Infinite crossed products

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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.

ACADEMIC PRESS, INC. 1250 Sixth Avenue, San Diego, CA 92101

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.

This Page Intentionally Left Blank

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


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


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


(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


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


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


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


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


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


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


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).


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


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


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


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


]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


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


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


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


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


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


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


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


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


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


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 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


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


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


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


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) 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,


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

78


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,. . . , 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 , . . . , 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


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


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.


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


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


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


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


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


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


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


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


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


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

+


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


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


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


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


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


( 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


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


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


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.

~ ~


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


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.


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 ,


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


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


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.


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


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.


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


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


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]


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.


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


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


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


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


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


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[ 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

188

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.

189

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

190


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 } ,


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


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


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,,

194


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


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.

196


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.


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


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


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


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]


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.


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


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


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


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


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 ) .


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


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


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].


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


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


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


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


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


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.


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


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


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


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+
Ib-a) and form the ring extension S = S[C]/(l+C+. . .+ 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 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]( 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(-' 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[ 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.

324

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


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


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


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


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


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 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


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

370


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


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


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 .

373


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

374


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


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.

376


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


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).

378


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.

379

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

380


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


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.

382


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


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


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


(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


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


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


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


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


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


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,


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


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


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,

398


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.


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


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 )


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


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


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


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.


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


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


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


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


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


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 ; 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


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])


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

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


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


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


+ +

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


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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[190] J. J . Rotman, “Notes on Homological Algebra,” Van Nostrand Reinhold, New York, 1970. [191] L. H. Rowen, “Polynomial Identities in Ring Theory,” Academic Press, New York, 1980. [192] S . Schanuel, see [lo,Proposition 2.l(b)]. [1931 J. P. Serre, Cohomologie des Groupes Discrets, in “Prospects in Mathematics, ” Annals of Math. Studies No. 70, Princeton, 1971, pp. 77-169. [194] M. K. Smith, Group algebras, J. Algebra 18 (1971), 477-499. [195] R. L. Snider, The zero divisor conjecture for some solvable groups, Pacific J. Math. 90 (1980), 191-196. I1961 J. T. Stafford, A simple Noetherian ring not Morita equivaEent to a domain, Proc. Amer. Math. SOC.68 (1978), 159-160. [197] J. T. Stafford Stably free, projective right ideals, Compositio Math. 54 (1985), 63-78. [198] H. Tominaga and T. Nagahara, “Galois Theory of Simple Rings, ” Okayama Math. Lectures, Okayama Univ., 1970. [199] K. H. Ulbrich, Vollgraduierte Algebren, Abhandlung Math. Sem. Univ. Hamburg 51 (1981), 136-148. [200] M. Van den Bergh, A duality theorem for Hopf algebras, in “Methods in Rang Theory, ” AS1 Series C, No. 129, Reidel, Dordrecht, 1984, pp. 517-522. [201] 0. E. Villamayor, O n the semisimplicity of group algebras, Proc. Amer. Math. SOC.9 (1958), 621-627. [202] 0. E. Villamayor and D. Zelinsky, Galois theory with infinitely many idempotents, Nagoya Math. J. 35 (1969), 83-98. [203] R. G. Walker, Local rings and normalizing sets of elements, Proc. London Math. SOC.(3) 24 (1972), 27-45. [204] D. B. Webber, Ideals and modules of simple Noetherian hereditary rings, J. Algebra 16 (1970), 239-242. (2051 B. A. F. Wehrfritz, see [57,Theorem 2.11. [206] C. Welsh, Prime length in crossed products. [207] J. Wiegold, Groups with boundedly finite classes of conjugate elements, Proc. Royal SOC.London Ser. A 238 (1957), 389-401. [a081 A. E. Zalesskii and 0. M. Neroslavskii, There exist simple Noetherian rings with zero divisors but without idempotents

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(Russian with English summary), Comm. Algebra 5 (1977), 231-244. [209] A. V. Zhuchin, Cross products with identity, Soviet Math. (Izv. VUZ) (no. 9) 29 (1985), 28-34.

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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Infinite crossed products

Infinite crossed products

Infinite Crossed Products (Pure and Applied Mathematics, Vol 135)

Crossed Products of C^* Algebras

Crossed products of C*-algebras

The Algebraic Structure of Crossed Products

The algebraic structure of crossed products

Crossed products of C-star algebras

Crossed

Crossed

Crossed

Continuous crossed products and type III Von Neumann algebras

Duality for Crossed Products of Von Neumann Algebras

Duality for crossed products of von Neumann algebras

Forever Crossed

Bone Crossed

Crossed Bones

Bone Crossed

Crossed Wires

Bone Crossed

Bone Crossed

Bone Crossed

Crossed Bones

Crossed Bones

Bone Crossed

Crossed Swords

Crossed Bones

Crossed Bones

Crossed Bones

Bone Crossed

Bone Crossed

Infinite crossed products