with W ess l$R. Then W contains a nonzero S-submodule of V . ProoJ Since V # 0 we have W # 0. Note that w E W\O implies that w R 2 W . Now choose X a subset of G of maximal size such that there exists w E W \ 0 with w R ( X ) W. We claim that X = G . If not, let g E G \ X . If w R ( g ) = 0 then w R ( X U ( 9 ) ) 2 W , a contradiction. Thus w R ( g ) # 0. Now wR(g) is a nonzero R-submodule of V and W ess Y R .Thus there exists w‘ E w R ( g )n W with w’# 0. Observe that w’R(g-lX) G wR(g)R(g-’X)
wR(X)
cW
and w‘ E W so w’R(g-lX U (1)) 2 W , again a contradiction. Thus X = G and w R ( G ) is a nonzero S-submodule contained in W . I
4. Maschke’s Theorem
35
Proposition 4.10. [68] Let S = R(G) be G-graded with G finite and let V be a completely reducible S-module. Then 1 / / ~is completely reducible.
Proof: We can assume that V is irreducible. The preceding lemma then implies that VIR has no proper essential submodules and hence is completely reducible. We now begin our work on J(S).
Definition. If S is a G-graded ring we let JG(S)denote its graded Jacobson radicu2, that is JG(S)is the intersection of the annihilators of all graded simple right S-modules. It is clear that JG(S)is a graded ideal of S. By standard arguments (see Exercise 7) JG(S) is the intersection of the maximal graded right ideals of S ; it is the largest graded ideal with quasi-regular identity component, and its definition is right-left symmetric.
Lemma 4.11. Let S
= R(G) be G-graded with G finite. i. J(S) f l R = J(R). ii. J(S#G*) n S C J(S).
Proof: (i) Let R I M be an irreducible R-module. Then M S is a graded right ideal of S with M S n R = (MS)l = M so M S # S. Choose M’ a maximal right ideal of S containing MS. Then M’nR = M so R I M S/M‘. Since J(S) annihilates S/M’ , it follows easily by varying the maximal right ideal M that J(S ) n R C J(R ). Conversely, let V be an irreducible S-module. Then is com-
-
pletely reducible by the preceding proposition so VJ(R) = 0 and J(R) C J(S) nR. (ii) If M is a maximal right ideal of S , then M(S#G*) # S#G* by the freeness of S#G* over S. The argument now proceeds as in (i) above. I Part of the above proof could of course be replaced by a quasiregularity argument. Indeed suppose S 2 R are any rings with the same 1 and with the property that any element of R invertible in S
36
1. Crossed Products and Group-Graded Rings
is invertible in R. Then J(S) f l R is a quasi-regular ideal of R and hence is contained in J(R). We now come to the graded analog of Theorem 2.2. Here part (iii) is an observation of [68].
Theorem 4.12. [39] Let S
= R(G) be a G-graded ring with G finite. i. JG(S) is the largest graded ideal contained in J(S). ii. J(S)IGl C JG(S) G J(S). Furthermore JG(S)= J(S) if 1Gl-l E S. iii. J G ( S ) ~C~SI J ( R ) S 2 JG(S).
Pro05 (i) Let L be the largest graded ideal of S contained in J(S). If V is a graded simple S-module, then V is a cyclic S-module so V L # V by Nakayama’s lemma. But V L is graded, so V L = 0 and we conclude that L 2 JG(5’). For the converse we use the one-to-one correspondence between the graded S-modules and the S#G*-modules given by Lemma 3.9. It follows that the graded simple S-modules must correspond to the simple S#G*-modules and therefore that J(S#G*) f l S = JG(S). Since JG(S) is a graded ideal, Lemma 4.1l(ii) yields the result. (ii) Let S = S/JG(S)so that is G-graded and J(S) contains no nonzero graded ideal. Since J(S#G*) is a characteristic ideal of S#G* it follows from Lemmas 4.7 and 4.11(ii) that J(S#G*) = 0. Now (S#G*)G MG(S) by Theorem 2.5. Hence, since J(S#G*) = 0, Theorem 2.2 implies that J ( M ~ ( 3 ) ) l= ~ l0 and that J(MG(S)) = 0 if [GI-’ E S#G*. Since J(MG(S)) = MG(J(S)),we conclude from the definition of S that J(S)lGIC_ JG(S)and that J(S) JG(S)if /GI-’ E S. (iii) Set I = S J ( R ) S so that I is a graded ideal of S. By Lemma 4.11(i) and part (i) above, I S JG(S)G J(S). Moreover J(R)CInRGJ(S)nR=J(R) so JG(S)/Iis a graded ideal of S / I which has identity component 0. By Lemma 4.5 we have J G ( S ) ~ ~I I. This completes the proof. I We close this section with the graded version of Maschke’s theorem.
37
4. Maschke’s Theorem
Proposition 4.13. [148] [lSO] Let S = R ( G ) be a G-graded ring with G finite and let W c V be S-modules with no (GI-torsion. Assume that W ess V . If 0 # v E V , then for some y E G we have wSy n W # 0. If in addition f!v(S,) = 0 for all 2 E G , then WIR ess V~R.
ProoJ We use duality and recall that RowG(V) is an S(1)G-module. Moreover, by Lemma 3.6, it follows that RowG(W) is an essential S(1)G-submodule of RowG(V). Thus since RowG(V) has no ]GItorsion, Proposition 4.3 implies that
Now let 0 # E V and let V E RowG(V) have 1-component v and all other entries zero. Then the essentiality implies that VS(1) n RowG(W) # 0; say 0 # VQ E RowG(W) with Q E S(1). By definition of V and a , the z-component of Va is given by (Va)(z)= WQ(Z,1) E vS,-1. Thus for some y E G we have wSy f l W # 0 as required. Finally suppose l?v(S”)= 0 for all z E G and choose 0 # w E wSy n W . Then
and hence Wp ess
YR. I
We remark that the condition l?v(Sz)= 0 is of course a module analog of component regularity.
EXERCISES 1. Let S = R ( G ) be a strongly graded ring and let V be an Smodule. If U is an R-submodule of V, define U” = U S , C_ V. Show that this yields an inclusion preserving permutation representation of G on the R-submodules of V . In particular U ess W if and only if U” ess W”.
38
1. Crossed Products and Group-Graded Rings
2. Let R be a semiprime ring with ideals A and B. Show that the conditions AB = 0, B A = 0 and A n B = 0 are equivalent. Now assume that A = !R(B). Prove that RIA is semiprime and that if R has no n-torsion, then neither does RIA. For any ring R, let N(R) denote the sum of its nilpotent ideals and let P(R) be its prime radical. Thus P(R) is the intersection of all prime ideals of R. Furthermore P(R) is the last term in the upper N-series defined transfinitely by No = 0, N,/N,_l = N(R/N,-l)if a has a predecessor a - 1, and N , = U4
&,
2
Delta Methods and Semiprime Rings
5. Delta Methods It’s time to move on and consider crossed products R*G of infinite groups G. Our goal is to discover when these rings are semiprime or prime. Of course this is not completely settled even when G is finite. Thus our answer will be a reduction to R*N for certain finite subgroups N of G. Specifically, the subgroups N to be checked depend upon the action of G on the ideals of R. Various aspects of this problem have been studied for over 25 years. The first results concerned ordinary group algebras and appeared in [155] and [44]. These papers handled the semiprime and prime problems respectively using a coset counting technique known as the A-method to effectively reduce these questions to the finite normal subgroups of G. Furthermore, the same techniques handled twisted group algebras with little additional difficulty [157]. However, the crossed product situation is considerably more complicated. Thus it is appropriate to begin by briefly discussing the A-method in the case of ordinary group algebras K [ G ] .
39
40
2. Delta Methods and Semiprime Rings
A linear identity is an equation in K[G] of the form
which holds for all z E G. Such identities arise in a number of different contexts, most notably in studying when K[G] is semiprime, prime or satisfies a polynomial identity. Perhaps the simplest example here is the equation az - za = 0 for all z € G which of course merely says that a is central in K[G]. Amazingly, it turns out that all linear identities are intimately related to Z(K[G]), the center of K[GI* Now it is easy to see (Exercise 1) that Z(K[G]) has as a K basis the set of all finite class sums of G. In other words, these basis elements are of the form = CZEK.xE K[G] where K: is a conjugacy class in G of finite size. It follows that the elements of G which appear in the support of elements of Z(K[G]) are precisely those which belong to
k
A(G) is called the f.c. center (finite conjugate center) of G and any group G with G = A(G) is called an f . c . group. It turns out that the elements z of finite order, o(z), in A(G) are of particular importance. Thus we define A+ = A+(G) =
{ z E G I IG : CG(Z)[<
00
and o(z) < m } .
Key properties of these subsets are listed below (see [161, Section 4.11).
Lemma 5.1. Let G be any group. i. A and A+ are characteristic subgroups of G. ii. A/A+ is a torsion-free abelian group. iii. A+ is generated by the finite normal subgroups of G. This explains the first half of the name “A-method”. The second half is apparent in the following proof.
41
5. Delta Methods
Lemma 5.2. Let
be a linear identity in K[G].Then
and
Proof:We show first that
xT
T A ( c Y ~ =)0.~ To ~
this end, suppose the expression is not zero, fix u in its support and, for each i, write ai = 7ra(ai) Cii with Supp @ n A = fl. Let W denote the intersection of the centralizers C,(y) over all y E Supp n ~ ( a ifor ) all i. Then, by definition of A, we have IG : WI < 00 and we will restrict our attention t o group elements II: E W . Note that each such x centralizes each ~ ~ ( a i ) . Let x E W and multiply the given identity for z on the left by z-'to obtain Ci(ai)zpi= 0. This yields ra(ai)@i= @pi, and since u is in the support of the left-hand term, it must occur on the right. Thus there exist a E Supp Cii, b E Supp pi, for some i, with u = axb. Hence ax = ub-' and z E Cw(a)w,,b, a fixed right coset of Cw(a)depending only on a, b and u. We conclude that W is the finite union W = Ua,bC~(a)w,,b. As we will see in the next section (Lemma 6.2), this implies that (W : Cw(a)(< 00 for some a and we have a contradiction since JG : WI < 00 and a 4 A. Thus 7ra(ai)Pi= 0 and by applying T A to this expression we obtain ~ a ( a i ) r a ( P=i )0. I
+
xi
xi
x; x;
The second conclusion above is from [155].The first is a particularly useful formulation contained in 11941. The relevance to the primeness or semiprimeness of K[G]is as follows. Suppose A and B are nonzero ideals of K[G]with AB = 0 and choose elements a E A and P E B . If x is any element of G, then ax E A so azp E AB = 0 and we obtain the linear identity
azp = 0
for all x E G.
42
2. Delta Methods and Semiprime Rings
We conclude from the above that ..A(.) . = 0 and therefore that T A ( A ). T A ( B )= 0. This reduces the various questions to K[A] where they are easily solved. The results are:
Theorem 5.3. If char K
= 0, then K[G] is semiprime.
Theorem 5.4. [155] If charK and only if Ap(G) = (1).
=p
> 0, then K[G] is semiprime if
Here AP(G) is the subgroup of A+(G) generated by all elements of order a power of p .
Theorem 5.5. [44] K[G] is prime if and only if A+(G) = (1). Now let us move on to crossed products R*G. As will be apparent, when G acts nontrivially on the ring R, a new dimension is added to the problem. Here the first result, due to [9], asserts that if R is a simple ring and G is outer on R, then R*G is a simple ring (see Exercise 6). This was extended in [127]where it was shown that if R is prime (or semiprime) and if G is X-outer on R , then R*G is prime (or semiprime). Furthermore, as we observed earlier, [58] essentially settled the semiprime question for G finite. The final attack on infinite groups began in [132] where the A-method and the techniques of [58] combined to handle the case where R is a prime ring. This was extended in [162] to semiprime coefficients and then the problem was completely solved in [166] and [167]. Somewhat later, the results were generalized in [180], using duality, to component regular group-graded rings. In this and the next three sections, we will discuss the results of [167]. Following that paper, we will work in the context of strongly group-graded rings. The proofs are no harder in that generality and in fact they are sometimes more natural. We begin by introducing some notation which will enable us to state the main result and prove its easy direction.
Lemma 5.6. If S is strongly G-graded, then it is component regular. Furthermore if ay is a nonzero element of S, and xyz = 1, then S,a,S, is a nonzero ideal of R = S1.
43
5. Delta Methods
Proof: This is clear since 1 E S1 = S,S,-l and since S,a,S, nonzero ( R ,R)-subbimodule of R = S1. I
is a
We will use this lemma, and its obvious generalizations, freely throughout the remainder of this chapter. Now let S = R(G) be G-graded and let N Q H G. For x E H and I an ideal of R ( N ) ,we define I” = S,-I IS,. In this way H acts on the ideals of R ( N ) and basic properties are as follows.
Lemma 5.7. Assume that S = R ( G ) is strongly graded and let N a H 5 G. If x,y E H and I , J are ideals of R ( N ) ,then i. I” is an ideal o f R ( N ) , ii. (I”), = I”9 and I’ = I , iii. I J implies that I” C J“, iv. ( I J ) ” = P J ” .
Proof: Since N
Q H and z E H , it follows easily that R ( N ) S z = S,R(N). From this we conclude first that I” R ( N ) and then that I” is an ideal of R ( N ) . Now for (ii) and (iv) we have
and
since I R = I . Finally (iii) is obvious, so the lemma is proved. I
If we let Z denote the set of ideals of R ( N ) in the above situation, then (i) and (ii) assert that the map x H ” is a homomorphism of H into Sym(Z). Furthermore, by (iii), these are inclusion preserving permutations and in particular they preserve the lattice operations of arbitrary intersections and sums. Part (iv) says that finite products are also preserved. Continuing with this notation, an ideal I of R ( N ) is said to be H-inwariant if I” = I for all x E H . Since SxSx-l = R, this clearly occurs if and only if I S , = S,I for all x E H .
44
2. Delta Methods and Semiprime Rings
Note that if N = (l),then R ( N ) = R and, in this way, G acts on the ideals of R. In fact, if S = R*G is a crossed product and I is an ideal of R, then
I” = sx-lmx = ~ ( R I R =) 3-93 ~ and this is precisely the action we discussed earlier. We can now state the main result for stongly groupgraded rings.
Theorem 5.8. [166] [167] Let S = R(G) be strongly G-graded with base ring R. Then S contains nonzero ideals A, B with AB = 0 if and only if there exist i. subgroups N a H C G with N finite, ii. an H-invariant ideal I of R with IxI= 0 for all x E G \ H , iii. nonzero H-invariant ideals A, of R ( N ) with AB = 0 and
A, B c I
R(N). Furthermore A = B if and only if A = B. *
As we will see in Section 8, this yields appropriate analogs of Theorems 5.3, 5.4 and 5.5 in the case of strongly group-graded rings. It is convenient now to record the following elementary
Lemma 5.9. Suppose that S = R(G)is strongly graded, H is a subgroup of G and I is an ideal of R with I x I = 0 for all x E G \ H. Then i. I S X I= 0 for all x E G \ H, ii. I . R ( G ) . I C I . R ( H ) E R ( H ) .
Proof. Part (i) is clear since S is component regular. Moreover I . R ( G ). I = e
C IS,I xEG
=e
C IS,I c I . R ( H ) 2 R ( H ) xEH
and (ii) is proved. 8 We can now offer the
Proof of Theorem 5.8 (Easy Direction). Here we assume that S = R(G)is given and that H , N , I , A and B exist and satisfy the appropriate properties. We set A = SAS and B = SBS so that these are
45
5. Delta Methods
nonzero ideals of the strongly G-graded ring S. The goal is to show that AB = 0 or equivalently that A S x B = 0 for all z E G. If x E H then, since A is H-invariant, we have &,B = S,AB = 0. On the other hand, if x E G \ H , then since A, 8 C I . R ( N ) we have ASxB G I * R ( N ) S X I * R ( N ) . a
S, and N x G \ H so Lemma 5.9(i) implies But R ( N ) S , = that I.R(N)S,-I = 0. Therefore AS,B = 0 in this case also, and we have shown that A B = 0. Since = B implies A = B, this direction is proved. I
A
The converse proof is considerably more difficult because of problems with the A-method. As an indication of this, suppose that S = R*G is a crossed product and that we are given the identity
azp = 0
for all z E G.
Following our earlier group algebra argument, we begin by defining W = C G ( ~where ) the intersection is over all y E Supp n ~ ( a ) . Then we have IG : WI < 03 and if x E W we might suspect that Z centralizes ~ a ( a )However, . this fails for two distinct reasons. First, although x commutes with all elements y E Supp n ~ ( a we ) , cannot conclude that 3 centralizes each ?j because of the twisting. Moreover, even if there were no twisting, Z would still not necessarily commute with the coefficients in .n(a) because of the action of 3 on R. In some sense, we avoid the first difficulty by assuming that T A ( ~= ) rf and .rra(P) = si for some r , s E R. We then obtain an interesting relationship between cosets of centralizers and the action of G on R in the following way. Write a = ~ ~ ( t rCU ) = ri d and p = T A ( ~ ) ,6 = si as before and let x E G = W (in this case). If r x s # 0, then the identity coefficient r x s # 0 in a"P = 0 must be cancelled. We conclude from this that x E U a , b C ~ ( a ) w , , bwhere the union is over suitable a E Supp li and b E Supp 0. In other words we have shown that rxs = o for all IC E G \ CG(a)%,b.
ny
+
+
+
u a,b
+p
46
2. Delta Methods and Semiprime Rings
This relation is probably hopeless to study, but with a little more care, T and s can be replaced by arbitrary elements in some nonzero ideal of R. That condition then becomes quite manageable and indeed the next section shows how to translate it t o an extremely useful form. The proof of Theorem 5.8 is finally completed in Section 7 where most of the work involves finding an appropriate a and /?with T A ( ~ and T A ( ~ relatively ) well behaved. The formal A-method argument shows up in Lemma 7.5. We close this section with two elementary observations which will be needed at the end of the proof. They require the following graded-ring definitions which are the obvious analogs of their crossed product counterparts. Suppose S = R(G) is G-graded and let a = C x E G a x - Then, S u p p a = { x 6 G 1 a,
ZO}.
Similarly if H is a subgroup of G, then the map T H : R ( G ) R ( H ) given by --f
TH(a)=TH
(c
ax)
xEG
=
a x xEH
is an ( R ( H ) R(H))-bimodule , homomorphism. Furthermore, suppose A is a nonzero ideal of the strongly W graded ring R ( W ) and let N a W . Then we denote by minN A the additive span of all elements a # 0 of A whose support meets the minimum number of cosets of N .
Lemma 5.10. Let S = R(G) be given, let H s G and suppose N a W are subgroups of G normalized by H . If A is a nonzero H-invariant ideal of R ( W ) ,then i. minN A is a nonzero H-invariant ideal of R(W), ii. ‘ I T N ( A is)a nonzero H-invariant ideal of R ( N ) .
Proof: (i) By definition, minNA is nonzero and let a E A be any generator of this set. If w E W , then Supp asw C (Supp a ) w and Supp Swa w(Supp a ) so, since N a W , it follows that minN A
)
47
5. Delta Methods
is an ideal of R(W). Finally if h E H, then Supp Sh-IaSh h-l(Supp a ) h so, since A is H-invariant and N is normalized by H , we see that minN A is H-invariant. (ii) We know that TN is an (R(N),R(N))-bimodule homomorphism so 7 r ~ ( Ais) an ideal of R ( N ) . Next if a E A and h E H then, since H normalizes N , we have ~ ~ ( S h - ~ a=s Sh h) - ~ r ~ ( a ) and Sh therefore 7 r ~ ( A is ) H-invariant. Finally choose 0 # a = aw E A. If J: E Supp a , then aS,-1 C A and clearly r ~ ( a S , - i )# 0. I
xwEW
Lemma 5.11. Let R ( W ) be given, let N a W and assume that WIN is a unique product group (for example, an ordered group). If A and B are nonzero ideals of R(W) with AB = 0, then
-
T N (minN A ) T N (minN B )
= 0.
Proof: In view of the graded analog of Lemma 1.3, we may assume that N = (1).In particular T N a,) = a1.
(c,
Let a = CaZ and ,i3 = Cpy be generators of minNA and mins B , respectively. Since W is a unique product group, we can Iet zoyo be a unique product element in (Supp a)(Supp p). From a,i3 = 0 we deduce that a,,,py0 = 0. Then a&, E A has smaller support size than that of a. Hence by the minimal nature of a , we have apy0 = 0 and therefore a,py0 = 0 for all z E W . Similarly we now see that a,P E B has smaller support size than that of p. Thus a,,B = 0 so a,& = 0 for all z,y E W . In particular W ( Q ) W ( P )= QlPl = 0. I
EXERCISES 1. Show that a E Z ( K [ G ] )if and only if a" = a €or all z E G and hence if and only if the coefficients of a are constant on the conjugacy classes of G. Deduce that Z ( K [ G ] )has as a K-basis the finite class sums of G . Find an analogous result in the case of twisted group algebras Kt[G].
2. Delta Methods and Semiprime Rings
48
2. Let K[G] satisfy the linear identity Ciaixpi = 0 for all x E G. Prove that ra(ai)zpi = 0 and 7ra(ai)x7ra(pi)= 0 for all x E G. This generalizes Lemma 5.2. 3. If (1) # N is a finite normal subgroup of G, set a = fi = EXEN x E K[G].Show that a is a central element of K[G]satifying a(a - IN/) = 0. Conclude that K[G]is not prime and that K[G] is not semiprime if (NI = 0 in K . This yields the easy direction of Theorems 5.4 and 5.5. 4. Complete the hard direction of the proofs of Theorems 5.3, 5.4 and 5.5 by using Lemma 5.2 to reduce the problem to K [ A ]and then Lemma 5.11 to further reduce it to K [ A + ] .Note that A/A+ is a torsion-free abelian group and hence an ordered group. 5. Suppose R is a simple ring, u is an automorphism of R and 0 # u E R. If ur = r‘u for all T E R, show that u is a unit of R and 0 is the inner automorphism of R induced by the unit u-’. 6. Let R*G be given with R a simple ring and assume that for all 1 # z E G the automorphism ” is outer on R. Use a minimal length argument to show that R*G is simple ([9]). To this end, let I be a nonzero ideal of R*G and choose 0 # a E I of minimal support size. Show that we can assume that 1 E Supp a and then, since R is simple, that the coefficient of 1 is 1 E R. Deduce that a centralizes R and then use the result of the preceding exercise to conclude that a = 1.
xi
xi
6. Coset Calculus Let H be a group and let R be a ring. We say that H strongly permutes the ideals of R if for all x,y E H and ideals I , J of R we have i. I” is an ideal of R, ii. (I”)v = 1”s and 1’ = I , iii. I J implies that I” J ” , iv. ( I J ) ” = I”J”. Thus, for example, if we are given the strongly H-graded ring R ( H ) ,
49
6. Coset Calculus
then Lemma 5.2 asserts precisely that H strongly permutes the ideals of R. As we observed previously, the first three conditions above imply that this action comes from a homomorphism of H into the group of inclusion preserving permutations on the set of ideals of R. In particular, each also preserves the lattice operations of arbitrary sums and intersections. It follows that 0" = 0 and RZ!= R for all x EH. Now let J be a nonzero ideal of R. We are interested in studying those h E H with J h J = 0. As we have seen, the A-method usually implies that something occurs for all elements of H except for those in a finite union of cosets of varying subgroups. In this section we show how to reformulate such a conclusion so that it in fact holds for all elements in H \ L where L is a single subgroup of H suitably determined by the situation. Suppose A and B are subgroups of H and that the left cosets X Aand yB are not disjoint. If z E ZAnyB, then z A = zA, yB = z B and hence Z A n y B = z A n z B = z ( An B ) . In other words, the intersection of two left cosets is either empty or a coset of the intersection. Property (iv) above will be crucial in the next few lemmas.
Lemma 6.1. Let H strongly permute the ideals of R and let J be a nonzero ideal of R such that
u n
JhJ = 0
for d l h E H
\
hkHk.
1
uy
Here hkHk is a fixed finite union of left cosets of the subgroups Hk of H . Then there exists a subgroup L of H and a nonzero product 0 # K = JY1JY2 - . . JYT of H-conjugates of J , with some yi = 1, such that K h K = 0 for all h E H \ L. Furthermore IL : L n H k I < 00 for some k.
Proof: In the course of the proof we will replace { H I , H z , . . . , Hn} by certain other finite sets A of subgroups of H with the property that A E A implies that A C Hk for some k . We note that if the
50
2. Delta Methods and Semiprime Rings
result is proved for such a set A, then from IL : L fl A\ < 00 for some A and L f l A L n Hk we obtain ( L : L n H k ( < 00. In other words, the result will then follow for the original subgroups H I ,H2, . . . ,H,. If A is the set of all proper (that is, nonempty) intersections of the Hk’s, then A is finite and closed under intersections. Thus without loss of generality we can now assume that the Hk’s are contained in a finite set d closed under intersections and we prove the result by induction on /A/2 0. If (dl= 0 then the hypothesis and conclusion both assert that J h J = 0 for all h E H . Assume now that I d1 2 1, let A be a maximal member of A and set A‘ = A \ { A } . Then Id’[ < Id1 and A’ is closed under intersections. We will be concerned with finite unions of left cosets S = Uhi,,Ai with Ai E A. By the support of S we mean those Ai’s which occur in this representation. Suppose K = JY1J Y 2 . * - JYr # 0, some yi = 1, and K h K = 0 for all h E H \ S. If A $ Supp S , then Supp S C A’ and induction applies. Thus there exists a finite product I = KU1 K U 2. . .K U 8# 0 with some u j = 1 and IhI = 0 for all h E H \ L. Since I is a suitable product of conjugates of J and since IL : L n At1 < 00 for some At E A’, the result follows in this case. Thus we can assume, for all such pairs K and S as above, that A E Supp S. Of course there is at least one such pair by hypothesis and now we choose K and S so that S has the smallest number, say m 2 1, of cosets of A occurring in its representation. Then
where T is a finite union of cosets of groups in A’ and we define L bv
Suppose K”K # 0 for some x E H . Then K x K is a nonzero finite product of H-conjugates of J with some conjugating element equal to 1. Furthermore the symbolic formula K(H\S)K = 0 yields K“‘”-l(H\S)K= 0 so ( K ” K ) h ( K ” K= ) 0 for all h with
h E ( H \ S )Ux-l(H
\ S ) = H \ ( S fl z-lS).
51
6. Coset Calculus
Since A is closed under intersections, it is clear that S I I X - ~ Sis also a finite union of left cosets of members of A. Indeed, since clearly Supp z-lS = Supp S , we see that S n zW1Sis a union of cosets of groups of the form B n C with B , C E Supp S. Furthermore, since A is a maximal member of A, it is clear that A = B f l C can occur if and only if B = C = A. By the definition of m, we note that S n z-'S must contain m' 2 m cosets of A . Since
the A-cosets of S f l z-lS come from
But cosets of A are either disjoint or identical so this intersection has m' 5 m terms. Thus we must have m' = m, so
and hence z E L. In other words, we have shown that I d H j L ) K= 0. Now H permutes the set S2 of left cosets of A by left multiplication and L is the stabilizer of the finite set A = { zlA, zzA,, . . , zmA } of 0. Thus L is a subgroup of H . Moreover if a = zA E A, then we have Ha = { h f H I ha = a } = zA2-l and it follows that IL : L n (zAz-')l < 00. We can eliminate the conjugating element z-l by conjugating both L and K by z. However in so doing we lose the property of K that some yi = 1. Thus we must take a different approach. Suppose first that L n ziA = 0 for all i = 1 , 2 , . . . ,m.Then KhK = 0 for all h with
h E ( H \ S ) u (H\ L)= H
\ (Ln S )
and by the above assumption, L n S = LnT is a finite union of cosets of the groups B = L n A I A E d'}. Since 1231 5 ld'l < Id1 and
{
52
2. Delta Methods and Semiprime Rings
since B is clearly closed under intersections, induction applies here. Thus, as before, there exists I = K U 1 K U* 2- KuB# 0 with some uj = 1 such that IhI = 0 for all h E H \ L where IL : r l BI < 00 for some B E B. Since I has the appropriate form, the result follows in this case. Finally, if L n ziA # 8 for some i, then we may take z = zi E L. Since IL : L n (zAz-')l < 00, conjugating this expression by z E L then yields IL : L n A1 < 00 and the lemma is proved. I We remark that the same result holds with J h J replaced by J J h . Indeed, in the proof, merely replace all occurrences of K h K by KKh. Furthermore both of these hold if left cosets are replaced by right cosets. Merely note that J h J = 0 is equivalent to JJh-l = 0 and that replacing h by h-' effectively interchanges right and left cosets. The form of the lemma proved here is the one we will use. Moreover, suppose that H = hkHk is given, let H act trivially on the ideals of any ring R with 1 and take J = R. Then the hypothesis of the above lemma is vacuously satisfied and the conclusion implies that RhR = 0 for all h E H \ L. Thus we must have L = H and IH : HkI < 00 for some k. In other words, Lemma 6.1 generalizes the well-known result given below concerning the existence of subgroups of finite index.
u:
Lemma 6.2. Given the set theoretic union G=
Uxi,j Hi i, j
where H I ,H 2 , . .. ,H, are finitely many subgroups of the group G. If the union contains finitely many (or countably many) terms, then for some k, the index IG : Hkl is finite (or countable).
Proof. We proceed by induction on n. If a full set of cosets of Hn occurs in the above, in particular if R, = 1,then the result is clear. So assume that the coset x H , is missing. Then XHn G G = Ui,j x*,jHi and xH, n xn,jHn = 0, SO XHn
xn,rHn G
Uifn,jXijHi.
U xn,Tx-lxi,jHi i#n 3
Thus
53
6. Coset Calculus
and G can be written as a finite (or countable) union of left cosets of the n - 1 subgroups HI,Hz, . . . ,Hn-l. The result follows by induction. I We will require a slight strengthening of Lemma 6.1.
Lemma 6.3. Let H strongly permute the ideals of R and let J be a nonzero ideal of R such that n
J h J d
forall h E W \ U W k H k . 1
u:
Here W is a subgroup of H of finite index and wkHk is a fixed finite union of cosets of the subgroups HI, of W . Then there exists a subgroup L of H and a nonzero product 0 # K = JY1JY2 - . JYr of H-conjugates of J , with some yi = 1, such that K h K = 0 for all h E H \ L. Furthermore, IL : L n Hkl < co for some k.
Proof: Since ( H : WI < 00 and J # 0 we can choose the sequence A = { h l , h2,. . . , h, } H to be of maximal size subject to i. 1 E A and the hi are in distinct right cosets of W in H , ii. K = J h l JhZ . . . J h s # 0.
Now let x E H and suppose that K”K # 0. Then by considering the right cosets of W , we have WAX= WA. Indeed, if this were not the case then, for some j , the element hjx would be in a new right coset of W . We would then have Jhjx
Jh’
Jh2
. . . Jhs
2KxK # 0
contradicting the maximality of the size of A. Since 1 E A and WAX = W R , this implies that hix E W for some i. Furthermore J h i x J 2 K”K # 0 so, by hypothesis, since h p E W , we must have hix
E
uy
WkHk.
Thus K”K # 0 implies that x E h i 1 Equivalently, we have shown that
KhK = 0
for all h E H
\
(uy
ZvkHk)
u
hF’WkHk.
i,k
for some i.
54
2. Delta Methods and Semiprime Rings
We can now apply Lemma 6.1 t o this situation to obtain a nonzero product
I=KY'KYz . . . K Y r # O with some yj = 1 and a subgroup L of H with IL : L n Hkl < 00 for some k and with I h I = 0 for all h E H \ L. Since 1 E A, we conclude that I is an appropriate product of H-conjugates of J , thereby completing the proof. I Again there are three other forms of this lemma which are also valid. We close this section with some definitions and minor observations. We assume in the remaining three lemmas that G strongly permutes the ideals of R.
Lemma 6.4. Let I be an ideal of R. i. IG = CxEG I" is the smallest G-invariant ideal of R containing I . ii. If I is G-invariant, then so are rR(I) and ! R ( I ) .
Proof: Part (i) is clear since G preserves arbitrary sums and (ii) follows from the formula (IJ)" = I" J". I Now let I be a G-invariant ideal of R. Then I is said to be G-nilpotent-free if I contains no nonzero G-invariant nilpotent ideal of R. Similarly I is said to be G-unnihihtor-free if for all nonzero G-invariant ideals A , B G I we have AB # 0. Obviously the latter property implies the former. When I = R these conditions are called G-semiprime and G-prime respectively.
Lemma 6.5. Let I be a G-invariant ideal of R. i. If I is G-nilpotent-fkee then rR(I)= rR(12). ii. I f H is a subgroup of G of finite index, then I is G-nilpotentfree if and only if it is H-nilpotent-free.
Proof: (i) Obviously r R ( I )
r R ( 1 2 ) = J and these are G-invariant ideals of R. Since 12J = 0, we see that I J is a G-invariant ideal contained in I with ( I J ) 2= 0. Thus I J = 0 and J C rR(I).
55
6. Coset Calculus
(ii) If I is H-nilpotent-free then it is obviously G-nilpotent-free. Conversely suppose I is G-nilpotent-free and let J be an H-invariant nilpotent ideal of R contained in I . Then since IG : HI < 00, we see that JG is a finite sum of the nilpotent ideals J" and hence J G is nilpotent. Thus JG = 0 and J = 0. I
Lemma 6.6. Let H be a subgroup of G and let I be an ideal of R. Suppose that I"I = 0 for all x E G \ H. i. I H is an H-invariant ideal of R with
( I H > " ( I H= ) 0
for all
II: E
G \ H.
ii. Let I J with I an H-invariant ideal and J a G-invariant ideal. If J is G-nilpotent-free (or G-annihilator-free), then I is H nilpotent-free (or H-annihilator-free) .
ProoJ (i) If a,b E H and z E G \ H , then Ia"Ib = (Iazb-lI)b =0 since azb-l 4 H . (ii) Let A and B be H-invariant ideals of R contained in 1 with AB = 0. By considering the cases II: E H and z E G \ H separately, we see that A"B = 0 for all II: E G. It then follows that AG and BG are G-invariant ideals contained in J with AGBG = 0. If J is Gannihilator-free, then AG or BG is zero and hence I is H-annihilatorfree. By taking A = B we obtain the analogous result for nilpotentfree ideals. 1 EXERCISES 1. Find an example of a set S and a sequence of subsets S1 C_ S2 C_ . . . such that S = Si and with each S \ S, uncountable. Then find an example of a group G and a sequence of subgroups GI C_ G2 ... such that G = Gi and with each IG : G,( uncountable. Compare this with Lemma 6.2.
lJF
uy
In the following, let G be a group and let H I ,H z , . . . ,H , be finitely many (not necessarily distinct) subgroups of G .
2. Delta Methods and Semiprime Rings
56
0;
2. If D = Hi show that every left coset of D is an intersection of cosets of the various Hi’s. Deduce that (G : D (5 (G : H i ( .
ny
Now suppose that G = U; xi&.
3. Show that G = U ’ x i H i where the union is over those Hi of finite index in G. To this end, suppose H I ,H 2 , . . . ,Hk are the subgroups of finite index in G and set D = Hi. Then UF xiHi is a finite union of left cosets of D. If G # UF H i , then some coset of D is missing. 4. Prove that IG : H k ( 5 n for some k. For this we can assume that all H j have finite index in G. Then each xjHj is a finite union of left cosets of the subgroup D = Hi and one can then count the number of cosets of D which occur. 5. Finally suppose that the union Uy q H i is irredundant so that all Hi have finite index in G. Suppose D = HI n H2 n . . n Hk with k < n. Then U; xiHi is a finite union of left cosets of D properly smaller than G because of the irredundancy. Thus some coset of D is missing. Deduce that ID : D n Hj I 5 n - k for some j > k. Conclude by induction that IG : 0:H i / 5 n!.
n!
ny
7. Minimal Forms The goal here is to complete the proof of Theorem 5.8. We begin with some notation and observations. First, suppose S = R ( G ) is G-graded and let Q = CIEGaz. Then we recall that Supp a = { z E G I a, # O } Also if H is a subgroup of G, then the projection map 7 r ~R(G) : + R ( H ) is an ( R ( H ) ,R(H))-bimodule homomorphism. Next, if H is a subgroup of an arbitrary group G, then the almost centralizer of H in G is defined by
{
D ~ ( H=) x E G
I
(H:c H ( X ) l <
-> .
7. Minimal Forms
57
It is clear that D G ( H ) is a subgroup of G normalized by H . Furthermore, H n D G ( H ) = DH(H) = A ( H ) , where A ( H ) is the f.c. center of H . Finally) if S = R ( G ) is strongly G-graded then, by Lemma 5.7, G strongly permutes the ideals of R. Thus the resuIts and definitions of Section 6 apply here and we will freely use this observation without further comment. In particular, recall that R is said to be G-semiprime if it contains no nonzero G-invariant nilpotent ideal. When dealing with infinite groups, we obviously cannot proceed by induction on the size of the group. Thus it is frequently necessary to produce another numerical parameter to study. In the case of the following crucial result, we introduce a structure called a form. We can then consider forms of minimal size.
Proposition 7.1. Let S = R(G) be a strongly G-graded ring and assume that the base ring R is G-semiprime. Suppose that A and B are nonzero ideals of S with AB = 0. Then there exists a subgroup H G , a nonzero H-invariant ideal I of R and an element ,B E B such that i. I“I = 0 for all x E G \ H , ii. I T A ( A ) # 0 and 17ra(,B)# 0 where A = A(H), iii. I.rra(A). I,B = 0. The above conditions motivate the following definition. Suppose
A and B are nonzero ideals of S with AB = 0. We say that the 4tuple ( H ,D , I,,L?) is a form for A, B if i. H is a subgroup of G and D = DG(H), ii. I is a n H-invariant ideal of R with I”I = 0 for all z E G \ H , iii. ,h’ E B , I 0 # 0 and I A # 0.
The proof of Proposition 7.1 proceeds in a sequence of lemmas, the final one being Lemma 7.6. Throughout the argument we assume that the hypotheses of this proposition are satisfied.
Lemma 7.2. Forms exist.
58
2. Delta Methods and Semiprime Rings
Proof: Take H = G, D = A(G) and I = R. Since A , B # 0 and 1 E R we have I A # 0 and I p # 0 for any p E B \ 0. I We define n = ( H , D , I , p ) # , the size of the form, to be the number of right D-cosets meeting Supp p. We now assume for the rest of this argument that ( H ,D , I , 0) is a form whose size n is minimal. In the next lemma we make a slight change in p. Afterwards, no additional changes in the form will be made. Set A = A(H).
Lemma 7.3. With the above notation we have i. I is H-nilpotent-free and rs(I) = rs(12), ii. I T A ( A )# 0 and we may assume that ITA(P)# 0, iii. for all y E R ( D ) we have I y P = 0 if and only if I ~ T D ( @ = 0. )
Proof: (i) By assumption, R is G-semiprime. Since I is H-invariant and P I = 0 for all x E G \ H , it follows from Lemma 6.6(ii) that I is H-nilpotent-free. Hence, by Lemma 6.5(i), rR(I) = rR(12). Finally r s ( I ) C rs(12) and suppose y = x,y, is contained in the latter ideal. Then for all x we have 12y, = 0 so y,S,-1 rR(12) = rR(I). This yields Iy,Sz-l = 0 so Iy, = 0 as required. (ii) Since I A # 0, we have IAS, # 0 for all x. Thus since AS, A it follows that I T A ( A )# 0. Now write P = C, pz. Since I p # 0 we have I& # 0 for some x E G. Thus Ip,S,-1 # 0 and we can choose U,-I 6 S,-I with Ip,u,-1 # 0. It is now clear that ( H ,D , I,pu,-1) is also a form with the additional property that I T A ( D C T - 1 ) # 0. Furthermore we have
so (H,D,I,Pa,-1) also has minimal size n. We now replace ,8 by pa,-^ for the remainder of the proof of the proposition, or equivalently we can assume that I.ira(P) # 0. This implies in particular that T A ( ~#) 0 and hence that TO(^) # 0 since D 2 A. (iii) If IyP = 0, then applying T D yields I ~ T O (=~0.) Conversely suppose I y ~ o ( p= ) 0. Then for any T E I we have rrP € B and Supp ryp meets less than n right cosets of D since y E R(D) and ~ ~ ( r y=p 0) while TD(P) # 0. By definition of n, this implies
59
7. Minimal Forms
that (H,D,I,ryp) is not a form. Thus IrrP = 0 for all r E I so 1 2 y p = 0. We conclude from (i) above that IyP = 0.
It follows from the above that H,I and /3 satisfy (i) and (ii) of the conclusion of Proposition 7.1. If in addition they satisfy (iii), then the result is proved. Thus we will assume throughout the remainder of the proof that I;lra(A)- I p # 0 and we derive a contradiction.
Lemma 7.4. With the above assumption, there exist W a subgroup of H of finite index, a = C , a, E A n R ( H ) and d E Supp 7 r ~ ( a ) such that i. W centralizes Supp K D ( ( Y ) and Supp K ii. for some u E W
D ( ~ ) ,
iii. for all y E W
PruuJ (i) By assumption I7ra(A) . I p # 0 and hence, since r s ( I ) = rs(12j we also have
A with I T A ( ~# ) 0.~ Observe that Thus there exists a E IAI R ( H ) by Lemma 5.9(ii). Thus since D n H = A we have 7 r ~ ( a= ) 7ra(a)and I 7 r ~ ( a )# p 0. We can now assume that a is chosen so that [Supp 7 r ~ ( a )is I minimal subject to a E A n R ( H ) and I7ro(a)/3# 0. Let W be the intersection of the centralizers in H of the elements of Supp 7 r ~ ( c r ) and Supp T O ( @ . Since Supp no(a)U Supp 7 r ~ ( / 3 is ) a finite subset of D = DG(H),it is clear that IH : Wl < cm. Note that I is Hnilpotent-free, by Lemma 7.3(i), and hence it is also W-nilpotent-free by Lemma 6.5(ii). This completes the proof of (i). (ii) This part does not use the minimal nature of Supp .rr~(a). Set y = r ~ ( a ) and P write a = C , a,, P = C , p, and y = C , 7.,
a E IS1
60
2. Delta Methods and Semiprime Rings
Then 7 r ~ ( a=) &D a d and we let J be the W-invariant ideal of R given by J = & D ( R a d s d - l ) W . Notice that for all d E D, y E G we have a'dPySy-ld-l
& RffdSd-1
*
SdPySy-ld-l
cJ
'
R =J
and from this it follows that y,S,-l & J for all z E G . Now suppose that I J y = 0. Then IJy,S,-1 = 0 so yzS,-l & rR(IJ) and hence, by the above, Iy,S,-1 C_ I J fl rR(IJ) = 0 since the latter is a W-invariant nilpotent ideal contained in I . This yields Iy, = 0 and therefore I n ~ ( a ) = p I y = 0, a contradiction by the choice of a. Thus we have I J n ~ ( a ) f#l 0 and hence, and u E W with by definition of J , there exists d E Supp no(.) I(RadSd-l)"TD(a)P # 0. Since u E W H and I is H-invariant, this part is proved. (iii) Let y E W and choose any ay-1 E Sy-l and & - I y E s d - l y . We study the element
c
6 = Uy-lQdbd-iya! Clearly 6 E A CzEG
- Uy-lQbd-iyQd.
n R ( H ) and since a
=
CzEC a,
we can write 6 =
44 where a ( z )= a y - l a d b d - l y a ,
- ay-l(Y,bd-lyad.
Observe that, since y centralizes d , the summands in a(z)have grades x and y-lzy, respectively. In particular, if x 4 I), then neither of these grades is in D since y E H normalizes D. On the other hand, if z E D then z E Supp n ~ ( aso) y commutes with z and hence both these summands have grade z E D. It follows that n ~ ( 6=) CsEDu(z) and that ISupp n ~ ( 6 ) I5 JSupp~ ~ ( a )Inl fact, . this inequality is strict since clearly ~ ( d=) 0. The minimality of [Supp n ~ ( a )now I implies that Ino(6)P = 0 and hence, by applying T D , that I n ~ ( b ) n ~ (= / 30.) Now as we observed above, TD (6) comes precisely from the D-homogeneous components of a so we have
61
7. Minimal Forms
and hence
Notice that this formula holds for all ay-l E Sy-i and that ISy-l = Sy-1I since I is H-invariant. We can therefore cancel the S,-1 factor and obtain
and since this holds for all
bd-Iy
E
Sd-lY, the lemma is proved.
The next result is proved by a variant of the A-method. Fix
a = C , a,, d, W and u as in the preceding lemma for the remainder of the argument.
Lemma 7.5. With the above notation,
for all y E W \ Ui x i H i . Here U; xiHi is a fked finite union of left cosets of the subgroups Hi and each Hi is the centralizer in W of some element in (Supp p) \ D.
Proof. We freely use the fact that I is H-invariant and, in particular, that S h I = ISh for all h E H. Let y E W and suppose that
Then
I(RadSd-1 )%(a>P # 0 so Lemma 7.3(iii) yields
2. Delta Methods and Semiprime Rings
62 so Lemma 7.4(iii) yields
and therefore finally
Write a = T
+ & and P = TO(@)+ p. Since
D ( ~ )
we have
We consider the supports of each of the four summands obtained from the above expression to see how cancellation can occur. Observe that y E W so y normalizes D and centralizes d E Supp T D ( Q ) . In particular, we have Sd-lyad & S, and from this it follows easily that the sets
have supports disjoint from D. On the other hand,
by the work of the preceding paragraph so it follows that this expression must be cancelled by terms from the fourth summand
In particular, the latter two summands must have a support element in common. Thus there exist f E Supp r5, g E Supp a E Supp T D ( C Y )and b E Supp T,(P) with ayby-' = fygy-l. Since y E W centralizes -1 b E Supp T D ( ~ ) this , yields g y = ygy-l = f-lab so y E zCw(g), some fixed left coset of C w ( g ) depending only on the finitely many
p,
63
7. Minimal Forms
parameters f , g , a , b. Since g E Supp is proved.
p = (Supp @)\ D ,the lemma
We remark that the truncation from p to .rro(p),using Lemma 7'.3(iii), in the above argument is crucial. Otherwise the subgroups Hi turn out to be centralizers of elements of (Supp a ) \ D.
Lemma 7.6. Contradiction.
Proof. We use the notation of the preceding two lemmas and we set y = .~ro(a)@ = x z E G y z . Then by Lemma 7.4(ii), there exists z E Supp y with (IadSd-~)~y~ # 0 and hence
J = (I~dSd-i)~^/~S~-i is a nonzero ideal of R contained in I since I is H-invariant and -1 u E W E H . Furthermore, since J C (1adSd-1)~ we have J" C IadSd-1 and hence J"-'Y C ( I a d S d - l ) Y for all y E W . -1 It follows from the above and Lemma 7.5 that J" Y . ~ r o ( a ) @= 0 for all. y E W \ U; z& or equivalently that JY.rro(a)@= 0 for all y E \ u - l z i ~ isince u E W . In particular, J Y ~ =, o for all those y so JY(R?;Sz-l) = 0 and hence JYJ = 0. Since ( H : WI < co, Lemma 6.3 applies and there exists a subgroup L of H and a nonzero product K = J"1 Jv2. . J v s such that K h K = 0 for all h E H \ L. Furthermore, wi = 1 for some i, so K C_ J & I and ( L : L fl Hk( < 00 for some 1 5 k 5 t. We claim that ( L ,DG(L),K L ,xo(cr)P) is also a form. To start with, we have K h K = 0 for all h E H \ L and then, since K I, it follows that KgK = 0 for all g E G \ L. Thus, by Lemma 6.6(i), K L is a nonzero L-invariant ideal of R with ( K L ) g K L= 0 for all g E G \ L. Furthermore, since R is G-semiprime, Lemma 6.6(ii) implies that K L is L-nilpotent-free and in particular ( K L ) 2# 0. Suppose K L y = K L r ~ ( a )=p 0. Then K L y z = 0 so K L ( R y z S z - i )= 0 and hence K L J = 0. But rR(KL) is L-invariant, by Lemma 6.4(ii), so this yields K L J L = 0, a contradiction since K L 5 J L and ( K L ) 2# 0. Thus KL.rro(a)P# 0. This then implies that KL.rro(a)# 0 so, since a E A , we have K L A # 0. Finally .rro(a)@E B so ( L ,D G ( L )K , L ,T D ( a ) @is) indeed a form.
w
2. Delta Methods and Semiprime Rings
64
It remains to compute the size of this new form. Since H 2 L we have DG(L) 2 DG(H) = D. Thus since ~ r g ( aE) R ( D ) , it is clear that Supp ~ , ( a ) p meets at most n right cosets of DG(L). But observe that IL : L n Hkl < 00 and that Hk = Cw(g)for some g E (Supp p) \ D. Thus IL : CL(g)I< 00 so g E DG(L) and in fact Dg g DG(L). Since TO@) # 0, the two D-cosets D and Dg, which meet elements of Supp p, merge to a single coset of DG(L) and therefore Supp .~r,(a)p meets less than n right cosets of DG(L). In other words,
contradicting the minimal nature of ( H ,D,I , p). I
As we observed previously, the contradiction of Lemma 7.6 is based on the assumption that IT~(A).IP # 0. Thus ITA(A).IP = 0 and Proposition 7.1 is proved. With this in hand, it is now a simple matter to prove the main result.
Proof of Theorem 5.8 (Hard Direction). Here we assume that S = R(G) is a strongly G-graded ring and that A and B are nonzero ideals of S with AB = 0. Suppose first that R is not G-semiprime. Then there exists a nonzero G-invariant ideal of R with = 0. The result now follows with H = G, I = R, N = (1) and fi = A. We can therefore assume that R is G-semiprime so Proposition 7.1 applies. Thus there exist a subgroup H 2 G, a nonzero H-invariant ideal I of R, and an element p E B such that
a
a2
(i) I”I = 0 for all z E G \ H , (ii) I?ra(A)# 0, Ixa(,B)# 0 where A = A ( H ) , (iii) ITA(A)- I p = 0. We have therefore found an appropriate H and I . It remains to find N , A and Set A1 = ITA(A)and B1 = I - ( S T A ( ~ ) S By ) ~ .Lemma 5.9 and (ii) above, A1 is a nonzero H-invariant ideal of R ( A ) . Since
B.
65
7. Minimal Forms
I T A ( A ) I. p = 0 we have I r a @ ) .Ira@) = 0 and, again by (ii), it follows that B1 is a nonzero H-invariant ideal of R ( A ) with AlBl = 0. Note that A l , B1 E I . R(A) and, since I T A ( / ~5) ITA(B), we have B1 C I T A ( B ) . By Lemma 5.1(ii), A ( H ) / A + ( H )is torsion-free abelian and we set A2 = rninAt A1, A3 = ?rA+(A2),B2 = minA+ B1 and B3 = T A + ( B in ~ )the notation of Lemma 5.10. Then that lemma implies that A3 and B3 are nonzero H-invariant ideals of R(A+) both contained in I R ( A + ) . Furthermore, since A/A+ is an ordered group and AlBl = 0, Lemma 5.11 implies that A3B3 = 0. Since A3 and B3 are nonzero, it follows from Lemma 5.l(iii) that there exists a finite normal subgroup N C A+ of H with A4 = A3 f~R ( N ) and B4 = B3 r l R ( N ) both nonzero. Certainly A4 and B4 are H-invariant ideals of R ( N ) contained in I . R ( N ) and they satisfy A4B4 = 0. For general A and B , the result now follows by taking N as above, A = A4 and B = B4. Finally if A = B , then since B1 C I T A ( B )= I . T A ( A )= AI we have B: = 0. It then follows as above that B? = 0 for all i so we can take A = B = B4. This completes the proof. I +
Using duality, and in particular Proposition 2.6, Theorem 5.8 can be extended to component regular group-graded rings. Suppose S = R(G) is such a ring. If N a G and I a R ( N ) we can still define I” = R(Nz-’)IR(zN)a R ( N ) but we no longer have a strong action of G on these ideals. Here we say that I is G-invariant if I“ C I for all 3: f G. We state the following result without proof; details of the proof will be considered in the exercises.
Corollary 7.7. [180]Let S = R ( G ) be a component regular groupgraded ring. Then S contains nonzero ideals A and B with AB = 0 if and only if there exist i. subgroups N a H G with N finite, ii. an H-invariant ideal I of R with P I = 0 for all z f G \ H , iii. nonzero H-invariant ideals A and B of R ( N ) both contained in R ( H ) I R ( H )fl R ( N ) and with A8 = 0. Furthermore A = B if and only if 2 = 8.
66
2. Delta Methods and Semiprime Rings
EXERCISES 1. In Section 5 the A-method example used right cosets; here we used left. How did this reversal occur? 2. Consider the proof of Lemma 7.5 and suppose we were not able to truncate p to no(/3).In other words, assume that we only have the weaker expression InD(a)Sd-lya&Sy-l # 0. What conclusion can we obtain from this?
Let S = R(G) be G-graded. 3. If I a R and z E G, then by the above 1' = S z - ~ I S z Show . that the equalities of Lemma 5.7 become inclusions in general. Prove that I is G-invariant if and only if I = J n R where J = S I S a S. 4. If A a S and G is finite, define A" = AS,. Show that A" is an ideal of S . Furthermore if A is contained in the graded ideal I , then A" 11s. Finally suppose 2 1 , z2, . . . ,x, E G are chosen so that the products yi = xixi+l.. -x, for i = 1 , 2 , . . . ,n contain all elements of G. Show that AS,, SZ2- SZnC A" and deduce that A # 0 and S component regular imply that A" # 0.
nxEG
-
a
We recall the notation of Proposition 2.6. Thus let H be a subgroup of G and let I be an ideal of S{l}H C_ MG(S). Then IeaR(H) is given by I t = { a ( 1 , l )I a E I}so that Itel,l = e1,1Ie1,1.
5. If g E G, prove that Sg-lel,lg and g-'e1,lSg are both contained in S(1). Next, if I is as above and g E N G ( H ) use the inclusion Sg-lel,lg - IS . ij-le1,lSg 2 I S to deduce that (Ie)9 (IS)<.Conclude that if I is g-invariant then It is g-invariant and if IS1 = 0 then (1t)gIt = 0. 6. Suppose I = J H a S{ 1)H where J is an H-invariant ideal of S(1). Prove that I t is a graded ideal of R ( H ) with ( I t ) ,= J t . Now 1 and H finite. Conclude from suppose that A a S(1)H with A Exercise 4 that (AC)" JC.R(H) and hence that A% [JE.R(H)]# 0 if R ( H ) is component regular.
8. Sufficient Conditions
67
7. Prove Corollary 7.7 using Proposition 2.6, Theorem 5.8 (in the case of skew group rings) and the preceding four exercises.
8. Sufficient Conditions The goal now is to use Theorem 5.8 to obtain sufficient conditions for the strongly graded ring S = R(G) to be either prime or semiprime. In particular, we will obtain results which are clearly analogous to Theorems 5.3, 5.4 and 5.5 on ordinary group algebras. To avoid trivialities we will assume that R is G-semiprime. We begin with two improvements on Theorem 5.8. The first yields a better understanding of the ideal I 2 R. The second sharpens the relationship between N , H and I . In particular we show that H can be taken to be the normalizer of N , thereby giving an indication of how close N is to being normal in G. We require some definitions. Let G strongly permute the ideals of R and let I be an ideal of R. Then we denote by GI = { x E G I I” = I } the stabilizer of I in G. Clearly GIis a subgroup of G. The nonzero ideal I is said to be a trivial intersection ideal if for all x E G either I” = I or I” nI = 0. Note that I” n I = 0 implies that I”I = 0.
Lemma 8.1. Let R be a G-semiprime ring, let H be a subgroup of G and let I be a nonzero H-invariant ideal of R. Suppose that I”I = 0 for all x E G \ H . Then i. GI = H , ii. I is a trivial intersection ideal, iii. i f X is a right transversal for H in G, then IG = @ CzEX I”.
Proof. Since R is G-semiprime, Lemma 6.6(ii) implies that I is H nilpotent-free and hence I n ! R ( I ) = 0. In particular, I 2 # 0 so (i) is immediate. Next observe that
68
2. Delta Methods and Semiprime Rings
and this clearly yields the direct sum in (iii). Finally, part (ii) is immediate from (iii). I
Corollary 8.2. Let S = R(G) be a strongly G-graded ring whose base ring R is G-semprime. Then S contains nonzero ideals A and B with AB = 0 if and only if there exist i. a trivial intersection ideal I of R, ii. a finite group N with normalizer NG(N)= GI, iii. nonzero Gpinvariant ideals A and B of R ( N ) with A, B I R ( N ) and AB = 0. Furthermore, A = B if and only if A = B. Proo$ If I, N , A and B exist, then so do A and B by Theorem 5.8. Conversely suppose A and B are given. We apply Theorem 5.8 and use its notation. In particular, N is a finite group and N G H G NG(N)= H . Since I"I = 0 for all z E G\H 2 G\H, Lemma 6.6(i) implies that I = I f i is an H-invariant ideal of R with fxI = 0 for all x E G \ H . By Lemma 8.1, is a trivial intersection ideal with stabilizer G,- = H = NG(N). Now we consider the action of H 2 H on R ( N ) . Set 3 = A H and B = BH so that these are H-invariant ideals of R ( N ) clearly contained in f , R ( N ) .Since IxI = 0 for all x E H\ H and since N C H, it follows easily that [I.R ( N ) ]" . [I - R ( N ) ] = 0 for all z E H \ H and hence that A x B =-0- for - all such z. But A is H-invariant and Ab = 0 so this yields A H B = 0. Hence since the right annihilator of AH is H-invariant, we conclude that AB = AHBH = 0. I
To paraphrase the above, we see that S = R(G) is prime (or semiprime) if and only if for all trivial intersection ideals I C R and all finite subgroups N C G with NG(N)= GI we have I . R ( N ) a Gpannihilator-free (or GI-nilpotent-free) ideal of R ( N ) . The prototype example here is as follows. Let G be a group, H a subgroup and let K be a field. Then G transitively permutes the set R of right cosets of H in G by right multiplication and we let wo E R correspond to the coset H itself. Define R = K,, the complete direct product of copies of K indexed by the elements of R, so that R is a ring and in fact a K-algebra. Note that the permutation
nwER
69
8. Sufficient Conditions
action of G on R extends to a group homomorphism G AutK(R) and we form the skew group ring S = RG. It follows that R is Gprime, I = K,, is a trivial intersection ideal of R with GI = H and I . RH K [ H ] . In particular, if K [ H ]is not prime or semiprime, then neither is S. Moreover, it is easy to see (Exercise 2) that the converse to this is also true, even though R has trivial intersection ideals other than I . Thus Theorems 5.3, 5.4 and 5.5, applied to K [ H ] ,completely settle the question. Under certain circumstances the primeness or semiprimeness of R(G) depends only on the finite normal subgroups of G. Some of these circumstances are obvious; one is slightly less so. Let K ( G , 5 2 , . . .) be the polynomial ring over the field K in the noncommuting indeterminates 5 1 , 5 2 , . , . . An algebra E over K is said to satisfy a polynomial identity if there exists a nonzero polynomial f(C1, 5 2 , . . . ,L ) E K ( h ,5 2 , . . . ) with --$
for all al,a 2 , . . . ,aa E E. For example, any commutative algebra satisfies f ( & ,52) = 5152 - 5251 and note that this is a special case of the standard identity defined by
The well known Amitsur-Levitzki theorem [5] asserts that the matrix ring Mm(K)satisfies s2,. A simple linearization argument (see [161, Lemma 5.1.1 3) shows that if E satisfies a polynomial identity of degree n, then it satisfies a multilinear one of the form
with all k, E K and
k1
= 1.
Proposition 8.3. Let S = R(G) be a strongly G-graded ring whose base ring R is G-semiprime and suppose that either i. G acts trivially on the ideals of R,
70
2. Delta Methods and Semiprime Rings
ii. R is prime, iii. R is Noetherian or a t least contains no infinite direct sum of nonzero two-sided ideals, or iv. S is a K-algebra satisfying a polynomial identity. Then S is prime (or semiprime) if and only if, for all finite normal subgroups N of G, the subring R ( N ) is G-prime (or G-semiprime).
Proof. We use Corollary 8.2 and its notation. In particular we let N , I , A and B satisfy (i), (ii) and (iii) of that result. If G acts trivially on the ideals of R then certainly GI = G. Furthermore if R is a prime ring, then I # 0 implies P I # 0 and again we have GI = G. Thus in either case we must have N a G. It remains to consider assumptions (iii) and (iv). If R is Noetherian or at least contains no infinite direct sum of nonzero two-sided ideals, then it follows from Lemma 8.l(iii) that IG : G I ( < 00. Now suppose S satisfies a polynomial identity of degree n which we can take to be the multilinear polynomial
with all k, E K and Icl = 1. We claim that (G : GII 5 n. Indeed, if zO,q,.. . ,zn are in distinct left cosets of GI in G and if cti E S -1 I S z i , then it follows from Lemma 5.9(i) that if u # 1 then Xi-1 a,(l)a,(2) . = 0. Thus the polynomial identity implies that slag . . a, = 0 and we conclude that S,,I"S,, = 0, a contradiction by Lemma 6.6(ii)In other words, assumptions (iii) and (iv) each imply at least that 1G : GI[ < 00. Since N is a finite normal subgroup of GI it A+(G) and then, by Lemma 5.l(iii), that N C M follows that N where M is a finite normal subgroup of G. Since A = S A S and B = SSS are ideals of S with AB = 0, we conclude that A f l R ( M ) and B R ( M ) are nonzero G-invariant ideals of R ( M ) with product zero. This completes the proof. I In particular if S = R*G with R prime, then 5' is prime (or semiprime) if and only if, for all finite normal subgroups N of G, the
8. Sufficient Conditions
71
subring R*N is G-prime (or G-semiprime). The latter result, due to [132],will be considered in more detail in the next chapter (see Theorem 12.7 and the remarks following its proof). Now let us return to a general strongly graded ring S = R(G) with base ring R which is G-semiprime. Then it is clear from Corollary 8.2 that if R ( N ) is prime, for all relevant N , then S is prime. In Section 17, we will consider when finite crossed products are prime. Here we just finesse the problem by assuming that all such I? are trivial. The following is formulated in a manner analogous to Theorem 5.5.
Corollary 8.4. [167]Let S = R(G) be a strongly G-graded ring whose base ring R is G-prime. Suppose that, for every trivial intersection ideal I of R, we have A+(Gl) = ( 1). Then S is prime.
Proof:We use Corollary 8.2. Suppose 1 is a trivial intersection ideal of R and N is a finite subgroup of G with NG(N)= GI. Then N E AS(Gl) = ( l),by assumption, so clearly N = (l),R ( N ) = R and GI = G . Since R is G-prime, it follows from Corollary 8.2 that S is prime. I Observe that if G is torsion free, then the hypothesis A+(G,) = (1) is clearly satisfied. Thus we obtain
Corollary 8.5. (1661 I1671 Let S = R(G) be a strongly G-graded ring whose base ring R is G-prime. If G is torsion free, then S is prime. We can of course handle the semiprime problem in a similar manner. It follows from the work of Section 4 that the appropriate subrings R ( N ) will be semiprime if R is semiprime with no INI-torsion. However, this yields a rather weak result which we must strengthen in two ways. First we want to assume that R is G-semiprime rather than semiprime and second we wish to consider the torsion of I rather than that of R. This is achieved via the following variant of Theorem 4.8. We use the proof of Theorem 4.4.
72
2. Delta Methods and Semiprime Rings
Lemma 8.6. Let S = R ( H ) be a strongly H-graded ring, let N be a finite normal subgroup of H and let I be an H-invariant ideal of R. If I is H-nilpotent-free with no IN]-torsion, then I - R ( N ) is an H-nilpotent-free ideal of R ( N ) .
Proof: We first observe that I - R ( N ) has no ”1-torsion.
Indeed suppose 7 = C , 7, E I . R ( N ) with IN17 = 0. Then for all z E N we have INly,S,-l = 0 and ~,S,-I I . Since I has no (NI-torsion, we conclude that ~,S,-I= 0 and hence y = 0. Now suppose A is an H-invariant ideal of R ( N ) contained in I - R ( N ) with A2 = 0. Then L = !I.R(N)(A) is a two sided ideal of R ( N ) which is essential in I . R ( N ) as a right R(N)-submodule. Since I - R ( N ) has no INl-torsion and R ( N ) is strongly graded, it follows from Proposition 4.13 that L p ess I - R ( N ) I R .Therefore L n I = !z(A) is essential in I as a right ideal . Now observe that L‘ = !z(A) is an H-invariant ideal of R contained in I . Thus since I is H-nilpotent-free we have L‘ r l rI(L’) = 0 and since L’ ess I this yields rz(L’) = 0. Finally if a = C , a, E A I . R ( N ) ,then L’a = 0 implies that L’a,S,-1 = 0 for all z. Thus since a,S,-1 E I we have a,S,-I = 0 and a = 0. We conclude that A = 0 as required. I It is clear that the above proof only needs H to strongly permute the ideals of R ( N ) and of R in a suitably compatible manner. If V is an additive abelian group and G is an arbitary group, we say that V has no [GI-torsion if, for all finite subgroups N of G, V has no IN/-torsion. We can now obtain the analog of Theorems 5.3 and 5.4.
Corollary 8.7. [167]Let R(G) be a strongly G-graded ring whose base ring R is G-semiprime. Suppose, for every trivial intersection ideal I of R, that I has no IA+(Gz)I-torsion. Then R(G) is semiprime.
Proof. Let I be any trivial intersection ideal of R and set H
= GI. Then I”I I” n I = 0 for all z E G \ H so Lemma 6.6(ii) implies that I is H-nilpotent-free. Suppose N is a finite normal subgroup of H . Then N A+(Gz) so, by assumption, I has no IN[-torsion. We
8. Sufficient Conditions
73
therefore conclude from Lemma 8.6 that I - R ( N ) is an H-nilpotentfree ideal of R ( N ) and Corollary 8.2 yields the result. I Observe that if R has no JGJ-torsion,then certainly I has no lA+(Gr)l-torsion. Thus we obtain
Corollary 8.8. [lSS][167]Let R(G) be a strongly G-graded ring whose base ring R is G-semiprime. If R has no (GI-torsion, then R ( G ) is semiprime. This completes our combinatorial approach, via the A-methods, to the problem of determining when R(G) is prime or semiprime. As we will see in the next section, the A-methods help in characterizing group-graded rings satisfying a polynomial identity. Furthermore, they are of use in computing certain rings of quotients and in describing the prime ideals in ordinary group algebras of polycyclic-by-finite groups.
EXERCISES 1. Verify the details of the skew group ring example S = RG given immediately after Corollary 8.2. Show that S can fail to be prime or semiprime even if G has no nonidentity finite normal subgroup. 2. In the above example, find all trivial intersection ideals of R. If A is a nonzero ideal of RG, show that 0 # IAI I RH. Determine precisely when S is semiprime or prime. 3. The main results of this section apply equally well to component regular group-graded rings. Verify, in particular, the analogs of Corollaries 8.5 and 8.8. 4. Show by example that there exists a G-graded ring S = R(G) with R a field, G a torsion-free group and with S not semiprime. Obviously S cannot be component regular. 5. Let R be an integral domain with field of fractions K and let a E R. Assume that the polynomial f(z)= zn - a E K [ z ] is irreducible. If a is a root of f(z),show that the integral domain
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2. Delta Methods and Semiprime Rings
R[a]is graded by the cyclic group of order n. Conclude that a crossed product can be prime even when the group has a nontrivial normal subgroup. Furthermore show that a component regular groupgraded ring need not be strongly graded.
9. Polynomial Identities We close this chapter by briefly considering group-graded rings and crossed products which satisfy a polynomial identity. We start with the case of finite groups. For this we require certain sufficient conditions for an algebra to satisfy a polynomial identity. Recall that the Amitsur-Levitzki theorem guarantees that the matrix ring Mm(.K) satisfies the standard identity sq,. More generally suppose R = M,(T). If T satisfies a polynomial identity of degree d then, by [176], R satisfies (s2,d)' for some integer t? 2 1. Moreover if T is semiprime, we can take t? = 1. Notice that R 2 E = M,(K) and that T = C R ( E ) . Furthermore, E is a finitedimensional separable K -algebra. That is, E remains semiprimitive under all field extensions. This explains the following result which we quote without proof.
Theorem 9.1. (1391 Suppose R is a K-algebra and E is a finitedimensional separable K-subalgebra. I f C R ( E )satisfies a polynomial identity of degree d then, for some integer e 2 1, R satisfies (sdn)' with n 5 dimK E. Furthermore if R is semiprime, then t? = 1. Notice that by taking E = K we see that if R satisfies a polynomial identity of degree d, then it satisfies (sd)' for some !2 1. This is an old result of [l]. As a consequence we have
Theorem 9.2.[14] Let S = R ( G ) be a G-graded K-algebra with G finite and K C R. I f R satisfies a polynomial identity of degree d then, for some integer t? 2 1, S satisfies (sdn)' with n 5 IGI. Furthermore if S is graded semiprime, then we may take e = 1.
9. Polynomial Identities
75
ProoJ Form T = S#G* and let E T be the K-subalgebra E = @ C x E G p x K .Then E is surely separable with dimK E = IGl. Furthermore it follows easily (see Exercise 1) that C T ( E ) = @ C x E G p xisR isomorphic to the ring direct sum of [GI copies of R. Thus C T ( E )also satisfies a polynomial identity of degree d. We conclude from Theorem 9.1 that T satisfies (sdn)' and hence so does T . Finally if S is graded semiprime, then T is semiprime, by S Theorem 4.8, and we can take != 1. I In particular, if G is finite, then R satisfies a polynomial identity if and only if S = R(G) does. Now let us move on to infinite groups. In the case of ordinary group algebras K [ G ] ,the polynomial identity condition essentially depends on the structure of G. Indeed if c h a r K = 0, then it was shown in [79] that K[G] satisfies a polynomial identity if and only if G is abelian-by-finite. This was proved, in its more precise version, using the character theory of finite groups and simple reductions from the infinite to the finite case. In characteristic p > 0 more ring theoretic methods were required. Here the first result, due to [194], handled prime group rings and obtained the same characterization as above, That proof used localization and brought the A-methods into play. These ideas were pushed further in [158]and the problem was finally solved in [160]. Let p be a prime. We say that the group A is p-abelian if A', the commutator subgroup of A , is a finite p-group. For convenience, O-abelian will mean abelian. Then the result is
Theorem 9.3. [79] [160] Let K be a field of characteristic p 2 0. Then K[G] satisfies a polynomial identity if and only i f G has a p-abelian subgroup A o f finite index. Furthermore there are bounds relating the degree of the polynomial identity and the product IG : A \ . [A'\. Now it turns out that most of the proof in [160]carries over essentially verbatim to the context of component regular group-graded rings. One reason for this is that in the necessary A-lemma, namely Lemma 9.5, the various ~ ~ ( a arei either ) 0 or 1. We will offer the,
2. Delta Methods and Semiprime Rings
76
suitably modified, ring theoretic aspects of the proof here. We will just quote the group theoretic facts which are required. To start with, let G be a group and let T be a subset of G. We say that T has finite indea: in G if there exist XI,x2,. . . ,X k E G for some finite Ic with
We then define the zndex IG : TI to be the minimum such k. Of course, if no such k exists, then IG : TI = 00. Observe that if T is a subgroup of G , then this agrees with the usual definition of index. We remark that this definition is not right-left symmetric. If G = U I Txi, then taking inverses yields G = U ; x i 1 T - l and we conclude that the index of T is equal to the left index of T-l. On the other hand, it is quite possible for T to have (right) index equal to 2 and left index equal to co (see Exercise 4). The following lemma is based on the fact that distinct cosets of subgroups of G are disjoint. Proofs can be found in [161, Lemmas 5.2.2 and 5.2.11. See also Exercise 4 of Section 6.
Lemma 9.4. Let G be a group, let H I ,H2,. . . , H I , be subgroups and let S = U k1 Higi for some gi E G. i. If IG : Hil > k for all i, then S # G. ii. If S # G , then there exist ~ 1 ~ x 2. .,,xt . E G with t = (Ic + l)! such that ni Sxi = 0. In particular, if G = S U T for some subset T of G , then IG : TI 5 (k + l)!. For any group G and integer k 2 1 we define
Then A k is a normal subset which is closed under taking inverses, but it need not be a subgroup of G. Note that A l ( G ) = Z(G), A ( G )= Ah(G) and that A,&, & &,. The next result is again a coset counting argument.
UT
Lemma 9.5. Let R ( G ) be a component regular G-graded ring, let a1, a2,. . . ,at, PI, 0 2 , . . . ,,&,y E R(G)and let Ic be a fixed positive
77
9. Polynomial Identities
integer. Suppose that (Supp ai) n Ak = 0 for all i and that
with r s < k. Let T be a subset of G and suppose that for all x E G\T and X E R(x) we have
Then either y = 0 or IG : TI 5 k!.
Proof: Let USuPPaa = {Yl,Y2,..4T) a
USUPPPZ = {z1,zz,...,zs). a
We assume that y # 0 and let u E S u p p y so that y, # 0. If yi is conjugate to vz3T1 in G for some z , j , choose h,,j E G with h7fy.h. . = -'. a , j a a,3 zj Let x E G \ T and, since R ( G ) is component regular, choose X E R(z) with Xy, # 0. Then by hypothesis we have
Because xu E Supp Xy, it follows that xu occurs in the support of the above left-hand side, and thus for some i , j we have yixzj = zu. Therefore, z-lyiz = uz3T1, so yi and uz3T1 are conjugate in G and we have x -1 yix = e z-1 . = h71y.h- .. 3
a,j
2
It follows that x E CG(Yi)hi,j. We have therefore shown that G = S U T where S is the set S = Ui,j C ~ ( y i ) h i , jNow . (Supp a t ) n Ak = 8 by assumption so we have IG : C G ( Y ~ >)k~for all i. Moreover there are at most rs < k cosets in the union for S. Thus we conclude from Lemma 9.4(i) that S # G and hence Lemma 9.4(ii) yields IG : TI 5 ( r s + l)! 5 k! as required. I
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2. Delta Methods and Semiprime Rings
Now let K be a field and consider the noncommutative polynomial ring K(C1,&, ...,G ) .A linear monomial is an element p E K(C1,C2,. . . ,
+
Theorem 9.6. [160] Let S
= R(G) be a component regular group-
graded ring which is an algebra over the field K & R. If S satisfies a polynomial identity over K of degree n, then
IG : A,(G)I < (k + l)! where k = (n!)2.
Proof. We assume by way of contradiction that IG : A,[ 2 (k + l)!. As we mentioned earlier, S must satisfy the multilinear polynomial
f(C1,C 2 7 . e , Cn) = *
c
%Cu(l)Cu(2)*
* *
6(n)
oESym,
with a, E K and a1 = 1. Clearly n > 1. Now, for j = 1 , 2 , . , . ,n define f j E K((j,Cj+l,.. . , Ca) by
Thus clearly f1 = f, fn = Cn and f j is a multilinear polynomial of degree n - j 1. In particular, for all j, the variable Cj occurs in each monomial of fj. Furthermore we have
+
fj
= Cjfj+l
+ terms not starting with C j .
9. Polynomial Identities
79
For each j = 2,3, . . . ,n let M j denote the set of all linear monomials in K(<j,< j + 1 , . . . , <), and let M I be empty. Then, by the remarks preceding this theorem, we have (Mjl 5 lM2l 5 n! for all j . We show now by induction on j = 1 , 2 , . . . ,n that for any ~ j , x j +. .~. ,x, , E G we have either p ( x j , xj+l,. .. ,x,) E Ak(G) for some p E M j or f j ( X j , X j + l , . . . ,X n ) = 0 for all Xi E R(xi). Because f = f1, the result for j = 1 is given. Suppose the inductive result holds for some j < n. Fix the elements xj+l,xj+2,. . . ,x, E G and let x play the role of the jth variable. Let p E Mj+l. If p ( z j + l , x j + 2 ,. . . ,zn)E A,(G), then we are done. Thus we may assume that for all such p E Mj+l we have ~ ( x j + i , x j + 2 , . xn) $! &(G). Set M j \ Mj+i = 5 . Now let p E 5 so that p involves the variable < j . Write p = p’cjp” where p’ and p” are monomials in K(<j+l,< j + 2 , . . . , <,}. Then we have p ( x ,~ j + . .~. ,,z), E Ak(G>if and only if
a fixed right translate of Ak since A, is a normal subset of G. It therefore follows that for all z $! T = UpEI,Akh, we have p(z,xj+l,. . . ,x,) $! Ak(G), and this holds for all p E M j because M j C_ Mj+l U 5 . Because the inductive result holds for j,we conclude that if x $! T then f j ( X , Xj+l,. ..,X,) = 0 for all X E R(z) and X i E R(xi). Write
where ai,j3i E K ( < j + l (, j + 2 , . . . , Cn) and ai is a linear monomial. Hence ai E Mj+l. Now let X j + l ,Xj+z,. . . ,X , be fixed elements of S with X i E R(xi). Then by the above
for all x E G \ T and X E R ( z ) .
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2. Delta Methods and Semiprime Rings
We apply Lemma 9.5 to the above with y = fj+l(Xj+l,. . . ,X,). Now f has at most n! monomials in its expression and thus clearly, in the notation of that lemma, we have T , s < n!. Hence T S < (n!)' = Ic. Moreover, observe that Supp a i ( X j + l , . . . ,X,) is disjoint from Ak(G) since ai E Mj+l implies that a i ( z j + l , .. . ,z,) E G \ Ak(G). Thus, since S is component regular, the hypotheses of Lemma 9.5 are satisfied and there are two possible conclusions. If IG : TI 5 Ic! then since T = UpEqA,+(G)h, and 5 ( M j (5 n! we see that ( G : Ak(G)I 5
131 (G : TI 5 (n!)(Ic!)< (Ic + l ) ! *
a contradiction by assumption. Thus we conclude that
and the inductive step is proved. In particular, the inductive result holds for j = n. Here f,([,) = and M , = { Cn }. Thus we conclude that for all x E G either z E Ak(G) or R ( z ) = 0, a contradiction since G # Ak(G) and S is component regular. Therefore the assumption IG : A,(G)I 2 (k+l)! is false and the theorem is proved. 1
cn
To obtain a more understandable conclusion from the above, we require some additional group theoretic facts. The first is a simple exercise (see [161,Lemma 5.2.31 or Exercise 6). The second is a more substantial result of [207] (see [161,Theorem 5.2.91).
Lemma 9.7. Let T be a subset of G with J G : TI 5 k and set T* = T U ( l } U T - l . Then
is a subgroup of G.
Theorem 9.8. [207] Let G be a group and let Ic be a positive integer. i. If IG'l 5 Ic then G = Ak(G). ii. If G = Ak(G) then IG'( 5 (/c')'~.
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9. Polynomial Identities
The bound in (ii) above is certainly not sharp. Indeed it is a question of some group theoretic interest to obtain a bound of the correct order of magnitude. As a consequence we have
Corollary 9.9. [lSO] Let S = R(G) be a component regular Ggraded ring which is an algebra over the field K C R . If S satisfies a polynomial identity over K of degree n, then setting k = ( n ! ) 2we have i. IG : A(G)( < ( I c + l)! and A(G)’ is finite, ii. G has a characteristic subgroup A with IG : A1 and ]A‘] bounded above by a finite function of n.
< (k + l)!
Proof. For convenience set m = (k + l)!. Then by Theorem 9.6 we have (G : Ak(G)l < m and hence (G : A(G)l < m since Ak C A. Moreover every translate of A, is either contained in A of disjoint from it, so we have IA : < m. Say A = Akyi with t < m. Then there exists an integer !with y i E Ae for all i, so A AkAe 2 Ake. We conclude that A(G) = &(G) = Ake((A(G)) so, by Theorem 9.8(ii), A(G)’ is finite and (i) is proved. For (ii) let A be the characteristic subgroup of G generated by &(G). Again A 2 Ak implies that JG : A1 < m. Moreover since 1 E Ak and A, is closed under taking inverses, we see that Ak = (A,)* in the notation of Lemma 9.7. Thus, by that lemma, A = (Ak)4k A,(G) with q = k4’. But then A = A,(A) so, by Theorem 9.8(ii), /A’]5 (q 4)q4 and the latter is a finite (though very large) function of n. I
ui
We remark that lA(G)’l need not be bounded by a function of n (see Exercise 7 ) . In particular all this applies t o crossed products. But it certainly does not solve the problem. If S = R*G satisfies a polynomial identity, then G has the structure given above. But this, in itself, is not sufficient. We must also know something about the twisting and how G acts on R. This can be achieved by studying prime homomorphic images of R*G and is more appropriately left for a later section (see Section 23).
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2. Delta Methods and Semiprime Rings
EXERCISES 1. Let T = S#G* be as in Theorem 9.2. If E = C z E G p z K , show that C T ( E )= C z E G p z T p z= @ C z E G p z R and that this ring is isomorphic to a direct sum of IGl copies of R. This can be proved directly or one can use the isomorphism T S(1) & MG(S). 2. Let E be an infinite dimensional exterior algebra over the field K . Thus by definition there are elements e l , e2, e3,. . . E E such that 1 and all finite products ei, ei, . eis with il < i 2 < . < i, form a K-basis for E. Furthermore multiplication in E is determined by the relations e: = 0 and eiej = -ejei for all i # j. Let El be the subspace of E spanned by all basic products of even degree and let E, be spanned by the odd degree ones. Show that E = E1@E, is graded by the group G = { 1,x } of order 2. Furthermore if char K # 2 prove that El = Z ( E ) . 3. Continuing with the above notation, compute the polynomial expressions s,(el, e2,. . . , e n ) and sz(e1 . . . ee, ee+l . e~e)~. Deduce that if charK = 0 then E satisfies no standard identity or ( ~ 2 ) ‘ . What does Theorem 9.2 say about the polynomial identities satisfied by E? 4. Let G = ( z , y 1 y 2 = 1 , x - l ~=~y-’) be the infinite dihedral group and let T = { x n , x-ny 1 n 2 0). Notice that yT = T so ziyjT fails to contain the elements xm for all sufficiently small rn. Conclude that T has infinite left index in G but that IG : TI = 2. 5. Both Theorems 9.2 and 9.6 assume that K C R and this hypothesis is used differently and subtly in each case. Where is it used and what happens if this assumption is dropped? 6. Let T be a subset of the group G with 1 E T and T = T-l. Suppose IG : TI 5 b and say G = UFTxi with 2 1 = 1. If T 2 T then T 2n T x i # 8 for some i > 1. Deduce that T 4 2 T z l U T x i so IG : T41 5 b - 1 and hence that T4kis a subgroup of G. 7. Suppose H is the finite dihedral group H = ( A , z )where A is an abelian group of odd order, z2 = 1 and ax = a-1 for all a E A . Show that the group algebra K [ H ]satisfies the standard identity s4 but that H’ = A can be arbitrarily large. Conclude that lA(G)’l is not bounded by a function of n in Theorem 9.9(i).
--
+ +
+ + +
3
The Symmetric Ring of Quotients
10. The Martindale Ring of Quotients To go further with our study of crossed products, it is necessary to localize. Fortunately there is a rather easily constructed and well behaved ring of quotients, the Martindale ring of quotients, which suffices for our purposes. This ring was introduced in [116]as a tool in the study of prime rings satisfying a generalized polynomial identity. The concept was extended to semiprime rings in [4]. As we will see, a good deal of the remainder of this book is a consequence of the existence of this quotient ring. It is, in fact, an absolutely necessary ingredient in numerous aspects of ring theory including crossed products, Galois theory, differential operator rings and normalizing extensions. In this section, we define the quotient ring, describe its elementary properties and compute some examples. For simplicity, we will restrict our attention to prime rings. We begin with the definition. Let R be a prime ring and consider the set of all left R-module homomorphisms f:R A + RR where A ranges over all nonzero two-
83
84
3. The Symmetric Ring of Quotients
sided ideals of R. Two such functions are said to be equivalent if they agree on their common domain which is a nonzero ideal since R is prime. That this is an equivalence relation follows from
Lemma 10.1. Let f : R A + RR with A a R. If B f B with 0 # B A , then Af = 0.
= 0 for some ideal
€ A. Since Ba G B and f is a left R-module homomorphism, we have 0 = ( B a )f = B - (af). Thus since R is prime, af = O . I
Proof. Let a
We let j denote the equivalence class off and let Qe = Qe(R) be the set of all such equivalence classes. The arithmetic in Qe is defined in a fairly obvious manner. Suppose f : R A --t RR and g : RB --t RR are given. Then f + i j is the class of f g: R ( Afl B ) + RR and fg is the class of the composite function f g : R ( B A )-+ RR. It is easy to see (Exercise 1) that these definitions make sense and, by Lemma 10.1, that they respect the equivalence relation. Furthermore, the ring axioms are surely satisfied so &e is a ring with 1. Finally let rp:RR RR denote right multiplication by T E R. Then the map T H .^p is easily seen to be a ring homomorphism from R into Qe. Moreover if T # 0 then Rrp # 0 and hence T^p # 0. We conclude therefore that R is embedded isomorphically in Qe with the same 1 and we will view Qe as an overring of R. It is the left Martindale rang of quotients of
+
---f
R. Suppose ~ : R + A R R and a E A. Then apf is defined on RR and for all T E R we have
Hence $f = (af), and the map f translates in Qe to right multiplication by f. This leads to the following abstract characterization of
Qe. Proposition 10.2. Let R be a prime ring. Then Qe = Qe(R)satisfies i. Qe(R) 2 R with the same 1, ii. if q E Qe then there exists 0 # A a R with Aq
E R,
85
10. The Martindale Ring of Quotients
iii. if q E Qe and 0 # A a R with A q = 0, then q = 0, iv. i f f : R A -+ RR is given with 0 # A a R, then there exists q E Qe with aq = af for all a E A . Furthermore Qe is uniquely determined by these properties.
Proofi In view of Lemma 10.1 and the preceding paragraph, it is clear that Qe satisfies the four properties. We need only show that if Q and Q’ satisfy (i)-(iv) then they are R-isomorphic. For this we define a map ~7:Q -+ Q‘ as follows. Let q E Q and choose 0 # A Q R with Aq C R. Then the map a H aq is a left R-module homomorphism from A to R so by (iv) there exists q‘ E Q’ with aq = aq’ for all a E A. If follows easily from (iii) that q’ is uniquely determined by q independent of the choice of A and we set q’ = qa. Clearly q’ = q6 is the unique element of &’ with aq = aq‘ for all elements a in some nonzero ideal of R. From this we conclude easily that u is a ring homomorphism which is the identity on R. Now we reverse the roles and define r:Q’ ---t Q in a similar manner. Then (q’)‘ satisfies aq’ = a(q’)7 and therefore we see that CTT = rg = 1. I
A slight modification of the above shows that if 7:R R1 is a ring isomorphism, then 7 extends to an isomorphism 7 : Qe(R)-+ Qe(R1). Furthermore, let S contain the prime ring R with the same 1 and assume that S satisfies (ii) and (iii) above. That is for all s E S we have As R for some nonzero ideal A of R and A s = 0 implies s = 0. Then it follows that S embeds isomorphically in Qe(R). One can of course define Qr(R), the right Martindale ring of quotients of R in a similar manner. It is obtained from the set of all right R-module homomorphisms g : BR RR with 0 # B a R and it satisfies a result analogous to the above. We briefly consider two well-known examples. --f
--f
Lemma 10.3. Let R be a domain, that is a ring without zero divisors. i. If R is commutative then Qe(R) is the field of fractions of R. ii. If R = K ( x ,y> is the free algebra over the field K in the noncommuting indeterrninates x and y, then x and y are zero divisors
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3. The Symmetric Ring of Quotients
Proof. (i) Let Q be the field of fractions of R. We show that Q satisfies (i)-(iv) of the preceding result. Since (i), (ii) and (iii) are obvious, we need only consider (iv). Thus suppose f : R A -+ R R and let a,b be nonzero elements of A. Then
a(bf) = (ab)f = (ba)f = b(af) yields a-'(af) = b-'(bf), an equation in Q. This implies that the fraction a-'(af) is a constant for all nonzero elements of A and if q E Q is this constant value, then af = aq for all a E A . (ii) Let I be the augmentation ideal of R = K ( z ,y ) , that is the set of all polynomials in z and y with zero constant term. Then I = Rx Ry is clearly free as a left R-module with basis { z, y }. Thus we can define ~ : R + I R R by xf = 1 and y f = 0 and there exists q E Qe(R)with zq = 1 and y q = 0. I
+
It is because of (i) that Qe is a ring of quotients and it is because of (ii) that Qe is in some sense too big. Fortunately there is a smaller, better behaved ring which suffices for the applications; it is the symmetric Martindale ring of quotients Q s ( R ) . We choose to define it abstractly by its properties and then prove its existence later on.
Proposition 10.4. Let R be a prime ring. Then the ring Qs = Q,(R) is uniquely determined by the properties i. Q,(R)2 R with the same 1, ii. if q E Qs then there exist 0 # A , B a R with Aq, qB R, iii. if q E Q , and 0 # I a R, then either Iq = 0 or qI = 0 implies q=o,
iv. let f : R A + RR and g: BR -+ RR be given with 0 # A, Ba R and suppose that for all a E A and b E B we have ( a f ) b = a(gb), then there exists q E Q,(R) with af = aq and gb = qb for all a E A , b E B.
Proof.We proceed as in the proof of Proposition 10.2. Let Q , Q' both satisfy (i)-(iv) and let q E Q. Choose 0 # A, B a R with Aq, qB C R
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10. The Martindale Ring of Quotients
and define f : R A + R R and 9 : BR + RR by a f = aq and gb = qb. Then ( a f ) b = (aq)b = a(qb) = a(gb) so f and g satisfy the balanced condition of (iv). Thus there exists q' E Q' with aq = af = aq' and qb = gb = q'b for all a E A and b E B. With this observation, the earlier proof can now apply. 1 We view the formula ( a f ) b = a(gb) in (iv) as either a balanced or an associativity condition. We remark that given (ii) above, part (iii) can be weakened to a one-sided condition.
Lemma 10.5. In the above context, (iii) is equivalent to either of the conditions iii'. if q E Q9 and 0 # I a R , then I q = 0 implies q = 0, iii". if q E Qs and 0 # I a R, then qI = 0 implies q = 0.
Proof: Certainly (iii) implies (iii'). Conversely assume (iii') is satisR. fied and say qJ = 0 with 0 # J a R. Let 0 # I a R with I q Since R is prime, J # 0 and 0 = I ( q J ) = ( I q ) J , we have I q = 0. Thus by (iii'), q = 0. I We are now ready to prove the existence of Q,(R). One approach of course is to modify the proof of the existence of Qe by considering equivalence classes of ordered pairs (f,g ) of balanced functions. However, we can avoid this by identifying Q s as a specific subring of the quotient ring Qe.
Proposition 10.6. I f R is a prime ring, then Q,(R) exists. Indeed Q,(R) = { q f Qe(R) 1 qB C_ R for some 0 # B a R } and
Qs(R= ) {q
E
I
QT(R) Aq
R for some 0 # A a R } .
Proof:Let S=
{q E
Qe(R) I qB E R for some 0 # B a R } .
88
3. The Symmetric Ring of Quotients
We will show that S is a ring satisfying the four conditions of Proposition 10.4. Let 4 1 , E~ S with q1&, q2B2 G R. Then (q1 q2)(B1 n B2) and qlq2B2B1 are both contained in R. Thus S is a ring satisfying condition (i). By definition of S and properties of Qe, S also satisfies (ii) and (iii’) and therefore also (iii). Finally let f:R A 4 R R and g : BR -+ RR be balanced maps with 0 # A, B a R. By properties of Qe applied to f , there exists q E Qe with af = aq for all a E A . By the balanced condition we then have
+
a ( g b ) = ( a f ) b = (aq)b = a ( q b )
so A ( g b - qb) = 0. Since g b - qb E Qt,this yields g b = qb for all b E B. In particular, qB = g B R so q E S. Since af = aq and gb = qb, the ring S satisfies (iv). I
As we mentioned above, Qe(R) was defined in [116].Its subring Qs(R),as described in Proposition 10.6, was used in the Galois theoretic studies of [87] and [88]. The formulations given in Propositions 10.2 and 10.4 are from [171]. It follows, as in the remarks after Proposition 10.2, that if R and R1 are prime rings with q:R + R1 a ring isomorphism, then q extends to an isomorphism q: Qs(R)-+ Qs(R1).Furthermore, let S contain the prime ring R with the same 1 and assume that S satisfies (ii) and (iii) of Proposition 10.4. Then S embeds isomorphically in Q8
(R).
We observe now that, in comparison to Lemma 10.3(ii), Qs(R) is quite close to R.
Lemma 10.7. Let R be a prime ring. i. If R is a domain, then so is Qs( R ) . ii. If X 5 R is right (or left) regular in R, then it is right (or left) regular in Qs (R). (i) Let qi, 42 E Q s ( R with ) q1q2 = 0 and choose nonzero ideals A i , A2 of R with A i q i , q2A2 C R. Then (Alql)(qzAz) = 0 and since R is a domain either Alql = 0 or q2A2 = 0. Thus q1 = 0 or qz = 0.
89
10. The Martindale Ring of Quotients
(ii) Let X q = 0 for some q E Q,(R) and choose 0 # A Q R with qA R. Then X ( q A ) = 0 and, since X is appropriately regular in R, we have qA = 0 and therefore q = 0. I Let us compute a few more examples.
Lemma 10.8. Let R be a prime ring. i. ZfR is simple, then Qe(R)= Qs(R)= R. ii. Qe(M,(R)) = M,(Qe(R)) and Qs(Mn(R))= Mn(Qs(R)). iii. Let R 2 S with the same 1 and assume that S 2 I where I is a nonzero ideal of R. Then Qe(S) = Qe(R)and Q s ( S )= Qs ( R ) .
ProcrJ:(i) This is clear since any left R-module homomorphism R R -+ R R is right multiplication by some element of R. (ii) If Q = Qa(R), then M,(&) 2 M,(R) and it is clear that this overring satisfies (ii) and (iii) of Proposition 10.2. It suffices to verify (iv). To this end, let f:M,(A) -+ M,(R) be a left M,(R)-module homomorphism with 0 # A&. For each i,j let ei,j denote the usual matrix unit and let ri,j:M,(R) -+ R denote the projection into the (i,j)th-entry. Then the map
A
--f
e+A
-
f M,(R) % R
is a left R-module homomorphism and hence is represented by an element gi,j f Q. It now follows that the matrix [qi,j] represents f since for all subscripts u,21 and all a E A (eu,va>f = eu,v . ( e v , v a ) f = ezl,u .
c
etJ,w(%,w) = ( e , , , 4 . [4i,j].
W
Thus Qe(M,(R)) = M,(Qe(R)). The result for Q,(M,(R)) follows from Proposition 10.6. (iii) We prove the result for the symmetric ring of quotients; the proof for Qe is similar. Let 0 # J S. Then 0 # I J I 5 J and I J I is an ideal of both R and S. It follows that S is prime. Note that S C R C Q,(R) and we use the characterization of Q s ( S )given by (i)-(iv) of Proposition 10.4 to show that Qs(R) = Q,(S).
90
3. The Symmetric Ring of Quotients
c
Let q E Q,(R) and say 0 # A , B a R with Aq, qB R. Then I A and BI are nonzero ideals of S with IAq I C S and qBI E I S. Next suppose 0 # J a S with J q = 0 or qJ = 0. By the above, J contains a nonzero ideal of R and therefore q = 0. Thus Q,(R) satisfies (i), (ii) and (iii) for the ring S. Finally suppose A, B are nonzero ideals of S and that f : sA + sS and g:Bs + Ss are balanced maps. Let A and B be nonzero ideals of R with A A and B E B. We show that f : A -+ R is an R-homomorphism. To this end, let T E R, i E I and a E A . Since i,ir E S and ~a E A we have
c
i ( ( r a ) f )= (ira)f = ir(af) so I ( ( r a ) f - r ( a f ) ) = 0 and hence (ra)f = ~ ( a f )Similarly . g: B R is a right R-module homomorphism and the balanced condition is surely satisfied. Thus there exists q E Q,(R) with af = aq and g b = qb for all a E A, b E B. We must show that iif = aq and g b = qb for all si E A and b E B. For the first, let a E A . Then a E S and aii E A so --f
a(6.f) = (aa)f = (aa)q = a(aq)
and A(af - siq) = 0. Since A is a nonzero ideal of R we conclude that iif = siq. Similarly g b = qb so the result follows. I Basic properties of these quotient rings are as follows.
Lemma 10.9. Let R be a prime ring and set S = Qe(R) or S = Qs(R)i. S is a prime ring; in fact if q1, q2 E S with q1Rqg = 0 then q1 = 0 or q g = 0.
ii. Every automorphism (or derivation) of R extends uniquely to an automorphism (or derivation) of S. iii. If C = C s ( R ) ,then C is a field which is the center of both Qe(R) and QS(R)-
Prooj (i) As usual, q1Rq2 = 0 yields (Aq1R)qz = 0 so either AqlR = 0 or
q2
= 0.
91
10. The Martindale Ring of Quotients
(ii) Assume that S = Qe(R).If u is an automorphism of R, then the isomorphism u:R -+ R extends to an isomorphism u:S --t S . Alternately, if f : R A + RR is given, we can define fa: R A + ~ RR by aafa = (af)". Now let 6: R + R be a derivation. If s E S, choose 0 # A a R with As C_ R and define f,:A' -+ R by bf, = 6(bs) - S(b)s. Note that & ( A ZC ) A so S(b)s E R and bf, E R. Certainly f, is additive and since 6 is a derivation of R we have ( T b ) f s = S ( T b S ) - S(rb)s
= TS(bS)
+ 6 ( T ) b S - T 6 ( b ) S - S(r)bs = r(bf,).
It follows that there exists q, E S with bf, = bq, for all b E A'. Furthermore, it is clear that q, is uniquely determined by s and this formula, independent of the choice of A . Notice that if T E R, then bf, = 6(br)-b(b)r = b6(r)so qr = 6(~) in this case. We can therefore define S(s) = qs for all s E S. By definition of S(s) = q, we then have 6(bs) = bfs+6(b)s = bS(s)+6(b)s for all b E A'. It remains to show that 6 is a derivation of S . To this end let s, t E S and choose 0 # A a R with As, At C_ R. Since A3s C A2 and A2st R it follows that for all b E A4
b&(st)= 6(bst)- 6(b)st = G(bs)t + bsG(t) - S(b)st
+ G(b)st+ bsS(t) - 6(b)st = b(S(s)t+ s6(t)) and thus 6 ( s t ) = S ( s ) t + s6(t). = b6(s)t
It is easy to see that u and 6 both map Q,(R) to Q,(R). For uniqueness it suffices to assume that u is the identity on R and that 6 is the zero map. If s E S with As C R then for all a E A we have as = (as)a = ads'' = asa so A(s - s a ) = 0 and u is the identity on S. Similarly 0 = S(as) = S(a)s aS(s) = aS(s) so A6(s) = 0 and S(s) = 0 as required. (iii) Again let S = Qe(R) and fix q E Cs(R). If s E S with As R, then for all a E A
+
(aq)s = (qa)s = q(as) = (as)q
so A(qs - sq) = 0. Thus qs = sq and we conclude from Proposition 10.6 that Z(Qe)= Z(Qs) = CsfR).
92
3. The Symmetric Ring of Quotients
Now if q # 0 choose 0 # B u R with Bq G R. In this case 0 # qB = Bq a R so . l R ( q ) = 0 and we can define f : Bq + R by ( b q ) f = b. It follows that f represents an element q' E S with qq' = q'q = 1. Thus q' = q-l E Z(S) and Z(S) is a field. I The field C above is called the extended centroid of R and the subring RC of Qe(R) is the central closure of R. It is a result of [116](see Exercises 7 and 8) that RC is a prime ring equal to its central closure. In other words, this is a closure operation. On the other hand, none of the quotients Qe, QT or Qs is a closure operator. Indeed, let K be a field and define
R = K [t ] [ x y, 1 zy = tyz]. Then it was shown in [171, Section 41 that
and Qs(R)= K ( t ) [ zY, I ZY = t ~ x l Furthermore if S denotes any of the above rings then
is properly larger than S. In the next section we will continue with additional examples and computations. The true importance of the symmetric Martindale ring of quotients will not begin to emerge until Section 12.
EXERCISES
+
1. Let R be a prime ring. Show that the definitions of .f g in Qe(R)and flj make sense and respect the equivalence relation. Furthermore verify that the ring axioms are satisfied. To this end,
93
10. The Martindale Ring of Quotients
note that two functions are equivalent if they agree on any nonzero ideal contained in both of their domains. 2. Let R be any ring with 1 and let F be a nonempty multiplicatively closed family of two-sided ideals of R each with zero left and right annihilator. Consider the set of all ~ : R A RR with A E 3 and let two such functions be equivalent if they agree on any C E F which is contained in both their domains. Show that the set, RF,of equivalence classes f^ of functions is an overring of R and determine its basic properties. 3. If R is semiprime, show that 3 = { A a R 1 AR ess RR } satisfies the hypotheses of the preceding exercise. In this case, RF is the Martindale ring of quotients defined in [4]. 4. Let K be a field, let M,(K) denote the set of all infinite matrices over K and let I be the set of finite matrices in M,(K). If R = K I , show that R is a prime ring with unique nontrivial ideal I. Furthermore prove that ---$
+
Qe(R)= {row finite matrices},
QT(R)= { column finite matrices}, and
Q,(R) = { row and column finite matrices } . The proof of Lemma 10.8(ii) should be helpful. 5. In the above example, let S = Qs(R).Show that I a S and determine Qt(S ) , QT ( S ) and Qs(S).Is I an ideal of Qt(R)? 6. Let R be a prime ring. An element 0 # a E R is said to be normal if a R = Ra. Show that any such a is regular in R, invertible in Q,(R) and that r H a-'ra is an automorphism of R. 7. Suppose R is a prime ring and f:R A -, RR is given. Show that f is contained in the extended centroid of R if and only if f is a bimodule homomorphism, that is if f:RAR + RRR. 8. Again let R be a prime ring and let S = RC be its central closure. Show that S is prime and choose z in its extended centroid C'. If Az S for some 0 # A a S , prove that J = { a E A n R 1 az E R } is a nonzero ideal of R and that right multiplication by z yields a bimodule homomorphism RJR + RRR. Deduce that there exists
94
3. The Symmetric Ring of Quotients
c E C with J ( z - c ) = 0 and, since z - c belongs to the field C’, that z = c. Conclude, as in [116], that S is its own central closure.
11. Separated Groups In our later studies we will discover that the elements between R and Qs(R)seem to cause the most difficulty. Thus it is of interest to obtain conditions which guarantee that no such elements exist. Specifically, if R is a prime ring, we say that R is symmetrically closed if R = Qs(R).Our goal in this section is to obtain reasonable sufficient conditions for a crossed product R*G to be symmetrically closed. To start with, we will assume that R is a prime ring which is itself symmetrically closed. Then we will assume that A S ( G ) = (1) to guarantee that R*G is prime. But this is not sufficient. We will require conditions on G which assert that certain types of separation occur via conjugation. Furthermore we will have to avoid anomalous behavior in the ring R. We begin with the group theoretic aspects.
Lemma 11.1. Let K[G] be a prime group algebra. If K[G]is symmetrically closed, then A ( G ) = ( 1 ) .
Proof: If K[G] is symmetrically closed then, by Lemma 10.9(iii), 2 = Z ( K [ G ] is ) a field. Now K[G]is prime so we have A+(G) = (1) and hence A ( G ) is torsion free abelian. Thus all units in K(A] are trivial, that is of the form kx for some 0 # k E K and x E A. In particular, 2 K [ A ]must consist of 0 and trivial units. But 2 is close under addition, so this yields 2 = K . Finally if x E A, then its class sum a is central so Q E 2 = K and x = 1.
Therefore it is appropriate to assume that A ( G ) = (1) so that G has no finite conjugacy classes other than the identity class. As will be apparent, it is the existence of countable conjugacy classes which keep R*G from being symmetrically closed.
95
11. Separated Groups
Lemma 11.2. Let G be a group. The following are equivalent. i. Every nontrivial conjugacy class of G is uncountable. ii. Every nontrivial normal subgroup of G is uncountable. iii. If H is a countable subgroup of G, then coreG(H) = (1). iv. If H is a countable subgroup of G and A is a finite subset of G, then there exists t E G with At n H C { 1}.
Pruuf. Since any countable normal subgroup is a union of countable conjugacy classes and since any countable conjugacy class generates a countable normal subgroup, it is clear that (i), (ii) and (iii) are equivalent. Here of course coreG ( H ) is the largest normal subgroup of G contained in H . Thus if 2 E coreG(H), then { x } ~C H for all t E G , so (iv) implies (iii). Finally assume (i), let A and H be given as in part (iv) and let ul,u 2 , . . . ,a, be the nonidentity elements of A. Observe that
is a (possibly empty) union of countably many right cosets of C G ( U ~ ) .
If
uu n o o
CG(ai)xi,j= G
i=l j = 1
then we conclude from Lemma 6.2 that IG : C G ( U is ~ countable )~ for some i, contradicting part (i). Thus there exists t E G not in this union and, for this t , we have At n H { 1}. a We say that G is separated if it satisfies any of the four equivalent conditions above. It follows from (ii) that every uncountable simple group is separated. We also consider a somewhat weaker condition. A group G is weakly separated if for every finitely generated subgroup H of G and every finite subset A of G there exists t E G with At n H C { 1). Note that this notation is slightly different from that of [171].Some additional examples not covered by Lemma 11.2 are as follows.
96
3. The Symmetric Ring of Quotients
Lemma 11.3. If G is a free product of infinitely many nontrivial groups, or if G is a locally finite group with A ( G ) = 1, then G is weakly separated.
* F2 * is a countably infinite free product. Given H , A we may assume that H, A C PI* F2 *. - - * F,. If t E F,+1 with t # 1,then At n H 5 { 1}. In the second case, since finitely generated subgroups of G are finite, the same proof as the implication (i) + (iv) of Lemma 11.2 applies.
Proof. In the first case, we may assume that G = F1
It follows that if F is free of infinite rank and G is arbitrary, then the free product F * G is weakly separated. Furthermore, if G is weakly separated then A(G) = (1) since G has no nontrivial finitely generated normal subgroups. Now let us turn to the ring theoretic aspects. It will be apparent that we must assume a condition for R somewhat stronger than primeness. Since the precise condition needed is not completely clear, we will choose one which is at least familiar. A ring R is said to be (right and left) strongly prime if for every 0 # a E R there exists a finite subset T of R such that t ~ ( T a=) 0 = rR(aT). We call T a (finite) insulator for a. A strongly prime ring is, of course, necessarily prime.
Lemma 11.4. Let R*G be a prime crossed product and let q E Q,(R*G). Suppose there exist 0 # A , B a R with Aq, qB C R*G. If R is strongly prime and symmetrically closed, then q E R*G.
Proof. Fix a group element x
G. We consider the %-coefficients in Aq and qB. For each a E A , b E B write aq = ( a f ) Z + . - . and qb = Z ( g b ) + Then clearly ~ : R A RR and ~ : B R --t RR. Furthermore, (aq)b = a(qb) yields ( ~ f ) =~ a'(gb) b where c - . a .
E
--f
is the automorphism '. Now by Proposition 10.2(iv), there exists s E Qe(R)with af = as for all a E A . Hence since c extends to an automorphism of Qe ( R )we have a'(gb) = ( a f ) " b = (as)"b = a'sub
11. Separated Groups
97
so A'(s"b - gb) = 0. Thus s a g = g B & R so sa E Q,(R) = R, since R is symmetrically closed, and hence s E R. We have therefore shown that for each x E G there exists r, E R with aq = CIEG ar,3. Of course, since the right hand term here is in R*G, it follows that for any a E A the coefficients ar, are almost all zero. We show that the r,'s are almost all zero. To this end, let 0 # a E A and let T be a finite insulator for a. Then aT G A and, for each t E T , almost all atr, are zero. Since T is finite this implies that, for almost all z E G , aTr, = 0 and the claim is proved. We can now set a = CzEG r,3 E R*G. Then the above yields aq = aa so A(q - a ) = 0. Since R is prime, it follows from the freeness of R*G over R that rR*G(A) = 0. Hence, by Lemma 10.7(ii), A is right regular in Q8(R*G)and we conclude that q = a E R*G.
We will need the following elementary A-lemma for two different applications. The proof is easy since we assume A(G) = (1).
Lemma 11.5. Let R*G be given with G # (I) and A(G) = (1). Suppose S is a subring of R, 0 # a , p E R*G and F is a finite subset o f G . Assume that some (right) coefficient a ofa satisfies r R ( a S ) = 0. Then there exist s E S , g E G such that 6 = a(sg)p # 0 and has support disjoint from F .
Proof: Set X = Supp a and Y = Supp ,l and ?let g E G. Note that X g Y n F # 8 implies that g E X-lFY-' and the latter is a finite set which we view as a finite union of cosets of the identity subgroup. Next if zlgyl = z2gy2 for distinct elements X I ,2 2 E X and y1,y2 E Y , then g-l(z,lzl)g = y2yT1 so g is contained in a fixed right coset of C ~ ( z F ~ zdepending 1) only on 2 1 , x2,7 ~ 1and y2. Since G # (1) and A(G) = (1) it follows that IG : (1)l= 00 and (G : Cc(z,'zl)l = 00. Thus, by Lemma 6.2, we can choose g E G not in any of the above right cosets. For this g , we have Xg Y f l F = 0 and zlgyl = x2gyz implies z1 = z2 and y1 = 92. It follows that for all s E S , the element S = asgP has support disjoint from F and that there is no cancellation between terms in the product. Finally if a is the coefficient of 3 1 in Q and b is the (left) coefficient of y1 in 0,then 6 contains the term 3 1 a . sg . by1 = 31. asc. gy1,
98
3. The Symmetric Ring of Quotients
where c = gbg-' E R \ 0, and this term can be chosen to be nonzero since, by assumption, aSc # 0. I The next lemma explains how the subring S comes into play.
Lemma 11.6. Let R*G be given with A ( G ) = (1) and with R strongly prime. Suppose q E Q,(R*G) and 0 # a E R*G with a(R*G)q G R*G. Then there exists a countable subring S R and a countable subgroup H of G such that i. S*H is a sub-crossed product of R*G, ii. a E S*H and A ( H ) = (l), iii. a(S*H)q C S*H, and iv. for some coefficient a of a we have rR(aS) = 0.
Proof. Let a be a nonzero coefficient of a and let T be a finite insulator. Then rR(aT) = 0 and, since S will be chosen to contain T , (iv) follows. Set HO = (Supp a ) and let SO G R be generated by T , the coefficients of a , all twistings of pairs of elements in HO and all conjugates of these elements under the action of Ho. Since Ho is countable, SOis countably generated and hence countable. Certainly So*Ho is a sub-crossed product of R*G and a E SO*Ho. We construct an ascending sequence of countable sub-crossed products S,*H, so that each element of Hn# = H , \ { 1) has infinitely many conjugates in H,+l 2 H, and with a(S,*H,)q G S,+l*H,+l. To this end, suppose S,*H, is given. Since A(G) = (l),for each h E H# we can find a countably infinite subset Wh of coset representatives for C G ( ~in) G. Now let Hn+l be generated by H,, the sets Wh for all h E Hn# and the countably many supports in a(S,*H,)q. It is clear that Hn+l is countable. In the same way, we let Sn+l be generated by S,, the countably many coefficients in a(S,*H,)q, all twistings of pairs of elements of Hn+l and all images of these elements under the action of &+I. Thus Sn+1*H,+1 is a countable crossed product. Finally, let H = U,"=, H , and S = U,"==,S,. Then it is clear that S*H has the appropriate properties. It is now a simple matter to prove
99
11. Separated Groups
Theorem 11.7 [171] Let R*G be given with R strongly prime and symmetrically closed and with G separated. Then R*G is symmetrically closed.
ProoJ: We may assume that G # (1). Since G is separated, A ( G ) = (1) so R*G is prime by Proposition 8.3(ii) (or by Lemma 11.5 with R = S and F = 0). Let q E Q8(R*G)and let 0 # I a R*G with q l , I q C R*G. If a E I \ 0, then a(R*G)q C R*G so we can apply Lemma 11.6 to obtain the sub-crossed product S*H with appropriate properties. In particular, a E S * H , A ( H ) = (l),a(S*H)q C S*H and H is countable. Choose 6 E I with 1 E Supp b. Since Supp t'-'bt' = t-l(Supp 6)t and G is a separated group, it follows from Lemma 11.2(iv) that there exists t E G with Supp t'-l6t'n H = (1). But t'-l6i? E l so we can clearly assume that Supp 6n H = (1). Let 0 # b be the identity coefficient of 6, let r E R and note that qr6 E qI & R*G. If y E a( S*H ) S*H then yq E S*H and (yq)rS = y(qr6). Applying the projection map TH:R*G + R*H yields ( Y q ) r r H ( S ) = ynH(qr6). Note that T H ( ~ = ) b since Supp 6 r l H = (1) and that a, = n ~ ( q r 6 E) R*H. Thus we have yqrb = ya, for all y E a ( S * H ) and r E R. In particular, for fixed T we have a(S*H)(qrb- a,) = 0. Now by Lemma 11.5 applied to a and S*H R*H we see that rR*H(a(S*H))= 0. This of course uses the fact, by Lemma 11.6(iv), that rR(aS) = 0 for some nonzero coefficient a of a. Thus since R*G is free over R*H we have rR*G(a(S*H)) = 0 and hence, by Lemma 10.7(ii), a(S*H) is right regular in Q3(R*G). We conclude from the above that qrb = a, E R*H. In other words, if B = RbRaR then B # 0 and qB R*G. By symmetry we also obtain a nonzero ideal A of R with Aq 5 R*G and Lemma 11.4 yields the result. I
c
In a similar manner we prove
Theorem 11.8. [171] Let R*G be a strongly prime ring. Suppose R is strongly prime and symmetrically closed and that G is weakly separated. Then R*G is symmetrically closed. choose 0 # l a R*G with q I , I q R*G. a E I and let T = { 71, ~ 2 , .. . ,T~ } be an insulator for a.
Proof: Let q Let 0
#
E Qs(R*G)and
100
3. The Symmetric Ring of Quotients
Then aqq E R*G for all i and we set H = (Supp a ~ iSupp , aTiq I i = 1 , 2 , . . . , n). Since H is finitely generated and G is weakly separated, there exists 6 E I with Supp S n H = (1). Let 0 # b be the identity coefficient of S and let T E R. By applying the projection map T H to (aTiq)rS = aq(qr6) we obtain a.riqrb = a7iaT where aT = T , y ( q T S ) . Thus aT(qrb - a,) = 0 and since (YTis right regular in Qs(R*G), by Lemma 10.7(ii), we conclude that qB C R*G where B = RbR is a nonzero ideal of R. Similarly there exists 0 # A a R with Aq E R*G and Lemma 11.4 implies that qER*G. I In particular, suppose R is a symmetrically closed domain and G is a free group of infinite rank. Since G is an ordered group, we see that R*G is also a domain. Hence Lemma 11.3 and Theorem 11.8 imply that R*G is symmetrically closed. We will discuss the case of finitely generated free groups later on. If we take a close look at the proof of Lemma 11.4 we arrive at a useful, but unfamiliar, definition. Let R be a ring and let G permute its ideals. We say that R is not G-cohesive if there exists nonzero G-invariant ideals A, B of R and an infinite sequence T I , ~ 2 . ., . of nonzero elements of R such that for all a E A, b E B there are at most finitely many subscripts i with ari # 0 or rib # 0. If the above situation can never occur then, of course, R is G-cohesive. In case G = (I), all ideals of R are G-invariant, and we use cohesive instead of (1)-cohesive. Note that a cohesive ring is necessarily prime (Exercise 5) and that cohesive implies G-cohesive for all G. It turns out that Lemma 11.4 only requires that R be cohesive and symmetrically closed. Indeed it was essentially shown in the proof that strongly prime implies cohesive. More to the point, we have
Proposition 11.9. Let R*G be a prime crossed product. i. If R*G is symmetrically closed and G is infinite, then R is G-cohesive. ii. Assume that every nonzero ideal of R*G meets R nontrivially. If R is prime, symmetrically closed and G-cohesive, then RxG is symmetrically closed.
101
11. Separated Groups
Proof. (i) Suppose R is not G-cohesive. Then there exist nonzero G-invariant ideals A and B of R and a sequence q , r g , .. . of nonzero elements of R such that, for all a E A, b E B , there are at most finitely many subscripts i with ari # 0 or rib # 0. Since G is infinite, we can let z1,22,.. . be a sequence of distinct group elements. Let y be the formal infinite sum y = C z , risilciri where each si is chosen in the prime ring R so that ris&ri # 0. Notice that for each a E A we have a y E R*G. Hence since A is G-invariant, formal right multiplication by y defines a map f:A*G R*G which is certainly a left R*G-module homomorphism. Similarly, formal left multiplication by y defines a right RxG-module homomorphism g: B*G -+ R*G. Since the balanced condition is surely satisfied, we deduce from Proposition 10.4(iv) that there exists q E Q,(R*G) with aq = a y for all a E A. Finally if q E R*G, then Supp Aq 5 Supp q is finite. On the other hand, since R is prime, we see that Supp Ay = { 1c1,z 2 , . . . } is infinite. Thus q 4 R*G and R*G is not symmetrically closed. (ii) Let q E Q,(R*G) and choose nonzero ideals I,JaR*G with Iq,qJ C_ R*G. By assumption, A = I r l R # 0 and B = J n R # 0 and clearly these are G-invariant ideals of R. Since Aq,qB R*G and R symmetrically closed, we deduce as in Lemma 11.4 that there exist elements r, E R such that for all a E A, b E B we have --f
qb =
r,zb = xEG
3rkb xEG
where T; = 3-'rX3. The goal is to show that r, = 0 for almost all group elements z. If this were not the case, let T I , 7 - 2 , . . . be an infinite sequence of nonzero coefficients and choose si E R so that Si = r i s i ~ # i 0. Then S1,S2,. . . is an infinite sequence of nonzero elements of R with the property that, for all a E A , b E B , there are at most finitely many subscripts i with a6i # 0 or Sib # 0. This is a contradiction since R is G-cohesive and we can now proceed as in Lemma 11.4. I
102
3. The Symmetric Ring of Quotients
Corollary 11.10. [171]Let K[G] be a prime group algebra with G locally finite. Then K[G]is symmetrically closed if and only if G is separated.
Proof. If G is separated, then K[G] is symmetrically closed by Theorem 11.7. Conversely suppose G is not separated and let N be a countably infinite normal subgroup of G. Note that K[G] = R*(G/N) where R = K"]. Since N is locally finite and countable, we can write N = Hi, an ascending union of finite subgroups, and we let ~i = H i denote the sum of the elements of Hi in R = K"]. If I is the augmentation ideal of K [ N ]and a E I , then a E K[H,] for some m and then 0 = a f i k = f i k a for all k 2 rn. Thus since I is (GIN)-invariant, we see that R is not (GIN)-cohesive. By Proposition 11.9(i), K[G] = R*(G/N) is not symmetrically closed. I
Uf"
It is tempting to try to modify the proof of Proposition 11.9(i) to show that if R is a symmetrically closed prime ring, then it must be cohesive. However this is not the case. For a counterexample, merely take R to be the ring of row and column finite matrices in M,(K) (see Exercise 4). On the other hand, as we see below, crossed products are not counterexamples.
Theorem 11.11. [171]Let R*G be a symmetrically closed crossed product with R prime. If G # (1) and A(G) = (l),then R*G is cohesive.
Proof. Suppose by way of contradiction that R*G is not cohesive. Then there exist nonzero ideals A and B of R*G and a sequence of nonzero elements 71,7 2 . . . of R*G such that, for all a E A and p E B , there are at most finitely many subscripts i with ayi # 0 or
# 0. We proceed in a series of steps. Step 1. We may assume that there exist al, a2,. . . E A such that aiyi # 0 and ai'yj = 0 i f j > i. a1, a2, . . . , a,-1 E A satisfy# 0 and R*G is prime, there exists a, E A
Proof. Suppose we have already found ing the above. Since ?,
103
11. Separated Groups
with any, # 0. Now there are at most finitely many i with a,yi # 0 so we can delete those yi with i > n and a,yi # 0. We then renumber the remaining yi with i > n and continue the process. 1
Step 2. We can assume that
Proof: We will replace each yi in turn by a suitable yiSiyi with ai(yi6iyi) # 0. This will certainly maintain the earlier properties of the gamma sequence. Suppose y1,72, ...,?,-I have already been modified. The goal is to find 6 so that
+ +
72 . + m - 1 ) is fixed, we need has large support. Since a,(yl only find 6 so that a,y,Sy, has large support. This is achieved using Lemma 11.5 with S = R as follows. Suppose 6‘ is given. Then by Lemma 11.5 there exists r E R and g E G such that a,y,rgy, # 0 and has support disjoint from that of any,6’y,. Thus setting 6 = 6‘ rg we see that a
+
In other words, we have an inductive procedure for enlarging the support of such products and thus the required element 6 will certainly exist. 1
Step 3. The formal sum s = C r l yi defines an element of Q,(R*G) not contained in R*G. Hence R*G is not symmetrically closed.
Proof: As usual formal right and left multiplication by s yield balanced module homomorphisms f:A -, R*G and g : B --t R*G. Thus there exists an element Q E Q,(R*G) with aq = as for all a E A . Suppose by way of contradiction that g E R*G. Then for all n
104
3. The Symmetric Ring of Quotients
by Step 1. Since
by Step 2, we obtain a contradiction when n 2 JSupp 4).This completes the proof of the theorem. I There are additional group algebra examples computed in [171]. Of particular interest is the following which we state without proof.
Theorem 11.12. [171] Let K[G] be a prime group algebra with G polycyclic-by-finite. Then Q s ( K [ G ] )= K[G]Z-' where 2 = Z ( K [ G ] ) Moreover, . this is the central closure of K[G]. We will consider free rings in Section 13.
EXERCISES 1. Let G be a n algebraically closed group and let a l , a2, . . . ,a, and b l , b 2 , . . . , b, be nonidentity elements of G. Show that there exist x,y E G such that x centralizes all b j , but z centralizes no a:. Deduce that G is weakly separated. 2. Prove that R is strongly prime if and only if every nonzero ideal contains a finite subset F with & R ( F = ) 0 = rR(F). 3. Let K ( ( 1 ,( 2 , . . .) be the free K-algebra on infinitely many variables and let J be the homogeneous ideal generated by all products [&ck with i > j > k. Show that R = K ( c 1 ,c 2 , . . . ) / J is right strongly prime but not left strongly prime. This is an example of paper [70]. 4. Let S = {row and column finite matrices in M,(K) }. In Exercise 5 of the preceding section it was shown that S is symmetrically closed and that I = {finite matrices} is an ideal of S. Prove that S is not cohesive.
105
12. X-Inner Automorphisms
5 . Prove that cohesive implies prime. To this end, if AB = 0 first consider a constant sequence with all ri = ba to conclude that B A = 0. Then consider a constant sequence with all ri = b. Show by example that cohesive does not imply strongly prime. Here we can let R = K[G]with G a locally finite, uncountable simple group. 6. Let R be prime but not cohesive and let A , B and r l , r2,. . . be given. Show by induction on n that, by deleting terms if necessary, we have i. S n , , = Eiri # 0 for all ci = 0, fl which are not all zero, ii. there exist an,€ E A with an,sSn,e # 0, ... 111. an,€rj = 0 for all j > n. For each set I of positive integers, let = CiEr ri. Prove that the various 01 determine distinct elements of Q , ( R ) and conclude that Q , ( R )is uncountable. 7. Let R*G be given with R prime, countable and symmetrically closed. If G is separated, deduce that R * G is symmetrically closed.
xy
12. X-Inner Automorphisms We now come to the key property of the symmetric ring of quotients. It appears in [116] in the special case u = 1. The same proof yields this more general observation of [87].
Lemma 12.1. Let u be an automorphism of the prime ring R and let a , b, c, d be fixed nonzero elements of Q e ( R ) . If
arb = crOd for all r E R, then there exists a unit q E QB(R)with c = aq, d = q - l b and r' = q-lrq for all r E R.
PruuJ Choose a nonzero ideal J of R with J a , Jc
R and set A = J a R and C = J c R . Then A and C are nonzero ideals of R and we define f : A C and g:C A by ---$
-
i
i
3. The Symmetric Ring of Quotients
106
and 9:c x i c y i H Cxeayg-l i
a
with xi E J and yi E R. To see that f is well defined, suppose xiayi = 0. Then for all T E R the formula atb = ct"d yields
xi
and hence Cixicyg = 0 since R is prime and d # 0. Similarly g is well defined and since both are clearly left R-module homomorphisms, there exist q,ql E Qe(R) which represent f and g respectively. Since f g = 1 on A and gf = 1 on C , it follows that q' = q - l . Now for all 2 E J we have (za)q = (xa)f = xc so J(aq - c) = 0 and aq = c. Similarly for all x E J and y , r E R we have (xcy)q-'rq = ( " a y q r q = [..(y"-'r)]q
= (2cy)r"
so C(q-lrq - r") = 0 and q-lrq = r". Moreover this shows that Q,(R) since l q R implies ql" R. Finally
c
qE
c
-
arb = crud = aq q-lrq
a
d = ar(qd)
so aR(b - qd) = 0 and b = qd. I
A somewhat disguised version of this result is Lemma 12.2. Let u be an automorphism of the prime ring R. S u p pose A is a nonzero ideal of R and f:R A -+ R R is a nonzero map satisfying (tar)f = (af)r' for all a E A and r E R. Then there exists a unit q E Q,(R) with af = aq for all a E A and r" = q-lrq for all T
E R.
Proof: We know that there exists
q E Qe(R) with af = aq for all
a E A . Hence
(ar)q = ( a r )f = ( a f ) r C= (aq)r"
107
12. X-Inner Automorphisms
0 and rq = qra. Since q # 0, we conclude from the preceding lemma that there exists a unit u E Q,(R) with 1 . u = q and u-lru = r U . I
so A(TQ- Q
T ~= )
It turns out that the above two results guarantee that Q,(R) is large enough to deal with most problems in crossed products and Galois theory. Thus the automorphisms a which occur there are of particular importance.
Definition. Let u be an automorphism of the prime ring R. Then (7 is said to be X-inner if there exists a unit q E Q,(R) with ra = q-lrq for all r E R. When this occurs, then 's = q-lsq for all s E Q,(R) since any two automorphisms of Q,(R) which agree on R must be equal. We let XinnfR) be the set of X-inner automorphisms of R. Then it is clear that this is a subgroup of Aut(R) containing Inn(R), the group of ordinary inner automorphisms. Furthermore, let u E Aut(R) and let q be a unit of Q,(R) which induces an X-inner automorphism on R. Since o extends to an automorphism of Qs(R),we have (q-lrq)" = (qa)-'raqa for all r E R. This shows first that q' induces an X-inner automorphism of R and then that, as automorphisms, qo = aqa. Thus Xinn(R) a Aut (R). Since the center of R remains central in Q,(R), it follows that Xinn(R) acts trivially on Z ( R ) . In particular, if R is a K-algebra, then Xinn(R) acts as K-automorphisms. Other basic properties are as follows.
Lemma 12.3. Let R be a prime ring. i. If (7 E Aut(R), then o is X-inner if and only if there exist nonzero elements a, b, c, d E R with arb = crud for all T E R. ii. If 7:G + Aut(R) is a group homomorphism, then Ginn = { IC E G 1 x7)is X-inner on R } is a normal subgroup of G. iii. If R*G is a crossed product, then Ginn
= { IC E G
1*
is X-inner on R }
108
3. The Symmetric Ring of Quotients
is a normal subgroup of G.
Proof. (i) If a , b , c , d are given, then
is X-inner by Lemma 12.1. Conversely suppose ra = q-lrq for q a unit of Q8(R). Let 0 # A, B a R with Aq, qB R and choose a E A, b E B with aq, qb # 0. Then qr' = rq yields (aq)r"b= ar(qb) as required. (ii) This is clear since Xinn(R) Q Aut(R). (iii) Here we recall that conjugation determines a group homomorphism from 9 , the group of trivial units of R*G, to Aut(R). Thus Ginn a G and the result follows since Qinn 2 U , the group of units of R, implies that Ginn = Sinn/U. I In the context of (ii) or (iii) above, we say that G is X-inner on R if Ginn = G. Similarly, G is X-outer on R if Ginn = (1). We remark that not every unit of Q,(R) induces an X-inner automorphism of R. For example if R = M,(Z) where 2 is the ring of integers then, by Lemma 10.8(ii), Q,(R) = M,(Q) where Q is the rationals. Since 2 is a principal ideal domain, it is easy to see (Exercise 1) that the only units of M,(Q) which normalize R are of the form Q' - GL, ( 2 )and this is properly smaller than GL, (Q). Let R be a prime ring and let N be the subgroup of units of Q,(R) which normalize R . Then, following [128],the normal d o sure of R is defined to be R N , the R-linear span of N . It follows (Exercise 3) that RN is a prime subring of Q,(R) but that, unlike the central closure, it does not yield a closure operation. Now it will be apparent that the ring RN is sufficiently large to handle problems in crossed products and Galois theory. However, once derivations come into play, for example in the study of differential operator rings or enveloping rings (see [15,90,173,174]), one needs the larger ring Qs(R).Thus for the most part, we will restrict our attention to the symmetric ring of quotients.
Proposition 12.4. [58] Let R*G be given with R prime and let S = Qe(R) or Q s ( R ) . i. R*G extends uniquely to a crossed product S*G. .. 11. S*Ginn = S @c E where E = CS*G(S) = C S + G ( Rand ) C = Z(S) is the extended centroid of R.
12. X-Inner Automorphisms
109
iii. E Ct[Ginn],some twisted group algebra of Ginn over C. iv. Let a E S*G, o E Aut(R) and suppose that T Q = TO for all T E R. I f g E Supp a , then go-' induces an X-inner automorphism on R and = aoij with QIO E S*Ginn.
ProoJ (i) This uses the fact that every automorphism of R extends uniquely to one of S. Thus we can define S*G to be the set of all formal finite sums CzEG%s, with sx E S and with the usual addition. Multiplication is defined distributively using
where
T:
G x G + U(R) E U(S) is the given twisting of R*G, and
where a ( a ) E Aut(S) is the unique extension of ~ ( xE)Aut(R). The associativity of S*G now follows immediately from Lemma 1.1 and the uniqueness of extension of automorphisms. (iv) Here it is convenient to write a = x , a s l c with a, E S. Then for all T E R we have u
-
X
Comparing coefficients yields ra, = U , T ~ ~ - ' for all r E R and it follows from Lemma 12.1 that if a, # 0 then a??-' induces an Xinner automorphism on R. In particular, Supp Q G (Ginn)g and a = aog as required. (ii) It follows from (iv) above with a = 1 that E = C S + G ( RC_) S*Gin,. Now for each 2 E Ginn choose a unit u, E Q,(R) 2 S such that T % = u,~u;' and set 2 = lcu,. Then 5 centralizes R and hence also S by the uniqueness of extension. Furthermore, we have made a diagonal change of basis, so every element of S*Ginn is uniquely of the form CZb,. It now follows easily that p = C2b, E E if and
110
3. The Symmetric Ring of Quotients
only if each b, E C = C s ( R )= Z(S). Thus E is a C-algebra with basis G, E centralizes S and S*Gin, = S @c E . (iii) If ? denotes the twisting for G, then @j = @+(x, y) implies that ?(z,y) E E . Thus ?(x,y) E C and we conclude that E is a twisted group algebra of Ginn over the field C. We remark that part (iv) above with r~ # 1 is contained in [123]. Furthermore, it is easy to see that a can be written as a = qpg for some unit q E S and some p E E . The next proposition is the crossed product interpretation of a Galois theory result of [87]. If M E R*G we let Supp M = UaEM SUPP a*
Proposition 12.5. Let R*G be a crossed product with R prime, let S = Qe(R) or Q,(R) and extend R*G to S*G. If M is a nonzero ( R ,R)-subbimodule of S*G with 1 E Supp M , then there exists an element a y E M n R*G such that a E R \ 0 and y E C S + G ( Rwith ) identity coefficient equal to 1.
ProoJ For convenience we write all coefficients on the left, If a = Z S ,E~M with s1 # 0, then since R is prime, there exists r E R with rs, f R for all x E Supp Q and rs1 # 0. Thus M‘ = M n R*G has the same property as M and we may assume that M 5 R*G. Next there exists a finite subset X of G of minimal size with 1 E Supp ( M n R * X ) . Replacing M by the smaller bimodule M n R * X , we can now assume that every a E M is contained in R*X and if Q
x
EXEX
= a,a: satisfies uy = 0 for some y, then al = 0. For each E X set
there exists a =
CavyE M with a, = r
Since M is an (R,R)-bimodule, it follows from the definition of X that each A, is a nonzero ideal of R. Fix z E X. If a = Ca,y E M , we claim that a, uniquely determines al. Indeed if a’ = xaby E M satisfies a, = a:, then a’ - a E M has zero %coefficient and hence zero identity coefficient. This means that we have a well defined map f,: A, A1 2 R given --f
12. X-Inner Automorphisms
111
by a,f, = a1 for all a = xayjjf M . Since r a = x r u y j j E M it follows that f, is a left R-module homomorphism. Furthermore, from QT = za,(jjry-l)jj E M we have (a,r)f, = (a,f,)r". We conclude from Lemma 11.2 that either f, = 0 or there exists a unit u, E Q,(R) C S with a,f, = a , ~ ; and ~ rz = u,Tu;'. In particular, a: commutes with us.Note that f1 = 1 so 211 = 1. Finally choose cy E M with 1 f Supp a. Then a = x u , % and al # 0. Since a1 = a,f, we see that x E Supp a implies that f, # 0 and then that al = a,f, = a,u;'. Thus a, = alu, so
with a = a1 and y E CS*G(R)as required. Of course, the above applies if M is any nonzero ideal of R*G or S*G. Hence since Supp a y E Ginn we have
Corollary 12.6. 11271 Let R*G be a crossed product over the prime ring R. If I is any nonzero ideal of R*G, then I n R*Gi,n # 0. In particular, if R*Ginn is prime (or semiprime), then so is R*G.
Theorem 12.7. 11321 Let R*G be a crossed product over the prime ring R. Then R*G is prime (or semiprime) if and only if R*N is G-prime (or G-semiprime) for every finite normal subgroup N of G with N C G i n n . Proof: By Proposition 8.3[ii), R*G is prime (or semiprime) if and only if R*N is G-prime (or G-semiprime) for every finite normal subgroup N of G. Now we need only observe that Ninn = NnGinnaG and that, by the previous corollary, if I is a nonzero G-stable ideal of R*N, then I n R*Ninn is a nonzero G-stable ideal of R*Nin,. I We remark that for R and N as above, R*N is G-semiprime if and only if it is semiprime. This follows from Theorem 16.2(iii) since R*N has a unique maximal nilpotent ideal. Another application of Proposition 12.5 is as follows. Here, as usual, n~ denotes the projection from R*G to R+A(G).
112
3. The Symmetric Ring of Quotients
Lemma 12.8. Let 0 # l a R * G with R prime and GinnnA+(G)= (1). Then there exists 01 E I such that n ~ ( a ) isR right regular in R*G and Rxa(01)is left regular in R*G.
Proof. Extend R*G to S*G with S = Qs(R).Since I is a nonzero (R,R)-subbimodule of S*G with 1 E Supp I , Proposition 12.5 implies that there exists an element a = ay = ya E I such that a E R\O and y E E = C S + G ( Rwith ) identity coefficient 1. Since y E S*Ginn and 1 E Supp y we have 0 # .rra(r)E S*(Ginn n A). Furthermore, .rra(y) is also in E . By assumption, Ginn n A’ = (1)SO Ginn n A(G) is torsion free abelian. This implies that Ct[Ginn n A] E Ct[Ginn] = E is a domain and hence na(y) is regular in Ct[Ginn nA]. By freeness, we conclude in turn that T ~ ( Yis) regular in E = Ct[Ginn], S*Ginn = S @ cE and S*G. Finally .rra(a) = u . R A ( ~= ) .na(y)a, so since rs(aR) = 0 = !s(Ra),it follows that na(01)Ris right regular in S*G and R.rrn(ct) is left regular. 1 Suppose R*H is given and G a H with R*G prime. Then R*H = (R*G)*(H/G) and it is appropriate to consider (H/G)inn. But the action of H / N is special, in that it normalizes both R and the group Q of trivial units of R*G. In particular, there is an induced action on G E G/U(R). This explains the hypothesis of the next result. The proof in [136] is amusing, since it requires that one extend RaG to three different rings, namely S*G, QB(R*G)and R*G where the latter group is an extension of G. All these rings actually live in some large quotient ring, but we will deal with them separately here. In fact, the first extension already occurred in the previous lemma and the group extension is somewhat finessed.
Theorem 12.9. [136] Let R*G be a crossed product with R prime and Ginn n A+(G) = (1). Then R*G is prime. Now suppose q is a unit of Qs(R*G) and IS is an automorphism of G with q-lR%q = Rz” for all x E G. Then IS = 1 ~ 1 0 2where (TI centralizes a subgroup of G of finite index and 0 2 is an inner automorphism of G.
113
12. X-Inner Automorphisms
ProoJ If N is a finite normal subgroup of G, then N C_ Af(G). Thus since Ginnn Af(G) = (l),by hypothesis, it follows from Theorem 12.7 that R*G is prime. Now let q and c be as given and choose 0 # A, B Q R*G with Aq,qB R*G. By Lemma 12.8 there exists a E A with n ~ ( a ) R right regular in R*G. Since q is a unit, aq # 0 and then, as in the proof of Lemma 12.3(i), there exist 0 # p, y, S E R*G with
for all T E R and x E G. Write a = ) . ( a . a’ where Supp a’ n A = 0 and fix x E G. Since na(a)Ris right regular and 3Cp # 0, there exists T E R with ~ ~ ( a ) r#z0.p Now, by hypothesis,
+
for some s E R. Thus by considering a group element in the support of n ~ ( a ) ~# z0owe see that either axb = a‘xb’
for some a E Supp ;rra(a),a’ E Supp a‘ and b, b’ E Supp
p or
axb = cx‘d
for some a E Supp ;rr~(a), b E Supp 0,c E Supp y and d E Supp 6. To better understand the last equation, form the semidirect product G = G XI ( t )where t is an element of infinite order acting like v on G. Then xu = t-lxt and we conclude from the above that 2 belongs to a right coset of C ~ ( a - ~ a ’ or) of CG(~C-’U) depending on finitely many parameters. In other words, we see as usual that G is a finite union of cosets of the subgroups CG(U-’U’) and C G ( ~ C - ’ U as) above. Hence, by Lemma 6.2, one of these subgroups must have finite index in G. But observe that a E Supp n ~ ( aE) A and a’ E Supp a’ so a-la’ f
114
3. The Symmetric Ring of Quotients
A(G). Thus for a suitable g E G we see that C G ( t g - ' ) has finite index in G. Since t = tg-' - g and tg-' centralizes a subgroup of G of finite index, the result follows. 1 Paper [136]then goes on to describe precisely what q looks like and to discuss the group of all such X-inner automorphisms. The material becomes technically quite complicated. Let G be a group and let H be a subgroup of G. Then we recall, from Section 7, that
is a subgroup of G normalized by H . The group algebra version of the following result is due to [62].
Corollary 12.10. Let R*G be given with R prime and let HaG. SupposeHi,,nA+(H) = (1).If0 # laR*G, thenInR*(HDG(R)) # 0. In particular, if R * ( H D G ( H ) )is prime, then so is R*G.
Proof. We have R*G = (R*H)*(G/H)and R*H is prime by Theorem 12.9. Furthermore, if 3 E G induces an X-inner automorphism on R*H, then by Theorem 12.9 again, 2 E D G ( H ) H = H D G ( H ) . In other words, (G/H)inn H D G ( H ) / H and Corollary 12.6 yields the result.
I
We can now combine this fact with the methods of the last section t o obtain some more symmetrically closed crossed products. For example
Corollary 12.11. Let R*G be given with R prime and let H a G with D G ( H )= (1).I f R*H is symmetrically closed, then so is R*G.
Proof. We may assume that G # (1). Then D G ( H ) = (1) implies that H # (1) and that A ( H ) = (1). Hence, since R*H is symmetrically closed, it is cohesive by Theorem 11.11. Since H D G ( H ) = H and R*G = ( R * H ) * ( G / H ) ,it follows from Corollary 12.10 and Proposition 11.9(ii) that R*G is symmetrically closed. I
12. X-Inner Automorphisms
115
Proposition 12.12. [171] Suppose G is a nonabelian free group and R is strongly prime. Then R*G is symmetrically closed.
Proof. Since R is strongly prime and G is an ordered group, it follows (Exercise 4) that R*G is also strongly prime. If G has infinite rank, the result follows from Theorem 11.8 and Lemma 11.3. In case G has finite rank, G has a normal subgroup H with G / H infinite cyclic and with H free of infinite rank. Furthermore, DG(H)= (1). Then R*H is symmetrically closed by the above and Corollary 12.11 yields the result. I We close this section with
Theorem 12.13. [123]Let R*G be a crossed product over the prime ring R. If u is a unit of R*G which normalizes R, then u = 210s for some g E G and some unit uo E R*Ginn which induces, by conjugation, an X-inner automorphism on R. In particular if all units of R*Ginn are trivial, then u is trivial.
Proof. If c E Aut(R) with U-'TU = T " , then ru = ura. Thus by Proposition 12.4(iv), u = uog with uo a unit in R*Ginn and with acting in an X-inner fashion on R. Since uo = ug-I induces 0g-l on R, the result follows. I 0g-I
As a consequence we have
Corollary 12.14. 11231 Let R*G be given with R prime and Ginn = (1).I f a is an automorphism of R*G with R" = R, then a normalizes the group of trivial units of R*G.
ProuJ This follows from the preceding theorem since Ginn = (1) implies that the group of trivial units of R*G is the set of those units of R*G which normalize R. I This result can then we combined with Theorem 12.9 to yield
Corollary 12.15. Let R*G be a crossed product with R prime and Ginn = (1). Suppose q is a unit of Q,(R*G) which normalizes
3. The Symmetric Ring of Quotients
116
both R*G and R. Then there exists an automorphism LT of G with q - l E q = R F for all x E G. Furthermore u = u1u2 where L T ~ centralizes a subgroup of G of finite index and 0 2 is an inner automorphism of G.
EXERCISES 1. Let R be a commutative unique factorization domain. Prove that Xinn(M,(R)) = Inn(M,(R)). To this end, let A E GL,(K), where K is the field of fractions of R, with A-lM,(R)A = M,(R). We may assume that A E Mn(R) with relatively prime entries and we write A-' = B/d where B E M,(R) and d E R has no prime factor in common with all entries of B. Since d divides all entries of BM,(R)A, conclude that d is a unit of R. 2. On the other hand, suppose R is a commutative domain having a nonprincipal ideal I = aR + PR with I 2 = dR principal. Prove that M2(R) has an X-inner automorphism which is not inner. Here let y,6 E I with a6
- By = d and show that
A =
[rj
1631
induces the appropriate automorphism. 3. Prove that R N , the normal closure of the prime ring R, is a prime ring. Now suppose R = K [ t ] [ zy, 1 z y = tyz] is the example at the end of Section 10. Using the results given there, show that S = RN = K ( t ) [ zy, 1 xy = tyz] and that S N = K(t)[z-l,z, y, y-I I zy = tyz] . This is an example of [19]. 4. Suppose G has the unique product property so that for any two nonempty finite subsets A, B G there is at least one uniquely represented element in the product AB. For example, any ordered group is a unique product group. Let R*G be given. If R is a domain, or prime or strongly prime, show that the same is true of R*G. Let K[G]be a prime group algebra. 5. If c is an automorphism of G, note that extends to an algebra automorphism of K[G].Suppose 0 centralizes a subgroup of finite index, let T be a right transversal for Q(c) in G and define
117
13. Free Rings a = CtETt-lt" E
K[G].If g E G , prove that g - l a g = a. Conclude that 0 # a is a normal element of K[G](see Exercise 6 of Section 10) and that conjugation by a induces the X-inner automorphism a on K [ G ] .Following [133], such automorphism are said to be of central type. 6. Let A: G K be a linear character of G and note that X determines an automorphism A#: K[G]-, K[G]given by X#(C, a,g) = C, a,A(g)g. Suppose Ker(X) C G ( X for ) some z E A ( G ) ,let T be a right transversal for CG(Z)in G and set p = CtET X(t)zk E K [ G ] . If g E G, prove that g - l p X ( g ) g = p. Conclude that 0 # p is a normal element of K[G]and that conjugation by p induces the X-inner automorphism A# on K[G]. Such automorphisms are said to be of scalar type. --f
It is shown in [133]that every X-inner automorphism of K[G] normalizing the group of trivial units is of the form a = a1a203 where ~1 is the extension of an inner automorphism of G, a2 is an autamorphism of central type and 03 is an automorphism of scalar type. 7. Let G = (2,y I y2 = 1, y - l z y = z-') be the infinite dihedral group and let X:G + K be given by X(z) = -1 and X(y) = 1. If charK # 2, prove that X# is not an X-inner automorphism of K[G]. For this, consider the action of A# on Z ( K [ G ] ) . 8. Let R*G and &G be two crossed products and let a : R*G + k*G be a ring isomorphism with R" = If R is prime, prove that G/Ginn G/Ginn. This follows from Theorem 12.13 and is a result of [123].
a.
13. Free Rings In this section, we continue with our computations, stressing in particular symmetrically closed rings. As we will see, one such example is the free K-algebra S = K ( z ,y, . ..) on at least two generators. Note that S = K [ F ]is the semigroup algebra of the free semigroup
118
3. The Symmetric Ring of Quotients
F = (z,9,.. .). More generally, we begin by considering the semigroup crossed product R*F as defined in Section 1. In this case, we use a variant of our earlier techniques to obtain the appropriate result. The following is clearly the direct analog of Lemma 11.4 and Proposition 11.9(ii). The proof is the same.
Lemma 13.1. Let R*G be a prime semigroup crossed product and let q E Q,(R*G). Suppose there exist 0 # A, B a R with Aq, qB E R*G. If R is prime, cohesive and symmetrically closed, then q E R*G. Now let F = (z,9,.. .) be the free semigroup on the variables 2,y, . . .. A subset D of F# = F \ { 1) is said to be separated if for all elements 1 # w E F , if w is an initial segment of a E D and a final segment of b E D , then we must have a = w = b. We will use la1 to denote the length of a. All this notation will remain in force until Theorem 13.4 is proved.
Lemma 13.2. Let D be a separated subset of F and let a , b E D and f , g E F. i. Ifaf = bg then a = b and f = g . ii. I f f a = g b t h e n a = b a n d f = g . iii. If af = g b and either f # 1 or g # 1, then f = wb and g = aw for some w E F .
Proof: (i) Say la1 5 Jbl. Then af = bg implies that b = ab’. But then w = a # 1 is an initial segment of b and a final segment of a. We conclude that a = w = b and then that f = g . Part (ii) is similar. (iii) Suppose af = gb. If 191 2 la1 then g = aw so a f = awb and f = w b . Now suppose (91 < la\. Then a = gw with w # 1 and g b = a f = g u l f so b = w f . Thus w is an initial segment of b and a final segment of a and w # 1. This yields a = w = b and then f=g=l. I
Lemma 13.3. Let D be a finite nonernpty subset of F . If F has at least two generators, then there exist f , g E F with f o g separated.
119
13. Free Rings
ProoJ: Let F be generated by x, y, . . . and choose integer n with n - 2 larger than the lengths of all elements of D . We claim that x n y D x y n is separated. Thus suppose w # 1 is an initial segment of xnyaxyn and a final segment of xnybxyn with a , b E D. If 120) 5 n then w must be both a power of x and a power of y, a contradiction. Thus (wI > n. From the xnyaxyn term we see that w starts with xn and from the xnybxyn term we see that w ends with yn. But n is suitably larger than Jal so the only yn segment in xnyaxyn OCCUTS at the end and thus w = xnyaxyn. Similarly w = xnybxyn. We can now prove the analog of Theorem 11.8.
Theorem 13.4. [171]Let R t F be a crossed product with F the free semigroup on a t least two generators. Assume that R is prime, symmetrically closed and cohesive. Then R*F is prime and symmetrically closed.
Proof: Since F is ordered, it follows that for all 0 # a E R*F we have aR right regular in R*F. Hence R*F is prime and, by Lemma 10,7(ii), a R is right regular in Qs(R*F). Let J denote the augmentation ideal of R*F so that J is the set of all elements with identity coefficient zero. R*F. Let q E Q,(R*F) and let 0 # I Q R*F with I q , q I Replacing I by J I J # 0 if necessary, we can assume that Iq, qI 2 J . Choose 0 # a E I. If f , g E F , then fag E I and Supp fag = f(Supp a)g. Thus by Lemma 13.3 and the fact that F has at least two generators, we can assume that D = Supp a is separated. Now fix r E R. Then for all r’ E R we have
ar’(qra) = (ar’q)ra and note that qra,ar’q E J c R*F since ra,a‘~’E I . Set p = qra and let f E Supp 0. Since R is prime and D is separated, Lemma 13.2(i) implies that af E Supp ar’p for some T’ E R and a E D. Hence af E Supp (ar’q)ra so af = g b for some g E Supp (ar’g) and b E D = Supp a. Note that f # 1 since ,L? = qra E J . Hence, by Lemma 13.2(iii) and the fact that D is separated, we have f = f’b
3. The Symmetric Ring of Quotients
120
for some f‘ E F. We have therefore shown that every f E Supp p has a final segment in D . We can now write Q = CaEDr,U and p = CaEDpuii with 0 # r, E R and pa E R*F. The equation ar’P = (ar’q)m then yields cur’p,a = (ar’q)rr,a.
C
C
aED
aED
Now Lemma 13.2(ii) asserts that if an element of F has a final segment in D , then that segment is unique. We therefore conclude from the above that QT’fla = (Qr’q)rT,for all a E D SO ~R(qrr,-fl,) = 0. Since QR is right regular in Q,(R*F), this yields q w , = pa E R*F. In other words, we have qB G R * F where B = RrbR is the nonzero ideal of R generated by T b for some b E D. Similarly we obtain Aq 5 R*F for some 0 # A a R and Lemma 13.1 yields the result. I
Of course, if R is not cohesive, then (Exercise 1) R [ F ] is not symmetrically closed. In view of the above, Xinn(R*F) = Inn(R*F). Furthermore, since F is ordered, it is clear that U(R*F) = U(R). Thus we have Corollary 13.5. Let R*F be a crossed product with F the free semigroup on a t least two generators. Assume that R is prime, symmetrically closed and cohesive. Then Xinn(R*F) = Inn(R*F) is generated by the action of the units of R .
In the case of free algebras, the preceding two results are due to [89] (see Theorem 13.11).
Corollary 13.6. [SS] If S is the free K-algebra S
= K ( z ,y, . . .), then
Xinn(S) = (1). This follows from Corollary 13.5 in case (z, y, . . .) has at least two generators. Otherwise Lemma 10.3(i) applies. Note that, by Lemma 10.3(ii), Qe(S) > S. Observe that free algebras are filtered via total degree. More generally a ring R is filtered if R = Ur=, R, is the ascending union
121
13. Free Rings
of additive subgroups R, with RnR, 2 Rn+m. Furthermore, 1 E Ro so Ro is a subring of R. The associated graded ring R of R is given by n=O
with R-1 = 0.
Theorem 13.7. [130]Let R be a filtered ring with associated graded ring R a domain. Then R is a domain and if 0 is an X-inner automorphism of R, then u preserves the filtration and hence acts on R. Furthermore, u acts trivially on the center of R .
a
E R, \ R,-1, let E R,/Rn-l R be its leading term. Since R is a domain, it follows that a b = ab. Hence R is a domain with additive degree function. Now let u be an X-inner automorphism of R. Then, by Lemma 12.3(i), there exist a,b,c,d E R \ 0 with arb = crud for all r E R. We therefore have
Proof: If a
dega + d e g r
+ degb = degc + degr" + degd
and since deg 1 = deg 1" we conclude that deg r = deg r". Thus u preserves the filtration and hence acts on R. Finally we have i b = c r' d for all r # 0. In particular, setting r = 1 yields b = d. Since u stabilizes Z(R) and R is a domain, we conclude that = rb = i" for all i E Z(R).
a
c
a
This of course does not apply to free algebras, but it does apply to U = U ( L ) ,the universal enveloping algebra of a Lie algebra L over K . Indeed we know that U is filtered and that U is a commutative polynomial ring. Also if L is finite dimensional, then U is an Ore domain. Let q be a unit in Q s ( U )which gives rise to the X-inner automorphism (T.Then the above implies that q - l t q = ! A(!) for all !E L where A: L --t K . Thus A E HomK(L, K ) and ,![ q] = !g - q! = A(!)q so Q is a semi-invariant for L with g E Q s ( U ) x . Conversely, suppose 0 # Q E Q s(U ) with [!,q] = !q - g! = A(!>q for all !E L. It follows
+
3. The Symmetric Ring of Quotients
122
that U ( L ) q = qU(L) and hence that q is a unit of Q s ( U ) (see Exercise 6 of Section 10). Multiplying by q-l then yields q - l t q = t + A ( t ) and again we have an X-inner automorphism. We conclude that
Corollary 13.8. [130] Let L be a Lie algebra over the field K and let U = U ( L ) be its enveloping ring. i. The group of X-inner automorphisms of U ( L ) is isomorphic to the additive subgroup of HomK(L, K ) consisting of those X with Q s ( U ) x # 0. ii. The semi-invariants for L in U are precisely the normal elements of U . Hence the semicenter is a characteristic subring of U . Filtered rings occur naturally in the study of coproducts. Let R1 and R2 be rings containing a common division ring D. Then R = R1 Rz, the coproduct over D , is filtered by Fo = D and F" = (R1 Rz)" = Ri, . . Ri,. The X-inner automorphisms of such rings have been studied in a series of papers, most notably
+
[119,101,118].The best result is the following, which we offer without proof.
Theorem 13.9. [118] Assume that each Ri > D , at least one of the four dimensions over D is larger than 2, and that one-sided inverses in R, are two-sided. Then every X-inner automorphism of R = R1 R2 is inner unless one of the following occurs. i. Each Ri is primarj that is R, = D + Ti with T: = 0. ii. One R, is primary and the other is Pdimensional. iii. char D = 2, one Ri is not a domain, and one is quadratic. In the course of the proof, one shows that the X-inner automorphisms u are strongly bounded, that is there exists an integer k 2 0 with deg rU 5 deg r Ic for all r E R. Now let R = 03 be an arbitrary filtered ring. Then, as above, the degree of r E R is defined to be deg r = rnin{n r E 22"). This function clearly satisfies
+
I
1. degr 2 0 for all T # 0 and, by convention, degO = -00, 2. deg(r - s) 5 max{deg r, deg s},
13. Free Rings
123
3. deg rs 5 deg r 4. deg 1 = 0.
+ deg s ,
Of course deg(r - s ) = max{deg T , deg s} if deg T # deg s. Conversely, by defining R, = { r E R I deg r i n }, any such degree function gives rise t o a filtration on R. Definition. Let R be a filtered ring with degree funtion as above. With respect to this function, we say that { al, a2,. . . ,an } R is right dependent if either some ai = 0 or there exists b l , b p , . . . , bn E R not all zero with
Otherwise the set is right independent. We say that a E R is right dependent on { al, a2,. . . , a, } if either a = 0 or there exist bi E R such that deg(a - x a i b i )
< dega
i
+
while deg ai deg bi 5 deg a for all i. Otherwise, of course, a is right independent of { al, a2,. . . ,a, }. Finally we say that R satisfies the n-term weak algorithm if given any right dependent set { a l , a2, . . . ,a , } with m 5 n and degal 5 dega2 5 ..- 5 dega,, some ai is right dependent on { a l , a2,. . . ,ai-1). Notice that i = 1 can occur here only if a1 = 0 since otherwise we have deg(a1 - 0) < degal, a contradiction. If R satisfies the n-term weak algorithm for all n, we say that R satisfies the weak algorithm. To familiarize ourselves with these concepts, we prove the following few elementary properties.
Lemma 13.10. Let R be a filtered ring satisfying the 2-term weak algorithm. i. For all a , b E R, we have deg a is a domain.
+ deg b = deg ab and hence R
3. The Symmetric Ring of Quotients
124
ii. Ro is a division ring. iii. R satisfies the (left) 2-term weak algorithm. iv. Let a1,a2 E R with 0 5 degal 5 dega2 and suppose that albl = a2b2 for some bl, b2 with b2 # 0. Then a2 = alr + s for some r , s E R with degs < degal.
ProoJ: (i) We may assume that b # 0. If degab < dega + degb, then { a } is a right dependent set. The 1-term weak algorithm yields a = 0. (ii) Let 0 # a E &. Then a l - l a = 0 implies that { a, 1} is right dependent. Hence since dega = deg 1 = 0, we see that 1is dependent on the set { a } so there exists b E R with deg(1- ab) < deg 1 = 0 and deg a deg b 5 deg 1. Thus ab = 1 and b E Ro. Since R is a domain, = b E &. (iii) Assume that { b l , b2 } is a left dependent set with 0 5 deg b l 5 deg b2. Then there exist a l , a2 not both zero with
+
deg(albl+ a2b2)
< max{deg a1 + deg b l , dega2 + deg b2}.
In particular, this forces deg a1
+ deg b i = deg albl = deg azb2 = deg a2 + deg b2
so deg a1 2 deg a2 2 0. Observe that { al, a2 1 is right dependent so the 2-term weak algorithm implies that a1 = a2c d with degd < deg a1 = degaz+degc. Thus setting e = albl+azbz and substituting in for a l , we have az(cb1 b2) = e - dbl and
+
+
We conclude from (i) that deg(cb1+b2) < deg b~ with deg c+deg bl = deg b2 and this fact is proved. (iv) We proceed by induction on deg az. Since { a l , a2 } is dependent and 0 _< degal 5 dega2, the 2-term weak algorithm yields r’, s’ with a2 = alr’ s‘ and degs’ < dega2. If degs’ < degal we are done, so we may suppose that degal 5 degs’. Moreover
+
125
13. Free Rings
so al(b1 -r’bz) = s’b2. Since 0 5 degal 5 degs’ < dega2, induction implies that s’ = alr” + s” with degs” < degal. Hence we have a2 = air' s’ = q ( r ’ T I ’ ) s“ as required. I
+
+
+
Part (iii) above is a special case of the fact [43, Section 2.31 that the n-term weak algorithm is right-left symmetric. Part (iv) is related to the division algorithm.
Theorem 13.11. 1891 Suppose R = Uz=, R, is a filtered ring satisfying the 2-term weak algorithm. Then either Q,(R) = R or R is a generalized polynomial ring in one variable over the division ring &.
Proof. By the previous lemma, we know that R is a domain and hence, by Lemma 10.7(i), Q,(R) is also a domain. We assume that Q,(R) > R and proceed in a series of steps.
Step 1. There exists q E Q,(R) with q-’ = z E R \ Ro.
Proof. Among all elements of Q,(R)\ R choose q so that 0 # rq E R has smallest possible degree for any r E R. Now let s f R with 0 # qs E R and consider the relation (rq)s = r ( s q ) in R. By Lemma 13.10(iv), this yields two possibilities. First if degr 5 degrq, then rq = rt u for some t , u E R with deg u < deg r 5 deg rq. But then q - t E Q,(R)\ R and r(q - t ) = u # 0 has degree smaller than that of rq, a contradiction. Thus we must have degrq < degr and then r = (rq)z v for some z , v E R with degv < degrq. Now r(1- q z ) = and deg v < deg rq, so this implies that 1- qz E R and we conclude from Lemma 13.10(i) that 1-qz = 0 since degv < deg r. Finally, Q,(R) is a domain so q-’ = z E R and, since Ro is a division ring and q 4 R, we have degz 2 1.
+
+
Step 2. Every two element subset of R is right dependent.
Proof: Let al, a2 E R and choose integer n with deg zn > deg ai for i = 1,2. Let 0 # C Q R with qnC C R and choose c E C \ 0. Then zn(qnaic) = aic and degz”
> degai so, by Lemma 13.10(iv),
126
3. The Symmetric Ring of Quotients
zn = aibi +ei for some bi, ei E R with deg ei
Now albl
< deg ai.
Clearly bi
# 0.
+ el = zn = a 2 b 2 + e2
so albl - a 2 b 2 = e2 - el. Since deg(e2 - el) < max { degal, dega2 }, we conclude that { al,a2 } is right dependent. I
Step 3. R is a generalized polynomial ring over the division ring &.
Proof: Let D = & so that D is a division ring. By Step 1, R > & so we can choose x E R to be of minimal positive degree. Let d E D. Then { dx, x} is right dependent, by Step 2, and degdx 5 degx so the 2-term weak algorithm yields
for some dl,d2 E R with degd1,degda SO. Thus dl,d2 E D and, since degx > 0, it follows easily that dl and d2 are uniquely determined by d. Furthermore, the map a: D -+ D given by 0:d H dl is clearly a monomorphism and 6: d H d2 is a a-derivation. In addition, by Lemma 13.lO(iii) and the left analog of Step 2, { x d , x } is left dependent. Thus xd = dix da for suitable di, dh E D so it follows that a is onto and hence is an automorphism of D. It remains to show that R is generated by x and D and we proceed by induction on the degree. Let r E R \ Ro. Then { x,r } is right dependent, by Step 2, and degr 2 degx. Thus, by the 2term weak algorithm, T = 2s t with deg t < degr. Since clearly degs < degr, the result follows by induction. We have therefore shown that every element of R is a polynomial in x with coefficients in D and, since deg x > 0, we conclude that R = D [ x ;a, 61. I
+
+
The ring D[x;a,6]will be considered in more detail in Exercises 5, 6 and 7. The above theorem applies to free algebras because of the following result. For a proof, see [43, Section 2.41 or the graded special case given in Theorem 32.2.
Theorem 13.12. [41]Let R = U,"==,R, be a filtered ring with Ro = K a central subfield. Then R is the free associative K-algebra on a
127
13. Free Rings
right independent generating set if and only if R satisfies the weak algorithm. We close by briefly discussing a related ring theoretic property. Let R be a ring and let n be a positive integer. Then R is said to be an n-fir if all n-generator right ideals of R are free right R-modules of unique rank. A basic property is as follows.
Lemma 13.13. Let R be an n-fir and let I be a right ideal of R which is free of rank Ic < n. Let a, b E R with ab E I . Then either b = 0 or I + a R is free of rank 5 Ic-
Proof: Note that I
+ aR has Ic + 1 5 n generators and hence is free
of unique rank. Suppose this rank is > k. Then R has a right ideal which is free of rank Ic 1. By uniqueness it follows that every free R-module of rank k 1 has unique rank. Let F = I @ R so that F is free of unique rank Ic 1and map F onto I a R via the homomorphism g:i @ r H i - ar. Since I aR is free, the map splits so F E ( I aR) @ L where L = Ker(a). Note that L 2 { r E R I ar E I }, a right ideal of R, and L has at most Ic 1 generators since it is a homomorphic image of F . Thus L is also free of unique rank and
+
+
+
+
+
+
+
k + 1 = rank F = rank ( I + aR) + rank L > k + rank L. Thus rankL = 0 so L = 0 and b = 0. I In particular, by taking I = 0 and Ic = 0, we conclude that any n-fir is a domain. In fact, by Exercise 8, R is a 1-fir if and only if it is a domain. Now let R be any ring. An element b E R is said to be bounded if bR n Rb contains a nonzero two-sided ideal of R. The relevance of this is as follows.
Lemma 13.14. Let R be a prime ring and let b
E
R. Then b-l E
Q,(R) if and only if b is bounded.
Proof: Suppose b-' = q E Q,(R) and choose 0 # I a R with q l , Iq C_ R. Then I C bR n Rb.
128
3. The Symmetric Ring of Quotients
Conversely suppose b is bounded and let I bR n Rb. Since R is prime, it follows that b is a regular element of R. Thus the maps f : RI + RR and g : IR RR given by ( r b ) f = r and g(br) = r are well defined. Furthermore, these maps are balanced and hence determine an element q E Q,(R) which is clearly the inverse of b. I ---$
Now by [43,Theorems 2.2.4 and 2.2.51, if R is a filtered ring satisfying the ( n 1)-term weak algorithm, then R is an n-fir. Furthermore, if R satisfies the weak algorithm, then it is a fir, that is all right ideals of R are free of unique rank. In particular, a free algebra is a fir.
+
[loo] Let R be a 2-fir. If s E Q,(R), then there exists a bounded element b E R with bs E R. In particular, R is symmetrically closed if and only if all bounded elements are units of the ring R.
Theorem 13.15.
PruuJ Note that R is a domain and hence so is Q 3 ( R ) .Let s E Q3(R) and choose 0 # i , z E R with i s , s z E R. Set sz = y and observe that (is)% = iy. Now I = iR is a free right ideal of rank 1 and z # 0. Thus by Lemma 13.13, I + (is)Ris free of rank 5 1 and hence ZR + (is)R= r R for some r E R. Viewing this as an equation in Qs(R),we see that r = iq for some q E Qs(R).Hence since Qsis a domain, we can cancel the i factor and conclude that R+sR = qR. Now 1 E R C qR so 1 = qb for some b E R and, since Qsis a domain, we have q = b-l. Finally s E qR so bs E R and Lemma 13.14 yields the result.
I
It is also shown in [loo] that if R is a 2-fir, then Q,(R) is a symmetrically closed 2-fir.
EXERCISES 1. Let F # (1)be a free semigroup and let R be a prime ring. If R is not cohesive, show that R [ F ]is not symmetrically closed. If R*F
129
13. Free Rings
is symmetrically closed, prove that R*F is cohesive by modifying the argument of Theorem 11.11. In the next three problems we consider an example from [118]. Let K be a field of characteristic not 2, and define the rings R, = K [ z i1 = ai f K ] for i = 1 , 2 so that, depending on the choice of ai, the ring R, is either primary, a quadratic field extension of K or Ri % K @ K . Let R = R 1 u R2 be the coproduct of R1 and R2 as K-algebras. 2. Show that R = K ( z l , z 2 I ( ~ 1 = ) ~ al, ( ~ 2 = ) ~a2) has as a K-basis { yn,z2yn,ynzl,z2ynzlI n = 0 , 1 , . . . } where y = z1x2. Prove that R is prime. 3. Show that a : R -+ R given by a(zi) = -zi defines an automorphism of R. Furthermore, prove that c is not inner by considering the natural homomorphism of R onto the commutative ring
I
K[x1,z2 ( E d 2
= a17
(.212
= a21.
4. Let y = 21x2 - z2z1 E R. Show that ry = yr" for all r E R. Deduce that y is a normal element of R and that c is an X-inner automorphism induced by the unit y of Q,(R). In the next three problems, let S = D [ z ; a , S )be a generalized polynomial ring in the variable z over the division ring D so that S is filtered by the usual degree function.
5. If f(z),g(x) E S with g(x) # 0, show that f(z)= g(z)q(z)+ r ( z ) with degr(z) < degg(z). Deduce that S satisfies the weak algorithm and that every nonzero one-sided ideal of S is generated by a unique monic polynomial of minimal degree. 6. Let 0 # I a S and let f(x) be the monic generator of I as a left ideal. Show that fD = Df and that xf - f z = df for some d E D. Conclude that f is a normal element of S. 7. Let R be a prime ring, like S above, with the property that every nonzero ideal contains a nonzero normal element. Prove that Qe(R) = Q,(R) = Q T ( R )= RN is equal to R localized at the multiplicatively closed set of its nonzero normal elements. Conclude
130
3. The Symmetric Ring of Quotients
that Q,(R) is simple and that R is symmetrically closed if and only if it is simple.
8. Prove that R is a l-fir if and only if it is a domain. One direction has already been noted. 9. Let F be a free semigroup on at least two generators and let R be a domain. Prove that every bounded element of R*F is contained in R. To this end, if q = b-l E Qs(R*F),use the argument of Theorem 13.4 to conclude that qB G R*F and hence that B C b(R*F).
4
Prime Ideals The Finite Case
14. G-Prime Coefficients
In previous sections we discussed when R*G is prime or equivalently when 0 is a prime ideal of R*G. Now we would like to know, more generally, the nature of all prime ideals of crossed products. This is obviously a more difficult problem and success has been limited to the cases where G is a finite group or where G is a polycyclic-by-finite group with R a right Noetherian ring. We must, of course, begin by considering ordinary group algebras. Here, if G is finite, then K[G] is a finite dimensional algebra and we will assume that such rings are reasonably well understood. On the other hand, if G is polycyclic-by-finite, then the determination of the primes of K [ G ]is a major achievement of [185]. Since this is group algebra material which has been adequately expounded in [185] and [168], we will not offer proofs here. However, we will state the necessary results and we will discuss at least one aspect of the proof where crossed products come into play. This will all be considered in later sections.
I31
132
4. Prime Ideals - The Finite Case
In this section and the next, we show how primes in crossed products are determined by primes in certain twisted group algebras. For this we will follow the arguments of [log],[lll]and [172].We begin by recalling some basic definitions and observations. Let R*G be given and, for each I Q Rand x E G, set 1” = 3T-lIZ. Then we know that this yields a permutation action of G on the ideals of R. Furthermore, if I is G-stable we set
By Lemma 1.4, if J a R*G, then J n R is a G-stable ideal of R with J 2 ( J n R)*G. Conversely, if I is a G-stable ideal of R, then i*G a R*G with (I*G) n R = I and R*G/I*G 2 (R/I)*G. Let I a R be G-stable. We say that I is a G-prime ideaE of R if for all G-stable ideals A, B a R, the inclusion AB G I implies A I or B C I . In particular, R is a G-prime ring if and only if I = 0 is a G-prime ideal.
Lemma 14.1. Let R*G be given. i. If P is a prime ideal of R*G, then P n R is a G-prime ideal of R. ii. If I is a G-prime ideal of R, then there exists a prime P of R*G with P n R = I . Proof: (i) Let P be given and let A , B be G-stable ideals of R with A B G P n R. Then (A*G)(B*G) C (Pn R)*G P and, since P is prime, one of A*G or B*G is contained in P. But then, for example, if A*G 2 P we have A = (A*G)n R P n R. (ii) Since (I*G) n R = I , we can apply Zorn’s lemma to find an ideal P of R*G maximal with P n R = I . Suppose A, B are ideals of R*G containing P with AB 2 P. From ( A n R ) ( B n R ) P n R = I and the fact that I is G-prime, we conclude that one of A r l R or B n R is contained in I . By the maximality of P we have A = P or B = P and P is prime.
If P is a prime ideal of the crossed product R*G, then P 2 ( P n R)*G and P/ [(PnR)*G] is a prime ideal of R*G/ [(PnR)*G] 2
133
14. G-Prime Coefficients
( R I P n R)*G. Since the homomorphism R*G -, R*G/ [ ( Pn R)*G] is well understood, we see that to describe P it suffices to replace P and R*G by their images under this map. Equivalently, we can assume that P n R = 0 and hence that R is a G-prime ring by Lemma 14.1(i). In this section, we study the situation of G-prime coefficients rings and show how to reduce the problem to the case of prime coefficient rings. We therefore assume throughout the proof that R*G is given with R a G-prime ring. Furthermore, we suppose that either G is finite or that G is polycyclic-by-finite and R is right Noetherian. In the latter case, R*G is also right Noetherian by Proposition 1.6.
Lemma 14.2. Let Q be a minimal prime of R. i. Q" = 0 so R is semiprime. ii. { Q" 1 x E G } is the finite set of minimal primes of R. iii. Let H denote the stabilizer of Q in G and set N = annR(Q). Then H is a subgroup of G of finite index,
nxEG
N =
n
Q"#O
"6-
and for all x E G \ H 0 = NZN = N
n N"
=N
iv. If A is a nonzero ideal of R with A
n Q.
c N , then annR(A) = Q.
Proof. (i)(ii) Suppose first that G is finite. Then by Zorn's lemma,
nxEG
Q" = 0. The goal is to show we can choose QaR maximal with that Q is prime. To this end, let A , B 2 Q with AB E Q. Then
Since R is G-prime, one of the first two intersections must be zero. But if A" = 0, then the maximality of Q yields A = Q. On the other hand, suppose R is right Noetherian. Then R has finitely many minimal primes, say Q = Q1, Q2,. . . , Qn and
nxEG
4. Prime Ideals - The Finite Case
134
(0; Qi)m = 0.
Clearly G permutes these ideals and we let C = Q" and D equal to the intersection of the remaining primes if any. Then C and D are G-stable and (CD)m= 0 so either C or D is zero. But D = 0 implies Q 2 D contradicting the fact that Q is a minimal prime of R. Thus C = 0 as required. Part (ii) is immediate in either case. (iii) Since R is semiprime, right and left annihilators of twosided ideals are equal. Thus N = annR(&) is unambiguously defined and it is clear from (i) that N 2 nX$HQx. On the other hand, if z 4 H then Q" 2 NQ = 0 and Q" 2 Q so Q" 2 N and we have N = n X g H Q x . It follows from this formula that if z 4 H then Q 2 Nx so N n N" N n Q = 0. Finally 0 = N x N = Z-lNZN yields N3N = 0. (iv) If 0 # A N then certainly Q E annR(N) E annR(A). On the other hand, Q 2 A . annR(A) and Q 2 A so we conclude that Q > a n n ~ ( A ) .I
nxEG
The above notation will be in force for the remainder of this section. Thus Q is a minimal prime of R, N = annR(Q) and H is N" so that the stabilizer of Q in G. In addition we set M = M is a nonzero G-stable ideal of R. Part (ii) of the following lemma is a crucial observation; part (i) is needed for its proof.
xxEG
Lemma 14.3. Let H and N be as above. i. Let V be a nonzero right R-submodule of NG and let T be a finite subset of G. Suppose V n (R*T)# 0 but that V n (R*T') = 0 for all TI c T . Then T is contained in a right coset of H . ii. Let I be an ideal of R*G. Then there exists a nonzero Gstable ideal E of R (depending on I ) with
G N ( I n R*H)G.
EI
Proof. (i) Fix s E T . By assumption there exists 0 # Q = Cat?E V n (R*T). Since V C NG we have at E N for all t E T and the minimality condition on T implies that at # 0 for all t . Now for any q E Q" we have QQ
=
c t
t-1-
atq
t
E
v n (R*T)
135
14. G-Prime Coefftcients
E N Q = 0. Therefore by minimality again, and note that since s E T, we have a&' = 0 for all t so (at)' E a n n ~ ( Q " )= N". Thus 0 # (at)fE N 3 n N t = ( N n Nts-' )" so ts-l E H by Lemma 14.2(iii). (ii) We first show that there exists a nonzero ideal B of R with
and this is clear if N I = 0. Thus we may suppose that V = N I # 0. Note that V satisfies the hypothesis of (i) and that V is a right ideal of R*G. Let 7 denote the family of all finite subsets T of G such that V n (R*T)# 0 but V n (R*T') = 0 for all T' c T . By (i) above, each such T E 7 is contained in a right coset of H . For convenience we choose a canonical element y = y(T) E T for each T E 7 and we let there exists ,O =
b t f E V with r = b, t€T
Since V is an (R,R)-subbimodule of N G , it is clear that each AT is a nonzero ideal of R contained in N . For convenience we also arbitrarily linearly order the elements T E 7. If S is a finite subset of G we can then define Bs = N AT where the product is taken with the T's in 7 in increasing order. Now let S be a finite subset of G with IS1 = m. We show by induction on m that Bs(V nR*S) E N ( I nR*H)G. This is clear for m = 0 so assume that m > 0 and that the result holds for all smaller size subsets. Let 0 # Q E V n R*S. By definition of 7, there exists T C S with T E 7. We assume T is largest possible in the ordering of 7 and we set y = y(T). Let Q = ajj and let d E AT. Then by definition there exists p E V n R*T C V n R*S with p = dy . Since T S it follows that = d a - pug E n (R*s).
nTcs
+ a
a
+
a.
v
But the y-coefficient of y is zero so y E R*S' where S' = S \ {y} is a proper subset of S. By induction, B p y N(In R*H)G. Furthermore since T H y , it is clear that 0 = (,Oy-')y E ( I nR*H)G
136
4. Prime Ideals
- The Finite Case
s
so since Bst N we have Bstpag N ( I n R*H)G. This all implies that Bstda N ( I n R*H)G. Note that y 6 S’ so T S‘. Thus if B’ denotes the obvious product with Bs = P A T , then Bst 2 B‘ so B’da N ( I n R*H)G. But this holds for all d E AT so Bsa N ( I n R*H)G as required. Set E = { T E R I T I CGN(InR*H)G}.
c
s
If G is finite, then 0 # BGN E so E # 0. On the other hand, if R*G is right Noetherian, then N I = aiR*G for suitable a;E V. Note that S = Uy Supp ai is a finite subset of G and 0 # BsN E. Thus again E # 0. Finally with the additional G factor in front, G N ( I n R*H)G is an ideal of R*G so it is clear that E must be a G-stable ideal of R and the result follows. I
zy
c
Definition. We continue with the above assumptions. In addition, if I a R*G we let I I H = { a E R*H I N a C I } and if L Q R*H we set
Since JHand IG are defined on ideals here, this notation cannot be confused with the restriction and induction of modules IR*H and lR*G previously considered (see Exercise 1). Note also that LIG can be defined for any subgroup H of G and that a number of the basic properties we prove apply to this more general situation. Recall that TH:R*G+ R*H is an (R*H,R*H)-bimodule homomorphism.
Lemma 14.4. With the above notation we have i. IIH aR*H and i f I n R = 0 then I I Hn R = Q, ii. LIG is the unique largest two-sided ideal of R*G contained in LG and if L n R = Q then LIG r l R = 0, iii. LIG is the unique largest ideal of R*G with r ~ ( L l ~L ),
137
14. G-Prime Coefficients
iv. 0lH = Q*H and (Q*H)IG = 0 .
Proof: (i) Since I a R*G and N a R, it follows that I I H is a left Rmodule and a right R*H-module. Furthermore, both N and I are H-stable under conjugation so I I H is also H-stable. This along with the above implies that IIHaR*H. Finally if I n R = 0 then T E I p n R if and only if N r C_ I n R = 0 so I p f l R = annR(N) = Q. (ii) If I is an ideal of R*G contained in LG, then since I is n,,,(LG)’ = LIG, On G-invariant under conjugation we have I the other hand, LIG is clearly a right R*G-module, a left R-module and it is G-invariant. Thus LIG a R*G. Finally if L n R = Q , then LIG n R L f l R = Q. But LIG n R is G-stable so LIG n R n X c&”~ = 0. (iii) Let I a R*G. If I LG then clearly ~ F H ( C I ) L. On the other hand, if .rr~(1) C L , then I T H ( I ) G LG. Thus (ii) above yields this result. (iv) The first statement follows since annR,H(N) = Q * H . For the second, set I = (Q*H)IG.Then I G (Q*H)G = Q G so .rr(l)(I)5 Q. But . r r ~ ) ( Iis) a G-stable ideal of R so T ( ~ ) ( = I ) 0 and I ~ ( 1 ) ( 1 ) *= G0. I
c
c
c
c
c
Note that part (iii) of the previous lemma asserts that the definition of LIG is right-left symmetric. This however is not the case for IIH (see Exercise 4). Now it is clear that the maps IG and p are monotone. They also have the following multiplicative properties.
Lemma 14.5. (i) I f Il and 12 are ideals of the crossed product R*G then ( I I ) ~ ~ ( N * H )S( (1112)IH. ~)I~ (ii) I f L1 and L2 are ideals of R*H then L1IGL2IGE (L1L2)IG.
ProoJ (i) By definition N I p C I so N ( I I ) -, N ~ (12)lH G 1 1 1 2 . Thus ( 1 1 ) I H N ( 1 2 )C_I H (1112)1H and the result follows since (I1),Ha R*H. (ii) Since L21G aR*G we have GL21G & L21G and hence
L
~
I
~ E LL ~~ G ~ L J~ ~ L
~
LL~ L ~ G . ~
~
Thus since LllGL21Ga R+G, Lemma 14.4(ii) yields the result. I
4. Prime Ideals - The Finite Case
138
Lemma 14.6. (i) Let L a R*H with L n R 2 Q. Then GNLG E LIG c LG and L c (LIG)IH.Furthermore N ( L I G ) I H C L. (ii) rf I a R*G, then M ( I I * ) I ~ c I. Moreover there exists a nonzero G-stable ideal E of R with E I E ( 1 , ~ ) ' ~ .
Proof: (i) If z z $ H then
E H , then
3NLG
c LG since L is an ideal of R*H. If
ZNLG = N ~ - ' L ~ - % E Q(R*G) c_ LG since N"-' 2 Q C L. Thus GNLG LG and since GNLGaR*G we have GNLG C LIG LG. In particular N L LIG so L E (LIG)IH. In the other direction, N(LIG)'* C LIG LG so this clearly yields N ( L ~E ~L. ) ~ ~ (ii) We have N(I1H)lG C NIIHG IG = I , where the second ' ~both inclusion holds by definition of I I H . Since 1 and ( 1 1 ~ )are G-stable it then follows that M(IlH)IG I. In the other direction, we know by Lemma 14.3(ii) that E I G N ( I nR*H)G for a suitable nonzero G-invariant ideal E of R. Furthermore I n R*H I I Hand I I H a R*H with Ip 2 Q = annR(N). Thus by (i) above
c
c
EI
c
c G N ( I n R+H)G c_ G N ( I ~ ~c) G(+)IG
as required. I We now come to the main result of this section.
Theorem 14.7. [111] [172]Let R*G be given with R a G-prime ring. Assume that either G is finite or G is polycyclic-by-finite and R is right Noetherian. Let Q be a minimal prime of R and let H be its stabilizer in G so that (G : HI < 00. Then the maps P H Pp and L H LIG as described above yield a one-to-one correspondence between the prime ideals P of R*G with P n R = 0 and the primes L of R*H with L n R = Q.
Proof: We start with an observation on a form of cancellation. Let L be a prime ideal of R*H with LnR = Q and suppose E I & LIG where
139
14. GPrime Coefficients
I Q R*G and E is a nonzero G-stable ideal of R. Then E I G LG so by applying the projection map T H we have E T H ( I )G L and hence ( E * H ) T H ( I )2 L. But certainly E*H L since E is G-stable, L n R = Q and 0,Q" = 0. Thus since L is prime we deduce that T H ( I )2 L and hence f 2 LIG by Lemma 14.4(iii). Now let P be a prime ideal of R*G with P n R = 0 and set L = q ~ By. Lemma 14.4(i), L n R = q~ n R = Q. Let us first observe, by Lemma 14.6(ii), that P 2 M(q,)IG = (M*G)(PlH)IG.
c
Thus since P is prime and M*G P , we see that P 2 (Pp)IG= LIG. Next we show that L is prime. Indeed if L1 and L2 are ideals of R*H containing L with L 2 LlL2, then Lemma 14.4(ii) yields
Since P is prime, P 2 LilG for some i and then, by Lemma 14.6(i), since Li f l R 2 L n R = Q , we have L = q~ 2 (LilG),H 2 Li. Thus L is a prime ideal of R*H with L n R = Q. Finally by Lemma 14.6(ii) there exists a nonzero G-stable ideal E of R with E P C ( q ~= ) LIG. Hence by the cancellation property of LIG mentioned above we have P LIG so P = LIG and this half of the correspondence is proved. In the other direction let L be a prime ideal of R*H with Q = L n R and set P = LIG. By Lemma 14.4(ii) we have P n R = 0 and, by Lemma 14.6(i), L E ( 1 5 1 ~=) q~ ~ ~ and L 2 N(LIG)IH= (N*H)+. But L is a prime ideal of R*H and L 2 N*H since N n Q = 0, so the latter yields L 2 PIHand hence L = P I H .Next we show that P is a prime ideal of R*G. To this end, let 11, I2 QR*G with 1 1 1 2 P. Then by Lemma 14.5(i)
and thus since L is prime and L
2 N*H we have L 2 (Ii)lH for some
i. Now applying Lemma 14.6(ii) to Ii we obtain Eli C LIG for some nonzero G-stable ideal E of R. Hence, by the cancellation property for LIG we have Ii C_ LIG = P and P is prime. Since L = P j H , the result follows.
~ ~
4. Prime Ideals - The Finite Case
140
We observe, with the above notation, that there is an obvious one-to-one correspondence between the prime ideals L of R*H with L n R = Q and the prime ideals & of
( R * H ) / ( Q * H )= (R/Q)*H = R*H with i n k = 0. Since f? is prime, we have therefore reduced the study of prime ideals in R*G with R a G-prime ring to those of R*H with fi a prime ring. We will consider the latter situation in the next section. In view of Lemma 14.4(iv) we have
Corollary 14.8. Let R*G be given with R a G-prime ring. Assume that either G is finite or G is polycyclic-by-finite and R is right Noetherian. Let Q be a minimal prime of R and let H be its stabilizer in G so that IG : HI < 00. Then R*G is a prime ring if and only if ( R * H ) / ( Q * H )= (R/Q)*H is prime. We close with a generalization of Theorem 14.7. We begin by proving the transitivity of induction using Lemma 14.4(ii)(iii).
Lemma 14.9. Let R*G be given and let H & A be subgroups of G. If L is an ideal of R*H, then (LIA)lG= LIG. PmoJ Since LIA 5 LA, we have (LIA)lG5 LIAG LG and hence (LIA)lG LIG. Conversely set I = T A ( L I ~so) that I a R*A. Then 7 r ~ ( I=) T H ( L ~E ~L )so I C LIA and LIG E (LIA)lG. I Corollary 14.10. [114] Let R*G be given and assume that either G is finite or G is polycyclic-by-finite and R is right Noetherian. Let Q be a prime o f R with finitely many G-conjugates and let H be its stabilizer in G. Suppose A is a subgroup of G containing H and set J = n,@A
Q".
i. The map 1; H LIG yields a one-to-one correspondence between the prime ideals L of R*A with L i l R = n a E A Q" and the primes P of R*G with P n R = n x E G Q x . ii. If P = LIG with P and L as above, then J L P n (&A) C_ L and L is the unique minimal covering prime of P n (R*A) not containing J .
141
14. G-Prime Coefficients
Proof: (i) By Theorem 14.7 applied to R*G, the map T
TIG yields a one-to-one correspondence between the primes T of R*H with T n R = Q and the primes P of R*G with P f l R = n z c G Q". Similarly, by Theorem 14.7 applied to R*A, the map T H TIA yields a one-to-one correspondence between the primes T as above and the Q". Since LIG = (TIA)lG= primes L of R*A with L n R = TIG = P , by transitivity, this fact is proved. (ii) If P = LIG, then P 2 LG so, by freeness, P n (R*A) E L. Furthermore n a E A Q" E L so J L' if x 4 A. Since P = LIG = LzG, it follows that J L Pn(R*A). Finally if L' is a minimal covering prime of P n ( R t A ) with L' n R 2 J , then L' 2 J L yields L' = L as required. 8 ++
s
nzEG
EXERCISES 1. Let R*G be given and let H be a subgroup of G. Suppose V is a right R*H-module and L = annR,H(V). Show that annR+G(VIR*G)= LIG. This shows that induced ideals and induced modules are intimately related. In particular, deduce that LIG = ann~,c([R*H/L]lR*G) .
2. Suppose R*G is given, H a G and L a R*H. Prove that (nz,G L') * (R*G). 3. L e t R = K ( z l , z 2 , y , , y 2 l x i y j = O = y j x i f o r a l l i , j ) a n d l e t G = (a),a group of order 2, act on R by (zi)"= yi and (yi)" = xi. Form the skew group ring RG and let Q = (21, x2) = RzlR+Rx2Ra R. In the notation of this section, show that Q is a minimal prime of R, N = Q" = annR(Q) = (y1,y2) and H = (1). 4. Continuing with the example of the previous problem, let I = (21x2, ( ~ 2 )y1y2, ~ , ( ~ 2 ) so ~ )that 1 is a G-stable ideal of R and I G a RG. Prove that
LIG
=
y2 E (IG)IH = {CX E RH I NCXG I G }
but that Y2
f { P E R H I PN E I G } .
This shows that the definition of
IH is
not right-left symmetric.
4. Prime Ideals - The Finite Case
142
5 . Let J be a right ideal of R*G and let H be a subgroup of G. Show that ( J n R*H)G E J C_ T H ( J ) G . Conclude that ( J n R*H)G = J if and only if T H ( J )G J . 6. With the above notation, prove that there exists a unique minimal subgroup W of G (depending on J ) such that (JnR*W)G= J . W is called the controller of J .
15. Prime Coefficients We continue with the task of describing prime ideals in crossed products. Specifically we study primes P in R*G with P n R = 0 and now we assume that the coefficient ring R is prime. Furthermore we will assume, at some point, that either G is finite or that G is polycyclic-by-finite and R is right Noetherian. In the latter case, of course, R*G is right Noetherian. We will follow the approach of [172]which is a modification of the finite arguments in [log]. We note that both of these proofs ultimately depend on ideas in [58] and require that R be localized to its symmetric Martindale ring of quotients Qs(R). We begin with two technical lemmas. Here we view S 8 T as the algebra freely generated by its commuting subalgebras S and 7'.
Lemma 15.1. Let C be a field, let S and T be C-algebras and let I' = ( S 8~T)*H be a crossed product. Assume that H normalizes both S and T and let I be an H-stable ideal of T . Then rim = ( s8 T ) * H / ( s8 I)*H E (S' 8 T')*H = where S' 2 S and T' = T / I . Furthermore Cp(S') = C,(S)', where the latter is the image of Cr(S).
Proof: Since I a T is H-stable, it is clear that S B I a S 8 T is H-stable. Thus IF = ( S 8 I)*H a I' and it follows from Lemma 1.4(ii) that
rim = ( sB T ) * H / ( s
I)*H = (5'8T / S 8 I ) * H .
143
15. Prime Coefficients
But ( S @ T ) / ( S @ IS) S @ ( T / I )so we have obtained the appropriate structure of r’ = I’/II’. It remains to consider the centralizers and certainly Cr(S)’ Cp(S’),that is the centralizer of S in r maps into the centralizer of S’ in I?’. For the other direction, let { t l ,t 2 , . . . } be a C-basis for a complement for I in T and let a‘ E Cp(5”). Then we may assume that a , an inverse image of a’, is of the form
with az.x E S. Let s E S. Since H normalizes S we have
sa - a s =
c[
(SUZ,,
-U
i , , P )
€4 ti]3.
i,X
On the other hand, sa - as E ( S €4 I ) * H since a’ E Cp(S’). By definition of { ti } it therefore follows that sa - a s = 0. Since this is true for all s E S we have a E Cr(S) and hence a’ E Cr(S)’. I
Lemma 15.2. Let
r
= (S € 4 T~ ) * H be as in the previous lemma. In addition assume that R is a prime ring, S = Q,(R) and that H normalizes R, S and T. If I is a nonzero (R, R)-subbimodule of r
with 1 H - l = I , then there exists 0 # a E Cr(S) and 0 # A a R with A a C_ I .
PmoJ Let { t o , t l , . . . } be a C-basis for T . Then every element p E I? is uniquely writable as
with bi,z E S. Choose 0 # P E I with a minimal number, say n, of nonzero coefficients b ~ , We ~ . may suppose b ~ # ,0. ~Furthermore, by multiplying P on the left by an appropriate element of R, we may assume that b ~E ,R. ~ Let T E R. Since I is an (R, R)-subbimodule of r we have 7 = bO,yTP - P T q b O , y ) y E I ,
4. Prime Ideals - The Finite Case
144
Furthermore
Y=
c[
(bo,yrbi,z - bi,xrgz-l (bo,y)y*-l) 63 t i ] Z
i.X
and the (0,y)-term here is zero. Thus the minimality of n implies that y = 0 and hence that
for all r E R. It follows from Lemma 12.1 that there exists a unit qi,z E S with 1 ---l-
2s- ryx
and
bi,,
-r
yz-1
- -1 - Pi,xTQi,z
= b O , y q i , z for all i , x which occur in the support of
0.
Set i,X
Since q i , x Z y - l centralizes R and normalizes S , it follows from the uniqueness part of Lemma 10.9(ii) that q i , z l c y - l centralizes S and hence that Q E Cr(S). Furthermore since b+ = b O , y q i , x we have
bo,ya=
C(bZ,, ti)& @J
= py-l E I
i,x
using 1R-l = I. Finally since 0 # b ~ E ,R,~I is an ( R ,R)-bimodule and a commutes with R we have
and the lemma is proved. We fix some notation for the remainder of this section. Let I? = R*G be a crossed product with R a prime ring. By Proposition 12.4(i), if S = Q s ( R ) ,then I’ extends uniquely to a crossed product I” = S*G. Set C = Z(S) = C s ( R )so that C , the extended centroid of R, is a field. Recall that an automorphism u of R is
145
15. Prime Coefficients
said to be X-inner if its unique extension to S becomes inner. By Lemma 12.3(iii) Ginn = { z E G
I ' is X-inner on R )
is a normal subgroup of G . The next lemma is essentially all notation.
Lemma 15.3. If T = Cp(S) then T = Ct[Ginn], some twisted group algebra of Ginn over C , and G acts on T normalizing its group of trivial units. Furthermore, S*Ginn = S @ T and r' = (S*Gi,n)*H = ( S @c T ) * H , a suitable crossed product of H = G/Ginn over S @ T .
Proof:Most of this follows from Proposition 12.4 and Lemma 1.3. All that remains is to observe that G acts on T . For this we note that G, the group of trivial units of RIG, normalizes S and hence acts by conjugation on T = Cp (S). Furthermore U = U(R) centralizes T so we obtain an action of G + G/U on 7'. It is clear that G normalizes the group of trivial units of T . I For the remainder of this section we write T = Cp(S). In addition, if I is a (T',r)-subbimodule of I?' = S*G we set
I = { a E T 1 Aa C I
for some 0 # A a R } .
Lemma 15.4. With the above notation, Furthermore if P E
7 is a G-stable ideal of T .
k'then there exists 0 # B a R with BP 5 I .
+
with A a , BP 5 I , then ( A fl B ) ( a P) C I so a + p E I. Now let y G T and choose 0 # C a R with C-y C_ r. Since Q and y centralize R we have
Proof: If a , P E
and
C A ? ~= ( c T ) ( A ~ )E ri = I so a?, y a E j . Thus I a T . Furthermore since I and R are G-stable, so is f.
-
4. Prime Ideals The Finite Case
146
Finally if p E ?I” then p = ~ ~ with a aj E f ~ and --yi ~ E r’. i Choose 0 # A , CaR with Aai C I and Cyi C I? for all i = 1,2,.. . ,n. Then
and the lemma is proved.
a
The following result is crucial. It applies when I a r and I a r‘.
Lemma 15.5. If I is a (I?, I?)-subbimodule of I?’, then ?I” a I” and I gP.
ProoJ: By Lemma 15.3, I?’ = (S & T)*H where H = G/Gjnn and, choosing E G, we see that R, S and T are H-invariant. Furthermore ? is an n-stable ideal of T . By Lemma 15.1, fl?’ a I“ and
Moreover T = Cp(S) maps onto the centralizer of S in I?”. fr”. Then the image I“ of I in rr’ is a nonzero Suppose I (R,R)-subbimodule of r’‘with I”H-’ = I” and Lemma 15.2 implies that there exists 0 # a” E C y ( S ) with Aa” 5 I“ for some 0 # AQR. As we observed, a” is the image of some element a E T and certainly A a C I ?I?’. Choose a E A with a # 0 and write a a = --y P with --y E I and P E f i t . Then, by Lemma 15.4, there exists 0 # B a R with BP C I. Since B y is certainly contained in I , we have Baa I and hence, since a centralizes R,we obtain (BaR)a = (Baa)R 5 I R = 1. But 0 # BaRaR implies that (Y E f, by definition of ?, and therefore the image a” of a is zero, a contradiction. We conclude that I 5 fr’. a
+
+
Recall that G acts on T . If Q # T is a G-stable ideal of T , then Q is G-prime if for all G-stable ideals I,J of T the inclusion I J C Q Q. We can now set up a correspondence implies I 5 Q or J between the primes of r = R*G disjoint from R and the G-prime ideals of T. With little additional work we can add to this link the primes of I” = S*G disjoint from S.
147
15. Prime Coefficients
Lemma 15.6. Let Q be a G-stable ideal of T and set P = &I” n r and PI = Q P . Then P a r and PIaI’l with P n R = PI n S = 0 and P = = Q. Furthermore if Q is G-prime then P is a prime ideal of I? and P’ is a prime ideal of I?‘.
Proof: We apply Lemma 15.3 and write I?’ = ( S @c T)*H where H = G/Gi,,. Since Q is a G-stable ideal of T , it follows that S 8 Q is an ideal of S @ T which is H-stable. Thus by Lemma 1.4 we have
) S@Q. and P’n(S@T)= S@Q. Thus P = P’flr’aI?and P f l ( S @ T 2 This implies that PI n S = P n R = 0. Let a E p or Then by definition there exists 0 # A a R with
F.
and hence QI E Q. Conversely if p E Q then there exists 0 with BP E I’ and thus
BP = ,LIB 2 Qr‘flr= P
# BaR
PI.
We conclude therefore that P = = Q. Finally assume that Q is G-prime. We show that P is prime, the proof for PI being identical. Let I , J a R with I J 2 P and let a E j , P E j. If 0 # A , B a R with Aa I and B,LI J then
ABaP = (Aa)(BP)E I J 2 P so ap E = Q. In other words, ij 2 Q. But 1 and j are G-stable ideals of T , by Lemma 15.4, and Q is G-prime. It follows that one of or J is contained in Q , say C Q. We conclude from Lemma 15.5 that I r“rln r 2 Qr’ n r = P
r
and the lemma is proved. The last step requires the full assumptions on G. If G is finite, we will be able to restrict our attention to the G-stable ideals of R.
4. Prime Ideals - The Finite Case
148
If G is infinite, the ideals of R cannot be taken to be G-stable, so we compensate by using the Noetherian hypothesis.
Lemma 15.7. Assume in addition that either G is finite or that G is polycyclic-by-finite and R is right Noetherian. Let P be aprime ideal of with P nR = 0 or let P’ be a prime ideal of r‘ with P‘ r l S = 0. Then P and are G-prime ideals of T with P = Pr‘ f l r and PI= m i . ProoJ It is convenient to proceed in a series of steps. Step 1. P = Prl n r
ProoJ Let W = PI‘’r l I‘so that W a r and W 2 P by Lemma 15.5. Suppose first that G is finite. If a E W then by Lemma 15.4 there B” we see that A exists 0 # B 4 R with Ba C P . Letting A = is a nonzero G-stable ideal of R with Aa P. Thus (A*G)a E P. Finally P is a prime ideal and P f l R = 0 so A*G g P and hence, for all a E W , we have a E P .
nzEG
Now assume instead that is right Noetherian and write W = Since ai E W E PI”, it follows from Lemma 15.4 that there exist 0 # Bi q R with Biai P. Letting A = fly Bi we have 0 # A u R with AW C P. Finally P is prime and P n R = 0 so A g P and hence W P as required. I
CyaJ.
Step 2. PI = Frl.
Proof: Set W’ = Fr’ so that W’ q r’ and W’ 2 P’. If either G is finite or polycyclic-by-finite, then T is right Noetherian. Thus is a finitely generated right ideal of T and hence W’ is a finitely generated right ideal of I”. We can now proceed as in the previous paragraph. In other words, if W’ = Xy air’ then there exists 0 # B a R with BW’ C_ PI. Since P’ is a prime ideal of r’ and B P’ we conclude that W’ C P’. I
Step 3. P and
are G-prime ideals of T .
149
15. Prime Coefficients
Proof. Since the proofs are identical in the two cases, we wilI only consider p . Let I and J be G-stable ideals of T with I J c p. Since IF’ and JI” are ideals of I”, by Lemma 15.6, we have F’J Jr’ and hence (Ir’)(Jr’)c IJr’ 2 p!?. It follows that
and thus, since P is prime, one of these factors is in P , say Ir’nI‘ P. By Lemma 15.6, this yields I = (Ir’n I?)- E p so p is G-prime. Similarly I;; is G-prime. This completes the proof. I The above two lemmas now combine to form the main theorem of this section. At this point, it is essentially all notation and requires no additional proof.
r = R+G be a crossed product with R prime. Assume that either G is finite or that G is polycyclic-byfinite and R is right Noetherian. Let S = Qs(R), C = Z(S) and let I” = S*G be the natural extension of r. Then T = Cp(S) = Ct[Ginn] is a twisted group algebra of the group Ginn a G and G acts on T as automorphisms normalizing the group of trivial units. Furthermore there exist one-to-one order preserving correspondences between the primes P of r with P n R = 0, the primes PI of I” with P‘ n S = 0, and the G-prime ideals Q of T . Specifically these maps are given by
Theorem 15.8. [log]1172) Let
Q
H Qr‘ = P’,
P’
H
P’nr = P,
P
H
P =Q
where
p = { Q E T I there exists 0 # A a R
with A a
P }.
We remark that the assumption above that G is finite can be replaced by the weaker hypothesis that any nonzero ideal of R contains a nonzero G-stable ideal (see Exercise 1).
Corollary 15.9. With the above notation, the following are equivalent .
150
4. Prime Ideals
- The Finite Case
i. R+G is prime. ii. S*G is prime. iii. Ct[Ginn]is G-prime.
ProoJ In view of the preceding theorem, it suffices to show that 6 = 0 and this is immediate from Lemma 15.6 with Q = 0. 1 Finally we remark that the G-prime ideals of T are closely related to the primes of that ring. Indeed, since T is Noetherian, it follows from Lemma 14.2(i)(ii) with G acting on T / Q that
nzEG
Lemma 15.10. Q is a G-prime ideal of T if and only if Q = Qx where { Qx I x E G } is the finite set of minimal covering primes of Q. In particular, there is a one-to-one correspondence between the G-prime ideals of T and the finite G-orbits of primes of T .
EXERCISES 1. Show that the assumption of Theorem 15.8 that G is finite can be replaced by the weaker hypothesis that any nonzero ideal of R contains a nonzero G-stable ideal. This requires taking a closer look at Lemma 15.7, proving Step 2 along the lines of Step 1. In the remaining three problems let R be a centrally closed prime ring with extended centroid C = Z(R)and let T be any C-algebra. We obtain some results of [116]. 2. If I is any nonzero ideal of R @JcT , prove that there exist 0 # P E R and 0 # t E T with P 8 t E I . Conclude that if I is prime, then either I n R # 0 or I f l T # 0. 3. Suppose S is a prime C-algebra generated by the commuting subalgebras R and T . Prove that S R a c T by mapping the tensor product onto S. 4. Show that the prime ideals P of R 8~ T with P f l R = 0 are all of the form R 8 Q with Q a prime ideal of T .
16. Finite Groups and Incomparability
151
16. Finite Groups and Incomparability Let R C S be a finite integral extension of commutative domains. In algebraic number theory one studies, among other things, the relationship between the prime ideals of these two rings, obtaining in particular (see [34,95]) the classical properties known as Lying Over,Going Up, Going Down and Incomparability. In noncommutative ring theory, we are also interested in the relationships between the prime ideals in certain finite extensions R C S. Here there are various candidates for such extensions. For example we have: (i) crossed products S = R*G with G finite; (ii) groupgraded rings S = R(G) with G finite; (iii) fixed rings R = SG where G is a finite group of automorphisms of S; (iv) finite centralizing, normalizing and intermediate extensions; (v) triangular or finite subnormalizing extensions; and (vi) extensions analogous to (i) and (iii) determined by finite dimensional restricted Lie algebras or more generally by finite dimensional Hopf algebras. In this section, we will consider crossed products in detail and we will briefly discuss the situation for finite normalizing extensions. In later sections we will show how the theorems obtained here translate to analogous results on group-graded rings and on fixed rings. We start with crossed products where the main tools are of course the one-to-one correspondences given in the preceding two sections. These ultimately relate the prime ideals of R*G to those of a finite dimensional twisted group algebra. The following lemma lists the basic multiplicative and intersection properties of these correspondences. It uses the specialized notation of the previous two sections which we do not repeat.
Lemma 16.1. Let R*G be a crossed product satisfying the hypotheses of Section 14 or 15. i. In the notation of Section 14, if I , J Q R*H then IIGJIG C (IJ)IG and IIG n JIG = ( I n J)IG. ii. In the notation of Section 15, if I , J are G-stable ideals of T = Ct[Ginn]then we have (Ir’ n I?)( Jr’ n I?) C ( I n J)r” n r and (rrl
n r) n (Jr‘n r) = ( ~ J)r’ n n r.
4. Prime Ideals - The Finite Case
152
Proofi (i) The first part follows from Lemma 14.5(ii) and since IG is order preserving we have ( I n J)IG C_ IIG n JIG. For the other direction, note that, by Lemma 14.4(iii), IIG nJIG is an ideal of R*G I n J so IIG n JIG C_ ( I n J ) ' ~ . with T H ( P n J I G ) (ii) The first part was noted in the proof of Lemma 15.7, Step 3. For the second, since we have I" = (S*Ginn)*(G/Ginn) and S*Ginn = S @C Ct[Ginn],it follows that I" is a free right and left Ct[Ginn]module. Thus the map I H II" preserves intersections. I The next result is clearly an analog of the Wedderburn theorem.
Theorem 16.2. [lo91 Let R*G be a crossed product with G finite and with R a G-prime ring. i. A prime ideal P of R*G is minimal if and only if P n R = 0. ii. There are finitely many minimal primes, say P I ,P2,. . . ,Pn, and in fact n 5 IGl. iii. N = PI nP2 f l - nPn is the unique maximal nilpotent ideal of R*G and NIGl = 0. iv. If Q is a minimal prime ideal of R, then { Q" I z E G } is the set of all minimal primes of R and n x E G Q" = 0.
Proof. Part (iv) is just Lemma 14.2(i)(ii). We show below that R*G has n 5 [GI primes Pi with Pi n R = 0. Furthermore, these are incomparable and satisfy Pi = N with NIGl = 0. Once this is obtained, the result follows immediately. Indeed if P is any prime ideal of R*G, then P contains the nilpotent ideal N , so P contains some Pi. Thus since the Pi are incomparable, (i) and (ii) are proved.
ny
Finally (iii) follows since every nilpotent ideal is contained in each Pi. There are two cases to consider.
Case 1. R is prime.
Proofi We use the notation and conclusion of Theorem 15.8. Thus T = Ct[Ginn] is a finite dimensional C-algebra and hence has at most d i m c T = lGinnI prime ideals all of which are minimal. It therefore follows from Lemma 15.10 that T has finitely many Gprime ideals Q1, Q2,. . ., Qn with n 5 IGinnI. Furthermore, these
153
16. Finite Groups and Incomparability
ny
ideals are incomparable and Qi = J is the Jacobson radical of T . Thus JIGinnI = 0. Set Pi = r’Qi n and N = r‘J n I?. Then it follows from Theorem 15.8 that PI, P2,. . . ,P, are precisely the prime ideals of I’ = R*G with Pi f l R = 0. Since the Qi’s are incomparable, so are the Pi’s. Furthermore, by Lemma 16.1(ii), we have Pi = N and NlGinnl = 0. I
n:
Case 2. R is G-prime.
ProoJ Here we use the notation and conclusion of Theorem 14.7. By the prime case, R*H/Q*H = ( R / Q ) * H has n 5 IN(incomparable prime ideals El, L 2 , . . . ,L, with Ei n (R/Q) = 0. Furthermore nyii = J is nilpotent with jlHl = 0. Lifting these to ideals of R*H, we see that R*H has n 5 (HI incomparable prime ideals L1,L a , ... ,L, with Lif l R = Q. firthermore 0; Li = J and JIHlE Q*H. Set Pi = and N = JIG. Then Theorem 14.7 implies that P I ,Pz,, . . ,P, are precisely the prime ideals of R*G with Pi nR = 0. Since the La’s are incomparable, so are the Pi’s. Finally NIHl C_ Q*H and (Q*H)IG = 0 by Lemma 14.4(iv). Thus by Lemma 16.l(i) we have 0; Pi = N and NIHl = 0. Since (HI 5 (GI, the result clearly follows. I There is obviously a good deal of additional information contained in the above. Aside from a precise description of the primes, there are sharper bounds on the number of primes in certain special cases. To start with, we have
Lemma 16.3. Let Kt[G] be a twisted group algebra with G a finite p-group and with K a field of characteristic p . i. Kt[G]/J(Kt[G]) is a purely inseparable field extension of K of finite degree. ii. Kt[G] is commutative if and only if G is abelian. iii. If J ( K t [ G ] )# 0 then there exists a central elementary abelian subgroup H of G with J(Kt[H]) # 0.
154
4. Prime Ideals - The Finite Case
Proof: (i) Let I? denote the algebraic closure of K and embed Kt[G] in &[GI = I? @ K Kt[G]. Then by [161, Lemma 1.2.101, &[GI is isomorphic to the ordinary group algebra I?[G].Furthermore, by [161, Lemma 3.1.61, J = J ( k [ G ] is ) a nilpotent ideal with I?[G]/J= K and, via the isomorphism, the same is true of I?‘“[G].Thus if P = J ( k t [ G ]fl) Kt[G],then P is a nilpotent ideal of Kt[G]with K t [ G ] / Pa K-subalgebra of k.It follows that K t [ G ] / Pis a finite dimensional field extension of K which is purely inseparable since it is generated by the images of the elements 3 with x E G and these satisfy 3”E K for some n. Obviously P = J(Kt[G]). (ii) It is clear that Kt[G]is a commutative ring if and only if I? @K Kt(G] E K[G]is commutative and the latter occurs if and only if G is abelian. (iii) Suppose first that G is nonabelian. By (i), it follows that the group of trivial units of Kt[G’]maps to K in K t [ G ] / J ( K t [ G ]Thus ). if W is any subgroup of G‘, then J(Kt[G])f l K t [ W ]is a nilpotent ideal of codimension 1 in K t [ W ] .Now take W to be a subgroup of order p in G’ n Z(G) # (1). On the other hand, suppose G is abelian so Kt[G]is commutative by (ii). Let H = { x E G I x p = 1} so that H is a central elementary abelian subgroup of G. Since J ( K t [ G ]# ) 0 is nilpotent, we can choose a E J ( K t [ G ]with ) a # 0 but a p = 0. Furthermore, by commutativity, we can assume that 1 E Supp a and say a = C, k,Z with k, E K . Then by commutativity again 0 = a p = C,(k,)”3 and hence if p = r ~ ( athen ) p # 0 and P P = 0. Thus P generates a nonzero nilpotent ideal of K t [ H ] . In view of the proof of Theorem 16.2, the preceding lemma yields
Proposition 16.4. Let R*G be given with G a finite p-group and with R a G-prime ring of characteristic p . Then R*G has a unique minimal prime P which is necessarily nilpotent. Now it turns out that Theorem 16.2 contains within it all the basic relations between the prime ideals of R and of R*G except for one aspect of Going Up. The original proof of that latter fact in [110]
16. Finite Groups and Incomparability
155
used integrality methods. However here we use a simpler argument based on the following lemma of [23].
Lemma 16.5. Let R
S be an extension of rings and assume that S=CpIXa with Rxi = x i R for all i. Let Q be an ideal of R which intersects nontrivially all nonzero ideals of R. Then there exists I a S with 0 # I n R Q. C_
Proo$ In this proof we consider the (R,R)-subbimodules of S so the hypothesis asserts that Q ess R. Let M be a maximal complement for R in S so that R @ M ess S. It follows that L = Q @ A4 ess S. For each i, j set Li,j = { s E S 1 xiszj E L }. It is then easy to see that Li,j is an ( R ,R)-subbimodule of S. Furthermore, Li,j ess S. Indeed, let U be a nonzero ( R ,R)-subbimodule of S. If xiUxj = 0, then U Li,j and Li,j rl U # 0. On the other hand, if xiUxj # 0, then ziUxj is a nonzero (R, R)-bimodule and L ess S so z&YzjflL # 0 and again Li,jf l U # 0. It now follows that I = Li,j is essential in S. Since L is an (R,R)-bimodule and S = C ziR = C Rxi,it is clear that I = { s E S I SsS C_ L}. Thus I is clearly the largest twoR@M, sided ideal of S contained in L. Next, since I C_ L = &@A4 we see that I r l R C_ Q. Finally since I ess S, we have I r l R # 0. I
ni,,
s
We can now list the Kru21 rehtions in R*G and it is convenient to do so diagrammatically. Thus for example the diagram in (iii) below indicates that PI 2 P2 are primes of R*G, Q1 1 Q2 are primes of R and that Pi lies over Qi. Furthermore, one reads the first diagram of (iv) as follows. Given Q1 2 Q2 primes of R and a prime Pz of R*G which lies over Q 2 , there exists a prime P1 of R*G such that PI 2 P2 and PI lies over Q1. The other three charts have similar interpretations.
Theorem 16.6. [lo91 Ill01 Let R*G be a crossed product with G finite. The following basic relations hold between the prime ideals of R and of R*G. Here PI and P2 denote primes of R*G while Q1 and Qz are primes of R.
4. Prime Ideals - The Finite Case
156
i. Cutting Down. If P is a prime ideal of R*G, then there exists a prime ideal Q of R, unique up to G-conjugation, with Q minimal over P n R. Indeed, we have P n R = n s E G Q". When this occurs, we say that P lies over Q . ii. Lying Over. If Q is a prime ideal of R, then there are primes Pl,P2, ...,P, ofR*G with 15 n 5 /GI such that Pi lies over Q . iii. Incomparability. If
and PI # Pz,then Q 1 # Q z . iv. Going Up. With the above notation we have
pi
Pl p2
Qi
I
/
:I
p2
Qi
:
.
/
Q2
Q2
v. Going Down. Similarly we have
Pl
/ p2
Qi
I Q2
Qi
I p2
.
*:
Q2
ProuJ (i)(ii)(iii) These follow from Theorem 16.2 once we mod out by ( P n R)*G, Q"]*G and [ n z E G ( Q 2 ) z ] *respectively. G (iv) In the first part, note that P2 n R G Q2 2 QI and thus we can choose PI a R*G maximal subject to PI 2 P2 and Pi n R E Q1. Since &1 is a prime ideal of R, it follows easily that PI is a prime of R*G. The goal is to show that &I is minimal over Pi n R. To this end, let 3 = S/P1and let R = R / ( R n Pi). Then h E S satisfies
[n,
157
16. Finite Groups and Incomparability
the first hypothesis of the preceding lemma by taking { xi } to be the image of G. Furthermore, by definition of Pl, it is clear that i f f is any nonzero ideal of S, then i n R Q1 = Q 1 / ( Rn P I ) . Thus by Lemma 16.5, Q1 is not an essential ideal of fi, say Q1 n J' = 0 with J' # 0. Finally, by (i) above, fi is semiprime and if 2, is a minimal prime of not containing j,then 2 0 = QlJ' implies that 2, 2 41. Thus equality occurs and Q1 is a minimal prime of fi as required. For the second part, let PI lie over &I. Then
2 ( Q 2 ) x for some z. Now replace Q1 by ( & 1 ) " - l . (v) In the first part, let L1, L z , . . . ,L, be the primes lying over Q2. Then by Theorem 16.2, the intersection L1 n L2 n ... n L, is nilpotent modulo [nxEG(Q2)x]*G PI. It follows that PI contains some Li. For the second part, we merely note that Q1 PI n R P2 n R. Thus Q1 contains a minimal covering prime of P2 n R. I
so
Q1
>
>
In particular we have obtained a one-to-one correspondence between certain finite subsets of prime ideals of R*G and the G-orbits of primes of R. It is natural now to see which ring theoretic properties are inherited via this correspondence. One such is primitivity.
Proposition 16.7. Let P be a prime ideal of R*G which lies over Q. Then P is primitive if and only if Q is primitive.
n R = nxEGQx= 0. Suppose first that Q is a primitive ideal of R so that Q = annR(W) for some irreducible R-module W . Let V be the induced module V = WIR*G= CxEG W 8 3. If I = annR*G(V), then it is easy to see that n ( l ) ( I ) annR(W 8 1) = Q. Thus since 7r(1)(I)is a G-stable ideal we have .rr(l)(I)= 0 and hence I = 0. Now, by Lemma 3.3, each W 8 3 is an irreducible R-module so V has finite length as an R-module and hence as an R*G-module. If L1, Lz, . . . , Lt are the annihilators of the composition factors of V, then these are primitive ideals of R*G with L1L2 . .Lt I = 0. Since P 2 0 = L1L2. . .Lt
Proof: We may assume that P
4. Prime Ideals - The Finite Case
158
we conclude that P _> Li for some i and the result follows from the minimality of P . In the other direction, let P be primitive with P = annR,G(V’) for some irreducible R*G-module V’. Since P n R = 0, R acts faithfully on V‘ and, by Proposition 4.10, V’p is completely reducible. Furthermore, since V is cyclic and G is finite, V ’ p is finitely generated and hence has finite length. By taking the annihilators of the composition factors of V’p, there exist primitive ideals L i , Lh, . . . ,L‘, of R with L‘, n La n - n L‘, = 0. Since Q is a minimal prime of R, we conclude that Q = LI for some j. I Now it is obvious that Going Up, Going Down and Incomparability allow us to compare the prime lengths of R and of R*G. Recall that the prime length of R is the maximal n such that R has a chain of primes Qo C Q1 C c Qn of length n. Of course if no such maximum exists then the prime length is infinite. Similarly, the primitive Zength of R is defined to be the maximal n with all Qi primitive. If Q is a prime of R, then the height of Q is the maximal n such that there exists a chain of primes as above with Q = Qn.Similarly, the depth of Q is the maximal length of chains of primes with Q = Qo. The following is an immediate consequence of Theorem 16.6 and Proposition 16.7.
Corollary 16.8. [lo91 [llO]Let R*G be a crossed product with G finite. Then the prime (or primitive) length of R is equal to the corresponding length of R*G. Furthermore if P is a prime of R*G which lies over Q , then P and Q have the same height and the same depth. We close this section by briefly considering finite normalizing extensions.
Definition. Let R
S be an extension of rings with the same 1. We say that S is a finite centralizing extension or a liberal extension of R if S = Rxi and each xi centralizes R. For example we could have S = R[G]with G finite, or S = M,(R), or S = F @K R where R is a K-algebra and F is a finite extension field of K .
xy
16. Finite Groups and Incomparability
159
More generally we say that S is a finite normalizing extension of the ring R if S = Rxi with Rxi = x i R for all a. For example we could have S = R*G with G finite since 2R = R% for all x E G. Unlike crossed products, these extensions are closed under homomorphic images. In other words, if I a S and if S 2 R is a finite normalizing (or centralizing) extension, then the same is true of SII 2 R / ( R n I ) .Note that the beginning hypothesis of Lemma 16.4 asserts precisely that S is a finite normalizing extension of R. Finally one also considers intermediate extensions. Here S is a finite normalizing extension of R and T is an intermediate ring. For example we could take R E S = M,(R) with T the ring of upper triangular matrices. One can of course study either extension R T or T C S.
xy
We will restrict our attention here to normalizing extensions; in particular, we will state the appropriate analog of Theorem 16.5. This result is the product of a series of papers. To start with, [23]and [lo21contain Cutting Down, Lying Over and Going Up. Some of this is also proved in [164] in a rather nonstandard manner. However, by far the most difficult relation to prove is Incomparability. Here a special case appears in [102];the general result is in [72].We state the following without proof and, since most of the difficulty concerns Incomparability, we credit it to [72].A proof and many additional details can now be found in the book [121].
xy
Theorem 16.9. [72]Let S = Rxi be a finite normalizing extension of the ring R with x i R = &i for all i. i. Cutting Down. If P is a prime ideal of S, then P f l R has only finitely many minimal covering primes. In fact, if these are Q1, Q 2 , . . . ,Qt then t 5 n, Qi = P n R and R / Q i si R / Q 1 for all i. We say that P lies over each Qi. ii. Lying Over. If Q is a prime ideal of R, then there are finitely many primes PI,P 2 , . . . ,P, of S such that Pi lies over Q. Here 1 5 s 5 n. iii. Incomparability. Let P c I be ideals of S with P prime. Then P f l R C I f l R; indeed no prime ideal of R minimal over P fl R contains I n R.
ni
4. Prime Ideals - The Finite Case
160
iv. Going Up and Going Down. If P I ,P2 are primes of S and if Q 1 ,Q2 are primes of R, then we have at least
Pl
Pl
/ p2
Qi
I
I
&I
p2
.
/ Q2
Q2
Unlike the crossed product case, the above does not yield a oneto-one correspondence between appropriate finite sets of prime ideals of R and of S. To be precise, if P and P' are prime ideals of S , write P P' if they lie over the same prime of R and extend this transitively to an equivalence relation. Similarly we obtain an equivalence relation on the primes of R and, in either case, equivalent primes are said to be linked. The following clever example of [72] exhibits some of the pathology which is present here. Suppose A is a ring and let a1, ~ 2 , . .. ,Q, be finitely many epimorphisms ai:A -+ A. Set S = M,(A) and define p : A -+ S by p: a H diag(al(a), a*(a),. . . ,a,(a)). If R = p ( A ) , then using the matrix units of S , it follows easily (Exercise 5) that S is a finite normalizing extension of R. Let us assume for convenience that Ker(ai) = 0 so that p is an isomorphism from A to R. Now every prime P of S is of the form P = M,(P) with P a prime ideal of A and we have N
n;
p-'(P
n R ) = { a E A 1 a i ( a ) E P for all i } n
1
Thus the prime Q of R contains
P n R if and only if n
Q = ,+(Q)
2 p - l ( P n R)= nail(.) 1
and hence if and only if Q 2 a i l ( P ) for some i. Thus the minimal covering primes of P r l R correspond to the minimal members of the set { Q;'(P) }.
161
16. Finite Groups and Incomparability
Conversely suppose we are given Q and Q = p-l(Q>. If p lies over Q, then Q = a i l ( P )for some i so we must have Q 2 Ker(ai) and P = ai(Q) since ai is onto. This limits the possibilities for P and allows for easy computations. We specialize now to a concrete situation. Let A = K[<1,52,.. .] be the polynomial ring in infinitely many variables, let a1 = 1 and let a2 be the K-algebra epimorphism determined by a z ( f i + l ) =
cannot be completed at either corner. Thus the missing Going Up and Going Down relations do indeed fail in general. [34] contains numerous commutative examples. Finally for each i 2 1 let Pi = (1,2,. .. ,i , i + 2 ) . Since c.T'(Pi) = Pi+l and Pi and Pi+l are incomparable, it follows that pi lies over both P: and This yields the infinite diagram
/
\
where the slanted lines represent Lying Over. Thus we see that both R and S contain infinite equivalence classes of linked primes. It is
4. Prime Ideals - The Finite Case
162
amusing to note that the above diagram is actually an upwards spiral since Pi+z 3 Pi for all i .
EXERCISES 1. Let K = F ( ( ) where F is a field of characteristic p > 0 and let K t [ G ]be the twisted group algebra of the group G = (x)x (3) of order p 2 given by Zfj = yZ, ZP = and j j p = 1 + I . If z is a nonidentity element of G, show that 3' is not a pth power in K and conclude that K t [ ( z ) ]is a field. On the other hand, show that Q = 1+I - fj generates a nilpotent ideal of Kt[G].Hence K t [GI is not semiprime even though K t [ H ]is semiprime for all cyclic subgroups H of G. 2. Prove that the two versions of Incomparability given in Theorems 16.6 and 16.9 are equivalent. To this end, if Q is a prime of R and if I is an ideal of S 2 R maximal subject to I n R Q, note that I is a prime ideal of S. 3. Let S = h i be a finite normalizing extension and suppose that I # 0 is an ideal of R of square 0. Set I0 = I and define IiaR inductively for i 5 n as follows. If Ii-1 is given and Ii-1xiIi-1 = 0 take Ii = & - I . Otherwise let Ii = { T E Ii-1 1 zir E I i - l x i } . Verify in either case that Ii is an ideal of R contained in Ii-1, that IixjIi = 0 for all j 5 i and that Ii # 0. For the later, use Ii-1xiIi-1 E li-lxi fl xJi-1. Deduce that SI,S is a nonzero ideal of S of square 0. This argument is from [102]. 4. Let S 2 R be a finite normalizing extension. If Q is a prime ideal of R show that there exists a prime P of S with Q minimal over P f l R. For this use the argument in the proof of Theorem 16.6(iv) along with the result of the preceding problem. Now verify the assertions in Theorem 16.9(iv). 5. Consider the example discussed at the end of this section. First show that S = M,(A) is a finite normalizing extension of R = p(A). Then verify that the diagram cannot be completed at either corner. For this, note that (1,2,3)- only lies over (1,2,3)P. Furthermore since (2,4) 2 Ker(az), (2,4)P can only lie under (2,4)".
xy
17. Primeness and Sylow Subgroups
163
6. If I is an ideal of the ring A, form S = A @ ( A / I ) and map A into S via p : a I+ a @ si. Show that S is a finite centralizing extension of R = p(A). Use this to construct an example with R a commutative domain and with S a commutative ring having no maximal nilpotent ideal.
17. Primeness and Sylow Subgroups There are numerous consequences of the preceding work, some of which we study in this and the next section. First let R*G be a crossed product with G finite and consider the possibility that R*G is prime or semiprime. We wish to see to what extent these conditions are inherited by subgroups or conversely inherited from analogous subgroup information. Of particular interest of course are the Sylow subgroups of G. Next let R(G) be a G-graded ring with G finite. Then via duality, we expect that certain crossed product results should translate directly to this new context. We exhibit a few more examples of this phenomenon. We actually start by considering a completely different topic, namely von Neumann regularity. Recall that a ring R is said to be won Neurnann regular if for every T E R there exists r' E R with rr I r = T . This is directly equivalent to the fact that every cyclic right or left ideal of R is generated by an idempotent. Furthermore it is equivalent (see Exercises 1 and 2) to the fact that every finitely generated submodule of a free R-module is a direct summand. We will need the following subgroup version of Maschke's theorem.
Lemma 17.1. Let W 2 V be R*G modules and let H be a subgroup of finite index in G. Suppose that W is a direct summand of V as an R*H-module and that JG : H1-l E R. Then W is a direct summand of V as an R*G-module.
Proo$ By hypothesis there exists an R*H-projection x : V -+ W. Let { z1,22,. . .,z, } be a right transversal for H in G and, since n E R, define p: V W by p(v) = xix(v%F1)3i.Then p is certainly additive and maps to the R*G-submodule W . Furthermore
4. Prime Ideals - The Finite Case
164
let a E Rij for some g E G. Then there exists a permutation i of the subscripts such that ZiazCi,' E R*H for all i. Thus
H
'i
so p is an RIG-homomorphism. Since p is the identity on W , we conclude that V = W @ U where U = Ker(p) is an R*G-submodule ofV. I
As a consequence we have
Proposition 17.2. [122][182]Let R*G be given and let H be a subgroup of G. i. If R*G is von Neumann regular, then so is R*H. ii. Suppose H has finite index in G and (G : H1-l E R. If R*H is von Neumann regular, then so is R*G. '
Proof. (i) Let TH:R*G -+ R*H be the natural projection and let a E R*H. Since R*G is von Neumann regular, there exists a' E R*G a a so R*H is with aa'a = a. Applying T H we obtain a r ~ ( a ' ) = von Neumann regular. (ii) Let W be a finitely generated submodule of the free R*Gmodule V and observe that V ~ R * His also free. Furthermore, since is finitely generated. Since R*H IG : HI < 00, it follows that WIR*H is von Neumann regular, WIR*H is a direct summand of YR*.. Thus using IG : H1-l E R and the preceding lemma, we conclude that W is a direct summand of V and therefore that R*G is von Neumann regular. We remark that if R is Artinian, then semiprimitive, semiprime and von Neumann regular are equivalent conditions. Thus the above
165
17. Primeness and Sylow Subgroups
translates to a result on the semiprimeness of the twisted group algebra Kt[G] with G finite. Note that the hypothesis )G : H1-l E R of (ii) is certainly needed. Indeed for G finite, K[G] is von Neumann regular if and only if 1GI-l E K . Now we turn to the primeness of R*G with G finite. For this we will follow the route of the proof of Theorem 16.2. Namely we first consider twisted group algebras, then prime coefficient rings and then G-prime coefficients. Let Kt[G] be a twisted group algebra and for the moment s u p pose that K is an algebraically closed field of characteristic 0. Then the condition that Kt[G] is prime is closely related to what group theorists call groups of central type. To be precise, the group H is of central type if it has a faithful irreducible K-representation of degree ( H : Z(H)Ih and this is equivalent to a certain twisted group algebra K t [ H / Z ( H ) ]being simple. It is known [46] that H is of central type if and only if (in a certain specified manner) all its Sylow subgroups are and that [78] a group of central type is necessarily solvable. The proof of the latter result requires, at the present time, the classification of all finite simple groups. In the next two lemmas we use the following notation and observations. Let K t [ H ]be a semiprime twisted group algebra of the finite group H . Then
is a direct sum of full matrix rings over division algebras. As a consequence, if p ( H ) denotes the right regular module of K'[H], then p ( H ) = '& nix where V, is the unique irreducible module of Mni(Di). We call ni the multiplicity of V , in p ( H ) and this integer is uniquely determined by the Jordan-Holder theorem. Computing dimensions over K yields ni dimK V,.
[HI = dimK K t [ H ]= a
Let L be a subgroup of H so that K t [ L ]is semiprime by Proposition 17.2(i). Since K t [ H ]is a free Kt[L]-module of rank IH : L ( ,
166
4. Prime Ideals - The Finite Case
it follows that p ( H ) l ~ t [ = q IH : Llp(L). In the other direction, = W mKt[L] K t [ H ] is a if W is a Kt[L]-module, then WIKt[[Hl = 1H : L(dimK W and right Kt[H]-module. Clearly dimK WIKt[IH] p(L)l'c"Hl = p ( H ) . Suppose u is an automorphism of K t [ H ]so that Q permutes the Kt[H]-modules as in Section 3. Indeed if V is given, then V u = {' w I w E V } with module action defined by w'au = ( ~ afor) all ~ a E Kt[H].Assume in addition that u stabilizes Kt[L].Then clearly (V')lKt[[q = ( V i ~ t p ] ) ~ Conversely . if W is a Kt[L]-module, then
(WIKtl[H])u Z (W')IKt[H1 via the isomorphism (w@a)OH w'@)ou. Observe that if 5 is the unique irreducible module of Mn,(Di), then v,b is the unique irreducible of M,,(Di)'. Now let A be a finite group of automorphisms of K t [ H ]normalizing both K and the group 3-1 of trivial units. Then A acts on X/K' E H . Furthermore, K t [ H ]is A-simple if and only if there is precisely one A-orbit of irreducible Kt[H]-modules. Note that, in any case, all irreducible modules in a fixed A-orbit occur with the same multiplicity in p ( H ) . Finally let p be a prime and let A, be a Sylow psubgroup of A. Then A, is a p-group which permutes the Sylow psubgroups of H . Since the number of Sylow subgroups is congruent to 1 modulo p , it follows that A, must normalize one such subgroup. Let H, be a fixed Sylow psubgroup of H normalized by A,. Thus A, acts as automorphisms on Kt[H,]. For any positive integer n, we use lnlp to denote its p p a r t ; when vertical lines are already present, we will not repeat them so for example [HI, = IHpl.
Lemma 17.3. With the above notation, if K t [ H ]is A-simple, then Kt[H,] is A,-simple.
PmoJ Since K t [ H ]is A-simple, we have p ( H ) = aV where V is the sum of an A-orbit of irreducible modules. Write p ( H p ) = bW + U where W is the sum of an A,-orbit of irreducible modules and U contains the remaining irreducibles. Now
167
17. Primeness and Sylow Subgroups
1
I
so a IH : Hplb and hence \alp b. Furthermore, let T be a right transversal for A, in A. Then WIKt[H] is A,-stable, so &T(WlKt[H1)7 is A-stable and hence equals CV for some integer c. Computing dimensions then yields
1
so since IT( = IA : A,I is a p'-number, I dimK VIP dimK W . Thus
But a d i m V ~ = [HI and bdimK W 5 lHpl = IHlp. We conclude therefore that bdimK W = lHpl so U = 0 and hence Kt[H,] is A,simple. I
Lemma 17.4. With the above notation, write p ( H ) = aV + U where V is the sum of an A-orbit of irreducible modules and U contains the remaining irreducibles. If Kt[H,] is A,-simple, then lHpl a dimK V . In particular, if Kt[H,] is A,-simple for all primes p dividing IHI, then Kt [HI is A-simple.
1
Proof. By assumption, p(H,) = bW where W is the sum of an Aporbit of irreducible modules. Now V is Astable so I / i ~ t is p ~A,-~ stable and hence y ~ t [ = ~ cW , ] for some integer c. Thus dimK W dimK V . Furthermore
I
so clearly b [ a. Combining these yields
1
bdimK W a dimK V and, since b dimK W = lHpl, the first fact follows. Finally if the above holds for all p (HI,then (HI a dimK V so U = 0 and K t [ H ]is A-simple. 1
1
1
4. Prime Ideals - The Finite Case
168
Note that if R*G is prime, then R must be G-prime. It is now a simple matter to prove
Theorem 17.5. 11131 Let R*G be a crossed product with G finite and with R a G-prime ring. Suppose Q is a minimal prime of R and let H be its stabilizer in G. For each prime number p, Jet Hp be a Sylow p-subgroup of H , let Gp be a Sylow p-subgroup of G containing H p and let
be the naturally obtained crossed product. Then RIG is prime if and only if RI,*Gp is prime for each p (GI.
I
Proof: There are two cases to consider.
Case 1. R is prime. Proof: Here we have Q = 0, H = G and RI, = R. Thus the goal is to
1
show that R*G is prime if and only if R*Gp is prime for all p ]GI. In the notation of Section 15, if T = CS*G(S)= Ct[Ginn], then G acts on T as automorphisms normalizing C and the group of trivial units. Furthermore, by Corollary 15.9, R*G is prime if and only if Ct[Ginn] is G-prime. Let Gp be a Sylow p-subgroup of G so that (Ginn)p = Gp n Ginn = (Gp)inn is a SylOW psubgroup of Ginn Q G. Suppose R*G is a prime ring so that Ct[Ginn] is G-prime. Since Ct[Ginn]is a finite dimensional algebra, it follows immediately that Ct[Ginn]is also semiprime and G-simple. Thus, by Lemma 17.3, we see that Ct"(Ginn)p]is Gp-simple and, by Corollary 15.9 again, R*Gp is prime. Conversely suppose R*Gp is a prime ring for all p ]GI. Then by Corollary 15.9, Ct[(Ginn)p] is Gp-prime for all primes p . Again this implies that each Ct[(Ginn)p] is semiprime and Gp-simple and, by Proposition 17.2(ii) applied to p = char C, we see that Ct[Ginn] is semiprime. Now Lemma 17.4 applies and we conclude that Ct[Ginn] is G-simple and hence that R*G is prime. I
I
Case 2. R is G-prime.
169
17. Primeness and Sylow Subgroups
Proof: Here we use the notation of the statement of the theorem. Then by Corollary 14.8, R*G is prime if and only if ( R / Q ) * His prime and hence, by the above, if and only if ( R / Q ) * H , is prime for all p . But note that Gp acts on Rp = R / [ n Z E G Q"] , and H p = H f l G, is the stabilizer in G, of the minimal prime Q/[nxEG,Q"] . Thus by Corollary 14.8 again, R,*G, is prime if and only if ( R / Q ) * H , is prime. This completes the proof. I We remark that primeness is not in general inherited by subgroups even in the case of twisted group algebras (see Exercise 5). We now change topics and consider group-graded rings. The goal is to obtain appropriate analogs of Theorems 16.2 and 16.6. Let S = R(G) be a G-graded ring with G finite. Recall from Section 2 that the smash product S#G* is given by
S#G* = @
pxS xEG
with 1 = C x E G p xa decomposition of 1 into orthogonal idempotents and with PxSz-1y
= P X S P , = Sx-lyPy
for all z,y E G and s E S . Furthermore, G acts on S#G* via (p,s)g = pg-lZs for all z, g E G and s E S and duality, Theorem 2.5, asserts that (S#G*)G 2 MG(S). Here MG(S) is the JGIx ]GI matrix ring over S . A precise realization of this isomorphism is given in Lemma 2.4(ii). A graded ideal I of S is said to be graded prime if for all graded ideals A , B of S, the inclusion AB S I impIies that A or B is contained in I. In particular, S is a graded prime ring when 0 is a graded prime ideal. If L is any ideal of S , we let LG denote the largest graded ideal contained in L. Clearly LG = CxEG(Ln SZ). Again if I is a graded ideal of S , then the graded epimorphism S -+ S / I gives rise to an epimorphism S#G* -+ (S/I)#G* whose kernel is clearly (S#G*)I = C x E G p x I .Thus I ( S # G * ) = (S#G*)I is an ideal of S#G* which we abbreviate as I#G*. In view of the homomorphism S + S / I , it is clear that (I#G*) n S = I . Note
170
4. Prime Ideals - The Finite Case
that I#G* is G-stable and conversely, by Lemma 4.7, every G-stable ideal of S#G* is of this form.
Lemma 17.6. Let I be a graded ideal of S = R(G). The following are equivalent. i. I is graded prime. ii. I = PG for some prime ideal P of S. iii. I#G* is a G-prime ideal of S#G*.
Proof. (i) e (ii) Let I be graded prime. Since I = IG, we can choose, by Zorn’s lemma, P a S maximal with PG= I. If A, B are ideals of S properly larger than P, then AG,BG 3 I so AGBG g I = PG and hence AB P. Thus P is prime. Conversely let I = PG with P prime. If A’, B’ are graded ideals not contained in I , then A’, B’ g P so A’B’ g P and hence A’B’ g PG = I. (i) (iii) Suppose first that I is graded prime and let A , B be Gstable ideals of S#G* with AB C I#G*. Then by Lemma 4.7, A n S and BnS are graded ideals of S with (AnS)(BnS) I#G* nS = I. Thus since I is graded prime, say A n S E I. By Lemma 4.7 again, A = ( A n S)(S#G*) C I(S#G*) = I#G*. Finally let I#G* be a G-prime ideal of S#G* and let A’, B’ be graded ideals of S with A’B’ 2 I . Then A’#G* and B’#G* are G-invariant ideals of S#G* with product contained in I#G*. Since I#G* is G-prime, say A’#G* C I#G*. Then A’ I#G* n S = I and I is graded prime. We can now obtain the analog of Theorem 16.2.
Theorem 17.7. [39]Let S = R(G) be a G-graded ring with G finite and assume that S is graded prime. i. A prime ideal P of S is minimal if and only if PG= 0. ii. There are finitely many minimal primes, say P I ,Pz, . . . ,P n with n _< [GI. iii. N = PInPz n . - - nPn is the unique maximal nilpotent ideal of S and NIGl = 0.
171
17. Primeness and Sylow Subgroups
with
iv. There exists a prime Q of S#G* unique u p to G-conjugation Qx= 0.
nzEG
Proof: By Lemma 17.6, S#G* is G-prime. We consider the skew group ring (S#G*)G S MG(S) and we note that (iv) follows from Theorem 16.2(iv). Furthermore, by Theorem 16,2(ii)(iii), there are n 5 IGI minimal primes PI,Pz,. . . ,Pn of MG(S) such that PI n Pz n n P, = JV satisfies NIGI = 0. Setting Pi = M,(Pi) and JV = MG(N), we see that P I ,P2,. . . ,P, are the n 5 [GIminimal primes of S and that PI r l P2 n . n P, = N satisfies NIGl = 0. This yields (ii) and (iii). For (i) we require the precise realization of the isomorphism MG(S)S (S#G*)G given by Lemma 2.4(ii). Here
--
S#G* = S{l} =
{ a E MG(S) I a(2,y) E R ( 2 - l ~ for ) all z, y E G } .
Now by Theorem 16.2(i), P = M G ( P ) is a minimal prime if and only if P n S{ 1) = 0. Since this is clearly equivalent to P n R(g)= 0 for all g E G, we see that P is minimal if and only if PG = 0. I We are actually more interested in the relationship between the primes of S = R(G) and those of R. For this we require the following classical result.
Lemma 17.8. Let e be a nonzero idempotent of the ring R. Then the map 'p: P H ePe = P r leRe determines a one-to-one correspondence between the set Reof prime ideals of R not containing e and the set of all prime ideals of eRe. Moreover if P , PI and PZ are in Re,then Py P z if and only if Pl Pz and P'+'is primitive if and only if P is primitive.
Proof: Observe that if A is an ideal of eRe, then RAR is an ideal of R with RAR n eRe = e ( R A R ) e = ( e R e ) A ( e R e )= A . Suppose that P is a prime ideal of R not containing e. We show that P'+'= ePe is prime in eRe. First since e 4 P , we have e 4 P+'
172
4. Prime Ideals
- The Finite Case
and hence Pq is a proper ideal of eRe. Now suppose A1 and A2 are ideals of eRe with AlAz ePe. Then
c
(RAlR)(RAzR) = RAl(eRe)AzR = RA1A2R
P
so RAiR c P for some i. Thus Ai = e(RAiR)e ePe and ePe is prime. Pz. Then eP1e c Now let P1,P2 E Reand assume that PT ePze so (ReR)Pl(ReR) G PZ and since e $ P 2 we have PI E Pz. Conversely if PI Pz, then PF Pz. In particular, we see that cp is injective. Next we show that cp is onto. For this let Q be a prime ideal of eRe and observe that e(RQR)e = Q. We can now let P be the unique largest ideal of R with Q = ePe = P i l eRe and it follows easily that P is prime. Furthermore e $ P since e $ Q so Q = P p as required. It remains to verify the primitivity assertions. Suppose first that P E Re is a primitive ideal of R and say P = annR(V) with V an irreducible right R-module. Then Ve is a nonzero eRe-submodule of V which is irreducible. Indeed if 0 # U C Ve is an eRe-submodule, then from U R = V we have Ve = URe = U(eRe) = U . Since ePe is easily seen to be the annihilator of Ve in eRe, it follows that ePe is primitive. Conversely let Q be a primitive ideal of eRe and say that Q is the annihilator of the irreducible module W . Write W E eRe/X where X is a maximal right ideal of eRe. Then X1 = X R @(1- e)R is a right ideal of R with
c
c
X1n eRe = X R n eRe = XRe = X and thus XI is contained in a maximal right ideal Y of R. Clearly X = Y n eRe since X is maximal and e $ Y. Hence, if V is the W . Let P = irreducible R-module R/Y,then v e R e 2 eRe/X annR(V). It follows that P is a primitive ideal of R with P n eRe = ePe C ameRe(W) = Q and, in particular, e 4 P . For the reverse inclusion, observe that eRQ = (eRe)Q C X Y and (1 - e)RQ c (1 - e)R Y . Thus RQ C Y so RQR is a two-sided ideal of R
173
17. Primeness and Sylow Subgroups
contained in Y . This yields RQR annR(V) = P and hence Q C ePe = PV. We have therefore shown that Q = PV and since P is primitive, the result follows. I Finally we have the analog of Theorem 16.6.
Theorem 17.9. [39] Let S = R(G) be a groupgraded ring with G finite. The following basic relations hold between the prime ideals of R and of S . i. Cutting Down. If P is a prime ideal of S , then there are n 5 (GI primes Q1, Q2, . . . , Qn of R minimal over P fl R and we have P n R = Q1 n Q2 n - . Qn. We say that P lies over each Q i . ii. Lying Over. If Q is a prime ideal of R, then there exists a prime P of S such that P lies over Q . Indeed there are m 5 IGl such primes Pi which lie over Q and these are precisely the primes satisfying (Pi)G = PG. iii. Incomparability. Given the lying over diagram
I f p i # P2, then Q i # Q 2 . iv. Going Up and Going Down. We have a t least
Q2
Q2
Proof: We will use the smash product S#G*
Q2
= CxEGpzS. Let I be a graded ideal of S. Since I#G* = C x E G p x we I have pl(l#G*)pl = p l I p l = Ipl where I1 = I n & = I n R . In particularpl(S#G*)pl = R p l and since p l centralizes R we see that Rpl is ring isomorphic to R. By the previous lemma there is a one-to-one correspondence
4. Prime Ideals - The Finite Case
174
between the primes Q of S#G* not containing pl and the primes Q of R. Specifically the map p: Q H Q is given by pl Qp1 = Qp1 = Qppl. (i)(ii) Let P be a prime ideal of S so that PG is graded prime by Lemma 17.6. It follows from Theorem 17.7(iv), applied to the graded prime ring S/PG,that there exists a prime ideal Q of S#G* with PG#G* = Q”. Since PG n R = P n R , we conclude that
nxEG
n’(
Thus, deleting those Q” containing pl, we obtain P n R = Qx)p. This shows that P n R is a semiprime ideal of R with finitely many minimal covering primes. Furthermore since the Q” are incomparable, so are the ( Q ” ) q . This proves (i). Conversely let Q be a prime of R and let Q be the prime of S#G* with Q’+’ = Q . Set I = s n n z E G Q”. Then n x E G QZ = I#G* by Lemma 4.7 and I is graded prime by Lemma 17.6. In view of the work of the preceding paragraph, a prime P of S lies over Q if and only if PG = I . Again by Lemma 17.6, at least one such P exists. Furthermore by Theorem 17.7(ii) applied to S / I , there are at most IGI such primes. Thus (ii) is proved. (iii)(iv) The third diagram of (iv) is trivial since Q1 2 PI nR 2 P2 f l R implies that Q1 contains a minimal covering prime of P2 n R. For the rest let Q i be the prime of S#G* with ( Q i ) p = Qi and define the graded ideal 1i of S by Ii = S n n x E G ( Q i ) Z . Then Q1 2 Q2 so 11 2 I2 and, by (ii) above, P i lies over Qi if and only if (P.)G= Ii. Suppose P2 is given. Then ( P 2 ) G = 12 C I1 so, by Zorn’s lemma, we can find PI 2 P2 11 maximal subject to ( P 1 ) G = 11.Then PI is prime and the first diagram is satisfied. One the other hand suppose PI is given. Then PI 2 I1 2 I2 so Pl 2 P2 where P2 is a minimal covering prime of 12. By Theorem 17.7(i) applied to S/I2 we have ( P 2 ) G = 1 2 and part (iv) is proved. Finally suppose both PI and P2 are given and that Q1 = Q 2 . Then I1 = I2 so ( P ~ ) G = ( P 2 ) G . By Theorem 17.7(i) applied to
+
175
17. Primeness and Sylow Subgroups
S/(P~)G we conclude that PI = Pz and (iii) follows. This completes the proof. I We remark that the above yields a one-to-one correspondence between certain finite sets of primes of R and of S. Furthermore, the missing Going Up result does not hold in general. An example will be offered immediately after Theorem 28.3, a result on primes in fixed rings.
Corollary 17.10. 1391 Let S = R ( G ) be given with G finite and let .P be a prime of S lying over Q. If P C I a S with I n R Q , then I = P.
Proof: Choose P' 2 I maximal with P' n R Q. Then it follows easily that P' is a prime of S and that P' lies over Q. By Theorem 17.9(iii), P' = P and hence I = P. I EXERCISES 1. Let R be a von Neumann regular ring. Show that every cyclic right ideal aR is generated by an idempotent. Now let aR+PR be a two generator right ideal and let e be an idempotent with a R = eR. Prove first that aR PR = eR (1 - e)PR. Now let f be an idempotent with (1- e ) P R = f R and set g = f(1- e ) . Prove that g is an idempotent orthogonal to e and that f R = gR. Conclude that aR+PR=(e+g)R. 2. Again let R be von Neumann regular. Prove first that every finitely generated right ideal of R is a direct summand of R and hence is projective. Now suppose that W is a finitely generated submodule of a free R-module V . Prove that W is a direct summand of V . For this it suffices to assume that V is finitely generated. Proceed by induction on the number of generators of V by projecting W into the last summand. 3. Let K t [ G ] be a twisted group algebra of the finite group G over the field K . For z f G define the map A Z : C ~ ( z-+) K* by jj-'?jj = X,(y)? for all 7~ E CG(Z).Prove that A, is a homomorphism
+
+
176
4. Prime Ideals - The Finite Case
with kernel C&(z)= { y E CG(Z)1 e&j= @ } and that (A,)m = 1 if x m = 1. Show that Z ( K t [ G ] has ) as a K-basis the set of conjugacy class sums for all elements z E G with CG(Z)= C",x). 4. Let Kt[G]be as above and assume when necessary that K is algebraically closed. Prove that Kt[G] is simple if and only if Z(Kt[G]) = K . For this, first show, using the homomorphism A,, that if Z(Kt[G]) = K and if charK = p > 0, then G is a p'-group. Observe in addition that if Kt[G] is simple then IGI is a square integer. 5. Assume that the field K contains a nonidentity pth root of unity 6 for some prime p. Let G = (XI,.. . ,z, y1,. . . , y,) be an elementary abelian pgroup of order p2n and define K t [ S ]by the relations ( Z i ) P = ( & j i ) P = 1 and ?i? = ~& l J i d for i 1 5 i 5 n. All other pairs of generators are assumed to commute. Prove that Kt[G]is simple. Find a subgroup H of G with K t [ H ]not simple. 6. Let S = R ( G ) be G-graded with G a finite group. Discuss the analog of Corollary 16.8 in this context. In addition, discuss the one-to-one correspondence between finite sets of prime ideals of R and of S.
18. Semiprimeness and Sylow Subgroups Now let us move on to consider RtG and the semiprime condition. As we will see, this property is inherited by all subgroups and inherited from at least certain subgroups. In addition, the results translate quite readily to groupgraded rings via duality. We start interestingly enough by considering twisted group rings over commutative von Neumann regular rings. If C is a central subring of a ring R and M is a maximal ideal of C, then we denote by RM the central localization by the multiplicatively closed set C \ M . Note that C \ M need not consist of regular elements here, so the homomorphism from R to RM is not necessarily an embedding.
Lemma 18.1. Let C be a central subring of R and let M be a maximal ideal of C.
18. Semiprimeness and Sylow Subgroups
177
i. If C is von Neumann regular, then CM is a field. ii. If R is semiprime and a finitely generated C-module, then RM is semiprime. iii. R is embedded in RM.
nM
Proof: (i) If c E M , then there is an idempotent e E Cc C M with ce = c. Hence c(l - e) = 0 and, since 1 - e E C \ M , the image of c in CM is zero. Thus all nonzero elements of CM have numerators in C \ M and are invertible. (ii) Let I be an ideal of Rn/i of square zero and let a E R map to a numerator in I . By assumption we can write R = C i r i C , a finite sum. Now for each i, we have aria = 0 in RM so there exists ci E C \ M with ariaci = 0. Set c = ci E C \ M . Then (ac)ri(ac) = 0 for all i so, since R = riC, we have (ac)R(ac) = 0. But R is semiprime so a c = 0 and thus the image of a in Rl\(i is zero. In other words, I = 0. (iii) If 0 # T E R then r c ( r ) # C and the result follows. I
xi
ni
This of course applies to twisted group rings @[GI with C commutative and G finite. Observe that if M is a maximal ideal of C, then the freeness of Ct[G]over C implies that C'[G]M= ( C M ) ~ [ G ] . Moreover if H is a subgroup of G , then C t [ [ H ]=~( C M ) ~ [ is H ]the naturally embedded subring of (C M )[GI. ~
Lemma 18.2. Let Ct[G]be a twisted group ring with C a commutative von Neumann regular ring and with G finite. i. If @[GI is semiprime and H is a subgroup of G, then C t [ H ] is semiprime. ii. If Ct[G]is not semiprime, then there exists a prime p and an elementary abelian p-subgroup P of G such that C t [ P ]is not semiprime and C has p - torsion.
Proof: (i) Let M be a maximal ideal of C. Since Ct[G]is semiprime, Lemma 18.l(ii) implies that C t [ G ]=~(CM)'[G] is semiprime. Thus since CM is a field, Proposition 17.2(i) implies that ( C M ) ~ [ H =] C'[H]Mis semiprime. Lemma l$.l(iii) now yields the result.
4. Prime Ideals
178
- The Finite Case
(ii) Suppose I is a nonzero nilpotent ideal of Ct[G]. Then by Lemma 18.l(iii) there exists a maximal ideal A4 of C such that the image of I in C t [ G ]is~ nonzero; thus ( C M ) ~ [ G is ]not semiprime. Since C M is a field, it follows that CM must have characteristic p for some prime p > 0 and that, by Proposition 17.2(ii) applied to a Sylow psubgroup o f G and Lemma 16.3(iii), there exists an elementary abelian psubgroup P of G with ( C M ) ~ [= PC ] t [ P ]not ~ semiprime. We conclude from Lemma 18.l(ii) that Ct[[P]is not semiprime. Moreover, since the image of p E C is zero in C M ,it follows that some element of C \ 1M has ptorsion. fl We now discuss certain crossed product reductions. Let R*G be given with G finite. Then G permutes the ideals of R and we recall that 0 # A a R is a trivial intersection ideal if for all x E G either Ax = A or A" n A = 0. These occur as follows.
Lemma 18.3. Let 0 # A be an ideal of R . Then A contains a trivial intersection ideal A.
Proof. Since A # 0 and G is finite, we can choose a subset X G maximal with A = A" # 0. Then As = A" so A9nA # 0 implies that X g E X . Thus X g = X and = 2. fl
nzEX
nzEXg
Lemma 18.4. Let R*G be semiprime and let G be finite. i. If N a G , then R*N is semiprime. ii. Suppose A is a trivial intersection ideal of R with stabilizer H C G. Then A*H contains no nonzero nilpotent ideal of R*H. iii. Let A and H be as above and set B = rR(A). Then (R/B)*H = ( R * H ) / ( B * H )is semiprime. Proof. (i) If I is a nilpotent ideal of R * N , then the finite sum L = CzEGI" is a G-stable ideal of R*N which is also nilpotent. Hence L(R*G) = L*G is a nilpotent ideal of R*G and therefore L*G = 0. (ii) Let I a R * H with I A*H and I2 = 0. If g E H , then IgI E I2 = 0. On the other hand, if g E G\H then from AgA E A W A = 0 it follows that I g I ( A * H ) g ( A * H ) = 0. Thus I ( R * G ) I = 0 so I = 0 since R*G is semiprime.
18. Semiprimeness and Sylow Subgroups
179
(iii) Observe that R is semiprime by (i) so we have B n A = 0 and ( A B ) / B is essential in RIB. If (R/B)*H has a nonzero nilpotent ideal, then this ideal meets ( ( A B ) / B ) * H nontrivially. Since A n B = 0, the inverse image of this intersection in A*H is a nilpotent ideal of R*H, contradicting (ii). Thus (R/B)*H is semiprime. I
+
+
We remark that (iii) above applies when A # 0 is a G-invariant ideal of R and H = G. The main results of this section are proved by techniques which originated in [58] and were further developed in the work of [lo91 and [172]as described in Section 15. Therefore some of the preliminary lemmas here are just variants of earlier work and we will be somewhat skimpy with their proofs. For example, the following is essentially contained in Proposition 12.5.
Lemma 18.5. Let I be an ideal of R*G and let A be a subset of G of minimal size with 1 E A and I n (R*A) # 0. For each x E A define
I
A,={~ER x ~ ~ y ~ I a n d r = r , VEA
i. Each A , is a nonzero ideal of R. ii. For each x E A there is an additive bijection jx: A1 A, with ( r a s ) f , = r(af,)s'-' for all r,s E R and a E A1. Here f1 = 1. iii. I n (&A) = { CxEA(afz)Z 1 a E Al }. --f
Proof: Part (i) is clear since I n (&A) is an (R,R)-bimodule. Now we note that the 1-coefficient of a E I n (RxR) uniquely determines a. Indeed if a,a' E I fl (R*A) have the same 1-coefficient, then a - a' E I n (R*A) has smaller support and hence must be zero. This allows us to define f x : A 1 -+ A, with f l = 1 so that (iii) is satisfied. Furthermore, fx is onto by definition of A, and it is oneto-one by the minimality of A. Finally
X
and (ii) follows. I
X
X
4. Prime Ideals - The Finite Case
180
We will also need the symmetric Martindale ring of quotients, this time applied to semiprime rings. As we mentioned in Section 10, the construction and basic properties of Q,(R) in this context is a routine generalization. Of course it requires that we restrict our attention to essential ideals of R, that is ideals A with !R(A)= r R ( A ) = 0. With this understanding, Q,(R) exists and is characterized as in Propositions 10.4 and 10.6. Furthermore, most of the proof of Lemma lO.g(ii)(iii) goes through so that if S = Qe(R)or Q,(R),then every automorphism of R extends uniquely to one of S. Also if C = Cs(R),then C is the center of both Qe(R)and Q,(R). The key difference here is that C , the extended centroid of R, is no longer a field in general. Indeed we have
Lemma 18.6. [4] If R is a semiprime ring, then its extended centroid
C is a commutative von Neumann regular ring. ProoJ: Let c E C and let A‘ be an essential ideal of R with A’c R. If B = !A, (c) then B a R and we can choose A a R so that A 6 B is essential in A’ and hence in R. Notice that Ac Q R and choose D a R so that Ac 6 D is essential in R. Since multiplication by c is one-to-one on A, we can define a left R-module homomorphism f : Ac @ D + R by (ac+ d )f = a. Of course f determines an element c’ E Qe(R)satisfying (ac + d)c’ = a so, for all a E A and b E B , we have
(a
+ b)cc’c = (ac)c’c = ac = ( a + b)c
and thus cc’c = c. Finally if r E R then, for all a E A and d E D ,we have (ac d)rc’ = (arc dr)c’ = a?-= (ac d)c’r
+
+
+
so rc‘ = c’r and c‘ E Z(Q,(R))= C as required. I This explains our concern with commutative von Neumann regular rings in Lemmas 18.1 and 18.2. The next result is a variant of Lemma 12.2.
Lemma 18.7. Let R be a semiprime ring, CT an automorphism of R and A and B ideals of R. Suppose f : A -+ B is an additive bijection satisfying ( r a s ) f = r(af)s‘ for all r, s E R and a E A.
18. Semiprimeness and Sylow Subgroups
181
i. The image B' = A'f of any ideal A' & A is an ideal contained in B. Furthermore ~ R ( A '=) !R(B') is a-invariant. ii. If A is an essential ideal, then there exists a unit q E Qs(R) such that af = aq for all a E A . Moreover, CJ extended to Q,(R) is the inner automorphism induced by q .
ProoJ (i) It is clear that B' = A'f is an ideal of R. Now for all r E R we have (rA')f = T(A'f ) = rB'. Hence since f is a bijection, it follows that l,(A') = ~ R ( B ' )Similarly . since (A'r)f = (A'f)r" = BIT", it follows that rR(A')" = rR(B'). Now use the fact that right and left annihilators are equal in semiprime rings. (ii) Suppose A is essential so that f determines an element q E Qe(R) satisfying af = aq for all a E A . Then (as)q = ( a s ) f = ( a f ) s " = UQS" implies that sq = qs" for all s E R. In particular, qAb = Aq C R so q E Qs(R). Finally since f is a bijection, its inverse g exists and satisfies g: B -+ A with (rbs)g = r(bg)s"-'. Furthermore, B is essential by (i) and hence g determines an element q' E Qs(R).Clearly q' = q-l and sq = qs" yields q-lsq = s" for all s E R. But two automorphisms which agree on R must also agree on Q,(R), so the result follows. The final lemma follows as in Proposition 12.4 and Lemma 15.4. It requires minimal proof here.
Lemma 18.8. Let R be a semiprime ring, S = Qs ( R )and let C be the extended centroid of R. Given R*G, there exists a uniquely defined crossed product S*G which contains R*G. Assume in addition that the action of G on R becomes inner when extended to S. i. S*G = S @c E where E = C S * G ( S= ) Ct[G],some twisted group ring of G over C. ii. If R*G is semiprime, then so is CtfG]. G , then S*H = S @C ( E n ( S * H ) ) and E f l ( S * H ) = iii. If H C t [ H ]where the latter is the natural subring of Ct[G]. iv. If I is an ideal of R*G then
I=
{ a E E I Aa
5I
forsome essential A a R }
is an ideal of E . Furthermore, I2 = 0 implies ( j ) 2= 0.
182
4. Prime Ideals
- The Finite Case
Proof: (ii) If I # 0 is an ideal of Ct[G]of square 0 , then S @cI is a nonzero ideal of S*G of square 0. Now observe that any nonzero ideal of S*G meets R*G nontrivially. (iv) Conversely let I a R*G with I2 = 0. Let a,p E i and say Aa, BP E I . Then (AB)aP = (Aa)(BP)5 I2 = 0 and, since AB is = 0. Thus (f)’ = 0. I essential in R, we conclude that We can now obtain the first main result of this section.
Theorem 18.9. [169] Let R*G be a crossed product with G finite and let H be a subgroup of G. If R*G is semiprime, then so is R*H.
Proof: We proceed by induction on (GI + IHI. Observe that R is semiprime by Lemma 18.4(i). Suppose by way of contradiction that I is a nonzero ideal of R*H of square 0. If N denotes the normal N a G and R*N is semiprime by closure of H in G , then H Lemma 18.4(i) again. Thus by induction, G = N is the normal closure of H . We apply Lemma 18.5 (with G = H ) to the ideal I of R*H and we let A C H and the nonzero ideals A, a R be as in that lemma. If H1 = (A), then H I is a subgroup of H and I f l ( R * H l ) # 0. By induction, H = €€I = (A). By Lemma 18.3, A1 contains a nonzero trivial intersection ideal D with stabilizer GI. Since D E A1 and R is semiprime, we observe that 0 # D ( I n ( R * h ) ) DE I n (&A). It follows by the minimality of A that the supports in D ( I n (R*A))Dcannot be properly smaller than A. Hence for each II: E A we have 0 # DA,TD C DZD. Thus since D is a trivial intersection ideal, we have II: E G1 and hence GI 2 (A) = H . Furthermore if B = ! R ( D ) then, by Lemma 18.4(iii), (R*Gl)/(B*Gl)E (R/B)*G1 is semiprime. On the other hand, since B n D = 0, the image of I in (R/B)*G1is a nonzero nilpotent ideal of ( R / B ) * H . Induction now implies that G1 = G. Let ’ denote the natural map R RIB onto the semiprime ring R’ = R / B and for each z E A let D, = Df,where f, is the bijection given by Lemma 18.5. Then by Lemma 18.7(i), each D, is an ideal of R with left annihilator !R(D,) = !R(D) = B. Thus B n D, = 0 and ---f
18. Semiprimeness and Sylow Subgroups
183
it follows that each fz gives rise to an additive bijection f;: D’ Clearly each Dl is essential in R‘ and (r’d’s’)fk = r ’ ( d ’ j ~ ) ( s ’ ) ’ - for all T ’ , s‘ E R’ and d’ E D’. It follows from Lemma 18.7(ii) that each 3 with z E A acts as an inner automorphism on S = Qs(R’). Hence since (A) = H and G is the normal closure of H , we see that G is inner on S. Lemma 18.8(i)(ii) now implies that R’*G C S*G = S@cCt[G] where C is the extended centroid of the semiprime ring R’. Furthermore, C is von Neumann regular, by Lemma 18.6, and Ct[G]is semiprime. By Lemma 18.2(i), C t [ H ]is also semiprime. Finally I 1 { zzEA(df,)3 I d E D } so I’, the image of I in R’*H contains { EzEA(d’f&)Z I d’ E D’ }. By Lemma 18.7(ii), there is a unit qz E S with d’f; = d’q, for all d‘ E D‘ and such that the action of 3-l on S is the inner automorphism induced by q,. Thus in S*H we see that I’ 2 D’y where y = XzEAqz3 E C S * H ( S= ) C t [ H ] . In other words, in the notation of Lemma 18.8(iv) (with G = H ), y is a nonzero element of the ideal of Ct [HI. But (1’)= 0 so (1’)2= 0 and this contradicts the fact that C t [ H ]is semiprime. The result follows. We now consider the converse direction. For convenience we say that a crossed product R*G with G finite is weakly semiprime if R*P is semiprime for P = (1) and for every elementary abelian psubgroup P of G such that R has ptorsion. In view of the preceding theorem, if R*G is semiprime then it is weakly semiprime. Furthermore, it is obvious that the weakly semiprime condition is inherited by subgroups. The second main result of this section is
Theorem 18.10. [l69l Let R*G be a crossed product with G finite. If R J G is weakly semiprime, then it is semiprime. Proof. We proceed by induction on /GI. If GI is a proper subgroup of G, then R*G1 is weakly semiprime and hence, by induction, R*GI is semiprime. Suppose by way of contradiction that I is a nonzero ideal of R*G of square 0. We apply Lemma 18.5 and use its notation; in addition we set A = A1 and B = & R ( A ) .Since I n ( R * h ) # 0 , it is clear that
4. Prime Ideals - The Finite Case
184
(A) = G. Moreover, by Lemma 18.7(i), B = !R(A,) for all x E A and B is 2-invariant. Thus B is G-invariant and hence B = t ~ ( A 9 ) for all g E G. It follows easily (see Exercise 4), since R is semiprime and B f l As = 0, that B is the left annihilator of A9, where the product is taken in any order. Thus D = n,,-G A9 is a G-stable ideal of R contained in A = A1 with !R(D) = B. For all x E A set D, = Df,.Then by Lemma 18.7(i), each D, is an ideal of R with !R(D,) = f R ( D )= B. Let denote the natural map R --t R I B onto the semiprime ring R’ = R / B . Clearly each DL is essential in R’. Furthermore, each f, gives rise to an additive bijection f:: D‘ + DL satisfying (r’d’s’)f; = r’(d’f;)(s’)’-’ for all T ’ , s’ E R’ and d‘ E D’. It follows from Lemma 18.7(ii) that each 5 with z f A acts as an inner automorphism on S = Qs(R’). Hence, since G = (A), we have G inner on S and Lemma 18.8 applies. Thus R’*G G S*G = S @C Ct[G] where C is the extended centroid of R’. Furthermore if R’ has p torsion, then so does R and it follows from Lemma 18.4(iii) and the hypothesis that R’*G is weakly semiprime. In addition, if C has ptorsion then so does R’. Thus we see from Lemma 18.8(ii)(iii) that Ct[G] is weakly semiprime. Since C is a commutative von Neumann regular ring, we conclude from Lemma 18.2(ii) that @[GI is semiprime. Finally I 2 { xzEA(df,)31 d E D } so I’, the image of I in R’*G contains { C,EA(d’fA)ji.I d‘ E D’ }. By Lemma 18.7(ii), there exists a unit q, E S with d’f; = d‘q, for all d‘ E D’ and such that the action of 2-l on S is the inner automorphism induced by qz. Thus in S*G we see that I‘ 2 D’y where y = CIEA q,3 E C S * H ( S = ) Ct[[H]. In other words, in the notation of Lemma 18.8(iv), y is a nonzero element of the ideal ? of Ct[G].But = 0 so (1’)2= 0 and this contradicts the fact that Ct[G] is semiprime. Therefore, the result follows. 1
ngEG
‘
The above of course generalizes Theorem 4.4. It has numerous crossed product corollaries, but we mention just two.
Corollary 18.11. [169] Let R*G be a crossed product with G finite and R semiprime. Then R*G is semiprime if and only if, for each
18. Semiprimeness and Sylow Subgroups
185
prime p such that R has p-torsion, we have R*G, semiprime where G, is a Sylow p-subgroup of G.
Proof. One implication is immediate from Theorem 18.9. For the other, suppose R has p-torsion and P is an elementary abelian p subgroup of G. Then P is conjugate to a subgroup of G, and R*G, is semiprime by hypothesis. It follows that R*P is semiprime, so R*G is weakly semiprime; Theorem 18.10 yields the result. I
Corollary 18.12. [169] Let R*G be a crossed product with G finite and let H be a subgroup of G. If R*H is semiprime and R has no )G : HI-torsion, then R*G is semiprime. Proof. Since R*H is semiprime, so is R. Furthermore, if R has p torsion, then p is prime to JG: HI so H contains a Sylow psubgroup G, of G. By Theorem 18.9, R*G, is semiprime; Corollary 18.11 now yields the result. a We close this section with the following consequence of duality.
Theorem 18.13. [180]Let S
=
R ( G ) be a G-graded ring with G
finite.
i. If R ( G ) is semiprime and H is a subgroup of G, then R ( H ) is semiprime. ii. Conversely suppose R ( P ) is semiprime for P = (1) and for all elementary abelian p-subgroups P of G such that S has p-torsion. If S is component regular, then S = R ( G ) is semiprime.
Proof. We use Corollary 2.7 and its notation throughout. (i) If R ( G ) is semiprime, then so is S{l}G by Corollary 2.7(ii). Thus Theorem 18.9 implies that S{l}H is semiprime and Corollary 2.7(i) yields the result. (ii) Note that the torsions of S and of S{l} are identical. Hence Corollary 2.7(ii) and the hypothesis imply that S(1)G is weakly semiprime. Thus S{ l)G is semiprime by Theorem 18.10 and Corollary 2.7(i) again yields the result. I
186
4. Prime Ideals - The Finite Case
EXERCISES 1. Let C be a central subring of R. Prove that R is von Neumann regular if and only if RM is for all maximal ideals M of C. F’urthermore, if C is von Neumann regular and if R is a finitely generated C-module, show that R is von Neumann regular if and only if it is semiprime. 2. Let P be a Sylow psubgroup of G and let K have characteristic p > 0. If Kt[G] is semiprime, show that P is abelian and has a normal complement in G. For this, observe from Lemma 16.3(i) that K t [ P ]is a purely inseparable field extension of K . Thus for all g E NG(P) the automorphism induced by g on K t [ P ]is trivial. Deduce that P is in the center of its normalizer. 3. Suppose C is a commutative von Neumann regular ring and that G is a finite group. If Ct[G] is semiprime, prove that G’, the commutator subgroup of G, is a p’-group for all primes p for which p-’ $ C. For this, merely observe that if p-’ $ C, then CM is a field of characteristic p for some maximal ideal M . 4. Let R be a semiprime ring and let A l , A z , .. . , A , and B be ideals of R. Suppose that B = ! R ( A ~for ) all i . Prove that B = t ~ ( A 1 A .2. - A , ) and then that B = t ~ ( A n 1 A2 n f l An). Furthermore show that A1 @ B is essential in R. 5. Let R*G be a crossed product with G finite and R semiprimitive. Use Theorem 4.2 to show that R*G is semiprime if and only if it is semiprimitive. Now translate Theorems 18.9 and 18.10 into results on semiprimitivity. 6. Find an example of a group graded ring R(G) and two Sylow psubgroups PI and P2 of G such that R(P1) is semiprime while R(P2) is not. Thus show that direct analogs of Corollaries 18.11 and 18.12 fail in this context. Where explicitly is the crossed product structure used in the proofs of those results? 7. Let S = R(G) be G-graded with G finite. Assume that S is component regular and that R is prime. Use Corollary 2.7, Theorem 17.5 and the argument of Theorem 18.13 to show that S is prime if and only if R(G,) is prime for each Sylow psubgroup G, of G. You should first observe that S{ 1) is prime.
5
Prime Ideals The Noetherian Case
19. Orbitally Sound Groups This chapter is concerned with prime ideals in Noetherian crossed products. Its theme is the interplay between the crossed product theory and that of polycyclic group algebras. The story begins with the work of [185]which comes tantalizingly close to completely answering the group algebra question. The remaining finite index problem is then settled in [114]using crossed product methods. Finally these results on K[G]lift to analogous ones on Kt[G]and then, via Theorems 14.7 and 15.8, to results on R*G with R Noetherian. We devote this section to the work of [185]on group algebras K[G]with G polycyclic-by-finite. The proofs here are interesting but they are quite long and would really take us too far afield. Fortunately this material is already nicely presented in the original paper I1851 and in the survey [168]. Thus, with two brief exceptions, we content ourselves here with merely discussing these results. Even this will take some time since it is necessary to first understand the key concepts and definitions.
187
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5. Prime Ideals - The Noetherian Case
The goal, of course, is to describe the primes of K[G]and apriori there are two inductive ways to do this. First, suppose (1) # N a G. Then the homomorphism G + G / N extends to the natural algebra epimorphism K[G]4K [ G / N ] .In particular, if Q is a prime of K [ G / N ] then , its complete inverse image P is a prime of K[G].Thus if Q is adequately described in the “smaller” group algebra K [ G / N ] , then P is certainly well understood in K[G]. Second, let Z be the center of G so that Z is a finitely generated abelian group and K [ Z ] is a finitely generated commutative K-algebra. We assume that the commutative theory tells us all we need to know about the primes of this ring. Suppose Q‘ is such a prime and that P’ = Q’K[G],its extension to K [ G ] ,is also prime. Then certainly we understand P‘. The main result of [185]essentially asserts that these two schemes describe all the primes of K [ G ] . More precisely, one must use the f.c. center A(G) instead of the center Z(G). Furthermore, this description does not apply to K[G]but rather to K[Go]where Go is a certain characteristic subgroup of finite index in G. We begin our formal arguments by discussing this subgroup.
Definition. In this section, G will always be a polycyclic-by-finite group. Note that his property is inherited by subgroups and factor groups. In particular, all subgroups of G are finitely generated and therefore G satisfies the maximal condition on subgroups. We will use r to denote an arbitrary group. Suppose I? acts as permutations on a set R. An element w E s1 is said to be orbital, or more precisely I?-orbital, if its r-orbit is finite. In particular, when I‘ acts on itself by conjugation then
q r ) = { x E r 1 Ir : cr(z)l < 0 0 ) is the subset of orbital elements. Furthermore, if V is a group on which acts, we can refer to the I?-orbital elements and subgroups of v. Now let the polycyclic-by-finite group G permute its subgroups by conjugation. Then H 5 G is an orbital subgroup if (G : NG(H)(< 00. Furthermore H is said to be an isolated orbital subgroup if it is orbital and if for any other orbital subgroup H I 3 H we have ( H i : HI = 00.
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19. Orbitally Sound Groups
Lemma 19.1. Let H be an orbital subgroup of G and define its isolator iG ( H ) by iG(H) = ( L I L 2 H is orbital and IL : HI
< 00).
Then iG(H) is the unique isolated orbital subgroup of G containing H and having H as a subgroup of finite index.
Proof: We first show that if L1, L2 are orbital subgroups of G with (Li : HI < 00, then I(L1,Lz) : HI < 00. For this it suffices to assume that G = (&,La) and that coreGH = H” = (1). Since these subgroups are all orbital and 1Li : HI < 00, we can find a normal subgroup A of finite index in G which normalizes H and satisfies [A,Li] C H for i = 1,2. Since A a G, this yields [ An H, Li] C A n H so A n H is normalized by (L1,L2) = G. Thus A r l H core& =
nzEG
(1) and H is finite. We conclude that L1 and L2 are finite orbital subgroups of G and hence, by Lemma 5.l(iii), G = (L1,L2) is finite. Finally, since G satisfies the maximal condition on subgroups, we can choose L maximal subject to L being orbital and ( L: HI < 00. By the above, L = iG(H) has the appropriate properties. I
We note that if z E G normalizes the orbital subgroup H , then z normalizes iG(H) so NG(iG(H)) 2 NG(H). In other words, the isolated orbital subgroups have the largest normalizers among nearby orbital subgroups.
Definition. We say that G is orbitalty sound if all its isolated orbital subgroups are normal. Since G is polycyclic-by-finite, this is equivalent (see Exercise 2) to any of the following three conditions applied to all orbital subgroups H of G. 1. There exist NI, N2aG with N1 C H C N2 and IN2 : N1J < co. 2 . There exists N1 a G with N1 C H and IH : NII < 00. 3. There exists NZa G with H N2 and IN2 : HI < co.
We note that a group of Hirsch number 5 1 is orbitally sound since every subgroup of such a group is either finite or of finite index. Furthermore the orbitally sound condition is inherited by all
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5. Prime Ideals - The Noetherian Case
homomorphic images and by subgroups of finite index. On the other hand, it is not inherited by arbitrary subgroups. For example, let G = A >Q D be the semidirect product of A by D where A = ( a ,b) is free abelian of rank 2 and D = (z,y I z-'yz = y-',z2 = 1) is infinite dihedral. The action here is given by UY = a3b, b y = a-', ax = b and b" = a. Then G is orbitally sound, but W = ( A ,z) is not (Exercises 3 and 4). For convenience, we record the following standard fact about finitely generated groups.
Lemma 19.2. Let H be a subgroup of finite index in G. Then M c H c G with M a characteristic subgroup of finite index in G.
Proof. Let (G : HI = n < 00. Since G is finitely generated, there are only finitely many homomorphisms of G into the symmetric group Sym,. If M is the intersection of all their kernels, then M is clearly a characteristic subgroup of finite index in G. Finally, the permutation action of G on the right cosets of H yields such a homomorphism, so H 2 M as required. I For any polycyclic-by-finite group G we define nio(G) to be the intersection of the normalizers of all isolated orbital subgroups of G. Thus, by definition, G is orbitally sound if and only if G = nio(G). More generally we have
Theorem 19.3. [185]Let G be a polycyclic-by-finite group. Then nio(G) is a characteristic subgroup of finite index in G. Furthermore, it is the unique largest orbitally sound subgroup of finite index in G.
The main thrust of this result is that G has some orbitally sound subgroup of finite index. Once this is proved, the subgroup found can be easily related to nio(G) with the help of Lemma 19.1. Later in this section we will state a useful sufficient condition for G to be orbitally sound.
Definition. If I is an ideal of K[G], we let It = { z E G I 1 - z E I } ,
19. Orbitally Sound Groups
191
Notice that G K[G]and that It is the kernel of the homomorphism of G into the group of units of K [ G ] / I .Thus I t Q G. We say that I is faithful if I t = (1)or equivalently if G is faithfully embedded in K [ G ] / I .Furthermore, I is almost faithful if It is finite. Note that
where the latter is the augmentation ideal of K [ I t ] . Hence I 2 w ( K [ I t ].) K [ G ] ,the kernel of the natural epimorphism K[G] + K [ G / I t ] and , therefore I is the complete inverse image of an ideal f of K [ G / I t ]which is clearly faithful. Let P be a prime ideal of K[G].Then P is said to be a standard prime if P = ( P n K [ A ( G ) ].)K [ G ] ,and P fl K [ A ( G ) ]= n x E G Q” where Q is an almost faithful prime of K [ A ( G ) ] .Note that G acts like a finite group on K [ A ( G ) ]since A ( G ) being finitely generated implies that G / C G ( A ( G )is ) finite. Thus { Q” 1 z E G } is necessarily the finite set of minimal covering primes of the ideal P nK [ A ( G ) ]in K [ A ( G ) ] It . is easy to see that a standard prime is almost faithful. We say that P is virtually standard if the image of P in K [ G / P t ] is a standard prime. If P is almost faithful and virtually standard, then it follows easily that P is standard. The main result is then
Theorem 19.4. [185]If G is an orbitally sound polycyclic-by-finite group, then the almost faithful prime ideals of K[G]are standard. Hence all prime ideals of K[G]are virtually standard. Thus all the primes of K[G]can be described via the two reductions mentioned in the introduction to this section. As we will see in Theorem 21.2(ii), the above property of prime ideals actually characterizes orbitally sound groups. One consequence of the preceding theorem is that the prime and primitive lengths of K[G]can be computed. For this we require some more terminology.
Definition. Let Q denote the field of rational numbers and 2 the ring of integers. A finite dimensional &[I‘]-moduleV is said to be
5. Prime Ideals - The Noetherian Case
192
a rational plinth for I? if V is an irreducible &[r]-module for all subgroups f' of finite index in r. Now let r act on a finitely generated torsion free abelian group A . Then A is a plinth for r if, in additive notation, V = A 8 2 Q is a rational plinth. Thus A is a plinth if and only if no proper pure subgroup of A is r-orbital. Notice that if A is infinite cyclic, then \I'/Cr(A)\5 2 so in this case A is called a centric plinth. Otherwise, when rankA 2 2, A is said to be an eccentric plinth. A normal series
for the polycyclic-by-finite group G is called a plinth series for G if each quotient GilGi-1 is either finite or a plinth for G. It is not necessarily true that every G has a plinth series. However we have (see Exercise 7)
Lemma 19.5. Any polycyclic-by-finite group G has a normal s u b group H of finite index with a plinth series. With G and H as above, let
be a plinth series for H . Then the number of infinite factors HilHi-1 is called the plinth length of G and is denoted by pl(G). Furthermore the number of infinite factors of rank 2 2 is called the eccentric plinth length of G and is denoted by epl(G). It is easy to see, by taking common refinements, that these parameters are independent of the choice of H and of the particular series for H . Note that epl(G) 5 pl(G) 5 ti(G), where the latter is the Hirsch number of G. Moreover, G is nilpotent-by-finite if and only if either pl(G) = h(G) or epl(G) = 0. Finally we distinguish two types of fields according to the nature of their multiplicative group K'. We say K is absolute if K' is periodic or equivalently if K is algebraic over a finite field. Otherwise, K is nonabsolute. With this we have
19. Orbitally Sound Groups
193
Theorem 19.6. [184] [185]Let K[G] be a group algebra over the field K with G polycyclic-by-finite. i. The prime length of K[G] is equal to pl(G). ii. If K is a nonabsolute field, then the primitive length ofK[G] is equal to epl(G). iii. If K is absolute, then all primitive ideals of K[G] are maximal.
Parts (i) and (ii) are first proved for K[nio(G)] using Theorem 19.4. Then one can apply Corollary 16.8 since
K [GI = K [nio(G)]c (G/nio( G) ) . Actually the equality of the prime lengths of K[G] and of K[nio(G)] is quite easy in this case since the rings involved are Noetherian. Observe that (iii) asserts that K[G]has primitive length 0 if K is absolute. This is the earlier result of [184]. In addition, it is shown in [185]that if G is orbitally sound then K[G] satisfies the saturated chain condition. In other words, if P is a prime ideal of K[G], then all saturated chains of primes with largest prime P have the same length. It is not known whether this property lifts to arbitrary polycyclic-by-finite group algebras. Theorems 19.3, 19.4 and 19.6 are the key facts we need about prime ideals in K[G], but there are other results which should be mentioned. To start with, we offer a sufficient condition for G to be orbitally sound. Suppose (1) = Go GI & G, = G is a plinth series for G. Then for each i with GilGi-1 a plinth we obtain the rational plinth V, = (Gi/Gi-1) @ &. Again, via common refinements, it follows that the collection (with multiplicities) of these rational plinths is independent of the particular series and hence is an invariant of G.
Proposition 19.7. [185]Let G be a polycyclic-by-finite group and assume that i. G has a plinth series with rational plinths VI, V2,. . . , V,,
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5. Prime Ideals - The Noetherian Case
ii. G/CG(V,)is abelian for each i, iii. V , V, if and only if V,~Q[HI V,,e[Hlfor any subgroup H of G with IG : HI < 00. Then G is orbitally sound. It is not difficult to see that any polycyclic-by-finite group has a normal subgroup of finite index with the above properties. Thus Proposition 19.7 supplies the main ingredient in the proof of Theorem 19.3. An alternate proof of this ingredient can be found in [205] where it is observed that a solvable connected linear group over the integers is orbitally sound. Now suppose r is a group of operators (that is automorphisms) on G so that acts on K[G]. We assume for convenience that r contains the inner automorphisms of G . Then the set of I'-orbital elements of G is
and this is a normal subgroup of G contained in A(G). Moreover, I? permutes the ideals of K[G]and we can speak about r-orbital primes. W e now briefly discuss some operator analogs of the preceding results. To begin with, we have the important
Theorem 19.8. [16] Let A be a finitely generated free abelian group which is an eccentric plinth for the group I?. If P is a faithful, rorbital prime of K [ A ] then , P = 0. This is proved by an extremely clever valuation theoretic argument. Next comes a far reaching generalization which is crucial to the proof of Theorem 19.4. We remark that it is really a result about finitely generated abelian groups; the extension to the f.c. case follows from the fact that such groups are center-by-finite.
Theorem 19.9. [185]Let A be a finitely generated f.c. group and let l? be a group of operators on A. I f P is an almost faithful, r-orbital , P = ( P n K [ D A ( ~ )- ]K) [ A ] . prime of K [ A ] then
19. Orbitally Sound Groups
195
In other words, the almost faithful prime P is r-orbital if and only if it has r-orbital generators. As an indication of the power of this theorem, we note that it quickly yields an affirmative answer to a multiplicative analog of Hilbert's 14th problem (see Exercise 8).
Corollary 19.10. [53] Let R be a commutative integral domain generated as a K-algebra by a finitely generated group of units A . Suppose r is a group of algebra automorphisms of R which stabilizes A . Then the fixed ring Rr is a finitely generated K-algebra. Finally, a second application of Theorem 19.9 to Theorem 19.4 yields
Corollary 19.11. [185] Let I? be an operator group on the polycyclicby-finite group G and let P be a I?-orbital almost faithful prime of K[G]. IfG is orbitally sound and I? contains the inner automorphisms ofG, then P = ( P ~ K [ D G ( ~ ) ] ) . K and [ G ]PnK[D,(r)] = Q" where Q is an almost faithful prime of K[DG(I')].
nzEG
The interested reader should consult [186) for other approaches to Theorem 19.9. We note that a generalization of that theorem appears in [206]. In the latter result, the hypothesis is weakened so that l? is merely assumed to act on K [ A ] ,stabilizing K and the group of trivial units. We remark that Theorem 19.4 can be used to characterize the primitive ideals of K[G]. Indeed, suppose K is a nonabsolute field and P is a standard prime of K[G] with P n K[A] = n x E G Q x . Then ([185])P is primitive if and only if dimK K[A]/Q < 00. Furthermore, we say that a K[G]-module M is finite induced, if there exists a subgroup H C_ G and a finite K-dimensional K[H]-module V with M = VIKIG].While it is rarely true that all irreducible K[G]-modules are finite induced, the finite induced irreducibles are nevertheless ample in the following precise sense.
Theorem 19.12. [57] Let P be a primitive ideal of the group algebra K[G]with G a polycyclic-by-finite group. Then P is the annihilator of an irreducible finite induced K[G]-module.
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5. Prime Ideals - The Noetherian Case
EXERCISES 1. Let G be a polycyclic-by-finite group. Prove that every subgroup and factor group of G is also polycyclic-by-finite and hence finitely generated. Deduce that A+(G) and G / Q ( A ( G ) >are finite. 2. Prove the equivalence of the four definitions of orbitally sound given immediately after Lemma 19.1. For this, use the previous exercise along with Lemma 19.2. 3. Show that the orbitally sound condition is inherited by factor groups and by subgroups of finite index. Furthermore show that any group of Hirsch number 5 1 is orbitally sound. 4. Let G = A >Q D be the semidirect product given immediately before Theorem 19.3. Prove that W = ( A , s )is not orbitally sound. Now view A additively and describe the action of D on A via 2 x 2 matrices. Conclude that A is a plinth for G. 5. Continuing with the above example, let H be an isolated orbital subgroup of G. If H n A = (l),prove that H centralizes A and hence that H = (1). Otherwise show that H 2 A. Conclude that G is orbitally sound since h ( G / A )= 1. 6. Let Go be an orbitally sound subgroup of finite index in G and let H be an isolated orbital subgroup of G. Use Lemma 19.1 to conclude that HO= H n G o is an isolated orbital subgroup of Go and that H = iG(H0). Since HOa Go conclude that Go normalizes H and hence that GO nio(G). 7. Let G be an arbitrary polycyclic-by-finite group and consider all subnormal series
with IG/HJ < 00, Hi a H and HilHi-1 either torsion free abelian or finite. Prove Lemma 19.5 by choosing such a series with a maximal number of infinite factors. Furthermore, by taking common refinements of two such series, prove that pl(G) and epl(G) are well defined. 8. Prove Corollary 19.10. To this end, first observe that there is a natural I?-epimorphism u:K [ A ] R and that Ker(o) is a I?-stable and faithful prime P. Thus if D = DA(I'), then P = ( PnK[ D] ) .K[ A] ---f
20. Polycyclic Group Algebras
197
hence R = S * ( A / D ) where S = a ( K [ D ] ) Now . show that Rr = Sr and that I' acts like a finite group on S.
20. Polycyclic Group Algebras In this section we return to our past practice of offering complete proofs. Specifically we consider an application of crossed product theory to polycyclic group algebras. Let K[G] be a group algebra of the polycyclic-by-finite group G over the field K . If Go = nio(G), then Go is a characteristic subgroup of finite index in G and all primes in K[Go] are well understood. The goal is to describe the primes of K [ G ] . This is what is known as a finite index problem because we need to lift information from a subgroup of finite index in G to all of G. Notice that K [ G ] = K[Go]*(G/Go) and that G/Go is finite. Thus Theorems 14.7 and 15.8 must surely apply here. In fact the former result (as Corollary 14.10) does apply directly, but the latter seems to require that a certain group of X-inner automorphisms be computed. Fortunately this computation turns out to be unnecessary. What we actually need is to show that a certain ideal is prime. In view of our previous work on primeness and on computing X-inner automorphisms, it is not surprising that A-methods come into play in this part. For convenience we will assume throughout this section that K [ G ]is given with G polycyclic-by-finite. Furthermore, we will write A = A(G).
Lemma 20.1. Let I be a G-stable ideal of K [ A ]which is an intersection of almost faithful primes of that ring. Suppose further that A C N a G with N / A E A ( G / A ) . I f J is a G-stable ideal of K [ N ] and J 3 1.K [ N ] ,then J n K [ A ]3 I .
PrmJ Let Z = Z ( A ) so that Z a G. Since A / Z is finite and N / A A ( G / A ) , it follows easily that N / Z E A ( G / Z ) . In particular there exists a subgroup H of finite index in G which centralizes both A and NIZ.
5. Prime Ideals - The Noetherian Case
198
Since J 3 I - K [ N ] ,we can choose a E J \ ( I - K [ N ] )with support meeting the minimal number, say n, of cosets of A. Thus a= aizi with ai E K[A] and with ~ 1 ~ x 2 .,.,x, . E N in distinct cosets of A. Multiplying by a group element if necessary, we can assume that z1 = 1. Clearly ai E K[A] \ I for all i. The goal is to show that n = 1. Suppose by way of contradiction that n > 1 so that 2, E N \ A. Then x, has infinitely many G-conjugates and hence infinitely many H-conjugates since IG : HI < 00. Note that H centralizes both K [ A ] and N / Z . Thus for any h E H we have ah = h-lah = c y a i ( z i ) h E Zzi C Asi. Setting 0 = ah - a E J, we see that and ,f3 = a i [ ( ~ i ) ~z 1I.id l has support meeting less than n cosets of A. It follows that ,f3 E I . K [ N ]and hence that ~ , [ ( z , ) ~ z ;~ 11 E I . Now write I = Q j , an intersection of almost faithful primes. Then for each h E H and subscript j we have ~ , [ ( z , ) ~ z ;-~11 E Qj. Since (Qj)t is finite and z, has infinitely many H-conjugates, we can choose h so that (z,)hz;l 4 (Qj)t. In other words, (z,)hz;l - 1 4 Q j . But (~,)~z;' - 1 is central in K [ A ] and Qj is prime, so we conclude that a, E Qj. This yields a, E Qj = I , a contradiction. Thus n = 1 and a E (J r\ K[A]) \ I as required. I
xy
x;
nj
nj
The next result is the necessary replacement for Theorem 15.8. It actually follows easily from Lemma 20.1 and Proposition 8.3(iii). However, instead of quoting the latter proposition, we offer a reasonably brief independent argument.
Proposition 20.2.[108] Let K[G] be a group algebra of thepolycyclicby-finite group G and let I be a G-stable ideal of K[A]. If 1 is an intersection of almost faithful primes of K [ A ] ,then I - K[G] is a semiprime ideal of K[G]. If, in addition, I is G-prime, then I.K[G] is a prime ideal of K[G].
Proof. Let A C N 2 G with N / A = A(G/A) and suppose that A, B are ideals of K[G] containing I.K[G] with AB E I.K[G]. Set A' = 7 r ~ ( Aand ) B' = 7 r ~ ( Bso) that A' and B' are G-stable ideals of K[N] containing I . K [ N ] .Here of course T N : K[G] -+ K [ N ]is the natural projection. We first show that (A' f l K[A])(B'n K [ A ] )g I .
199
20. Polycyclic Group Algebras
To this end, let a’ E A’ fl K [ A ] and p’ E B’ f l K [ A ] . Then by definition there exist a E A, /3 E B with a = a‘ aixi and j3 = p’ piyi. Here { 1,x1,. . . ,x, } and { 1,y1,.. . ,ym } are sets of distinct coset representatives for N in G and cxi,pi E K [ N ] . Now let H be a subgroup of finite index in G which centralizes A . Since each xiA f A(G/A), the usual coset counting in G / A shows that there exists h E H such that ( ~ i ) ~ 4y jA for all i , j . Thus since h centralizes K [ A ] ,it follows that a’/3‘ = xa(ahp).But ah E A and AB C I - K [G ]so
+ xy
+ xy
as required. Note that I is a semiprime ideal of K [ A ] .Thus if A’ 5 I . K [ G ] , then (A’ fl K[ A ])’ C I by the above and we have A‘ n K [ A ] E I. Lemma 20.1 now yields A’ = 1.K [ N ]so
A
c ~ N ( A. K) [G ]= A ’ . K [ G ]= I . K[ G]
and I - K[ G ] is a semiprime ideal of K [ G ] . I - K [ G ] ,then Finally suppose I is a G-prime ideal. If AB we have (A’ n K [ A ] ) ( B fl ’ K [ A ] ) I so one of these two G-stable factors is contained in 1, say A’ f l K [ A ]C I . As above, this yields A I - K[ G ]and therefore 1.K [G ]is a prime ideal of K [ G ] . I
c
c
Recall from Section 14 that if H induced ideal LIG is defined by
c G and if L a K [ H ] ,then the
It follows from Lemma 14.4 that LIG is the unique largest two-sided ideal of K[G]contained in L.K[G]. Furthermore, it is unique largest ideal with x ~ ( L l ~L.) Note that if H a G, then
5. Prime Ideals - The Noetherian Case
200
Now recall from the previous section that a prime ideal P of K[G]is standard if P = ( P n K [ A ] ) - K [ G and ] P n K [ A ]= n x E G Lx where L is an almost faithful prime of K [ A ] .The following is merely a reformulation of Proposition 20.2 since A a G and since the ideal Lx is clearly G-prime.
nxEG
Lemma 20.3. Any standard prime P of K[G] can be written ils P = LIG with L an almost faithful prime of K [ A ] .Conversely if L is an almost faithful prime of K [ A ] ,then LIG is a standard prime ideal of K[G]. The next lemma describes the behavior of standard primes under restriction to normal subgroups of finite index. Note that K[G]is right and left Noetherian by Proposition 1.6. Thus if N a G and P is a prime of K [ G ] ,then Lemmas 1.3 and 14.2 imply that P n K [ N ]= n x E G Q xwhere Q is a G-orbital prime of K [ N ] unique up to Gconjugacy. Indeed { Qx I IC E G } is the finite set of minimal covering primes of the ideal P n K [ N ] .
Lemma 20.4. Let H be a normal subgroup of finite index in G and let P be a prime ideal of K[G].Write P n K [ H ]= n x E G Q" with Q a prime ideal of K [ H ] . i. P is standard if and only if Q is standard. ii. Assume that P is standard and write P = LIG with L an almost faithful prime of K [ A ] . If J is a minimal covering prime of L n K [ A ( H ) ]then , JIH is a minimal covering prime of P nK [ H ]and it is standard.
ProoJ We first prove (ii). Set D = A ( H ) so that D = H n A since (G : HI < 00. By assumption, P is standard so P = ( P n K [ A ] ) X [ G ] and hence TA(P) & P. Therefore D = H n A implies that T D ( Pn K [ H ] = ) T A ( Pn K [ H ] ) P
n K[H].
Setting I = P n K [ H ] ,it follows from I T D ( I ). K [ H ]and the above that I = ( I n K [ D ] -) K [ H ] .Moreover, since
P n K I A ]= ~l~
n K [ A ]=
n
xEG
L"
201
20. Polycyclic Group Algebras
we see that
n
~nK I D ]= P n K I D ]=
( L n~K [ D ] )=
xEG
n
( Ln K ~ D ] ) ~ .
xEG
,n,
Since D a A we have L n K [ D ] = J Y where J is any minimal covering prime. Hence ~t n D = n,,,(Jt)Y. In particular, if 2 = Z(A) n D ,then J t n 2 = Lt n 2 C Lt. But L is almost faithful and [ D: 21 < 00 so we conclude that J is almost faithful. It now follows from Lemma 20.3 that JIH is a standard prime of K [ H ] .Moreover, since D a G,
n
P n K [ H ]= I = ( ~Kn [ D ] .) K [ H ]=
( Ln K [ D I ) K ~ [. H ]
xEG
n
=
( F I X .
xEG
Thus since (G : HI < m we see that JIH is a minimal covering prime of P n K [ H J . This proves fii) and one implication of (i) since the uniqueness of Q implies that Q is G-conjugate to JIH and hence is clearly standard. Conversely assume that Q is standard and write P n K[A] = Sx where S is a prime ideal of K[A]. Since Q = TIH for some almost faithful prime T of K [ D ] we , have Q f~ K[DJ = TY and hence
nxEG
nyEH
( sn K[D])"= ( P n K [ A ] )n K [ D ]= ( P n K [ H ] n) K I D ] xEG
=
n zfG
(Q"
n K p ] )=
n
TX.
x€G
Since G acts like a finite group on K [ A ] ,these intersections are finite. Thus since T is prime, we have S n K [ D ] T" for some 2 E G and it follows that St n D (Tt)". But the latter group is finite and IA : Dl < 00, so S is an almost faithful prime of K[A]. By Lemma 20.3,
P' = SIC =
( n 55) . K[G]= ( P n K[A]) XEG
K[G]
5. Prime Ideals - The Noetherian Case
202
is a standard prime ideal of K[G]which is clearly contained in P. Finally since Q = T Y ) . K [ H ]we have
(nyEH
and thus ( P n K [ H ] )- K[G]C P' C P. Since K[G]= K [ H ] * ( G / H ) , it now follows from Incomparability, Theorem 16.6(ii), that P' = P. Thus P is standard and the lemma is proved. I We require some additional terminology. Let N be a subgroup of G and let I be an ideal of K[G].We say that I is almost faithful mod N if It 2 N and / I t : NJ < m. In particular if NaG, then this occurs if and only if I is the complete inverse image of an almost faithful ideal of K [ G / N ] .More importantly, we say I is almost faithful sub N if I t C N and IN : It1 < 00. In addition we let VG(N) denote the complete inverse image in NG( N ) of A (NG( N )/ N ). We can now obtain the first main result of this section.
Theorem 20.5. [114]Let G be a polycyclic-by-finite group and let P be a prime ideal of K[G].Suppose H is a normal orbitally sound subgroup of finite index in G and write P n K [ H ]= (IIEGQ" with Q a prime of K [ H ] .Define N = iG(Qt), A = N G ( N ) and let B = { z E G J Q" = Q }. i. H Z B E A . ii. There exists a unique prime ideal T of K [ A ]with P = TIG and T n K [ H ]= Q". Indeed T is the unique minimal covering prime of P n K [ A ]not containing Q". iii. There exists an almost faithful sub N prime ideal L of K[vG(N)] with T = LIA and hence with P = LIG. Indeed L is a minimal covering prime of T n K[vG(N)].
naEA
Proof: We use the notation given in the statement. We note that such a normal orbitally sound subgroup H of finite index in G does exist; indeed we could take H = nio(G) by Theorem 19.3. Moreover P I I K [ H ]has the above structure with Q unique up to G-conjugation.
20. Polycyclic Group Algebras
203
Also Qt a H so Qt is orbital in G and, by Lemma 19.1, N = iG(Qt) is an isolated orbital subgroup of G with IN : Qtl < 00 and A = N G ( N ) 2 N G ( Q ~2 ) H. Observe that if Q" = Q, then x certainly normalizes Qt and hence we have A 2 NG(Qt) 2 B 2 H . Thus by considering K[G]as the crossed product K[H]*(G/H), we see from Corollary 14.10 that there exists a unique prime ideal T of K [ H ] * ( A / H= ) K[A]with T n K [ H ]= n a E A Q" and with P = TIG. This proves parts (i) and (ii) since Corollary 14.10(ii) describes T appropriately. By Lemma 19.2 we can let M be a characteristic subgroup of N of finite index contained in Qt so that M Q N G ( N ) = A. We use - : K [ A ]+ K [ A / M ]to denote the natural map. Since M a H and IQt : M ) < 00, it follows that Q, the image of Q is an almost faithful prime of K[I?].Furthermore, fi is orbitally sound so Q is a standard prime by Theorem 19.4. Note that T n K [ H ]= n a E A Q" and hence ? n K [ H ]= Q'. We can therefore conclude from Lemma 20.4(i) that f is a standard prime of K [ A ] .In other words there exists an almost faithful prime 2, of K [ A ( A ) with ]
niLEA
By lifting this expression back to K [ A ] we , see that there exists an almost faithful mod M prime ideal L of K [ V A ( M ) ]with
Observe that N , M a A and IN : MI < 00. This implies easily (see Exercise 4) that V A ( M ) = V A ( N ) = V G ( N )since A = NG(N). Also iG(M) 2 N 2 M so iG(M) = N . Since Lt is orbital in A , it is orbital in G , and from Lt 2 M and (Lt : MI < 00, we iG(M) = N and that IN : Lt( < m. Thus conclude that Lt we see that L is almost faithful sub N . Finally we have T = LIA and P = TIG so Lemma 14.9 yields P = (LIA)lG = LIG. Since T ~ K [ V G ( N=) ] La, a finite intersection of primes, L is clearly a minimal covering prime of T n K[VG(N)] and the result follows. I
naEA
5. Prime Ideals - The Noetherian Case
204
Definition. Let P be a prime ideal of K [ G ] , N an isolated orbital subgroup of G, and L an almost faithful sub N prime ideal of K [ V G ( N ) ]If. P = LIG, then N is said to be a vertex of P and Furthermore, for this N , the prime L is said we write N = VXG(P). to be a source of P. The previous theorem describes how to find at least one vertex and source for P. This is all the more useful once we prove uniqueness in the next main result.
Theorem 20.6. [114]Let N be an isolated orbital subgroup o f G and let L be an almost faithful sub N prime ideal of K [ V G ( N ) ] .Then P = LIG is a prime ideal of K[G].Furthermore, let H = nio(G) and write P nK [ H ]= QI n Q 2 n n Qn, an intersection of G-conjugate primes. i. For some Q = Qi, we have N = iG(Q+) so N is uniquely determined by P up to conjugation in G. ii. For this N , the ideal L is uniquely determined by P up to conjugation by A = NG(N). iii. I f J is a minimal covering prime of L nK [ H n V G ( N ) ]then , JIH = Q" for some a E A . Furthermore, H n V G ( N )= V H ( Q ~ ) . a
Proof. Set A = NG(N)and let H = nio(G) so that, by definition, H C A . Since ( N : N n HI < 00, and ( N : Lt( < 00, there exists, by Lemma 19.2, a characteristic subgroup M of finite index in N N n H and M C L t . Thus M a A and therefore also with M M Q H . Furthermore, since IN : MI < 00, we have iG(M) = N , so NG(M)= A and then clearly V G ( M )= V G ( N )(see Exercise 4 again). Let - : K [ A ]-+ K [ A / M ]denote the natural map. Then is an , V G ( M )= V G ( N ) and , we almost faithful prime of K [ A ( A ) ]since conclude from Lemma 20.3 that
is a standard prime of K [ A ] . Furthermore Lemma 20.4(i) implies that T n K [ f i ] = n a E A Q a is a finite intersection of A-conjugate
205
20. Polycyclic Group Algebras
standard primes of K [ f i ] .Lifting this information to K [ A ] ,we see immediately that
naEA
is a prime ideal of K[A] and that T n K [ H ]= Q" is a finite intersection of A-conjugate primes of K [ H ]which are almost faithful mod M . Set B = {x E G Q" = Q } so that H E B E G and surely B C NG(Qt). Observe that IQt : MI < 00 and Qt is orbital in G so iG(Qt) = iG(M) = N and hence B E N G ( Q ~C) NG(N) = A . We now view K [ G ]as the crossed product K [ H ] * ( G / H ) .Then, since B C A, it follows from Corollary 14.10(i), that P = TIG is a prime ideal of K [ G ]and, since TIG = (LIA)lG= Ll', by Lemma 14.9, the first assertion is proved. Furthermore, by Corollary 14.10 again, Q is a minimal covering prime of P n K [ H ]= Q1 n Qz n . n Qn so Q = Qi for some i. Thus since N = iG(Qt), (i) is proved. Now suppose both N and P = LIG are known. We consider those Qi with iG(Qf) = N. Since Qi = Q" for some x E G, we have
I
N = iG(Qf) = iG(Qt)" = N", and hence this occurs if and only if x E A. In particular, the ideal &AQ" is the intersection of all the minimal covering primes Qi of P K [ H ]with iG(Qf) = N and this is surely determined by N and P. It now follows from Corollary 14.10(ii) that the prime ideal T of K [ A ] is uniquely determined by the conditions TIG = P and T n K [ H ]= Q". But T = LIA and L is certainly a minimal L", so L is unique up to covering prime of T n K[VG(N)] = conjugation by A . This proves (ii). Finally, recall that 2= LIA is a standard prime of K [ A ] .Thus Lemma 20.4(ii) asserts that if j is a minimal covering prime of L n K [ A ( B ) ]then , is a minimal covering prime of 5? n K [ B ]= n k E A Q' and hence jIH= Q' for some zt E A. Lifting this information back to K [ A ] ,we see that for any minimal covering prime J of L n K [ V H ( M ) ]we , have JIH = Q" for some a E A . Since clearly V H ( M )= V H ( Q ~= ) V G ( N )n H , part (iii) follows. I
naEA
naEA
206
5. Prime Ideals - The Noetherian Case
The preceding two theorems are extremely technical in nature and actually yield more information than we require. Therefore we close this section by offering simpler, more understandable, versions of these results. The formulation divides naturally into three parts.
Theorem 20.7. (Existence) [185] 11141 Let G be a polycyclic-byfinite group and let P be a prime ideal of K [ G ] . Then there exists an isolated orbital subgroup N of G and an almost faithful sub N prime L of K [ V G ( N ) ]with P = LIG. This is Theorem 20.5(iii). Note that V G ( N ) / Nis a torsion free abelian group so V G ( N ) / Lis~ finite-by-abelian and hence an f.c. group. Thus L essentially corresponds to a prime ideal of a commutative group algebra. Recall that N above is a vertex for P and L is a source.
Theorem 20.8. (Uniqueness) [114] Let G be polycyclic-by-finite and let P be a prime of K [ G ] . Then the vertices of P are unique u p to conjugation by G. Furthermore, if N is a vertex, then the sources for this N are unique up to conjugation by N G ( N ) . This is of course Theorem 20.6(i)(ii). Finally the beginning of Theorem 20.6 yields
Theorem 20.9. (Converse) [114] Let N be an isolated orbital subgroup of the polycyclic-by-finite group G. If L is an almost faithful sub N prime of K [ V G ( N ) ]then , LIG is a prime ideal of K [ G ] . In the next section we will consider some corollaries and extensions of these results.
EXERCISES 1. Let I be a G-stable ideal of K [ A ] which is an interseetion of almost faithful primes of that ring. Let A N a G with
207
20. Polycyclic Group Algebras
N / A E A(N/A) and let ments of IS”]. If
&,ti2
,...,ti,
and
$I,,& ,...,fin
be ele-
for all z E G, show that
and hence that
To this end, let H be a subgroup of finite index in G centralizing both A and N / Z where 2 = Z(A). Now proceed by induction on the number of A-cosets which meet the supports of the various &. 2. Let I be as above and let Q I , ~ 2 , . .. , (lin and P I , P 2 , . . . , Pn be elements of K[G]. If
for all 2 E G, show that
For this, let N/A = A(G/A) and set tii = . r r ~ ( c x i and ) ,& = .rrp~(Pi). Now use A-methods to reduce to the hypothesis of the preceding problem. This is a result of [108]. 3. If N is a finite normal subgroup of G, show that Q G ( N )= A(G). Conclude that a virtually standard, almost faithful prime of K[G]is standard. 4. Let N be an isolated orbital subgroup of G and let M be a characteristic subgroup of finite index in N . Prove that N G ( N ) = N G ( M )and then that V G ( N )= V G ( M ) .
5. Prime Ideals - The Noetherian Case
208
5. Let N be a nonidentity torsion free polycyclic-by-finite group, let W be a finite group and set G = N1 W . Thus G is the semidirect product G = H >Q W where H = N , is the direct product of copies of N indexed by the elements of W . Moreover W permutes the factors of H in the natural manner. Show that each N , is an isolated orbital subgroup of G and deduce that nio(G) H .
nWEW
21. Polycyclic Crossed Products We now consider some consequences and extensions of the results of the previous section. We start with group algebras and then move on to twisted group algebras and polycyclic crossed products. Again we let G denote a polycyclic-by-finite group and K will be a field. We first list some elementary properties of isolators.
Lemma 21.1. Let N E H be orbital subgroups of G. i. iG(N) c iG(H) and iG(N) n H = iH(N). ii. If N a H, then iH(N)/N = A + ( H / N ) . iii. If N is an isolated orbital subgroup of G, then V G ( N ) / Nis torsion free abelian.
Proof. (i) Let A = NH(N) 2 N . Then A is a subgroup of finite index in H and it is G-orbital. In particular, iG(A) = iG(H). Since A normalizes N, it normalizes iG(N) so AiG(N) is an orbital subgroup of G containing A . Moreover
Thus AiG ( N ) & i G (A) = iG ( H ). The second part follows easily since iG(N) n H is H-orbital and iH(N) is G-orbital. (ii) If M / N = A + ( H / N ) ,then we have M a H and (MINI < 00 by Lemma 5.l(iii). Thus M iH(N). Equality follows since every H-orbital subgroup of HIN is contained in A + ( H / N ) . (iii) Set H = VG(N). Then by the above, A+(H/N) = (1) so H/N is torsion free abelian by Lemma 5.1(ii).
c
21. Polycyclic Crossed Products
209
The following is a converse to Theorem 19.4. It shows that the latter result cannot be extended beyond the class of orbitally sound groups.
Theorem 21.2. [114]Let K[G]be given with G a polycyclic-by-finite group. i. Every isolated orbital subgroup N of G is the vertex of a prime ideal of K [ G ]. ii. I f all primes of K[G]are virtually standard, then G is orbitally sound.
Pmo$ (i) Let N be an isolated orbital subgroup of G and let L denote the kernel of the natural epimorphism K [ V G ( N ) ]+ K [ V G ( N ) / N ] . Since V G ( N ) / Nis torsion free abelian, by Lemma 2l.l(iii), its group algebra is a domain and L is a prime ideal of K [ V G ( N ) ] Further. more, Lt = N so we conclude from Theorem 20.8 that P = LIG is a prime ideal of K [ G ]with vertex N . (ii) In view of the above, it suffices to show that every vertex is normal in G. To this end, let P be a prime ideal of K[G]. If N = iG(Pt) then, by Lemma 21.l(iii), N / P t = A + ( G / P t ) so N a G. This also implies that V G ( N )= V,(Pt). Finally, since P is virtually standard, there exists an almost faithful mod Pt prime L Of K [ v G ( P t ) with ] P= L,) . K [ G ] .Thus P = LIG and since L is almost faithful sub N , we have VXG(P) = N a G . It follows from Theorem 20.9 that N is the unique vertex of P.
(n,
Next we consider an operator version of Theorem 20.7 and for this we need a variant of V G ( N ) . Thus suppose r is a group of operators on G, N is a subgroup of G, and let
; Then ro normalizes N , NG(N)and V G ( N )and we define V G ( N r) to be the set of IC E V G ( N )having only finitely many I'o-conjugates modulo N . It is clear that V G ( N I?) ; is a ro-invariant subgroup of V G ( N )containing N . Furthermore if G acts on G by conjugation, then V G ( N G) ; =VG(N).
210
5. Prime Ideals - The Noetherian Case
Proposition 21.3. Let G be a polycyclic-by-finite group, let be a group of operators on G and let P be a r-orbital prime of K [ G ]with VXG(P) = N . Then there exists a r-orbital almost faithful sub N prime L of K [ v G ( N ; r)]with P = d G .
Proof: We proceed with a series of special cases. Case 1. Assume that G is an f.c. group and P is almost faithful.
Proof: By Lemma 21.1(ii), if N = A f ( G ) , then N is a finite isolated orbital subgroup of G. Since P is almost faithful sub N and G = V G ( N ) ,it therefore follows that N = VXG(P). Moreover since P is r-orbital, we conclude from Theorem 19.9 that P = ( P n K[D&)]) - K[G]. Now it is clear that &(r) contains the finite group A + ( G ) = N so DG(r) a G and it follows easily that DG(r) = VG(N;r). Furthermore, P fl K [ D G ( ~= ) ]n x E G Lx where L is a prime ideal of K [ & ( r ) ] so we have P = LIG. Finally if 2 = Z ( G ) ,then IG : 21 < 0;) and P t n Z = n,,-G(Lt)” n 2 = Lt n 2 so Lt n 2 is finite and L is an almost faithful ideal. Certainly L is I?-orbital. \
Case 2. Assume that G = V G ( N ) . ProoJ By definition, P is an almost faithful sub N prime ideal of K [ G ]and, by Lemma 19.2, we can choose M to be a characteristic Pt. Since N / M is finite, subgroup of finite index in N with M G / M is an f.c. group. Furthermore, the image of P in K [ G / M ]is an almost faithful I?-orbital prime. The preceding case, applied to K [ G / M ]quickly , yields the result. I
Case 3. Assume that G is arbitrary.
Proof: By Theorem 20.7, P = JIG where J is an almost faithful sub N prime of K[VG(N)]. Since P is I?-orbital, it follows from uniqueness in Theorem 20.8 that N is r-orbital. Thus (I? : rol < 00 where ro = Nr(N). Furthermore, by uniqueness again, J is a ro-orbital prime of K[VG(N)].By the previous case, J = LlvG(N)with L a
211
21. Polycyclic Crossed Products
ro-orbital almost faithful sub N prime of K [ V G ( N ; ~ ) ]Hence, . by the transitivity of induction in Lemma 14.9, we have P = JIG = LIG. Since L is clearly I?-orbital, the result follows. 4 At this point it is convenient to consider a few properties of poly-{infinite cyclic} groups. A group G is said to be p o l y - 2 if it has a finite subnormal series (1)= Go a G1 a .
a
- a G,
=G
with each quotient GilGi-1 infinite cyclic. It is clear that such groups are polycyclic-by-finite and also torsion free. In addition we have
Lemma 21.4. i. A polycyclic-by-finite group has a normal poly-Z subgroup of finite index. ii. Any poly-Z group is a unique product group. iii. Let R*r be a crossed product with a unique product group. If I is a r-prime ideal of R, then I*r is a prime ideal of R*r.
Pvoaf. (i) This is [161,Lemma 10.2.51. (ii) By induction we need only show that if H a G is a unique product group and if G / H is infinite cyclic, then G is unique product. For this, let G = ( H , z ) and let A , B be finite nonempty subsets of G. Choose n,m maximal with A0 = A n H x n # 0 and Bo = B n H x m # 0. Then a unique product element in z - ~ A. ~ easily determines one in AB. (iii) We may clearly assume that I = 0 so that R is r-prime. Suppose 0 # A , B a R*r with AB = 0. Then by Lemma 5.12 we have the product 7 r - ((~m i q l ) A ) (min(1)B ) = 0. But these factors are nonzero r-stable ideals of R, by Lemma 5.11, so we have a contradiction. I Of course (iii) follows directly from Corollary 8.5 since a unique product group is torsion free. The next result offers an affirmative answer to a conjecture of [185].
5. Prime Ideals - The Noetherian Case
212
Theorem 21.5. [27] Let H be an isolated orbital subgroup of the polycyclic-by-finite group G and let L be a G-orbital prime of K [ H ] . Then LIG is a prime ideal of K[G].
Proof: Let Go = NG(H)so that IG : Go1 < 00. Then Go is a group of operators on K [ H ]and L is Go-orbital so the previous proposition ) D = VH(N;GO), then applies. In particular if N = v x ~ ( L and there exists a Go-orbital almost faithful sub N prime I of K [ D ]with L = IIH. Furthermore, N is Go-orbital and hence also G-orbital. By Lemma 21.1(i), iG(N) iG(H) = H so iG(N) = iH(N) = N and N is an isolated orbital subgroup of G. Moreover, IG : Go1 < 00 so it follows easily that VG(N) fl H = V H ( N ; G ~= ) D. Note that V G ( N ) / Nis torsion free abelian, by Lemma 2l.l(iii), so DaVG(N). Since D is G-orbital, we conclude from Lemma 21.l(i) that iG(D) E H and then, from Lemma 21.l(ii) applied to D C VG(N), that VG(N)/D is torsion free abelian. But a finitely generated torsion free abelian group is poly-2, so Lemma 21.4(ii)(iii) implies that J
= IlvG(N) =
(
n
IZ)
.KPG(WI
xE VG (N)
is prime. Finally the above formula for J shows first that J t E D and then that J t = n r E V c ( N )(It)”. Thus since IN : I t ( < 00, Lemma 19.2 yields ( N : Jtl < 00. We have therefore shown that J is an almost faithful sub N prime of K[VG(N)] so, by Theorem 20.9, JIGis prime. But by transitivity
so the result follows. I We now move on to twisted group algebras. For this it is necessary to briefly discuss finitely presented groups. We recall that a group r is finitely presented if it is determined by finitely many generators and relations. We write such a group as
21. Polycyclic Crossed Products
213
where z1,x2, . . . ,z, generate r and where all relations among these generators can be derived from w1 = 1,202 = 1, . . . ,w, = 1. Formally this means that the homomorphism from the free group F = ( X 1 , X 2 , . . . , X,) onto I?, given by X i H xi, has as its kernel the normal subgroup (W1,W,, . . . , W,) generated by the finitely many words WI, W2,. . . , W, E F with wi = W i ( z 1 , 2 2 , .. . ,x,). The next result follows from [161, Lemma 12.3.12(ii)(iv)]. We just briefly sketch its proof.
Lemma 21.6. Let G be a polycyclic-by-finite group. i. G is finitely presented. ii. If H is a finitely generated group, Z is a central subgroup of H and H f Z Z G , then Z is finitely generated and hence H is polycyclic-by-finite.
ProoJ (i) Here, by induction, we need only show that if H is finitely presented and G I N is infinite cyclic or finite, then G is finitely presented. In either case, we start with the generators and relations of H . If G / H is infinite cyclic, then G = (H,x) and we add J: to the generating set. Furthermore we add the finitely many relations which describe the x-conjugate of each generator of H . On the other hand, if G / H is finite, then G = Uy Hxi and we add z ~ , z ~ . . ,x, ,. to the generating set. As above, the action of each xi on H can be described via a finite number of additional relations. Finally for each i , j we have zizj E Hxk for some k and we add one more relation for each of these containments. (ii) We know that G is finitely presented, say
.. ,x,). Let H be finitely generated with where wi = Wi(x1,xz,. H / Z E G for some central subgroup 2 of H . By adding generators if necessary, we can assume that the generating set includes h l , h 2 , . . . ,h, with hi2 = gi. Then H = ( h l ,h2,. . . ,h,)Z so we can assume that the remaining finitely many generators z1,z2,. . . , zt are in 2. Replacing H by H / ( ~ 1 ~ 2.2..,,zt ) if necessary, we can suppose that H = ( h l ,ha,. . . , h,). With this, we claim that Z is generated by
5. Prime Ideals - The Noetherian Case
214
Wi(h1,h2,. . . , h,) for 1 5 i 5 s. To this end, let z E 2 and write z in terms of the generators of H so that z = W(h1,hz, . . . , h,) where W is a word in the free group F = ( X I ,X Z ,. . . ,X,). Hence, by mapping z to G, we have W ( z l , 2 2 , .. . ,zn) = 1 and, by definition, W is in the normal closure of (W1,W2,.. . , W , ) .Replacing each X i by hi and using the fact that Wi(h1,hz,. . . , h,) E 2 is central in H now yields the result. I
As an application we have
Lemma 21.7. Let Kt[G] be a twisted group algebra with group of trivial units 6 and let H be a finitely generated subgroup o f 6 with HK' = 6. Then 2 = H n K' is a central subgroup of H and H / Z S G. Thus H is a polycyclic-by-finite group and the inclusion map H
-+
8 extends to a K-algebra epimorphism K[H] -+ Kt[G].
Proof. Since 6 / K * G is finitely generated, we know that H exists, 2 = H n K' is central in H and H I 2 2 G. By Lemma 21.6, H is polycyclic-by-finite and the remaining facts are clear. I In particular, if P is a prime ideal of Kt[G], then it lifts to a prime of K [ H ]and the results of the previous section apply. We can then map the information back to Kt[G]. For example we have
Proposition 21.8. If G is polycyclic-by-finite, then the prime length of Kt[G] is at most equal to the plinth length of G.
Proof: By Lemma 21.4(i), let Go be a normal poly-2 subgroup of finite index in G. Since Kt[G] = Kt[Go]*(G/Go), the rings Kt[G] and Kt[Go] have the same prime lengths by Corollary 16.8. F'urthermore, it is easy to see that pl(G) = pl(G0). Thus, replacing G by Go if necessary, we can assume that G is poly-2. Let H and 2 be as in Lemma 21.7 and suppose that h ( 2 ) = n. Then 2 has a series 20 c Z1c - . -C 2, = 2 with 20 torsion and with ZilZi-1 infinite cyclic for i = 1 , 2 , .. . ,n. By Lemma 19.5, G has a normal subgroup of finite index with a plinth series. By lifting this series to H and adjoining it to the above series for 2, we see that pl(H) = pl(G) n.
+
215
21. Polycyclic Crossed Products
Finally let PO C PI C C P k be a chain of primes in Kt[G] and let Pi c Pi C C Pi be its inverse image in K[H]. Note that under the epimorphism K [ H ] Kt[G], we have 2 K'. Thus if Qi denotes the kernel of the homomorphism K[&] -+ K , then Qi is an H-stable prime ideal of K[&]. Hence, since H/Zi is poly-2, Lemma 21.4(ii)(iii) implies that Q!, = Qi K [ H ]is a prime ideal of K [ H ] .We therefore obtain, in K[H], the chain of primes --f
+
of length at least n k. By Theorem 19.6(i), n n pl(G) so k 5 pl(G) as required.
+
--f
+k
5 pl(H)
=
We remark that equality need not occur above. For example (see Exercise 2) there is a simple twisted group algebra Kt[G] with G free abelian of rank n for any n 2 2. Thus pl(G) = n but Kt[G] has prime length 0. We can of course go further with Lemma 21.7 and precisely describe the primes of K t f G ]in general, This is done in [172], but it is too technical to include here. However, one aspect of uniqueness is worth discussing. Notice that the choice of H in Lemma 21.7 is quite arbitrary. Therefore finding a normal subgroup of finite index in G whose inverse image in H is orbitally sound might also appear to be arbitrary. This turns out not to be the case as we see below. The proof of the following result is a simple variant on the proof of Proposition 19.7. It uses the alternate characterizations of the orbitally sound property which are given in Section 19.
Lemma 21.9. Let G be polycylic-by-finite, let Z be a central subgroup of G and assume that G/Z is orbitally sound. Define W a G to be minimal with W 2 Z and G / W an elementary abelian 2-group. Then (G : W (< oc, and any automorphism of G which normalizes Z also normalizes W. i. Let N be a W-orbital subgroup of W. If N contains no nonidentity normal subgroup of G, then IN : N n Z(W)l < 00. ii. W is orbital1.y sound.
216
5. Prime Ideals - The Noetherian Case
ProoJ Since G is finitely generated, its homomorphic images which are elementary abelian 2-groups are of bounded order. It follows that there exists a unique minimal W with G 2 W 2 Z and G / W an elementary abelian 2-group. Thus (G : WI < 00 and the uniqueness of W yields the result on automorphisms. (i) We proceed by induction on h ( N ) . Since f G : WI < co,N is G-orbital so there exists H a G with (G : HI < 00 such that H normalizes N . We may assume that W 2 H 2 2. Since G / Z is orbitally sound and Z N / Z is an orbital subgroup, it follows that there exists C a G with Z N 1C 2 Z and IZN : CI < 00. Replacing C by C n H , we may assume that C H . Furthermore, C = Z ( N n C) and ( N : N n C ( < 00. Thus since N n C is also orbital in W and normalized by H , we can replace N by N n C , if necessary, and assume that C = Z N . Since N contains no nontrivial normal subgroup of G, we have Z f l N = (1). Also H a G and N a H , since N C C H , so we have [C,H] a G and [ C , H ] = [ Z N , H ] = ",HI C N . Thus [C,HI = [ N ,H] = (1) so N is central in H and in particular N is abelian. If h ( N ) 2 2 we can write N = N1 x NZ with h(N1),h(N2) < h ( N ) . Since Nl,N:! are central in H , they are orbital in W . It follows by induction that INi : Ni n Z ( W ) ( < 00 for i = 1 , 2 so ( N : N n Z ( W ) (< 00 as required. It remains to consider h ( N ) = 1. Choose an integer n so that both 2" and N" are torsion free abelian. Then C" = 2" x N n a G so replacing N by N" we can assume that D = T x NaG with T = 2" a torsion free central subgroup and N infinite cyclic. Let us now think of D as being additive, so the finite group G / H acts linearly on D. Furthermore, if Q denotes the field of rational numbers, we can let G / H act on D @ &. Since &[G/H]-modules are completely reducible and T c3 & is a central subspace of codimension 1, we see that G / H can be diagonalized. In fact each element acts like diag(1, 1,. . . ,1,A) where A:G/H -+ &' is a linear character. This implies that A2 = 1 so, by definition of W , we have W E Ker(A) and hence W centralizes D c3 Q. Thus W centralizes N 5 D @ Q and (i) is proved. (ii) Let N C W with N orbital in W . Set M = Ng a G and consider G = G/M. We have G 2 2 = Z M / M so Z is central
ngEG
217
21. Polycyclic Crossed Products
in G. Also G / Z 2 G / Z M is a homomorphic image of G/Z so GI2 is orbitally sound. Since W 2 Z M , it is clear that is minimal in G subject to 2 2 and G/W being an elementary abelian 2-group. Furthermore n,,~fi9 = A?f = (1) so (i) applies and we conclude that : N n Z(W)l < co. If T denotes the complete inverse image of n Z ( w ) in G, then clearly T a W and IN : TI < m. Thus W is orbitally sound. I
w
w
We remark that G above need not be orbitally sound. For example let G = C 1 C2 where C is infinite cyclic and lC2l = 2. If Z = Z(G), then G / Z has Hirsch number 1 and hence is orbitally sound. But G itself is not orbitally sound. Let G be polycyclic-by-finite. By Theorem 19.3, Go = nio(G) is a characteristic orbitally sound subgroup of G of finite index. Let G1 = nio2(G) be the subgroup of G generated by the squares of all elements of nio(G). Then Go/G1 is an elementary abelian 2-group and clearly the largest such homomorphic image of Go.
Proposition 21.10. [172]If G is a polycyclic-by-finite group, then nio2(G) is a characteristic subgroup of finite index. Furthermore let X be any polycyclic-by-finite group such that X / Z 2 G for some central subgroup Z of X . If Y is the complete inverse image of nio2(G) in X, then Y is orbitally sound.
PmuJ Write Go
= nio(G) and GI = nio2(G). Then IG : Go1
< 00
and [Go : GI1 < 00 so GI is clearly a characteristic subgroup of G of finite index. Now let X / Z 2 G and let Xi be the complete inverse image of Gi. Then X o / Z 2 Go is orbitally sound so, by Lemma 21.9 and the definition of GI, we have X I orbitally sound. 1 In view of the work of Sections 14 and 15, a description of the primes in twisted group algebras leads to a similar result in Noetherian crossed products. Thus for example we have
Theorem 21.11. [172]Let RlrG be a crossed product with R right Noetherian and G polycyclic-by-finite of Hirsch number n. If Po c PI C . . . c P,+1 is a chain ofprimes in R*G, then PonR c P,+l nR.
5. Prime Ideals - The Noetherian Case
218
Proof. We may clearly assume that POn R = 0 so that R is G-prime. Suppose by way of contradiction that Pn+l n R = POn R = 0. Then Pi n R = 0 for all i. We proceed with the usual reductions. Case 1. R is G-prime.
Proof: We use the notation and results of Theorem 14.7. Thus Q is a minimal prime of R with stabilizer H and there is a one-toone correspondence between the primes of R*G disjoint from R and certain primes of R*H. Moreover, IG : HI < 00 so h ( H ) = h(G) = n. Now let Li be the prime ideal of R*H corresponding to Pi. Then Li n R = Q and LO c L1 c ... c Ln+l. Furthermore, if - : R * H ( R * H ) / ( Q * H )= ( R / Q ) * H denotes the natural map, then we obtain a chain of primes LOC L1 c - . C with Linfi = 0 and fi prime. ---f
a
Thus we have reduced the problem to the case of prime coefficient rings. I
Case 2. R is prime. Proof. Here we use the notation and results of Theorem 15.8. In particular, we have Gi,,aG and a correspondence between the primes of R*G disjoint from R and the G-prime ideals of a certain twisted group algebra C'[Ginn]. If Pi is the G-prime corresponding to Pi, then we obtain the chain c c c pn+l. In addition it follows easily from Lemma 15.10 that there exists a chain QO C Q1 C - - .C Qn+l of primes in Ct[Gin,] with each Qi minimal over Pi. But pl(Ginn) 5 h(Ginn) 5 h(G) = n, so this contradicts Proposition 21.8 and the result follows. I
PO
We remark that if H a G then we do not have pl(H) 5 pl(G) in general. Thus the last line of the above proof prevents us from replacing n = h(G) by the smaller parameter pl(G). On the other hand, the result is in fact true with n = pl(G). This is proved in [206] using the extension of Theorem 19.9 mentioned at the end of Section 19.
219
21. Polycyclic Crossed Products
EXERCISES 1. Let A = (ul,uZ,. . . , a n ) and B = (b1,bz ,..., b,) be free abelian groups of rank n and set R = K [ A ] . If X is an element of infinite multiplicative order in K', define the K-action of B on R by (ai)*,= ai for i # j and ( u ~ )=~ Aui. z Show that B acts faithfully on R and that R is B-simple. 2. Continuing with the above, suppose (A, p 1 , pz, . . . , pk) is a free abelian subgroup of K'. If C = (c1,cz, . . . , c k ) is a free abelian group, define the action of C on R by ( a i ) c ~= pjai. Show that B x C acts faithfully on R and hence is X-outer since R is commutative. Conclude that the skew group ring R(B x C) is simple and observe that R ( B x C) E K t [ Ax B x C ] . 3. Let 3 be a finite field extension of the rationals Q with (F: Q ) = n and let 0 be the ring of algebraic integers in F. Then 0+ is a free abelian group of rank n and if a is a unit in 0 then, via multiplication, 0 is a module for the cyclic group ( a ) .Prove that 0 is a plinth for ( a )if and only if F = Q[ak]for all integers k 2 1. 4. Continuing with the above, let p be a prime and let a be a root of the polynomial xp+px-l. If F = &[a]show that (F: Q ) = p so that there are no intermediate fields. If 0' = 2 [ a ] where , 2 is the ring of ordinary integers, show that a is a unit of 0' which is not a root of unity. Conclude that the additive group of 0' is a plinth for ( a ) of rank p . Note that 0' & 0 but we do not claim that equality occurs. 5. Let A be a free abelian group of rank n which is a plinth for the infinite cyclic group (x)and let G be the semidirect product G = A >a (x). Show that h(G) = n 1 but that pl(G) = 2. Use the result of the preceding problem to construct such examples. 6. Let I a Kt[G]and define
+
I t = { x E G 1 % - k E I for some k E K } . Prove that I t is a normal subgroup of G. Furthermore show that there exists a homomorphism -: Kt[G] + K t [ G / I t ]where , the latter is some twisted group algebra of the group G / I t , such that I 2 Ker(-) and f t = (1).
220
5. Prime Ideals - The Noetherian Case
7. Let R*G be given with R right Noetherian and G polycyclicby-finite. If h ( G ) = n and R has prime length m, show that R*G has prime length less than ( n+ l)(m + 1). This follows from Theorem 21.11.
22. Jacobson Rings A ring R is said to be a Jacobson ring if, for every prime ideal P of R, the factor ring RIP is semiprimitive. This is of course equivalent to the assertion that every prime (or semiprime) homomorphic image of R is semiprimitive. Thus, for example, the Hilbert Nullstellensatz implies that K [ q ,z2, . . . , xn] is a Jacobson ring. In fact, every finitely generated commutative K-algebra is a Jacobson ring, since this property is clearly inherited by homomorphic images. The obvious noncommutative analog concerns crossed products R*G with G polycyclic-by-finite. Namely we ask whether the Jacobson property for R implies the same for R*G.As we will see, the answer is ((yes” if R is right Noetherian, but “no” in general. In the course of this work we will obviously have to consider J(R*G). Furthermore, suppose P is a prime of R*G and let J / P = J(R*G/P). Since we wish to show that J = P , an incomparability result for J 3 P would be most welcome. This is the approach we follow and we keep the Noetherian hypothesis to a minimum here. It is clear, by induction, that we need only consider G infinite cyclic or finite. We start with the latter case.
Lemma 22.1. Let R & S be an extension ofrings. i. I f S = R @ U where U is a complementary R-submodule of S , then J(S) n R J(R). ii. Suppose S is a finitely generated right R-module and that J ( R ) S SJ(R). Then J(R) J(S). iii. Let S = R ( G ) be a G-graded ring and let H be a subgroup of G. Then J(R(G)) n R ( H ) J(R(H)).
Proof: (i) Suppose r E R has an inverse s E S and write s = T’ + u E R @ U . Then 1 = sr = r‘r
+ ur implies that ur = 0 so u = 0 and
221
22. Jacobson Rings
R. It now follows that J(S)n R is a quasi-regular ideal of R and hence is contained in J(R). sE
(ii) Let V be an irreducible S-module. Then l)~ is cyclic and
SIRis finitely generated so V is a finitely generated R-module. By Nakayama's Lemma, VJ(R) C V. In addition, J(R)S C SJ(R) implies that VJ(R) is an S-submodule of V. Thus since V is irreducible, we have VJ(R) = 0 and hence J(R) J(S). (iii) Here we need only observe that U = R(G \ H ) is a complementary R(H)-submodule of R ( G ) so (i) applies.
Lemma 22.2. (681 Let S = R(G) be a G-graded ring with G finite and let I a R(G). Set R = R / ( Rn I ) C_ S = S / I . i. Iff E R has an inverse in S, then its inverse is in R. ii. J(S) n R = J(R). T E R with f = T + ( R n I ) . We show by inverse induction on )A/ that if A is a subset of G with 1 E A , then there exists s E R(A) with rs s 1 (mod I ) . By hypothesis, s exists if A = G. Now suppose 1 E A C G and that the result holds for all subsets of G of larger size. Choose g E G \ A. Then A U {g} is a larger subset of G and R ( A U { g } )= R ( A ) + R ( g ) .Thus by induction there exists s = s ( A ) s ( g ) E R ( A ) R ( g ) with T S 1 (mod I ) . Furthermore g-'A U { 1) is also a subset of G of larger size so again, by induction, there exists t = t(g-'A) + t(1) E R(g-'A) R(1) with rt = 1 (mod I ) . Now l - r . s ( A ) z r . s ( g )and l - r - t ( l ) = r-t(g-'A) so multiplying these equations yields
PruuJ (i) Let
+
+
+
+
1 - T [ s ( A ) t(l)-s(A) 3
-T
*
t(1)]
r [ s ( g ). T . t(g-'A)] (mod I ) .
But the factors of r in the square brackets both belong to R ( A ) since T E R = R(1) and 1 E A . Thus the inductive statement is proved. The result follows by taking A = (1). (ii) By (i) above, J(S)f l R is a quasi-regular ideal of A and thus J(S) n R C J(R). In the other direction, let V be an irreducible Smodule. Then V is an irreducible S-module and hence a completely reducible R-module, by Proposition 4.10. Thus V is a completely reducible R-module so VJ(R) = 0 and hence J(R) C J ( S ) fl R. I
222
5. Prime Ideals - The Noetherian Case
We remark that the above is false in the case of infinite groups (see Exercise 1). We can now prove
Theorem 22.3. Let S = R(G) be a G-graded ring with G finite. Then R is a Jacobson ring if and only if S is. Proof. Suppose first that R is a Jacobson ring and let P be a prime ideal of S = R(G).By moding out by the graded ideal of S generated by P n R, we may assume that P n R = 0. It follows from Theorem 17.9(i) that R is semiprime and hence semiprimitive. Now define J a S by J / P = J(S/P). Then J n R c J(R) = 0, by Lemma 22.2(ii), and we conclude from Corollary 17.10 that J = P. Conversely suppose that S is a Jacobson ring and let Q be a prime of R. By Theorem 17.9(ii), there exists a prime P of S with Q minimal over P n R. Again we can assume that P n R = 0. Since S / P is semiprimitive, it follows from Lemma 22.2(ii) that R is semiprimitive. Finally we define J’ a R with J’/Q = J(R/Q) and we let N be the intersection of the other minimal primes of R. Then it follows from Theorem 17.9(i) that N # 0 and N n Q = 0. In particular J’N c N is an ideal of R which embeds isomorphically in J’/Q. Hence J‘N is quasi-regular and therefore contained in J(R) = 0. Finally Q 2 J‘N = 0 and Q 2) N , so we conclude that Q _> J‘. I This completes the finite group argument. Now suppose S is a G-graded ring with G infinite cyclic. In this case, we normally think of G as the additive group of integers 2 and say that S is Z-graded. Then S has the structure S = @ Si with SiSj Si+j. As usual, if 0 # s E Si for some i, then s is said to be homogeneous and degs = a. Furthermore if s = si is a nonzero element of S written in terms of its homogeneous components with sm, s, # 0, then we let s+ = s, be the leading term of s and s- = sm be its trailing term. Moreover degs = degs+ = n and we let br(s) = n - rn 2 0 denote the breadth of s. Notice that br(s) = 0 if and only if s is homogeneous. For convenience we set br(0) = -co. The next goal is an incomparability result for such rings.
xiEz c:=,
Lemma 22.4. Let S be a Z-graded ring and assume that S is prime. Let 0 # I a S and let u be an element of minimal breadth in I \ 0. If
223
22. Jacobson Rings
0 # w E I, then there exist 0 # t E S and homogeneous h E S such that tsu = whsu+ for all s E S.
Proof: We first consider u. Note that for any homogeneous j E S the element uju+ - u+ju E I has smaller breadth than u.Thus this expression must be zero and by linearity we have
(*)
for all s E S.
usus. = u+su
Now we study 0 # w E I and we proceed by induction on br(w). Let j be any homogeneous element of S and form wju+ - w+ju = w‘ E I . Since br(w) 2 br(u) and since the term w+ju+ cancels, we see that br(w’) < br(w) and that deg w‘ < deg w d e g j degu. If w’ = 0 for all j , then by linearity wsu+ = w+su holds for all s E S and the result follows with t = W+ and h = 1. Now suppose w‘ # 0 for some j. Then we conclude by induction that there exist 0 # t‘ E S and homogeneous h’ E S with
+
+
for all s E S.
t’su = w’h’su+
(**I
Notice that (*) yields w+j[uh’su+]= w+j[u+h’su]and thus by substituting the definition of w’ into (**) we have
[t’ + w+ju+h’]su = wlju+h’]su+
for all s E S.
Thus the result follows with t = t’+w+ju+h’ and h = ju+h’ provided we show that t # 0. We do this by computing degrees. If w+ju+h’ = 0, then it is clear that t = t’ # 0. Thus we may assume that w+ju+h‘ is not zero so deg(w+ju+ h‘) = deg w
+ deg j + deg u + deg h’.
Since S is prime, there exists a homogenous element s E S with # 0. Using this s in (**) yields
t’+su+
deg t’
+ deg s + deg u = deg ~
S
U
= deg w‘h’su+ <
deg w‘
+ deg h‘ + degs + degu
5. Prime Ideals - The Noetherian Case
224
so deg t' 5 deg w' deg w' yields
+ deg h'. Finally plugging in the upper bound for
deg t' < deg w so clearly t = t'
+ deg j + deg u + deg h' = deg(w+ j u + h')
+ w+ju+h' # 0.
I
We can now prove
Theorem 22.5. [18] Let S be a 2-graded ring and let P be a nonzero prime ideal. If I is an ideal of S properly containing P , then I has a nonzero homogeneous element. In particular, if S is component regular, then I r l So # 0. Proof: Suppose first that S is not prime and let a, b E S \ 0 with aSb = 0. Then clearly a+Sb+ = 0 c P so at least one of a+ or b+ is contained in P . Thus P contains a nonzero homogeneous element and hence so does 1. We can now assume that S is prime and that P itself has no nonzero homogeneous elements. We are given 0 C P C I and hence we choose w to be an element of minimal breadth in P \ 0 and u an element of minimal breadth in I \ P. Since w+ is homogeneous, w+ 4 P so w+Su P and there exists a homogeneous element j f S with w+ju 4 P. It follows that w' = w+ju - wju+ is an element of I \ P and clearly br(w') < max{br(w), br(u)}. But br(u) 5 br(w'), by definition of u, so br(u) < br(w). It follows that u is an element of minimal breadth in I \ 0 so the previous lemma applies to yield appropriate t ,h E S with tsu = whsu+
for all s E S,
Since S is prime, we can find homogeneous a, b E S with t+au+ # 0 and t-bu- # 0. Letting s = a or b and computing degrees as before yields deg t+ deg u+ 5 deg w+ deg h deg u+
+ deg t- + deg u-
2 deg w-
+ + + deg h + deg u+
225
22. Jacobson Rings
+
so br(t) br(u) 5 br(w). Moreover tSu C P and u $- P so t E P and br(t) 2 br(w) by definition of w. We conclude that br(u) = 0 so that u E I is homogeneous as required. Finally if is component regular and u E sk, then 0 # U s - k I n So and the result follows.
s
Next we consider the Jacobson radical of a 2-graded ring. The first result along this line appeared in [2]where the polynomial ring R[z]was studied. Somewhat later, crossed product algebras were considered in [157].The techniques used in those two papers are sufficient to prove
Theorem 22.6. Let S be a 2-graded ring. Then J(S) is a graded ideal and J(S) f l S, is nilpotent for all m # 0.
Proof. We first prove that J(S) is graded. Specifically, the goal is to show that if s = Cisi E J(S),then each si E J(S). We proceed by induction on the size n of the support of s over all choices of s and S. If n = 0 or 1 the result is clear so assume n 2 2. Say sa # 0 and fix S b # 0 with b # a. Choose any rational prime p > Ib-a) and form the ring extension S = S[C]/(l+C+. . .+
Ci
~ = C E ~ S ~ - E ~ ~ S ~ i
i
and this is an element of support size less than n. We conclude by induction that t , = ( 6 , - eb)sa E J(S) so ( 1 - ebPa)s, E J(S) and note that p does not divide b - a by the choice of p . It follows that P-1
psa = n(1-e j ) . sa E J(S) j=1
226
5. Prime Ideals
- The Noetherian Case
so p s , E J(S) f l S = J(S). Finally since this is true for at least two distinct p , we conclude that sa E J(S) as required. For the second part, let um E J(S) n Sm and assume for convenience that m > 0. Then 1 - urn is invertible in S with inverse w = c i Z k u i and say W k # 0. Since the lowest degree term in 1 = (1 - um)u is 1 wk = wk, it follows that b = 0 and wo = 1. Furthermore using u,wi = wi+,, we conclude by induction that w j m = ( u m ) j . Since w j m is eventually zero, the result follows. I
Corollary 22.7. 111) i. Let R*G be a crossed product with G a poly-2 group. Then J(R*G) C J(R)*G. ii. J(R[z;o]) = I0 + I ~ z R [ z ; owhere ] I0 C 1 1 are a-invariant ideals of R. In fact, I0 = 11 r l J(R). Proof. (i) By induction, it suffices to assume that G is infinite cyclic. In that case, by the previous theorem J(R*G) is a graded ideal and hence we have J(R*G) = I*G where I = J(R*G) n R. It follows from Lemma 22.l(iii) that I C J(R). (ii) Again J(R[z;o]) is a graded ideal and hence it is equal to C ~ = o I , x nwith each I, a R. Moreover, it is clear that I, E I,+l. Now let n 2 1 and observe that
(I,zR[z;4) C InznR[z; a]C J(R[z; 4). J(R[z;o]) and I, E 11. We conclude that all Thus I,zR[z;o] I, with n 2 1 are equal. Finally note that I0 2 I1 n J(R) by CT] is a right ideal Lemma 22.1(i). Furthermore, [I1f l J(R)] IlzR[z; of R[z; u] whose image in R[z; o]/J(R[s; 4)is quasi-regular. Thus 2 the image is zero and I1 n J(R) C Io. In addition, (R[z;c~]zI1)~ J(R[z;o]) so zI1 5 I1z and, by symmetry, these sets are equal. Therefore I1 is a-stable and the result follows. I
+
We remark that equality need not occur in (i). Furthermore, in (ii) above it is quite possible to have 10# 11 (see Exercise 5). The next result is a consequence of the division algorithm.
Lemma 22.8. Let R*G begiven with G infinite cyclic, let 0 # IaR*G with I fl R = 0 and let JII = J(R*G/I). Then there exist nonzero G-stable ideals A, B of R with ( J n R)AB C J(R).
227
22. Jacobson Rings
Proof: If G = ( x ) then, by replacing 5F by 2" if necessary, we can assume that R*G = R ( x ) is a skew group ring. As usual, any R-linear combination of those z* with n 2 0 will be called a polynomial. Since I # 0 it is clear that there exists y = r o + r l z + - . .+T,x, E I with TO # 0 and we choose y so that n is minimal. Note that n 2 1 since I f l R = 0. For this n we define A = { T E R I cg + c l x + +c,xn E I and T = co} B = { T E R I co + c l x + ..-+c,xn E I and T = c n } . Then A and B are both nonzero G-stable ideals of R. Suppose q E R(z). We show that there exist integers u,v 2 0 such that for any c E A"B" we have qc p (mod I ) where p is a polynomial of degree less than n. To this end write 77 = eixi with m >_ 0. If a E A, then the x - coefficient ~ of qa is contained in A . Thus there exists a polynomial a: of degree 5 n in I such that
=
xi=-,
i=-m+l
=
or equivalently qa Ci=-m+l eixa (mod I ) . Continuing in this manner, using various a's we get qalaz - a, p (mod I) where p is a polynomial of degree at most max{degq, n - 1). Write p = fixi with t 5 maxtdegq, n - 1). If b E B and t 2 n, then there exists a polynomial p E I of degree 5 n such that 4
=
c;=,
t-1
i=O
or equivalently p b f f:xi (mod I ) . Continuing in this manner, using various b's we get pblbz .. . bt-,+l p (mod I ) where p is a polynomial of degree less than n. Thus this observation is proved withu=max{-degq-,O} and v=max{degq-n+l,O}. Finally let M be any maximal right ideal of R. The goal is to show that ( J f l R)AB 2 M . If J 17R C M this is clear, so assume J n R M . Then R = M + ( J n R ) so 1 = m + j with m E M and j E J n R. By definition of J , 1 - j = m is invertible modulo ~~~~
=
5. Rime Ideals - The Noetherian Case
228
I, say m 9 EE 1 (mod I ) . We apply the observation of the preceding paragraph to 7. Thus there exist integers u,v 2 0 such that for any c E A"B" we have 9c E p (mod I ) where p is a polynomial of degree less than n. Thus c = m9c
= rnp
(mod I ) .
But mp- c is a polynomial of degree < n since n 2 1so, by definition of n, we have mp - c = 0. By considering the identity coefficient in this expression, we conclude that c E mR E M . In other words, we have shown that AuB" M . But the largest two-sided ideal of R contained in M is a primitive and hence a prime ideal. Thus we conclude that either A or B is contained in M and hence ( J f l R)AB & M as required. Since this holds for all such M , the result follows. I
It is now a simple matter to prove
Theorem 22.9. [ 6 5 ] Let R*G be a crossed product with G infinite cyclic and assume that every G-prime ideal of R is a semiprimitive ideal. Then R*G is a Jacobson ring.
Proof: Let P be a prime ideal of R*G. Moding out by ( P n R)*G if necessary, we can assume that P n R = 0. Thus R is a G-prime ring and it is semiprimitive by hypothesis. Let J a R*G with J / P = J(R*G/P). The goal is to show that J = P. By Corollary 22.7(i), R*G is semiprimitive, so the result follows if P = 0. On the other hand, if P # 0, then by Lemma 22.8 there exist nonzero G-stable ideals A, B of R with ( J n R)AB C J(R) = 0. But R is G-prime and A , B # 0, so this implies that J n R = 0. Theorem 22.5 now yields the result. 4
Corollary 22.10. [65] Let R*G be a crossed product with R right Noetherian and G polycyclic-by-finite. If R is a Jacobson ring, then so is R*G.
Proof: By induction it suffices to assume that G is either infinite cyclic or finite. If G is finite, then Theorem 22.3 yields the result. If
22. Jacobson Rings
229
G is infinite cyclic and P is a prime ideal of R*G then, since R is right Noetherian, Lemma 14.2(i) implies that P n R is a semiprime ideal of R. Since R is a Jacobson ring, P n R is therefore a semiprimitive ideal and the result follows from Theorem 22.9(i). We remark that certain analogs of these results were known earlier. In particular, it was shown in [66,96] that if R is a commutative Jacobson ring then so is R[z].Furthermore, [185]considered R[G] with R Noetherian and G polycyclic-by-finite. We close this section by sketching three examples of interest. First, it is quite possible for R*G to be a Jacobson ring even though R is not. For example, let R = K [ [ ( ]be ] a power series ring over the field K and let X be an element of infinite multiplicative order in K'. Define the action of the infinite cyclic group G = (9)on R via the obvious extension of Cg = A<. Note that R is an integral domain with J(R) = ( R # 0 and thus R is not a Jacobson ring. On the other hand, let P be a prime ideal of RG and let J / P = J(R/P). If P = 0, then J = 0 by Theorem 22.6 since RG is obviously a domain. On the other hand, if P # 0, then P n R # 0 by Lemma 10.3(i) and Corollary 12.6. Thus P n R = ("R for some n 2 1. But then ((RG)n C P so ( R G 2 P and P corresponds to a prime of the Jacobson ring RG/
5. Prime Ideals - The Noetherian Case
230
since YoRG # 0 is nil. To see this, let y = CFkrigi f RG with k 2 2. Then (@-,ij~y)~ is an RG-linear combination of monomials of the form
(y0)”1
(yo>g”Z
...(yo)”k
- Yn,Yn, .
* *
Ynk
withnj = d l + d 2 + - a . + d j , d l = Oand ldjl 5 k. But ynlynz E I since this monomial has k factors with subscripts at most k2 apart. ~ 0 and we conclude that RG is not a Jacobson ring. Thus ( 9 0 7 )= Finally it is clear that the prime radical and the Jacobson radical of any Jacobson ring coincide. In particular, this implies that the Jacobson radical of such a ring is nil. It was thought for awhile that every finitely generated K-algebra S also had J(S) nil. Indeed this wits shown to be true in [3]if K is nondenumerable. However the result is false in general ([12]).To see this, let K be a countable field and let D be the polynomial ring K[y]localized at the prime (9). Then D is a countable local domain and J(D) = yD # 0. Now let T= Di be the complete direct product of copies of D indexed by the integers and let ei be the idempotent in T with 1 in the ith entry and 0’s elsewhere. Furthermore, since D is countable, we can write D = { di I i E 2 } and define S E T to have ithcomponent equal to di. As usual T admits the shift operator g and we let R be the finitely generated subalgebra (or the finitely generated subring) of the skew group ring T ( g )generated by 1,eo, 6,g and 9-l. Note that . e . y n k
ni,,
e&eo 5 eoT(g)eo = eoT = eoD and that eoSgieo = eodi. Thus eoReo = eoD 2 D. Since JD 2 J(eoRe0) E J(R) and J(D) is not nil, we conclude that J(R) is not nil. We remark that R is in fact the skew group ring ( Rn T ) ( g ) .
EXERCISES 1. Let R be a commutative integral domain which is not a field ] the group ring of the infinite cyclic group (x) and let S = R [ ( z )be over R. If 0 # a is a nonunit of R, let P be the ideal of S generated by x - a. Show that P is prime and that ti = a P is invertible in SIP but not in R S R.
+
231
22. Jacobson Rings
2. Let G be a group. A ring S is called a G-system if S = CgEG S, is a sum, not necessarily direct, of additive subgroups satisfying SzSy 2 Szy.Show that S is a G-system if and only if it is the natural homomorphic image of a G-graded ring. 3. Let S be a 2-graded ring. If u , w are nonzero elements of S and j is homogeneous, prove that
br(u+jw - ujw+)
< max{br(u), br(w)}.
Furthermore if S is prime and asb = csd holds for all s E S , prove that br(a) br(b) = br(c) br(d). 4. Let S = R(G) be given with G a finitely generated free abelian group. Assume that S is an algebra over the infinite field K C R. Prove that every characteristic ideal of S is graded. 5. Suppose S = K[yi I i E 21 is the polynomial ring over K in infinitely many variables subject to the relations yigj = 0 for i # j and consider the skew polynomial ring S[z;o] where a is the shift operator. Show that J(S[z;a])= I1zS[z;a]where 11 = CiyiS. 6. Let P be a prime ideal of R[z;a]with z $! P. Prove that P n R is a-stable and then a-prime. For this, note that (R[z; a]~)(PrlR)~ P. Similarly, if
+
A = {T E R
+
I a0 + alz +
+ anzn E P and
T
=a ~ }
prove that A is a a-stable ideal of R and that A # 0 if P # 0. 7. Apply the ideas of the preceding problem to prove the following result of [65]. Let R[z;a]be a skew polynomial ring with R a right Noetherian Jacobson ring. Then R[z;a]is a Jacobson ring. This also uses the fact that every essential ideal of a semiprime right Noetherian ring contains a regular element. 8. Let S be a finitely generated algebra over the nondenumerable field K . If s E J S , use the fact that the uncountably many elements (1 - ks)-' with k E K are K-linearly dependent to conclude that s is algebraic over K . Then show that s is nilpotent so that J(S) is a nil ideal. This is the argument of [3].
232
5. Prime Ideals - The Noetherian Case
23. P. I. Algebras We close this chapter by completing earlier work on two topics of interest, namely incomparability and crossed products satisfying a polynomial identity. We start with incomparability and here the goal is to extend Theorem 22.2 to more general polycyclic-by-finite groups. To proceed, we first need an operator version of the latter result. This could be proved directly by suitably modifying the original argument. Instead we take the opportunity to exhibit a trick first used in [166]. Let I' act as automorphisms on the Z-graded ring S. We say that I' respects the grading if each component Si is I?-stable.
Corollary 23.1. 1301 Let S be a Z-graded ring and let I' be a group of operators on S which respects the grading. Suppose I is a rstable ideal of S which properly contains the nonzero I?-prime ideal P . Then I has a nonzero homogeneous element. In particular, if S is component regular then I n SO# 0.
PruuJ If f' is a free group which maps onto r, then the action of I' on S lifts to a corresponding action of f' clearly satisfying the same hypotheses. Thus without loss of generality we may assume that = is a free group and hence an ordered group (see [161, Corollary 13.2.81). Form the skew group ring SI'. Since each Si is r-stable, it follows that SI' is also Z-graded with (SF), = (&)I?. Furthermore 0 c PI' c IF and PI? is a prime ideal of Sr by Lemma 21.4(iii). We can now conclude from Theorem 22.2 that II' contains a nonzero homogeneous element of SI' and, by considering the r coefficients of this element, we see that I contains a nonzero homogeneous element of s. The following is the best crossed product incomparability result to date without a Noetherian assumption. Recall that a finitely generated nilpotent group G is necessarily polycyclic-by-finite. Furthermore Z(G) is finite if and only if G is finite (see [161, Lemma 11.4.31 with N = G).
23. P. I. Algebras
233
Theorem 23.2. [30]Let R*G be a crossed product with G a finitely generated nilpotent group of Hirsch number n. If PO C PI c . C P, is a chain of primes of R*G with m = 2n, then POn R c P, n R. Proof: Replacing R*G by R*G/(Po n R)*G, if necessary, we may clearly assume that POn R = 0. We proceed by induction on n. If n = 0 then G is finite and Theorem 16.6(iii) yields the result. Now suppose n > 0 so that, by the above comment, Z(G) is infinite and hence contains an infinite cyclic subgroup 2. Then Z a G, R*G = (R*Z)*(G/Z)and fi(G/Z) = n - 1 so induction implies that
Po n ( R * Z )c Pm/2 n (R*Z) c Pm n (R*Z). Observe that S = R*Z is a 2-graded ring and, since Z Z(G), the group 6 of trivial units of R*G acts on S and respects the grading. Furthermore, each P i n s is a 6-prime ideal of S by Lemma 14.l(i). In particular P,12 f l S is a nonzero 5;-prime ideal and P, f l S is properly larger. Thus, by the previous corollary, Pm n R # 0 = Po n R and the result follows. 1 Finally we transfer this back to group-graded rings via duality.
Corollary 23.3. [31] Let S = R(G) be a group-graded ring with G a finitely generated nilpotent group o f Hirsch number n. I f Po c PI c . - c P, is a chain ofprimes of S with m = 2n, then CxEG Pons, C C x E G p7n n sz.
Proo$ By moding out by the graded ideal CzEG POf l S, if necessary, we can assume that POcontains no nonzero homogeneous elements. Now, by Proposition 2.8, there is a chain of primes PA C Pi C . . C Ph of the skew group ring S(1)G where P,! = MG(P,) n S(1)G. It therefore follows from the previous result that Ph n S(1) # 0. But Ph C MG(P,) and the entries in the matrices of S(1) are all homogeneous elements of S . Thus CxEG Pm n S, # 0. 1 This completes our discussion of incomparability. Now we consider crossed products satisfying a polynomial identity. Recall that a
5. Prime Ideals - The Noetherian Case
234
K-algebra S satisfies a polynomial identity of degree n if there exists a polynomial f(C1, ( 2 , . . . ,(k) E K(C1,( 2 , . . .>of degree n such that f ( s l , s 2 , .. . , s k ) = 0 for all si E S. It is a standard fact that f can be taken to be multilinear, that is of the form
with a, E K and a1 # 0. Group algebras satisfying a polynomial identity have been completely classified (see [161,Chapter 51) and p. i. crossed products are at least reasonably well understood. The latter study began with the work of [71]on skew group rings. More generally [209]considered crossed products with a central twisting. The approach we take here avoids any concern with the nature of the twisting. We start by quoting an omnibus result from the theory of p. i. algebras. We refer the reader to the books [191]or [161]for complete details.
Theorem 23.4. Let S be a prime K-algebra satisfying a polynomial identity of degree n and let Z = Z(S). i. The central localization SZ-I is a simple ring with center F = 22-l. ii. dimF SZ-I = m2 where m 5 [n/2] and S2-l C M,(T) for some extension field T of F . iii. Q s ( S )= SZ-I and cr E Aut(S) is X-inner if and only if Q is trivial on 2.
Proof: (i) This seems to be due to a number of people independently. See for example [191,Theorem 1.7.91 or [161,Theorem 5.4.101. (ii) By taking a multilinear identity, it follows that SZ-' also satisfies a polynomial identity of degree n. But SZ-' is simple and hence primitive so [191,Theorem 1.5.161 or [161,Theorem 5.3.41 yields the result. (iii) By Lemma 10.9(iii) we have Q s ( S )2 S2-l. Conversely let q E Qs(S)and let 0 # A a S with qA C S.By [191,Theorem 1.5.331 or [161,Theorem 5.4.91 there exists 0 # z E A n 2. But then qz = s E so q = sz-l E sz-l.
s
235
23. P. I. Algebras
Finally if u is X-inner on S , then u acts trivially on Z(Q,(S)) and hence on 2. Conversely if u is trivial on 2, then u extends to an automorphism of the finite dimensional simple ring SZ-' which is trivial on its center F = 22-l. The Skolem-Noether Theorem ([73, Theorem 4.3.11) implies that u is inner on SZ-' and hence X-inner on S . I The next lemma is the heart of the matter
Lemma 23.5. Let R*G be given with R a G-prime ring. Furthermore, suppose R*G satisfies a polynomial identity of degree n over the field K
R and set m = [n/2]. i. There exist k 5 m minimal primes Q1, Q 2 , . . . ,Q k of R. These are G-conjugate and satisfy Qi= 0. ii. If Q is any minimal prime of R, there exists a subgroup H of G with IG : HI 5 m2 such that H stabilizes Q and acts as X-inner automorphisms on the prime ring R/Q.
ni
Prm$ By Lemma 14.l(ii) there exists a prime P of R*G with P n R = 0 and we let 7 : R*G t R*G/P = S be the natural homomorphism. Then S is a prime algebra satisfying a polynomial identity of degree n. If Z = Z(S), then the previous theorem implies that SZ-' is a simple algebra with center F = 22-' and dimF SZ-' I m2. It follows that SZ-' = Mj(D) where D is an F-division algebra and j I m. Note that S is generated by RV = R / ( P n R ) E R and G V where G is the group of trivial units of R*G. Furthermore, G V normalizes both R V and F. It follows that RVF is a Gq-prime ring. Indeed if A and B are GV-stable ideals of RqF with A B = 0, then AG? and BGq are ideals of SZ-' with product zero. Thus since SZ-I is simple, one of these factors must be zero. Since RQF is finite dimensional over F , we conclude that RVF is semisimple and that G V transitively permutes its simple summands. If ei denotes the centrally primitive idempotent correspondk ing to the ith summand, then from 1 = CiZ1eiand the structure of S = Mj(D), it follows that k I j 5 m. Furthermore, since
5. Prime Ideals - The Noetherian Case
236
the simple summands eiRqF are all F-isomorphic, dimFeiRqF 5 (dimF S)/k5 m 2 / k . Let La = (1 - ei)RqF be the prime ideal of RqF corresponding to the ith summand. Since F is central, it follows that Li = L: n R” is a prime of RQ. Moreover, Gq permutes these primes transitively Li = r),(L: fl Rq) = 0. We conclude that L1, L2, . . . ,Lk are and precisely the minimal primes of Rq N= R and (i) is proved. Note that we do not claim that all Li are distinct. Thus there may be less than k such minimal prime ideals of R. Finally let Q1, Q2,. . . , Qk be the minimal primes of R with QI = Li. If Gi is the stabilizer of Qi is 6 , then 1 9 : Gil 5 k and 6; acts on the simple ring eiRqF. Note that eiRqF is generated by R, = eiRq E R/Qi and the central subfield Fi = e i F F and that dimFi &Fi 5 m2/k. Thus if Zi denotes the center of &, then ZiFi is a central subfield of &Fi. Moreover G: acts as field automorphisms on ZiFi fixing Fi. We conclude from the Galois theory of fields that
ni
IG? : Cq(Zi)J5 dimFi ZiFi 5 m 2 / k . In particular, if 9 i i denotes the complete inverse image of CG:( Z i ) in G, then
Furthermore ‘Hi acts on R/Qi fixing its center. Theorem 23.4(iii) now implies that ‘Hi is X-inner on R/Qi and (ii) follows with Hi =
WW). I We now require some fairly simple observations.
Lemma 23.6. Let R*G be given. i. R is G-semiprime if and only if the G-prime ideals of R intersect to zero. ii. If R*G satisfies a polynomial identity, then R is G-semiprime if and only if it is semiprime. iii. Let H be a subgroup of G of finite index k and let I be an H-stable ideal of R. Then there is a homomorphism from R*G into Mk(R*H/I*H) with kernel I”)*G.
(n,
237
23. P.I. Algebras
Proofi Part (i) is standard (see Exercise 2 or 3) and part (ii) follows from (i) and the fact that every G-prime ideal is semiprime by Lemma 23.5. We consider (iii) and for this we may clearly assume that n z E G I" = 0. Let { 1 = X I ,22,.. ., x k } be a right transversal for H in G. Then V = R*G is a free left R*H-module with basis { 1 = Z1,32,. . . ,Zk }. Furthermore V is a faithful right R*G-module. Since right and left multiplication commute as operators, it follows that R*G is isomorphic to a subring of EndR,H(V) Mk(R*H). To be precise, let a E R*G and for each i write Z i a = ai,j?j with ai,j E R*H. Then the map Q I+ [ai,j]E Mk(R*H) is the appropriate embedding. Finally consider the composite map
'&
If Q E Ker(O), then since 31 = 1 we have
We can now offer the necessary and sufficient conditions for R*G to satisfy a polynomial identity. As is apparent, the result is quite tedious to state and not particularly useful; but it is the correct answer. Note that it seems preferable here to use a lower case q to denote a prime ideal.
Theorem 23.7. [71][209] Let R*G be a crossed product with R a G-semiprime ring. Assume that R*G is a K-algebra with K & R and that R satisfies a polynomial identity over K . If q is a prime of R, write S, = Qs(R/q) and let C, = Z(S,) be the extended centroid of R/q. Note that these are all K-algebras. Then R*G satisfies a polynomial identity over K if and only if there exists a constant k and a multilinear polynomial f(G, 5 2 , . . .) E K(C1,&, .. .) such that for each minimal prime q of R there is a subgroup H , of G with i. IG : H,[ 5 k and H, stabilizes q, ii. R*H,/q*H, S,*H, = S, @ Ci[H,] where Ci[H,] is a twisted group algebra, iii. Ci[[H,] satisfies the polynomial identity f .
5. Prime Ideals - The Noetherian Case
238
Proof. Suppose first that R*G satisfies a polynomial identity f of degree n which we may assume to be multilinear. Let q be a minimal prime of R. By replacing R by R / ( n Z E G 4”) if necessary, we can clearly assume that nzEGqz = 0 so that R is G-prime. By Lemma 23.5(ii), there exists a subgroup H , of G with IG : If,[ 5 [n/2I2such that Hq stabilizes q and acts as X-inner automorphisms on the prime ring R / q . We conclude from Proposition 12.4(i)(ii)(iii) that R*H,/q*H, & S,*H, = S, €4 C;[H,]. Since each Ci[H,] satisfies the polynomial identity f , the result follows in this direction. Conversely suppose k, f and the subgroups Hq are given and write m, = 1G : H,l. Then for each minimal prime q, it follows from (i), (ii) and Lemma 23.6(iii) that we have a homomorphism
8,: R*G +M,,
(R*H,/q*H,) C Mk(R*H,/q*H,)
c M k ( S , c?J C p q I ) . Note that this combined map need nut send the element 1 to 1. By Lemma 23.6(iii) we have Ker(8,) = q”)*G. Furthermore, since R satisfies a polynomial identity of degree n, so does S, and indeed, by Theorem 23.4(ii)(iii), S, C Mm/q(F,) for some field F, 2 C,. Thus
(n,
s, @ qH,1 c M[n/2](F, @ C p q I ) and 8, extends to a map
8,: R*G
--f
@ Ci[H,])
M,(F,
where rn = k [ n / 2 ] . Finally by combining all of these maps, we obtain a ring homomorphism (not taking 1 to 1)
8: R*G
+
M,
(n
F, €4 Ci[H,])
P
with
23. P. I. Algebras
239
since R is semiprime by Lemma 23.6(i). Moreover, by hypothesis, each Ci[H,] satisfies the multilinear polynomial identity f and hence so does n , F , ~3Ci[H,]. It now follows from Theorem 9.1 with E = Mm(K) that the m x m matrix ring over this ring also satisfies a polynomial identity and hence, since Ker(0) = 0, so does R*G. I Thus, as with other crossed product problems, the question of when R*G satisfies a polynomial identity is reduced to the case of twisted group algebras. Here the result is a simple analog of the ordinary group algebra situation, so we just state it without proof. In fact, given Corollary 9.9, the proof is not a t all difficult and we discuss some aspects of it in the exercises.
Theorem 23.8. [159]Let Kt[G]be a twisted group algebra over the field K . i. If Kt[G] satisfies a polynomial identity of degree n over K , then G has normal subgroups A C B with JAJand IG : BI bounded by functions of n and with Kt[B]/J(Kt[A])Kt[B]commutative. ii. Conversely suppose G has normal subgroups A C B with IAJ = a < 00 and IG : B ( = b < 00 and with Kt[B]/J(Kt[A])Kt[B] commutative. Then Kt[G]satisfies ( ~ 2 b where ) ~ S2b is the standard identity of degree 2b. Thus we see that this result meshes precisely with Theorem 23.7 to settle the polynomial identity problem for R*G with R a Gsemiprime ring.
EXERCISES 1. Let G be an infinite elementary abelian 2-group with generators z i , y i for i = 1 , 2 , 3 , .. .. Define the twisted group algebra R = Kt[G] so that Zi& = -&Zi for all i but that all other generators commute. If charK # 2 show that K'[[(zl,yl)]"= M2(K) and hence that R E Mz(R). 2. Let R*G be given. Prove that R is G-semiprime if and only if the G-prime ideals of R intersect to zero. To this end suppose R is Gsemiprime and let O # T E R. Define a sequence T = { TO, T I , ~ 2 . ., . }
5. Prime Ideals - The Noetherian Case
240
of nonzero elements of R inductively by T O = T and T,+I E T,RT,G~ for some gn E G. If P a R is maximal subject to being G-stable and disjoint from T , show that P is G-prime. 3. Give an alternate proof of the above using the same trick as in Corollary 23.1. Namely, reduce to the case where G is a free group and use the fact that if R is a G-semiprime ring, then R*G is semiprime. 4, Suppose R is prime, G is X-inner on R and R*G satisfies the multilinear polynomial identity f. If S = Q,(R) and C = Z(S), prove that C is central in S*G and that S*G = C (R*G). Deduce that S*G also satisifies f. This is used in the proof of Theorem 23.7. 5 . Let Kt[G] and A 2 B be given as in Theorem 23.8 with \ A \ = a and IG : BI = b. Show that Kt[G]/J(Kt[A])Kt[G] is isomorphic to a subalgebra of Mb(Kt[B]/J(Kt[A])Kt[B]). Since the coefficient ring of this matrix ring is assumed to be commutative and since J(Kt[A])Kt[G]is nilpotent of degree 5 a, conclude that Kt[G] satisfies ( S Z b ) a . 6. Let Kt[G] be given with G an abelian group and define W = { 2 E G I if E Z(Kt[G])}. If I is any ideal of Kt[G], prove that I = ( I nKt[W]) . Kt[G] and that Kt[W] maps onto the center of Kt[G]/I. In particular, if T is a transversal for W in G, conclude that (T I ) / ] is linearly independent over Z(Kt[G]/I). 7 . Let Kt[G] and W be as above. If Kt[G] satisfies a polynomial identity of degree n, show that IG : W1 5 [n/2I2. For this, take I to be a maximal ideal of Kt[G] in the preceding exercise. How close does this come to proving Theorem 23.8?
+
6
Group Actions and Fixed Rings
24. Fixed Points and Traces Let G act on the ring R so that we have a group homomorphism G -+ Aut(R). Then we can form the skew group ring RG which is an associative ring containing R, G and the fixed ring
RG = { r E Rl rg = T for a l l g E G}. In other words, RG contains all the ingredients of the Galois theory of rings. Thus we might hope to use crossed product results to obtain Galois theoretic information. Indeed there are such applications, but not an overwhelming amount for a number of reasons. First, it is necessary to assume that G is finite. Only certain classes of infinite groups are allowed in the theory, and their description does not readily translate into crossed product terms. Second, the structure of RG is best understood when R has no JGJ-torsion. Thus the route through crossed products usually requires this assumption. Third, certain natural Galois theory questions do not translate into natural questions about RG and vice versa.
241
242
6. Group Actions and Fixed Rings
In this chapter we consider skew group ring applications to the Galois theory of rings. As indicated above, this represents only a small part of the theory. Thus the material here is necessarily very selective. For many more details the reader should consult the books [129]and [92]and the survey paper [60]. We start by considering the existence of fixed points, that is elements of RG. If G acts on R,then obviously 1 E R is a fixed point so nonzero fixed points exist. On the other hand, if I # 0 is a G-stable (right) ideal of R, then it is not so obvious that nonzero fixed points exist in I . For example, let S = K(z,y) be the free algebra on z,y over the field K of characteristic p > 2, let R = Mz(S) and let G be the group of units of R generated by the matrices
Then (GI = 2p2 and G acts on R by conjugation so that RG = CR(G). To compute the latter, first note that the centralizer of
(i1:)
is precisely the set of diagonal matrices since char K
# 2.
It then follows easily that RG = K , embedded as scalars. But there is a natural homomorphism S -+ K obtained by mapping z, y H 0 and this extends to a ring homomorphism R -+ Mz ( K )with kernel I # 0. Since I a R, we see that I is G-invariant but I n RG = I n K = 0. We note three properties of this example. (1) R is a prime ring but not a domain (and hence has nontrivial nilpotent elements). (2) G is inner on R. (3) R has JGJ-torsion. As we will see in this and later sections, each of these properties is in some sense a necessary ingredient of the example. One way to construct fixed points is with a truce map. If G is finite and acts on R, define trG: R + R by
Then clearly trG(T) E RG and if T is a fixed point then tTG(T) = (Glr. Thus we have
243
24. Fixed Points and Traces
Lemma 24.1. If G is finite, then trG is an RG-bimodule homomorphism from R to RG. Moreover, (GI . RG C_ trG(R) a RG. Notice that if 1 a R, then I is itself a ring (without 1). Thus, in this section and the next, it is convenient to allow these more general rings. Of course a ring without 1 is merely an ideal in a ring with 1. Indeed if R is any ring (without l),we let R# = Re2 be its natural extension to a ring with 1. The main result is
Theorem 24.2. [22]Let G be a finite group of automorphisms of the ring R (without 1) and set n = (GI. If trG(R) = 0, then nR is nilpotent of degree 5 2n2.
Proox It is convenient, although not necessary, to formulate this argument in terms of the skew group ring RG. Notice that R is naturally embedded in this ring but that, since R need not have a 1, there is no natural embedding of G. As in the case of ordinary group rings, there is an augmentation map p : RG -+ R given by p(CSEGrgg)= CgEGrg. However here this map is certainly not a ring homomorphism; rather it is just a left R-module homomorphism. Define A = { a E RG 1 p(aR) = O } .
Then A is clearly both a right and left R-submodule of RG. Observe that if 1 E R, then it follows immediately that p(A) = 0. On the other hand, even if R does not contain a 1, we can still show that p ( A ) is nilpotent. To this end, recall that n = IGI and, for each integer m 5 n, define Am to be the linear span of all elements of A of support size 5 m. Then clearly
A = A, 2 A,-
1
2 .. 2 *
A1
2 A0
=0
and these are both right and left R-submodules of RG. Suppose X = { zl, x2,.. . , x, } is a subset of G of size m and let a , ,B E Am have support in X so that a = C;l"aizi and ,O = bixi for suitable a { ,bi E R. For each subscript Ic we observe that
x;l"
m
a&
- a ( b / J X k = x ( a k b i - ai(bk)x , x , l )xi E Am-1 i= 1
244
6. Group Actions and Fixed Rings
since the i = k summand vanishes. Thus
But p ( a ( b k ) " l ) = 0 since Q! E A so akp(P) = p(akP) E p(Am-l). Summing over k = 1,2,. . .,m, we obtain p(a)p(P) E p(Am-l). Now let us consider ( P ( A ~ ) R # ) ~This " . subset of R is clearly spanned by all products of the form
with 0 # ai E Am and ri E R#. Since there are less than 2n nonempty subsets of G of size 5 rn, it follows that in this expression at least two of the ai terms must have identical support. Thus for some subset X of G of size m we see that
with a', p' both having support in X and with T ; , T ; , T; E R#. Because y = p ( ~ ~ a ' ) p ( @ ' ) r = ~ p(a)p(P)r&,we conclude from the result of the preceding paragraph that y E p(Am-l)R#. In other words, we have shown that
It follows that if t = (2n)n then
Hence p(A)R# is nilpotent. But p(A) C p(A)R# so P ( A )=~ 0 . ag E RG. If Finally, let a E R and form the element a = CSEG T E R, then we have
by assumption. Since p ( m ) = 0 for all T E R, it follows that a E A . In particular, since p ( a ) = IGlu = nu, we have p(A) 2 nR. Thus (nR)t= 0 and the result follows. I
24. Fixed Points and Traces
245
We note that the nilpotence bound given above is not sharp. In fact, a closer look at the proof shows that it could be improved [ 11. However this is presumably also much too slightly to large since, for G solvable, the bound is n. As we will see, there are other proofs of this result, but none of them yield better information on the bound.
nz=,(G)+
Lemma 24.3. Let the finite group G act on R. i. If H a G, then G I H acts on RH. Moreover RG = (RH)G/H and trG = tTG/H 0 trH. ii. Suppose p is a prime, pR = 0 and G is a p-group. If R # 0 then RG # 0.
ProoJ: (i) Most of this is obvious. For the last part, let T be a transversal for H in G. If T E R, then
since HT = G. (ii) In view of (i), it suffices to assume that G = (x) is cyclic of order p. In this case, we forget the multiplicative structure of R and just view R as a module for the group algebra GF(p)[G]. Since z p = 1, we have (1 - x)” = 0 so I - IC is a nilpotent linear transformation on R # 0. It follows that RG = anna(1 - x) # 0.
A finite group G is said to be p-ndpotent for the prime p if G has a normal p’-subgroup H with G / H a pgroup. In particular, any pgroup is pnilpotent.
Corollary 24.4. [22] Let G be a finite group of automorphisms of the ring R (without 1) and assume that RG = 0. If either R has no IGI-torsion or G is p-nilpotent and pR = 0, then Rk = 0 where k = 2nZ and n = IGI.
246
6. Group Actions and Fixed Rings
Pro05 We know that (nR)k= 0. Thus if R has no n-torsion, then
Rk = 0. Now suppose pR
= 0 and let H be a normal p’-subgroup
of G with G / H a pgroup. If RH # 0 then, since G / H is a pgroup, # 0, a it follows from the preceding lemma that RG = ( R H ) G / H contradiction. Thus RH = 0 and hence, since R has no (HI-torsion, we conclude that Re = 0 where C = 2m2 and rn = 1HJ5 n. I We consider two examples of interest. The first shows that we can have R # 0 and RG = 0 even when R has no JGI-torsion. To this end, let K be a field and let R be the ring of strictly upper triangular n x n matrices over K . Suppose in addition that K contains n distinct nth roots of unity, say €1,€ 2 , . . . ,E,. This implies that one of these elements has order n and also that the characteristic of K does not divide n. Let G be the cyclic group of units of Mn(K) generated by the diagonal matrix g = diag(e1, € 2 , . . . ,en). Then IGI = n, G acts on M,(K) by conjugation and Mn(K)G is the ring of diagonal matrices. Furthermore, G acts on R and RG = 0. Note that R is nilpotent of degree n = IGJ. The second example, from [22], is a partial converse to the preceding corollary. We state it as
Lemma 24.5. Let K be a field of characteristic p > 0 and let G be a finite group which is not p-nilpotent. Then there exists a finite dimensional, nonnilpotent K-algebra R (without 1 ) on which G acts with RG = 0.
ProoJ Let H be the normal subgroup of G generated by all its p’elements, Then G / H is a pgroup so that, by assumption, p IHI
I
and therefore T = ChEH h is contained in the augmentation ideal w(K[H]). Let R be the right K[G]-module R = K[G]/TK[G]. We show first that RG = 0. To this end, let a E K[G] correspond to an element in the fixed submodule RG. Then for all h E H we have a(h - 1) E TK[G] and hence a ( h - 1)2 = 0 since T is central in K[G] and ~ ( -h1) = 0. Furthermore charK = p 2 2 so a(hP - 1) = a ( h - 1)”= 0. Now H is generated by p’-elements and thus H = ( h p I h E H ) . We conclude that a is contained in the left
247
24. Fixed Points and Traces
annihilator of w ( K [ H ] so ) a E T K [ G ]and a corresponds to the zero element of R. Now let p : K[G]+ K denote the usual augmentation map with kernel w(K[G]).Since 7 E w ( K [ G ] )we , see that p gives rise to a nonzero linear functional f : R + K . Furthermore for all a E R and z E G we have f ( a ~=) f ( a ) . We now define a ring structure on R by setting a x /3 = a f ( P ) . It is trivial to verify that this is distributive with respect to addition and it is associative since
Thus R is a ring and in fact a finite-dimensional K-algebra (without 1). Moreover P E R is a right identity if and only if f(P) = 1. Thus R has numerous right identities and is therefore not nilpotent. Finally because f(Pz)= f(0)we see that G acts on R as ring automorphisms and the result follows. I
As we indicated earlier, the nilpotence bound in Theorem 24.2 can be appreciably sharpened in case G is solvable. This is based on the following important observation.
Lemma 24.6. Let G be an abelian group of period n and let K be a field containing a primitive nth root of unity. Suppose G acts as K-automorphisms on the K-algebra R (without 1). For each linear character X E G = Hom(G, K') define Rx = { r
I
E R r9 = X(g)r
for all g E G }
In particular, R1 = RG. Then R = @ xxEe Rx is a G-graded ring. Specifically, if r E R, then its X-component rx is given by
Proof: It is clear that each Rx is an additive subgroup of G and, since G acts as K-automorphisms, we have easily RxR, C Rxp. The goal is to show that R = @ ExEdRx. This is based on the fact that if
248
6. Group Actions and Fixed Rings
1 # X E G then x g E G X ( g ) = 0 and the dual fact that 1 # g E G yields CAE& X ( g ) = 0. Let T E R and define T A as above. If z E G, then
so r~ E RA.Moreover
Finally if r E R,, then T A = 0 for X # 1.1 and r p = T . This implies easily that the sum R = CAE- RAis direct and the result follows. I Of course, not every ring is a K-algebra, but sometimes we can work with a generic model. Let G be a group, let A be an integral domain and form the free A-algebra
where I is some index set. Then G acts as A-automorphisms on this ring by permuting the variables in regular orbits. Specifically ((i,g)"= (i,gz for all z,g E G. Write
for the augmentation ideal of T , namely the ideal generated by all < i , g . Then T' is a A-algebra (without 1) and it is G-stable. Now let A = 2 be the ring of integers. If R is any ring (without 1) acted upon by G and if ri E R for all i E I, then the map
24. Fixed Points and Traces
249
given by ci,g H ( ~ i ) g is a G-homomorphism of rings. Furthermore, for sufficiently large I , this map can be made an epimorphism. It is convenient to record the following elementary observation.
Lemma 24.7. Let G act on both R and S and let 0: S R be a Ghomomorphism of rings. Then t9(SG) RG and e(trG(S)) C trG(R) --f
if G is finite. kforeover, if 8 is onto then 8(trG(S)) = trG(R).
As an application we prove
Lemma 24.8. Let G be a finite abelian group of order n acting on the ring R (without 1). Then for any integer d 2 1 we have
Proof: Let
be a primitive complex nth root of unity, set A = Z [ E ] and K = Q [ E ]We . work with the generic models E
By Lemma 24.6 we know that T = C A E &is T G-graded. ~ Furthermore, by that lemma,
In particular, nd
(A) k = l
where the latter sum is over all nd-tuples (XI, X2, . . . ,),A, of elements of G. Fix such an nd-tuple and consider the nd 1 partial products pi = X l X 2 ... X i for i = 0, 1,.. . ,nd. Since (dl = n, the pigeonhole
+
6. Group Actions and Fixed Rings
250
principle implies that at least d + l of these pi are equal. But pi = p j for i < j implies that
k=i+l
Thus we conclude that
Next observe that for suitable rn m
m
S = @ x R t i and SG = i=O
i=O
Thus
(nR)ndc ( n q n dc S#(SG)dS#
c CB
2
R#(RG)dR#€i
i=O
and by reading off the coefficient of
6'
we have
Finally, by choosing the index set sufficiently large, we can map R onto any ring on which G acts. It therefore follows, by Lemma 24.7, that the above formula holds for all such rings. I
Theorem 24.9. [22] Let G be a finite solvable group of order n acting on the ring R (without 1). Then for any integer d 2 1 we have
Proof. We proceed by induction on JGJ.If G is abelian, the result follows from the previous lemma. Otherwise, G has a nontrivial normal subgroup H with 1 < M = (HI < n. By induction (using (n/m)dfor d ) (mR)nd
c R#(RH)(+%#.
251
24. Fixed Points and Traces
Furthermore G / H acts on RH and (RH)GIH = RG so induction yields ((,/,)RH)(~/~)~
5
(RH)#(RG)~(RH)#
5 R#(RG)~R#.
Finally n 2 n / m so combining these yields
(nR)ndC R # ( ( ~ L / ~ ) R ~ ) (G~R#(RG)dR# ’~)~R# and the result follows. In particular, if RG = 0 and d = 1 we obtain (nR)n= 0, the appropriately sharper version of Theorem 24.2 for solvable groups. We close this section with another nice observation from [22] and then a consequence.
Proposition 24.10. [22] Let G = (x)be a cyclic group ofprime order p acting on the ring R (without 1 ) . If R has no nonzero nilpotent elements, then either G acts trivially on R or trG(R) # 0.
Proofi Assume by way of contradiction that trG(R) = 0 and RG # R. Then Theorem 24.2 implies that pR is nilpotent so, by assumption, pR = 0. We can therefore view R as a right GF(p)[G]-module. With this notation, if T , S E R then (TS)(l-
x) = f S - ( T Z ) ( S Z ) = ( T - T Z ) S + T Z ( S - sx) = T ( 1 - x) s + T 2 . s(1- z) *
and hence by induction
Since (1- x)”-l = 1+ IC +. . . + xp-l and trG(R) = 0, it follows that 1- x is nilpotent on R of degree t 5 p - 1. Moreover t 2 2 since RG # R. Now put k = t , T E R(l - z ) ~ - ’ and s E R in the formula of the preceding paragraph. Since R ( l - x ) = ~ 0, it follows that 0=
(:)+
- Z)Zt-l *
s(1-
2y-l
6. Group Actions and Fixed Rings
252
and hence ~ - ( l - x ) x ~ - ~ . ~ ( l - = x )0~since - ~ t < p . But r(l-z)zt-' and s ( 1 - z ) ~ - ' are typical elements of R(1- z ) ~ - ' so we conclude 2 that (R(1- x ) ~ - ' ) = 0. Since R has no nonzero nilpotent elements, it follows that R(1- z ) ~ - ' = 0 and this contradicts the definition of the degree t. I Let G act on the ring R and let A be a nonempty finite subset of G. The map trA: R 4 R given by
is called a partial trace if trA(R) 5 RG. It is nontrivial if 0 # trA(R). Certainly any partial trace is an RG-bimodule homomorphism from R to RG. In particular, trA(R) is an ideal of the fixed ring.
Corollary 24.11. [38]Let G be a finite solvable group acting on the ring R # 0 (without 1). If R has no nonzero nilpotent elements, then a nontrivial partial trace exists.
Proof. We proceed by induction on IGl, the case J G )= 1 being clear. Let H a G with G / H cyclic of prime order p . By induction, there exists a subset fl C H with 0 # trn(R) c RH. Note that G need not act on the ideal trn(R) = S , but it does act on S" G RH and
zzEG
then so does G / H . If G / H acts trivially on CsEG S", then 0 # trn(R) RG and trQ is a nontrivial partial trace. On the other hand, if G / H acts nontrivially, then by the previous proposition 0 # trG/H(xzEGSz) C RG. In particular, 0 # trG/H(Sz) RG for some x E G and the result follows with A = flxT where T is a transversal for H in G. I
c
We offer some examples in the following exercises to show that the above result is essentially best possible without additional assumptions on the ring.
253
24. Fixed Points and Traces
EXERCISES 1. Let R be a prime ring. Prove that R is a domain if and only if it has no nonzero nilpotent elements. 2. Let K[G] be a group algebra of the finite group G and let H be a subgroup of G. If 7 = ChEH h E K [ H ] ,prove that r K i ~ l ( 7 = ) w(K[H])K[G] and eK[G](w(K[H]) = K[G]T* 3. Suppose G is a cyclic group of prime order p acting faithfully on the ring R with pR = 0. Show that the only possible partial trace here is the ordinary trace. To this end, observe that every element of GF(p)[G] not in the augmentation ideal is a unit. 4. Let R = M2(K) where K is a field of characteristic p > 0 and let g =
(i :)
E
R. Then G = (9) acts on R by conjugation.
Viewing R as a right GF(p)[G]-module,show that R ( l - g) consists of upper triangular matrices, R(1 - g)2 consists of strictly upper triangular matrices and R(l - g)3 = 0. Conclude that for p 2 5 there are no nontrivial partial traces even though R is a simple ring. This example is from [38]. The remaining exercises are based on [69]. Let G be a finite group of order divisible by p , let K be a field of characteristic p and let fl be the set of all right cosets of subgroups of G of order p. Then G permutes fl by right multiplication, each point stabilizer G, for w E fl is a subgroup of G of order p and all such subgroups occur in this manner. Form 4 = @ EWER K,, the direct sum of copies of K indexed by the elements of fl, and let G act on the reduced ring Rp by permuting these summands. Suppose trA(Rp) (Bp)Gfor some nonempty subset A of G.
5. Let 0 # T E K,. Show that trA(r) = 0 if and only if A is a union of right cosets of G, or equivalently if and only if G,A = A. On the other hand, if trA(r) # 0 show that A is a disjoint union of 1 5 n 5 p - 1 right transversals of G, and hence that IAl = nlGl/p. To this end, write A = U iAiti where { t i } is a right transversal for IAiJrti. G, in G and hi G,. Then note that trA(r) =
Xi
254
6. Group Actions and Fixed Rings
6. Suppose G is generated by its subgroups of order p . Show that trh(Rp) = 0 if and only if A = G and conclude that IGl/p divides ]A( in general. 7. Suppose p and q are distinct primes and that G is generated by its subgroups of order p and by its subgroups of order q. Show that there is no nontrivial partial trace for the action of G on Rp@R,. 8. Suppose G is generated by its subgroups of order 2 and that G has no subgroup of index 2. Show that there is no nontrivial partial trace for the action of G on R2. To this end, if G, = (x)has order 2, observe that either XA = A or A is a right transversal for G, and hence xA = G \ A. Conclude that G permutes the subsets A and G \ A by left multiplication.
25. Integrality
It is possible that Theorem 24.2 is trying to tell us something more. It may be that R is integral, in some sense, over the fixed ring RG. For this we need some definitions. Let R be a ring (possibly without 1) and let T be a subring. If r1, r2,. . .,r, E R, then a T-monomial in the ri's is a product of these elements in some order with elements of T such that at least one factor from T occurs. For example, if t l ,t 2 E T , then r?tlr2t2rlr3 is a T-monomial but rfr2rlr3 is not unless 1 6 T . The degree of such a monomial is the total degree in all the Ti's. We say that R is filly integral over T of degree m if for any rl,r2, . . . ,r, E R we have
r1r2 * . . rm = Cp(r1,r2,. . . ,r,) where tp is a sum of T-monomials in the T ~ ' Sof degree less than m. In particular, setting r1 = ~2 = . . = rm = T E R we see that rm = $ ( r ) where 4 is a sum of T-monomials in r of degree less than m. This says, by definition, that R is Schetter integral over T of degree m. In particular, if R is fully integral over RG of degree rn and if RG = 0, then R" = 0. Thus such an integrality result is an appropriate generalization of Theorem 24.2. As a simple example, a
255
25. Integrality
let G = { 1 , x } haveorder 2 andact on R. Ifr E R, t h e n t = T + T ” = trG(r) E RG and t - T = T” so
+
In other words, r satisfies 2r2 - rt - tr s = 0 and 2r is Schelter integral over RG. To see what might be involved in a general proof of this result, we begin with a crucial special case. Let S be an algebra over the field K and set R = M,(S). Furthermore, let G be a finite absolutely irreducible subgroup of GL,(K) so that, by definition, the K-linear span of G is M,(K). For example, if K has characteristic 0, then the symmetric group Sym,+, has such an irreducible representation. Then G acts on R by conjugation and RG = S, embedded as scalars. Thus at the very least we expect M,(S) to be integral over S. Fortunately, this is a result of the extremely clever paper [154].Variants of that theorem were later obtained in [110],[lo21and [163]using the same proof. The formulation of the following is from [163]. If A is a ring with 1, we let { ei,j } denote the usual set of matrix units of M,(A).
Theorem 25.1. [154]Let A be a ring with 1 and let R 2 T be subrings of M,(A) (without 1). Assume that for all k = 1 , 2 , . . . ,n we have i. (C!ez,i)R(C: e2,z) G R, ii. T consists o f diagonal matrices and ek,kRek,k = ek,kTek,k. Then there exists an integer m = m(n) 2 1 such that R is fully n+l-2 integral over T of degree m. Here m < 22 ProoJ For ease of notation we first consider Schelter integrality. As we will see at the end of the proof, full integrality is an easy consequence of this. For each k = 1 , 2 , . . . , n we embed &,(A) into M,(A) as the k x k upper left corner. The goal is to show by induction on k that RrlMk(A) is Schelter integral over T of degree m(k)< 22k++‘-2. Furthermore, the monic polynomials involved will all have constant term zero.
6. Group Actions and Fixed Rings
256
First let k = 1 and choose
T
(ii) implies that there exists t = m(1) = 2 < 22*--2. Now let k 2 1 and for any
T
=
(g
(g
n)
e)
E
R f l M , ( A ) . Then
E T . Thus r2 = tr and
E R n Mk+l(A) write
r= 0
0
0
where r' is the k x k corner block. Furthermore, for this r set ?=("
0
x e)
0
0
so that r' E R by assumption (i). Given T ~ , T ZE R n M k + l ( A ) we say that r1 = r2 if and only if ry = rg. We note two properties of =. (1) Let T ~ , T ~ E , T R f l Mk+l(A) and t E T. If r1 = r2, then tr2. i5-1= Frz and trl (2) Let r1,r E R n Mk+l(A). Then there exists t E T with Flr = T I T - rlt. For this, note that there exists t E T with
r--t= and then
o o *
(; ; i) rir''
r1(r-t) = By induction on j
=FIT-.
> 0, if r1, 7-2,. . . ,~ jr ,E R fl Mk+l(A),then
where T is a sum of nonconstant T-monomials in the ri's of degree smaller than j 1. Indeed if
+
257
25. Integrality
then by the above we have
where t is a suitable element of T and T = rj7 - rjt is an appropriate sum of T-monomials of degree less than j 1. Now for the inductive part of the proof. Let T E R n Mk+l(A). Then by induction, ? satisfies a monic polynomial Fm - $(T) = 0 over T of degree m = m(lc). Hence certainly (T"" - $(T"))T E 0. We now apply the observation of the preceding paragraph to each of the monomials in this expression. Since TMk+1(A), Mk+1(A)T C Mk+I(A), each nonconstant T-monomial in T is a product of factors all belonging to Mk+l(A). With this it follows easily from the above that (Trn - @ ( T ) ) T T ( T ) = (T"" - $(T"))T = 0 ,
+
+
where ~ ( ris) a suitable sum of nonconstant 7'-monomials in T of degree less than m 1. In other words, we have shown that there exists s = r m+l - @(.>. + T ( T ) E R n Mk+l(A),
+
+
a monic polynomial over T in T of degree m 1,with s =- 0. Note that S E R n Mk(A) also satisfies a monic polynomial say P - O ( i ) = 0 over T. Hence since s = 0, we see that v=sm--O(s)=
0 0 0 (I a 0) 0 0 0
for some a E A. Furthermore if b E T with b = diag(*, a, *), then we have ( u ~ - ~ u=)0.~ Since the latter expression is a monic polynomial in s of degree 4m and therefore a monic polynomial in T of degree 4m(m + l),the induction step follows with m(k + 1) = 4m(lc)(m(k)+ 1) < 22 . 22
k+l-2
This proves that R is Schelter integral over T.
22k+fl-2
- 22b++2-2
6. Group Actions and Fixed Rings
258
Finally form the free ring A = A(&,(2, . . .) and let k and T be the subrings of M,(A) = Mn(A)((l,(2, . . .) given by k = R((1,(2,. . .) and T = T(C1,(2, . . .). It follows easily that T E k C M,(A) satisfy hypotheses (i) and (ii). Thus k is Schelter integral over of deeee rn = m(n). In particular if T I , T Z , ...,r m E R then r" = rl(1 rzcz - r m c m satisfies an appropriate monic polynomial of degree rn over ?. By considering the coefficient of * - - (m, we conclude that R is fully integral of degree m over 7'. I
+
+ +
The above theorem has numerous consequences. To start with, let R be a ring with 1and let 1= el +ez+. - -+en be a decomposition of 1 E R into orthogonal idempotents. If & = { e l , e2,. . . ,en }, then it is trivial to see that C R ( € )= elRe1 e2Rez . e,Re,. If & is central in R, then an &-transversal for R is a subring T such that eiR = eiT for all i = 1 , 2 , .. . ,n. In the following, m(n)denotes the particular function given by Theorem 25.1.
+
+ +
Corollary 25.2. [163] Let R be a ring, let 1 = el + ez + - + e, be a decomposition of 1 E R into orthogonal idernpotents, and set & = { e l , e2,. . , ,e, }. If T is an &-transversal for C,(&),then R is fully integral over T of degree m(n).
ProoJ For each T E R, let ri,j = eirej. Then the map a : R -+ M,(R) given by D ( T ) = [ri,j]is easily seen to be an isomorphism into. We show that a ( T ) & D ( R )5 M,(R) satisfy the hypotheses of the preceding theorem. To start with, since ri,j E eiRej and a(ei) = ei . ei,i, it follows that ei,io(r) = a(eir), a(r)ei,i = o(rei) and hence that k
k
k
k
( C e z , i ) a c R ) ( C e i , i ) = a ( ( C e i > R ( c e e ) ) 5 o(R) i=l
i=l
i=l
i=l
so (i) holds. eiRei so a ( T ) G ei,iM,(R)ei,i and Next, T g C R ( € )= therefore consists of diagonal matrices. Also, since ri,i = eirei we have
xi
xi
ei,ic(r)ei,i = o(ei)o(r)o(ei)= o(eirei).
25. Integrality
259
But if r E R, then eirei E C R ( & )so, by assumption,
for some t E T. Hence
and (ii) holds. We conclude from Theorem 25.1 that o(R) is fully integral of degree m(n) over o ( T ) . Applying IT-clearly ' yields the result. I Note that if R = M,(A) and if & = { e1,1, e2,2,.. . ,en+ }, then the scalar matrices A are an &-transversal for C R ( & )the , diagonal matrices. Thus Corollary 25.2 recaptures much of Theorem 25.1. Another consequence of that result is
Corollary 25.3. 1211 Let G be a finite group and let S be a G-graded ring (without 1). Then S is fully integral of degree rn(lG1) over the identi ty component S1.
ProoJ Write S = R(G) where R = S1 and form S# = S @ 2, the natural extension of S to a ring with 1. In the notation of Section 2, let MG(S#) be the ring of /GI x IGl matrices over S# and let
If T{l} = R, embedded as scalar matrices in MG(S#), then it follows immediately that T{l} C S{1} C_ MG(S#) satisfy the hypotheses of Theorem 25.1. We conclude that S{l} is fully integral over T { 1 } of degree rn(lG1). Finally if s E S , recall that the matrix S E MG(S) is defined by S(z,y) = s,-lY so that S E S{1}. Moreover, by Lemma Z.l(ii), the map s H S is a ring embedding of S into S{1}. Since the image of R under this map is T{1}, the result clearly follows from the above. I With this result in hand, it is not surprising that we can obtain full integrality for abelian group actions.
6. Group Actions and Fixed Rings
260
Theorem 25.4. [163]Let G be a finite abelian group of order n acting on a ring R (without 1). Then nR is fully integral over RG of degree m(n).
Proof: Let
be a primitive complex nth root of unity, set A = Z[E] and K = &[el. We work with the generic models E
as described in the preceding section. By Lemma 24.6 we know that T = C x E e Tis~G-graded. Furthermore, by that lemma,
X€G
Note that Cx,e(Tx f l S ) is G-graded so, by Corollary 25.3, this ring is fully integral of degree m(lG1)= m(n)over the identity component TI n S = T G n S = SG. Thus nS is fully integral over SG of degree m(n).Next observe that for a suitable integer q 0
4
S = @ C R t i and S G = @ z R G e i i=O
i=O
with E Q + ~ a Z-linear combination of 1,E , . . . ,E P . It then follows easily by reading off the coefficient of E’, that nR is fully integral over RG of degree m(n). Finally, by choosing the index set I suffciently large, we can map R onto any ring on which G acts. Lemma 24.7 therefore yields the result. I The original proof of the above in [163] used skew group rings directly (see Exercises 2-5). Note that the case of solvable group actions does not follow directly from this result. An example of [20], on Schelter integrality, shows that an integral extension of an integral extension need not be integral.
261
25. Integrality
Now we move on to consider group actions by arbitrary finite groups. We remark that if S and T are merely subsets of a ring A, then it still makes sense to consider whether S is fully integral over T. It is with this understanding that we prove the following lemma since the set eTe need not be a subring of A.
Lemma 25.5. Let T
S
A be rings (without 1) and assume that S is fully integral over T of degree m. If e is an idernpotent of A with eSe C S , then eSe is fully integral over eTe of degree m. C_
Proof: Let s1,SZ,.. ,,s E S. Since eSe C S and S is fully integral over T of degree rn, we have m
k
i=l
where each a k is a T-monomial in the esie of total degree less than m. Certainly m
i=l
k
Now T is a subring of A, so adjacent T factors in eake can be merged. Once this is done, each T factor t in eake has an e on either side and hence can be replaced by ete E eTe. Thus eake is an eTe-monomial in the esie of total degree less than m and the lemma is proved. I
We can now obtain
Theorem 25.6. f l 8 l ) Let G be a finite group acting on the ring R (without 1 ) and suppose that (GI R = R. Then R is fully integral of degree m(lG1) over the fixed ring RG.
P~oo_f: Let n = fGl and m = m(n). We first consider the generic models
3 = z[l/nI(Ci,, I g E G,1 5 5 m) 5’ = Z[1/n](
6. Group Actions and Fixed Rings
262 Let B = M,(S) CcBby
E A = Mn(S) and if G
C = { diag(s"l,
=
s Z 2 , . . . ,sZn)
{z1,z2,
...,z n }
define
1 s E S }.
It then follows immediately from Theorem 25.1 that B is fully integral over C of degree rn. Now let e = ;a E A where Q is the matrix with all entries equal to 1. Then e is an idempotent of A and if ,B = [ b i , j ] E A , we have e,Be = (i bi,j)e. Thus eBe B and the previous lemma implies that eBe is fully integral of degree m over eCe. But eBe 2 Se, with S embedded as scalars, and eCe = $trG(S)e SGe. Thus Se % S is fully integral over SGe E SG and this case is proved. Now set
T = Z(Ci,g By the above, we have (1
7
where 8 is a sum of SG-monomials in the C ~ Jof degree less than m. It then follows by clearing denominators that, for some integer k >_ 0, nkO is a sum of TG-monomials in the Hence we see that
where cp is a sum of TG-monomials of degree < rn in the variables n'[i,l. Finally if T I ,7 - 2 , . . . ,r , E R, then there is a G-homomorphism T -+ R given by Ci,g H r: and we conclude from Lemma 24.7 and (*) that nkR is fully integral over RG of degree rn. Thus since nR = R, we have nkR = R and the result follows.
As we mentioned earlier, this yields a far reaching generalization of Theorem 24.2 with however a poorer bound. We close this section by obtaining an integrality result for normalizing extensions. Let R S = Cr.l Rxi be a finite normalizing extension of rings (with 1). An ideal A of R is said to be normal if z i A = Azi for all i = 1 , 2 , . . . ,n. It follows that AS is an ideal of S and hence a subring (without 1) containing A .
c
25. Integrality
Theorem 25.7. [154] [110] Let R C S =
263
C ~ = = lbef i ia finite
normalizing extension of rings (with 1 ) and let A be a normal ideal o f R. Then A S is fully integral of degree m(n)over A.
ProoJ Let F = R" be the free left R-module of rank n and let T :F --t S be the left R-module epimorphism given by ( T I ,~ 2 ,... ,T,)" C;,' ~ i x i .If I = Ker(n) & F , then U = { cp E End(RF) I I 9 c- 1 ) is a subring of End(RF) = M,(R). Furthermore, there is a natural homomorphism -: U -+ End(RS) such that if cp E U then
Routine verifications show that (p is well defined and that - is a ring homomorphism. Indeed (p = n-lym with the understanding that -1 S" is any inverse image of s. Let T be the set of all diagonal matrices in M,(R) = End(RF) of the form cr. = diag(a1, a 2 , . . . ,a,) with ai E A and aixi = xia for some a E A . Then T is a subring of M,(A) and the normality of A implies easily that ei,+M,(A)ei,i = ei,iTei,i for all i. Furthermore if f = ( T ~ , T B.,.. , T , ) E F , then f a = ( ~ l a l , r 2 a 2 ,.. . ,T,u,) so
It follows immediately that a E U and that 6 is right multiplication by a E A. Now let s E AS be given. Since AS is a two-sided ideal of S , we have xis E A S so xis = C j a i , j x j for suitable ai,j E A. Of course the elements ai,j need not be uniquely determined. Let P = [ a i , j ] E M,(A). Then it follows easily from the above that if f = (TI, 7 - 2 , . . , ,r,) E F then f"" = (f")s. Hence we see that P E U and that the endomorphism 0 is right multiplication by s E A S . By Theorem 25.1 applied to T M,(A) M , ( R ) we see that M,(A) is fully integral of degree m(n)over T . Hence U n M,(A) is also fully integral over T and therefore (U fl M,(A))- is fully integral over T . But as we have shown above, (U n M,(A))" 2 AS, viewed as right multiplication, and T = A. Thus the result follows. I
264
6. Group Actions and Fixed Rings
EXERCISES 1. Let A = Z(x,y) be the free ring on x , y and let a = diag(x,y) E Mz(A). Show that a is not integral, in the ordinary sense, over the scalar matrices A. Indeed, show that the powers of a are (right or left) independent over A .
In the next four problems, let G be an abelian group of order n and let K be a field containing n distinct nth roots of unity. Set G = Hom(G, KO). 2. For each 1 # X E G prove that CgEGX(g) = 0 and for each 1 # g E G prove that CAE& X(g) = 0. Show that G G. 3. For each X E G define eA E K [ G ]by ex = CgEG A(g-l)g. Show that 1 = Exex is a decomposition of 1 into orthogonal idempotents. 4. Suppose G acts as K-automorphisms on the K-algebra R with 1 so that R G 2 KG = K [ G ] .If & = { e x I X E G}, show that RG is an &-transversal for CRG(&). Conclude that R G and hence R is fully integral over RG of degree m(n). 5. Now suppose G acts on any ring R (without 1). Use the above and generic models to obtain an alternate proof of Theorem 25.4. 6. Let G be a group of order n acting on a ring R with E R. If I is a G-stable ideal of R, prove that ( R / I ) G = (RG I ) / I . Lemmas 24.7 and 24.1 apply here. 7. Let G be a group of order n acting on a ring R (with 1) and form the skew group ring RG. As in Lemma 23.6(iii) we have RG EndR(RG) S M,(R) using the natural left R-basis G. Show that the element a = CzEG x E RG corresponds to the matrix with all entries equal to 1. In some sense, this underlies the proof of Theorem 25.6.
i
i +
26. Finiteness Conditions We now move on to more direct skew group ring applications. For this we will assume throughout that G is finite. Furthermore we
265
26. Finiteness Conditions
return to our usual assumption that all rings have a 1. For the most part we will be concerned in this section with finiteness conditions, in particular the Artinian, Noetherian and Goldie properties of rings. We will consider just a small sample of what is known. The reader should consult [129]for the complete story. Let G act on the ring R with 1. We recall that the trace map trG:R --+ RG is defined by trG(T) = C z E Gand ~ ”satisfies part (i) below. In addition if A is a right ideal of RG, then A G (AR)G and trG(AR) = AtrG(R) & A so (ii) follows.
Lemma 26.1. Let G act on R. i. trG is an (RG,RG)-bimodule homomorphism from R to RG. ii. If A is a right ideal of RG, then (GI A 5 (GI . (AR)G C trG(AR) C A .
-
As usual, if G acts on R we can form the skew group ring RG. Recall that G is embedded isomorphically in RG so we do not need the overbars. Define G = CTEG2 E RG and observe that for all g E G we have G g = G = gG. We remark that the element G is sometimes denoted by G. We have chosen this alternate notation to avoid confusion with the dual group.
Lemma 26.2. Let G E RG be as above. i. G(RG) = eR E R as right RG-modules. Here R acts on R by right multiplication and G acts in the given manner. Furthermore EndRG(R) = RG, acting by left multiplication. ii. If T E R, then GrG = G trG(r). Hence
iii. If I and J are G-invariant right ideals of R, then RGI a RG and (RGI)(RGJ) = RG trG(I)J. Furthermore, for all n 3 0 we have (RGI)”” = RG . trG(I)nI.
Proof: (i) Since RG = GR, we have G:(RG) = G(GR) = G R R where the latter isomorphism is given by O(Gr) =
T.
%
This is
266
6. Group Actions and Fixed Rings
certainly a right R-module map and for g E G we have O(G:Tg) = O(Gg-'rg) = : ~ g= O ( G : T ) ~ . Finally, EndR(R) = R, acting by left multiplication, and therefore we have EndRc(R) = RG, also acting by left multiplication. (ii) This is clear since
(iii) RGI is surely an (R, R)-bimodule. Furthermore, since both R and I are G-invariant and since G absorbs factors fiom G, we have R G I a R G . Finally (RGI)(RGJ)= RGtrG(I)J since GIRG = G trG(f) by (ii) a,bove. The result on (RGI)n now follows easily by induction. I We can now obtain
Theorem 26.3. 1221 112'71 Let G act on R and suppose that the skew group ring RG is semiprime (or prime). This occurs, for example, if R is semiprime with no /GI-torsion (or if R is prime and G is X-ou ter) . i. If I # 0 is a G-invariant right or left ideal of R, then trG(I) is not nilpotent and hence IG # 0. ii. RG is semiprime (prime).
PrwJ The sufficient conditions for RG to be semiprime or prime are contained in Theorem 4.4 and Corollary 12.6. Now let 0 # I be a G-invariant right ideal of R and consider the nonzero ideal RGI of RG. By assumption, RG is semiprime so RGI is not nilpotent. Thus Lemma 26.2(iii) implies that trG(I) is not nilpotent. In particular, t T G ( I ) # 0 and (i) is proved. For (ii), let A and B be nonzero right ideals of RG and set I = AR and J = BR. Since trG(I) A, Lemma 26.2(iii) implies that I J RGAB. Thus clearly if RG is semiprime (or prime), then so is RG. I The following is a special case of Theorem 24.2 on the existence of fixed points. We include it to see how close ring theoretic methods
267
26. Finiteness Conditions
can come to proving the general result. Additional comments are contained in Exercise 6.
Theorem 26.4. [22] Let G act on R, a ring with no IG(-torsion, and suppose that 0 # I a R is G-invariant. If trG(I) is nilpotent, then I n rR(I) # 0 and I is nil.
Proof. By Lemma 26.2(iii), RGI is a nilpotent ideal of RG. Thus J = r R G ( R G I ) is a two-sided ideal of RG which is essential as a left ideal. By the left analog of Proposition 4.3 it follows that R J ess RRG and hence R ( J fl R) ess RR. But surely J f l R = r R ( I ) so, since I # 0, we have I n r R ( 1 ) # 0. For the second part, let N be the nil radical of R and set R = R / N . Then G acts on the semiprime ring R and it is easy to see that R has no !GI-torsion. Furthermore, trG(f) = trG(I) is nilpotent. It therefore follows from Theorem 26.3(i) that I = 0 and hence I E N is nil. I Let V be an R-module. Then the Goldze rank of V , written rankVR, is defined to be the largest integer k such that V contains V1@V2 @ . . . @ Vj, a direct sum of k nonzero submodules. If no such maximum exists, then rankVR = m. Basic properties of this rank can be found in [64]. In particular it is clear that W & V implies that rankW 5 rankV and if W ess V then the ranks are equal. Modules of rank 1 are called uniform so that U # 0 is uniform if and U we have X n Y # 0. only if for all nonzero submodules X , Y Furthermore rank (V CB W ) = rank V rank W and if U,, U2,. . . , Uk are uniform and (Ul CB U2 @ . . . CB Uk) ess V , then rank V = k .
+
Lemma 26.5. Let V be a module for the crossed product R*G. Then
Proof. The first inequality, namely rank VR*G5 rank VR,is obvious. We consider the second. For each R-submodule
AR of
VR define
A = n,.,AZ. Then A = { v E V I v3-l E A for all z E G } and it
268
6. Group Actions and Fixed Rings
is easy to see that A is an R*G-submodule of V . In fact, A is the largest R*G-submodule of V contained in A. Since G is finite, Zorn’s lemma applies and there exists A maximal with A = 0. Suppose that (&/A) @ ( & / A ) @ . . @ (Bk/A) is a direct sum of nonzero R-submodules of VIA. Then Bi is properly larger than A, so & # 0 and it is easy to see that @ @ @ & is a direct sum of nonzero R*G-submodules of V . Thus we see that rank (V/A)RI rank VR*G.Finally each A3 is also an R-submodule of V with the same property so rank (V/A?)R I rank VR*G.Thus, since A = 0, VR is embedded isomorphically in @ CxEG VIA3 and
& &
This completes the proof. I Now return to the action of G on R. The following lemma shows that there exists a strong relationship between the essential right ideals of R and those of RG. In part (i) below, the RG-module structure of R is as described in Lemma 26.2(i). Furthermore, the rank of any ring R is its rank as a right R-module.
Lemma 26.6. Let G act on R, a semiprime ring with no IGI-torsion. i. rank RG = rank RRG. ii. rank RG I rank RR 5 IGI - rank RG. iii. If E is an essential right ideal of R, then ( E n RG) ess RG. iv. If A is an essential right ideal of RG, then AR ess R.
Proof. We will freely use Lemma 26.l(ii) and Theorem 26.3(i). (i) If 11 @ 1 2 @ . * - @ 1k is a direct sum of nonzero RG-submodules of R, then IF @ IF @ - - .’@ IF is a direct sum of nonzero right ideals of RG. Conversely if Al @ A2 @ . . . @ Ak is a direct sum of nonzero right ideals of RG,then AIR @ A2R @ - @ AkR is a direct sum of nonzero RG-submodules of R . Indeed the latter sum is direct since, for each i, Bi = ( A I R A2R . . . + Ai-lR) n AiR is a G-invariant right ideal of R with trG(Bi) C (A1 A2 Ai-1) n Ai = 0. Thus Bi = 0 and the ranks are clearly equal.
+
+
+
+
+
26. Finiteness Conditions
269
(ii) This is immediate from (i) above and the previous lemma. (iii) If E ess R then, (n,,,E”) ess R, since G is finite, and this intersection is G-invariant. Thus we may assume that E is Ginvariant. Now suppose X is a nonzero right ideal of RG. Then E n X R # 0 and hence 0 # trG(E fl X R ) C ( E fl RG)f l X . (iv) If A ess RG then it is clear that ARRG ess RGRG. Indeed if X is a nonzero G-invariant right ideal of R, then X G # 0 so A n X G # 0 and hence AR f l X # 0. Proposition 4.3 now yields ARR ess RR and the result follows. I Note that a ring R is semisimple Artinian if and only if it has no proper essential right ideal. Thus we have
Theorem 26.7. [98] [36]Let G act on R and suppose that R is semiprime with no (GI-torsion. Then R is semisimple Artinian if and only if RG is.
Proof: Suppose RG is semisimple Artinian and let E ess R. Then ( E n RG)ess RG,by Lemma 26.6(iii), so 1 E E fl RG and E = R. Conversely if R is semisimple Artinian, then the center of R is a sum of fields and hence (GI-’ E R. If A ess RG then AR ess R, by Lemma 26.6(iv), so 1 E AR. Thus )GI = trG(1) E trG(AR) C A and A = RG.In view of the above remarks, the theorem is proved. I R is said to be a Goldie ring if and only if rank RR < 00 and R satisfies the maximum condition on right annihilators of subsets of R. Goldie’s theorem [64, Theorem 1.371 asserts that a ring R has a classical right quotient ring Q(R)which is semisimple Artinian if and only if R is a semiprime Goldie ring. These rings can be characterized in terms of their essential right ideals and therefore a correspondence analogous to the above theorem is to be expected. For convenience, we quote the following lemma which merely isolates the last few steps in the proof of Goldie’s theorem (see Exercises 7).
Lemma 26.8. Let T be a multiplicatively closed subset of regular elements of R and suppose that i. t E T implies tR ess R, ii. E ess R implies E n T # 8.
6. Group Actions and Fixed Rings
270
Then T is a right divisor set in R, the ring RT-l is semisimple Artinian and R is a semiprime Goldie ring with classical quotient ring Q(R)= RT-l. With this we can now prove
Theorem 26.9. [85] Let G act on the ring R and suppose that R is semiprime with no (GI-torsion. Then R is Goldie if and only if RG is. Furthermore if this occurs, then Q(R)= RT-l where T is the set of regular elements of RG and Q(R)G= Q(RG).
Proof: Observe that RG is semiprime by Theorem 26.3(ii). Suppose first that R is Goldie. Then RG C R 2 Q(R)and Q(R) is semisimple Artinian, so RG surely satisfies the maximum condition on right annihilators of subsets. Furthermore rankRG is finite, by Lemma 26.6(ii), and hence RG is semiprime Goldie. Conversely suppose that RG is Goldie and hence that Q(RG)= RGT-' exists, where T is the set of regular elements of RG. Certainly T is a multiplicatively closed subset of R. Furthermore, if t E T and if X is either the right or left annihilator of t in R, then X is a G-invariant right or left ideal of R and X n RG = 0. Hence, by Theorem 26.3(i), X = 0 and T consists of regular elements of R. We show now that T satisfies (i) and (ii) of the previous lemma. Indeed if t E T , then tRG ess RG since RG is a Goldie ring. But then, by Lemma 26.6(iv), tR = (tRG)Ris essential in R. On the other hand, if E ess R , then ( E n RG) ess RG, by Lemma 26.6(iii), and hence E n T = ( E n RG)n T # 8 since RG is semiprime Goldie. Lemma 26.8 now asserts that R is a semiprime Goldie ring with Q(R) = RT-l. Finally, since T C RG, we see immediately that Q(R)G= (RT-')G = RGT-' = Q(RG)and the result follows. I We remark that the above theorem was discovered independently in papers [35]and [59]. To proceed further, we assume that 1Gl-l E R and define e E RG by
271
26. Finiteness Conditions
Lemma 26.10. Let G act on R with \GI-' E R. Then e = JG/-'G is an idempotent of RG with i. e(AG)e = eAG for any G-invariant ideal A of R, ii. e(RG)e = eRG 2 RG where the latter is a ring isomorphism.
Proof: Since gG = G, it follows that (G)2 = [GIG and hence that e is an idempotent. Now let A be a G-invariant ideal of R. Then AG = trG(A) since 1GI-l E R so (i) is immediate from Lemma 26.2(ii). Furthermore, since RG clearly commutes with e, the isomorphism eRG RG does indeed preserve the ring structure. 1 In view of part (ii) above, the following lemma is surely of interest. Let S be a ring. Recall that if W is an S-module, then L(Ws) denotes the lattice of S-submodules of W .
Lemma 26.11. Let W be a right S-module and let f be an idempotent of S. Then there exist inclusion preserving maps
such that i. or = 1 so o is one-to-one and r is onto, ii. r preserves direct sums.
Proof. Define o:L(W f Ifsf) L(Ws) and r in the other direction by A' = A S and BT = B f . Then A"' = ASf = AfSf = A and the result clearly follows. 1 -+
Theorem 26.12. [113] Let G act on the ring R with 1GJ-l E R and let VR be a right R-module. If W = VIRGis the induced RG-module, then there exist inclusion preserving maps
such that i. or = 1 so o is one-to-one and r is onto, ii. r preserves direct sums.
272
6. Group Actions and Fixed Rings
Proo_f:Let S = RG. In view of the previous two lemmas, it suffices to show that ~ R 2 G WeleSe. But this is clear since W = VIRG= @ CzEG V 8 x implies that the map TJ H CzfG TJ 8 x gives the necessary RG-isomorphism. I
Corollary 26.13. Let G act on R with \GI-'
E R and let
VR be a
right R-module. i. If VR is Noetherian or Artinian, then so is ~ R G . ii. If R is right Noetherian, then R is a finitely generated RGmodule and RG is right Noetherian. iii. If R is right Artinian, then so is RG. iv. If VR is completely reducible of composition length n, then ~ R is G completely reducible of compcsition length 5 n - ]GI.
Proo_f:Let W be the induced RG-module W = VIRG.Then W I R= @ CzEG V 8 J: with each V @ x an R-module conjugate to V (see Lemma 3.3). In particular, the lattice of R-submodules of V 8 x is isomorphic to ~ ( V R ) . (i) If VR is Noetherian or Artinian, then so is each V 8 2 and hence W p . It therefore follows that WRGis Noetherian or Artinian and hence so is V ~ R Gby Theorem 26.12 since u is one-to-one. (ii) By (i) applied to V = R, we conclude that RRG is Noetherian and hence finitely generated. Since RG R, it follows that RG is a right Noetherian ring. Part (iii) is similar. (iv) If VR is completely reducible of composition length 5 n, then the same is true of each V @ x .Thus W p is completely reducible of length 5 n - IGI. By Theorem 4.1, WRGis completely reducible and hence so is YRGby Theorem 26.12 since 7- is onto and preserves direct sums. I Part (ii) above is essentially due to [56]. Analogous results hold for (S,R)-bimodules.
Corollary 26.14. [113] Let G act on the ring R with 1GI-l E R and let H act on the ring S with 1HI-l E S . Suppose SVR is an (S,R)-bimodule. i. If SVR is Noetherian or Artinian, then so is S~ I ~ R G .
273
26. Finiteness Conditions
ii. If the ideals of R satisfy the ascending or descending chain condition, then the same is true of the ideals of RG. iii. If ~ V isRcompletely reducible of composition length n, then the restriction S H I I / ( R G is completely reducible oflength _< n.IHJ.IGI.
ProoJ We may suppose that V # 0. Since V is a unitary right module for the ring SopC3 R, it follows that the latter is a ring with 1. Also H x G acts on this ring and JHx GI-' E S o p 8 R. Now it is easy to see that
and thus
With this observation, (i) and (iii) follow from Corollary 26.13(i)(iv). Part (ii) follows by viewing R as an (R,R)-bimodule. In particular, if R is a direct sum of n simple rings, then (iii) implies that RG is a direct sum of at most n /GIzsimple rings. This bound, however, can be sharpened to n IGl, a fact first observed in 11511. We have a
Theorem 26.15. 1861 Let R be a ring wliich is a direct sum of n simple rings and let G act on R. If R has no (GI-torsion, then RG and RG are direct sums of at most n . JGJsimple rings. ProoJ The structure of R implies immediately that \GI-' E R. Let V = RG and view V as an (RG,RG)-bimodule. Then ~ ( = c3
xzEG Rx is a direct sum of IG(bimodules, each of which is clearly a
direct sum of n simple bimodules. Thus ~ 1 V is p completely reducible of composition length n. IGJ.By viewing V as a module for the skew group ring ( B o p 18R)(G x G), as in the previous result, Theorem 4.1 implies that V is also a direct sum of at most n. (GIsimple (RG, RG)bimodules. In other words, RG is a direct sum of at most n.lGJsimple rings. Since e(RG)e RG, Lemma 17.8 now yields the result. I
y ~
274
6. Group Actions and Fixed Rings
Finally we prove
Theorem 26.16. [56] Let G act on R and suppose that R is semiprime with no /GI-torsion.If RG is right Noetherian, then so is R.
ProoJ In view of Corollary 26.13(ii), we expect R to be a finitely generated RG-module. Observe that RG is Goldie and hence, by Theorem 26.9, R is semiprime Goldie and G acts on the semisimple Artinian ring Q(R). Note that R has no /GI-torsion and hence (GI-'E Q(R). Thus since Q(R) is left Noetherian, the left analog of Corollary 26.13(ii) implies that Q(R) is a finitely generated left Q(R)G-module. We can of course assume that the generators have a common denominator. Thus there exist T I , 2-2,. . .,T n E R and t regular in R with Q(R) = Q(R)Grit-l so that
xT
xy
We now define a map 8: R + (RG)nby 8 ( r ) = @ trG(rir). It is clear that 8 is a right RG-homomorphisminto the free right RGmodule (RG)".The goal is to show that 8 is one-to-one. To this end, suppose r E Ker(0). Then trG(rir) = 0 for all i so trG(Q(R)Grir)= 0 and thus, by the above, trG(Rtr) = 0. If I is the G-invariant left ideal I = x 2 E G ( R t r ) Sit, follows that tTG(I) = 0. But R is semiprime with no \GI-torsion, so Theorem 26.3(i) implies that I = 0. We conclude that tr = 0 and hence, since t is regular, that r = 0. Thus 8 is one-to-one as required. In other words, RRG is isomorphic to a submodule of a finitely generated free RG-module. Since RG is right Noetherian, we conclude that RRGis Noetherian and therefore RR is also Noetherian. I An interesting special case is as follows. Suppose G acts as automorphisms on the polycyclic-by-finite group H . Then G acts on the Noetherian ring K [ H ]and, by the above, K[HIGis Noetherian if 1GI-I E K . It is an open question as to whether the hypothesis (GI-'E K is required here.
275
26. Finiteness Conditions
EXERCISES
1. Find an example of a prime ring R and a group G such that R is not Artinian, Noetherian or Goldie, even through RG is a field. An appropriate example occurs in Section 24.
z)
2. Let K be a field of characteristic not 2 and let F be an and let infinite dimensional extension field. Form R =
(f
G = { 1,g} act on R with g = diag(1, -1). Show that \GI-' E R and that RG is Artinian, Noetherian and Goldie, but that R is not.
3. Let D be a K-division algebra generated (as a division algebra) by q,z2,. . . ,z., Assume that D does not satisfy a polynomial identity and that charK = p > 0. Let R = M2(D) and let G be the group of units generated by
(i i)
and
(i y )
for
i = 1 , 2 , . . . ,n. Show that G is a finite pgroup and that the fixed ring R~ equals
(t i)
where 2 =
Z(D).Conclude that R is
Artinian, Noetherian and Goldie, but that RG is not.
4. Construct an appropriate division ring D for the preceding problem starting with the group algebra of a poly-2 group.
R = M,(K) where K is a field of characteristic not 2 and let G = { l,g } act on R with g = diag(1, -1, -1,. . . , -1). Prove 5. Let
that R R G requires at least n generators even though (GI = 2. To this end, note that RG = MI(K) CB M,-l(K) and that the first column of R is an RG-submodule annihilated by Mn-l(K). 6. Let G act on R and let I be a G-stable ideal. If R has no ]GI-torsion and IG = 0, prove that I is nil of bounded degree by applying Theorem 26.4 to a generic model. Then use the action of G on the free ring R(<1,&,.. .) to conclude that I is nilpotent.
7. If E is an essential right ideal of a ring R and T E R, show that ralE = { s E R 1 rs E E } is essential in R. Use this to show that (i) and (ii) of Lemma 26.8 imply that T is a right divisor set. Then observe that RT-' has no essential right ideals and complete the proof of that lemma.
6. Group Actions and Fixed Rings
276
27. Rings With No Nilpotent Elements We return to the question of the existence of fixed points. Recall that the beginning of Section 24 contains an example where fixed points do not exist. In that example, R is a prime ring but not a domain and hence has nontrivial nilpotent elements. Furthermore G is inner on R and R has (GI-torsion. We have already seen in Theorem 26.3(i) that the latter two conditions are necessary ingredients for this example. Now we show that the presence of nilpotent elements also matters. Note that a ring with no nilpotent elements, that is a reduced ring, is necessarily semiprime. The proof of this result requires the use of the symmetric Martindale ring of quotients for semiprime rings (see Section 18 for details). In particular, we need the following observations.
Lemma 27.1. Let R be a semiprime ring. i. Q,(R) is semiprime. ii. If R has no nilpotent elements, then neither does Qs(R). iii. If G acts on R, then Ginn={x€GI
isinneronQ,(R)}
is a normal subgroup of G. iv. If R has no nilpotent elements and B is an annihilator ideal of R, then RIB has no nilpotent elements.
Proof. (i) This is clear since every nonzero ideal of Q,(R) meets R nont rivially. (ii) Let q E Q,(R) with q2 = 0 and choose essential ideals A, B of R with gA, Bq E R. Since qABq is a nilpotent subset of R, we have qABq = 0. But then qAB is also nilpotent so qAB = 0. Since AB is essential in R, it follows that g = 0. (iii) This is the obvious analog of Lemma 12.3(ii). (iv) Say B = !R(A)with A a R and let z E R with x2 E B. Then x 2 A = 0 so z A z is nilpotent and hence x A x = 0. But then xA is nilpotent so xA = 0 and x E !R(A)= B as required. I
27. Rings With No Nilpotent Elements
277
The next result is a key ingredient.
Theorem 27.2. [51][87] Let R be a ring with no nilpotent elements and let G be a finite group acting faithfully on R. IfpR = 0 for some primep, then the commutator subgroup [Ginn,Ginn] has no elements of order p .
Proof. We may assume that G = Ginn. Let H be the set of units of S = Q,(R) which act by conjugation on R like some element of G. Then H is clearly a subgroup of the unit group of S and, since G acts faithfully on R, there is a well-defined homomorphism 0: H -+ G given by 0(u) = g if and only if u-lru = rg for all T E R. Furthermore, 0 is onto since G = Ginn. If 2 = Ker(0), then 2 centralizes R and hence S . Thus 2 is central in H and H is center-by-finite. By Theorem 9.8(ii) (or see [161,Lemma 4.1.41) the commutator subgroup [ H ,HI of H is finite. Finally let u E [ H , H ]with up = 1. Since pR = 0, it follows that pS = 0 and hence (u - 1)P = 0. But S has no nilpotent elements by Lemma 27.l(ii) so u = 1. We conclude that [ H , H ]is a finite p'-group and, since 8 ( [ H , H ] = ) [G,G], the same is true of [G,G]. 1 The proof of the existence of fixed points proceeds via a series of reductions. The next lemma handles the final step.
Lemma 27.3. Let R be a ring with no nilpotent elements and with pR = 0 for some prime p. Let G be a finite nonabelian simple group acting on R and let 0 # L a R be G-stable. Suppose that every L of R contains a nonzero G-stable ideal. Then nonzero ideal A LG # 0.
Proof. Assume by way of contradiction that LG = 0. Form the skew group ring RG and let I = RGL so that 0 # I a RG, by Lemma 26.2(iii), and I C_ LG. If I2 # 0 then, by that same lemma, trG(L) # 0 and hence LG # 0, a contradiction. Thus I 2 = 0 and then clearly I n R = 0. We apply Lemma 18.5 to I and use its notation. In particular, L for each 2 E A there is a subset 1 E A C G, nonzero ideals A,
6. Group Actions and Fixed Rings
278
and additive bijections f,: A1
-+
A, with
fl
= 1 satisfying
( r a s ) f , = r(af,)s,-’ for all r , s E R and a E A l . Note that A # { 1} since I n R = 0. By assumption, A1 contains a nonzero G-stable ideal B and for each x E A we set B, = Bf,. Then it follows from Lemma 18.7(i) Since B is Gthat each B, is an ideal of R and that !R(B,) = !R(B). stable, so is D = &R(B).Moreover, since R is semiprime, B, n D = O and B, @ D is essential in R. Let ‘ denote the natural map R -+ R/D onto the ring R’ = R / D . Then each B; is clearly essential in R‘. Furthermore, each fZ gives rise to an additive bijection f:: B’ --t BL satisfying (r’b’s’)f; = r’(b‘f:)(~’)~-’ for all r’,s‘ E R’ and b‘ E B’. Since R‘ is semiprime, it follows from Lemma 18.7(ii) that each x E A acts as an inner automorphism on Qa(R’).In particular, in its action on R’, we have Ginn # (1) since A # { 1}. But G is simple and Ginna G so we have in fact Ginn = G. Furthermore, since B embeds in R’ and B E L , we see that G acts nontrivially on R’. But again, since G is simple, this implies that G acts faithfully on R‘. Finally, R’ has no nilpotent elements by Lemma 27.l(iv) and pR‘ = 0. Thus the preceding theorem implies that [G,G] has no elements of order p. But G is nonabelian simple so G = [G,G] and we conclude that G is a p’-group. Since R is semiprime, Theorem 4.4 now implies that RG is also semiprime and this certainly contradicts
r2=o. I
With this we can prove the first main result of this section.
Theorem 27.4. [87] Let G be a finite group acting on R a ring with no nilpotent elements. If L is a nonzero G-stable (right or left) ideal of R, then LG # 0.
Proof: We proceed by induction on IGl,the case 1GJ= 1being trivial. We require three reductions. First we can assume that pR = 0 for some prime p dividing (GI and that L a R . Indeed if L has no JGJ-torsion,then LG # 0 by
27. Rings With No Nilpotent Elements
279
1
Theorem 26.3(i). Thus we may assume that for some prime p IGl we have L’ = annL(p) # 0. Then G acts on L‘ and hence on the ring L’ @ GF(p) = R’. Clearly R‘ has no nilpotent elements, pR’ = 0 and L‘ a R‘. Since L’ C L , it suffices to show that # 0. Next we can assume that G is a nonabelian simple group. Indeed suppose N is a proper normal subgroup of G. Then, by induction, LN is a nonzero ideal of R N . But GIN acts on RN so, by induction again, LG = ( L N ) G / N# 0. Thus G is simple. Now if G is abelian, then G must be cyclic of order p and Lemma 24.3(ii) implies that LG # 0. Thus we may take G to be nonabelian. Finally we can assume that every nonzero ideal of R contained in L contains a nonzero G-stable ideal. Indeed let 0 # A a R with A 2 L and suppose that A contains no nonzero G-stable ideal. By Lemma 18.3, A contains a trivial intersection ideal A # 0 and, for convenience, we may take A = A. By definition this means that for all x E G either A” = A or A ” n A = 0. Let H = GA be the stabilizer of A in G and let T be a right transversal for H in G. Then H # G, since A cannot be G-stable, so AH # 0 by induction. Furthermore since R is semiprime, Lemma 8.1 implies that the sum CtET At is direct. But then any nonzero a E AH gives rise to CtET at a nonzero element of L clearly fixed by G. Thus we may suppose that no such A exists. In other words, the hypothesis of Lemma 27.3 is now satisfied and it follows from that result that LG # 0.
This completes our work on fixed points; however we continue to study rings with no nilpotent elements. Now we take a closer look at the relationship between the nontriviality of the trace map trG and the semiprimeness of the skew group ring RG. The main result in this direction is contained in [128] which considers group actions on domains. Here we combine the techniques of that paper with those of Section 18 to obtain a slightly more general result. Let R[G] be an ordinary group ring. Then the map p: R[G]-+ R given by p(CgEG T 9 g ) = CgEG rg is a ring homomorphism called the augmentation map. In the case of crossed products, p is at best an R-module homomorphism, but it is still useful. For example it is
6. Group Actions and Fixed Rings
280
a key ingredient in the proof of Theorem 24.2 and it is crucial for Theorem 27.7 below. Let us recall how the centralizer structure in RG comes about (see Proposition 12.4). Let G be a group, finite or infinite, which acts on the semiprime ring R and suppose that G = G i n n . By definition, this means that for each g E G we can choose a unit ug E S = Q,(R) such that rg = ZL;'TZL, for all T E R. It is clear that ug is unique up to a factor of a unit in C,the extended centroid of R. Setting g = u;'g E SG, it then follows that E = CSG(S)is a free C-module with the elements g as a basis and in fact that E = C t [ G ] .Note that the C-linear span of the units ug is a well-defined C-subalgebra of S. It is denoted by B ( G ) and called the algebra of the group. With this notation we have
Lemma 27.5. Let G act on the semiprime ring R with G = G i n n and let S = Q,(R). Then the augmentation map p:
c
xEG
sxx H
c
sx
xEG
defined on E = C S G ( S )C SG is a ring antihomomorphism from E onto B(G). Furthermore it is a C-homomorphism and if g E G then p ( g ) is a unit of S with rg = p(g)rp(g>-' for all r E R.
Proof. The map p is certainly a C-module homomorphism. For multiplication, we first note that C z E G s 2 xE E if and only if each s x x E E. Thus we need only consider p((sz)(ty)) with s , t E S , x,y E G and sx,ty E E . But then sx commutes with t so (sx)(ty) = t(sx)y = (ts)(xy) and we have P((SX)(tY>)= P ( W ( X Y ) ) = t s = p(ty) p(sz)
as required. Finally since g = u;'g E E , we have p(g) = u;' and hence r g = p ( g ) ~ p ( g ) - ' for all T E R. Moreover since E is the Clinear span of the elements g, we conclude that p ( E ) = B(G) and the result follows. a
If I is an ideal in a semiprime ring R,then t ~ ( 1=)rR(I). Thus the annihilator of I is unambiguously defined. The following is a special case of Theorem 27.7.
27. Rings With No Nilpotent Elements
281
Lemma 27.6. Let G act on R, a ring with no nilpotent elements, and let S = Q,(R). Assume that G is an elementary abelian p-group for some prime p and that G = Ginn. If C S G ( S )is not semiprime, then there exists a subgroup H of G such that trH(R) annihilates a nonzero ideal of R.
ProoJ We know that C S G ( S )= Ct[G]where C is the extended centroid of R and that, by the previous lemma, there is a ring antihomomorphism p: C'[G]+ B where B is the algebra of the group. Furthermore, p is a C-homomorphism, C is a commutative von Neumann regular ring by Lemma 18.6 and B has no nilpotent elements by Lemma 27.1(ii). Since Ct[G]is not semiprime by hypothesis, Lemma 18.l(iii) implies that there exists a maximal ideal M of C such that the central localization C t [ G ]is~not semiprime. Here we are localizing at the multiplicatively closed set C \ M . Since Ct[G]is free over C, we have C'[G]M= ( C M ) ~ [ GFur]. thermore, by Lemma 18.1(i), CM = K is a field. Now Kt[G]= C'[G]Mis not semiprime and G is an elementary abelian pgroup. Thus it follows from Theorem 4.4 that charK = p and then from Lemma 16.3(ii) that Kt[G]is commutative. Note that the map p extends to a K-algebra epimorphism p ~Kt[G] : = C'[G], + B M . Furthermore BM has no nilpotent elements. Indeed suppose =0 (in B,> with b E t3 and t E C \ M . Then there exists tl E C \ M with b2tl = 0 (in B ) . But btl E B has square zero, so btl = 0 and we conclude that bt-' = 0 (in a,). Let W be a subgroup of G maximal with K t [ W ]being a field and say K t [ W ]= F . Since KtfG] is not semiprime, we can choose g E G \ W and we set H = (W,g ) = W x (9). Again since G is elementary abelian, we have g p = a1 E K F and then clearly K t [ H ]E F[[]/([P- a l ) . But K t [ H ]is not a field so the polynomial [P-al E F[<]is reducible and hence [P-al = ( [ - a ) P for some a E F . In other words, (9 - u)P = 0. Since BM has no nilpotent elements, we conclude that p ~ ( g a) = 0 so that p ~ ( i j = ) p ~ ( a in ) B M . It follows easily from this that there exists a nonzero element e E C and an element a E C t [ W ]with p(g)e = p ( a ) in B. Furthermore, since C is von Neumann regular, we can assume for convenience that e is an idempotent.
282
6. Group Actions and Fixed Rings
We can now relate this to the trace map. Suppose r E RW. Then r is centralized by p ( a ) and hence by p(ij)e. Since 7-9' = ~ ( g ) ~ r p ( i j > - * and e is a central idempotent, we see that erg' = er and hence etr(g,(r)=peer = 0. Note that tTH = tr(g)otrwand that trw(R) C RW. We conclude therefore that etrH(R) = 0. Finally there exists an ideal A of R with 0 # Ae & R. Then ( A e )tTH(R) = 0 and, since e E C centralizes R, Ae is a nonzero ideal of R which annihilates trH (R). I Now we come to the second main result of this section.
Theorem 27.7. Let G be a finite group acting on a ring R with no nilpotent elements. Then the skew group ring RG is semiprime if and only if, for all elementary abelian p-subgroups H of G , tlH(R) annihilates no nonzero ideal of R. Proof:Suppose first that RG is semiprime and let H be any subgroup of G . Then R H is semiprime by Theorem 18.9. Furthermore if J is any nonzero ideal of R, then R H J is a nonzero left ideal of RH and 0 # ( R f i J ) 2= R f i t r H ( J ) J . Thus 0 # t r H ( J ) J 5 trH(R)J and trH(R) annihilates no nonzero ideal of R. Conversely suppose that, for all elementary abelian psubgroups H of G, trH(R) annihilates no nonzero ideal of R. The goal is to show that RG is semiprime. By Theorem 18.10 we need only show that RW is semiprime for every elementary abelian psubgroup W of G. In other words, it suflices to assume that G itself is an elementary abelian pgroup. Furthermore, by induction we can assume that RGo is semiprime for all properly smaller subgroups Go. Suppose by way of contradiction that I is a nonzero ideal of RG with I' = 0. We apply Lemma 18.5 to I and use its notation. In particular, there is a subset 1 E A E G, nonzero ideals A, C L for each x E A and additive bijections f z : A A, satisfying ---f
( r a t ) f , = r ( af x ) t " - l for all r,t E R and a E A. Here A = A1 and
f1
= 1. In addition
27. Rings With No Nilpotent Elements
283
Note that I n R ( A ) # 0 so we must have (A) = G since the skew group rings of all proper subgroups are semiprime. Let B = !?R(A). Then it follows from Lemma 18.7(i) that B = !R(A,) for all x E A and that B is stabilized by (A) = G. Moreover, since R is semiprime, A , n B = 0 and A, CB B is essential in R. Let ’ denote the natural map R -+ RIB onto the ring R’ = R / B . Then each A; is clearly essential in R’. Furthermore, each f, gives rise to an additive bijection f;: A’ A’, satisfying (~’a’t’)fL = ~’(a’fL)(t’)~-’ for all r‘,t’ E R’ and a’ E A‘. Since R‘ is semiprime, it follows from Lemma 18.7(ii) that each x E A acts as an inner automorphism on Qs(R’).Hence since G is generated by A, we have G = Ginn in its action on R‘. Now the map ‘ extends to an epimorphism RG R’G and 1’, the image of I , contains { C z E A ( a ’ f L )Ixa E A }. By Lemma 18.7(ii) there exists a unit q, E S = Q,(R’) with a’f; = a’gz for all a‘ E A’ and such that the action of x-l on R’ is the inner automorphism inq,x E duced by 4%. Thus in SG we see that I‘ 2 A’y where y = CSG(S)= Ct[G]. In other words, in the notation of Lemma 18.8(iv), y is a nonzero element of the ideal I’ of C’[G]. But = 0 so (1’)2= 0 and we conclude that Ct[G] is not semiprime. Lemma 27.6 now implies that there exists a subgroup H of G and a nonzero ideal J’ of R’ with J’trH(R’) = 0. Thus if J is the R B. ) complete inverse image of J‘ in R, then J 3 B and J ~ ~ H ( E It follows that AJtrH(R) C AB = 0. Since J 3 B = rR(A), we see that A J is a nonzero ideal of R. But trH(R) annihilates A J , so this contradicts the hypothesis and the result follows. 8 --f
--$
The formulation of the original result on domains is much more satisfying.
Corollary 27.8. [128]Let G be a finite group acting on a domain R. The following are equivalent. i. trG(R) # 0. ii. trG(I) # 0 for all nonzero right ideals I of R. iii. The skew group ring RG is semiprime. Proofi (iii)
+ (ii) is immediate from Lemma 26.2(iii) and (ii) + (i)
6. Group Actions and Fixed Rings
284
is obvious. We prove that (i) + (iii). Thus suppose that trG(R) # 0 and let H be any subgroup of G. If T is a right transversal for H in G and if T E R, then trG(T) = CtET trH(r)t. Thus it follows that trH(R) # 0. But R is a domain so trH(R) annihilates no nonzero ideal of R and Theorem 27.7 implies that RG is semiprime. I These results cannot be extended beyond rings with no nilpotent elements. For example, let K be a field of characteristic p > 0 and let R = M,(K). Then R contains an element u of order p which is a full p x p Jordan block with 1's on the main diagonal and the super diagonal and with 0's elsewhere. The group G = (9) of order p then acts on R with g acting as conjugation by u. We first observe that trG(R) # 0 and indeed that trG(e1,l) # 0. To this end, note that u-iel,l = e1,l for all i and hence that el,^)"' = el,lui. Thus P-1
P-1
i=O
i=O
= el,l(l - u)p-l = el,lel,p = el,p.
Now R is simple so trG(R) annihilates no nonzero ideal of R. F'urthermore, R = &,(It), C = K and RG = R @ K K ~ [ with G ] g = u-lg. But g p = 1 so Kt[G]= K[G]is not semiprime and therefore neither is RG.
EXERCISES 1. Let R be semiprime and let A be a nonzero ideal of R. If B = !R(A)show that ( A @ B ) / B is essential in RIB. 2. Let f: R A -+ R R be a left R-module map with A Q R and let I a R. If I A = 0, prove that Af E rR(I). Now suppose R is semiprime with finitely many minimal primes P I ,P2, . . . ,P, and set A%= ! R ( P ~ )Prove . that A; # 0, Pi = ! R ( A ~and ) that the sum CZ1Ai is direct and yields an essential ideal of R. 3. Suppose R is a semiprime ring with finitely many minimal primes PI, P2,. . . ,P,. Prove that n
i=l
285
28. Prime Ideals and Fixed Rings
To this end, use the preceding exercise to show that the above right hand side satisfies the semiprime analog of the defining properties given in Proposition 10.4. In the remaining problems, R is a ring with no nilpotent elements and P is a prime ideal of R.
4. If a , b E R with ab = 0, show that ba = 0 and then that aRb = 0. Deduce that R is prime if and only if it is a domain. 5. Let a , b E R \ P and suppose abz = 0 with z E R. Show that cz = 0 for some c E R \ P. Conclude that the multiplicatively closed subset A of R generated by R \ P does not contain 0. 6. Let Q be an ideal of R maximal with respect to Q n A = 8. Show that Q is prime and that Q P. Finally suppose P is a minimal prime of R so that Q = P. Deduce that A = R \ P and that RIP is a domain. This is a result of [7]. 28. Prime Ideals and Fixed Rings Probably the most successful application of skew group rings to Galois theory concerns the behavior of prime ideals in the ring extension R 2 RG.To start with, let us recall some basic properties of prime ideals in crossed products of finite groups. Let R*G be given with G finite and with R a G-prime ring. Then Theorem 16.2 asserts that 1. A prime ideal P of R*G is minimal if and only if P n R = 0. 2. There are finitely many such minima1 primes P I ,Pz,. . . ,P, with n 5 [GI. 3. If Q is a minimal prime of R, then { Q” I z E G } is the set of all minimal primes of R and Q” = 0.
nzEG
Furthermore if R has no IGI-torsion, then (3) and Theorem 4.4imply that R*G is semiprime and thus that Pi = 0. These results then give rise to appropriate Incomparability, Going Up and Going Down properties as described in Theorem 16.6. In
n:=l
286
6. Group Actions and Fixed Rings
addition, Proposition 16.7 asserts that P i is a primitive ideal of R*G if and only if Q is a primitive ideal of R. As a consequence of all of this we have
Corollary 28.1. Let R*G be given with [GI-' E R and let A be a G-prime ideal of R. Then A*G = PI n P2 n - - - n P,, an intersection of n 5 JGI minimal covering primes. Furthermore if P is a prime ideal of R*G, then P = Pi for some i if and only if P n R = A. Proof: Let -:R*G
+ (R*G)/(A*G)= (R/A)*G denote the natural epimorphism. Then k = R / A is a G-prime ring with no ]GItorsion since /GI-' E R. By the above observations, we see that &G has n 5 IGl minimal primes which intersect to zero and hence, if PI, P2,. . . , P, are their complete inverse images in R*G, then each Pi is a prime ideal and Pi = A*G. Finally if P is a prime of R*G then, by the above again, P = Pi for some i if and only if P 2 A*G and P n k = 0. Since the latter conditions are easily seen to be equivalent to P n R = A , the result follows. I
Now suppose that G is a finite group acting on R with \GI-' E R. Then we recall from Lemma 26.10(ii) that e = CzEG 2 is an idempotent in the skew group ring RG and that eRGe = eRG is ring isomorphic to RG. In this situation, Lemma 17.8 applies to relate the primes of RG to those of RG.Specifically if 1 is an ideal of RG, we define I p = ele = I n (eRGe) so that I" is an ideal of eRGe. Then Lemma 17.8 asserts that cp yields an inclusion preserving (in both directions) bijection between the primes of RG not containing e and all the primes of eRGe. Note that Iq = eJ for a uniquely determined ideal J of RG. By combining all of the above observations, it was shown in [113]that R and RG have equal prime lengths. However that paper did not specifically look at the primes of RG. It remained for [131] to complete the formalities. The following result uses the function cp as defined above.
&
Lemma 28.2. Let G act on the ring R with /GI-' E R and let A be a G-prime ideal of R. Then AG = Q1n Qz n n Q k , a finite - a -
28. Prime Ideals and Fixed Rings
287
intersection of k 5 [GI minimal covering primes. firthermore, if Q is a prime ideal of RG,then Q = Qi for some i if and only if eQ = P9 for some prime P of RG with P f l R = A .
Proof: By Corollary 28.1 we have AG = PI n P2 n - - .n P, with n 5 IGI. Thus if P: = eQi, then Lemma 26.10(i) yields
and hence AG = Q1 n Q2 n . . . n Qn.We can of course delete those Qi's equal to RG or equivalently those Pi's containing e. When this is done and the primes are suitably relabeled, we then have, by Lemma 17.8, AG = Q 1 f l Qz n.. . n Q k an intersection of k 5 n 5 IGI primes. Furthermore, since the Pi's are incomparable, so are the Qi's. Hence Q l , Q 2 , .. . ,Qii: are precisely the minimal covering primes of the ideal AG of RG. Finally if Q is a prime of RG,write eQ = P'f' for some prime P of RG. Then Q = Qi if and only if P = Pi and hence if and only if P n R = A . Note that this P necessarily satisfies e !$ P. I With this result in hand, we can relate the primes of RG to those of R. To start with, if T is a prime ideal of R and Q is a prime of RG,we say that T lies over Q or equivalently Q lies under T if Q is a minimal covering prime of T f l RG. As usual we will describe basic relations between the primes diagramatically. Thus €or example the middle diagram in (iv) below is read as follow. Suppose Q1 _> Q2 are primes of RG and TZis a prime of R lying over Q2. Then there exists a prime TI of R such that TI lies over Q1 and TI 2 T2. The key result on primes is
Theorem 28.3. [131] Let G act on R and suppose that (GI-' E R. The following basic relations hold between the prime ideals of R and of RG. i. Cutting Down. If T is a prime ideal of R, then there are k 5 IGI primes Q1, Q2,. . . ,Qk of RG minimal over T n RG and we have T n RG = Q1 n Q2 n .. - n Q k .
6. Group Actions and Fixed Rings
288
ii. Lying Over. If Q is a prime ideal of RG, then there exists a prime T of R, unique up to G-conjugation, such that T lies over Q. Furthermore, the distinct T" with 2 E G are incomparable. iii. Incomparability. Given the lying over diagram
Tl
/
I T2
Qi
I
/
Q2
Then TI = T2 if and only if Q1 = Q2. iv. Going Up and Going Down. We have at least
Ti
/ Qi
Ti
I T2
/
T2
Qi
I Q2
Tl
Q2
/
I
:
T2
Qi *:
Q2
Proof: (i) Let T be a prime of R and observe that for any 2 E G we have T" n RG = (T n RG)"= T f7 RG. Hence if A = T", then A is a G-prime ideal of R and AG = A n RG = T n RG. Now apply
nxEG
Lemma 28.2. (ii) Let Q be a prime ideal of RG and define the prime P of RG and A 4 R by P" = eQ and A = P n R. Then A is a G-prime ideal of R so, by our earlier observations, A = T" for some prime T . Since T fl RG = A n RG,we see from Lemma 28.2 again that T lies over Q. Furthermore that lemma shows that A is the unique G-prime ideal of R with Q minimal over AG. Since A uniquely determines { T xI IC E G}, its set of minimal covering primes, we conclude that T is unique up to G-conjugation. In addition, the distinct T" are clearly incomparable. (iii) If TI = T2, then since Q1 is minimal over TI n RG we have &I = Q2. Conversely if QI = Q2, then the uniqueness of T in (ii) above, along with the incomparability of the various T", implies that Ti = T2.
nxEG
28. Prime Ideals and Fixed Rings
289
(iv) The first relation is obvious. We have Q1 2 TI n RG 2 T2nRG so Q1 contains a minimal covering prime Q2 of T2 nRG. For the second and third, define Piprime in RG and Ai a R by Pr = eQi and Ai = Pi n R. Then Q1 2 Q2 implies, by Lemma 17.8, that Pl r> P2 and hence that Al 2 A2. It is now a simple matter to compare the corresponding Ti's. We consider two examples of interest. First let R = M,(K) where K is a field of characteristic not 2 and, for each 1 5 i 5 n - 1, let di be the diagonal matrix di = diag(1, 1,.. . ,-1,1,. . . ,1) with -1 in the ith entry and 1's elsewhere. Let G be the group of automorphisms of R generated by g1,92,. . . ,gn-l where each gi acts like conjugation by di. Then IGl = 2,-' and G is faithful and inner on R. The latter implies that RG = R 8~ E where E = Kt[G]. But E is semiprime and is generated by the commuting elements digi of order 2, so E "= K, a direct sum of ]GI copies of K . Thus RG cBC[~'R and therefore RG has IGl = 2,-' minimal primes. On the other hand, RG is the ring of diagonal matrices, so RG has precisely n minimal primes. Thus for n 2 3 we see that there are primes of RG lost under the cp map; that is, there are primes containing e. This extends an example of [131]. Second, let A be a simple domain over the field K with char K # 2 and assume that A is not a division ring. Then we can choose I # A to be a nonzero left ideal. For example we could take A to be the Weyl algebra A l ( K ) = K[x, y 1 xy - yx = 11 with char K = 0 and I = Ax. Now define
@xiG'
It follows easily that R is a prime ring, so
T2
of R. Also observe that T I =
is a maximal two-sided
(; i)
= 0 is a prime ideal
ideal of R with R/T1 = K . Let G be the group of automorphisms of R generated by conjugation by diag(1,-1). Then IG( = 2 and RG = diag(K I , A ) . Thus RG has two minimal primes, one of which is Q2 = diag(K+I,O). But RG/Q2 2 A so Q 2 is also maximal.
+
290
6. Group Actions and Fixed Rings
Thus we see that there exists no prime Q1 of RG which completes the diagram
Tl
In other words, the missing Going Up result in Theorem 28.3 does indeed fail. It does not fail in RG, but the prime we get may contain e and hence not correspond to a prime of R". This is from [138]. Thus the missing Going Up result in Theorem 17.9 also fails. Indeed, we need only observe, by Lemma 24.6, that R is a G-graded ring with identity component R1 = RG. Now it is clear that Theorem 28.3 yields a one-to-one correspondence between the G-conjugacy classes of prime ideals of R and certain finite subsets of primes of RG. To be precise, if Q1 and Q2 are prime ideals of RG, write Q1 Q2 if and only if Q1 and Q2 lie under the same prime T of R. This then defines an equivalence relation whose classes are finite of size IGI and there is a one-to-one correspondence between these classes and the G-conjugacy classes of primes of R. It is natural to consider which ring theoretic properties are shared by equivalent primes. Obviously these include those properties which are inherited by lying over and lying under primes. Some examples are as follows. Recall that the height of a prime P of R is the largest n such that Po C PI C . . C P, = P is a chain of primes in R. If no such maximum n exists, then the height is infinite. The depth of P can be defined similarly by looking at primes containing P. Equivalently, it is the prime length of R I P . N
<
Proposition 28.4. [113] [131] Let G act on R with [GI-' E R and let T be a prime ideal of R which lies over Q. i. T is primitive if and only if Q is primitive. ii. T and Q have the same height. iii. R and RG have the same prime lengths and the same primitive lengths.
291
28. Prime Ideals and Fixed Rings
Proof: (i) This is a consequence of Proposition 16.7, Lemma 17.8 and the argument in the proof of Theorem 28.3(ii). (ii)(iii) These are immediate from Going Up, Incomparability and (i) above. 1 In the case of Goldie rings, we are also concerned with the relationship between the ranks of R/T and of the corresponding factor rings RG/Qi. For this we require the following special case of the additivity principle of [84]. If e and f are idempotents of a ring R, we write e f if and only if eRR E fRR. As is well known (see Exercise l), this occurs if and only if there exist elements u,v E R with e = uv and f = vu.
-
Lemma 28.5. Let S 2 R be rings with S semisimple Artinian and write S = @ C: Si, a direct sum of simple rings. If f i is a primitive idempotent in Si,then rank R = Cf=,(rankSi)(rank fiRR). Proof: We know that Si
2 Mn$(Di), a full matrix ring over a di-
vision ring, and we let { fi,,, f i , z , . . . ,fi,ni 1 be a family of primitive orthogonal idempotents summing to the identity of Si. Then 1= fi,j is an orthogonal decomposition of 1 E S 2 R and hence R = @ fi,jR. Observe that fi,j fi in S and hence in R, by the above comments, so fi,jRR G! f&. Thus computing ranks yields rankR = ni . rank fiRR and the lemma is proved since ni = rank Si. 1 N
x:
As a consequence we have the additivity principle applied to fixed rings.
Theorem 28.6. [lo71Let G act on R, a ring with no (GI-torsion, and suppose that R is G-prime and Goldie. If T is a minimal prime of R with H = GT, the stabilizer of T in G, and if Q 1 , Q2,. . . , Q k are the minimal primes of RG,then k
zi . rank (RG/Qi)
rank ( R / T )= i=l
for suitable integers xi satisfying 1 5 xi 5 \HI.
292
6. Group Actions and Fived Rings
Proof. Since R is G-prime, it is semiprime and Theorem 26.9 applies. In particular, RG is also semiprime Goldie and Q(RG) = Q(R)G. Furthermore, it is known (see Exercises 2-4) that the minimal primes of R correspond in a one-to-one manner to the primes of Q(R) and indeed that their factor rings have equal Goldie rank. In view of these remarks, we may clearly replace R by Q(R) and assume that both R and RG are semisimple. Observe that IGl is now invertible in R. Write R = R1@R2 @ . . - @ Rj, a direct sum of simple rings, and say T = R2@. [email protected] R is G-prime, these factors are permuted transitively by G and hence it is clear that the projection map onto R1 defines an isomorphism RG S RF. Thus since R1 2 R/T, we may replace R by R1 and G by H and then assume that T = 0. Finally write RG = s = & @ & @ - - - @ & as in the previous lemma. Then by that lemma we have k
rank(R/T) = rankR = z(rankS,)(rankfiR). i= 1
Since Si = RG/Qi and zi = rank f i R is ~ a positive integer, it remains to bound these integers. To this end, we see from Lemma 26.l(ii) that (fiR)G = fiRG is an RG-module of rank 1. Hence Theorem 26.3(i) implies that fiR is an RG-module of rank 1 and Lemma 26.5 yields the result. Let us consider two more examples, this time with /GI-' 4 R. First, in the example at the beginning of Section 24 we have R = M2(K(z,y)) with K a field of characteristic p > 2 and with G a group of automorphisms of R of order 2p2. Furthermore, RG G K . Thus RG has one prime ideal, namely 0, while R has infinitely many. This certainly precludes a reasonable correspondence between the primes of R and of RG. Second, suppose B is a simple domain over a field K of characteristic p > 2 and let R = M2(B) so that R is also a simple ring. Define G to be the group of units of R generated by
1 0 ( 0 -1)
(;
:>
(; T)
293
28. Prime Ideals and Fixed Rings
where z is some noncentral element of B . Then \GI = 2p2 again, G acts by conjugation on R and RG 2 C,(x). Suppose in addition that C B ( ~is )the group ring K[(z)]= K[z,z-l]. Then we see that R has only one prime, namely 0, but RG has infinitely many primes and again no correspondence can exist. Note that a candidate for B is B = K [ z ,z - l , y, y-l I zy = Xyz] where X is an element of infinite multiplicative order in K' (see Exercise 5 ) . Finally we briefly consider the relationship between J(R) and J(RG). The first result in this direction is based on the well known fact that i f f is an idempotent in a ring S, then fJ(S)f = J ( f S f ) .
Theorem 28.7. [126] Let G act on R with 1GI-l E R. Then J(RG)= J(R) n RG = J(R)G.
Proof: Form the skew group ring RG and let e = IGl-lG. Then by Lemma 26.1O(i)(ii) and Theorem 4.2 we have
eJ(RG) = e(J(RG))e = e(J(R)G)e = eJ(R)G. Thus we see that J(RG) = J(R)G. I The above conclusion fails if 1GI-l $ R. In fact it is easy to construct examples (see Exercise 6) with R simple but RG not semiprime or with RG simple but R not semiprime. Even so, there is more that can be said about this situation. Indeed, we offer a generalization of the preceding theorem from [117] using the recent clever approach of [178].
Lemma 28.8. Let I a R. i. Suppose L1 and L2 are right ideals ofR with I+L1 = I+L2 = L 1 f L2 = R. Then I + (L1 n L2) = R. ii. Let Ml, M2,. . . ,Mk be maximal right ideals of R. If I Mi for all i, then I IM~= ) R.
+ (n:
Proof: (i) First I Next, since I
+
+
= RI = ( L 1 L2)I so L1 = R we see that L1
+
I = (In L I ) ( I n L2). (I n L2) = R and hence
+
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6. Group Actions and Fixed Rings
+
+
L2 = (L1n L2) ( I n L2). Finally using I La = R and the above we obtain I (L1n L2) = R. (ii) This follows easily from (i) by induction on I;. I
+
Proposition 28.9. [178]Let G act on R and let I be a G-stable right or left ideal. I f trG(I) J(RG), then /GI . I c J(R).
Proof:Write n = \GI. We proceed in a series of three steps, assuming throughout that I is G-stable.
Step 1. If I a R and M is a maximal right ideal of R, then there exists i E I with ( n - trG(i)) I C M . +
Proof: Since I a R, we may assume that I M . Then I + M = R and hence, for all g E G, I + Mg = R. Part (ii) of the previous lemma now shows that I + i@ = R where l@ is the G-stable right ideal M = n g E G M g . Thus there exists i E I with 1 - i E &f. Furthermore, for all g E G , 1 - is E l@,so summing over g yields n - trG(i) E M M . We conclude that (n- tTG(i)) I M I M as required.
c
c
step 2. I f l a R and trG(I) C_ J(RG), then nI C J(R).
Proof: If M is a maximal right ideal of R then, by Step 1 applied to nI a R,there exists ni E nI with (1 - trG(i)) * n21= ( n - trG(ni)) . n I & M .
Since tTG(z) E J(RG) by assumption, 1- tTG(i) is a unit in the ring and hence (1- trG(i)) n21 = n21. Thus n21 M for all such M so n21 J(R). But then ( n I ) 2G J(R) so clearly n I 5 J(R).
c
c
Step 3. I f I is a right or left ideal o f R and tTG(I) G J(RG), then nI E J(R).
ProoJ Say I is a right ideal of R and let S = I @ 2 be the natural extension of I to a ring with 1. Then clearly G acts on S and I is a G-stable two-sided ideal. Furthermore, since quai-regularity is
295
28. Prime Ideals and Fixed Rings
an internal condition and trG(I) is a right ideal of RG contained in J(RG),we see that trG(I) is a quasi-regular ideal of SG. Thus trG(I) J(SG)and, by Step 2 applied to I Q S, we have nI J(S). But again this means that nI is quasi-regular, so nI J(R) and the result follows. I
c
We can now prove
Theorem 28.10. [117]Let G be a finite group acting on R. Then \ G I . J(RG)2 J(R) n RG E J(RG).
PmoJ If u E RG is invertible in R, then clearly u-' E RG. Thus J ( R )nRG is a quasi-regular ideal of RG and J ( R )nRG & J (RG). For the other inclusion, let I = J(RG)R. Then I is a G-stable right ideal of R with tr,(I) C J(RG)by Lemma 26.1(ii). We conclude from the previous proposition that [GI I C J(R) so [GI . J(RG)E J(R). I It was observed in [178]that Proposition 28.9 can be used to give yet another proof of Theorem 24.2. Indeed suppose I is a Gstable right ideal of R with trG(I) nilpotent. We extend the action of G to the polynomial ring R[C]in the natural manner and note that t r ~ ( I [ < ]is ) still nilpotent. Thus Proposition 28.9 implies that IGI . I[C] J(R[<])so IGI . I 2 J(R[C]) n R. But, as is well known, J(R[C])nR is a nil ideal, so 1GI.I is nil. Now we can use the argument of Exercise 6 of Section 26 or certain more exotic ring extensions of R (as in [178]) to conclude that IGI . I is nilpotent.
c
EXERCISES 1. Let e and f be idernpotents in a ring R. Suppose first that e = uv and f = vu for some U , Z I E R and define a : e R + R by a ( e T ) = veT. Prove that a determines an R-isomorphism from e R to fR.Conversely suppose that there exists an R-isomorphism a:e R -+ fR.Prove that e = uv and f = vu for some u , v E R.
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6. Group Actions and Fixed Rings
2. If R is a prime Goldie ring, prove that rankR = rankQ(R). 3. Suppose R is a semiprime Goldie ring. Let T be the set of regular elements of R and let A be an essential right ideal of R. Prove that Q(R) = RT-' = A ( A n T)-'. To this end, f i s t observe by Lemma 26.8 that Q(R) = R ( A f W ) - l . Then note that A(AnT)-l is an essential right ideal of Q(R). 4. Let R be a semiprime Goldie ring and let Pl,P,, . . . ,P, be its minimal primes. Prove that Q(R) = CBCy=lQ(R/Pi). This follows fairly easily from the previous exercise along with Exercise 2 of Section 27. 5. Let B = K [ z , ~ - ~ , y , y I- 'zy = Xyz] where X is an element of infinite multiplicative order in K'. Show first that Cg(z) = K[z,z-'] and hence that (y) is X-outer in its action on this ring. Next show K [ z , z - ' ] is (y)-simple. Conclude from Corollary 12.6 and Lemma 1.4(i) that B = K[z,z-l](y) is a simple domain. 6. Construct examples of G acting on R with R simple but RG not semiprime or with RG simple but R not semiprime. Suitable 2 x 2 matrix rings will do the trick. 7. Let A = Z(2) be the ring of integers localized at the ideal (2), let R = M2(A) and let G be the group of automorphisms of R
generated by conjugation by the matrix
( y t).
Show that
and that
is a quasi-regular ideal of RG not contained in J(R). This example is from [153]and is certainly relevant to Theorem 28.10.
7
Group Actions and Galois Theory
29. Traces and Truncation The Galois theory of noncommutative rings is a natural outgrowth of the Galois theory of fields. It began with the work of [147] on inner automorphisms of central simple algebras and continued, for a number of years, to be concerned with rather special rings. Thus, for example the Galois theory of division rings was initiated in [28,75,80,81]. Furthermore, complete rings of linear transformations were investigated in [ 47,1461 and paper [187]studied continuous transformation rings. Much of this can now be found in the book [82]. In addition, simple Artinian rings were considered in [76,145] and in a series of papers leading to the monograph [198]. Somewhat later, the Galois theory of separable algebras was investigated in detail. We note, in particular, the papers [29,93,124,202]. The best results to date are now in [88]and concern group actions on semiprime rings. In that generality, the proofs are long and technical and would be inappropriate for this book. Thus the goal here is merely to convey the flavor of the subject. We do this by
297
298
7. Group Actions and Galois Theory
restricting our attention to important special cases. In particular, in this section we assume that R is prime and in the next section we assume in addition that the algebra of the group is a domain. This includes, for example, the case of X-outer actions and the case where R itself is a domain. For more information the reader should consult the papers [88]and [134]or the forthcoming book [92]. The exposition here comes from [134]and [135]. Throughout this section we assume that R is prime, Q = Q,(R) and that G acts on R. As we will see below, it is frequently necessary to assume that G is infinite. So to start with, we must describe the allowable groups. As usual, we let Gin, be the subgroup of G consisting of those elements which act as X-inner automorphisms; thus Gi,,aG by Lemma 12.3(ii). Furthermore recall that the algebra of the group B = B(G) = BR(G) is defined to be the linear span of all units q of Q such that conjugation by q gives rise to an automorphism of R contained in Ginn. It follows easily that L3 is an algebra over C , the extended centroid of R. Note also that if h E Ginn, g E G and q-lrq = rh for all r E R, then ( q g ) - l r g q g = rhg so conjugation by qg gives rise to the automorphism induced by g - l h g E Ginn. We conclude therefore that B is stable under the natural action of G on the ring Q.
Definition. We say that G is an M-group if i. IG : G i n n l < 00, ii. B is a semisimple finite dimensional C-algebra. These are the finiteness conditions required for most results. In addition, note that B(G) is spanned by units of Q acting like elements of G on R and hence B C C Q ( R ~ )In. particular, conjugation by any unit of 23 will fix RG and it is appropriate to add these units to G. To be precise, we say that G is an N-group if it is an M-group satisfying the saturation condition iii. If b is any unit of belongs to G.
B with b-'Rb
= R, then conjugation by b
299
29. Traces and Truncation
This definition differs from that of [88] and [134].The latter more restricted version of N-group, which we call a n N*-group, will be discussed in the next section. Note that N-groups tend to be infinite since, if G is not outer, then the group of units of B will usually be infinite. By definition, any M-group can be extended to an N-group having the same fixed ring. Furthermore, if G is a finite group and R has no IGI-torsion, then it follows from Lemma 27.5 and Theorem 4.4 that G is an M-group. In this section, we develop some necessary machinery. In particular, we show how to construct and truncate trace forms. These trace forms replace the trace map trG which was used so effectively in the case of finite groups. As we see below, appropriate trace forms exist for M-groups. Let A be a finite dimensional algebra over any field C. If A* = Homc(A, 6 )is the d u d of A, then A* can be given a right A-module structure by defining the functional Xa to be
Aa(z) = X(az)
for all z E A.
A,: then A is said to be a Here X E A* and a E A . If A A Frobenius dgebru. The first part of the following result asserts that A: is isomorphic t o the transpose of the left regular representation of A. For the second part, if V is a vector space over C,we say that a basis for V is compatibEe with the decomposition V = Vl@V2@. . .@Vk if it is a union of bases of the subspaces Vi. The third part contains necessary and sufficient conditions for A to be a Frobenius algebra.
Lemma 29.1. Let { a l , a2,. . . ,a, } be a C-basis for the algebra A and Jet { u;, a;, . . . , a: } be its dual basis in A*. i. If a E A with aai = C j ajci,j, then a3a = Xici,jaZ. Here of course ci,j E C . ii. If e is an idempotent in A, then { ai } is compatible with A = eA @ (1- e ) A i f and only i f { a t } is compatible with A* = A*e @ A*(l - e ) . Furthermore, when this occurs then ai E eA if and only if at E A*e.
300
7. Group Actions and Galois Theory
iii. AA 2 A2 if and only if there exists X E A* whose kernel contains no nonzero right ideal of A. Furthermore if A is semisimple, then A is Frobenius.
Proof: (i) Write aai = C j u j ~ , and j $a = ments C ~ J d, i , j E C.Then
xidi,jai with the ele-
(ii) Take a = e in the above. Then { a l , a2, . . . , an } is compatible with A = eA @ (1 - e)A if and only if the matrix [ci,j] is diagonal with 1 and 0 entries on the diagonal. Furthermore, by (i), this is precisely the same criterion for { u;, a$, . . . ,a*, } to be compatible with A* = A*e @ A*(l - e). Finally when this occurs, then ai E eA if and only if ci,i = 1 and hence if and only if a: E A*e. (iii) Observe that any module homomorphism f : A + A* is determined by f(1) = A. Moreover f ( u ) = Xu is the zero map Ker(X). Thus f is one-to-one and hence an if and only if aA isomorphism if and only if the kernel of X contains no nonzero right Ai, a ring direct sum ideal. Now if A is semisimple, write A = @ of simple rings. Since A* = @ x i A Z , it clearly suffices, in view of (ii), to show that Ai Z A: as Ai-modules. But this is trivial since dim Ai = dimA5, both modules are completely reducible and Ai has a unique irreducible module.
xi
We remark that the condition on the existence of X in (iii) above is actually right-left symmetric (see Exercise 1). Furthermore, if A has a 2-dimensional right ideal all of who subspaces are right ideals, then it is clear that no such X exists and A is not Frobenius. We will use the above lemma to construct certain trace forms. Specifically a trace form is a formal expression in the variable z given bY n i=3
where ai, bi E Q and oi E Aut(R) 2 Aut(Q). The various ai need not be distinct here. Of course T clearly gives rise to a linear function
29. Traces and Truncation
301
from Q to Q . The goal is to find suitable T , depending on G, such that T ( Q )& QG. The main result in this direction is
Proposition 29.2. 1881 Let G be an M-group and let A be a transversal for Ginn in G with 1 E A. Then there exist trace forms
QG. Furtherwith ~ i ,bi,,~ ,E B = B(G), n = dimc B and T ( & ) more we have } and { bi,g } are C-bases of B. i. For each g E A, the sets { ii. Either basis { ai,l } or { b i l l } may be prescribed beforehand. iii. If e E B is an idempotent, then { ai,l } is compatible with B = eB @ (1 - e)B if and only if { b i , l } is compatible with B = Be @ B(l - e). When this occurs, ai,l E eB if and only if bi,l E Be.
Proof: Let O:B* B be a right B-module homomorphism and let { b l , bz, . . . , bn } be a C-basis for B. If { b:, b;, . . . ,b i } is the dual basis for B*, we show that --f
QG. To this end, let us first satisfies ~ i , ~ , bEi ,B~ and T ( Q ) consider T~ = C i b i x O ( b t ) . If b E B and bbi = C j b j c i , j then, by Lemma 29.l(i) and the fact that C is central in &, we have
for all z E Q. Moreover since 0 is a right B-module homomorphism, the last term above equals (& bjzO(b;))b. Thus for all x E Q, T ~ ( x ) commutes with B and in particular is fixed by Ginn. Since A is a
302
7. Group Actions and Galois Theory
transversal for GinnaG,it is now immediate that ~ ( z=) CgEh ~1(~)g maps Q to QG. Since G is an M-group, B is semisimple, so a right module isomorphism 8:B* -+ B exists by Lemma 29.l(iii) and we use this 8 in the above construction. With any choice of basis { bi }, the trace form ~ ( z so ) obtained satisfies T(Q) C QG and ~ i ,bi,g ~ ,E B. Now ai,l = bi and b i , ~= 8 ( b t ) so both { ai,l } and { b i , }~ are bases for B. Moreover the basis { a i , }~ may clearly be prescribed beforehand by taking bi = a i , ~and the basis { b i , l } may be prescribed by choosing { b i , }~ to be the dual basis to { 8-'(bi,l) } in B** = B. Indeed if bi = F 1 ( b i , l ) * , then bt = 8-'(bi,l) so B(b;) = bi,l. Thus we have (i) and (ii) since ai,g = ( a i , l ) g and = (b+)g. Finally let e be an idempotent in B. Since 8 is an isomorphism and b i , ~= O(b;), it is clear that { b;, b;?. . . ,b;*,} is compatible with B* = B*e @ B*(1- e) if and only if { b i ~ b, ~. . ., , bi,n } is compatible with B = Be @ B(l - e). Therefore since a i , ~= bi, part (iii) now follows immediately from Lemma 29.1(ii), I
If T is a trace form, then T ( Q ) E Q, but T ( R ) need not be contained in R. Nevertheless we have
Lemma 29.3. Let T ( z ) =
xiaiz"2bi be a trace form. Then there
exists a nonzero ideal I of R with T ( I ) R. Indeed, i f J is any nonzero ideal of R, then there exists a nonzero ideal L C J with T(L) J .
Proof: Let 0 # J a R be given. Since R is prime, it suffices to assume that T ( z ) = m U b . Now choose nonzero ideals A , B of R with aA, B b C R and set L = J n (AJB)"-'. Then L clearly has the appropriate properties.
Definition. Let S be a subring of R. We say that S has the bimodule property if every nonzero ( R ,S)- or ( S ,R)-subbimodule M of Q contains a nonzero ideal of R and satisfies M n S # 0. It follows immediately from Proposition 10.4 that S = R has the bimodule property; we will use this observation freely throughout this and the next section. We remark that this definition differs from that of [88]
303
29. Traces and Truncation
and [134].We are able to use this simplified version here because of the restricted goal of our presentation. Now let T ( z )= Cy==luizutbi be a trace form. For any finitely many elements r k , sk E R we let
where bi = x k ( r k ) a i b i s k E Q. We call any such F obtained in this way a right truncation of T . More generally, if we insist that all S k above belong to S , then T is a right (R,S)-truncation of T . Similarly the trace form
where ui =
Skf&(?-k)‘a
is called a left ( S ,R)-truncation of T
The following result extends Proposition 12.5. The proof is similar, but somewhat more complicated because of the presence of the subring S.
Proposition 29.4. [88] Let S be a subring of R satisfying the bimodule property and let T ( z )= Cy=luizotbi be a trace form with bl # 0 and (TI = 1. i. There exist elements zi E Q with z1 = 1, zi either zero or a unit in Q and zis = s‘izi for all s E S . In particular, if R = S and zi # 0, then ~i is X-inner on R. ii. I f T(z) = Cy=lu i z a a z i ,then there exists a nonzero ideal J of R such that T(zj) is a right ( R ,S)-truncation o f T for all j E J . Furthermore, all right hand coefficientsin T ( z j ) belong to R. iii. There exists a right (R,S)-truncation T ( z )= uizuibi o f T with b l E S \ 0 and bi = zibl E R for all i.
cy=l
Proof: We begin with several general remarks. First, the ai’s merely play the role of place holders here. It is of no concern whether they are zero or not. Second, for any such T = ~ $ ‘ ~ b i , we let
cy=l
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7. Group Actions and Galois Theory
the support of T be the set of subscripts i with bi # 0. It is clear 5')-truncation of T, then Supp p E Supp T . that if T is a right (R, When Supp !f is empty, we say that 5? is trivial. Third, if T' is a right (R,S)-truncation off', then T' is also a right (R,S)-truncation of T . Finally if bj # 0 in the above, then the bimodule property implies that RbjS n S # 0. Thus there exists f' = uia"a bi with b j E s\o. The proof of the proposition proceeds by induction on the support size of T , which we may for convenience assume to be n. By the preceding remark we may further assume that bl E S \ 0. If n = 1 then, since 01 = 1, the result follows with z1 = 1 and J a nonzero > 1 and let 7 denote ideal contained in RblS. Now suppose the set of all right (R, S)-truncations of T. If p = uix"ibi E 7 with (Supp < n, then the result will follow by induction provided 61 # 0. Thus we may assume that all such f E 7 of support size less than n satisfy 61 = 0. One further reduction is necessary. For each i, there exists a -1 nonzero ideal Ii of R with Iibi C R. Thus if I = ni (I i )" t , then I # 0 and I"ibi C R for all i. In addition, we can choose T E I with rbl # 0. Then T = T(xr) is a truncation of T with all bi E R and b1 # 0. We can now assume that T itself has this property. As above, we may assume in addition that bl E S \ 0. We first show that if T' = uizUib: E 7 with JSuppT'J< n, then T' is trivial. We already know that b', = 0 but suppose bi # 0 for some j # 1. By truncating T' if necessary we may assume that b[i E S \ 0. Since SbiR contains a nonzero ideal of R, we have b1SbiR # 0 and we can choose s E S with blsbi # 0. For this s let
c:=l
Ci
xi
+(a) = T(x)sbi - T'(x(bjs)"y')
i=l
Here bi = bisbi - (bjs)"~'~""b;. Hence since b', = 0, we have bl = blsb[i # 0 by the choice of s. On the other hand, the above formula clearly yields b j = 0 and this contradicts the assumptions on 7. Thus all nontrivial elements of 7 have support size n.
305
29. Traces and Truncation
Now we return to T itself. Since RblS is a nonzero ( R , S ) bimodule, it contains a nonzero ideal J of R. Thus for each j E J there exists a right ( R ,S)-truncation T'(z) of T with n.
i=l
amd b i ( j ) = j. In fact, Tj is unique since if Tj and TY are two such truncations with the same 1-coefficient j , then Tj-T'' is a truncation of T with support size less than n and hence must be trivial. In other words, Tj' = Tj". Thus, for each i, we see that b6: J + R is a welldefined function. Furthermore, a similar uniqueness argument shows that each bi is one-to-one. Now b: is surely additive, but it is not a left R-module homomorphism. Indeed by comparing T j ( z r ) and T:j(z) we see that b$(rj) = r"ibi(j). But then the composite map (bi)'yl: J + R is a, left R-module homomorphism and hence represents, by Proposition 10.2(iv), an element gi of the left Martindale ring of quotients I a--1 Qe(R). In other words, for each j E J we have (bi) ( j ) = jgi, so b : ( j ) = j'iggi = j'izi where we have set zi = gg* E Qe(R). Of course z1 = 1 since b i ( j ) = j . Note also that the image b : ( J ) is a nonzero (R,S)-bimodule and hence contains a nonzero ideal of R. Thus, since bi is one-to-one, it follows, using the back map and the a,bove argument, that zi is in fact invertible in Qe(R). Let T(z) = C~=t=luizatzi. Then, for all j E J , we see that Z'(zj) = ~ i a i z u i j u i z=i Tj/(z)is a truncation of T ( z ) .Thus (ii) is proved. For (i), we must further study the elements zi. Let s E S. Then by comparing T',(z)sand T j s ( z ) ,we obtain the required identity zis = suizi for all s E S. With this we can now show that zi is a unit of Q = Qs(R).To start with, note that zi E Qe(R),Juazi G R a,nd ziS = Suizi. Next, z i ( J n S ) = ( J n S)uizi C R and J n S # 0 by the bimodule property. It therefore follows that 2
zi S ( J n S I R = sul. Z ~ ( Jn S ) . R a
G R.
E3ut S ( J n S ) R contains a nonzero ideal of R, so Proposition 10.6 iimplies that zi E Q,(R) = Q. Similarly, replacing Jui by the domain
306
7. Group Actions and Galois Theory
of definition of z i l , we see that z i l E Q and zi is a unit of Q. Thus (i) is proved. Finally for (iii), choose j E J n S with j # 0. Then j"izi = zij so n n
T ( z )= T(zj) =
c
uizbQj"izi=
i= 1
and the result follows with
bi
c
UiZ"iZ2j
i=l
= zij. I
Since the bimodule property and Q,(R) are both right-left symmetric, the left analog of the above applies to left (S,R)-truncations. We will not bother to state the result, but we will feel free to use it when necessary. Let T ( z ) = Ciuizaibi be a trace form. If any ui is X-inner induced by the unit qi of Q, then for any z E Q we have uizbibi = (aiq;')z(qibi). Because of this we can usually assume that no Xinner automorphism other than u = 1 occurs in T . Indeed we say that T is an outer truce form if ui X-inner implies ui = 1. Note that the traces constructed in Proposition 29.2 all have this property since 1 E A. If we allow elements of C to pass across the Z" factors in trace forms, then we have
Lemma 29.5. Let T ( z ) =
xiuizaibi be an outer trace form with
bl # 0. i. There exists a right truncation T ( z )= a& of T with bl E R \ 0 and (I! = C' aici. Here ci E C,c1 = 1 and the sum is over { i 1 02 = 1}. ii. Suppose that { ai I ui = 1} is C-linearly independent. If I is a nonzero ideal of R, then T ( I )# 0. (TI = 1 and
Proof. (i) This follows easily by applying Proposition 29.4(i)(iii) to T ( z )with S = R. If bi # 0, then ui is X-inner and hence cri = 1, by assumption. But z i l induces cri so zi E C. Thus bi = cibl with ci = zi E C and c1 = 1. Letting C' indicate the sum over {i I( ~ = i 1}, we conclude that
29. Traces and Truncation
307
as required. (ii) If T ( I )= 0, then certainly T ( I )= 0 for any right truncation T of T . In particular, if f’is as above, then 0 = ? ( I ) = a&. But bl # 0, so it follows (see Exercise 2) that 0 = a = C’aici and this contradicts the linear independence of the ai’s with ui = 1. I As an indication of the power of this machinery, we offer
Proposition 29.6. 1881 Let G be an M-group acting on the prime ring R . If I is a nonzero ideal of R, then I n RG # 0.
ProoJ: By Proposition 29.2 and Lemma 29.3 there is an outer trace form ~ ( z = ) ai,gzgbi,g and a nonzero ideal J 2 I with r ( J ) C I n QG = I n R< Now apply part (ii) of the previous lemma. I
xi
Note that the example at the beginning of Section 24 shows that some hypothesis on G is required for the existence of fixed points, even if G is finite. Additional consequences of the machinery will be considered throughout the next section.
EXERCISES 1. Let A be a finite dimensional C-algebra and let X E A*. Show that XA # A* if and only if there exists 0 # b E A with XA(b) = 0. Conclude that A A A2 if and only if there exists X E A* whose kernel contains no nonzero left ideal of A . 2. Let I be a nonzero ideal of the prime ring R and let a , b E Q,(R). If aIb = 0, prove that a = 0 or b = 0. To this end, choose nonzero ideals A , B of R with Aa, bB C R and observe that (AaI)(RbB)= 0. Now apply Lemma 10.7(ii). 3. Let G be a finite group acting on a ring R. Show that .RGR = (RG)G(RG)is equal to the skew group ring RG if and only if there exist elements a l , a2,. . . ,an and b l , b 2 , . . . , bn in R with n
1 ifg=h C ( a i ) g ( b i ) h= ( 0 i f g # h . i=l
308
7. Group Actions and Galois Theory
Here of course g , h E G. When these equivalent conditions occur, the action is said to be G-Galois. 4. Assume that the action of G on R is G-Galois and let { ai, bi } be as above. Define the left RG-module homomorphisms a : R -+ (RG)"and p: (RG)" -+ R by
Prove that cup = 1 and conclude that R is a finitely generated projective left RG-module. By symmetry, the same is true for R as a right RG-module. 5. We continue with the above notation, again viewing R as a left RG-module. Define the maps 7:EndRG ( R ) RG and 6: RG + EndRc( R )by --f
Thus S is the ring homomorphism given by Lemma 26.2(i). Show that 7 and 6 are inverse maps and conclude that RG is ring isomorphic to EndRG( R ). 6 . Let R = M2(K) with charK # 2 and let G = { 1 , g ) act on R with g acting as conjugation by diag(1, -1). Show that this action is G-Galois even though RG is not simple. 7. Suppose S is a G-graded ring with G finite and let R = S#G*. Show that the action of G on R, as described in Theorem 2.5, is G-Galois. Then use the conclusion of Exercise 5 to prove that (S#G*)G = RG %' MG(S). This is the alternate proof of duality contained in [152]and [200].
309
30. The Galois Correspondence
30. The Galois Correspondence We continue with the notation of the previous section. Thus R is a prime ring, Q = Q,(R) is its symmetric Martindale ring of quotients and C = Z(Q) is the extended centroid of R. Furthermore, G acts on R with algebra of the group B = B(G),a C-subalgebra of Q. The goal here is to obtain the usual sort of Galois correspondence between certain subgroups of G and appropriate intermediate rings containing RG. As we indicated earlier, these results will not be given in their full generality; we will make certain simplifying assumptions. These will still maintain the flavor of the proofs, but will avoid some of the unpleasant technicalities. We begin with
Proposition 30.1. [SS] Let G be an M-group acting on a prime ring R. Then C Q ( R ~=)B.
ProoJ: Certainly C,(RG) 2 B , since B is spanned by elements which induce the X-inner automorphisms of G. We consider the reverse inclusion. Let ,LI E c Q ( R ~ ) . Let e be a primitive idempotent of the algebra B and let ~ ( x=) Ci,gai,gxgbi,gand I be as in Proposition 29.2 and Lemma 29.3. Furthermore, we can assume that the C-basis { U ~ }J is chosen compatibly with the decomposition B = e B @ (1- e ) B . If T ( x ) is defined by T ( x ) = PeT(x) - .r(z)pe then, since Be E C Q ( R ~and ) ~ ( 1G)RG,we see that T vanishes on I . Furthermore, in the expression pe.r(x),we can delete all those ai,l in (1- e ) B and we use C’ t o denote such a deleted sum. Since T is an outer trace form, Lemma 29.5(ii) implies that the left-hand coefficients of
T ( x )=
c’
,Beai,gxgbi,g-
c
ai,gxgbi,g/3e
i,g
corresponding to g = 1 are C-linearly dependent. Thus there exist c i , d i E C , not all zero, with
a In particular, some ci must be nonzero. Furthermore, the U ~ Jin the left-hand sum all belong to e B and thus we have pa = diai,l
xi
310
7. Group Actions and Galois Theory
where Q = C’ciui,l is necessarily a nonzero element of eB. Since e is primitive and B is semisimple, e E aB and we conclude immediately that pe E B. Finally if 1 = el e2 . en is a decomposition of 1 into orthogonal primitive idempotents in B , then each pei E B and hence
+ + + a
-
PEB. I If S is a subring of R, let Gal(R/S) = { a E Aut(R) I a centralizes S} Thus Gal(R/S) is certainly a subgroup of Aut(R). We can now quickly prove the first main result on Galois theory.
Theorem 30.2. (Galois Group) [SS] Let G be an N-group acting OR the prime ring R. Then Gal(R/RG) = G .
Proof: Certainly Gal(R/RG) 2 G. In the other direction, suppose a E Gal(R/RG). Let ~ ( z = ) C~i,~zgb and i , ~I be as in Proposition 29.2 and Lemma 29.3. If T ( z )is defined by
then T vanishes on 1 since r ( I ) C RG. If ga is not an X-inner automorphism of R for any g above, then the only X-inner automorphisms in T ( z ) occur when g = 1 and in the first sum. However, this contradicts Lemma 29.5(ii). Thus g a is X-inner for some g E G and say this automorphism is induced by the unit q E Q. Since both g and a fix RG, it follows that q E Q ( R G )= B by the previous proposition. But then q is a unit of B normalizing R and hence q gives rise to an automorphism in G, since G is an N-group. In other words, ga E G and we conclude that aEG. I
It turns out that the N-group condition is not quite strong enough to prove the best possible results. We actually require N*groups.
30. The Galois Correspondence
31 1
Definition. Let G act on the prime ring R. We say that G is an N*-group if G is an M-group satisfying iv. If b is any unit of B ( G ) , then b-'Rb = R and conjugation by b belongs to G. Notice that an N*-group is necessarily an N-group, but the condition is definitely more restrictive. We will consider some examples after Theorem 30.4. Now we have already seen, in Proposition 30.1, how idempotents of B come into play in Galois theory arguments. Later proofs, in the presence of idempotents, become even more technical. To avoid this, we will make the simplifying assumption throughout the remainder of this section that B = B(G) is a domain and hence a division ring. Notice that this is always the case if G is X-outer, since then B = C , or if R is a domain, since then Q is a domain by Lemma 10.7(i). As we mentioned previously, the reader can consult [88], [134]or [92] for information on the general case.
Proposition 30.3. [88] Let G be an M-group acting on the prime ring R and assume that B is a domain. Let S be a subring of R containing RG. i. S is prime. ii. S has the bimodule property. iii. Suppose G is an N*-group and H = Gal(R/S). Then S contains a nonzero ideal of R H .
Pro05 (i) Suppose
s , t E S with sSt = 0 and t
# 0. Let ~ ( x =)
ui,gxgbi,gwith bl,l = 1and let 0 # I a R be as in Proposition 29.2 and Lemma 29.3. Then ~ ( 1C)RG S so s ~ ( I )=t 0. In particular, if T ( z )is the trace form T ( s )= S T ( Z ) ~ ,then T vanishes on I . Notice that T is outer and has as a right-hand coefficient b l , l t = t # 0. It therefore follows from Lemma 29.5(ii) that there exist suitable ci E C , not all zero, with Cisui,lci = 0. But { u i , ~ is } C-linearly independent, so ai,lci is a nonzero element of B and hence a unit of B. Since this element annihilates s, we conclude that s = 0.
Ci
312
7. Group Actions and Galois Theory
(ii) Let M # 0 be an (S,R)-subbimodule of Q and choose m E M\ 0. Let ~ ( x=) ai,,z9bi,, and I be as in Proposition 29.2 and Lemma 29.3. Since ~ ( 1E)RG S , it follows that ~ ( 1 ) m S M C M . Thus if T ( z ) = ~ ( z ) mthen , T ( I ) M . Furthermore, since M R C M , it follows that if F is any right (R,R)-truncation of T , then F ( I ) E M . Now let f ( z ) = azbl be given as in Lemma 29.5(i). Then bl E R \ 0 and Q is a nonzero element of B since { a a , l ) is C-linearly independent. Thus aIb1 M . Note that Q is invertible in B . Since a-'J G R for some nonzero ideal J of R, we have a-lJI 2 I and hence
c
Thus M contains the nonzero ideal JIblR. A similar argument, using the left analog of Lemma 29.5(i) works for (R,S)-bimodules. Finally, by Proposition 29.6, any nonzero ideal of R meets RG and hence S nontrivially. We conclude therefore that S has the bimodule property. (iii) Since H fixes S 2 RG, Theorem 30.2 implies that H 5 G. Thus GinnH = HGinn is a subgroup of G and we can choose H. a transversal A for Ginn in G with 1 E A and A n (GinnH) Using this A, we apply Proposition 29.2 and Lemma 29.3 as usual ) C i , g a i , g x g b i , gand a nonzero ideal I to obtain a trace form ~ ( z= of R with ~ ( 1C)RG. Since RG 2 S it follows that, for any right (R,S)-truncation 5? of T , we have F ( I ) S. By (ii) above, S has the bimodule property and thus Proposition 29.4(i)(ii) applies; we use its notation, allowing for double subscripts. Suppose zi,g # 0. Then zi,g is a unit of Q and Z ~ , ~=S s g ~ i for all s E S. Since S 2 RG and g fixes RG, we see that zi,g E C,(BG) = B , by Proposition 30.1. But then, since G is an N*group, conjugation by ~ iinduces , ~ an element w of Ginn so sgw = (~i,~)-~sg =zsi and , ~ gw E H. It follows that 9 E A n (HGinn) H, by assumption, and we conclude that w E H so zi,g E C Q ( R ~ ) . ~ i , ~ z 9 z i , ,Then, . by the above, we see Now consider T(z) = that T :Q + Q is a right RH-module homomorphism. Furthermore, by Proposition 29.4(ii), there exists a nonzero ideal J of R with T(zj) a right (R,27)-truncation of 7 for all j E J . Thus F ( I j ) C S for each
c
, ~
30. The Galois Correspondence
313
is a right RH-module homomorphism, j so T ( I J ) G S. Since I J # 0 and S R H ,it follows from Lemma 29.5(ii) that V = T ( I J ) is a nonzero right ideal of RH contained in S. Similarly, using the left analogs of these results, we find a nonzero left ideal U of RH contained in S . But S is prime, so U S V is a nonzero two-sided ideal of RH contained in S and the result follows. I We say that S is an ideal-cancellable subring of R if, for all nonzero ideals I of S and elements r E R, I r G S implies that r E S. We can now obtain the second main result on Galois theory.
Theorem 30.4. (Correspondence) 1881 Let G be an N*-group of automorphisrns of the prime ring R and assume that 23( G) is a domain. Then the maps H H RH and S H Gal(R/S) yield a one-to-one correspondence between the N*-subgroups H of G and the idealcancellable subrings S with R 2 S 2 RG.
Proof. Let H be a subgroup of G . We first show that S = RH is ideal-cancellable. Let I be a nonzero ideal of S. Since H C_ G we have S 2 RG so S satisfies the bimodule property by the previous proposition. In particular, R I contains a nonzero ideal of R and hence rR(I) = 0. Now suppose I r C S . If h E H and s E I , then sr E S so ST = ( ~ r=)shrh ~ = srh and I ( r - r h ) = 0. Thus r = r h for all h E H and r E RH = S . In addition, if we assume that H is an N*-group, then it follows that H = Gal(R/S), by Theorem 30.2. Conversely suppose S 2 RG is ideal-cancellable and let H = G by Theorem 30.2 again. Gal(R/S). Then H fixes RG so H Moreover, it is clear that H is an N*-group, since B is a division ring and G satisfies (iv). By Proposition 30.3(iii), S contains a nonzero ideal I of RH. Thus I is also an ideal of S and if r E R H , then Ir I S . We conclude from the ideal-cancellable property that r E S and thus S = RH as required. I We remark that the ideal-cancellable property can be restated in terms of the left Martindale ring of quotients. Indeed, if S 2 RG then it can be shown that Qe(S) is contained naturally in Qe(R).With this embedding, S is ideal-cancellable if and only if S = Qe(S) n R
314
7. Group Actions and Galois Theory
(see Exercise 3). Note that, if B ( G ) is not a domain, then the idealcancellable condition is not sufficient for 5' to be a fixed ring. The precise properties required can be found in [88],[134]or [92]. We now consider several examples of interest. Let D be a division algebra with center K and let D(z,y) be the free D-algebra on two generators z and y. This ring is of course a domain with group of units D' = D \ 0. Furthermore, it is symmetrically closed by Theorem 13.4, since domains are necessarily cohesive. In particular, K = Z ( D ( z ,3)) is the extended centroid. For convenience, we denote the augmentation ideal zD(z,y) yD(z, y) by D ( z ,y)'. Now let F be any subdivision ring of D and set R = F D ( z ,y)'. Then R is a domain containing an ideal of D ( z ,y) so, by Lemma 10.8(iii), we have
+
+
Thus we know the unit group of Q,(R) and it follows that the group G of all X-inner automorphisms of R is induced by conjugation by
{ d E D' I d-IFd
=F
}.
In other words, G % N p ( F ' ) / K ' since only elements of K will centralize D ( z ,y)'. For our first example take F = K and assume 1 < dimK D < m. Since d-lKd = K for all d E D', it follows that all such d induce elements of G. Thus B ( G ) = D and G is an N*-group. Notice that G is not inner on R since the only units of R are in K'. For a more interesting example take D = K Ki K j Kk to be the quaternion division algebra over the real field K and set F = K + Ki = K[i]. If d E D', then d-lFd = F if and only if d - l i d = &i and hence if and only if d E F U F j . Thus the elements of G are induced by all nonzero d E F U F j so G is an N-group on R but not an N*-group. Note that B(G) = F F j = D. Since RG is centralized by B(G) = D , it follows easily that RG = K ( z ,y). Now let E = K K ( i j ) = K [ i j ] D and let S be the intermediate ring S = K E ( z ,y)'. Note that d E D' centralizes S if and only if d E C D ( E ) = E ; But E n ( F U F j ) = K , so
+
+
+
+
+
+
+
+
30. The Galois Correspondence
315
H = Gal(R/S) = (1) and RH = R(l) = R. Clearly S contains no nonzero ideal of RH = R so we see that Proposition 30.3(iii) does indeed require that G be an N*-group. Furthermore, it is easy to see that S 2 RG is ideal-cancellable, but S is not a fixed ring. Thus Theorem 30.4 also requires that G be an N*-group. The latter example indicates a failure of the theory in its present state. Certain automorphisms which should be present are missing. We expect all of Do to act on R but, by accident, some of the elements do not normalize R. This problem can be eliminated by suitably redefining what we mean by an allowable automorphism. In the case of prime rings R, an appropriate candidate is the set of all automorphisms CT of Q,(R) with the property that 0 and cr-' both send some ideal of R into R. In other words, one weakens the con-1 dition R" = R to I" C_ R and I" R. This is presumably the approach being taken in [92]. We close this section by considering some applications of this work to the relationship between R and the fixed ring RG. First we have
Proposition 30.5. [88] Let G be an M-group acting on the prime ring R. If B( G ) is a domain, then Qe(R)G= Qe(RG)and Qs(R)G = Qs (RG).
c
Proof. Note that G acts on Qe(R)and Q,(R) and RG c Qs(R)G Qe(R)G.Moreover RG is prime by Proposition 30.3(i). We show first that Qe(R)G2 RG satisfies (i)-(iv) of Proposition 10.2. To this end, let q E Qe(R)G.If 0 # I Q R with I q S R, then I n RG is a nonzero ideal of RG,by Proposition 29.6, and clearly (In RG)q5 RG. This of course yields (ii). For (iii), suppose Y is a nonzero ideal of RG with Yq = 0. Then RYq = 0 and RY is a nonzero (R,RG)-bimodule. It follows from Proposition 30.3(ii) that RY contains a nonzero ideal of R, so q = 0 and (iii) is proved. For (iv), let f : Y + RG be a left RG-module homomorphism with 0 # Y a RG. We extend f to a map f*: RY + R, in the natural manner, by f*: rkyk H r k ( y k f ) where r k E R and yk E Y . To see that this is well defined, suppose Ckrkyk = 0 and choose T = Ci,gai,gxgbi,, and I as in Proposition 29.2 and Lemma 29.3
Ck
316
7. Group Actions and Galois Theory
with b l , ~= 1. Note that bi,, E B centralizes RG and hence Y . Furthermore, for all g E A we have 0 = & ~ k y k ) ~= z k ( r k ) g y k . It therefore follows that the right truncation T ( z ) = d,zrk)yk has all right-hand coefficients equal to zero and hence T ( I )= 0. In addition, since ~ ( 1E)RG and f is a left RG-module homomorphism, we see that
T'(4 =
c k
+-k)(Ykf)
=
c
ai,gzgbilg
i,g
also vanishes on I . But T' is an outer trace form, so Lemma 29.5(ii) . have therefore shown that f * is yields 0 = bi,l = C k r ~ ( y k f ) We well defined. Again, by the bimodule property, RY contains a nonzero ideal J of R and f * :J + R is a left R-module homomorphism. It follows that there exists q E Qg(R)with j f *= j q for all j E J. In particular, if y E Y and j E J , then j y q = ( j y ) f * = j ( y f ) so J ( y q - y f ) = 0 and yf = yq. Finally, since yq E RG for all y E Y , it follows easily that q E Qe(R)Gand we conclude from Proposition 10.2 that Qe(R)G= Qg(RG).The result for Q8(RG)is now immediate from the above, Proposition 29.6 and Proposition 10.6. Finally we specialize to the case of finite X-outer groups. Recall that any such group G is necessarily an N*-group. Furthermore, if R is a simple ring, then Q,(R) = R so outer and X-outer are equivalent properties.
Lemma 30.6. Let G be a finite group of outer automorphisms of the simple ring R. Then RG is simple if and only if R contains an element of trace 1, that is if and only if 1 E trG(R).
Proof: By Lemmas 26.1 and 29.5(ii), trG(R) is a nonzero ideal of RG. In particular, if RG is simple, then trG(R) = RG and 1 E trG(R). Conversely, suppose R is simple and 1 E trG(R). If I is a nonzero ideal of RG, then I R is a nonzero (RG,R)-subbimodule of R and hence contains, by Proposition 30.3(ii), a nonzero ideal of R. Thus I R = R and 1 E trG(R) = trG(1R) C I . We conclude therefore that I = RG and that RG is simple. 1
30. The Galois Correspondence
317
Theorem 30.7. Let G be a finite group of outer automorphisms of the simple ring R and suppose that 1 E trG(R). Then the maps H H RH and S H Gal(R/S) yield a one-to-one correspondence between the subgroups H of G and the intermediate rings S 2 RG. In particular, there are only finitely many intermediate rings and they are all simple.
Proof: If H is a subgroup of G and A is a left transversal for H in G, then it follows easily that trG(r) = trH(trA(r)) and hence that trG(R) trH(R). In particular, we now know that 1 E trH(R) and hence that RH is simple by the previous lemma. Conversely let S 2 RG be any intermediate ring and let H = Gal(R/S) E G. By Proposition 30.3(iii), S contains a nonzero ideal of its overring RH. But RH is simple, so we conclude that S = RH. The result follows from Theorem 30.4. I The hypothesis that R contains an element of trace 1 is trivially satisfied if )GI-' E R or if R is a division ring. Indeed if R is a division ring, then so is RG and Lemma 29.5(ii) yields this fact. Thus the Galois correspondence for fields is a consequence of the above, as is the analogous result for divisions rings in [80]. More generally, if R is simple Artinian and G is outer, then there always exists an element of trace 1, as the following lemma shows. Therefore the Galois correspondence of [76] and [144] can also be recovered from Theorem 30.7.
Lemma 30.8. Let R be a simple Artinian ring and let G be a finite group of outer automorphisms of R. Then RG is simple Artinian and 1 E tTG(R).
Proof. Set I = tTG(R) so that I is a nonzero ideal of RG by Lemmas 26.1 and 29.5(ii). According to Proposition 30.3(i)(ii), RG is prime and satisfies the bimodule property. In particular, I R contains a nonzero ideal of R so I R = R and ~ R ( I=) 0. We show first that if J is any nonzero right ideal of RG, then J contains a nonzero minimal right ideal. Since R = M,(D), the ring of n x n matrices over the division ring D , we can choose a E J \ 0 to
318
7. Group Actions and Galois Theory
have minimal rank as a matrix in R. Then a 1 is a nonzero minimal right ideal of RG contained in J . To see this, choose b E a 1 with b # 0. Then bR C a I R = aR, so a R = b R by the minimality of the rank of a. Applying the trace map yields
and a1 is indeed minimal. Now in any semiprime ring, a minimal one-sided ideal is generated by an idempotent. Thus if J1 is a minimal right ideal of RG,then J1 = elRG for some idempotent el and RG = elRG @ (1- el)RG = J1 @ J i . Assuming J i # 0, we can find a minimal right ideal ezRG = Jz contained in J i . Since JZ is a direct summand of RG,it is a direct summand of J[ so RG = J1 @ JZ @ J;. We continue in this manner and observe that the procedure must stop after at most n steps, since R = M,(D) cannot contain more that n mutually orthogonal idempotents. Thus RG is a finite sum of minimal right ideals and hence is Artinian. Finally, since RG is also prime, the ring is simple Artinian and 1 E trG(R) by Lemma 30.6. I The remaining aspects of the Galois theory will be considered in the next section.
EXERCISES 1. Let G be an M-group on R and let e # 0 , l be an idempotent of B . If 0 # I is an ideal of R with e1 E R, show that e1 is an nonzero (RG,R)-subbimodule of R which contains no nonzero ideal of R. Thus Proposition 30.3(ii) fails, according to the definition of the bimodule property given here, if B ( G ) is not a domain. 2. Show by example that Proposition 30.3(i) also fails if B(G) = B is not a domain. It is shown in [134]that if G is an M-group, then RG is semiprime with minimal primes in one-to-one correspondence with the G-centrally primitive idempotents of B. 3. Let G be an N*-group with B ( G ) a domain and let S 2 RG with H = Gal(R/S). Use Proposition 30.5 and Lemma 10.8(iii)
31. Almost Normal Subgroups
319
to show that Qe(S) = Qe(RH)C Qe(R)and Q s ( S ) = Q s ( R H )C Qs(R).Conclude that S = RH if and only if S = Qe(S)n R and that this is equivalent to S being ideal-cancellable. 4. Let R be a prime ring and let G be the set of all u E -1 Aut(Q,(R)) such that I" E R and I" C R for some nonzero ideal I of R depending on a. Show first that I' contains a nonzero ideal of R and then that G is a subgroup of Aut(Q,(R)). Furthermore, prove that G contains all inner automorphisms of Qs(R). 5. Let R = K [ z ' , ~ - ~ , y , y -1 ~z y = Xyz] be the simple ring described in Exercise 5 of Section 28 and let CT be the K-algebra automorphism of R determined by xu = z-' and y" = y-l. Show that (T is well defined, has order 2 and is X-outer. If G = (0)and charK = 2, prove that 1 4: trG(R) and hence that RG is not simple by Lemma 30.6. This is an example of [150].Are there any nontrivial intermediate rings? 6. Suppose u is a field automorphism of K of finite order 2 3 and let F be the fixed field F = K ( " ) . Set R = Mz(K) so that G = (GL2(K),a) acts on R. Show that G is an N*-group with B = M2(K) and RG = F , embedded as scalar matrices. Now define S = { diag(a,a") I a E K } so that S 2 RG and S E K . Show that S is ideal-cancellable, but not a fixed ring.
31. Almost Normal Subgroups In this section we complete our study of the formal aspects of the Galois theory of rings. Here we are concerned with normal subgroups H of the Galois group G and with the action of G / H on the fixed ring R H . The notation will be as before. We start with
Theorem 31.1. (Extension) [SS] [134]Let G be an N*-group acting on the prime ring R with B = B(G) a domain and let S 2 RG be an intermediate ring. If 'p: S -+ R is a ring isomorphism into, with 'p the identity on RG, then cp is the restriction of some g E G.
Proof. Let us recall some notation and observations from the beginning of the proof of Proposition 30.3(iii). First, H = Gal(R/S)
s
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7. Group Actions and Galois Theory
G and A is a transversal for Ginn in G chosen with 1 E A and A n (GinnH) C H . Next, T ( Z ) = ~ i , ~ z g band i , ~0 # 1 4 R are given as usual by Proposition 29.2 and Lemma 29.3. Third, there exist elements zi,g E Q such that if xi,g # 0, then g E H and qgis a unit of C , ( R H ) . Finally if T ( z )= ~ i , ~ z g z then i , ~ , there exists a nonzero ideal J of R with T ( z j )a right (R,S)-truncation of T for all j E J. Now for each i , g we construct a map I'&,~: J -+ Q as follows. Let j E J and write T ( z j )= C kT(2Tk)Sk with T k E R and S k E S. Then we set ei,g(j)
=c(Tk)gbi,g(sk)'k
We must first show that Oi,g is well defined. To this end, suppose T ( z j )is also equal to c k T ( z r k ) s k with Fk E R and s k E S. Then for any y E I we have
k
k
and, since ~ ( 1C)RG,we obtain by applying p
In other words, the trace form
k
k
vanishes on I . Note that F(z9-l) is an outer trace form for any g E A and that the left-hand coefficients corresponding to the automorphism 1 are C-linearly independent. Thus Lemma 29.5(ii) implies that 0 = &,g = c ( T k ) g b i , g ( s k ) ' - ~ ( ~ k ) g b i , g ( ~ k ) ' k
k
and Oi,, is indeed well defined. Now it is clear that 8i,g:J + Q is additive. Furthermore, since T(xcrj) can be obtained from 5?(zj) by replacing z by Z T , we have
321
31. Almost Normal Subgroups
easily &,,(rj) = r g B i , g ( j ) . Observe that & , ( J ) C Rbi,,R and that for some nonzero ideal L of R we have Lgbi,g C R. Hence & , ( L J ) = LgBi,,(J) & R and, by replacing J with L J if necessary, we can assume that 1 9 i , ~J : R. Since the composite map (Bi,g)g-l: J --t R is a left R-module homomorphism, there exists qi,g E Qe(R) with -1 & , , ( j )g = jiji,g. Equivalently, Si,,(j) = j g q i , g where we set qi,g = --f
E Qe(R). As we observed earlier, if zi,g# 0 then g E H and zi,gis a unit of Q centralizing RH. It follows that T ( z j s )= T(zj)s for s E S and hence that 0ilg(js)= Si,,(j)s". Thus (qi,g)g
3'9
9 .
Q2,g
= f%,g(js) = Bi,,(j)S'P
= jgqz,gs'p
and, since this holds for all j E J , we have sgqi,g = q i , g s p . With this, we can now show that qi,g E Q. Indeed, it follows from Jgqi,, & R and Sgqi,g = qi,gSp that
Q ~ , ~ S ' (nJ S ) p R = Sg . ( J n S)gqi,g R C_ R. But S and S p satisfy the bimodule property, by Proposition 30.3(ii), so J f l S # 0 and S p ( J f l S ) p R contains a nonzero ideal of R. Thus since qi,g E &!(It), Proposition 10.6 implies that qi,g E Q,(R) = Q. Note further that both cp and g fk RG and hence E C Q ( R ~=)B , by Proposition 30.1. It remains to find some i , g with qi,g # 0. To this end, observe that T does not vanish on I J , by Lemma 29.5(ii), and say T ( y j ) # 0 with y E I and j E J . If T ( z j ) = c k 7 ( z r k ) s k with r k E R and sk E then 0 # T ( y j ) = & ~ ( y r k ) s k . But T ( y r k ) E RG C and cp is one-to-one and fkes RG;thus
s
s,
0 # z7(yrk)'(src)" = k
T(Yrk)(sk)'* k
In particular, for some i , g we must have 0 # x k ( T k ) g b i , g ( S k ) p = 6$,g(j).It follows for this i , g that t9i,g # 0 and hence that qi,g # 0. But then, since G is an N*-group, qi,g is a unit of the division ring B which induces an automorphism go E Ginn. We conclude, therefore, from s g q i , , = q i , g s p that cp is the restriction of ggo to S and the result follows. I
322
7. Group Actions and Galois Theory
Now suppose H is an N-subgroup of G. Then RH is a prime subring of R and it is natural to consider Gal(RH/RG). We show below that this group is isomorphic to N / H where N = NG(H)and that N / H is an N*-group in its action. The proof of the latter is routine, but somewhat tedious.
Lemma 31.2. Let G be an N*-group in its action on the prime ring R and assume that B(G) = B is a domain. Let H be an N-subgroup of G and set N = NG(H). i. The restriction map yields an isomorphism between the groups N / H and Gal(RH/RG). ii. BRH(N/H) = B H C B and hence it is a domain. iii. N / H is an N*-group in its action on R H .
Proof. (i) It is trivial to see, from Theorem 30.2, that g
E G stabilizes RH if and only if g E N = NG(H). In particular, by way of restriction, we obtain a homomorphism from N to Gal(RH/RG). Since H is an N-group, the kernel of this map is H , by Theorem 30.2 again. Moreover, the map is onto by the previous theorem. Thus N / H 2 Gal(RH/RG). ) Qs(R)H= Q H . (ii) According to Proposition 30.5, Q s ( R H= Now suppose that q is a unit of QH which acts, by conjugation, like an element of N on RH. Then q centralizes RG and hence q E B , by Proposition 30.1. It follows that BR”(N/H) E B n QH = B H . Conversely, suppose q is a unit of B H . Since G is an N*-group, conjugation by 4 on R gives rise to an automorphism g E Ginn. Furthermore, q - l R H q E R n Q H = R H . Thus g E N and conjugation by q on RH gives rise to an automorphism in ( N / H ) i n n .We conclude in particular that BRH( N / H )= B H . Note also that G acts on the field C = Z ( Q ) and that Ginn acts trivially. Thus, since G/Ginn is finite, it follows from the Galois theory of fields that F = CG is a subfield of C with (C : F ) < 00. Hence dimp B H < 00. But F Z ( Q H ) and thus BH is finite dimensional over the extended centroid Z ( Q H )of R H . (iii) It is clear from (ii) above that B R N ( N / H )is a finite dimensional division algebra over the extended centroid of R H . Furthermore, every nonzero element of this division ring gives rise, via
31. Almost Normal Subgroups
323
conjugation, to an X-inner automorphism in N / H . Thus we need only prove that the subgroup of N / H consisting of X-inner auto) write 2 morphisms has finite index. For this, let 2 = C B ( R ~and for the multiplicative group 2'. Next, let N be the multiplicative group of all units of B which give rise to elements of Ninn in their action on R. Our goal is to show that 2 a N and that Z(N n B H ) is a subgroup of N of finite index. Once this is proved, the result will follow quickly. Note that 2 and BH commute elementwise, since 2 = B ( H ) is spanned by units which act by conjugation on Q like elements of the group Hi,, C H . Suppose h E H and y E N . If y gives rise to the automorphism a E N , then y-'yh is a unit of B which gives rise to the auto= a-lh-lah E H , since N normalizes H . Thus morphism y-'yh E C g ( R H )= 2 and we conclude that yh = yz for some z E 2. This observation is used twice in the proof. Since H is an N-subgroup of G and G is an N*-group, all units of Z give rise to automorphisms in H and hence in N . Thus 2 is a subgroup of N and in fact 2 a N since H a N . This implies that Z(Nf-123~) is a subgroup of N . Now N acts on 2 by conjugation and hence also on T , the center of 2. Since T is an extension field of C of finite degree and N centralizes C , we conclude that N1 = N n c B ( T ) has finite index in N . Thus it suffces to show that z ( N n B H )which , is clearly a subgroup of NI , has finite index in N1. Let y E N1. Then conjugation by y yields an automorphism of the finite dimensional division algebra 2 which fixes its center T . By the Skolem-Noether theorem, this automorphism must be inner on 2. Thus there exists z E 2 such that z-ly centralizes 2. In other z = NI n C , ( Z ) . Thus, we have words, N1 = 2NZ where N
N2n 2 ( N n BH)= ( 2n N2)(Nn BH)= T'(N n B H > and it suffices to show that IN2 : T*(Nn BH)I < 00. Note that H acts on B , 2 and T and that Hi,, acts trivially on T . Thus , L 2 Hi,, so IH/LI < 00. Furthermore, since if L = C H ( T )then H N , it is clear that H acts on N and then on N2. We show below that IN2 : N2 n BLI < 00 and then that Nz n BL = T'(N n B H ) . This will surely yield our goal.
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7. Group Actions and Galois Theory
Observe that H i n n centralizes Nz, so the finite group L/Hinn acts on N2. Fix h E L and, for each y E N2, write yh = yX(y) where X(y) E 2 by the observation of the second paragraph. Since y h , y E N2, we see that X(y) E 2 n N2 = T o and thus X maps N 2 + T o . Indeed, if y1, y2 E N 2 then, since NZ centralizes T , we have YlY2X(YlY2) = (Y1Y2)h = ( Y l ) h ( Y 2 ) h = YJ(Yl)Y2X(Y2) = YlY2X(Yl)X(Y2)
and X:N2 + To is actually a linear character. Furthermore, since h E L acts trivially on T', we have easily yh" = yX(y)" for any integer rn. But JL/HinnJ< 00, so hn E Hinn for some n 2 1 and thus X is a homomorphism from N2 into the finite group of nth roots of unity in T . We conclude therefore that h centralizes a subgroup of finite index in N2, namely the kernel of A. Since this is true for each element h E L and since L/Hi,, is finite, we deduce that
IN2 N2n
< 00.
BL 2 T0(Nn B H ) and fix an element y E E H we have, by the observation of the second paragraph again, yh = yp(h) where p ( h ) E 2. Again since yh, y E Nz, we see that p ( h ) E 2 nN2 = T o and thus p maps H to T o . Indeed, since L acts trivially on y E N2 f l B L , p is actually a map from the finite group H / L to T o . Next suppose h l , h2 E H. Then Note that
N 2
N 2 f l
n B L . Then for each h
so p satisfies Noether's equation p(hlh2) = p ( h ~ > ~ ~ p ( hTherefore, z). by the above remarks and the fact that H / L acts faithfully on the field T , we conclude from [83, page 751 that p is a trivial crossed homomorphism. In other words, there exists t E T o with p ( h ) = t/th for all h E H. But then yh = yp(h) = yt(th)-l implies that yt E N n B H . Hence, since yt = ty, we have y = t-' ty E T o ( Nn BH). Thus N2 n BL = T'(N n B H )and, as indicated above, this yields our goal. Finally since IN : Ninn(< m, it follows that the restriction of N to RH is a subgroup of finite index in N / H . Hence the restriction of 2(Nn B H )also has finite index. But 2 E C B ( R ~restricts ) to 1
31. Almost Normal Subgroups
325
and N n BH restricts to the subgroup of X-inner automorphisms in N / H . We conclude therefore that (N/H)innhas finite index in N / H and the result follows. I Since B(G) is a domain, it follows that any subgroup of G is an M-group in its action on R. In particular, this applies to N above. On the other hand, N need not be an N*-subgroup; we will consider some examples after the next theorem.
Definition. Let G be an N*-group in its action on the prime ring R. If N is an M-subgroup of G, then N can be completed to an N*subgroup fi of G by adjoining to N the action of all units of B ( N ) . Thus clearly B ( N ) = B ( f i ) and RN = R' since any element of R fixed by N is fixed by all units of B ( N ) . We say that H is almost normal in G if for N = NG(H) we have fi = G. In addition, we say that the extension RH/RG is N*-Galois if Gal(RH/RG) is an N*-group in its action on RH with fixed ring equal to RG. We can now quickly prove
Theorem 31.3. [134]Let G be an N*-group of automorphisms of the prime ring R and assume that B = B(G) is a domain. If H is an N-subgroup of G, then the following are equivalent. i. H is almost normal in G. ii. RH is N*-Galois over RG with algebra of the group a domain. iii. Gal(RH/RG) acts on RH with fixed ring RG. Proof: (i) + (ii) If N = Nc(H), then by assumption N = G. According to Lemma 31.2, N / H = Gal(RH/RG) is an N*-group in its action on RH with algebra of the group a domain. Furthermore, we have (RH)(N/H)= RN = Rfi = RG.
By definition, RH is N*-Galois over RG. (ii) + (iii) This is obvious. (iii) + (i) By Lemma 31.2 again, Gal(RH/RG) = N / H where N = NG(H) and note that fi is aq N*-subgroup of G. By assumption, we have G - RH G a l ( R H / R G )= ( R H ) N / H = RN = RN. R 4 )
326
7. Group Actions and Galois Theory
Thus by Theorem 30.4, N = G and H is almost normal in G. We remark that most of the Galois theory results we have considered extend to the case of arbitrary N*-action without any additional assumption on B(G). Of course, other properties do come into play. For example, in the characterization of fixed rings S 2 RG, S must satisfy more than just ideal cancellation. But the above theorem is different; it is more restrictive in nature and certain assumptions on both B(G) and B ( H ) are needed even in the general case. Now let us consider some examples to show that, in the previous theorem, N need not be an N*-group in its action on R. To start with, let D = K K i K j Kk be the quaternion division algebra over the real field K and let G be the group of all inner automorphisms of D. Then G is an N*-group with B(G) = D and DG = K. Suppose H is the subgroup of G consisting of all automorphisms induced by nonzero elements of F = K + Ki = K[i]. Then H is an N*-group and B ( H ) = F = D H . Note that F / K is Galois, but only admitting outer automorphisms. Furthermore, N = NG(H)= H U H j SO N # G. Of course B ( N ) = D and fi = G. This is actually part of a more general phenomenon. Namely, let D be any finite dimensional division algebra with center K and let G be the group of all inner automorphisms. Then, as above, B(G) = D and DG = K . Furthermore, if H is an N-subgroup of G, then B ( H ) = F is a division K-subalgebra. Now observe that if H a G, then F' a D' and the Cartan-Brauer-Hua theorem (see [82, page 1861) implies that either F = K or F = D. In other words, H a G if and only if H = (1) or H = G. The following is an immediate consequence of Lemma 31.2(ii) and Theorem 31.3.
+ +
+
Corollary 31.4. Let G be a finite group of X-outer automorphisms of the prime ring R. If H is a subgroup of G, then H a G if and only if the extension RH/RG is Galois with Gal(RH/RG) a finite group of X-outer automorphisms.
For our final application of traces and truncation we allow B(G) to be a general semisimple algebra. The additional work required for
327
31. Almost Normal Subgroups
this is fairly minimal.
Theorem 31.5. [88] Let G act as an M-group on the prime ring R. Then there exists a nonzero ideal A of R with the property that, for any a E A, aR is contained in a finitely generated right RG-submodule of R and Ra is contained in a finitely generated left RG-submodule of R. In particular, if R is simple, then R is finitely generated as a right and as a left RG-module.
ProoJ For each o E B , define RGri for some n < 00 and ri E R i= 1
It follows easily that A , is a two-sided ideal of R. The goal is to show that A1 # 0. To this end, define W = { o E B I A, # 0). Then 0 E W and W is closed under addition since clearly A, n Ap C_ A,+p for a,P E B. Moreover, if 0 # J a R with P J R, then the inclusion aPJr o R r implies that A,o 2 JA,. Thus W is a right ideal of B and hence a C-subalgebra of B = B(G). Suppose by way of contradiction that W # B. As usual, let 7= ~ i , ~ z g band i , ~0 # I a R be given as in Proposition 29.2 and Lemma 29.3. Furthermore, assume that the basis { a i , }~ is chosen compatibly with B = W’ @ W where W’ is a complementary Csubspace and with a1,l E W’. By Lemma 29.5(i), there exists a right (R, R)-truncation f’of T with
c
s
Here a = Ciciai,l E B with ci E C and c1 = 1. Moreover, b1,l E RGsk so 1&,1 C_ R \ 0. Since T(I) R ~we, see that a&,l A, and hence A, # 0. On the other hand, by the choice of the basis { a + } and the fact that c1 = 1, it follows that o $! W , a contradiction. We conclude therefore that W = B so 1 E W and A1 # 0. Similarly A:, the right analog of A1, is also nonzero and the result follows with A = Al n A:.
c
s
328
7. Group Actions and Galois Theory
As we mentioned above, if R is simple, then R is finitely generated as both a right and a left RG-module. However, if R is not simple, this need not be the case. For example, let R = K(-z,y)be the free algebra on two generators over the field K of characteristic not 2 and let u be the X-outer automorphism of order 2 determined by 2" = --z and yo = y. If G = { 1,u}, then (GI = 2 and gG is spanned by all monomials containing IC an even number of times. Suppose R is finitely generated as a right RG-module. Then for some n we have R = C,pRG where p runs over all monomials of degree 5 n. But ynz cannot belong to the right-hand side, so we have a contradiction. In case R is a simple Artinian ring and G is a finite group of outer automorphisms, we can sharpen the above result using skew group ring techniques. Indeed we have
Theorem 31.6. [146]Let G be a finite group of automorphisms of the simple Artinian ring R and assume that the skew group ring RG is simple. For example, this occurs if G is outer on R. Then there exists a division ring D and an integer k with RG Mk(D), RG % M k l ~ l ( D and ) 1 5 k 5 rank&. Furthermore, R is a free right or left RG-module on IGJgenerators.
Proof: If G is outer on R, then RG is simple by Corollary 12.6. Now assume that RG is simple. Since R is Artinian and G is finite, RG is Artinian and hence RG E M,(D) for some division ring D . In the following, all dimensions will be computed with D acting on the left. In particular, we have dimD RG = n2. Furthermore, if V is an irreducible right RG-module, then EndRG(V) = D , acting on the left, and dimD V = n. By Lemma 26.2(i), R is a cyclic RGmodule and thus RRG is isomorphic to V@', the direct sum of k copies of V. Clearly 1 5 k 5 rank RR. By computing dimensions we get En = dimD Vek = dimD R and hence kn(G(= dimD RG = n2. Thus LIGJ= n. Again, since RRG V e k ,it follows that
329
31. Almost Normal Subgroups
acting on the left. Furthermore, if W is an irreducible left MI,(D)module, then dimD W = k . Thus R, as a left Mk(D)-module, is a direct sum of dimD RldimD W = k n / k = n = klG( copies of W . But Wek is a free left Mk(D)-module, so we see that R = (W*'")*IGlis a free left Mk(D)-module on \GI generators. Finally, EndRG(R) = RG,by Lemma 26.2(i) again, so the result follows. I Note that the above yields an alternate proof of Lemma 30.8.
EXERCISES 1. Suppose D is a noncommutative division ring, finite dimensional over its center K , and let S = D ( X ) be the free D-ring on the variables in the set X . As usual, let D ( X ) ' be the augmentation ideal of S and set R = K D ( X ) ' E S. Then, as in the example following Theorem 30.4, G = D'/K' is the full group of X-inner automorphisms of R and it is an N*-group in its action on R. For which sets X is R a finitely generated RG-module? 2. Let G be a finite group of X-outer automorphisms of the prime ring R and, in the skew group ring RG, define I = RGR n R. Prove that 0 # l a R and that I is contained in the ideal A of Theorem 31.5. For the latter, let a E I so a = CrL1 T ~ G sNow ~ . compute GRu or uRG. 3. Let R = Mz(C) where C is the field of complex numbers and let u act on R by
+
g:
(z ;)
H
(-b
d-
- aF )
where - denotes complex conjugation. Prove that u is an outer automorphism of order 2 and determine R{ In particular, observe that k can be strictly less than rank RR in Theorem 31.6. 'lu}.
330
7. Group Actions and Galois Theory
The following examples, with K a field, indicate some of the limitations necessary in extending Theorem 31.1 to general N*-groups. Verify the details.
4. Let R = M4(K) and G = GL4(K) so that RG = K . Set S1 = { diag(a,a, b , b ) I a,b E K } and let SZ = { diag(a,b,b,b) I a,b E K } . Then the natural isomorphism cp: S1 --t Sz cannot be extended to an element of G. 5. Let u # 1 be an automorphism of K of finite order and set R = Mz(K) and G = (GLz(K), u) so that RG = F , the fixed field of u. If S is the subring of diagonal matrices, then the isomorphism 'p: S --+ S given by cp: diag(a, b) H diag(a", b) cannot be extended to an element of G. 6. Let T = K ( z ,y, z ) be the free algebra over K # GF(2) and let Sym, act on T by permuting the generators. Set R = Mz(T) and let G = GLz(K) x Sym, act on R. Then RG = TSym3,C = K and B = Mz(K). Now let H be the subgroup of G generated by GL1(K) x GL1(K) and
(! t)
u where u is the transposition (xy). It follows
that S = RH = { diag(a,a") 1 a E T }. If cp: 5' + S is defined by cp: diag(a, d) diag(a', aTu), where 7 is the transposition (y z ) , then cp cannot be extended to an element of G. ---f
32. Free Rings and Subrings We close this chapter by applying some of the Galois theory results of the previous sections to the special case of free rings. We begin, by working in the larger class of rings satisfying appropriate weak algorithms. These were considered at the end of Section 13 in the context of filtered rings. Here we are mainly concerned with graded rings and, in this case, the relevant definitions simplify. To start with, let R = CB CEO Ri be an arbitrary ring graded by Z', the semigroup of positive integers. In particular, 1 E & and &Rj & + j . Furthermore, R has a natural degree function and, by convention, deg 0 = -cm.We restrict our attention to Fl = UEoRi, the set of homogeneous elements of R. A set { a l , a ~. .,. ,a, } C 3t
331
32. Free Rings and Subrings
is said to be right dependent if there exist elements bi E 'H, not all zero, with C E l a t b i = 0. Moreover, the element a E 'H is said to be right dependent on { a l , a 2 , . . , , a , } if there exist bi E 'H with a = CE1aibi. Finally, R satisfies the n-term weak algorithm if given any right dependent set { a l , a 2 , .. ., a , } 'H with m 5 n and deg a1 5 deg a2 5 - - . 5 deg a,, then some ai is right dependent on { a l , a2, . . . ,a i - 1 ) . If R satisfies the n-term weak algorithm then, by definition, it satisfies the n'-term weak algorithm for all n' 5 n. Furthermore, it is easy to see that R satisfies the 1-term weak algorithm if and only if it is a domain. Of course, any Z+-graded ring R is filtered by the partial sums Cs=, R, and it is an easy exercise to verify that the above definition of the n-term weak algorithm agrees with that of Section 13. Indeed the degree inequalities of the original formulation merely indicate that certain leading terms vanish. Now let G be a group of automorphisms of the .%+ ' -graded ring R = @ C z oRi. We say that G is homogeneous if it stabilizes all the homogeneous components Ri. Note that this implies that RG = @ is a graded subring of R. With all this notation behind us, we can now prove
Theorem 32.1. [89] [97] Let G be a group of homogeneous automorphisms of the Z+-graded ring R. If R satisfies the n-term weak algorithm with respect to this grading, then so does RG. Proof: We know that RG is a graded subring of R and we show that it satisfies the n-term weak algorithm with respect to this grading. To this end, suppose m 5 n and { ul,u2,. . . ,a, } is a right dependent set consisting of homogeneous elements of RG with deg a1 5 deg a2 5 . . . 5 deg a,. Since RG is a graded subring of R, { ai } is also right dependent in R and we choose k 5 m minimal with { a l , a 2 , . . . ,a k } right dependent. By assumption, R satisfies the n-term weak algorithm so there exist homogeneous bi E R with ak = C:z$ aibi. Now let g E G. Since { ai } RG,we have k- 1
k-1
i=l
i=l
332
7. Group Actions and Galois Theory k- 1
ai (bi - (bi)”. But { a l , a2,. . . ,ak-l} is not and thus 0 = right dependent, by the choice of k, and bi - (bi)g is homogeneous, since g is a homogeneous automorphism. Thus bi = ( b i ) g for all i and all g E G. In other words, bi E RG,so Uk is dependent on { a l , a2,. . . ,a k - 1 } in RG and the result follows. I If R is a Z+-graded ring, we say that R satisfies the weak d g o rzthm if it satisfies the n-term weak algorithm for all n. The graded analog of Theorem 13.12 is given below. Notice that, if R = K ( X ) is the free K-algebra on the set X of variables, then R can be Zsgraded by making each x E X homogeneous of some arbitrarily assigned positive degree. For example, the usual grading is given by degx = 1 for all x E X. Furthermore Ro, the 0-component of R, is equal to the field K by our assumption that all generators have positive degree.
Theorem 32.2. [41] Let R = @ C z oR, be a Z+-graded ring with & = K a central subfield. Then R is the free associative K-algebra on a right independent homogeneous generating set if and only if R satisfies the weak algorithm.
PrmJ Suppose first that R = K ( X ) is graded as above and let { a l , a2,. . .,ak } 3-1 be right dependent. Say degal 5 dega2 5
. - 5 degak
and that C;=,aibi = 0 with bi E 3-1 not all zero. We can of course assume that all terms in the latter sum have the same degree. Thus after deleting zero terms if necessary, we have deg bk 5 degbk-1 5 5 degbl. Now let p be an X-monomial which occurs in bk and, for each z, write bi = bip by where p is not a trailing term of any monomial in the support of br. Then the dependence relation yields
+
k
k
i=l
i=l
Furthermore, deg bi 1 deg bk = deg p implies that p is not a right k factor of any X-monomial in Ci=luiby. Thus C!=laibi = 0. But bk is a nonzero element of the field K = Ro, so we can solve for a k and this direction is proved.
333
32. Free Rings and Subrings
Conversely, suppose R satisfies the weak algorithm. We first find X . To this end, note that Co<j
&
j5n I:
for some n, i with each pj,k homogeneous. In addition, we can assume that all summands satisfy deg Xj,kpj,k= n. In particular, if j = n then ,f?j,k E K while if j < n then xj,& E R j h - j . But then (*) yields a K-linear dependence of the set X , modulo Co.,j.,n RjR,-j and this contradicts the definition of X,. I
As an immediate consequence of the previous two results we have
Corollary 32.3. 1891 [97] Let R = K ( X ) be a free K-algebra graded with I& = K. If G is a group of homogeneous automorphisms of R, then RG is a free KG-algebra on homogeneous generators.
334
7. Group Actions and Galois Theory
It is not known whether the fixed ring of a free algebra is free without the assumption that the group acting is homogeneous. Again let R be a Z+-graded ring. The interesting cases of the n-term weak algorithm begin with n = 2. For example, we have the following correspondence theorem.
Theorem 32.4. Let R =
@xEo&
be a Z+-graded ring satisfying the 2-term weak algorithm and suppose G is a homogeneous N*-group of automorphisms of R. Then the maps H H RH and S H Gal(R/S) yield a one-to-one correspondence between the N*subgroups H of G and the intermediate rings S 2 RG which satisfy the 2-term weak algorithm with respect to any filtration.
Pro05 Since R is a domain, Theorem 30.4 applies. In particular, if H is an N*-subgroup of G then, by Theorem 32.1, S = RH is a graded subring of R satisfying the 2-term weak algorithm and H = Gal(R/S).In the other direction, if S 2 RG is given, then certainly H = Gal(R/S) is an N*-subgroup of G and the goal is to show that S = R H . By Proposition 30.3(iii), we know at least that RH 2 S 2 I where I is a nonzero ideal of R H .If S is symmetrically closed then, by Lemma 10.8@), Q s ( R H= ) Q s ( S )= S so S = RH as required. On the other hand, if S is not symmetrically closed, then Theorem 13.11 implies that S = D [ z ;m, 61 is a generalized polynomial ring in the variable z over the division ring D. Note that this ring clearly has a division algorithm and hence every right or left ideal is principal. In particular, I = a s for some nonzero a E S. Finally, if T E RH then o r E I = aS so a r = a s for some s E S. But R is a domain and a f 0, so T = s E S and hence RH = S in this case also. 1 Suppose G acts on R = K ( X ) . According to Corollary 13.6, R admits no nonidentity X-inner automorphism. In particular, G is an M-group in its action if and only if G is finite.
Theorem 32.5. 1891 Let R = K ( X ) be a free K-algebra graded so that its generators are homogeneous and Ro = K . Moreover, let G be a finite group of homogeneous automorphisms. Then the maps
335
32. Free Rings and Subrings
H H RH and S H Gal(R/S) yield a one-to-one correspondence between the subgroups H of G and the free algebras S 2 RG.
Proof: As we observed above, G is a finite X-outer group and hence an N*-group in its action on R. By Corollary 32.3, each RH is a free algebra and, by Theorem 32.2, each free algebra S satisfies the 2-term weak algorithm. Thus Theorem 32.4 now yields the result. I Notice that if S 1RG and S satisfies the 2-term weak algorithm, then S = RH so S is in fact a free algebra. Again, let G be a homogeneous group of automorphisms of K ( X ) . For example, if we take the usual grading in K ( X ) and assume that G consists of K-algebra automorphisms, then G homogeneous means that each g E G acts linearly. That is, for all z E X, we have z g = CyEX~,,yy with c , , ~ E K . On the other hand, an example of a homogeneous nonlinear action is as follows. Take X = { z, y } and define g: K ( X ) + K ( X ) by z g = -z y2 and yg = y. Notice that g has order 2 and that g is homogeneous if we define degz = 2 and degy = 1. In any case, since RG is free, it is natural to consider the size of its generating set and, in particular, whether it is finite or infinite. This has been settled at least for linear actions by finite groups. For the proof of this result, we first need some definitions. To start with, let R = K ( X ) have the usual grading. Then the component Rt is spanned by all monomials y1y2 - - - yt with yi E X . Moreover, if (T E S y q , then (T acts linearly on Rt by defining
+
In this way, Sym, acts as place permutations on Rt. Note that Sym, is not a group of algebra automorphisms; it only acts on the single component Rt . Next, if X = { z1,22,. . . ,Xd }, then an element of particular interest in R is the standard polynomial
336
7. Group Actions and Galois Theory
Note also that if G acts linearly on R,then each element of G is represented by a matrix in its action on R1 = K X and, in this way, G is a subgroup of GLd(K). In particular, we can speak about the determinant of each element of G. We record a few basic facts. In the following, part (i) is clear and part (ii) follows immediately by considering the effect of elementary row operations on 6.
Lemma 32.6. Let G act linearly on R = K ( x l , x 2 , .. . , x d ) . i. G commutes with all place permutations under the usual grading. ii. If g E G, then 69 = (det g)6, where 6 is the standard polynomial in 21,22,. .., x d . The proof we offer of the next result is fiom [48]. It significantly simplifies the original arguments.
Theorem 32.7. [49] [91] Let G be a finite group of linear automorphisms of the free algebra K(X). Then K(X)G is finitely generated if and only if X is finite and G acts as scalar matrices.
Proof. Write R = K ( X ) and consider the usual grading on R. S u p pose first that X is finite and G acts as scalar matrices. Then G = (9) is cyclic with x9 = cx for all x E X and some E E K' of order e. It follows that RG is spanned by all monomials of length divisible by e and therefore it is generated by the finitely many monomials of length precisely e . Now suppose that RG is finitely generated. If X is infinite, then RG K ( W ) for some finite subset W of X and there are infinitely many free intermediate rings S 2 RG. But this contradicts Theorem 32.5 and thus X must be finite; say = d. If I? is the algebraic closure of K , then certainly G acts on I ? ( X ) = I ? @ K K ( X ) , G GLd(I?) and l?(X)G = k € 3 K~ ( X ) G is finitely generated. Replacing K by I?: if necessary, we may therefore assume that K is algebraically closed. Fix g in G. The goal is to show that g is represented by a scalar matrix. Since K is algebraically closed, there is a basis Y of RI = K X with respect to which the matrix of g is in Jordan form.
1x1
337
32. Free Rings and Subrings
Note that Y also freely generates R as a K-algebra and with the same grading. Thus we may assume that Y = X . Let A denote the set of nonidentity X-monomials which occur in the supports of the finitely many generators of RG. Then RG is certainly contained in the K-subalgebra of R generated by A . Moreover, since A is finite, we can choose the integer n so that [GI divides n and n is larger than the degrees of the elements in A . Let 6 = &(XI, z 2 , . . . ,zd) E R d be the standard polynomial. If h E G then, by Lemma 32.6(ii), 6h = (deth)6 and hence (6n)h = (det h)n6n = 6" since (GI divides n. Thus 6" E ( R , d ) G . Furthermore, since the action of G commutes with the place permutations on R n d , it follows that every place permutation of Sn is also contained in ( R , d ) G . But RG is contained in the K-algebra generated by the monomials in A , so we conclude that every monomial of length nd occurring in the support of a place permutation of 6" is a product of elements of A. In particular, this applies to every place permutation of ( X I 2 2 ' - - Zd)". Suppose first that g is not a diagonal matrix. Then by considering the top of a nontrivial Jordan block and by appropriately labeling the variables, we can assume that z: = Xxl z 2 , z: = X z 2 z3 and that the remaining z: do not involve z 1 or z 2 . Here of course X E K' and the x3 term above may be missing if the block has size 2. Now x y x y - x: is a place permutation of ( 2 1 x 2 . . zd)" and hence is a product of monomials in A , each of which is different from 1 and has degree at most n. It follows that zy E A for some m with 1 5 m 5 n. In other words, x? occurs in a generator T for RG, which we may assume to be homogeneous of degree m; furthermore, we may take the coefficient of xy in T to be 1. Thus, for some k E K ,
+
+
where a involves the other monomials of degree m. Now it follows easily from the nature of the action of g that
338
7. Group Actions and Galois Theory
where is a sum of terms not involving xy or xy-1x2. But r g = r , so comparing coefficients yields Am = 1 and k = Am-' + ICX", a contradiction. Thus g must be diagonal and say x: = Xixi. Suppose that for every monomial v E R there exists an appropriate x E X with vx $ A . Then starting with = 1, we can construct a monomial u of degree n all of whose initial segments are not in A . But observe that there is a place permutation of ( ~ 1 x -2- .xd)= which starts with u and this monomial is a product of elements of A, a contradiction. In other words, there exists a monomial w with wxi E A for all i. Note that g sends any monomial to a scalar multiple of itself. Thus wg = pw for some p E K'. Moreover, since g fixes all generators of RG, it fixes all monomials in their supports and hence g fixes all members of A . We conclude that wxi = (wxi)g = pXiwzi for all i, so X i = p-' and g is indeed represented by a scalar matrix. This completes the proof. I
If G is allowed to be infinite, then the above is definitely false. For example, if just one g E G acts as a scalar matrix on X with eigenvalue of infinite order, then K(X)G= K. Suppose R = K(X) is any free algebra graded so that each component R, is finite dimensional over K . Then one can describe the degrees of the generators X by means of a Hilbert series. In the following, we just briefly sketch some of the basic ideas and we compute some examples. Define the two power series in C by 00
and
00
n=O
Note that H(R) has constant term 1 since Ro = K and, for the same reason, H(X) has constant term 0.
Lemma 32.8. T h e Hilbert series are related by H(R) =
1 1- H ( X ) '
339
32. Free Rings and Subrings
ProoJ For n > 0, R, has a basis consisting of the monomials w of degree n and every such monomial is uniquely writable as w = w'x with z E X. If degx = i, then w' is a monomial of degree n - i. Furthermore, for fixed i there are / X n Ri( choices for z and dimK &-i choices for the monomial w'. Thus
Since dimK Ro = 1, we conclude that H(R) = H ( R ) . H(X) the lemma is proved. I
+ 1 and
1x1
Suppose for example that = d and that R is graded in the usual manner. Then H(X) = d( so
and we obtain the not surprising result that dimK R, = d". More interesting examples arise when we consider the fixed ring RG. Let X = { x , y } and let G = { 1 , g ) act on R = K ( X ) with zg = x and yg = -3. Here of course charK # 2. We grade R as usual and consider the graded free ring RG. It is clear that RG is spanned by all monomials containing an even number of y-factors and thus RG is freely generated by the monomials Y = { x}
u { yzzy [ i 2 o } .
In other words, IY n (RG)n(= 1 for all n 2 1 so H(Y) = ( / ( l - () and hence
c;=,5" =
It follows that dimK(RG), = 2n-1 for all n 2 1. In case G is finite and charK = 0, there is another way to compute H(RG). Note that, since G E GLd(K), we can speak about the matrix trace of the elements of G. We have
340
7. Group Actions and Galois Theory
Theorem 32.9. [49] Let R = K ( X ) be a finitely generated free algebra over the field K of characteristic 0 and let G be a finite group acting linearly on R. Then
where tr(g) is the matrix trace of g E G.
Proof: We begin with two observations. Suppose first that V is any finite dimensional module for the group algebra K[G]and let e be the idempotent e = IGl-'G. Then V G= V e , so
Here, of course, trv denotes the matrix trace of the representation of G on V . Next it is clear that, as K-vector spaces, R, = ( R I ) @ ~ , the n-fold tensor product of RI over K . It then follows from this, by choosing appropriate bases if necessary, that trR,(g) = trRl(g)n = tr(g)" for all g E G. With these facts in hand, we can now quickly prove the result. Indeed, by the above,
and thus
as required. In particular, if char K = 0 in the example discussed immediately before this theorem, then tr(1) = 2 and tr(g) = 0 so 1
H(RG) = z(l +
32. Free Rings and Subrings
341
as before. Another example of interest occurs in characteristic p > 0. Let X = { z1,22,. . . ,xp } and let g act on R1 = KX as a Jordan block of size p with eigenvalues 1. Then gP = 1 and one can show that, for all n 2 1, the Jordan form of g on R, is a sum of these blocks of full size p . It follows that if G = (g), then d i m K ( k ) G = (dimK Rn)/p= pn-' for all n 2 1 and hence
If RG is freely generated by the homogeneous set Y ,then Lemma 32.8 yields
and thus (Y f l (RG),l = ( p - l)n-l for all n 2 1. Finally, we remark that the dichotomy of Theorem 32.7 also manifests itself in the prime correspondence of Theorem 28.3. This is apparent in the following two results which we offer without proof. Note that the assumption 1GI-I E K is required for the prime correspondence to exist.
Theorem 32.10. [137]Let G act as scalars on the free K-algebra R = K ( x l , x 2 , .. , ,z d ) with [GI-' E K. Let Q be a prime ideal of RG and let P be a prime of R lying over Q. If either RG/Q is finite dimensional over K or satisfies a polynomial identity, then the same is true of RIP.
Theorem 32.11. [137]Let G act linearly, but not as scalars, on the hee K-algebra R = K ( Q , 22,. .. ,zd) with 1GI-l E K. Then there exist uncountably many primes Q of RG with dimK RG/Q= 1such that if P is any prime of R lying over Q , then R I P does not even satisfy a polynomial identity. Some aspects of the proof are considered in 4-7 below.
EXERCISES 1. Show that the two definitions of the n-term weak algorithm agree in the case of ,%+ ' -graded rings.
342
7. Group Actions and Galois Theory
2. Let R = D[x;a,6]be a generalized polynomial ring over the division ring D and assume that R is Z+-graded with f i = D. Prove that R is in fact a skew polynomial ring over D in the variable y = x - d for some d E D. Now suppose that G is a group of homogeneous automorphisms of R. Show that G stabilizes each set Dy" and deduce that RG is either DG or a skew polynomial ring over DG. In either case, RG is Noetherian. 3. Let R be a Z+-graded ring satisfying the 2-term weak a l p rithm and let G act as an M-group on R. If G is homogeneous, prove that the division ring Ro is a finite module over hG. As a first step, show that G is an M-group in its action on & if R is symmetrically closed. In the remaining problems, G is a finite group acting on R with ]GI-' E R. We consider the prime correspondence of Theorem 28.3.
4. Let J be a G-stable ideal of R and set R = R / J . Show that the correspondence between the primes of I? and RG is consistent with the correspondence in R. 5. Let I a R G and define T(1) to be the unique largest ideal of R with tTG(T(I)) C I . Prove that T(1) is G-stable. Furthermore, if Q is a prime ideal of RG,show that T(Q) = P9 where P lies over Q. For the latter, go back to the original proof of the correspondence or apply the conclusion of Exercise 4 to I? = R / T . 6. Suppose R = K ( x l , x 2 , .. . ,z d ) and G acts as scalars on R so that we have G = ( 9 ) and x: = €xi where 6 has order e. Let I a RG. If a l , a z , . . . ,ae E I and T ~ , T z , .. . ,7e have degree 1, show that a 1 ~ 1 ~ 1 2 7* *2* a e T e E T(I). 7. Let R = K ( x l , x 2 , . . . , x d ) with K a countable field and suppose G acts linearly, but not as scalars, on R. Show that there exist uncountably many primes Q of RG with dimK RG/Q = 1 and only countably many primes P of R with dimK RIP < 00. For the latter, observe that dimKR/P < 00 implies that R I P is contained isomorphically in M,(K) for some n.
ngEG
8
Grothendieck Groups and Induced Modules
33. Grothendieck Groups The goal of this chapter is essentially to prove one marvelous theorem. This result of [140, 1421 asserts that the Grothendieck group of a Noetherian crossed product R*G is determined in a well defined manner by the sub-crossed products R*H for thy various finite subgroups H G . While this is very pretty in its own right, it also has far reaching and surprising applications. In particular, it solves two outstanding group ring problems concerning zero divisors and Goldie rank. Moreover, these results cannot be proved just for group rings; the proofs are inductive and make strong use of the elementary fact that RxG = ( R * N ) * ( G / N )for N a G . In other words, these are truly crossed product theorems. Our exposition, for the most part, is based on the papers [33] and [54] which considerably simplify the original argument. As we will see, the proof requires some familiarity with homological concepts, in particular with the projective dimension of a module and the derived functor Tor. To aid the reader in this, we will discuss
s
343
344
8. Grothendieck Groups and Induced Modules
the necessary background material, indicating without proof or with brief sketches the basic properties required. The latter are actually all quite elementary and can be found for example in [190]. Finally we remark that the study of Grothendieck groups becomes somewhat unnatural in the case of non-Noetherian rings. Thus we will usually assume that the rings being considered are right Noetherian.
Definition. Let R be a right Noetherian ring and let F be the free additive abelian group whose free generators are the isomorphism classes of the finitely generated right R-modules. Notice that every finitely generated R-module is isomorphic to R@"/N for some integer n and some submodule N of Re".Because of this, the collection of isomorphism classes of such modules is a set and we can choose representatives of the individual classes from among the modules Re"/N. In other words, the construction of F involves no set theoretic difficulties. If A is a finitely generated R-module, that is if A E modR, we let A denote the generator of F which is its equivalence class. Then we define the subgroup R of F to be generated by all expressions of the form - A - 6 if 0 -+ A + B + C -+ 0 is a short exact sequence with A , B , C E mod R. The group F / Ris called the Grothendieck group of R and is denoted by Go(R). Furthermore, we denote the image of A in Go(R) by [A].Thus Go(R) is generated by all such [A] subject to the relations [B]= [A] [C]for all 0 --t A ---t B -+ C + 0. Let us compute some examples.
B
+
Lemma 33.1. i. Let A be an additive abelian group and consider any map 8: modR + A. Suppose further that 8(B)= 6 ( A ) + 8 ( C )for all 0 + A -+ B -+ C + 0. Then 19 determines a group homomorphism 8:Go(R)+ A by defining 8 ( [ A ] = ) 8(A). ii. Suppose R is right Artinian and that VI ,V2, . . . ,vj are representatives of the isomorphism classes of the finitely many irreducible R-modules. Then Go(R)is the free abelian group on the generators [Vl],[V2),.. .,[V,].Furthermore, composition length determines a homomorphism p: Go(R) -+ 2.
Proof: (i) From the exact sequence 0
+
0
+
0
-,
0
--f
0 we see
345
33. Grothendieck Groups
that 8(0) = 0 and then from 0 -+ 0 -+ B -+ C -+ 0 we see that B C implies that 8(B) = 8(C). Since 3 is free abelian, we can now defined 8: 3 -+ A by B ( A ) = 8 ( A ) and, since B(R)= 0, this map factors through Go(R)= 3/R. (ii) Since every finitely generated R-module has a finite composition series, it follows that Go(R)is spanned by [Vl], .., [ V k ] . The goal is to show that these are independent. To this end, let A be the free abelian group on the generators vl, w 2 , . . . ,zlk and for each A E modR define B(A) = Cj mjwj E A, where mj is the multiplicity of V, as a composition factor of A . Then B(A) is well defined and respects short exact sequences. Thus by (i), 0 determines a homomorphism O:Go(R)-+ A. But O([K]) = vi and the vi are independent, so the same must be true of the elements [K]. Finally for p, we need only observe that the composition length is a well defined integer which respects short exact sequences.
[&I,.
Next we consider the behavior of Go under a change of rings. To start with, we recall that the tensor product is a right exact functor. That is, if 0 -+ A -+ B -+ C -+ 0 is any exact sequence of right Rmodules and if R M is a left R-module, then at least the abbreviated sequence A @ R M + B @ RM -+ C @R M -+ 0 is exact in general. We then say that R M is flat if for all 0 B -+ C -+ 0 we have
-+
A
-+
exact. Note that @R commutes with arbitrary direct sums and that A @ R 2 A via the map a @ r + ar. It then follows that every projective left R-module is flat. Furthermore, flatness is local so that every directed limit of flat modules is flat. Specifically, if M has a family { Mi } of flat submodules with the property that each finite subset of M is in some Mi, then M is flat (see [190, Corollary 3.29 and Theorem 3.301). Of course, there are analogous definitions and properties for flat right R-modules. Now suppose 8: R -+ S is a ring homomorphism, not necessarily onto. Then S can be viewed as a left R-module via r . s = O(r)s
346
8. Grothendieck Groups and Induced Modules
for r E R and s E S. Furthermore, this yields a map from right R-modules to right S-modules by A H A @ R S . Observe that if A = Cyrl a$ is finitely generated, then A @ R S is generated by the elements ai @ 1 and hence is a finitely generated S-module. We have
Lemma 33.2. Let 8: R + S be a ring homomorphism and assume that, in this way, RS is a ffat left R-module. Then A H A @R S determines a group homomorphism 8’:Go(R) + Go(S) given by [A]H [ A @ R S]. Furthermore, if 8 is onto, then so is 8‘ and we have A @R S E A / A I where I = Ker(8).
Proof. Note that if AR is finitely generated, then so is A @ RS. Thus we obtain a map 8’ from mod R to Go(S) by 8’: A I+ [ A@ R S ] . The flatness of RS and the relations in Go(S) imply that 8’ respects short exact sequences. We conclude from Lemma 33.l(i) that 8’: Go(R) + Go(S) exists. Finally suppose 8 is onto and that B is a finitely generated S-module. Then, by way of the homomorphism R + S -+ End(B), we can view B as an R-module and it will also be finitely generated. Since B @E S 2 B via the map b 8 s H bs, it follows that 8’ is onto. On the other hand, if A is an R-module, then A @ RS = A @ R( R / I )2 A / A I via the map a @ ( r + I ) H ar + A I . This completes the proof. I If R C S then, as in Section 3, we denote A @ R S by Als and call it the induced S-module. We will use the above result in some of the cases listed below.
Lemma 33.3. The following ring extensions are Aat, that is the overring is a flat left module for the subring. i. R R*G, where G is any semigroup. ii. RaH C R*G, where H is a subgroup of the group G. iii. R C RT-’, where T is a right divisor set of regular elements of R. Furthermore, the map Go(R) + Go(J2T-l) is onto.
Proof. Cases (i) and (ii) follow since the larger ring is a free module over the smaller one. For (iii), we observe that RT-l = Ute,,Rt-I and that each Rt-’ is a free, and hence flat, left R-module. Since
347
33. Grothendieck Groups
common denominators exist in this right ring of fractions, it follows that { Rt-' I t E T } is a directed family of flat R-submodules and hence RT-' is flat. Finally if B = biI2T-l is a finitely generated right RT-'-module, then A = biR is a finitely generated Rsubmodule with A 8~ RT-l E B. We conclude that the induced map G o ( R ) Go(l3T-l) is onto. 1
xy xT
---f
In particular, (iii) above applies to the classical right quotient ring Q(R)if it exists. On the other hand, the extension R Q,(R) is not flat, in general, as can be seen from the following example of [loo]. Let F = K(z,y) be a free K-algebra on two generators and let I be the ideal of F generated by x and y2. If R = K I F then, by Lemma 10.8(iii) and Theorem 13.4, Q,(R) = Q,(F) = F . On the other hand, RF is not flat (see Exercise 1). Another change of ring result occurs in the context of Z+-graded rings. To start with, let R = be such a ring. Then we can form the graded Grothendieck group gr Go ( R ) by restricting our attention to the finitely generated graded R-modules gr mod R. Specifically, gr Go(R) is the additive abelian group generated by the symbols [A] for each A E grmodR subject to the relations [B]= [A]+ [C]for all short exact sequences 0 -+ A -+ B -+ C -+ 0 with graded homomorphisms. The formal construction of gr G o ( R ) is of course analogous to the construction of G o ( R ) and it enjoys similar properties. Suppose again that R = CroRi is Z+-graded and form the polynomial ring R[t].Then R[t]can be Z+-graded by total degree, that is & t j has grade i + j . In particular, R[t]o= Ro. The following is a key result. Notice that the ring homomorphism considered cuts across the grade.
+
EFo&
Proposition 33.4. Let R = CEO R, be a right Noetherian Z+-graded ring and let R[t]be Z+-graded by total degree. Then the ring homomorphism R[t]-+ R given by t H 1 yields a group epimorphism g.Go(R[tl) Go@>. +
ProoJ Note that R[t]is right Noetherian and that the kernel of the homomorphism 8:R[t]-+ R is the principal ideal I = ( t - l)R[t]
8. Grothendieck Groups and Induced Modules
348
generated by the central element t - 1. As above, 8 determines a map from the finitely generated R[t]-modules to the finitely generated Rmodules given by A
HA
@qt] R
A/AI = A/A(t - 1).
In particular, we get a map from gr modR[t] to Go(R)by defining O’(A) = [A/A(t- l)] E Go(R). This will in turn yield a group homomorphism from grGo(Rft1) to Go(R)provided we show that 8’ respects graded short exact sequences. P To this end, let 0 + A + B + C t 0 be a short exact sequence of graded modules and graded homomorphisms. We may suppose that A is in fact a submodule of B. The assumption that the homomorphisms are graded then implies that A is a graded submodule of B . Now the combined map B + C --+ C / C ( t - 1) is certainly an epimorphism with B(t - 1 ) in its kernel. Thus we have an epimorphism B / B ( t - 1 ) C / C ( t - 1) with kernel U / B ( t- 1 ) and clearly U 2 A B(t - 1). We now show that equality occurs. For this, let u E U . Then P(u) E C(t-1) so, since pis onto, p(u) = P ( b ) ( t - 1 ) for some b E B. But then u-b(t-1) E Ker(P) = A and U C A+B(t-1). It follows that
p: +
--$
Ker(P) = U / B ( t - 1 ) = ( A + B(t - l ) ) / B ( t- 1 ) 2 A/ ( An ~
(- I)) t
and it remains to determine the intersection A fl B ( t - 1). Certainly A n B ( t - 1) 2 A(t-1). Conversely suppose ai E A and bi f B are elements written in terms of their homogeneous components with
xi
xi
Then by comparing homogenous components we have, setting b-1 = 0, a, = b,-lt - b, for all integers n 2 0. Thus bo = -a0 E A and, by induction, b, = b,-,t - a, E A for all n. Therefore b i ) ( t - 1) E A(t - 1) and we conclude that A n B(t - 1) = A(t - 1). Hence
(xi
Ker(P) 2 A / ( A n B(t - 1))= A/A(t - 1)
349
33. Grothendieck Groups
and we obtain a short exact sequence 0 --t A / A ( t - 1)--t B / B ( t - 1) -+ C/C(t - 1) --t 0.
In other words, in Go(R)we have 6’(B) = 6’(A)+ 6’(C>and thus, by the graded analog of Lemma 33.1(i), 6‘ determines a group ho-+ Go ( R ) as required. momorphism gr Go(R[t]) Now we show that this map is an epimorphism. Let V’ be a finitely generated R-module. Then, for some integer n, we can map f,lR onto V’ with kernel U’. Let F = the free R-module F’ = @ @ faR[t]be a free R[t]-module on the generators f ~fi,, . . . , fn, graded so that each f i is homogeneous of degree 0. Now each u‘ E U’ can be written uniquely as u’ = f,!ri = firi,j where ri = X j ri,j E Rj is the decomposition of ri into its homogeneous components with ri,j E Rj, If rn is the largest such j which occurs with ri,j # 0, we define q ( d ) E F by
xT
xy
xi
‘&
Then q(u’) is a homogeneous element of F of degree rn and we let U be the R[t]-submodule of F generated by all these q ( d ) for ZL’ f U‘. Clearly U is a graded submodule of F and it is finitely generated since R[t]is right Noetherian. Furthermore, if M = F / U then M is M --t 0 a finitely generated graded R[t]-module with 0 -+ U -+ F exact. It follows by the result of the preceding paragraph that -+
0 --t U / U ( t - 1) -+ F / F ( t - 1) --+ M / M ( t - 1) -+ 0
is an exact sequence of R-modules. Consider the additive group epimorphism F -+ F‘ given by f i r t j H f i r for all T E R. The kernel of this map is clearly F ( t - 1) and hence we obtain an R-isomorphism F / F ( t - 1) -+ F’. Under this map, the image of ~ ( u ’ ) F ( t - 1) is precisely u‘ and therefore U / U ( t - 1) = U / ( U n F ( t - 1)) maps onto U’.Thus V‘ Z F‘/U‘ S M / M ( t - 1) so [V‘]is in the image of 6‘:gr Go(R[t]) Go(R)and the result follows. I
+
--f
8. Grothendieck Groups and Induced Modules
350
It is apparent from the last part of the argument that the above result is an analog of the classical trick of adding a new variable to introduce homogeneous coordinates. Now let us return to arbitrary right Noetherian rings, but this time we restrict our attention to finitely generated projective modules. In this case, we get a further group of interest. Specifically, let R be a right Noetherian ring and let pr modR denote the family of finitely generated projective R-modules. Then KO( R )is the additive abelian group generated by the symbols [PI for each P E pr mod R subject to the relations [PI = [Q] [Q’]for all short exact sequences 0 -+ Q -+ P Q’ + 0. Again, the formal construction of this group is analogous to the construction of Go(R)and it enjoys similar properties. Furthermore, since Q’ is projective in the above, the sequence splits and is equivalent to P Q @ Q‘. Under certain circumstances, Ko(R) is naturally isomorphic to G o ( R ) . To prove this fact, we require some definitions. If AR is an R-module, then a projective resolution for A is a long exact sequence
+
--f
--+
Pz
+ Pi
+Po
-+
A
-+
0
with the Pi projective. For convenience, we abbreviate this by P --t A -+ 0 and we write P, for the nth module P,. Now it is clear that every module has such a resolution. Indeed, start with A and map a free module PO onto it, obtaining PO -+ A -+ 0. Then map a free module PI onto Ker(P0 + A ) to get PI -+ PO ---t A 0. Continuing in this manner, we clearly construct a free and hence projective resolution for A. Furthermore, if R is Noetherian and A is finitely generated, then this process yields a projective resolution for A consisting of finitely generated projective (or free) modules. A projective resolution is said to be finite if the Pn’s are eventually all zero. We record some basic properties. ---f
Lemma 33.5. Let
351
33. Grothendieck Groups
be exact sequences with Pi and Qi projective. Then
In particular, X is projective if and only if Y is.
Proof: We proceed by induction on n 2 0. Let B be the kernel of the map PO-+ A and let C = Ker(Q0
0 -+ B 0
+
C
--f
A ) . Then we have
+ Po + A -+ --+
0
Qo - + A + 0
and therefore B @ QO E C @ PO by Schanuel's lemma (see [190, Theorem 3.411 or Exercise 2). Furthermore we have
0
-+
Y
But B @ QO
+
Qn
+..*
+ Q2 +
Q1 @Po -+ C @ P o + 0.
C @ Po so the result now follows by induction. I
Notice that the above says that if P + A -+ 0 and Q -+ A -+ 0 are two projective resolutions for A , then Ker(P, -+ Pn-l) is projective if and only if Ker(Q, + Qn-l) is. Furthermore, when this occurs, then the truncated series
0 + Ker(P,
-+
Pn-l) -+ P,
-+
-+
P1
-+
Po
-+
A
-+
0
is a finite projective resolution for A . In particular, if A has some finite resolution, then any projective resolution can be suitably truncated to a finite one. As an immediate consequence we have
352
8. Grothendieck Groups and Induced Modules
Lemma 33.6. Let A be a finitely generated R-module with R a right Noetherian ring. If A has a finite projective resolution, then it has one consisting of finitely generated projective modules. The following is [190, Lemma 6.111.
Lemma 33.7. Let 0 A + B + C + 0 be a short exact sequence of R-modules and let P + A + 0 and Q + C + 0 be projective resolutions. Then there exists aprojective resolution P@Q+ B + 0 -+
such that the diagram 0
+
P
.+
1 O + A +
P@Q
1 B
+
Q
+
0
I
+ C + O
1
I
1
0
0
0
commutes. Here (P @ Q)n = P, @ Qn and the maps in 0 -+ P -+ P @ Q-+ Q -+ 0 are the natural ones determined by these direct sums. A module A is said to have finite projective dimension if it has a finite projective resolution. Indeed the projective dimension of A, pd,A, is the minimal n for which
exists. In particular, pd,A = 0 if and only if A is projective. Furthermore, if Q + A 0 is any projective resolution then, by our previous comments, pd,A is the minimal n with Ker(Q,-I + Q,-z) projective. We can now prove --f
Proposition 33.8. Let R be a right Noetherian ring. Then there is a natural map c: Ko(R) + Go(R) given by [PI H [PI. 16 in addition, every finitely generated R-module has finite projective dimension, then c is an isomorphism.
Proof. The existence of c is obvious from the projective analog of Lemma 33.1(i). Assume now that each finitely generated R-module
353
33. Grothendieck Groups
has finite projective dimension. We show that c is an isomorphism by constructing a back map. To this end, let A be a finitely generated R-module and, by Lemma 33.6, let 0 + P, + ... + PI + PO + A -+ 0 be a finite projective resolution consisting of finitely generated modules. We define O(A) = C ( - ) Z [ P i ]E Ko(R). i
In view of Lemma 33.5 with X = Y = 0, 8(A) is well defined, independent of the particular resolution chosen. Furthermore, by Lemma 33.7 and the relations in Ko(R), it follows that 6 respects short exact sequences. Thus by Lemma 33.1(i), 0 determines a group homomorphism 19:Go(R)-+ Ko(R). Now it is trivial to see that I9c is the identity on Ko(R). Moreover, since [A] = Ci(-)i[Pi] in Go@), we see that cI9 is the identity on G o ( @ . Thus 6 = c-l and c is an isomorphism. I The homomorphism c above is called the Cartan map. Now suppose R is Artinian with PI, P2,. . . ,Pk representatives of the isomorphism classes of principal indecomposable R-modules. Then, as in Lemma 33.1(ii), we see that Ko(R) is the free abelian group on the elements [ P I ] , . . . , [Pk].Thus since the Pi are in one-to-one correspondence with the irreducible R-modules V1, V2,.. . , V k , we see that Ko(R) and Go(R)are abstractly isomorphic. However, c itself need not be an isomorphism; indeed it is not even necessarily oneto-one (see Exercise 4).
[&I,
Lemma 33.9. i. Let 0
B + C --t 0 be given. If any two of these modules have finite projective dimension, then so does the third. ii. Let B = @ Ai and let k be an integer. Then pdRB 5 k if and only if pdRAi 5 k for all a . +
A
+
xi
proof. (i) If pdRA and p d R C are finite, then certainly so is p d R B by Lemma 33.7. Now suppose pdRA and pdRB are finite. Choose a finite projective resolution P + A 0 for A and any resolution Q C + 0 for C. Then by Lemma 33.7 again, P @ Q B +0 ---f
.--f
---f
354
8. Grothendieck Groups and Induced Modules
is a resolution for B. But pd,B < 00 so this resolution eventually has projective kernels. In addition, since P, = 0 for all sufficiently large n, it follows that Ker(Q, --t Qn-l) is eventually projective. Therefore C has finite projective dimension. The argument with pd,B, pd,C < 00 is similar. (ii) Now let Pi -+ Ai -+ 0 be projective resolutions. Then, since B =@ Ai, the obvious direct sum yields a projective resolution @ Pi + B -+ 0. Moreover, the kernel of the nth map in @ Pi is the direct sum of the corresponding kernels in the various Pi. Since a direct sum is projective if and only if each summand is, the result clearly follows. I
xi
Ci
We consider one last ring extension. Let R*G be a crossed product with G = ( 2 1 , 22,.. . ,q )the free abelian semigroup on the variables zi. In other words, R*G is an iterated skew polynomial ring and we have a homomorphism R*G -+ R given by Z i H 0 for all i. In particular, if M is an R-module, then via the combined map R*G -+ R -+ End(M), we see that M is also an R*G-module.
Lemma 33.10. Let R*G be as above with G = (XI,22,.. . ,xt) a free abelian semigroup. If M is any R-module of finite projective dimension, then M also has finite projective dimension when viewed as an R*G-module.
Proof. It suffices to assume, by induction on t , that G = (x)has one generator. Write S = R*G and consider the short exact sequence 0 -+ ZS S -+ R -+ 0 determined by 3 H 0. Since S and ZS S are free as right S-modules, we have pdsR 5 1 and indeed, by Lemma 33.9(ii), pdsP 5 1 for any projective R-module P. Finally let -+ Pi -+ Po -+ M --+ 0 0 -+ P, -+ --f
be a finite projective resolution for M as an R-module. Then as Smodules, each Pi has finite projective dimension and it follows from Lemma 33.9(i) and induction that the same is true for M . I We remark that Ko(R) is the first of a series of K-groups associated with the ring R (see [179]for details).
355
33. Grothendieck Groups
EXERCISES
+
F. 1. Let F be any K-algebra, let I a F and set R = K I Assume that 2 5 dimK F / I < 03 and 1 5 dimK 1/12 < 00. Show that RF is not flat. To this end, consider 0 -, I + R + R / I -, 0 and observe that ( R / I )8 F % F / I and R 8 F S F . Thus if 0 + I 8 F + R @ F + ( R / I ) @ F-, 0 is exact, then V = I 8 F % I. But V has as a homomorphic image
(1/12) 8F
2 (dimK 1 / 1 2 )
+
( R / I )8 F s (dimK 1 / 1 2 ). F / I
and this has larger dimension than 1/12,a contradiction. P 2. Let 0 + X + P 2 A + 0 and 0 + Y + Q + A -+ 0 be given with P and Q projective. Let W be the R-submodule of P @ Q consisting of all ( p , q ) with a ( p ) = p(q). Observe that the projection map W + Q is an epimorphism with kernel (X,O). Deduce that X @ Q r W SY@P. 3. Let R be a right Artinian ring with principal indecomposable modules P I ,P2, . . . , PI, and corresponding irreducible modules V,, V 2 , . . , VI,. The matrix of c: Ko(R) + Go(R) with respect to the bases { [Pi]} and { } is called the Cartan matria: of R. Describe this matrix in case R is semisimple or R = K [ x I xm = 01. In particular, note that c need not be surjective. 4. Let K be a field and write A ( a , b ) and B ( a , b ) for the sets of 4 x 4 upper triangular matrices
[&I
with the * entries arbitrary. Now let R and S be the 4- and 8dimensional subalgebras of M4 ( K ) given by
R = { A ( a , b )l a , b ~ K }
S={B(a,b) [ a , b ~ K }
Observe that R/J(R)2 K @ K 2 S/J(S) and describe the Cartan matrices of R and S . Conclude that c need not be injective in general.
356
8. Grothendieck Groups and Induced Modules
5. If 0 -+ A -+ B -+ C -+ 0 is given, then the projective dimensions of any two of these modules bound the dimension of the third. Obtain this information by sharpening the proof of Lemma 33.9(i). 6. Let R*G be a crossed product with G = (XI,2 2 , . . . , xt) the free abelian semigroup on the variables xi. If M is an R-module, show that pdR,,M 5 pdRM t.
+
34. Graded Rings In this section, we are concerned for the most part with Z+-graded R, is P - g r a d e d and right Noetherings. Specifically if R = @ CEO rian, our goal is to show that the induced map A H A @R,, R yields an isomorphism from Go(&) to Go(R). The proof of this key result relies on the derived functor Tor. Let R be any ring and let AR and R M be right and left Rmodules respectively. Then A and M determine a sequence of ad= Tor,(A,M) for ditive abelian groups denoted by Tor:(A,M) n = 0,1,2,. . .. These Tor groups enjoy many of the properties of the tensor product and indeed Toro(A, M ) = A @ R M. However, their construction is somewhat more complicated. To start with suppose
is a zero sequence or complex of additive abelian groups. By definition, this means that the maps d, are homomorphisms and that d,d,+l = 0 for all n. In particular, Im(d,+l) Ker(d,) and the homology groups H,(X) = Ker(d,)/Im(d,+l) are a measure of the failure of exactness of the complex X at the nth module X,. Now let
be any (left) projective resolution for RM. By tensoring this resolution with AR and ignoring the first term A 8 M, we obtain
... -+ AmQ,
-+
+
A@Q1- + A @ Q o - + O
34. Graded Rings
357
and the groups Tor,(A, M ) are precisely the homology groups of this complex. A few basic properties are as follows (see [190, Section 81 for details). First, Tor,(A, M ) is well defined, independent of the particular projective resolution for M . In fact, Tor can also be computed in a similar manner by using projective resolutions for A. Specifically, if
is such a resolution, then the groups Tor,(A, M ) are also the homology groups of the complex
In view of this, properties of Tor are clearly right-left symmetric. Morover, Tor, commutes with arbitrary direct sums and is functorial in the sense that any R-module homomorphism A + B gives rise to a natural group homomorphism Tor,(A, M ) -+ Tor,(B, M ) . Finally, we know that the tensor product is a right exact functor, but it need not be exact. The key property of Tor is that it takes up the slack. Indeed, we have the following long exact Tor sequence (see [190, Theorem 8.31).
Lemma 34.1. Let R M be a left R-module and let 0 --t A + B -+ C -+0 be it short exact sequence of right R-modules. Then there exists a long exact sequence
Here the maps Tor,(A, M ) -+ Tor,(B, M ) t Tor,(C, M ) are given by the functorial property of Tor and the various 6’s are called connecting homomorphisms. It follows easily from the construction of Tor that if R M is flat, then Tor,(A,M) = 0 for all n 2 1 and all
358
8. Grothendieck Groups and Induced Modules
right R-modules A . Conversely, if Torl(A, M ) = 0 for all A, then the preceding lemma implies that M is flat. Moreover, we have
Lemma 34.2. Let RM be a left R-module. i. Suppose 0 --t A B C --t 0 is a short exact sequence of right R-modules and that Tor,+l(C, M ) = 0 = Tor,(B, M ) for some integer n 2 0. Then Tor,(A, M ) = 0. ii. If pdRM = n, then Tork(A, M ) = 0 for all lc 2 n 1. --$
--f
+
Proof: Part (i) follows from Lemma 34.1 since Toro(A, M ) = A @ M . Part (ii) follows by constructing the groups Tork(A,M) from the finite projective resolution
This completes the proof. Now suppose that R = Ro @ I where I a R and Ro is a subring with the same 1. Then R = R / I is a right and left R-module and an &-module with R ~ SRa0RO. As usual, if A is any right Ro-module, then we are concerned with the induced right R-module A @ R ~R.
Lemma 34.3. Let R = Ro @ I with I a R, set R = R / I and let A be any right Ro-module. i. (A @R,, R ) @R R S A, where this is a functorial isomorphism as right Ro-modules. ii. If RoR is flat, then Torf(A 8 ' R, ~R ~ ) = 0 for all n 2 1.
ProoJ: (i) By the associativity of tensor products we have
as required. (ii) Here we compute the various Tor groups directly from the definition. Let
359
34. Graded Rings
be a projective resolution for the Ro-module A. Since R~R is flat, by assumption, the sequence
is also exact. Note further that Ro @ R R~ Z R and that @ commutes with arbitrary direct sums. It therefore follows that each P, @ R is a projective R-module and hence that the above sequence is in fact a projective resolution for A @ R. To compute Tor:(A,R), we tensor this resolution with R and find the homology groups. But by (i) above, we have the functorial isomorphism (P, @ R R) ~ @ R R "= P, and hence the new series is just
the original exact sequence. Since the abbreviated series
is clearly exact at all P, with n 2 1, the homology groups are trivial at those modules and therefore Tor,(A @I R, R) = 0 for all n 2 1. a
@xEol&
Now let R = be a Z+-graded ring and define I = R+ = @ Czl Ri. Then I a R and R = Ro @ I so the above result applies. We fix this notation throughout and also write I? = R / I = R/R+. Recall that A = @ Ai is a graded R-module if AiRj C Ai+j. The following is the graded version of Nakayama's lemma. It
xzo
is surprisingly powerful.
Lemma 34.4. Let R be a Z+-graded ring and let A be a graded R-module. I f A = AR+, then A = 0.
Proof. Since A = A(R+)" for all n 2 0, it follows that A has no elements of grade < n. I Observe that if A is an Ro-module, then
B
=
A
R~ = C ~B ~ ~ A C Z I R ~ R ~ i=O
8. Grothendieck Groups and Induced Modules
360
is a graded R-module with components Bi = A € 3 &. ~ ~Moreover, in this way, Ro @ R R ~ is graded isomorphic to R.
Lemma 34.5. Let R = @ CzoRi be a Z+-graded ring and let A be a graded R-module generated by its kth component A,+for some integer k. I f Torl(A,R) = 0, then A = AI,@ R ~R. Furthermore, if R ~ isRflat, then Tor,(A, I?) = 0 for all n 2 1.
Proof: Since A = AkR, it is clear that A has no elements of degree less than k. Let <:A,+ @ R R ~ -+ A& = A ’ b e the natural graded
<
epimorphism. The goal is to show that is one-to-one. To start with, let z f Ker(<). Since R = & @ I , we can write
with at, a: E A,+.Then we have
It follows that thus
Ctatrt = 0 so Ctat @ rt = (c,atrt) @ 1 = 0 and =
C a:
E A~ € 3 1 = ( A @ ~R R ) I .
(€1
In other words, Ker(<) C (A,+@ R)I. Now tensor the exact sequence 0 -+ Ker(J) -+Ak @ R + A --t 0 with R = R / I . Since B @ R ( R / I ) B / B I for any R-module B , Lemma 34.1 yields the exact sequence Torl(A, 8 ) + Ker(<)/Ker(J)I 5 (A,+@ R)/(A,+ €3 R ) I . But Torl(A,R) = 0 by assumption, so the map i is in fact an injection. On the other hand, we have shown that Ker(<) C (A,+@ R)I so i must also be 0. It follows that Ker(<)/Ker(J)I = 0 and the graded version of Nakayama’s lemma yields Ker(<) = 0 as required. In particular, A % A,+@ R R ~ and Lemma 34.3(ii) yields the result.
361
34. Graded Rings
We also need
Lemma 34.6. Let A be a graded R-module and let k 2 0 be an integer. If B = AiR, then B n AI = B I .
xi
Proof: Clearly B f l AI 2 B I . For the converse, first note that
i
i
(xi,,
Thus since B I A l , it clearly suffices to show that Ai)nAI B I . To this end, let z belong to the latter intersection. Then z E A I , so we can write z = Cj ajrj with aj E A and rj E I all homogeneous. Furthermore, since degz < k, we can assume that each summand satisfies degaj d e g r j < k. But then a j r j E B I , and the lemma is proved.
+
We now come to the first main result of this section and we follow the proof in [33].
Theorem 34.7. [179]Let R = @ CEO Ri be a right Noetherian Z+graded ring. Assume that R,,R is flat and that R = R/R+ has finite projective dimension as a left R-module. Then the map A H A @ RR~ induces an epimorphism G o ( & ) + gr G o ( R ) .
Proof: As we observed earlier, if E E mod&, then E @ R R ~ E grmod R. Thus since R,,R is flat, Lemma 33.l(i) implies that E I+ E @ RR~determines a group homomorphism 8: G o ( R 0 ) -, gr Go (R). The goal is to show that t9 is onto. Note also that TOr,(E@Ro R, R) = 0 for all TI 2 1 by Lemma 34.3(ii). Let A denote the family of all finitely generated graded Rmodules A such that Torn(A,R) = 0 for all n 2 1. We begin by showing that [A] E gr Go(R) is contained in the image of 8 for all such A E A. To this end, let A be given. Since A is finitely generated, we have A = C:=oAiR for some integer k and we proceed by induction on k. Set B = &kAiR = ‘&kBiR. If C = A / B , then 0 --t B -+ A -+ C + 0 is an exact sequence sequence with graded homomorphisms and by the preceding lemma
362
8. Grothendieck Groups and Induced Modules
we have 0 -+ B / B I -+ A/AI -+ C / C I -+ 0 exact. On the other hand, if we tensor the original sequence with R = R / I , then the long exact Tor sequence yields
.
.--t
Torl(A,R) -+ Torl(C,R)
-+
B/BI
+
A/AI
-+
C/CI
-+
0.
But Torl(A,R) = 0 and the map B / B I -+ A/AI is injective, so we conclude that Torl(C,R) = 0. Now it is clear that C = C k R and thus, by Lemma 34.5, we have C = Ck @ R R~ and Tor,(C, R ) = 0 for all n 2 1. Finally, by Lemma 34.2(i) applied to 0 -+ B A --t C -, 0, we see that Torn(B,R)= 0 for all n 2 1 so B E A. It therefore follows by induction on k that [B]is in the image of 8. But then [A]= [B] [C]E Im(0) and this part is proved. Next we show that the image of 8 contains [A]for certain additional modules A. Specifically, for each integer m 1, set & equal to the family of all A E gr mod R with Tor, (A,R) = 0 for all n 2 m. Thus for example A1 = A and we show by induction on m that if A E &, then [A]E Im(8); the case m = 1 has of course already been proved. Now suppose A E A, with m > 1 and let 0 -+ B -+ F -+ A 0 be a graded short exact sequence with F a finitely generated free R-module. Since R = & @ R R ~ and Tor commutes with direct sums, it follows that F E A and hence, by By induction, [B]E Im(0) and, Lemma 34.2(i), that B E &-I. since clearly [ F ] E Im(O), the same is true of [A]= [ F ]- [ B ] . Finally we use the hypothesis that R R has finite projective dimension; say pd,R = t . Then, by Lemma 34.2(ii), Tor,(A,R) = 0 for all right R-modules A and all n 1 t 1. In other words, gr mod R = At+l and, with this, the result follows. --$
+
>
-+
+
We remark that the map Go(R0) + grGo(R) is in fact an isomorphism. We will prove the one-to-one aspect after the next key result.
Corollary 34.8. 1179)Let R = @ CEOR, be a right Noetherian Z+graded ring. Assume that R~R is flat and that R = R/R+ has finite projective dimension as a lefc R-module. Then the map A H A @ RR~ induces an epimorphism GO(Ro)-+ Go ( R ).
363
34. Graded Rings
Proof. Form the polynomial ring S = R[t]and let S be Z+-graded by total degree. Then S is right Noetherian, since R is, and SO= Ro. Furthermore, R ~ isRflat so it is clear that soS is also flat and pdRR < so it follows from Lemma 33.10 that pd,S is also finite. We can now apply Theorem 34.7 to conclude that the map A H A @soS yields an epimorphism Go(Ro)= Go (So) gr Go ( S ) . Furthermore Proposition 33.4 implies that the ring homomorphism R[t]-+ R given by t H 1 yields an epimorphism gr Go(S) = gr Go(R[t]) -+ Go(R). Thus we obtain the epimorphism Go(&) + grGo(R[t]) -+ Go(&) and all that remains is to identify this combined map. To this end, let A E mod&. Then we see that 00
-+
where R is viewed as a left R[t]-module via the above ring homomorphism. Since the associativity of tensor product yields
the composite map is determined by [A] H [A @ R R] ~ and the result follows. It turns out that the epimorphisms of the preceding two results are in fact both isomorphisms. For this we require one more elementary property of Tor. Suppose AR and R M are R-modules and assume that M is an (R,S)-bimodule. Then Torf(A, Ad) is a right S-module and the maps of Lemma 34.1 are all S-homomorphisms. Indeed, for the most part, this is just a consequence of the functorial nature of Tor. Furthermore, if R is right Noetherian, AR E modR and M s E mod S , then Torf(A, M ) is a finitely generated right S-module (see Exercise 5).
Proposition 34.9. [179] The epimorphisms in Theorem 34.7 and Corollary 34.8 are both isomorphisms.
ProoJ The arguments are identical in the two cases, so we just consider Corollary 34.8. Here we have an epimorphism O:Go(Ro) -+
8. Grothendieck Groups and Induced Modules
364
Go(@ and we show that 6' is an isomorphism by constructing a back map. Since the module R R has finite projective dimension t , by ast + 1. sumption, Lemma 34.2(ii) yields TorF(A,R) = 0 for all k Furthermore, R is an (R,&)-bimodule and thus each Tor:(A, fz) is a right Ro-module. Since f z E~ (RO)R~, ~ it follows from the above comments that A E mod R implies that Torf(A, R) E mod Ri-, and we define the map modR + Go(&) by
>
t
A
H
x(-)k[Torf(A,
R)] .
k=O
Now it is an easy consequence of the long exact Tor sequence that this map respects short exact sequences and hence, by Lemma 33.1(i), we obtain a group homomorphism 'p: Go(@ -+ Go(&) given by t 'p:
[A] t+ x ( - ) k [ T o r z ( A , R)] . k=O
Finally if B E mod& then Lemma 34.3(ii) implies that we have Torf(B @ R R, ~ R ) = 0 for all n 1. Hence since Tor0 is the tensor product, we see that
>
(Po: [B]
[ ( B@Ro a)@ R R1 = rB1
I
by Lemma 34.3(i). Thus p6' is the identity on Go(&) and 8 is oneto-one as required. I We remark that an appropriate KO analog of Theorem 34.7 holds and is a simple consequence of the graded version of Nakayama's lemma (see Exercises 6-8). On the other hand, there is no KOanalog of Corollary 34.8 without additional assumptions on the ring. A counterexample can be found at the end of Section 36. We close with a few comments on Ko(R) which are relevant to that example. Let R be a ring. If A , B f m o d R , then A and B are said to be stably isomorphic if A @ R" B @ R" for some finitely generated free R-module R". Furthermore, A is stabZy free if A @ R" G Rm for some m,n 2 0.
365
34. Graded Rings
Lemma 34.10. Let P and Q be finitely generated projective Rmodules. i. [PI = [Q]in Ko(R) if and only if P and Q are stably isomorphic. ii. [PI is in the cyclic subgroup of Ko(R) generated by [R]if and only if P is stably free.
Proof: (i) If P and Q are stably isomorphic, then certainly [PI = [Q]in KO(@. For the converse, let F be the free additive abelian group on the isomorphism classes A of finitely generated projective R-modules. Furthermore, let R be the subgroup of F generated by the relations B - A - C with B E A @ C. Then Ko(R) S F / Rand [A]= A R. In particular, if [PI = [Q]then p - can be written as a finite sum and difference of relations, say
+
a
and this means that the summands on the left must match, term for term, with the summands on the right. Hence P @ V S Q @ V where V is the direct sum
But V is projective, so V C.B U 2 R" for some n and therefore P is stably isomorphic t o Q. (ii) If P is stably free with say P @ Rn S Rm, then [PI = (rn - n)[R]in Ko(R). Conversely suppose [PI is contained in the cyclic subgroup of Ko(R) generated by [R]and say [PI = n[R].If n 2 0, then P is stably isomorphic to Rn and hence P is stably free.
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8. Grothendieck Groups and Induced Modules
On the other hand, if n < 0, then P @ R(-,) is stably isomorphic to 0 and again P is stably free. I
EXERCISES 1. Let M be a left R-module. Show that M is flat if and only if Torl(A, M ) = 0 for all A and, when this occurs, that Tor,(A, M ) = 0 for all A and all n 2 1. 2. Suppose that R is a commutative domain with field of fractions Q. If A is a right R-module, prove that Torl(A,Q/R) is isomorphic to the torsion submodule of A , namely { a E A I ar = 0 for some r E R\O}.
3. Let 0 --t A, --t . . - --+ Al --t A0 -, 0 be an exact sequence of finitely generated R-modules. Show that Cy=o(-)2[Ai] = 0 in
GO(R). 4. Let X be a partially ordered multiplicative semigroup with the property that zy, y z > y for all z E X # = X \ (1) and y E X . If R*X is a right Noetherian crossed product, show that I = R * X # a R*X with R*X/I E R. Furthermore, prove that the obvious analog of Theorem 34.7 holds in this context. 5. Assume that R is right Noetherian, AR E modR and that RMS is finitely generated S-module. Use an appropriate projective resolution for AR to show that Torf(A,M) is a finitely generated S-module. 6. Let R be a Z+-graded ring and let A and P be graded R-modules with P projective. Suppose 8: A + P is a graded homomorphism which induces an isomorphism of the modules A/AI -+ P I P I . Use the graded analog of Nakayama’s lemma to show first that 8 is onto and then that 8 is an isomorphism. For the latter, note that Ker(8) is a graded submodule of A contained in A I and that A = P’ @ Ker(8) for some submodule (not necessarily graded) P f Fi P. 7. Again let R be a Z+-graded ring and let P be a graded Rmodule which is projective. Note that P I is a graded submodule of P
367
35. Group Extensions
xi
and set Qi = Pi/(Pi n P I ) so that PIPI = GI Qi. Show first that each Qi is a projective Ro-module and hence that Q = @ Q~@R~R is a projective R-module. Furthermore, if Q is graded so that Qi@Rj has degree i j,prove that Q / Q I is graded isomorphic to PIPI. 8. Continuing with the above notation, show that there exists a graded homomorphism 0 : Q + P which induces the isomorphism Q/QI PIPI. Conclude that 8 itself is an isomorphism.
Ci
+
--f
35. Group Extensions In this and the next section, we prove the key result of [142] on the Grothendieck group of Noetherian crossed products. As will be apparent, almost all the work involves the study of abelian-by-finite groups. Thus suppose A a I' with A abelian and r/A finite. Then the proof divides naturally into three cases according to whether the action of I'/A on A is (1) free, (2) rationally free or (3) arbitrary. Case (1) follows fairly easily from Corollary 34.8 once we observe that such groups split, and case (3) uses a Tor argument similar to that of the proof of Theorem 34.7. Case (2) is by far the most interesting; it is here that the finite subgroups miraculously appear as the stabilizers in a certain permutation action. In this section, we consider the first two cases. To start with, we need some general information on group extensions and we sketch the necessary background material in the special case of abelian normal subgroups. Let r be an arbitrary group, A a normal abelian subgroup and set G = r/A. Then I? acts on A by conjugation and A acts trivially, so we obtain a group homomorphism G -, Aut(A). In other words, A is a module for the integral group ring Z[G] (even though we continue to view A multiplicatively). Now suppose that we know G, A and the structure of A as a Z[G]-module. The goal is to understand the possibilities for r. One such, of course, is the sp2it extension r = A XI G, but there are certainly other groups which we now proceed to describe. For each z E G, choose Z E r to be a representative of the Acoset corresponding to x . Then I' = U z E G 3 A and Zy E ZyA for all z, E G. Specifically, let us write 30 = @ja(z,y) where a ( x ,y) E
368
8. Grothendieck Groups and Induced Modules
A. Notice that the map a : G x G --t A completely determines the structure of I?. Indeed if z, y E G and a , b E A, then
where of course UY denotes the image of a under the action of y. Furthermore, the associativity of this multiplication is easily seen (Exercise 1) to be equivalent to the equation
for all 5,y, z E G. Functions which satisfy this condition are called 2cocycles and we denote the set of all such a by C2(G,A). Note that, since A is abelian, C2(G,A) is a subgroup of the group of all functions from G x G to A with the operation being pointwise multiplication. We remark that the similarity between the 2-cocycle equation and the twisting relation for crossed products is certainly not surprising in view of the fact that Z [ r ]= Z[A]*(I’/A) = Z[A]*G. Now suppose that we choose different coset representatives, say I E I?. Then for each z E G we have 5 = %6(s)for some 6(z) E A and the 2-cocycle /3 associated with { 5 } is given by
Notice that the 2-cocycle (I! corresponding to a splitting of a ( z ,y) = 1 for all z, y E G and thus if ,O is defined by
r
has
for any function 6: G -, A, then is also a 2-cocycle, but a special one called a 2-coboundary. Now we let B2(G,A) denote the subgroup of C2(G,A) consisting of these 2-coboundaries and we define H2(G,A) to be the factor group H2(G,A) = C2(G,A)/B2(G,A). From the above discussion, it is clear that there is a one-to-one correspondence between extensions of A by G and the elements of this 2nd-cohomology group H2 (G, A). Thus we have
369
35. Group Extensions
Lemma 35.1. Let A be a normal abelian subgroup of r so that A is a Z[G]-module for G = r/A. If A‘ is a Z[G]-module containing A, then there exists a group I” 2 r with A’ a r’,I?’ = FA’, A’ n = A and r‘/A’= r/A = G acting appropriately on A’.
Proof: The structure of r gives rise to a 2-cocycle a : G x G -, A. By viewing a as a map to A’, it is clear that a E C2(G,A’) and thus a determines a group r’ with A’ a I?’ and I”/A’ % G. In fact, if { a : I x E G } is the set of coset representatives for A’ in I?’ obtained in this manner, then I?’ = UzEG?A‘ and I? E UzEG3A so the result follows. I Furthermore, we have
Lemma 35.2. Let A be a Z[G]-module with G finite. i. H2(G,A) has exponent dividing [GI. ii. If A is finitely generated, then H2(G,A) is finite. iii. If A is a free Z[G]-module, then H2(G,h) = (1).
Proof: (i) Let a E C2(G,A) be given and multiply the 2-cocycle equation over all z E G. Since A is commutative, we obtain S(z)S(y)’ = S(yz)a(y, z ) ~ where , 6(y) = a ( z ,y) and n = IGI. Thus an is
nzEG
a 2-coboundary. (ii) This follows from (i) and the fact that C2(G,A) is a finitely generated abelian group, being a subgroup of the direct product of IGI2 copies of A. (iii) Since A is a free Z[G]-module, we can write A as the direct product A = n z E G A Z where the A, are subgroups of A permuted regularly by G. In particular, if a, denotes the z-component of a E A, then a = n z E G u Zand ( a g ) , = (aZg-i)g for any z , g E G. Now let a E C2(G,A), replace z by z - l in the 2-cocycle equation and conjugate by z to obtain
a(zy,z-l), . a(x,y) = a ( x ,yz-’), Reading off the z-components then yields
. a(y, z - y
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8. Grothendieck Groups and Induced Modules
where, by definition, al(z,y)= a ( z , y ) l . Finally, since a ( z , y ) = n t E G c r ( z , y ) Zit follows, by multiplying the above over all z E G, that S(zy)cr(z, y) = 6(z)Y6(y) where S(Z) = Q I ( Z , z-')'. Thus cr is a 2-coboundary as required. I
nzEG
We can now handle case (1).
Proposition 35.3. [142] Let R*r be a crossed product with R right Noetherian and r a finitely generated abelian-by-finitegroup. Specifically, let A be a normal abelian subgroup of r with I'/A finite and assume that A is a free Z[I'/A]-module. Then r = A XI G for some finite subgroup G r and the map A H A @R*G R*r yields an epimorphism Go(R*G) --t Go(R*F).
s
Proof: The assumption that A is a free Z[r/A]-module is used in two different ways. First, by Lemma 35.2(iii), H2(r/A, A) = (1)and thus the extension splits; say r = A XI G for some finite subgroup G r. Second, since A is now a free Z[G]-module, we can choose a free generating set { 21, z2,. . . ,zn } for A which is permuted by G. It follows that if X is the subsemigroup of A generated by the Q'S, then X is the free abelian semigroup of rank n and R*X is a skew polynomial ring in n variables. Furthermore, if Xi denotes the set of monomials in X of degree i , then R*X is Z+-graded with ith component R*Xi. But observe that G normalizes X and in fact each Xi. Thus S = R*(XG) = R*(GX) is a subring of R*r and indeed S is Z+-graded with ith component Si = R*(XiG) = R*(GXi). In particular, since X O = { 1}, we have So = R*G. We show now that S = @ C,"=, Si satisfies the hypotheses of Corollary 34.8. First, S is right Noetherian since R*X is, according to Lemma 1.7, and since S = G . R*X is a finitely generated right R*X-module. Next we note that s0S is flat since, in fact, soS is free with X as a free basis. Finally, since R R is free, it follows from Lemma 33.10 that R*X/(R*X)+ = R has finite projective dimension as a left R*X-module and we can choose
35. Group Extensions
371
to be an appropriate finite (left) projective resolution. Now S is a free right R*X-module with basis G, so tensoring this resolution with SR*X yields the exact sequence
Furthermore, since S @R*X ( R * X ) z s S and @ commutes with arbitrary direct sums, we see that each S @R*X Qi is projective. In other words, the above is a finite (left) projective resolution for the module
and pds(S/S+) < 00 as required. We can now apply Corollary 34.8 to conclude that the map A H A @R*G S yields an epimorphism Go(R*G)= Go(So) + Go(S). Furthermore, since R*r = S X - l , it follows from Lemma 33.3(iii) that the map B I-+ B @s R*r yields an epimorphism Go(S) -+ Go(R*r).Thus the combined map
determines an epimorphism Go(R*G)+ Go(R*r)and the proposition is proved. I We now move on to case (2). Here the goal is to show that
G o ( R * r )is spanned by the images, under induction, of Go(R*Gi) for finitely many finite subgroups Gi of I'. The basic idea of the proof is to embed R*r in a larger, better behaved ring where the result is already known and then to translate this information back to R*r. The proof is in fact conceptually easy but it can get technically complicated. We simplify matters by doing most of the preliminary work in the context of arbitrary rings. We begin with a trivial observation. Suppose that T 5 S are rings and that u is a unit of S. Then T" = u-'Tu is also a subring of S and, since T 2 T", there is a natural correspondence between the modules of the two subrings. Specifically, if A is a right T-module, then A" = { a" 1 a E A } is an
372
8. Grothendieck Groups and Induced Modules
isomorphic copy of the additive abelian group A with the module structure given by a" . t" = (at)".
Lemma 35.4. Let T S be rings and let u be a unit of S . If A is a right T-module, then A" @ p S z A € 3 S. ~ Furthermore, A E mod T if and only if A" E modT".
ProoJ The appropriate maps between the two tensor products are given by a" €3 s H a €3 us and a @ s H a" €3 u-ls. The last part is clear.
I
Now we consider the classical version of the Morita correspondence. Let S be any ring and let e # 0 be an idempotent in S. If A is a right eSe-module, then a ( A ) = A Besee S is a right S-module, and if B is a right S-module, then P(B)= Be is a right eSe-module. For convenience, we call a and P the Morzta maps determined by e. Note that if S is right Noetherian, then so is eSe since I H I S is a one-to-one inclusion preserving map from the set of right ideals I of eSe to the right ideals of S. Thus, in this context, we can consider both Go(S) and Go(eSe). Part (iii) of the following lemma is not really required; we include it for the sake of completeness.
Lemma 35.5. Let e # 0 be an idempotent in the ring S and assume that SeS = f S where f is a central idempotent of S . i. The Morita maps yield a one-to-one correspondence between the right eSe-modules A and the right S-modules B with B = B f . ii. A E modeSe if and only if & ( A )E mods. iii. A is projective if and only if a ( A ) is projective. iv. If S is right Noetherian and f = 1, then G o ( S )and Go(eSe) are isomorphic with isomorphisms determined by the Morita maps.
Proof. Since e E fS,e(fS) = e S and e(fS)e = eSe, it clearly suffices to assume that f = 1. Thus S = SeS and we have 1 = ZTsies!, for suitable si,si E S. Note also that if N is a right S-module with Ne = 0, then 0 = N ( S e S ) = N S and N = 0. (i) We show that the composite maps pa and a/? are both the identity and the first is clear since ( A @ e S e eS)e = A @ e S e eSe A .
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35. Group Extensions
For the second, we need to prove that B e 8 e ~ e SeE B and at least we have a map 19: B e 8 e S -+ B given by 8:be 8 es (be)(es) = bes. Furthermore, Im(I9) = BeS = B(SeS) = B so I9 is onto and we have Ker(I9)e = 0 since (Be 8 eS)e = Be 8 eSe E Be. Thus Ker(8) = 0 and I9 is indeed an isomorphism. (ii) If A is a finitely generated eSe-module, then a ( A ) = A 8 e S is finitely generated since e S is a cyclic right S-module. On the other hand, if B = b j S is a finitely generated S-module, then P(B) = Be = bjSe and it suffices to show Se is a finitely generated right eSe-module. But 1 = C;” sies!, implies that Se = C;”si(eSe), so the result follows. (iii) Since a and p commute with direct sums, it suffices to show that a(eSe) and p(S) are projective and the first is clear since a(eSe) = eS. For the second, p(S) = S e and we define the right eSe-module homomorphisms u:(eSe)n --+ Se and 7: Se (eSe)n by
xy
xy
--f
c n
u:(tl,t 2 , . . . ,tn>H
sit2
1
and 7:se w
(esise, es’zse,. . . , esLse).
Since x y s i e s ! , = 1, it follows that r r ( s e ) = se and therefore that Se is projective. (iv) As we observed, if S is right Noetherian, then so is eSe. In view of (i) and (ii), we need only show that the maps A H [a(A)]E Go(S) and B H [P(B)]E Go(eSe) respect short exact sequences. The latter is clear since B1 C B2 implies that B2e n B1 = Ble. For the former, suppose A1 A2 and let N denote the kernel of the map A1 @ e S A2 8 eS. Since A @ eSe % A, it is clear that N e = 0 and thus that N = 0. This completes the proof. I --f
The next lemma is a combination of the previous two. It is slightly tedious to state, but it is precisely what is needed. Note that if T C S is an extension of rings and if e E T , then we also have the ring extension eTe C eSe.
Lemma 35.6. Suppose we are given
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8. Grothendieck Groups and Induced Modules
i. a right Noetherian ring S and an idempotent e E S with
S = SeS, ii. subrings Ti C S and idempotents ei E Ti such that TieiTi = fiTi with fi a central idempotent of Ti, iii. units ui o f S with (ei>.lli= uileiui = e. I f G o ( S ) is generated by the elements [Bifa @T~S] for all i and Bi E m o d Ti, then Go(eSe)is generated by the elements [Ai B e ( ~ , ) ueSe] ie for all i and all Ai E mode(Ti)"$e.
ProoJ By Lemma 35.4 we have
so these modules determine the same element of Go(S). Thus without loss of generality, we can now assume that each ui = 1 and ei = e. By Lemma 35.5(iv), p determines an epimorphism from G o ( S ) to Go(eSe). Thus since P(Bifi @ T ~S ) = Bifi @T%Se, we see that Go(eSe)is generated by the elements [Bif i @T~Se] for all i and Bi E m o d T i . In addition, by Lemma 35.5(i)(ii), each Bifi is isomorphic to Ai @ e ~ i eTi e for some appropriate Ai E mod eTie. In other words, G o ( e S e ) is generated by the elements [MI with modules M of the form ~ ~ Bifi @T%Se (Ai @eTie eTi) € 3 Se Ai @eTie (eTi @T%Se).
But eTi @Ti Se E eSe since @ commutes with direct sums and Ti = eTi @ (1- e)Ti. Thus
and the result follows. I Now we return to crossed products. Suppose R*r is given and I" is a group containing I? as a subgroup of finite index. Then we can not, in general, hope to extend R*r to some R*r' since, for example, the action of I? on R might not extend. Nevertheless, we have the following observation reminiscent of duality. As it turns out, we do
35. Group Extensions
375
not really care that R"*r' is a matrix ring, but it does offer a way to construct this ring without having to check associativity.
Lemma 35.7. Let I?' 2 F be groups with Ir' : I'l = n < 00 and let R*r be given. Then M,(R*r) is a crossed product Rn*I". Moreover i. R" = diag(R,R,. . . , R). ii. The group of trivial units of Rn*K" transitively permutes the matrix idempotents e1,1,e2,2,. . . ,en,n by conjugation. iii. If Q' is any subgroup of I?', then
with R*r embedded as scalar matrices.
Proof. Let 7 be a right transversal for r in I" and assume that 1 E 7. Since 1 7 1 = n, we can index the rows and columns of M,(R*r) by the elements of 7 with the first row and column corresponding to 1. In particular, for each t,t' E I we have the usual matrix units et,p E Mn(R*l?). We can now grade M,(R*F) by assigning to each expression rget,tt the grade t-lgt' f r'. Here of course T E R, g E r and t , t' E 7. It follows immediately that, in this way, M,(R*r) becomes at least a rl-graded ring. Furthermore, t-lgt' = 1 if and only if t = t' and g = 1. Thus the identity component in this grading is R" = diag(R, R, . . . ,R). Now let x E I?' and observe that for each t E 7 there exists t' E 7 with tx E rt'; say t x = xtt' with xt E r. Indeed, the map t I.+ t' corresponds t o the permutation action of x on the right cosets of r and thus ii = &T2tet,tj is clearly an invertible matrix which is homogeneous of grade x. It follows from this that M,(R*r) is not only I?'-graded, but it is in fact a crossed product Rn*r'. It remains to prove (ii) and (iii) and for the former we need only observe that the idempotents et,t are central in R" and that et,tii = 2tet,tt = 2et,,tt. Finally if s = e1,l . rget,tt . e1,l # 0, then t = t' = 1, s = rgel,l E R * r . e1,l and rget,tt has grade t-lgt' = g E I '. With this, (iii) follows and the proof is complete.
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8. Grothendieck Groups and Induced Modules
We can now quickly handle case (2). Note that if A is a finitely generated free abelian group and a Z[G]-module, then A @z Q is naturally a module for the rational group algebra Q[G]. We say that A is a rationally free Z[G]-module, if A@zQis free as a Q[G]-module.
Proposition 35.8. [142]Let R * r be a crossed product with R right Noetherian and I? a finitely generated abelian-by-finitegroup. Specifically, let A be a torsion free normal abelian subgroup of I? with r/A finite and assume that A is a rationally free Z[r/A]-module. Then there exist finitely many finite subgroups Gi of r such that Go(R*r) is generated by the images of the various Go(R*Gi) under the induced module map.
ProoJ: In this paragraph we temporarily view A additively. By assumption, A 632 Q is a free Q[I’/A]-module and thus it has a Q-basis { p l , p 2 , . . . ,pk } which is permuted in regular orbits by the group r/A. Note that each of the finitely many generators of A C A @ Q is a Q-linear combination of the pi’s. Thus if d is an appropriate A’ = C F Z p i / d . In other words, common denominator, then A A is contained in the finitely generated free Z[r/A]-module A’. In addition, since A‘ E A €3 Q, we see that A’/A is torsion and hence finite. We can now apply Lemma 35.1 to conclude that there exists a group I?’ 2 I’ with A’aI’’ and r’/A’ = r/A acting appropriately on A‘. In addition, I” = FA’ and A’ r l r = A so Jr’ : I’ =J JA’ : A1 = n < 00. Set S = Mn(R*r). Then, by Lemma 35.7, S = Rn*I” is a crossed product of r’ over the right Noetherian ring Rn.Thus since A’ is a free Z[r”/A’]-module, we conclude from Proposition 35.3 that I” = A’MG for some finite subgroup G C I”. Moreover, if T = Rn*G C S, then the map B t+ B @T S yields an epimorphism Go(T) -+ Go(S). Set e = e1,l E S and note that S = SeS. By Lemma 35.7, the trivial units of Rn*G permute the set { e j , j } of diagonal idempotents by conjugation. This action is not necessarily transitive so, for each orbit Oi,we choose ei E Oi and let fi denote the sum of the members of the orbit. Then 1 = fl + f2 + - - . is a decomposition of 1 E T = Rn*G into orthogonal central idempotents. Hence, since Go(T) + Go(S)is surjective, it follows that
35. Group Extensions
377
G o ( S ) is generated by elements of the form [Bifi @T S] for all i and Bi E modT. Moreover eiT fiT and in fact TeiT = fiT since fi is a sum of T-conjugates of ei. Finally, by Lemma 35.7(ii) again, for each i there exists a trivial unit & E Rn*P with xi E r’ and ZL1ei2i = e1,l = e. In other words, the hypothesis of Lemma 35.6 is satisfied with all Ti = T . We conclude therefore that Go(eSe) is spanned by elements of the form [Ai g e p z eSe] e for all i and Ai E m o d e T Z 2 e . But observe that eSe = R*r. e, by Lemma 35.7(iii), and that TZt = Rn*Gxz so eTZae= R*Gi . e where Gi = Gxa n I‘. Thus G o ( R * r .e ) is generated by the images of the various Go(R*Gi e ) under the induced module map. Since R*r . e 2 R*r and each Gi is a finite subgroup of r, the result follows. I We will deal with case (3) and prove the main result in the next section.
EXERCISES 1. Show that each element of H2(G,A) gives rise to a group r which is an extension of A by G. For this, one must prove not only the associativity of but also the existence of 1 and of inverses. 2. Let R be right Noetherian and let U , V E m o d R . Show that [U]= [V] in Go(R)if and only if there exist two short exact sequences 0 --t A + B -+ C -+ 0 and 0 --t A’ t B’ -+ C’ -+ 0 with U @ A‘ @ B @ C’ V @ A @ B’ @ C. This requires a slight modification of the proof of Lemma 34.10(i). 3. Suppose R[r] is a group ring with R right Noetherian and I’= H M G polycyclic-by-finite. Assume that R has finite projective dimension as a left R[H]-module. Use the argument of Proposition 34.9 to show that the induced module map G o ( R [ G ]-)+ Go(R[r]) is oneto-one. 4. Show that the proof of Proposition 35.8 from Proposition 35.3 works equally well for KO.In other words, if there is a KOanalog of the case (1)result, then there will also be an analog in case (2).
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8. Grothendieck Groups and Induced Modules
5. Let G be a finite group of X-outer automorphisms of the simple ring R and assume that (GI-' E R. What does Lemma 35.5 say about the relationship between R-modules and those of the fixed ring RG? 6. Let 1 = el e2 - . - e, be a decomposition of 1 E R into orthogonal idempotents and let G be a group of units of R which transitively permutes the set { ei } by conjugation. For each i choose gi E G with ei = gi'elgi and then define ei,j = gi'elgj. Show that the latter elements multiply like matrix units and then prove that R 2 M,(S) with S 2 elRel.
+ + +
36. The Induction Theorem The goal of this section is to prove the induction theorem. Specifically we show that if R*I' is a crossed product with R right Noetherian and r polycyclic-by-finite, then Go(R*F) is generated by the images under the induced module map of the Grothendieck groups Go(R*Gi) for finitely many finite subgroups Gi C_ F. As indicated earlier, most of the work here involves abelian-by-finite groups. Furthermore, such groups split into three cases and the first two have already been dealt with. Thus it remains to consider case (3) and for this we follow the argument of [33].Again the idea is to introduce a larger, better behaved ring, but this time R*F will not be a subring but rather a homomorphic image. We start with an observation which has already proved useful, namely transitivity of induction. Stated below in its full generality, it is still an immediate consequence of the (functorial) associativity of tensor product.
Lemma 36.1. Let 6 :R -, S and 'p: S -, T be ring homomorphisms and let A be a right R-module. Then ( A @R S ) @s T A @R T where RT is obtained from the ring homomorphism p6:R + T . Furthermore, if R S and sT are flat, then so is RT. Next we require a slight extension of Lemma 34.3(ii). Since the proof is essentially the same, we just briefly sketch it here.
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36. The Induction Theorem
Lemma 36.2. Let T be a subring of S with T S Aat. If A is a right T-module and M is a left S-module, then Torf(A
@JTS, M
)
TorF(A, M )
for all 2 0. In particular if T M is ffat, then for all i >_ 1.
@T
s,M ) = 0
ProoJ Let P A + 0 be a projective resolution for the T-module A. Since T S is flat, it follows that P @T S -+ A @T S -, 0 is a projective resolution for A @ S. Now, by definition, the groups Tor'(A,M) come from the homology of the complex P @T M -+ 0 while the groups Torf(A @J S, M ) come from the homology of the complex (P @T S ) @JSM + 0. But --f
(P,
@T
S ) @S M
P,
1M
for all n, so the two complexes are naturally isomorphic and hence have the same homology. I The following result is the analog of Lemma 35.6 in the context of homomorphisms.
Lemma 36.3. Suppose we are given i. a right Noetherian ring S and an epimorphism cp:S -+ S" such that sS9 has finite projective dimension, ii. a subring So of S with S = So @ Ker(cp) and soSAat, iii. subrings Ti C S with T,S and T , S ~ both Aat. If Go(S) is generated by the elements [Bi@ T ~S] for all i and Bi E modTi, then Go(S9) is generated by the elements [Ai @JT: S"] for all i and Ai E mod T ' , .
Proof. Notice that both S and S' are right Noetherian and that if A E mods, then TorS(A,Sp) E modSp. Furthermore, since sS9 has finite projective dimension, Lemma 34.2(ii) implies that, for each A, only finitely many of these Tor groups can be nonzero. In particular, it makes sense to define q: mod S -+ G o ( S V ) by q(A) = z(-)z[Tor?(A, S')] i=o
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8. Grothendieck Groups and Induced Modules
and it then follows from Lemma 34.1 that q respects short exact sequences. Thus q determines a group homomorphism 7: Go(S) -+ GO(S')* Now suppose T is any subring of S with T S and TS' both flat. If B E modT, then B @T S E mod S and we compute q([B@T S]). Using both flatness assumptions, it follows from Lemma 36.2 that Torf(B @T S,S') = 0 for all i 2 1. Thus since Tor0 is the tensor product, we have
Note that s0S is flat, by assumption, and that soS' E soSo since S = So@ Ker(cp). Thus, soS' is also flat and the above implies that q( [B@soS ] )= [B@soS'] for all B E mod SO.But cp restricts to a ring isomorphism cp:So + S' so it follows that every finitely generated S"-module is isomorphic to a suitable B @soS'. In other words, Go(SV) is spanned by the images under q of the various [B@soS] and thus 7 is certainly an epimorphism. Hence, since Go(S) is generated by the elements [Bi@T~S] for all i and Bi E modTi, it follows that Go(S") is generated by the images of these elements under q. But, by assumption, T~S and T~S' S ] )= [ B ~ @S'] T ~by the observation of the second are flat so q( paragraph. Thus since
and Bi
@ T ~Ti'
E modT,", the result follows.
I
Now we return to crossed products where we begin by constructing the necessary larger ring. Suppose r' = A'MI' is a split extension of A' by I?. Then r can be viewed as both a subgroup of I" and as a homomorphic image. To avoid this confusion, let us write To for the subgroup so that r' = A T 0 and reserve I' for the homomorphic image.
Lemma 36.4. Let I" = A' 4 I? be a split extension of A' by r and let R*r be given. Write T' = Afro where rois a complement for
381
36. The Induction Theorem
A' and let R*ro be the crossed product obtained naturally from the isomorphism I'o E I'. i. R*ro extends to a crossed product R*r" with R*A' = R[A'] an ordinary group ring. ii. The map p: I" + extends to an epimorphism 'p: R*r' -t R*T' with &*I" = R*ro @ Ker('p). iii. I f R has finite projective dimension as a left R[A']-module, then R*r has finite projective dimension as a left R*I"-module.
Proof. Since R*r is given, it is endowed with a twisting function T:I' x I' U(R) and an action u: I' -+ Aut(R) satisfying conditions (i) and (ii) of Lemma 1.1. In particular, if 'p:r" + I' is any map then, by way of composition, we obtain r':"' x I" + U(R) and 5': I?' Aut(R). Furthermore, if cp is a group homomorphism, then --$
--f
it is trivial to verify that (i) and (ii) of Lemma 1.1are also satisfied by these functions. It follows that r' and u' give rise to a crossed product R*r' and that 'p extends to a ring epimorphism R*r' + R*r. Observe that the restriction of cp to I'o yields the given isomorphism I'o r and hence that 'p:R*ro-+ R*I' is the natural isomorphism between these two crossed products. In addition, since i = 1 E R*r and A' = Ker(p), it follows that the twisting and action in R*A' 2 R*r' are trivial and hence that R*A' = R[A']. Finally if P --t R -+ 0 is a finite projective resolution for R as a left R[A']-module, then
is a finite projective resolution for completes the proof.
(m?) I'+@
R
!2
R*r. This
We also need an analog of Lemma 33.10 for group rings. Let R[A] be a group ring with A = (XI,~ 2 . ., . ,xt) free abelian on the generators xi. Then we have a homomorphism R[A]--t R given by xi H 1 for all i. In particular, if M is an R-module then, via the combined map R[A] -+ R t End(M), we see that M is also an R[A]-module.
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8. Grothendieck Groups and Induced Modules
Lemma 36.5. Let R[A] be as above with A = (x1,22,.. . ,xt) a free abelian group. If M is any R-module of finite projective dimension, then M also has finite projective dimension when viewed as an R[A]module.
Proof, It suffices to assume, by induction on t , that A = (x)is infinite cyclic. Write S = R[A] and consider the short exact sequence 0 + (x - 1)s+ S + R -+ 0 determined by x H 1. Since both S and (x - 1)s S are free right S-modules, we have pd,R _< 1 and indeed, by Lemma 33.9(ii), pd,P P. Finally let
5 1 for any projective R-module
be a finite projective resolution for M as an R-module. Then, as S-modules, each Pi has finite projective dimension so it follows from Lemma 33.9(i) and induction that the same is true for M . An analogous proof works for left modules. I We can now handle case (3).
Proposition 36.6. [142] Let R*r be a crossed product with R right Noetherian and J? a finitely generated abelian-by-finite group. Then there exist finitely many finite subgroups Gi of I? such that Go(R*I') is generated by the images of the various Go(R*Gi) under the induced module map.
Proof. By assumption, r has a finitely generated normal abelian subgroup A of finite index. Moreover, if the torsion subgroup of A has order n, then A" = { xn I x E A 1 is a characteristic torsion free subgroup of A of finite index. Thus we may replace A by An if necessary and assume that A is free abelian. In this paragraph, we will temporarily view A additively. Then A 8 2 Q is a finite dimensional module for the rational group ring Q[r/A]. Thus since the latter ring is semisimple Artinian, there exists a finite dimensional Q[I'/A]-module V with V @ (A @Z Q) free. Let B be a Q-basis for V and let A' be the 2-submodule of V
383
36. The Induction Theorem
generated by the finite set 23 - (r/A).It follows that A’ is a finitely generated Z[r/A]-module with A’ @Z Q = V and thus A’ @ A is a rationally free Z[I’/A]-module. Now revert to multiplicative notation and observe that I? acts on A’ with A acting trivially. Thus if I?‘ = A‘ >Q I‘,then I” has a normal abelian subgroup s2’ 2 A‘ x A of finite index and 0’ is a rationally free Z[I”/fl’]-module since I”/n’ S r/h. By Lemma 36.4, there exists a crossed product S = R*r’ such that the map p:I” --t I? extends to a ring homomorphism ‘p: R*r’ 4 R*r = Sp. Furthermore, S = So@Ker(cp)where So = R*ro. Since A’ is free abelian and RR is free, it follows from Lemma 36.5 that R has finite projective dimension as a left R[A’]-module. Thus, by Lemma 36.4(iii), S‘+‘ has finite projective dimension as a left S-module. The structure of I” and Proposition 35.8 imply that there exist finitely many finite subgroups Gi of I” such that Go(R*I”)is generated by the elements [Bi@ , R * G ~S] for all i and Bi E modR*Gi. In particular, setting Ti= R*Gi R*r‘, we see immediately that the hypotheses of Lemma 36.3 are all satisfied. Indeed S is right Noetherian, by Proposition 1.6, and certainly soS and T ~ are S free and hence flat. Furthermore, since Gi is finite, we have Gi n A‘ = (1) so 9:R*Gi -+ (R*Gi)V = R*GY is an isomorphism. It follows that S p = R*r is a free and hence flat left Ti-module. Finally, we conclude from Lemma 36.3 that Go(S’f’)is generated by the elements [Ai QT? Sp] for all i and Ai E modT,’. Thus, since T? = R*GY and the map Go(Tr) Go(S”) is well defined, it follows that Go(R*r) is generated by the images of the various Go(R*GT). This completes the proof. I ---f
At this point, it is a simple matter to prove the induction theorem. However, in order to give that result a more precise formulation, we require a few additional observations on group extensions. These will yield information on the conjugacy classes of finite subgroups of a polycyclic-by-finite group. Let I? be an arbitrary group which is the split extension I‘ = AG of a normal abelian subgroup A by some complement G. As usual, assume that we know the structure of A as a module over Z[G] Z[r/A]. The question now is to determine all possible complements
8. Grothendieck Groups and Induced Modules
384
for A in I?. Suppose that H is another such complement. Since each coset of A has a unique representative in H and in G, there exists a function a : G .--t A with H = { x a ( x ) 1 x E G}. Furthermore, since H is a subgroup we have ZQ(Z) - ya(y) = xya(zy) and hence (l-cocycle)
Q(x>YQ(?A = Q(4
for all z,y E G. Functions which satisfy this condition are called l-cocycZes and we denote the set of all such Q by C1(G,A). Since A is abelian, C1(G, A) is a group under pointwise multiplication. Now suppose that we replace H by a I'-conjugate to obtain a new complement for A. Since I? = AH = H A , this conjugation can be achieved by an element c-l E A and the l-cocycle associated with cHc-l is given by p ( x ) = a ( x ) ~ - ~ c "In . particular, l-cocycles satisfying
( l-coboundary)
P(.)
= c -1 cx
for some c E A are called l-coboundaries and we let B1(G, A) denote the subgroup of C1(G,A) consisting of these functions. It follows from the above discussion, that the elements of the lSt-cohomoZogy Qro'021P H1(G, A) = C1(G, A)/B1(G, A) are in one-to-one correspondence with the conjugacy classes of complements for A in I'. The next result is analogous to Lemma 35.2 with a similar proof which we just sketch.
Lemma 36.7. Let A be a Z[G]-module with 1GJ= n finite. i. H1(G, A) has exponent dividing (GI. ii. If A is finitely generated, then H1(G, A) is finite. iii. If A is a free Z[G]-module, then H1(G, A) = (1). iv. If G = (9)is cyclic, then
Proof: (i) Let a E C1(G,A) be given and multiply the l-cocycle equation over all z E G. Thus if c = ~ ( zE) A, then c Y ~ ( y = )~
nzEG
c and an is a l-coboundary. Part (ii) is clear.
385
36. The Induction Theorem
(iii) As in Lemma 35.2, we replace y by 9-l and conjugate the l-cocycle equation by y. Reading off the y-components then yields . a l ( y - ' ) y = al(xy-')y where, by definition, a l ( z ) = a(z)1. Now multiply over all y E G to obtain a ( x ) c = c" where c= al(y-')Y E A. (iv) Let a E C1(G,A) and write a(g)= a E A. Then it follows from the cocycle equation that, for all i >_ 0, a ( g 2 ) = al+"+".+""-'. But g n = go = 1, so we must have al+s+'.'+sn-l= 1. Since a is a coboundary if and only if a = c-'cg E Ag-', the result follows. I
ny
We remark that H1(G, A) and H2(G,A) are part of an infinite series of cohomology groups starting with Ho(G, A). However, only these first three have true group theoretic interpretations. We also note that part (iii) above explains why we can choose any complement G for A in Proposition 35.3. Indeed, all such complements are conjugate in that case. As a consequence of part (ii) we have
Lemma 36.8. If I' is a polycyclic-by-finite group, then
has only
finitely many conjugacy classes of finite subgroups.
r has a torsion free normal abelian If G is a finite subgroup of r, then finite group r/A and hence there are
Proo$ Let us first assume that
subgroup A of finite index. AG/A is a subgroup of the only finitely many possibilities for the group AG. Because of this, it suffices to assume that r = AG and to show that there are only finitely many conjugacy classes of finite subgroups H with AH = r. But observe that A is torsion free and H is finite, so A n H = (1) and H is a complement for A in r. Thus since H1(G,A) is finite, by Lemma 36.7(ii), the result follows in this case. For the general polycyclic-by-finite group r, we proceed by induction on the Hirsch number and we may clearly assume that I' is infinite. Then has a torsion free normal abelian subgroup A # (1) and, by induction, r/A has only finitely many conjugacy classes of finite subgroups, say with representatives ri/A for i = 1 , 2 , . . . ,t . If G is any finite subgroup of I?, then AG/A is conjugate to some I'i/A, so G is conjugate to a finite subgroup of I'i. But, by the above, ri
386
8. Grothendieck Groups and Induced Modules
has only finitely many conjugacy classes of finite subgroups and thus the result follows. 1 It is time to prove the induction theorem. Notice, in the following, how the simple crossed product property R * r = ( R * N ) * ( r / N ) with N a r allows for an almost immediate reduction to the abelianby-finite case.
Theorem 36.9. [140][142]Let R*r be a crossed product with R right Noetherian and r a polycyclic-by-finite group. Suppose G I , Gz, . . . ,Gt are representatives of the conjugacy classes of the maximal finite subgroups of r. Then G o ( R * r ) is generated by the images of the various Go(R*Gi) under the induced module map.
ProoJ We first show that Go(R*r)is generated by the images of the Grothendieck groups Go (R*Gi) for finitely many finite subgroups Gi C I?. The proof of this proceeds by induction on the Hirsch number of r and of course we may assume that is infinite. Then r has a normal torsion free abelian subgroup A # (1)and observe that
R*r = (R*A)*(r/A) = S*F with f' = r / A and with S = R*A right Noetherian by Proposition 1.6. Thus by induction, f' has finitely many finite subgroups f'i such that Go(S*f') is generated by the images of the various Go(S*f'i)under the induced module map. Note that f'i = ri/h for some subgroup I'i C I' and that S*Fi = R*ri. Thus, since S*f' = R*r, the above says that Go(R*r)is generated by the images of the various Go(R*ri)under the induced module map. But each ri is abelian-by-finite, so Proposition 36.6 applies. We conclude that there exist finitely many finite subgroups Gi,j I'i such that the induced module map @ Go(R*Gi,j) -+ Go(R*ri) is surjective. Thus the composite map
Cj
CB
C Go(R*Gi,j) @ C Go(R*ri)
-+
-+
i,j
Go(R*r)
i
is also surjective and, by transitivity of induction, this first key observation is proved.
387
36. The Induction Theorem
We have therefore shown that there exist finitely many finite subgroups Gi I'with the induced module map @ Go(R*Gi) -+ Go(R*r) surjective. Now observe that if G C_ H C_ I? then, by transitivity of induction again, the image of Go(R*G) is contained in the image of G,-,(R*H). In other words, we can replace each Gi by some Hi 2 Gi with Hi a maximal finite subgroup of I'. Furthermore, by Lemma 35.4, conjugate subgroups give rise to the same image in Go(R*l?). Thus we require only one representative from each conjugacy class of maximal finite subgroups and, with this, the result follows. I
xi
It is natural t o ask if an analogous result holds for Ko(R+r). The answer is ''no". A close look at the proof shows that all reductions from case (1) t o (2) to (3) to Theorem 36.9 do in fact hold for KO. The problem is that case (1) and Corollary 34.8 fail in this context. We close by constructing an appropriate commutative counterexample. To start with, we have
Lemma 36.10. Let R be a commutative domain with quotient field K and let P and Q be R-submodules of K . i. If PQ = R, then P and Q are both finitely generated projective R-modules. ii. If P is stably free, then P = cR for some c E P .
Proof. (i) We use the fact that PQ 5 R and that 1 = Cypiqi with pi E P and qi E Q. Define the R-module maps c : R n -+ P and T : P --t R" by n
i= I 7: P
( Q l P ,42P,*
. 4nP). * 7
Then m ( p ) = p for all p E P , so P is finitely generated and projective. (ii) Suppose P # 0 is stably free and that P @ R" = V = Rm. Since P K = RK = K , it follows, by computing the K-dimension of V K , that m = n 1. Now let { 210, 211,. . . ,vn } be an R-basis for V = Rn+' and let { w1,. . . , 20, } be a basis for W = Rn C V. Then,
+
388
8. Grothendieck Groups and Induced Modules
for each 1 5 j 5 n, we have wj = Cy=owiri,j with ri,j E R and we study the (n 1) x n matrix M = [ri,j].Notice that if I is any maximal ideal of R, then
+
V/VI=
G3
(P/PI)
so the elements w1,. . . ,w n are (R/I)-linearly independent modulo V I . It follows that if = [ F i , j ] , where F i , j = ri,j + I E R / I , then M must have rank n as a matrix over the field R / I . Hence at least one of the maximal minors of M is not zero. Since I is an arbitary maximal ideal, this implies that the set { mo, ml,.. . ,mn } of maximal minors of M is unimodular, that is Cy=omiR = R. In other words, we can find elements r i , E~ R so that the augmented ( n 1)x (n+1) matrix M* = [ri,j]has determinant 1 and hence is invertible. Finally, if n we define 200 = Ci=o wiri,~,then { W O ,w1,.. . ,w,} must also be an R-basis for V . Thus V = woR@ W and P 2 V/W E woR is free of rank 1.
+
With this we can prove
Proposition 36.11. [192]Let R be a commutative domain with quotient field K and suppose there exists u E K \ R with a2,u3 € R. If S is the polynomial ring S = R[t] or the group ring S = R[T] with T = ( t ) infinite cyclic, then S has a finitely generated projective module which is not stably free.
Proof. We will just consider the group ring R[T]. The polynomial ring proof is identical except at one point where it is in fact slightly simpler. Let f = at E K [ T ]and let P = (1 f,1 f f 2 ) be the R[T]submodule of K [ T ]generated by 1 f and 1 f f 2 . Similarly let Q = (1 - f , 1 - f + f 2 ) and observe that
+
+
+ + + +
+ f 3 , 1- f 3 , 1+ f 2 + f 4 ) C R[T] since u2,u3,u4 E R. Furthermore, we have 2 = (1+ f 3 ) + (1- f 3 ) PQ = (1 - f 2 , 1
E PQ a n d 3 = ( l - f 2 ) ( 2 + f 2 ) + ( l + f 2 + f 4 ) E PQ. Thus 1 E PQ and P is a finitely generated projective R[T]-module by Lemma 36.10(i).
389
36. The Induction Theorem
Now suppose, by way of contradiction, that P is stably free. Then Lemma 36.10(ii) implies that P = cR[T] for some c E K [ T ] . Moreover, we have P . K [ T ]= K [ T ] since , (1+f +f2)-(1+f)f = 1, and thus P = cR[T] implies that K [ T ]= cK[T]. In other words, c is a unit of K [ T ]so, since K is a field and T = ( t ) is infinite cyclic, c = kin for some k E K \ 0. But t" is a unit of R[T]so we have P = kR[T]and furthermore Q = k-lR[T] since PQ = R[T].Finally k(1 -at) E IcQ = R[T]so Ic E R and then 1+at E P = kR[T]G R[T] yields a E R, a contradiction. We conclude therefore that P is not stably free. Now for the example. Let F be a field and let R be the subring of the power series ring F [ [ x ] ]consisting of all elements with zcoefficient 0. Then R is Noetherian since R = 1- F [ [ x 2 ] ] x3 * F [ [ x 2 ] ] and R is certainly local. Thus all finitely generated projective Rmodules are free and Ko(R) is the cyclic group generated by [R]. On the other hand, if S = R[t]or R[T]as above, then S has a finitely generated projective module which is not stably free. This follows from Proposition 36.11 since x 2 , x 3 E R but x 4 R. Thus, by Lemma 34.10(ii), Ko(S) is properly larger than its cyclic subgroup generated by [S]and hence the induced module map Ko(R) -+ Ko(S) is not surjective. Finally notice that if S = R[t],then S is Z+-graded with So = R. On the other hand, if S = R[T]then, since T is torsion free, the only finite subgroup of T is G = (I) and R[G] = R. Thus we see that neither Corollary 34.8 nor Theorem 36.9 has a KO analog without additional assumptions on the ring.
+
EXERCISES 1. Show that the reductions from case (2) to case (3) and from case (3) to Theorem 36.9 work equally well for KO. 2. Let R*G be a crossed product with the property that every finitely generated right module has finite projective dimension. Show that the same is true for R*H with H any subgroup of G. To this end, use Lemma 3.10 to show that any R*H-module is a direct summand of the restriction of an R*G-module.
390
8. Grothendieck Groups and Induced Modules
3. Let R*r be a crossed product with R right Noetherian and J? polycyclic-by-finite. If all finitely generated R*r-modules have finite projective dimension, prove that a precise KOanalog of Theorem 36.9 holds. For this, use the previous exercise along with Proposition 33.8. 4. Let A be a Z[G]-module with G = (9)cyclic. If IG( = n, prove that H2(G,A) = AG/A1+g+'.'+gn-l where AG = Cn(G). If G is infinite, show that H1(G, A) = A/Ag-l. 5. Let A be a Z[G]-module and assume that G is a free group. Show that H2(G,A) = (1). 6. Let R be a commutative ring and let P and Q be ideals of R with P + Q = R. Prove that P @ Q E R @I ( P Q ) as R-modules. In particular, if R is a domain and PQ is principal, conclude that both P and Q are projective.
9
Zero Divisors and Idempotents
37. Zero Divisors and Goldie Rank We start this chapter by discussing the famous zero divisor problem. Let K[G]be a group algebra and suppose G has a nonidentity element z of finite order n. Then the equation
( 1 - z ) ( l + z + - . ’ + z n - l )= 1 - z n
=o
shows that K[G]has nontrivial zero divisors. On the other hand, if G has no such element 2, that is if G is torsion free, then K[ G] has at least no obvious divisors of zero. Because of this, and with frankly very little supporting evidence, it was conjectured that if G is torsion free then K[G]must be a domain. Amazingly, the conjecture has held up for over forty years. Progress on this problem divides naturally and historically into four stages. The first is the purely group theoretic approach concerned with ordered groups, unique product groups and their analogs. Specifically, one asks if the torsion free assumption on G implies other properties of G which are more relevant to the zero divisor question.
391
392
9. Zero Divisors and Idempotents
The answer here has been a hodge-podge of definitions and unsatisfactory results. In some sense, this era was finally brought to an end by the recent paper [183]where small cancellation theory was used to construct a torsion free group which does not have the unique product property. Indeed even polycyclic examples are now known to exist. The structure of the following group is well known. What is new here is the fact that the unique product property fails.
Proposition 37.1. [177] Let G be the group
Then G is a torsion free abelian-by-finite group which does not have the unique product property.
Proof: Set z = zy, a = z2, b = y2 and c = 2'. Then clearly ' a = a-1 and 6' = b-l. Moreover, we have (zy)-' = y-lz-' = b-' ayz so c-l - ( ~ y ) -=~yzyz and hence c" = cY - 1 = c-'. It follows that D = ( a ,b, c) is a normal abelian subgroup of G. Furthermore, G / D is generated by Z and g with Z2 = jj2 = (Zy)2 = 1 and hence G / D is at most a fours group. We next observe that D is free abelian of rank 3 and that JG/DJ= 4. To this end, consider the two rational matrices X , Y E GL4(Q) given by
0 1 0
x=(:
x:
0 0 1
0
0
0
1 0
y = ( 02 0 0 o1 ) 0 1/2 0 0
Then it is easy to see that X - l Y 2 X = Y-' and Y-lX2Y = X-2 so there is a well defined epimorphism a: G -+ (X, Y ) given by a(.) = X and a(y) = Y . If 2 = X Y , we have
a ( a ) = X2 = diag(2,2,1/2,1/2) a(b) = Y2 = diag(2,1/2,2,1/2) a ( c ) = 2' = diag(2,1/2,1/2,2)
37. Zero Divisors and Goldie Rank
393
and these matrices generate a free abelian group of rank 3. F’urthermore, since a ( D )is diagonal and 1,X , Y,2 are not diagonal multiples of each other, we have JG/DJ2 4 and hence G/ D is a fours group. Now we note that G is torsion free. To start with, if g E G has finite order then, since g2 E D and D is free abelian, we must have g2 = 1. Next, we consider the D coset containing g. Suppose for example that g E Dx and write g = dx with d E D. Then we have 1 = g 2 = dxdx = dd”a and this is a contradiction since a occurs with even exponent in dd”. Similarly g 4 D y or D z . Thus g E D and therefore g = 1. Finally for the unique product property, let S be the subset of G given by S = Az U B y U C with
A = { 1, a-’, a-lb, b, a-lc-’,
c}
C = { c, c - l }
B = { 1 , a , b - l , b-lc, c, a b - l c } . Then JSJ= 14 and S2 = S S has no unique product. For the latter, observe that A , B , C D and that D is commutative. Thus the intersections of S2 with the four cosets of D are given by
S2n D X = AzC U CAX = (AC” U A C ) Z
s2n DY = BYC u CBY = (BCYu B
C)~
S2n D X= AxBy U BYAX= (AB“ U AYBa-’bc-l)z S2 n D = AxAx U ByBy U C2 = AAZaU BBYbU C2. In particular, since C” = CY = G, the elements in S2 n D z and S2 f l Dy all occur with even multiplicities. Unfortunately, there is no such simple explanation for the other two cosets. The 72 and 76 products, respectively, must be checked by hand or computer to prove that all multiplicities are larger than 1. Since these computations all occur in the free abelian group D , there are no technical difficulties involved. I In view of this resuIt, the zero divisor problem for the group G above cannot be settled by group theoretic means. It was, however,
394
9. Zero Divisors and Idempotents
dealt with during the second stage of progress by the introduction of certain ring theoretic machinery. Specifically, this stage is based on the theorem of [42] which asserts that if R1 and R2 are both n-firs containing a common division ring D , then the coproduct R1 R2 over D is also an n-fir. This turns out to be relevant since a ring is a l-fir if and only if it is a domain (see Exercise 8 of Section 13) and indeed we have
uD
Theorem 37.2. [42] [99] Let K be a field, G a group and suppose that G has subgroups A, B and N a G with GIN = ( A / N )* ( B / N ) . If K [ A ] and K [ B ]are domains and ifK[N] is an Ore domain, then K[G]has no zero divisors. Here of course ( A I N ) * ( B / N )denotes the free product of the two groups. Now, in general, free products are quite complicated and are certainly not Noetherian. However, it is easy to see that C2 * Cz, the free product of two groups of order 2, is isomorphic to the infinite dihedral group and it turns out that this is a basic building block of torsion ftee supersolvable groups. Recall that a group G is said to be sdpersolvable if it has a finite normal series
with each quotient Gi+l/Gi cyclic. Thus any such group is necessarily polycyclic. As a consequence we have
Corollary 37.3. [Sl]If G is a torsion free supersolvable group, then the group algebra K[G]is a domain. In particular, this applies to the example of Proposition 37.1. The third stage of progress handled all polycyclic-by-finite groups. It began with the following general observation.
Theorem 37.4. [203] Let R be a ring satisfying i. R is right Noetherian and semiprime, ii. all finitely generated R-modules have finite projective dimension,
37. Zero Divisors and Goldie Rank
395
iii. all finitely generated projective R-modules are stably free. Then R is a domain.
In particular, if R = K[G]with G a torsion free polycyclicby-finite group, then R is certainly right Noetherian and it is also semiprime, by Theorems 5.3 and 5.4. Furthermore, a key result of [193](see [161,Theorem 10.3.121) asserts that K[G] has finite global dimension, that is all R-modules have finite projective dimension. Thus (i) and (ii) above are satisfied. While we now know that (iii) also holds (see Exercise 4), this fact was not available when the zero divisor applications were originally obtained. Indeed, (iii) was circumvented by either a suitable localization or by somehow dropping down to subgroup of finite index. The work began with 1261 where these ideas were applied, using localization, to handle all torsion free abelian-by-finite groups in characteristic 0 and abelian-by-(finite p ) groups in characteristic p. Then [55] used the fact that (iii) held for poly-2 groups to handle all torsion free polycyclic-by-finite groups in characteristic 0. That paper also considered some groups in characteristic p > 0, but it remained for [32]to complete the result. Combining all this, we have
Theorem 37.5. [26] [55] [32]Let K be a field and let G be a torsion free polycyclic-by-finite group. Then the group algebra K[G] is a domain. Some additional results on more general solvable groups were obtained by [195], but the above theorem essentially ended the third stage. The techniques used in its proof were powerful, but not powerful enough to handle Noetherian group rings D[GJ with D a division ring. We are now in the fourth stage of progress and this is all based on the upper bound for Go(R*G) given in Theorem 36.9. Indeed, the applications can be obtained quite quickly and elegantly by techniques which are not really new. What is new here and crucial is Theorem 36.9 itself. We begin with the necessary variant of Theorem 37.4.
Theorem 37.6. [203]Let R be a semiprime right Noetherian ring. I f G o ( R )is the cyclic group generated by [R],then R is a domain.
396
9. Zero Divisors and Idempotents
Proof. By Goldie's theorem, R has a classical right ring of quotients Q = Q(R) which is semisimple Artinian and, by Lemma 33.3(iii), the induced module map A H A @ R Q gives rise to an epimorphism 8:G o ( R ) -+ G o ( Q ) . In particular, the assumption on Go(R) translates to the assertion that G o ( Q )is generated by [Q].Lemma 33.l(ii) now implies first that Q has only one irreducible module and then that QQ is itself irreducible. Thus Q is a divison ring and R is a domain. We remark that it suffices to assume in the above that Go(R)is generated by [R]modulo its torsion subgroup. This follows since the torsion disappears in Go(Q(R)).We will need the following result on Ore localization. It actually follows as in Proposition 12.4(i), but we offer an alternate approach here.
Lemma 37.7. Let R*G be a crossed product and let T be a right divisor set of regular elements of R. If T is G-stable, then T is a right divisor set of regular elements of R*G and (R*G)T-' = (RT-')*G.
Proof. It suffices to show that T is a right divisor set in R*G; the remainder of the argument is then trivial. Thus suppose t E T and (1: = CiEca, E R*G. Since T is G-stable, we have t" E T for all z E Supp a and we can write the finitely many elements (t")-'a, E RT-' all with a common right denominator s E T. In particular, if (t")-'a, = b,s-', then t"b, = ass and hence t(C%b,)= QS as required. I It is now a simple matter to prove
Lemma 37.8. Let R*G be a crossed product with R an Ore domain and with G a locally polycyclic-by-finite group. Suppose that R*H is a domain for all finite subgroups H C_ G. Then R t G is an Ore domain.
ProoJ Since the property of being an Ore domain can be checked two elements at a time, it suffices to prove this result for the finitely generated subgroups of G. In other words, we may suppose that G
397
37. Zero Divisors and Goldie Rank
itself is polycyclic-by-finite. Furthermore, in view of the previous lemma, we can localize and assume that R is a divison ring. Thus R*G is right Noetherian and it is semiprime by Proposition 8.3(ii). Now let H be any finite subgroup of G. Then R*H is an Artinian domain and hence a divison ring, so Lemma 33.l(ii) implies that G o ( R * H ) is the cyclic group generated by [R*H].Hence since R*H @R*H R*G = R*G, the induced module maps sends Go(R*H) to the cyclic subgroup of Go(R*G) generated by [R*G]. Finally, by Theorem 36.9, Go(R*G)is generated by the images of finitely many such G o ( R * H ) and thus Go(R*G) is the cyclic group generated by [R*G].Theorem 37.6 now implies that R*G is a domain and then it is an Ore domain by Goldie’s theorem. I If G is torsion free, then the only possibility for H above is the identity subgroup. Thus R*H = R is a domain, and Lemma 37.7 already extends Theorem 37.5. But more can be done; we can consider arbitrary solvable groups. Indeed we study groups G having a finite subnormal series
with each quotient Gi+l /Gi locally polycyclic-by-finite. For convenience, we let p(G) denote the minimal such n for the group G. Thus p ( G ) = 0 if and only if G = (1) and p ( G ) = 1 if and only if G is a nonidentity locally polycyclic-by-finite group. We list a few basic properties of p.
Lemma 37.9. Let G be as above and let H be a subgroup of G. Then p ( H ) 5 p ( G ) and p ( G / H ) 5 p(G) i f H a G . Moreover if (1)# H U G and IG/HI
< 00, then p ( G ) = p ( H ) .
Proof: The first observations are clear. Now suppose (1) # H u G with (G/HJ < 00. Since n = p ( H ) _< p ( G ) , we need only obtain the reverse inequality. By assumption, n 2 1 so there exists N u H with H / N locally polycyclic-by-finite and with p ( N ) 5 n - 1. If T is a finite transversal for H in G, then M = N t a G and HIM is locally polycyclic-by-finite since H I M
c--t
ntGT ntETH / N t . Moreover,
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9. Zero Divisors and Idempotents
p ( M ) 5 p ( N ) 5 n - 1,so it clearly suffices to show that G = G / M is locally polycyclic-by-finite. To this end, let X be a finitely generated subgroup of G. Then l X / ( X n H)I < 00, so X n j? is also finitely generated (see Exercise 5 ) and hence X n H is polycyclic-by-finite. Thus X is also polycyclic-by-finite and the lemma is proved. I We now come to the main results of this section. Notice how the simple crossed product property R*G = ( R * N ) * ( G / N )with N a G allows for a fairly easy inductive argument.
Theorem 37.10. [94] (1411 Let R*G be a crossed product with R an Ore domain and assume that G has a finite subnormal series (1) = G o a G 1 a . . . a G , = G with each quotient Gi+l/Gi locally polycyclic-by-finite. If R*H is a domain for every finite subgroup H & G , then R*G is an Ore domain.
Corollary 37.11. (941 (1411 Let R*G be a crossed product with R an Ore domain and assume that G has a finite subnormal series (1) = G O a G l a . - . a G , = G
with each quotient Gi+l/Gi locally polycyclic-by-finite. If G is torsion free, then R*G is an Ore domain.
Proof: We prove the theorem by induction on p(G), starting with the trivial case p ( G ) = 0. Now if p ( G ) = n 2 1, then there exists N a G with p ( N ) 5 n - 1 and G / N locally polycyclic-by-finite. Thus, by Lemma 1.3, we have
R*G = ( R * N ) * ( G / N )= S * ( G / N ) where, by induction, S = R*N is an Ore domain. Furthermore, if H / N is a finite subgroup of G I N , then S * ( H / N ) = R*H and there are two cases to consider. If N # (1) then, by the previous lemma, p ( H ) = p ( N ) 5 n - 1 so S * ( H / N )= R*H is a domain by induction.
37. Zero Divisors and Goldie Rank
399
On the other hand, if N = (1) then H is a finite subgroup of G so S * ( H / N ) = R*H is a domain by assumption. We can now apply Lemma 37.8 to conclude that S * ( G / N ) = R*G is an Ore domain and the theorem is proved. The corollary is of course an immediate consequence. Paper [94] obtains a transfinite version of the above by allowing for more general types of subnormal series (see Exercises 7 and 8 for details). Now we move on to consider the Goldie rank problem. Specifically, let R*G be given with R right Noetherian and G polycyclic-by-finite and let us assume that R*G is prime. By Goldie’s theorem, the classical right ring of quotients Q(R*G) is isomorphic to M,(D), a full matrix ring over a division ring D. The integer n here is called the Goldie rank of R*G and the problem is to determine this rank. In the case of ordinary group algebras K [ G ] ,it was conjectured in [52] and [188] that n is the least common multiple of the orders of the finite subgroups of G. Special cases of this were verified in [103,170, 1891 before the conjecture was finally settled in the affirmative in [140]using, as usual, Theorem 36.9. More generally, the Goldie rank of R*G is the least common multiple of certain parameters determined by the finite subgroups of G and, for this result, we follow [1051. Let R be a semiprime Noetherian ring so that Q(R) exists and is semisimple Artinian. If A is a finitely generated R-module, then A @ R Q(R) is a finitely generated Q(R)-module and hence has finite composition length. The reduced rank of A is then given by
PR(A)= composition length of A @ R Q(R). We remark that the reduced rank can be defined for all Noetherian rings, but we only require the semiprime case here. Since RQ(R) is flat, by Lemma 33.3(iii), it follows that p~ respects short exact sequences and hence gives rise to a group homomorphism p ~Go : ( R )-+ 2. More generally, suppose S is an overring of R with SR finitely generated. Then any finitely generated S-module restricts to a finitely generated R-module and again this yields a group homomorphism p ~Go(S) : -+ 2. Finally the normalized reduced rank X R is defined
400
9. Zero Divisors and Idempotents
bY X R ( A ) = PR(A)/PR(R) so that X R ( A )is a rational number and X R ( R = ) 1. Again X R gives rise to a group homomorphism XR:G o ( S ) &. Now suppose R*G is a prime crossed product with R right Noetherian and G polycyclic-by-finite. If NaG, then we have R*G = ( R * N ) * ( G / N )and certainly R*N is also right Noetherian. Thus Lemmas 14.l(i) and 14.2(i) imply that R*N is G-prime and hence semiprime. In particular, it makes sense to consider X R * N . Note further that if \GIN] < 00, then R*G is a finitely generated right R*N-module. In the following lemma, we determine the normalized reduced rank of certain induced R*G-modules. --f
Lemma 37.12. Let R*G be a prime crossed product with R right Noetherian and G polycyclic-by-finite. Suppose N is a normal s u b group of finite index in G and that H C G is finite. i. If A E modR*G, then JG: NI XR*G(A)= X R * N ( A ) . ii. If B E mod& then X R * N ( B @R R * N ) = X R ( B ) . iii. IfC E m o d R * H , then XR*G(C@R*H R*G) = l H ( - l x ~ ( C ) .
Proof. (i) Since R*G = (R*N)*(G/N), it clearly suffices, in this part only, to assume that N = (1) and that G is finite. Moreover, since XR*G(A)and X R ( A )are obtained by localizing the rings involved, there is no harm in first doing a partial localization by a regular Ore set T provided T R. In other words, by Lemma 37.7, we may now assume that R = Q(R)is Artinian. But then, since G is finite, R*G is also Artinian; thus R*G = Q(R*G) and no further localization is required. Now R*G is prime, so it is a simple Artinian ring with a unique irreducible module M . Indeed, if A E modR*G, then A E n - M , the direct sum of n = ~ R * G ( A copies ) of M . Furthermore, A I R n.MlR so p R ( A ) = n’ p R ( M ) = PR*G(A) p R ( M ) In particular, this holds for A = R*G which, as an R-module, is free on /GI generators. Thus we have IGl p R ( R ) = pR(R*G) = PR*G(R*G)’ p R ( M )
37. Zero Divisors and Goldie Rank
401
and dividing the first equation by the second yields
as required. (ii) Again, there is no harm in localizing R since the equation
shows that this localization commutes with the induced module map. Thus we can assume that R is semisimple Artinian. Moreover, we note that the group 6 of trivial units of R*G acts as automorphisms on R, R*N and Q(R*N) and hence it permutes the modules of each ring. Indeed, if B E modR and g E 6, then we have
via the map bg 8 s g H ( b 8 s)g for all b E B and s E Q ( R * N ) . Since conjugate modules have the same composition length, we conclude that P R * N ( B g @R R*N) = P R r N ( B @R R*N). Now R*G is prime by assumption, so R is G-prime and hence Gsimple. This implies that all irreducible R-modules are G-conjugate and hence, when induced to R*N, yield the same reduced rank. Suppose M is a fixed irreducible R-module. If B E modR, then B is a direct sum of ~ R ( Birreducible ) R-modules and hence, by the above observation,
Furthermore, if B = R, then B 8~ R*N = R*N so
Finally, dividing the first of these equations by the second yields
402
9. Zero Divisors and Idempotents
and part (ii) is proved. (iii) Here we use the induced and restricted module notation of Section 3. Let N be a poly-Z normal subgroup of G of finite index and let C E m o d R * H . Then by (i) above
and we apply the Mackey decomposition to the latter module. Notice that N is torsion free so N n H = (1) and, since N a G, the ( H ,N ) double cosets in G are merely the right cosets of the subgroup H N . In particular, if D is a right transversal for HN in G, then Lemma 3.10 yields
d€D
since N n H d = (1) and C @ d 2 Cd. Thus since all C7R have the same reduced rank, we conclude from (ii) above that
But
ID1 = IG : HNI
= JG: NI/IHI, so the result follows.
We need one last bit of notation. Let R*H be a crossed product with R semiprime Noetherian and H finite. Since p ~Go(R*H) : t2 is a group homomorphism, we have Im(pR) = mZ for some positive integer m = m(R*H).Thus, m divides ~ R ( R * H=) p ~ ( R ) a[HI and we define the index of R * H to be the integer ind(R*H) = PR(R*H ) / m ( R* H).
Lemma 37.13. Let R*H be as above. i. ind(R*H) = ind(Q(R)*H). ii. If R is Artinian, then m(R*H) is the greatest common divisor of the ranks ~ R ( Aof) all irreducible R*H-modules A .
403
37. Zero Divisors and Goldie Rank
iii. R*H is a domain if and only if ind(R*H) = 1. iv. If R*H = R[H], then ind(R*H) = ~ R ( RIHI. ).
Proof: Part (i) follows since
p~ factors through the epimorphism Go(R*H) --f Go(Q(R)*H). In (ii), since R and R*H are both Artinian, Go(R*H) is generated by the elements [A]for all irreducible R*H-modules A. Thus 7n.Z
= Im(pR) =
c
PR(4
A
and it follows that m is the greatest common divisor of the various ~ R ( A )Part . (iii) is now immediate from the above and (iv) follows since R [ H ]has a module A with ~ R ( A=) 1. I We remark that there are many opportunities for ind(R*H) to equal ~ R ( R(HI. ) . For example, by Lemma 26.2(i), we can take R*H to be a skew group ring with R a domain. Alternately, R*H can be a twisted group algebra K t [ H ]with the degrees of its irreducible modules relatively prime. We now prove
Theorem 37.14. [lo51 Let R*G be a prime crossed product with R right Noetherian and G polycyclic-by-finite. If GI, GI,. . . ,GI,are representatives of the conjugacy classes of the maximal finite subgroups of G, then the Goldie rank of R*G is the least common multiple of the indices ind(R*Gi).
Proof: We compute the image of Go(R*G) under the map XR*G. To start with, if R*G has Goldie rank n, then Q(R*G) = M,(D) and it follows easily that Im(XR*G) = n-lZ. On the other hand, by Theorem 36.9, Go(R*G) = W iwhere Wi is the image of Go(R*Gi) under the induced module map. In particular, if ind(R*Gi) = w i , then it follows from Lemma 37.12(iii) that
xi
XR*G (Wi) =
IGil-lXR(GO (R*Gi)) = 1Gil-l. rn(R*Gi)/p~(R) * 2 = wC'Z.
xi
In other words, n-lZ = wT1Z. Now wtI1 E n-'Z implies that wiJnand thus we have wln where w is the least common multiple of
404
9. Zero Divisors and Idempoteats
the wi. On the other hand, since w / n E C i ( w / w i ) Z that n1w and the result follows. I
C 2, we
see
As an immediate consequence of this and Lemma 37.13(iv) we have
Corollary 37.15. [140] Let R[G]be a prime group ring with R right Noetherian and G polycyclic-by-finite. If GI, Gar.. . ,Gk: are representatives of the conjugacy classes of the maximal finite subgroups of G, then the Goldie rank of R[G]is equal to ~ R ( Rtimes ) the least common multiple of the orders of the subgroups Gi. Finally [94] extends aspects of this to more general groups. For example we quote the following result without proof.
Theorem 37.16. [94] Let G be a group having a finite subnormal series (1) = GOa GI a . . . a G, = G
with each quotient Gi+l /Gi locally polycyclic-by-finite. Then the group ring R[G]has a right Artinian quotient ring if and only if R has a right Artinian quotient ring and the finite subgroups of G have bounded order.
EXERCISES 1. Verify the group theoretic relations in the proof of Proposition 37.1. Write a computer program to show that all multiplicities in S2 are larger than 1. 2. Show that CZ* C, is isomorphic to the infinite dihedral group D. Prove that the group G of Proposition 37.1 can be embedded in the direct product D x D x D. Conclude that G is supersolvable. 3. Let R be a right Noetherian ring with finite global dimension. If Go(R) is the cyclic group generated by [R],prove that all finitely generated projective R-modules are stably free. For this, use the isomorphism of Ko(R) with G o ( R ) given by Proposition 33.8.
405
38. The Zalesskii-Neroslavskii Example
4. If G is a torsion free polycyclic-by-finite group, prove that all finitely generated projective modules for the group algebra K[G] are stably free. This is a result of [140]. 5. Let G be a group generated by X I , 22,.. . ,Q and let H be a subgroup of finite index with right transversal { 91, y2,. ..,yt }. For each i , j there exists hi,j f H with y i z j = hi,jyil and we let HO be the subgroup of H generated by all hi,j. If W = Hoyi, prove that W G = W and hence that W = G. Since HO C H , conclude that H = HO is finitely generated. 6. Suppose ( a / b ) Z = C i ( u i / b i ) Z with g.c.d.(u,b) = 1 and g.c.d.(ai,bi) = 1. Show that a = g.c.d.{ai} and b = l.c.m.{bi}.
ub,
Let A be a well ordered set. A subnormal series for G of type A is a collection { Gx 1 X E A } of subgroups such that i. if X is not a limit ordinal, then G x - 1 a Gx, ii. if X is a limit ordinal, then Gx = Ua<XGo, iii. Go = (1) and G = Ux,,, Gx. The quotients in the series are the groups Gx/Gx-1.
7. Let G be a group having a subnormal series as above with all quotients locally polycyclic-by-finite. Define a transfinite subnormal length p(G) and note that p(G) cannot be a limit ordinal if G is finitely generated. Show that the analog of Lemma 37.9 holds for finitely generated groups. 8. State an appropriate analog of Theorem 37.10 and prove it by transfinite induction. This is a result of [94].
38. The Zalesskii-Neroslavskii Example A question posed in [50] has given rise to some extremely interesting mathematics. Specifically, it was asked whether there exists a simple Noetherian ring having zero divisors but no idempotents. The affirmative answer given in [208] is a particular twisted group algebra Kt[G]of an abelian-by-finite group G over a field K of characteristic 2. This ring is now known as the Zulesskii-Nerosluvskii example. Most of its basic properties are quite easy to obtain; the
9. Zero Divisors and Idempotents
406
main difficulty is the absence of idempotents and the original proof of this fact was very computational. Somewhat later, it was shown in [196] that this same ring is not Morita equivalent to a domain. The recent paper [lo61 now contains a slick noncomputational proof of both of these results by studying the Grothendieck group of the ring. Moreover, [104]offers analogous examples in all characteristics p > 0. We follow the approach of [l06]. To start with, we study skew group rings S = RG which are algebras over a field K . By this we mean precisely that K Z(R) and that G acts trivially on K . In particular, S 2 K G = K[G], the ordinary group algebra of G over K . This structure allows us to copy a basic ingredient from the representation theory of finite groups, namely the tensor product module with diagonal action. Since S is a skew group ring, there is as usual no need to use overbars on the group elements. Suppose A is a right S-module and B is a right K[G]-module. Then A 8~ B can be made into an S-module by defining ( a 8 b )( Z T ) = a z 8 ~ bx
for all a E A,b E B,r E R and z E G. Indeed, it is clear that the above formula defines, for each zr E S , a K-endomorphism B(zr):A 8 B -+ A 8 B and hence we obtain an additive map 19:S -+ EndK(A 8 B ) . To see that I9 is multiplicative, we need only observe that B(Zr)O(yt) = O ( z r . y t ) = B(zyrYt) and this is trivial to check. Thus we have confirmed that A @K B is an S-module. To avoid confusion with the usual tensor product, we write this module as A 0 B = A O K B. In other words, A 0B has the K-vector space structure of A 8~ B , but its S-module structure is given by the diagonal action ( a 0b ) ( x r )= UZT 0bz. Furthermore, it is easy to see that if a : A -+ A’ is an S-module B’ is a K[G]-module homomorphism, homomorphism and p : B then a@P:A@B-+A’OB’ --$
is an S-module homomorphism. This is the content of Lemma 38.l(i) below.
38. The Zalesskii-Neroslavskii Example
407
If G is a finite group then, by Lemma 26.2(i), R is a right Smodule. Indeed this is true even if G is infinite; we merely define u . xr = uxr for all u,r E R and z E G. Then as above, this gives rise to an additive map 4:s+ End(R) and one checks easily that +(zr)+(yt)= +(zr y t ) = q5(zyrYt). It is this S-module structure for R which is used in part (iii) below. Part (ii) is an aspect of Frobenius reciprocity.
Lemma 38.1. Let S = RG be a skew group ring which is an algebra over the field K . i. I f A is an S-module and B is a K[G]-module,then A O K B is an S-module which is functorial in each factor. ii. I f A is an S-module, then A I R@R S E A OK K[G]. iii. R OK K[G] S S.
-
ProoJ Part (i) has already been discussed and for (ii) we need only observe that the map A I R@R S A OK K[G] given by a 18 TIC H arx 0 z is a n S-isomorphism with back map a 0 x H ax-' @ z. Finally for (iii), let A = R and observe that A I R= RR is the regular R-module. Thus by (ii) we have R O K K[G] 2 R p @R S 2 S and the lemma is proved. I If S is any ring and A is an abelian group, then an additive map 7 : S -+ A is called a truce if q(st) = ~ ( t sfor ) all s , t E S. It is clear that, for any such 7 , we have Ker(7) 2 [ S , S ] the , linear span of all Lie commutators [s, t] = st - t s . Thus 7 factors through the additive homomorphism 7: S -+ S / [ S ,S] which is itself a trace. The following key result uses a few simple properties of group algebras. First, K[G] has a one-dimensional irreducible module V1 called the principal modde. Specifically, each g E G acts like the identity on Vl, so V, 2 K [ G ] / I where I is the augmentation ideal of the algebra. Moreover, if G is a finite pgroup and char K = p > 0, then I is nilpotent by [161, Lemma 3.1.61 and hence V, is the unique irreducible K[G]-module. Thus, in this case, K [ G ] Khas ~ ~a composition series of length [GI with all composition factors isomorphic to V1.
Theorem 38.2. 11041 [lo61Let S = RG be a skew group ring which is an algebra over the field K. Assume that
408
9. Zero Divisors and Idempotents
i. char K = p > 0 and G is a nonidentity finite p-group, ii. R is right Noetherian and all finitely generated projective R-modules are stably free, iii. 1 4 [S,S ] . If A: Go ( S ) -+ Z is any group homomorphism, then A( [PI)E p Z for all finitely generated projective S-modules P.
Proof. Notice that S is a finitely generated free right R-module and that R is right Noetherian. Thus, since G is finite, S is also right Noetherian and it makes sense to consider Go(S). Also note that if A is any S-module then, by (ii) and the discussion above, the Smodule A 0 K[G]has a series of length IGl with factors isomorphic to A 0 V1 Z A. Hence if A is finitely generated, then in Go(S) we have [ A 0K [ G ] ]= /GI - [ A ] . On the other hand, if A is projective then the series splits so A 0K[G]3 AIGl. Now let P be a finitely generated projective S-module. Then PIRis a finitely generated projective R-module and hence it is stably free by assumption; say Pp @ Rm Z Rn with m,n 2 0. If R also denotes the S-module of Lemma 38.l(iii) then, by parts (ii) and (iii) of that lemma, we have
( P 0K [ G ] @ ) S"
3
( P 0K [ G ] @ ) ( R0K[G])" ( q R@
2 Rn @R
(RIR)") S 2 S" .
@R
s
Moreover, since P is projective, we have P 0 K[G]2 PIGl. Thus PIGl @ S" "= S" and, by adding additional S summands if necessary, we can assume that p I m. The first goal is to show that p I n. Since G is a nonidentity pgroup, p \GI and S" 2 PIG\ S" XP for some S-module X . In particular,
I
S' = M,(S)
Z
Ends(Sn) Z Ends(Xp) E M,(T) = T'
where T = Ends(X). If { ei,j } denotes the set of matrix units of MP(T), then we have 1 = C;='=,[ei,l, e l , i ] since charK = p. In other words, 1 E [T',T']and hence, via the above isomorphism, we have 1 E [S',S']. On the other hand, consider the combined map
S' 3 s
0:
L S / [ S ,S ]
409
38. The Zalesskii-Neroslavski Example
where tr denotes the usual matrix trace (that is, the sum of the diagonal entries) and T is the trace map of S discussed above. Then it is easy to see that u is also a trace map so a(1) = 0 since 1 E [S’,S’]. But u(1) = ~ ( n=)neT(1) and ~ ( 1#) 0 by assumption (iii). Thus, since char K = p > 0, we conclude that p 72 as required. Finally, since S E R 0 K[G]we have [S]= IGl . [R]in Go(S) and then PIG]@ S” 2 S“ yields [GI [PI = ( n- m)lGI * [R].Hence if A: Go(S) 2 is any group homomorphism then, since 2 is torsion free and p I ( n - m ) ,we have A( [PI)= ( n - m)A([R])E p Z and the result follows. I
I
--f
Two rings S and T are said to be Morita equivalent if their categories of modules are equivalent. The key theorem of [143] gives necessary and sufficient conditions for this to occur and in fact describes the equivalence. We will not need that result; we only require a few simple observations which follow directly from the existence of the category equivalence. These are all contained in the following lemma which is an immediate consequence of [6, Propositions 21.4, 21.6(2), 21.8(2) and Corollary 21.91. Here, we let Kg(S) G o ( S ) denote the image of Ko(S) under the Cartan map c.
Lemma 38.3. Let S and T be Morita equivalent rings with S right Noetherian. Then T is right Noetherian and there exists an isomorphism 8: Go(S)-+ Go(T) which sends K;(S) to Kg(T). With this, we can quickly prove
Corollary 38.4. [lo41 [lo61 Let S = RG satisfy the hypothesis of Theorem 38.2. Then S is not Morita equivalent to a domain. Furthermore, if the reduced rank of S satisfies ps(S) 5 p , then ps(S) = p and S contains no nontrivial idernpotents.
Pro05 Suppose T is a domain Morita equivalent t o S. Then T is right Noetherian and the regular module TT has reduced rank p ~ ( 2 ’ )= 1. In other words, there is a homomorphism p ~ : G o ( 7 ‘ ) 2 and a finitely generated projective T-module P with P T ( [ P ]=) 1. On the other hand, by Theorem 38.2, if A:Go(S) --t 2 is any map, then --f
410
9. Zero Divisors and Idempotents
A: Kg(S) p Z . Thus we have a contradiction, by Lemma 38.3, and the first fact is proved. Now suppose ps(S) 5 p . By Theorem 38.2 again, we know that p 1 p s ( P ) for all finitely generated projective S-modules P. In particular, this implies that p s ( S ) = p and that if P # 0 then p s ( P ) 2 p = ps(S). Thus S cannot have a nontrivial idempotent e since otherwise P = eS would be a projective module having smaller rank than S. 4 --f
To use this result, we obviously need some sufficient conditions for (ii) and (iii) of Theorem 38.2 to hold. Part (iii) is easy.
Lemma38.5. IfS = Kt[G]i s a twistedgroupalgebra, then 1 4 [S,S ] .
Proof: Since [S,S] is the K-linear span of the Lie commutators [Z,y] with z, y E G, it suffices to show that 1 $ Supp [Z, y]. In fact, since Supp [z, $1 C { z y , y z }, we need only consider the case y = 2 - l . But then ZGy = k E K’ so 3-l = Ic-ljj and hence Z and jj commute. Thus [%,GI = 0 here and the lemma is proved. 1 The next result predates Theorem 36.9 and was originally proved using the techniques of Section 33.
Theorem 38.6. [67][193] Let R*G be a crossed product with G a poly-Z group and with R a ring of finite global dimension. i. R*G has finite global dimension. ii. If R is right Noetherian, then the induced module map yields an epimorphism KO(R) -+KO(R*G) .
Proof: (i) By induction on the Hirsch number, it suffices to assume that G = (x) is infinite cyclic. Of course, this implies that S = R*(x) = R ( x ) is a skew group ring. Let B be an R-module and let P B 0 be a finite projective resolution for B which exists since R has finite global dimension. Then, since RS is free and hence flat, P @R S -+ B @ R S -+ 0 is a --f
--f
finite projective resolution for B @R 5’. In other words, we now know that any induced S-module has finite projective dimension.
41 1
38. The Zalesskii-Neroslavskii Example
Finally, let A be an arbitrary S-module and observe that there is an epimorphism a: A I R @ R S -+ A given by a 8 s H as for all a E A and s E S . Since R S is free with basis { xi I i E Z }, we have AIR 8 S = @ A 8 xi arid
xi
L = Ker(a) =
{
ai @ xi
a
I
aizz = 0 i
1
Furthermore, if AIR)^ is the conjugate R-module determined by the automorphism x , then the map , k l : ( A l ~ 8~ ) ~ S + L given by a" 8 xi H a 8 xi+' - ax @ xi is easily seen (Exercise 4) to be an S-module isomorphism. Thus
is a short exact sequence of S-modules. But both A I R8 S and ( A I ~8) S~ have finite projective dimension and hence so does A by Lemma 33.9(i). (ii) Here we note that S = R*G is Noetherian and that, by Proposition 33.8 and the above, the maps c:Ko(R) Go(R) and c-l: Go(S) + Ko(S) are isomorphisms. Furthermore, by Theorem 36.9, the induced module map yields an epimorphism Go(R)+ Go(S). This follows since (1) is the unique finite subgroup of G. Hence the combined map Ko(R) + Ko(S) is also an epimorphism and this is clearly the induced module map restricted to KO. I -+
Now we begin to construct the examples. Let K be a field containing an element X of infinite multiplicative order and let
be a free abelian group on the 2n generators xi and yi. Then we let R, = Kt[A,] be the twisted group algebra with Zi& = XjjiZi for all i and such that all other pairs of generators commute. Note that A, S A1 x A1 x . . . x Al and that R, S R1 8 R1 8 . . 8 R1. Note also that since A, is an ordered group, all units of R, are trivial. In particular, if G is a group of automorphisms of R,, then G acts on the group A, of trivial units and then on A, A,/K'. a
412
9. Zero Divisors and Idempotents
Lemma 38.7. Let A, and R, be as above and let G be a finite group of K-algebra automorphisms of R, such that the inherited action of G on A, is faithful. Then the skew group ring R,G is a simple Noetherian ring and it is a twisted group algebra Kt[A, >Q GI.
Proof. We first study R = h.Set X = ( Q , x ~. ,. . ,z,) and Y = (31,y2,. . . , y n ) . Then A = A, = X x Y and R is the iterated skew group ring R = ( K X ) Y . Note that K X = K[X] is an ordinary group algebra, so it is a commutative domain, and we consider the action of Y on this ring. To start with, KX is Y-simple. Indeed, suppose I is a nonzero Y-stable ideal of R and let a = x k z Z E 1 be a nonzero element of minimal support size. We may assume that 1 E Supp a and then, for all y E Y, we have Supp (a@-a)c Supp a. The minimality of Supp a now implies that a is centralized by Y and hence, since X has infinite multiplicative order, that cy E K'. Thus I = K X as required. Next, we note that Y acts faithfully on K X . Thus since KX is a commutative domain, Lemma 10.3(i) implies that Y is X-outer in its action. It now follows easily, from Corollary 12.6 and the comments preceding Lemma 14.1, that R is a simple ring; it is certainly a Noetherian domain by Proposition 1.6. Since R is a simple ring, we have Q B ( R )= R and hence all Xinner automorphisms are inner. But A is an ordered group, so all units of R are trivial. It follows from this, since A is abelian, that the X-inner automorphisms of R centralize A in the inherited action. In particular, if G is a finite group of K-algebra automorphisms of R acting faithfully on A , then G must be X-outer and hence, by Corollary 12.6 again, the skew group ring RG is simple. Since this ring is also clearly Noetherian and isomorphic to Kt[A>QG], the result follows. I We can now quickly construct examples in characteristic p no examples are known in characteristic 0.
> 0;
Theorem 38.8. (2081 [196]Let R = R1 = Kt[Al] be as above with K a field of characteristic 2. Suppose G = { 1 , a ) is a group of order 2 which acts on R as K-algebra automorphisms by ( 3 1 )= ~ --1 . Then the skew group ring S = RG is a simple 2 1 ,(Yl)' = YT1
413
38. The Zalesskii-Neroslavski Example
Noetherian ring with nonzero nilpotent elements. Furthermore, S has no nontrivial idempotents and it is not Morita equivalent to a domain.
Theorem 38.9. [lo41 Let K be a field of characteristic p > 0 and let R = Rp = Kt[A,] be a above. Suppose G is a group of order p which acts on R R1 @ R1 @ - . 63 R1 by cyclically permuting the p factors. Then the skew group ring S = RG is a simple Noetherian ring with nonzero nilpotent elements. Furthermore, S has no nontrivial idempotents and it is not Morita equivalent to a domain.
Proof. It is clear that G acts as K-algebra automorphisms in the second set of examples. For the first, since
31jjl = Xjj131,
we have
and again G AutK(R). In either case, since G acts faithfully on A = A1 or A = A,, Lemma 38.7 implies that S = RG is a simple Noetherian ring. Furthermore, since R is a domain and IGI = p , the reduced ranks satisfy ps(S) 5 ~ R ( S=)p . Now note that S e!K t [ A>Q GI so 1 4 [S,S ] by Lemma 38.5. Furthermore, since R = K t [ A ] ,Theorem 38.6 implies that the induced module map yields an epimorphism Ko(K) + Ko(R). But Ko(K) is the cyclic group generated by [ K ] so , Ko(R) is the cyclic group generated by [K @R R] = [R] and hence Lemma 34.10(ii) implies that all finitely generated projective R-modules are stably free. The results now follow from Corollary 38.4 and the fact that the augmentation ideal of K [ G ]C S is nilpotent. The first ring S = RG above is the Zalesskii-Neroslavskii example. It of course varies somewhat depending on the choice of the field K and the element X E K'. In the next sections, we will explicitly compute Go(S) along with other examples. To do this, we will need to know that the fked ring RG is Noetherian. We prove a more general result below.
414
9. Zero Divisors and Idempotents
Suppose A = Zkis the free additive abelian group on k generators. Then A can be ordered Zezicographically by defining
if and only if a1 = b l , a2 = bz, . . ., ai-1 = bi-1 and ai < bi for some subscript 1 5 i 5 k. In this way, A becomes an ordered' group. Alternately, we have the product ordering,where
if and only if ai 5 bi for all i . Note that this is only a partial order, but it again respects the group addition. Let N = { 0 , 1 , 2 , .. . } denote the set of nonnegative integers. Then N k & Zk= A and we have
Lemma 38.10. Any sequence { X I ,22,.. . } ofelements in N k contains , yi+l for all i . In particular, a subsequence { y1,y2,.. . } with yi 1 if M is a nonempty subset of N k ,then M has only finitely many minimal elements and each element of M contains at least one of these.
Proof. The natural ordering on N has the property that every sequence has a nondecreasing subsequence. Therefore, the same holds for Nk,by projecting the given sequence into each of the k summands and, at each step, taking an appropriate subsequence. It follows that N k cannot contain (i) infinitely many incomparable elements under
Proposition 38.11. [77]Suppose R = D*A is a crossed product with A a torsion free abelian group of rank k < 00 and with D a division ring. Let the group G = { l , o } act on R in such a way that D is centralized and the inherited action of (I on A is inversion, that is a" = a-l for all a E A. Then R is right and left Noetherian as a module over the fixed ring RG.
38. The Zalesskii-Neroslavskii Example
415
Proox We need mme definitions. First, let us think of A = Zk as being additive and define a linear ordering on A as follows. For each a = (ul,u2,. .. ,a k ) E A , let llall = \ail2 0. Then we say that a 5 b if and only if either (i) [ [ a ( < ( llbll or (ii) ( ( a (= ( llbll and a Se b where <e denotes the lexicographical ordering of A . Note that 5
xi
does not respect addition, but it is a well ordering. It will be necessary to keep track of the signs of the entries in a = ( a l ,a 2 , . . . ,ak). For this, we let E = ( E I , ~.,. . , ~ k )denote a multi-sign, so that each ~i = f,and we set
It is clear that the 2k sets A,, obtained in this manner, partition A . For convenience, we let { e(l), e(2), . . . ,e(k) } denote the canonical Z-basis of A . In other words, e ( i ) has a 1 in its ith entry and 0’s elsewhere.
Claim 1. Let M be a finite nonempty subset of A and let m = max M be the maximal element of M under the ordering 5 . If m E A, with E = ( E ~ , E Z , . . . , ~ k ) ,then m
+ Eie(i) = m a x ( M
e(i)).
+
Furthermore, if rn + Eie(i) = a de(i) for some a E M and 6 = f, then we have a = m and 6 = t i .
M , 6 = f and m + E i e ( i ) 5 a + 6e(i). The goal is to show that a = m and 6 = ~ i .To start with, we have
PrwJ Suppose that a
E
+
+
since a 5 m and m E A,. But, by assumption, a 6e(i) m t i e ( i ) so \la + 6e(i)II >_ Ilm Eie(i) 11 and we must have equality throughout. In particular, /la Se(i)II = llall 1 = Ilm cie(i)[(and therefore a 6e(i) >e m Eie(i). Furthermore, llall = llmll and hence a l e m since a 5 m. Now, if either ~i = or ~i = - and 6 = -, then a Se m clearly yields a 6 e ( i ) <e rn t i e ( i ) . Thus these two elements are
+
+
+
+
+
+
+
+
+
9. Zero Divisors and Idempotents
416
equal and it follows easily that a = m and 6 = ~ i . All that remains now is the case E = -, 6 = and we show that this cannot occur by comparing components. If aj < mj for some j < i, then certainly a + be(i) <e m + cie(i), a contradiction. Thus since a <e m, we must have aj = mj for all j < i and hence ai 5 mi. But then ai 5 mi < 0, since ei = -, and this contradicts the fact that 6 = and IIa Se(i)ll = ((all 1.
+
+
+
+
We continue with the proof of the proposition and introduce additional notation. If 0 # r E R,we let deg r = max (Supp r ) E A using the ordering 5 . In addition, if I is a nonzero RG-submodule of R, we let d e g I equal the set of degrees of the nonzero elements of I . In the following, N = { 0,1,2,. . . } will again denote the set of nonnegative integers.
Claim 2. Let I , J be nonzero RG-subrnodules of R. i. If I c J , then deg I C deg J . ii. I f O # r E I a n d d e g r E A , w i t h ~ = ( ~ 1 ,..., ~ 2 E, k ) , then
d e g I 2 degr
+ ( E ~ N , E Z N. .,,.E # ) .
Proof: (i) Certainly I c J implies d e g I G deg J . Now suppose, by way of contradiction, that deg1 = deg J and choose r E J \ I of minimal degree. Then degr E deg J = d e g I implies that there exists s E I with degs = degr. Furthermore, note that D E RG by assumption, so I is a D-vector space. Thus since D is a division ring, we can multiply s by an appropriate element of D so that the leading coefficients of r and s match. But then r - s E J \ I has smaller degree than that of r and this is a contradiction. (ii) Write Supp r = M and m = maxM E A,. If e(i> denotes the element of R corresponding to e ( i ) ,then we have
and Supp ti = { e ( i ) ,-e(i) } since, by assumption, the inherited action of 0 on A is inversion. Furthermore, rti E I - RG = I and Supp rti
2 (Supp r ) fe ( i ) = M f e(i).
417
38. The Zalesskii-Neroslavskii Example
+
Now, by Claim 1, m cie(i) = max (A4f e ( i ) ) and indeed) for all a E M and 6 = f,we have a se(i) = m eie(i) if and only if a = m and S = ~ i . Thus since D is a division ring, it follows that m eie(i) E Supp rti and then that m eie(i) = degrti. In other words, degr Eie(i) = degrti E degI. Furthermore, since m+Eie(i) E A, and rti E I , we can clearly continue this process and, in so doing, we obtain
+
deg r
+
+
+
+
+ 1niEie(i) = degr(tl)nl(tz)”’ - -
(tk)”’” E
deg I
2
as required. I It is now a simple matter to complete the proof of the proposition. Suppose by way of contradiction that 0 C I1 C 1 2 C ... is an infinite strictly increasing chain of RG-submodules of R. Then, by Claim 2(i), degI1 C deg12 c is strictly increasing, so we can choose elements a ( p ) E deg I p \ degIp-l. By considering a suitable subchain if necessary) we can assume that all a ( p ) belong to the same A, with E = ( E ~ , E Z , . . . , ~ k and ) then, for any a = ( a ~ , a z ,. .. , a k ) E A,, we write EU
= ( c l a l ,~ 2 a 2 ,.. . ) € k a l e ) E ( N ,N , . . . , N ) .
+
Finally, suppose p < q . Then a ( p ) ( E ~ N E z, N , . . E ~ NC_) degIp, by Claim 2(ii), so a(q) 4 a ( p ) + ( c l N , e 2 N , ..., e k N ) and hence ~ a ( q6) ea(p) ( N ,N , . . . ,N ) . In other words, in the product ordering on N k we have ea(p) g7Fca(q). But then the infinite sequence { ~ a ( l )4, 2 ) ) . . . } C N k cannot have a nondecreasing subsequence in the product order and this contradicts the previous lemma. )
+
EXERCISES 1. Let I = w ( K [ G ] )be the augmentation ideal of the group algebra K [ G ]of the nonidentity group G. Show that I is nilpotent if and only if charK = p > 0 and G is a finite p-group. For this, first observe that r ~ p l ( 1 # ) 0 if and only if G is finite.
418
9. Zero Divisors and Idempotents
2. Let S = RG be as in Lemma 38.1. Suppose A: G -+ K' is a linear character of G and let Vx be its corresponding l-dimensional K[G]-module. Prove that R OK Vx % R if and only if there exists a unit u of R with 219 = A(g)u for all g E G. To this end, note that any such S-isomorphism is an R-isomorphism and hence is determined by a suitable left multiplication. 3. Suppose S is a semiprime right Noetherian ring and that P is a nonzero projective (or flat) S-module. Use the inclusion 0 + S + Q ( S ) to show that p s ( P ) # 0. 4. Verify the details of Theorem 38.6. Specifically show that the map /3: AIR)^ @ R S + L is an S-module isomorphism. To see that /3 is onto, proceed by induction on the breadth of the elements of L. 5. Let G act on the crossed product D*H, normalizing the group 7-l of trivial units. If D is a division ring, prove that G normalizes D and hence acts on H E %/Do. 6. Let S = RG be the Zalesskii-Neroslavskii example and observe that the action of G on R is G-Galois (see Exercises 3, 4 and 5 of Section 29). Deduce that RRG is a finitely generated indecomposable projective module over the Noetherian domain RG. Note that indecomposability follows from the fact that EndRc(R) E RG = S has no nontrivial idempotents. 7. Continue with the above notation and observe, since charK = 2, that 0 + RG --+ R 3 T -+ 0 is an exact sequence of RG-modules with T = trG(R) RG. Deduce that ~ R G ( R = ) 2. This and the results of the preceding exercise are from 1771.
39. Almost Injective Modules In this section we develop some techniques for explicitly computing the Grothendieck group of the Zalesskii-Neroslavskii example and certain similar twisted group algebras. Since Theorem 36.9 already yields an upper bound for Go, the goal here is to obtain an appropriate lower bound. We do this by studying the restrictions of the modules of the larger ring to group algebras of finite subgroups. Since these restricted modules are most likely infinitely generated,
39. Almost Injective Modules
419
it is then necessary to isolate within them a certain finite parameter. A number of the preliminary lemmas below, and in particular Proposition 39.3, hold in the more general context of quasi-Frobenius rings, that is rings whose projective and injective modules coincide. However, we have opted to consider only group algebras here, since this allows us to proceed in a fairly self-contained manner. Let K[G] denote the group algebra of a finite group G. If A and B are K[G]-modules then they determine, as in the previous section, a tensor module A OK B with diagonal action. Specifically (a 0b)g = a g 0bg for all a E A, b E B and g E G. Recall that a ring R is said to be seZf-injective if the regular module RR is injective.
Lemma 39.1. Let K[G] be a group algebra with G finite and let A be a right K[G]-module. i. K[G] is self-injective. ii. A OK K[G] is a free K[G]-module on dimK A generators. iii. The map a:A 3 A O K K[G] given by &(a)= CgEG a O g, for all a E A , is a K[G]-isomorphism into. In particular, if A is irreducible, then A is isomorphic to a submodule of K[G].
Proof: (i) Let f:K[G]
-+ K
denote the map which reads off the identity coefficient. Then f is certainly a linear functional and, for all y E K[G], we have y = zgEGf(yg-')g. Now let A B be K[G]-modules and let a : A -+ K[G] be a K[G]-homomorphism. Then fa: A -+ K is a linear functional which can certainly be extended to ( f a ) * : B+ K . We now define u*:B -+ K[G] by a*@)= C S E G ( f ~ ) * ( b g - l ) g .It is easy to verify that a* is a K[G]-module homomorphism which extends u. (ii) If { ai 1 i E I }is a K-basis for A , then so is { aig 1 i E I } for any fixed g E G. Furthermore, since A O K K[G] = @ CgEG A 0g, it then follows that { aig 0g 1 g E G, i E I } is a K-basis for A 0K[G] and hence that { ai 0 1 I i E I } is a K[G]-basis for the module. (iii) It is clear that (Y is a one-to-one K[G]-homomorphism. Thus A 2 a ( A ) C A@KK[G]and the latter is isomorphic to a direct sum of copies of K[G]. In particular, if A is irreducible, then the projection of a ( A ) into one of the direct summands must be nontrivial and hence an embedding. I
9. Zero Divisors and Idempotents
420
In the following lemma we let E(A) = ER(A) denote the injective hull of the R-module A and we use the well known fact (see Exercise 2) that, in the case of Noetherian rings, a direct sum of injectives is injective. For convenience, we say that a module is injective-free if it contains no nonzero injective submodule.
Lemma 39.2. Let K[G] be the group algebra of the finite group G and let V be a K[G]-module. i. If A is an irreducible K[G]-module, then E(A) is isomorphic to a direct summand of K[G]. ii. If the socle of V is written as. SOC V 2 @ '&Ai with each Ai irreducible, then E(V) S 43 E(Ai). iii. V is projective if and only if it is injective. iv. V % Vl @ V2 where V2 is injective and V1 is injective-free.
Proof. (i) By the previous lemma, A is isomorphic to a submodule of K[G] and K[G] is injective. Thus E(A) is isomorphic to a submodule of K[G] and hence to a direct summand. (ii) Since K[G] is Artinian, SOCVis essential in V and hence E(V) = E(SOC V) €D E(Ai). (iii) Since K[G]K[GIis injective, it follows that any projective K[G]-module is also injective. Conversely, if V is injective and is described as in (ii) above, then V = E(V) 2 @ E(Ai). But each E(Ai) is projective by (i), so we conclude that V itself is projective. (iv) By Zorn's lemma, we can define V2 to be a maximal direct sum of injective submodules of V. Then V2 is injective, so V = V1$V2 for some submodule V1 and it is clear that VI is injective-free. I
xi
xi
We say that a K[G]-module V is almost injective if E(V)/V is finitely generated. For such a.2. modules V, [E(V)/V] is a well defined element of Go(K[G]) and we let O(V) denote the image of -[E(V)/V] in Go(K[G]) = Go(K[G])/KE(K[G]). Note that, if V is finitely generated, then so is E(V) by Lemma 39.2(i)(ii) and hence O(V) is the image of [V] - [E(V)] in Go(K[G]). But E(V) is projective, so [E(V)] E KC,(K[G])and O(V) is in fact the image of [V].We will not need all of the following proposition; it is included however for its potential usefulness.
42 1
39. Almost Injective Modules
Proposition 39.3. [1151 Let K[G] be a group algebra of the finite group G. i. Let V be a K[G]-module and write V = V1 @ V2 where V2 is injective and V1 is injective-free. Then V is a.i. if and only if V1 is finitely generated and, in this case, O(V) = [Vl] Kg(K[G]). ii. Let 0 -+ A -+ B C -+ 0 be a short exact sequence of K[G]-modules. If any two of these modules are a.i., then so is the third and we have O(B) = O(A) O(C).
+
-+
+
Proof: (i) Since Vz is projective and E(V) = E(Vl)
@ V2, it clearly suffices to assume that V2 = 0. In other words, we can suppose that V is injective-free. If V is finitely generated then, as we observed above, E(V) is finitely generated and V is certainly a.i. Conversely, suppose V is a.i. In the notation of Lemma 39.2(ii), we have V G E(V) = @ CiEr E(Ai) and we let 7ri denote the projection into the ithsummand. Then 7ri(V)# E(A,) since E(Ai) is projective and V, being injective-free, does not contain an isomorphic copy of E(Ai). Since the finitely generated module E(V)/V has as a homomorphic image the direct sum @ CiErE(Ai)/xi(V), we conclude that I is finite. It follows that E(V) is finitely generated and hence so is V. (ii) We may suppose that A E B and C = B / A . Let A2 be an injective submodule of A . Then A2 A B yields B = B’ @ A2 and A = A’ @ A2 where A’ = B’ n A. Since C E B’/A‘ and the module A2 is both injective and projective, it follows easily that we need only consider 0 -+ A’ -+ B’ -+ C -+ 0. In other words, we may assume that A2 = 0 and hence, by Lemma 39.2(iv), that A is injective-free, Next write C = C1 @ C2 as in Lemma ‘39.2(iv) and let B1 be the complete inverse image of C1 in B. Since C2 is projective, there exists a submodule B2 of B which maps isomorphically to C Z . It follows that B = B1 @B2and that A B1. Since B2 C2 are both injective and projective, it follows easily that we need only consider 0 -+ A -+ B1 -, C1 0. In other words, we may assume that C2 = 0 and hence that C is injective-free. Now if A and C are a.i. then, by (i) above, they are both finitely generated. Hence B is also finitely generated and a.i. Furthermore, [B]= [A] + [C]in Go(K[G]) and therefore O(B) = O(A) O(C) in
c c
---f
+
422
9. Zero Divisors and Idempotents
GJ(K[G]). Finally suppose B is a.i. so that E ( B ) / B is finitely generated. Now E(B) 2 E(A) yields E(B) = E(A) CB D for some injective submodule D and hence
where A = E(A)/A. Furthermore, ( A@ D ) / C E E(B)/B is finitely generated. In particular, if C is a.i., then C is finitely generated and hence so is A. On the other hand, if A is a.i., then A is finitely generated and hence so is E(A). But then (E(A) @ D ) / C is finitely generated and, since E(C) E(A) CBD ,we conclude that E(C)/C is finitely generated. This completes the proof. I
If A and B are K[G]-modules, then HomK(A, B ) is also a K[G]module with a conjugate action. Specifically, if f E HomK(A,B) and g E G, then f9:A -, B is defined by f g ( a ) = f(ug-l)g for all a E A. It is clear that f 9 E HomK(A,B) and that ( f 9 ) h = f g h for all g, h E G. Thus C = HomK (A, B ) is indeed a K[G]-module. Note that fg(a)g-' = f(ug-') and thus f g = f for all g E G if and only i f f E HornK[q(A,B). In other words, HomK[G](A,B)= CG = { f E C
I fg
= f for all g E G } .
Of course, if G is finite, then we have the trace map tTG:C given by trG(f) = C g ~ fg* G
+
CG
Lemma 39.4. Let G be a finite group and let V be a K[G]-module. i. Suppose X is an irreducible K[G]-module and T : X V is a K[G]-homomorphism. Then T E trG (HomK(X, V ) )if and only if T extends to a homomorphism T * : E(X) + V. ii. V is a.i. if and only if, for all irreducible K[G]-modules X , --f
HomKIG]
(x,V)/trG (HomK(X, v))
is finite dimensional over K.
Proof: (i) Suppose first that
CgEG fg.
is a trace and say T = trG(f) = Then we can define p:X OK K[G] --t V by P(z 0g) = T
423
39. Almost Injective Modules
f(zg-')g, for all x E X , g E G, and it follows easily that ,f3 is a K[G]-homomorphism. Furthermore, if a : X X OK K[G]is the map of Lemma 39.l(iii), then --f
so pa = trG(f) = r. Now X 0K[G]is free and hence injective, so a extends to a map a*:E ( X ) -+ X 0 K[G].Thus n extends t o a map r*:E(X) -+ V as required. Conversely if r extends to a map E(X) -+ V then, since E(X) is a direct summand of K[G], r extends to a map r*:K[G]-+ V. Let f : K[G]-+ K K[G]be the functional which reads off the identity coefficient. Then n*f maps K [ G ]to V and, for all y E K [ G ] ,we have r*(r) = r* f(rs-')9)
(c c SEG
=
r*f(?g-')
*
= (trG(T*f))(y)*
9EG
Thus r* = trG(r*f) and restricting these functions to X yields the result. (ii) Write V = V1 @ V, with V2 injective and V1 injectivefree. Then HomK(X, V) = HomK(X, V1) @ HomK(X, V2) is a K[G]module direct sum and it follows from (i) that H o m ~ p ] ( XV2) , = trG (HOmK (X,V2)). Thus
and, with this, it clearly suffices to assume that V2 = 0. In other words, we now suppose that V = V1 is injective-free. Furthermore, by (i) again, this implies that trG(HomK(X, V)) = 0. Indeed, if r : X -+ V is a nonzero trace, then r extends to a map r * : E ( X )+ V. But X is irreducible, so 7r is an embedding, and X ess E(X) implies that r* is also an embedding, a contradiction. Thus, by Proposition 39.3(i), we must show that V is finitely generated if and
424
9. Zero Divisors and Idempotents
only if H O ~ K [ G ] ( X V), is finite dimensional over K for all irreducible K[G]-modules X. If V is finitely generated, then dimK V < 00 and H o r n ~ [ ~ l (V) X, is certainly finite dimensional. For the converse note that, since X is (X, ] SOC V). In parirreducible, we have H O ~ K [ G ] V) = H o m ~ p(X, ticular, H O ~ K [ G ] ( X , Vis) finite dimensional over K if and only if socV does not contain an infinite direct sum of copies of X. Thus if H O ~ K [ G (X, ] V ) is finite dimensional for all the finitely many irreducible K[G]-modules X , then socV is a finite direct sum of irreducible K[G]-modules. In other words, soc V is finitely generated E(V) = E(socV), we conclude that V is finitely and, since V generated. 1 To proceed further, it is necessary to reformulate part (ii) above in terms of tensor products. If K = V1 is the principal K[G]-module, then X * = HomK(X,K) is called the contrugredient of X. Some properties are as follows.
Lemma 39.5. Let X be a finite dimensional K[G]-module. i. X** E X . ii. X is irreducible if and only if X* is. iii. If V is any K[G]-module, then HomK(X, V) E V OK X*.
Proof: (i) We know that, as vector spaces, X and X**are canonically isomorphic via z H u(x) where a ( z ) ( X ) = X(z) for all X E X*. Furthermore, it is easy to see that u is a K[G]-homomorphism. (ii) In view of (i), it suffices to show that if X is not irreducible, then neither is X*. For this, note that if Y is a K[G]-homomorphic image of X , then Y * is naturally embedded in X'. (iii) Define 0:V OK X* --t HomK(X, V) so that ~ ( Z0 I X)(z) = 2 1 . X(z) for all ZI E V, X E X * and x E X. It then follows easily that u is a K[G]-module homomorphism. Furthermore, suppose { 2 1 , x2,.. . ,z, } is a K-basis for X and let { x;,xa,. . . ,x; } be its dual basis in X*. Then we define T : HomK(X, V) + V OK X* by ~ ( f =) Cy=lf(z:i)0zz for all f E HomK(X, V). Since T = u-', it follows that u is a K[G]-isomorphism. 1
425
39. Almost Injective Modules
As above, if V is a K[G]-module then we let
V G = {w
E
V I wg = w for allg
E
G}.
In addition, the map trG: V V Gis given by trG(w) = CgEG wg. The following is an immediate consequence of Lemmas 39.4(ii) and 39.5(ii) (iii). -+
Proposition 39.6. [115]Let V be a K[G]-module with G a finite group. Then V is almost injective if and only if, for all irreducible K[G]-modules X , ( V 0q G / t r G ( V 0X ) is finite dimensional over K .
Recall that RG is said to be a K-algebra skew group ring if K E Z(R) and G acts as K-algebra automorphism on R. It then follows that RG 2 K[G].Furthermore, trG(R) is an ideal of the fixed ring RG and RG is a K-subalgebra of R. In the following we assume that RRG is a Noetherian module. Since RG C R, this clearly means that RG is a right Noetherian ring and that R is finitely generated as a right RG-module. If S is any right Noetherian ring then, as in the case of K[G], we let Kb(S) Go(S) denote the image of the Cartan map c and we set G o ( S ) = Go(S)/Kg(S).
Theorem 39.7. [115] Let RG be a K-algebra skew group ring with G finite. Assume that R is a Noetherian right RG-module and that RG/trG(R) is finite dimensional over K. Furthermore, let S 2 RG be an extension of rings with S right Noetherian and finitely generated as a right RG-module. i. If V is a finitely generated S-module, then y K [ G ] is an almost injective K[G]-module. ii. There is a homomorphism 8: Go ( S )-+ Go( K [G]) given by 8: [V] H -[E(V’)/V’]
+ KE(K[G])
426
9. Zero Divisors and Idempotents
iii. If in addition S K [ G ] is projective, then 8 factors through CO(S)so there exists a homomorphism I% C o ( S )+ e o ( K [ G ]given ) bY [V] K",S) H -[E(V')/V'] KE(K[G])
e: +
+
where V' = q K [ G ] .
ProoJ (i) We start with a few simple observations. Let V be any RG-module. Then it is easy to see that VG and trG(V) are both RG-submodules of V. Furthermore, if E VG and r E R, we have
-
Thus VG trG(R) C trG(V) and we conclude that VG/trG(V) is a module for the ring R/trG(R). Now assume in addition that V is finitely generated. Since G is finite and RRG is a Noetherian module, it follows in turn that ViR and are also finitely generated. Moreover, since RG is a Noetherian ring, we conclude that VG E ~ R isG finitely generated and hence, by the above, so is the RG/trG(R)-m0dUk VG/trG(V). But, by assumption, the latter ring is a finite dimensional K-algebra and therefore VG/trG(V) is finite dimensional over K . Furthermore, if X is any irreducible K[G]-module, then V O K X is also finitely generated since, as an R-module, it is a direct sum of dimlc X < 00 copies of V. Thus, by the above, (V@X)G/trG(VOX) is finite dimensional over K for all such X, and we conclude from the previous proposition that bjK[G] is a.i. Finally note that SRG is finitely generated. Thus if W is any finitely generated S-module, then V = WIRG is also finitely generated and the above implies that W [ K [ G ]= q I ( [ G ] is almost injective. (ii) If 0 -+ A + B -+ C + 0 is a short exact sequence of finitely generated S-modules, then 0 ---t A' ---t B' C' -+ 0 is also exact, where ' denotes the restriction to K[G]. But these modules are almost injective, so Proposition 39.3(ii) implies that
qRc
--f
39. Almost Injective Modules
427
O(B’)= O(A’)+ O(C’).It therefore follows from Lemma 33.l(i) that [A] H O(A’)determines a homomorphism from Go(S) to eo(K[G]). (iii) Now suppose in addition that S K [ G ]is projective. If V is any projective S-module, then it follows that V’ = V/K[G] is projective and hence injective. Thus O ( V ) = 0, so [V] H 0 and 13 factors through Go(S). fl
To use the above, we need some information about Go(K[G]). Suppose first that char K = 0. Then K[G] is a semisimple Artinian ring, so every K[G]-module is projective and eo(K[G]) = 0. On the other hand, we have
Lemma 39.8. Let char K = p > 0 and let G be a p-group. i. Go(K[G]) is the infinite cyclic group generated by [Vl], where V1 is the principal K[G]-module. ii. Go(K[G]) S Z / ( ( G I . 2 ) .
Proof: As we observed earlier, V1 is the unique irreducible K[G]module and thus (i) follows from Lemma 33.1(ii). Furthermore, since the augmentation ideal of K[G] is nilpotent, the regular representation of K[G]is indecomposable. Thus Kg(K[G])is the subgroup of Go(K[G]) generated by [K[G]] = (GI. [Vl]and (ii) is proved. fl
As a first application, we consider a simple but interesting example. Suppose that I’ = (z,y I y-lzy = z-’, y2 = 1) is the infinite dihedral group and that S = K [ r ]is a characteristic 2 group algebra. Now r has two conjugacy classes of maximal finite subgroups, and we can take G1 = (y) and G2 = (zy) as representatives of these classes. If Wi denotes the principal K[Gi]-module then, by the previous lemma, Go(K[Gi]) is generated by [Wi]and 2[Wi]= [K[Gi]]. Furthermore, by Theorem 36.9, Go(,!?) is generated by the images of Go(K[Gi]) for i = 1 , 2 under the induced module map. Thus setting ai = [Wi’”]E G o ( S ) ,we see that Go(S) is generated by a1 and (US. But these are not independent. Indeed, since 2[Wi]= [K[G,]] in Go(K[Gi]), we have
2ai = 2[WiIs]= [K[Gi]’”] = [S]
428
9. Zero Divisors and Idempotents
so putting ,f3 = a1 - a2 yields Go(S) = (a1,p) with 2a1 = [S]and 2p = 0. The goal is to show that a1 has infinite order and that p # 0 SO Go(S) 2 @ (2122). To start with, let R = K[(x)] C S so that S is the skew group ring S = RGi for i = 1,2. If p~ denotes the reduced rank as an R-module, then P R : G o ( S+ ) 2 and PR([S]) = 2. Thus p ~ ( a 1 = ) 1 and a1 certainly has infinite order. Next we consider p. Since Wl (1 y)K[G1] and W2 % (1 xy)K[G2], it follows that
+
+
wlIs B w2lS= (1 + y ) ~ [ re] (1 + ~ y ) ~ [ r ] . On the other hand, since y and xy generate r and charK = 2, the augmentation ideal I of K[r] is given by
and this is easily seen to be a direct sum (Exercise 4). Thus a1+a2 = [I]and, if V denotes the principal K[r]-module, then a1
+ a2 + [V] = [ I ]+ [V] = [S]= 2a1.
In other words, p = a1 - a2 = [ V ] . Now RG1 has as a K-basis the set (1) U { xi x-; I i > 0 }. It then follows easily that RG1 = K [ x x-'I is Noetherian and R is a finitely generated RG1-module with generating set { l , ~ }In. addition, dimK RG1/trG1 ( R ) = 1. Thus Theorem 39.7(ii) applies and we have a homomorphism 8:Go(S) -, Go(K[G1]) % 2 / 2 2 as described in that result. Since p = [V] and yK[G,] = w1 is a finitely generated K [G 1]-module, we have
+
+
{ [Wl]+ K W [ G 1 1 ) ,
e(p) = 1 + 22,
in Go(K[G11>; in 2 / 2 2 .
Thus e(p) # 0, so p # 0 and Go(S) 2 @( 2 / 2 2 ) as claimed above. By an accident of sorts, we are able to compute Ko(K[r])for fields of all characteristics. Indeed, as mentioned in Section 37, r z 2, * Z2, the free product of two groups of order 2, and hence K [ r ]z K [ Z z ] K[Z2].A result of [63] therefore yields
uK
KO(K[F]) 2 Ko(K[Z21) e K O ( K [ 2 2 1 )
429
39. Almost Injective Modules
where, for any ring S , K o ( S ) is equal to Ko(S) modulo its cyclic subgroup generated by [S].It then follows that
Ko(K[rl)
{ i7@
if charK = 2; 2 @ 2, if charK # 2.
In fact, when charK = 2, [17]proves that all projective K [ r ] modules are free. Finally, if char K # 2, then the group ring K [ r ]has finite global dimension, and hence Go(K[r])% Ko(K[r]) by Proposition 33.8. Thus we conclude that Z@(Z/22), i f c h a r K = 2 ; Z @ Z @ 2, if c h a r K # 2.
Notice, in particular, that Go(K[r]) Ko(K[r]) in the characteristic 2 case.
EXERCISES 1. Let R be a ring and let V be a right R-module. Baer’s criterion asserts that V is injective if and only if every R-homomorphism 0: I + V, with I a right ideal of R, extends to a map c*:R + V. Prove this. 2. Show that a finite direct sum of injective modules is injective. If R is Noetherian, use Baer’s criterion to prove that an arbitrary direct sum of injectives is injective. 3. Let R S be rings and assume that RS is flat. If I is a right ideal of R, prove that the induced module satisfies I @ R S I S S. 4. Let r = (z, y I y-lzy = z-l, y2 = 1) and let K be any field. If I is the augmentation ideal of K [ r ] prove , that
To show that the sum is direct, first observe that any element of (1 - y)K[r] is uniquely writable as ki(1 - y)zi with Ici E K . Then observe that any element of (1 - zy)K[r]is annihilated on the left by 1 + zy.
xi
430
9. Zero Divisors and Idempotents
5. Let R = K [ ( x ) ] K [ r ]and let GI = (y). Show that RG1 is the polynomial ring K [ x + z-'] and that R = RG1@ xRG1 is a free RG1-module on two generators. 6. Suppose that the finite group G acts on R. If R is right Noetherian and trG(R) = RG,prove that RG is Noetherian. For this, note that if A is a right ideal of RG, then trG(AR) = AtrG(R) = A . 7. A ring R satisfies Kdim R 5 1 if and only if, for every nonzero right ideal I, R / I is an Artinian R-module. Now let the finite group G act on R and suppose that RG is right Noetherian and Kdim R 5 1. If RG/trG(R) is a right Artinian ring, prove that KdimRG 5 1. This is a result of [77].
40. Stably Free Modules We close this book with two topics of interest. The first is concerned with Go computations. Specifically, we continue the work of the preceding section and explicitly determine the Grothendieck group of the Zalesskii-Neroslavskii example. The second is related to KO and concerns the existence of stably free modules which are not free. To start with, suppose G acts as automorphisms on the group N . Then we say that the action is fixed point free if, for all 1 # g E G, we have C,(g) = (1). In the case of finite groups, this turns out to be an extremely important concept. Indeed if both G and N are nontrivial, then N is called a Frobenius kernel, G is a Frobenius complement and N >d G is a Frobenius group. Furthermore, it follows that N is necessarily nilpotent and that the structure of G is fairly tight. For example, the Sylow subgroups of G are either cyclic or quaternion and G is almost always solvable. A detailed discussion of such groups can be found in the book [156].In the case of infinite groups, N need not be nilpotent. For example, the infinite dihedral group admits a fixed point free automorphism group of order 2 (see Exercise 1). We will not use any of these facts here; we just require an understanding of the notation. The following result can be slightly generalized. However, its statement is already quite tedious.
431
40. Stably Free Modules
Theorem 40.1. [115] Let K t [ r ]be a twisted group algebra with I? polycyclic-by-finite and with char K = p > 0. Furthermore, let N be a torsion free normal subgroup of finite index in r, set R = K t [ N ]and let G I ,G2,. . . , Gk, H I , Hz, . . . ,H , be representatives of the conjugacy classes of maximal finite subgroups of I?. Assume that, for all 1 5 i 5 k and 1 5 j 5 m, we have i. Gi is a nonidentity p-group which acts in a fixed point free manner on N , Kt[Gi]2 K[Gi]and R is a Noetherian right module over the fixed ring RGi7 ii. Kt[Hj]is a division ring. Then k
Proof. By Theorem 36.9, the induced module map determines an epimorphism k
m
i=l
j=1
Since K t [ H j ]is a division ring, by assumption, G o ( K t [ H j ]is ) generated by [K'[Hj]]and hence its image in G o ( K t [ r ] is ) generated by [ K t [ r ] ] .In particular, if k = 0 then G o ( K t [ r ] is ) the cyclic group generated by [ K t [ r ] and ] the result follows easily. Thus, for the remainder of this argument, we assume that k 2 1. Since the image of Go(K'[Gl])also contains [ K t [ r ] it ] , is now clear that the induced module map yields an epimorphism (: A -+ Go(K'[[r]), where A = @ Go(K'[Gi]).Furthermore, since projective modules induce to projective modules, we also obtain an epimorphism d -+ where A = @ Co(Kt[Gi]). Note that, by (i) above, Kt[Gi] 2 K[Gi] and Gi is a finite pgroup. Thus, by Lemma 39.8, Go(Kt[Gi])is the infinite cyclic
xf=l
6
co(Kt[r])
9. Zero Divisors and Idempotents
432
group generated by [Wi],where Wi is the principal module, and c o ( K t [ G i ]2 ) Z/(lGil . 2). In particular, k i=l
and the goal is to prove that ( is an isomorphism. We of course already know that it is surjective. To show that is injective, we need a back map of sorts and for this we use Theorem 39.7(iii). To start with, since Kt[Gi]% K[Gi], we see that K t [ N G i ] = RGi is a skew group ring of Gi over R = K t [ N ] . In addition, S = K t [ r ]is a Noetherian K-algebra which is finitely generated as a right RGi-module. It remains to consider the action of Gi on R and, by assumption, we know that R is a Noetherian module over the fixed ring RGi. Furthermore, the action of Gi on N is fixed point free. Thus, since R = @ CyEN Kjj and Gi permutes the summands with y # 1 in orbits of full size IGil, it follows easily that RGa = K trGi(R). Thus the hypothesis of Theorem 39.7 is satisfied and we conclude that there exists a map
<
+
as described in that result. Hence, by combining these maps, we obtain a group homomorphism k
It turns out that fj is not quite the inverse of 6, but it is close enough. To understand the combined map fjcd + A, we must first observe that each Gi is self-normalizing. To start with, since N is torsion free, we have N n G i = (1). Next since Nar, the commutator group ["(Gi), Gi] satisfies
["(Gi),
Gi] C N n Gi = (1).
Thus Gi centralizes " ( G i ) and hence " ( G i ) = (1)since Gi acts in a fixed point free manner on N . Finally, Nr(Gi)nN = " ( G i ) = (1)
433
40. Stably Free Modules
and )I? : NI < 00, so it follows that Nr(Gi) is finite. Thus, since N r ( G i ) 2 Gi and Gi is a maximal finite subgroup, we conclude that N r ( G i ) = Gi as required. With this, we can use the Mackey decomposition to compute
wz,j = (wiJK
)
,
K t [Gj1.
To be precise, let 23 be a complete set of (Gi, Gj)-double coset representatives in r. Then, by Lemma 3.10, we have Wi,j = G3 &-D V ( d ) where
and Wi @I d is a specific module for Kt[Gt]. In particular, if we let = dimK V ( d ) then, since dimK W i= 1, we see that
Now Wi,j is almost injective by Theorem 39.7(i) and
Thus it follows that almost all of the modules V ( d ) are injective and therefore have dimension divisible by lGjl and hence by p . Since any V ( d )has dimK V ( d )composition factors all equal to Wj, it then follows from all of the above that
where ai,j E 2 is congruent modulo p to the number of d E 27 with v ( d ) = 0. Notice that v(d) = 0 if and only if Gf f l G j = G j and hence, since G j is maximal, if and only if G t = Gj. In particular, if i # j then ai,j z 0 modp since Gi and G j are not conjugate in I?. On the other hand, if i = j then we must have d E Nr(Gi) = Gi and this corresponds to a unique (Gi, Gi)-double coset. In other words, ai.i G 1 modp. Now, by Lemma 39.8, G,(Kt[Gi]) is generated by
434
9. Zero Divisors and Idempotents
wi = [Wi]+ Kg(Kt[Gi])and, viewing this as an element of d,we conclude that k j=1
= wi
modpA.
In other words, fjf induces the identity map on and, since A is a finite additive pgroup, this implies that fjf is an automorphism of A. Hence (is injective. We have therefore shown that A -+ G 0 ( K t [ r ] ) is an isomorphism so
6
i=l
as required. Finally, since 5 is an epimorphism and Go(Kt[Gi]) is generated by [Wi],we see that Go(Kt[l?])is generated by the elements ai = [WilKt"rl]. Furthermore, lGil . [Wi]= [Kt[Gi]]in Go(Kt[Gi]), so (Gilai = [ K t [ r ]in ] G o ( K t [ r ] )In . particular, if ]GI(3 JGiJfor all i, then pi = ai-JG1J-JGiJ-lal E G o ( K t [ r ] )IGiJPi , = 0 and Go(K'[F]) is generated by a1, p2, p 3 , . . . ,p k . Now, if p~ denotes the reduced rank as an R-module, then p ~G:o ( K t [ r ] ) Z is a homomorphism and p ~ ( K ~ [#r 0. ] ) Thus p ~ ( a 1 #) 0 and a1 has infinite order. On the other hand, by mapping these generators to e,(Kt[r]), we see that pi has order precisely [Gil and that p 2 , p 3 , . . . ,p k are otherwise independent. Thus --f
and the result follows. Recall that the Zalesskii-Neroslavskii example, as described in Theorem 38.8, depends on the field K of characteristic 2 and the element X E K' of infinite multiplicative order. As a consequence of the above, we have
435
48. Stably Free Modules
Corollary 40.2. (1151 If S is the Zalesskii-Neroslavskii example determined by K and A, then
Furthermore,
€'roo$ We know that S = K t [ r ]where
Here A = ( a ,b) is a free abelian group of rank 2 and that, by Lemma 36.6(iv),
H'((c),A)
2 2
= (c}. Note
annA(l+ .)/AC-' = A / A 2
and this group has order 4. Thus I? has four conjugacy classes of maximal finite subgroups, all of order 2, and we can take as representatives of these classes the subgroups
G I = (c),
G2 = (uc),
G3 =
(bc),
G4
=(u~c).
In K t [ r ]we have Sib = XbSi, C-liiC = Si-l, C - l b C = b-l and C2 = 1. Thus 2, SiF and b C all have square 1, so Kt[Gi]= K[Gi] for i = 1 , 2 , 3 . On the other hand,
(--abc)2 - a-&-l&-l and hence a field
Kt[G41
{ K[G4~:
=A if X $ if X E ( K * ) 2 .
Finally observe that each Gi acts in a fixed point free manner on A and that R = K t [ A ]is a Noetherian module over the fixed ring RG1
436
9. Zero Divisors and Idempotents
by Proposition 38.11. Thus Theorem 40.1 applies and, in view of the dual nature of Kt [Gd], it yields the appropriate formulas for GO(S) and Go(S). We mention just one more consequence, this time for ordinary group algebras.
Corollary 40.3. [115] Let A a r with A a finitely generated free abelian group and with F/A a finite p-group acting in a fixed point free manner on A . If GI, Ga, . . . ,Gk are representatives of the conjugacy classes of the maximal finite subgroups of r and if K is a field of characteristic p, then
and
Proof: Certainly Gi n A
= (l),so Gi is a finite pgroup which acts in a fixed point free manner on A . Furthermore, since R = K [ A ]
is a finitely generated commutative K-algebra, it follows (see Exercise 3) that R is a Noetherian module over the fixed ring RGi. Thus Theorem 40.1 yields the result. Note that if some Gi = (l),then we must have k = 1 and GI = (1). In this case, just let HI = (1) in Theorem 40.1. 1 Now we move on to our last topic. Let S be a right Noetherian ring and let P be a finitely generated projective S-module. We recall that P is stably free if P @ Sn E S” for some m ,n 2 0 and that, by Lemma 34.10, this occurs if and only if [PI is in the cyclic subgroup of Ko(S) generated by [S]. Of course, any free module is stably free. The question then is whether stably free, nonfree modules exist. The answer is “yes” for large classes of noncommutative rings and
40. Stably Free Modules
437
in particular for group rings. A slick approach to this problem is contained in [197]and we offer a small sample of the results of that interesting paper. We start with two key special cases. The first is a generalized polynomial ring or Ore extension given by S = R[z;u, 61. Specifically, this ring has the additive structure of the ordinary polynomial ring over R in one variable, but with multiplication given by r z = zr' S(r) for all r E R. Here u E Aut(R) and 6 is a 0-derivation. Second, we consider the skew group ring S = R[z,z-I; u]= R(z) of the infinite cyclic group (z) with TIC = zr'. Notice that, in either case, if R is a domain then so is S and if R is right Noetherian then so also is S. A polynomial f (z) = Cr.oaizz E R[z;0,6] is said to be monic if a, = 1. Notice that this is independent of the side on which we write the coefficients since lo = 1. Similarly, deg f = n if a, # 0 and this is also right-left symmetric.
+
Lemma 40.4. Let R be a right Noetherian ring. Then the set of monic polynomials in S = R[z;u,61 is a right divisor set.
Proof: The set of monic polynomials is certainly multiplicatively closed. Now let f (z) be a monic polynomial in S. Then by long division, it follows that S/ f S is generated by the images of all polynomials of degree less than that of f. In other words, S/ f S is finitely generated and hence Noetherian as a right R-module. Thus if g(z) E S , then the R-module
is finitely generated, say by the images of gzo, g d , . . . ,gzn-'. Then gziR and we conclude that gh E f S where h ( z ) is a monic polynomial of degree n.
gzn E f S
+
~~~~
Notice that if R is a division ring, then both S = R[z;0 , 6 ] and S = R ( z ) are principal right ideal rings and hence every right ideal is a free S-module. Thus if S is to have a nonfree ideal, then R must certainly contain a nonunit r # 0.
438
9. Zero Divisors and Idempotents
Theorem 40.5. [197]Let R be a right Noetherian domain and let S = R[x;a,S] or R [x,x-l; a]. Suppose that there exists a nonunit a E R and some b E R with S = U S+ (x + b)S. Furthermore, if S is the skew group ring, assume that baa $ aR. Then I = { g E s 1 ag E (x + b ) S ) is a nonfree, stably free right ideal of S satisfying I @ S G S @ S.
Proof. The proofs of the two cases are quite parallel, but there are differences which have to be considered. Note that if S is the skew group ring R[z,x-'; a],then the subring of S generated by R and x is R[x;a,S] with 6 s 0. For obvious reasons, we call the elements of this subring the polynomials in S . Furthermore, since S = 0 in this case, the constant term of a polynomial is right-left symmetric. Now let S be either ring. Then the hypothesis of the theorem guarantees that a # 0 and that S is a Noetherian domain. Furthermore, in the skew group ring case, ba" $ a R implies that b # 0. Consider the natural epimorphism 'p: aS @
(x+ b)S
--f
aS
+ (X + b)S = S
and observe that, for each g E I, there exists a unique g* E S with ag = (x b)(-g*). Then ag @ (x b)g* E Ker(cp) and in this way we obtain a map from I to Ker(cp) which is easily seen to be an isomorphism. Since ass 2 Ss E (x b)Ss, this yields a short exact sequence 0 -+ I -+ S @I S -+ S -+ 0 which of course splits since S is projective. Thus I @ S S @ S, so I is stably free, and it remains to show that I is not free. Since S is a Noetherian domain, this is equivalent to showing that I is not cyclic. We first find some elements in I. To start with, we have
+ +
+
ux =
+ S ( U ) = (x+ b)aU+ ( S ( U ) - baa)
where again S = 0 in the skew group ring case. Since R is an Ore domain, there exist T ~ , T ZE R with T I # 0 and (6(a)- baU)r1= arz. This yields a(xr1 - T Z ) = (x b)aarl and hence zr1 - rz E I. Next, since R is right Noetherian, we can apply Lemma 40.4 to either
+
439
40. Stably Free Modules
R[z;0,S] or to the polynomial subring of R[z,x-*;01. In either case, since x b is monic, there exists a monic polynomial f(z)E S with af E (z b)S. Therefore f E I and I contains a monic polynomial. Suppose, by way of contradiction, that I = h ( z ) S is cyclic. We use the existence of the above elements of I to suitably narrow the possibilities for h. To start with, if S = R ( z ) ,then we can multiply h by an appropriate power of the unit z to make h a polynomial with nonzero constant term. Thus in either case, h is a polynomial. Next, since 27-1 7-2 E I with T I # 0, it is clear that degh(z) 5 1 at least if S = R[z;a,S]. On the other hand, if S = R[z,z-';a]and h(z)k(z) = Z T ~ 7-2 then, since h has a nonzero constant term, it follows that k(x) is also a polynomial and again we have deg h ( z )5 1. Thus h ( z ) = xX p for some A, p E R. Furthermore h E I implies that a h ( x ) E (II:+ b)S and hence X # 0 since b # 0 when S is a skew group ring. Finally since I = hS contains a rnonic polynomial, it follows easily that X is a unit of R. Multiplying by X-l, we can therefore assume that h ( z )= x p. Again h(x) E I implies that a h ( x ) E (z b)S so
+ +
+
+
+
+
a h ( z )= a ( z
+
+ p ) = xuu + S(a) + up = (x+ b ) t
(*I
for some t E S . We can now complete the proof, handling the two cases separately. If S is the skew group ring then, since b # 0, (*) implies first that t is a polynomial and then that t E R. Thus, comparing coefficients yieIds t = a" and then baa = bt = up f aR, a contradiction. On the other hand, if S is the Ore extension, then we use S = aS+(z+b)S to find elements u,'u E S with 1 = au+(z+b)v. In addition, by long division, u = (z + p)w + c for some w E S and c E R. Thus (*>yields 1 = uu = .(
+ (z + b)w = a(z + p)w + (z + b). + ac + b)(tw + ?+I)ac
and, by computing degrees, we have t w + v = 0 and then ac = 1. But R is a domain, so this implies that a is a unit of R, a contradiction, and the result follows. This yields a uniform and fairly simple approach to a number of diverse constructions. For example, we have
440
9. Zero Divisors and Idempotents
Corollary 40.6. [149]Let D be a Noetherian domain containing elements a,P such that [a,P] = a0 - pa = y is a unit of D. If S = D [ z ,y] is the polynomial ring over D in two variables, then S has a nonfree, stably free right ideal.
Proof: Let R = D[y] so that S
+
+
= R[5].Since [ c u , ~ = ] y, we see that
+
[y Q , X p] = y and y is a unit of S. Thus since y Q is not a unit of R, the result follows from Theorem 40.5 with a = y Q and b=p. I
+
Note that the existence of a,,Ll above is certainly guaranteed if D has a noncommutative division subring. Thus, if E is a noncommutative skew field, then the polynoniial ring S = E [ q ,52,.. . ,x,] in n 2 2 variables has a nonfree, stably free right ideal.
Corollary 40.7. [197]Let D be a Noetherian domain and let S = D [ z ,y I zy - yz = 11. Then S has a nonfree, stably free right ideal.
Proox If R = D[y], then S = R[z;S]is clearly an Ore extension. Thus since y is not a unit of R and [x,y] = 1, the result follows from Theorem 40.5 with a = y and b = 0. I In particular, let D be a Noetherian domain and define the nth Weyl algebra A,(D) inductively by Ao(D) = D and
Then A,(D) is a Noetherian domain and, by the above, it has a nonfree, stably free right ideal if n 2 1 (see [204]). Finally we have
Corollary 40.8. [8] Let G be a nonabelian poly-2 group and let D be a Noetherian domain. Then the group ring S = D[G]has a nonfree, stably free right ideal.
Proof: We proceed by induction on the Hirsch number of G. Let (1) = Go a GI a - .. a G, = G
441
40. Stably Free Modules
be a subnormal series for G with each Gi+l/Gi infinite cyclic and set H = G,-1. If R = D [ H ]and G = (H,x), then certainly S = R[z,x-'; a].There are two cases to consider. Case 1. H is nonabelian.
Proof. By induction, R = D [ H ]has a nonfree, stably free right ideal I . We show that the right ideal I S of S has similar properties. To start with, since S is a free left R-module, the embedding I r-t R yields I @R S R @ I R S. Thus since R @IR S S via the map r @I s H T S , we conclude that I @R S S I S . In other words I S 2 ' 1 and, since I R is stably free, it follows that 15's is also stably free. Now suppose, by way of contradiction, that I S is free and hence cyclic; say I S = f ( x ) S for some f(x) E S. Multiplying f by a suitable power of the unit IC E S , we can assume that f is a polynomial with nonzero constant term. Now f ( x ) S 2 I # 0 and say f(z)g(x) is a nonzero element of I & R. Then by considering the lowest degree L)
term of g , we conclude first that g is a polynomial and then that both f and g have degree 0. In other words, f,g E R and hence f R = I , a contradiction.
Case 2. H is abelian.
Proof. Since H is a poly-2 group having Hirsch number n - 1, it is free abelian with generators y1, y2, ... ,y,-l. Now suppose r E R commutes with rap' and observe that
+
(x r"
-1
)(. - r).-2
+ TT'-1x-2
= 1.
Thus the result will follow from Theorem 40.5 with a = r and b = -1 -1 r" provided that r" r U 4 rR. The goal then is to find such an element r . Since G is nonabelian, x does not centralize H and hence say -1 y; # y1. Suppose first that yf # ycl. Then y?, 9; 4 (yl) and we set T = 31- 1. Notice that T does indeed commute with its conjugate rb-' = yTP1- 1 and that r R is the kernel of the homomorphism from D [ H ]to the domain D[H/(31)]. Thus clearly r"-'r0 6 rR. On the
442
9. Zero Divisors and Idempotents
+ +
other hand, if yf = yT1 then we set T = 1 y1 9.: Again T -1 commutes with its conjugate ra = 1 y T 1 yT3 and it is easy to see that $! TR. Indeed iff-'^" = T S for some s f R then, by projecting this equation into D[(yl)],it follows first that s E D[(yl)] and then that s does not exist (see Exercise 4). Thus we have a suitable element T in all cases and the corollary is proved. I
+
+
Paper [197]then goes on to show that enveloping algebras of nonabelian, finite dimensional Lie algebras always have nonfree, stably free right ideals. Some aspects of this are considered in Exercises 5 and 6.
EXERCISES 1. Let r = (z,y 1 y-lzy = x-l, y2 = 1) be the infinite dihedral group. Show that A = (x2,y) is a normal subgroup of r which is isomorphic to I’ and that conjugation by zy induces a fixed point free automorphism of order 2 on A. 2. Let I? = A >a (z) where A = ( a ,b) is free abelian, z4 = 1 and a” = b, b” = a - l . If K is a field of characteristic 2, compute G o ( K [ r ]and ) Go(K[rI). 3. Let R = K[al,u2,. . . , a,] be a finitely generated commutative K-algebra and let G be a finite group of K-algebra automorphisms of R. Show that R is a Noetherian RG-module. To this end, let S be the finitely generated K-subalgebra of R generated by the coefficients of the polynomials fi(z) = - a:) for i = 1 , 2 , . . . ,n. Observe that S is a Noetherian subring of RG and that R is a finitely generated S-module. 4. If D[y] is any polynomial ring, show that 1 y y3 cannot divide (1 y2 y3)2. Use this to deduce the appropriate facts about T = 1 y1 y; in the proof of Corollary 40.8. 5. Let S = R[z;a,6]be an Ore extension with R a Noetherian domain and suppose that R contains a nonunit T with CEO Si(r)R= R. Show that S has a nonfree, stably free right ideal. For this, observe that rS zS contains all 6”~).
ngEG(z
+ +
+ + + +
+
443
40. Stably Free Modules
6. Now let R be a commutative K-algebra freely generated by the finite dimensional vector space L and let 6 be a nonzero Kderivation of R which stabilizes L. Use the previous exercise to prove that S = R[z;61 has a nonfree, stably free right ideal. To start with, choose a E L with S(a) # 0 and define ai = @ ( a ) . Since L is finite dimensional and 6-stable, there exists n minimal with a,+l in the K-linear span of ao, a l , . . . ,a,. If a,+l = 0, take T = 1 a,-la, in the above. On the other hand, if un+l # 0 then we can write a,+1 = CLjkiai with ki E K and kj # 0. Now take T = 1 a j .
+
+
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Index
0-abelian, 75 1-coboundary, 384 1-cocycle, 384 lSt-cohomology group, 384 2-coboundary, 9, 368 2-cocycle, 3, 368 2-fir, symmetric ring of quotients, 128 2-term weak algorithm, 125 2nd-cohomology group 4, 368-369 A
[ A ] ,344 A O K B, 406 A*, 299 Abelian group algebra, 194 Absolute field, 192 Action, 2 on ideals, 43 Additivity principle, 291 fixed ring, 291 a.i. module, 420ff Algebra of the group, 280, 298, 309 Algebraically closed group, 104 Almost centralizer, 56-58 Almost faithful ideal, 191 Almost faithful mod N ideal, 202 Almost faithful sub N ideal, 202 Almost injective module, 420ff characterization, 421-422, 425 existence, 421422, 425
Almost normal group, 325 Amitsur, S. A., 69, 74, 83, 93, 180, 225, 230, 231 Amitsur-Levitzki theorem, 69, 74 Ample, 195 Andrunakievitch, V. A,, 285 Annihilator-free, 54-55, 68 Artamonov, V. A., 440 Artinian fixed ring, 269 Artinian ring, Go, 344 KO,353, 355 Associated graded ring, 121 Associativity condition, 87 Augmentation ideal, 191, 248, 314, 407 Augmentation map, 243, 280-281 Automorphism, central type, 117 scalar type, 117 X-inner, 107ff Azumaya, G., 42, 48, 297, 328 B B(G), 280, 298, 309 B1(G,A), 384 B2(G,A), 368 Baer’s criterion, 429 Balanced condition, 87 Base ring, 10 Bedi, S. S., 226 Beidar, K. I., 230 Bell, A. D., 23, 26
459
460
Index
Bergen, J., 74, 108 Bergman, G., 116, 194, 224, 243, 245246, 250-251, 254-260, 266-267, 429 Bimodule property, 302, 311 Bit-David, J., 155, 159 Blather, R. J., 16, 28 Bounded element, 127 Brauer group, 4 br(s), 222 Breadth, 222 Brown, K. A., 212, 395 C X R , 399 C1(G,A), 384 C2(G,A), 368 Cartan, H., 297 Cartan-Brauer-Hua theorem, 326 Cartan map, c, 352-353, 409 Cartan matrix, 355 Case 1, induction theorem, 370 Case 2, induction theorem, 376 Case 3, induction theorem, 382 Center, group algebra, 40 Central closure, 92 Central localization, 176 Central simple algebra, 4 Central type, 117 Central type group, 165 Centric plinth, 192 Childs, L. N., 297 Chin, W., 18, 25, 232-233 Cliff, G. H., 343, 361, 378, 395 Cohen, I. S.,151, 161 Cohen, M., 12, 15-16, 26, 28, 32, 34, 36, 74, 170, 173, 175, 252-253, 264270 Cohesive ring, 100 Cohn, P. M., 125-126, 128, 332, 394 Compatible basis, 299 Completely reducible module, 30, 35 Complex, 356 Component, 12 Component regularity, 16, 32, 65 module analog, 37 Composition length, 344 Conjugacy class sum, 176 Conjugate action, 422 Conjugate module, Z(t21, 371-372 Connecting homomorphism, 357 Connell, I. G., 39, 42 Contragredient module, 424
Controller, 142 Coproduct, 122, 394 X-inner automorphism, 122 coreG(H), 95 Correspondence, prime, 138-140, 171 Correspondence theorem, 313 free algebra, 334 weak algorithm, 334 Countable index, 52 Crossed product, 2ff Goldie rank, 403 incomparability, 217, 233 Jacobson radical, 30 Jacobson ring, 228 Noetherian, 7-8 pgroup, 154 polynomial identity, 237 prime, 111, 168 prime correspondence, 149-150 prime ideal, 131ff, 142ff, 152 semiprime, 31, 111, 182-183 symmetrically closed, 99, 114-115 von Neumann regular, 164 weakly semiprime, 183 X-inner automorphism, 112 Cutting Down, 156, 159, 173, 287 D A(G), 40 &(GI, 76 Ap(G), 42 Af(G), 40 VG(N), 202 v ~ ( Nr), ; 209 A-lemma, 41, 61, 76, 97, 198 A-methods, 39ff, 61 crossed products, 45-46 group algebras, 39-42 DGW, 194 DG(H), 56-57 D(H), 13 D-'(H), 13 Dade, E. C., 11 degs, 222 Degree, 122, 254 Delta methods, 39ff DeMeyer, F., 165, 297 Depth, 290 prime ideal, 158 Derivation, 90 Derived functor, 356
Index
461
Diagonal, 406 Diagonal change of basis, 3 Dicks, W., 336, 340 Dieudonnb, J., 297 Dihedral group, 82 Division ring, 4 Domain, 85 not Morita equivalent, 409, 412 Dual, 299 Duality, 12ff, 33-34, 37, 65, 233 Duality theorem, 15 E &-transversal, 258 270 E R ( A ) , 420 Easy direction, Theorem 5.8, 44 Eccentric plinth, 192 Eccentric plinth length, 192 Enveloping ring, X-inner automorphism, 122 epl(G), 192 Essential ideal, 180 Essential Maschke’s theorem, 31 Exact functor, 20 Existence of fixed points, 243, 245 reduced ring, 278 Extended centroid, 92, 180 Extension theorem, 319 Exterior algebra, 82 e,
F
F,344 Faith, C., 277, 405 Faithful ideal, 191 Farkas, D. R., 195, 272, 274, 343, 395, 399 f.c. center, 40-42, 57, 188 f.c. group, 40 Filtered ring, 120ff X-inner automorphism, 121 symmetric ring of quotients, 125 Finite centralizing extension, 158 Finite class sum, 40, 47 Finite conjugate center, 40-42 Finite global dimension, 395 Finite index, 52 subset, 76, 80 Finite index problem, 197 Finite induced module, 195
Finite integral extension, 151 Finite normalizing extension, 155, 159 full integrality, 263 Krull relations, 159 Finite projective dimension, 352-354 Finite projective resolution, 350-352 Finite solvable group, 250, 252 Finite subgroup, polycyclic-by-finite group, 385 Finitely generated group, 190 Finitely presented group, 212 fir, 128 Fisher, J. W., 31-32, 42, 108, 142, 179, 242, 270 Fixed point free action, 430 Fixed points, 242 existence, 243, 245, 267, 307 Fixed ring, 241 additivity principle, 291 Artinian, 269 finitely generated, 336 free algebra, 336 full integrality, 260-261 Goldie, 270 Hilbert series, 340 Jacobson radical, 293-295 Krull relations, 287 Noetherian, 274, 414 prime, 266 prime ideal, 287 restricted bimodule, 272 restricted module, 272 ring of quotients, 315 semiprime, 266 simple, 273 weak algorithm, 331 Flat module, 20, 345, 357 directed limit, 345 projective, 345 Form, 57ff size, 58 Form of minimal size, 58 Formanek, E., 114, 336, 340, 394 Free abelian semigroup, 354 Free algebra, 127 prime correspondence, 341 Free group, 8, 100 Free product group, 394 Free ring, symmetrically closed, 119 X-inner automorphism, 120 Free semigroup, 118 Frobenius akgebra, 299
462
Index
Frobenius complement, 430 Frobenius group, 430 Frobenius kernel, 430 Frobenius reciprocity, 407 Full integrality, 254ff finite normalizing extension, 263 fixed ring, 26&261 graded ring, 259 matrix ring, 255 G I?-orbital element, 188 G,265 Gjnn, 107, 276, 298 GI, 67 Gal(R/S), 310 Go(R),344ff Artinian ring, 344 eSe, 372 group algebra, 436 Noetherian crossed product, 386 ring extensions, 345-346 skew group ring, 407 twisted group algebra, 431 ,%’+-graded ring, 361-363 Zalesskii-Neroslavskii example, 435 co(R),420, 425 grGo(R),347 Z+-graded ring, 361-363 gr mod R, 347 G-annihilator-free, 54-55 G-cohesive ring, 100 G-graded ring, l O f f G-nilpotent-free, 54-55 G-prime, 54 G-prime ideal, 132 G-prime ring, 132 G-semiprime, 54, 57 G-stable ideal, 5 G-system, 231 G-Galois action, 308 Galois group theorem, 310 Generalized conjugates, 13 Generalized polynomial ring, 125, 129, 437 Generic model, 248-249 Gersten, S., 428 Going Down, 151, 156, 160, 173, 288 Going Up, 151, 156, 160, 173, 288, 290 Goldie, A. W., 228, 231, 267, 269 Goldie fixed ring, 270
Goldie rank, 267, 399 crossed product, 403 group ring, 404 Goldie rank problem, 399ff Goldie ring, 269 Goldie’s theorem, 269 Goldman, O., 229 Graded Grothendieck group, 347 Graded homomorphism, 23 Graded ideal, 33 Graded Jacobson radical, 35-36 Graded module, 22ff Noetherian, 23, 25-26 simple, 23 Graded Noetherian module, 23, 25-26 Graded prime ideal, 169 Graded prime ring, 169 Graded ring, full integrality, 259 incomparability, 233 Jacobson ring, 222 Krull relations, 173 prime ideal, 170 semiprime, 185 Graded R-module, 359-362 Graded semiprime, 32-33 Graded simple, 23 Graded submodule, 23 Graded subring, 32 Graded version, Maschke’s theorem, 3637 Nakayama’s lemma, 359 Grothendieck, A., 410 Grothendieck group, 344ff Group, central type, 165 Group algebra, 1 abelian, 194 center, 40 Go, 436 prime, 42 semiprime, 42 Group of operators, 194-195, 210 Group of trivial units, 3 Group ring, 3 Goldie rank, 404 Group-graded ring, lOff, 169ff Jacobson radical, 36 nilpotent, 32 nondegenerate, 32-33 polynomial identity ring, 74, 78, 81 semiprime, 32-33 Grzeszczuk, P., 35-36, 221 Guralnick, R. M., 253
Index
463 H
A(G), 192 H1(G,A), 384 exponent, 384 free module, 384 H2(G, A), 368-369 exponent, 369 free module, 369 ( H , D,I , P), 57 ( H , D , I , P ) # ,58 H(R), 338 H(X), 338 Hn(X), 356 Handelman, D., 104, 234, 237 Hard direction, Theorem 5.8, 64 Height, 290 prime ideal, 158 Heinicke, A. G., 159-160 Higman, G., 32 Hilbert Basis Theorem, 7, 21 Hilbert Nullstellensatz, 220 Hilbert series, 338 fixed ring, 340 Hilbert’s 14th problem, 195 Hirsch number, 192 Hochschild, G., 297, 317 Hodges, T. J., 414, 418, 430 Homogeneous coordinates, 350 Homogeneous element, 222 Homogeneous group, 331 Homology group, 356 Howlett, R. B., 165
I
W ) 16 , iG(H), 189 I t , 190-191 I*G, 5, 132 Z#G*, 169 ,tI 136 i, 145 Ideal, trivial intersection, 67ff Ideal-cancellable subring, 313 Idempotent, 171, 372-374 Incomparability, 151, 156, 159, 173, 175, 288 crossed product, 217, 233 graded ring, 233 Z-graded ring, 224, 232 ind(R*G), 402
Index, 76, 402 countable, 52 finite, 52 Induced ideal, 199 Induced module, 19ff, 141, 346 Induction, 19 Induction theorem, 386 case 1, 370 case 2, 376 case 3, 382 KO,410 Infinite dihedral group, 117, 394, 430 Go, 427 KO,428 Injective hull, 420 Injective-free module, 420 Inn(R), 107 Insulator, 96 Integrality, full, 254ff Schelter, 254ff Intermediate extension, 159 Intersection theorem, 110-111, 114 Invariant ideal, 43, 65 Isaacs, I. M., 75, 165, 243, 245-246, 250251, 253, 266-267 Isolated orbital subgroup, 188 Isolator, 189, 208
J JG(S), 35 J(W, 30 Jacobson, N., 297, 317 Jacobson radical, 30 crossed product, 30 fixed ring, 293-295 graded, 35-36 group-graded ring, 36 2-graded ring, 225-226 Jacobson ring, 220ff crossed product, 228 graded ring, 222 Janusz, G. J., 165 Joseph, A., 291
K Ko(R), 350ff, 364-365 Artinian ring, 353, 355 induction theorem, 410 Kg(R), 409 Ko(R), 429
464
Index
Kdim R, 430 Kharchenko, V. K., 88, 105, 108, 110, 120, 125, 270, 273, 277-278, 297-299, 301-303, 307, 309-311, 313-315, 319, 327, 331, 333-334, 336 Kreimea H. F., 297 Kropholl'er, P. H., 398-399, 404-405 Krull, W.1 151, 229 Krull relations, 155 crossed product, 155 finite normalizing extension, 159 fixed ring, 287 graded ring, 173 L
L ( V ) , 271 LIG, 136, 199 LG,169 Lane, D. R., 331, 333 Lattice of submodules, 271 Lawrence, J., 104, 234, 237 Leading term, 121, 222 Left index, 76 Left strongly prime ring, 104 Levitzki, J., 69, 269 Lewin, J., 128, 347, 394 Lexicographical ordering, 414 Liberal extension, 158 Lichtman, A. I., 114, 122 Lie algebra, 11, 121-122 Linear automorphism, 335 Linear identity, 40-42 Linear monomial, 78 Linked primes, 160 Linnell, P. A., 343, 398-399, 404-405 Long exact Tor sequence, 357 Lorenz, M., 31, 132, 138, 140, 142, 149, 152, 154-155, 158-159, 162, 168, 179, 187, 198, 202, 204, 206-207, 209, 255, 263, 271-272, 286, 290-291, 399, 403, 406-407, 409, 413, 421, 425, 431, 435436 Lying Over, 151, 156, 159, 173, 288
M&(S), 12ff MIR, 19 M-group, 298 Mackey decomposition, 2&27, 402, 433 Martindale, W. S., 83, 88, 92, 94, 105, 122, 129, 150, 293, 295 Martindale ring of quotients, 83ff left, 84ff right, 85 Maschke, H., 29 Maschke's theorem, 29ff essential version, 31 graded version, 36-37 subgroup version, 163 Matrix ring, full integrality, 255 McLaughlin, J. E., 164 Michler, G. O., 228, 231 Mihovski, S. V., 110, 115, 117 minNA, 46-47 Minimal prime ideal, 133, 152 Miyashita, Y., 19, 297 Miyashita automorphism, 19 Module, completely reducible, 30 conjugate, 371-372 induced, 19ff rationally free, 376 restricted, 19 stably free, 364-365 stably isomorphic, 364-365 Monic polynomial, 437 Montgomery, S., 12, 15-16, 26, 28, 3132, 34, 36, 42, 71, 74, 108, 111-112, 114, 117, 121-122, 142, 170, 173, 175, 179, 252-253, 266, 269, 279, 283, 286287, 289-291, 293, 298-299, 303, 311, 314, 318-319, 325, 341 Moody, J. A., 343, 367, 370, 376, 382, 386, 398-399, 404405 Morita, K., 409 Morita correspondence, 372-374 Morita equivalent rings, 409 Morita maps, 372-374 Multi-sign, 415 Multilinear polynomial, 69, 234
M
N A G ) , 397 mod R, 344 m(n),255ff m(RtG),402 MG(S), 12ff
n-fir, 127-128, 394 Inlp, 166 n-term weak algorithm, 123ff, 331 n-torsion, 29
Index
465
N*-Galois extension, 325 N*-group, 311 N-group, 298 Nagahara, T., 297 Nakayama, T., 297, 317, 328 Nakayama’s lemma, graded version, 359 NhtLescu, C., 23, 37 Neroslavskii, 0. M., 405, 412 Nilpotent group-graded ring, 32 Nilpotent-by-finite group, 192 Nilpotent-free, 54-55, 68 nio(G), 190 nio2(G), 217 No (GI-torsion, 72 Noether, E., 297 Noether’s equation, 324 Noetherian crossed product, 7-8 Noetherian fixed ring, 274, 414 Noetherian module, 21 Nonabsolute field, 192 Nondegenerate, 32-33 Normal abelian subgroup, 367 Normal closure, 108 Normal element, 93 Normal ideal, 262 Normalized reduced rank, 399-400
0 Ojanguren, M., 440 Orbital element, 188 Orbital subgroup, 188 Orbitally sound group, 189-191, 209 sufficient condition, 193 Ordered group, 47, 391 Ordering, lexicographical, 414 product, 414 Ore extension, 437 Osterburg, J., 242, 270, 273, 308, 319, 414, 418, 430 Outer group, 317 Outer trace form, 306 nontrivial, 306
P 6, 46, 56 pr m o d R, 350 P(R), 38 pabelian, 75 p-group, crossed product, 154 twisted group algebra, 153 TH,
pnilpotent group, 245 Page, A., 296 Par6, R., 255, 263 Partial trace, 252 nontrivial, 252 Passman, D. S., 31-32, 39, 41-42, 44, 71-73, 75, 78, 81, 88, 92, 95, 99, 102, 104, 108, 111-112, 114-115, 117, 119, 131-132, 138, 140, 142, 149, 152, 154155, 158-159, 168, 179, 182-185, 187, 198, 202, 204, 206-207, 209, 215, 217, 225, 232, 239, 253, 255, 258, 260, 263, 271-272, 286, 290, 298-299, 303, 311, 314, 318-319, 325, 341, 399, 421, 425, 431, 435-436 pd,A, 352 Pearson, K. R., 229 Permutation module, 28 Permutes, strongly, 48ff p. i. algebra, prime, 234 PKG), 192 Place permutation, 335 Plinth, 192 centric, 192 eccentric, 192 Plinth length, 192 Plinth series, 192 Poly-{infinite cyclic} group, 211 Poly-z group, 211 Polycyclic-by-finite group, 6, 25-26, 104, 187ff, 197ff finite subgroup, 385 Polycyclic group algebra, 187ff, 197ff prime ideal, 191, 202-206, 210 prime length, 193 primitive ideal, 195 primitive length, 193 Polynomial identity, 69-70, 74ff, 233 crossed product, 237 group-graded ring, 74, 78, 81 twisted group algebra, 239 Power series ring, 389 Prime correspondence, 138-140, 171 crossed product, 149-150 free algebra, 341 Prime crossed product, 111, 168 Prime group algebra, 42 Prime ideal, 18, 194 crossed product, 131ff, 142ff, 152 existence, 198, 204, 206, 212 fixed ring, 287 graded ring, 170
466
Index
polycyclic group algebra, 191, 202206, 210 standard, 191 virtually standard, 191 Prime length, 158 crossed product, 158 polycyclic group algebra, 193 twisted group algebra, 214 Prime p. i. algebra, 234 Prime radical, 38 Prime strongly graded ring, 44, 68, 71 Primitive ideal, 157 polycyclic group algebra, 195 Primitive length, 158 crossed product, 158 polycyclic group algebra, 193 Principal indecomposable, 353 Principal module, 407 Procesi, C., 74 Product ordering, 414 Projective dimension, 352-354 finite, 352-354 Projective module, 350 not stably free, 388 Projective resolution, 350ff finite, 350-352 Promislow, S. D., 392 Puczlowski, E., 293-295
Q Qe(R1, 84 Q T ( R ) ,85 Q,(R),86ff Quasi-Frobenius ring, 419 Quillen, D., 354, 361-363 Quinn, D., 12-13, 16, 18, 25, 37, 42, 65, 185, 233, 261
R
R,344 R*G, 2ff RC, 92
R(G),11 RG, 241 R N , 108 (R, S)-truncation, 303 R ( X ) , 11 Ram, J., 226 rankV, 267 Rational plinth, 192
Rationally free module, 376 Reduced rank, 399-400 Reduced ring, 276ff existence of fixed points, 278 Reid, A., 164 Respects the grading, 232 Restricted bimodule, fixed ring, 272 Restricted module, 19 fixed ring, 272 Restriction, 19 Right dependent on, 123, 331 Right dependent set, 123, 331 Right exact functor, 345 Right independent of, 123 Right independent set, 123 Right strongly prime ring, 104 Ring, cohesive, 100 coproduct, 394 filtered, 120ff finite global dimension, 395 fir, 128 G-cohesive, 100 G-prime, 132 Goldie, 269 Jacobson, 220ff left strongly prime, 104 n-fir, 127-128, 394 quasi-Frobenius, 419 reduced, 276ff right strongly prime, 104 self-injective, 419 simple Noetherian, 412 strongly prime, 96 symmetrically closed, 94ff von Neumann regular, 163, 175 Z+-graded, 330, 347, 356ff Ring extension, Go, 345346 Ring of quotients, fixed ring, 315 Martindale, 83ff symmetric, 86ff Rings, Morita equivalent, 409 Rips, E., 392 Rjabuhin, Ju. M., 285 Robson, J. C., 155, 159-160 Roseblade, J. E., 131, 187-188, 190-191, 193-195, 206, 211, 229 Rosenberg, A., 297 Rosset, S., 399 b w G ( M ) , 24 Rowen, L. H., 32
467
Index S
S#G*, 14 S { H } , 13ff s-, 222 S n , 69 s+, 222 (S,R)-truncation, 303 Saturated chain condition, 193 Saturation condition, 298 Scalar type, 117 Schanuel, S., 388 Schanuel’s lemma, 351 Schelter, W., 234, 237, 255, 263 Schelter integrality, 254ff Segev, Y., 392 Seidenberg, A., 151, 161 Self-injective ring, 419 Semi-invariant, 11, 121 Semicenter, 11 Semigroup crossed product, 7, 118 Semiprime crossed product, 31, 111, 182-183 Semiprime graded ring, 185 Semiprime group algebra, 42 Semiprime strongly graded ring, 44, 68, 72-73 Separable K-algebra, 74 Separated group, 95 Separated subset, 118 Serre, J. P., 395, 410 Simple fixed ring, 273 Simple Noetherian ring, 412 Simple ring, Galois theory, 317, 327-328 Size of form, 58 Skew group ring, 4, 241, 265 Go, 407 Skew polynomial ring, 7, 354 Skolem-Noether theorem, 235, 323 Small, L. W., 74, 290-291 Smash product, 14-15, 26, 33-34, 169 Smith, M. K., 41, 74-75 Snider, R. L., 195, 272, 274, 395 socV, 420 Socle, 420 Source, 204 existence, 202, 206 uniqueness, 204, 206 Split extension, 367 Sridharan, R., 440 Stabilizer, 67 Stably free ideal, nonfree, 438, 440
Stably free module, 364-365, 436ff Stably isomorphic modules, 364-365 Stafford, J. T., 406, 412, 437438, 440, 442 Standard identity, 69 Standard polynomial, 335 Standard prime ideal, 191, 200 Stephenson, W., 229 Strongly graded ring, 10, 42ff prime, 44, 68, 71 semiprime, 44, 68, 72-73 Strongly permutes, 48ff Strongly prime ring, 96 Sub-crossed product, 5 Subnormal series, type A, 405 Supersolvable group, 394 Supp a , 5, 46, 56 Support, 5 Sylow subgroup, 168 Symmetric ring of quotients, 86ff a-fir, 128 domain, 88 filtered ring, 125 matrix ring, 89 semiprime ring, 180 Symmetrically closed, 94ff crossed product, 99, 114-115 free group, 100, 115 free ring, 119
T
O(V),420, 425 3 ( V ) , 426 T-monomial, 254 Tensor module, diagonal action, 406, 419 Tominaga, H., 297 Tor, 356ff, 363 Torsion, 29 Torsion free group, 391 Torsion submodule, 366 Total degree, 347 trG, 242, 265, 422, 425 Trace, nontrivial, 251, 282-283 Trace form, 300ff existence, 301 outer, 306 Trace map, 242, 265, 422 general, 407 Trailing term, 222 Transitivity of induction, 140, 378 Transversal, 258
468
Index
Trivial intersection ideal, 67ff, 178 Trivial units, 3, 115 Truncation, 303 Twisted group algebra, 214, 410 Go, 431 pgroup, 153 polynomial identity, 239 prime length, 214 Twisted group ring, 4 semiprime, 177 Twisting, 2, 368 U
W L ) , 11 Ulbrich, K. H., 16, 29 Uniform module, 267 Unique product, failure, 392 Unique product group, 47, 116, 391 Universal enveloping algebra, 11, 121122 Untwisting, 7-8 V
VG, 425 vls, 19 vc, 20 Van den Bergh, M., 308 Vertex, 204 existence, 202, 206, 209 uniqueness, 204, 206 Villamayor, 0. E., 30, 297 Virtually standard prime ideal, 191 Von Neumann regular crossed product, 164 Von Neumann regular ring, 163, 175 commutative, 176-177 VXGCP),204 W Walker, R. G., 394-395 Weak algorithm, 123ff, 332 Weakly semiprime crossed product, 183 Weakly separated group, 95 Webber, D. B., 440 Wedderburn theorem, 152 Wehrfritz, B. A. F., 194 Weiss, A., 343, 361, 378 Welsh, C., 195, 218 Weyl algebra, 289, 440
Wiegold, J., 80 Wreath product, 208, 217
X Xinn(R), 107 X-inner automorphism, 107ff coproduct, 122 crossed product, 112 enveloping ring, 122 filtered ring, 121 free ring, 120 X-inner group, 108 X-outer group, 108, 326
2 Z-graded ring, 222 incomparability, 224, 232 Jacobson radical, 225-226 Z+-graded ring, 330, 347, 356ff gr Go, 361-363 Go ( Ro),361-363 Zalesskii, A. E., 405, 412 Zalesskii-Neroslavskii example, 405ff, 412 Go, 435 Zelinsky, D., 297 Zero divisor, Go, 395 Zero divisor problem, 391ff polycyclic group, 395 solvable group, 398 supersolvable group, 394 Zero sequence, 356 Zhuchin, A. V., 234, 237
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