THE ALGEBRAIC STRUCTURE OF CROSSED PRODUCTS
NORTH-HOLIAND MATHEMATICS STUDIES Notas de Matematica (118)
Editor: Leopoldo Nachbin Centro Brasileiro de Pesquisas Fisicas Rio de Janeiro and University of Rochester
NORTH-HOLLAND -AMSTERDAM
NEW YORK
OXFORD .TOKYO
142
THE ALGEBRAIC STRUCTURE OF CROSSED PRODUCTS
G regory KARPILOVSKY Department of Mathematics University of the Witwatersrand Johannesburg, South Africa
1987
NOTH-HOLLAND -AMSTERDAM
NEW YORK
OXFORD .TOKYO
@
Elsevier Science Publishers B.K, 1987
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.
ISBN: 0 444 70239 3
Publishers:
ELSEVIER SCIENCE PUBLISHERS B.V. P.O. BOX 1991 1000 BZ AMSTERDAM THE NETHERLANDS
Sole distributors of the U. S.A. and Canada:
ELSEVIER SCIENCE PUBLISHING COMPANY, INC 52 VAN DER BI LT AVE NUE NEW YORK, N.Y. 10017 U.S.A.
Lihrav of Congress Cncalo~ngin-PubliationLhts
Karpilovsky, Gregory, 1940The algebraic structure of crossed products. (North-Holland mathematics studies ; 142) (Notas de matematica ; 118) Bibliography: p. Includes index. 1. Von Neumann algebras--Crossed products. I. Title. 11. Series. 111. Series: Notas de matematica (Rio de Janeiro, Brazil) ; no. 118. QAl.N86 no. 118 [QA326] 510 s (512l.551 87-9033 ISBN 0-444-70239-3 (U.S.)
PRINTED IN THE NETHERLANDS
FOR HELEN, SUZANNE and E L L 1 0 E t
This Page Intentionally Left Blank
vii
Preface
In the past
15 years, the theory of crossed products has enjoyed a period of
vigorous development.
The foundations have been strengthened and reorganized
from new points of view, especially from the viewpoint of graded rings. The purpose of this monograph is to give, in a self-contained manner, an upto-date account of algebraic structure of crossed products.
Although no claim to
completeness is made here, one of our goals is to tie together various threads of the development in an effort to convey a comprehensive picture of the current state of the subject. The present monograph is written on the assumption that the reader has had the equivalent of a standard first-year graduate course.
Thus we assume a famil-
iarity with basic ring-theoretic and group-theoretic concepts and an understanding of elementary properties of modules, tensor products and fields.
For the con-
venience of the reader, a chapter on algebraic preliminaries is included. chapter provides a brief survey of topics required later in the book.
This
A syste-
matic description of the material is supplied by the introductions to individual chapters.
There is a fairly large bibliography of works which are either direc-
tly relevant to the text or offer supplementary material of interest. A
word about notation.
As is customary, Theorem 3 . 4 . 2 denotes the second
result of Section 4 of Chapter 3 ; however, for simplicity, a l l references to this result within Chapter 3 itself are designated as Theorem 4.2. I would like to express my gratitude to my wonderful wife for the tremendous help and constant encouragement which she has given me in the preparation of this book.
I am also grateful to Nan Alexander, who smoothed over many technical
problems for me with her friendly advice.
My thanks to D S Passman for sending
viii
me some reprints of his fundamental work on crossed products.
Preface
Finally, I w i s h to
thank Lucy Rich for her excellent typing. Gregory Karpilovsky
ix
Contents
vii
PREFACE CHAP'FER 1.
PRELIMINARIES Notation and terminology 2. Projective, injective and flat modules 3. Artinian and Noetherian modules 4. Group actions Cohomology groups and group extensions 5. Some properties of cohomology groups 6. 7. Matrix rings and related results 1.
CHAPTER 2.
GROUP-GRADED ALGEBRAS AND CROSSED PRODUCTS: GENERAL THEORY 1. 2. 3. 4. 5. 6. 7. 8.
CHAPTER 3.
CHAPTER4.
1 11 19 25 28
41 46 59
59 71 78 92 103 119 125 147
THE CLASSICAL THEORY OF CROSSED PRODUCTS
151
1. Central simple algebras 2. The Brauer group Classical crossed products and the Brauer group 3.
161 169
CLIFFORD THEORY FOR GRADED ALGEBRAS
181
Graded modules Restriction to Al Graded homomorphism modules Extension from A , Induction from A ,
181
1. 2. 3. 4.
5.
CHAPTER 5.
Definitions and elementary properties Equivalent crossed products Some ring-theoretlc results The centre of crossed products over simple rings Projective crossed representations Graded and G-invariant ideals Induced modules Montgomery's theorem
1
151
186 191 196 205
PRIMITIVE AND PRIME IDEALS OF CROSSED PRODUCTS
223
Primitive, prime and semiprime ideals 2. Primitive ideals in crossed products 3. Prime coefficient rings 4. Incomparability and Going Down 5. A Going Up Theorem 6. Chains of prime and primitive ideals
223 225 230 246 259 267
1.
Contents
X
CHAPTER 6.
SEMIPRIME AND PRIME CROSSED PRODUCTS 1.
2. 3. 4.
5.
Coset calculus A-methods The main theorem and its applications Sufficient conditions for semiprimeness Twisted group algebras
2 17 285
294 305 3 11
BIBILIOGRAPHY
331
NOTATION
339
INDEX
345
1
1 Preliminaries
I n t h i s chapter we consider s e v e r a l p r i m a r i l y u n r e l a t e d b a s i c t o p i c s t h a t we s h a l l These i n c l u d e p r o j e c t i v e modules,
need i n v a r y i n g d e g r e e t h r o u g h o u t t h e book.
Later
group a c t i o n s , cohomology groups and group e x t e n s i o n s and m a t r i x r i n g s .
c h a p t e r s w i l l t r e a t v a r i o u s a s p e c t s of t h e s e t o p i c s i n g r e a t e r d e t a i l and d e p t h . Here w e s h a l l be i n t e r e s t e d o n l y i n d e r i v i n g e l e m e n t a r y p r o p e r t i e s , i n i n t r o d u c i n g v a r i o u s a u x i l i a r y c o n c e p t s , and i n developing some i m p o r t a n t n o t a t i o n a l conventions.
1. NOTATION AND TERMINOLOGY I t i s important t o e s t a b l i s h a t t h e o u t s e t v a r i o u s n o t a t i o n a l conventions t h a t
For convenience, we d i v i d e t h e r e l e v a n t i n f o r -
w i l l be used t h r o u g h o u t t h e book. mation i n t o s u b s e c t i o n s .
Maps and diagrams X
Let
Y
and
be a r b i t r a r y sets.
A
map
f :x-Y
x E X a unique element y
i s a f u n c t i o n t h a t a s s o c i a t e s w i t h each element
f
:
x
I-+
Let
y.
y,
This
Y
X+
f :
is denoted by
t h e image of
and
gof.
g
:
Y 4Z gaf
The map
(qof, Given
f
:
X
-f
Y
and
g
:
X
3:
-+
f,
be maps.
i s denoted by
The composite map
Y,
y = f(xc).
X+
Y--t Z
i s g i v e n by t h e r u l e (XI =
z, fzg
under
6
grf
(5) 1
t h e map :
x-
YXZ
for all
x
E
X
CHAPTER 1
2
i s d e f i n e d by
The
identity m p lX
:
x- x
i s d e f i n e d by
X (2)=
1
Y
If
f :
X
X'
of
X.
The
X,Y
a r e s e t s , we w r i t e
-+
f(X'
is any map, t h e n
i n c h s i o n map
i
:
X'+
for all
3:
denotes t h e
X
r e s t r i c t i o n of
i s d e f i n e d by
i(x)
f
= 3c
3: E
X
to a subset for a l l
x E X'. If
X is
elements of by
Y.
a proper s u b s e t of
1x1.
Y The
X If
n o t contained i n
5Y if X X 5Y , Y - X X.
i s a s u b s e t of
Y
and
X cY
if
d e n o t e s a s u s u a l t h e s e t of a l l
The c a r d i n a l i t y of t h e s e t
X
i s denoted
diagram
Z signifies that
X,Y,
and
Z a r e sets and t h a t f
:
X
+
Y, k
:
X - + Z,g
:
2
4Y
The diagram
commutes, o r i s s a i d t o be commutative, i n c a s e f = g o k .
An a r b i t r a r y diagram i s
commtative i f we g e t t h e same composite maps whenever w e
a r e maps.
f o l l o w d i r e c t e d arrows along d i f f e r e n t p a t h s from one s e t t o a n o t h e r s e t i n t h e diagram.
Groups Unless e x p l i c i t e l y s t a t e d o t h e r w i s e , a l l groups a r e assumed t o be m ultiplicat
NOTATION AND TERMINOLOGY
We use the symbol
yroup
n
yroup of order
is denoted by
N d G if N
We write
F*.
H
if there is a subgroup
=
X is a subset of G, C G ( X )
cyclic
A
G and say that G
splits
G such that
of
N - H and N n H
1
=
and N (X) will denote the centralizer and
G
By the normal closure of
X in G, respectively.
normalizer of
X is
If
‘n.
is a normal subgroup of
G If
is denoted by
G, < X > will denote the subgroup of G generated by X.
a subset of
over N
for both the identity element and identity subgroup of a
1
The multiplicative group of a field F
G.
3
understand the smallest normal subgroup of
G
X.
containing
As
X
G we
in
usual, Z ( G )
G.
denotes the centre of
G is finite, the least common multiple of the orders of the elements of
If
G is called the exponent of G. Given
Y are subsets of G , then
In particular,
X and
[X,Yl is defined by
G’ for
G.
As
is customary, we
[G,G].
G is of finite order, G is called a torsion group,
If every element of G
for the commutator ~ - ~ y - ~ qIf.
is the commutator subgroup of
IG,Gl
also frequently write
while
k,yI
z , y E G, we write
is torsion-free if all its elements, except for
1, are of infinite
order. Let H
be a subgroup of
from each left coset xH
G.
A
subset of G
containing just one element
is called a l e f t transversal for
H
in
G , and right
transversals are defined correspondingly. Let f
:
G-+
H
understand any map
be a surjective group homomorphism.
g
:
H
G
{Gi}iEI
be a family of groups and let
Then a typical element of
f
we
such that
f o g = lH and g(1) Let
By a s e c t i o n of
rIiEIGi is
(gi)
=
1
niEIGi
be their direct product set.
with
g
E
Gi
for all i E I and
CHAPTER 1
4
(9.)= ( g ; )
i f and o n l y i f
(gi)
two elements
and
gi
(g!)
=
g;
niEIGi @ G iEI i
d i s t i n c t from groups
1
of
Define t h e p r o d u c t of
(gig;)
=
d i r e c t product of t h e groups
becomes a group c a l l e d t h e
subgroup
I.
E
by t h e r u l e
( g i ) (g;) Then
i
f o r each
niE1Gi
c o n s i s t i n g of a l l
(g.1
Gi.
w i t h f i n i t e l y many
The
g
i
i s c a l l e d t h e d i r e c t sum ( o r r e s t r i c t e d d i r e c t product) of t h e
Gi
A sequence of groups and homomorphisms
i s s a i d t o be exact a t
Gi
if Kerfi
The above sequence i s c a l l e d
exact i f
Imfi-l
=
i t i s e x a c t a t e v e r y group.
L
1-G i s e x a c t i f and o n l y i f
In particular,
H
f i s i n j e c t i v e , while G - f t H - 1
i s e x a c t i f and o n l y i f
f
is surjective.
The e x a c t n e s s of
f i s an isomorphism.
i s equivalent t o the f a c t t h a t
I f w e a r e given a three-
term e x a c t sequence
l - G I L G also c a l l e d a
short exact sequence, w e see t h a t G1
Let
N
2
f(G1)
be a normal subgroup of
morphism.
A G - I
G
and
2
G and l e t f
G2/f(C1) : G+
G/N
G/N-+
1
Then t h e s h o r t e x a c t sequence
1-
N
A
G
i s a l s o c a l l e d t h e natural e m c t sequence.
b e t h e n a t u r a l homo-
NOTATION AND TERMINOLOGY
f
A homomorphism
:
G
-+
H
5
i s s a i d t o factor through a group
B
i n case
t h e r e i s a commutative diagram
G
f
\/" f
I n t h i s c a s e , one a l s o s a y s t h a t t h e homomorphism
G--t B .
morphism
G-+ G / N
homomorphism
f
Thus a homomorphism
G
H
-.+
f a c t o r s through t h e n a t u r a l
5 Kerf
N
i f and o n l y i f
:
f a c t o r s t h r o u g h t h e homo-
Kings and modules 1#
A l l r i n g s i n t h i s book a r e a s s o c i a t i v e w i t h
assumed t o have t h e same i d e n t i t y element a s assumed t o p r e s e r v e i d e n t i t y elements.
0 and s u b r i n g s o f a r i n g
R.
R
are
Each r i n g homomorphism w i l l be
R
w i l l b e denoted by
t-+
n - 1 i s a r i n g homo-
The c e n t r e of
Z(R).
R
Let
be a r i n g .
The map
-+
R
n
d e f i n e d by
morphism whose image i s c a l l e d t h e prime s u b r i n g of
d
m
f o r a unique
R
The r i n g
R
Let
be a r i n g .
n.
J
p o t e n t , while
of
e e 1
= 0, where
2
= e e 2
= 1
r
An element An i d e a l
J
6
R
Jn
i s t h e p r o d u c t of
e2
=
e.
J
Let
1c =
0
or
y = 0
for all
E
1
n
times.
e ,e
Two idempotents
r,y
n
with i t s e l f
A nonzero idempotent i s
0.
=
0
i s n i l i f e v e r y element of
R
in
rn
i s nilpotent i f
0 and
R.
f o r some
J
is nil-
such t h a t
e
An element
a r e orthogonal if
2
primitive i f it c a n n o t b e w r i t t e n as a
sum of two nonzero o r t h o g o n a l idempotents. implies
and denoted by c h a r R.
is nilpotent i f there is a positive integer
i s idempotent i f
R
R
c a l l e d t h e characteristic of
is s a i d t o b e simple i f t h e r e a r e no i d e a l s o t h e r t h a n
positive integer
J"
2 0,
i t s kernel i s an i d e a l
R;
We s a y t h a t
R
i s a domain i f
q = O
R.
{Rili 6 I} be a f a m i l y of r i n g s and l e t
R
be t h e d i r e c t p r o d u c t s e t
CHAPTER 1
6
ni&.
It is straightforward to verify that R direct product of the family R
of
R by the rules:
We can define addition and multiplication on
onto Ri
{Rili E I}.
is a ring homomorphism:
For all
i
as the
I, the projection pri
E
the injections
Xi
:
R preserve
Ri+
1 and s o are not ring homomorphisms.
addition and multiplication but not
Assume that I is a finite set, say I = {1,2, ...,n } . if and only if R
R
is a ring; we shall refer to
Then R
9
R1
x
... x R
contains pairwise orthogonal central idempotents I -
such that 1= e 1
+
...
ie
n
and
Rei
9
Ri qua ring (1 G i
G n)
~ l R-modules l in this book are left R-modules unless otherwise specified, and each
V
R-module
1.V
An R-module summand of
0 and If
...
is assumed to be unital, i.e.
V
=
v
for all v E V
is compZetely reducible if every submodule of
V.
A
V
is a direct
V is irreducible if its only submodules are
nonzero R-module
V. V
is an R-module, we shall write
End(V)
for the endomorphism ring of
R ! I . The
f,g
elements of
t End(V),
f+ g
End(V) are R-homomorphisms from R and fg are defined by
V to V ; given
R
(f+g, ( U ) = A
f(Vf
+
( f g f(V)
g(V)
{ V i l i E 11
direct sum of a family of modules
=
iv E V)
f(g(uf)
is written
@Vi
whenever the
index set I is not pertinent to the discussion. An R-module elements x. E V
'L
is said to have an R-basis such that each
a: E V 3: =
can be written as a finite sum C r .x
z i
with uniquely determined coefficients r . E R. R-modules.
Expressed otherwise, V
{xili E 11 if there exist
is free if and only if
a direct sum of copies of the left R-module
#.
V
Such modules V
are called free is isomorphic to
Here the additive group of
RR
7
NOTATION AND TERMINOLOGY
R,
coincides with t h a t of t h e product Let
r
and t h e p r o d u c t of
E
m
R with
E R
i s d e f i n e d as
rm i n t h e r i n g R.
V # 0
be an R-module.
A
composition s e r i e s f o r
V is
a series of sub-
modules of t h e form
v = v
3 v
0
such t h a t f o r each
i>
2
3...3vk=o vi-l/Vi
t h e f a c t o r module
1,
W
i r r e d u c i b l e R-module
3 v 1
is irreducible.
i s c a l l e d a composition factor of
V
w = vi-1/v i Let
R
be a r i n g .
V
be w r i t t e n a s a d i r e c t sum called
=
X@ Y
if
f o r some
V # 0
Then an R-module
is called
An
i
decomposable i f
of nonzero submodules;
i t can
P
otherwise,
is
indecomposable.
Given a p a i r of r i n g s
R,S,
V is
w e say t h a t
an
l e f t R-module and a r i g h t S-module, w i t h t h e a c t i o n s of
(R,S)-bimoduZe if V R
and
S
V
on
is a
commut-
ing
(r E R , v
E
V, s E
Tensors products Given a r i g h t R-module
V
and a l e f t R-module
W,
t h e a b e l i a n group
V@W R called the
tensor product of
f r e e &module w i t h
VxW
V
W,
and
as a basis;
i s defined a s follows.
t h e n e a c h element of
F
Let
F
be a
can be u n i q u e l y
w r i t t e n i n t h e form (Zij
CZij ( V i , W j ’
w i t h f i n i t e l y many
z
i.i
d i s t i n c t from
g e n e r a t e d by a l l e l e m e n t s of t h e form
0.
Let
T
E
z,vi
E V,W. E
be t h e subgroup of
s
F
W)
S)
CHAPTER 1
8
where
v,vi 6 V , w,w E W , i
group
F/T,
(v,w)
The image of V
R.
r
and
V 8W R
Then
under the natural homomorphism F
With this notation, the z-module V @ W R
8 w.
is defined to be the factor
-+
F/T
consists of all finite sums
zvi 8 wi and the elements V 8 U
is denoted by
v, w .
(Vi E
E
W)
satisfy the relations
(v+v)@w=v o w + v 8 w 1
2
V@(W+W) 1
v The map
f
:
v
w-+
X
I.'@ R
w
8
2
MJ =
=v@w
+v@w
vr 8 w
defined by
f(V,W) = V @ t:
satisfies the following
properties: (i)
f(vl+ V 2 , w )
=
~ ( u , 1w+ W 2 1
(ii)
(iii)
f(v
=
: I/ X
ism
f(v,wl)+ f(v,w2)
f(vr,w) = f ( v , r ~ )
Furthermore, if g g
,W) + f ( V 2 , W )
A
W4
is any baZanced map from
v
X
w
into A ,
i.e.
if
satisfies (i), (ii), and (iii), then there is a unique homomorph-
$ : V 8 W-+
A
which renders commutative the following diagram:
R
Assume that
V
is an (S,R)-bimodule.
Then
@ R
W can be regarded as a (left)
S-module by putting
In particular, if
V
and
W are
modules over a commutative ring
R,
then V 8 W R
NOTATION AND TERMINOLOGY
9
is an R-module. The following standard properties of tensor products are direct consequences of the definition. 1.1. PROPOSITION.
Let
be an (S,R)-bimodule.
The11
V @ R E V R 1.2.
PROPOSITION.
bimodule.
Let
1.3. PROPOSITION. (VA)
(vA PP w
I-r
If
v
V be an (S,R)-
V @ R
w1 0 ... a3 V
@
R
w,
(as S-modules)
are two free modules over a commutative ring R
W
and
(respectively, (w ) ) P
is an R-basis of
V
(respectively, W), then
S
and let
V @ W. R
Let R
1.4. PROPOSITION.
R-module with
1.5. PROPOSITION. module of
V
and
1
Then S @ W
as an R-basis.
R
2
{ l @ w w l , l @W
module with
be a subring of a ring
{ w ,w ,. . . ,w 1
and
... 0 W,’ =
is an R-basis of
)
be left R-modules and let
Then V @ (W1a3 R
and
WI,W*,...,W,
as S-modules
Let
V
2
,...,1 @ w 1
is a free left S-
as an S-basis.
be a right R-module, W
W’ a submodule of W.
W be a free left
If
V’
V’
a left R-module,
a sub-
is a direct summand of
V
W‘ a direct summand of W, then the canonical homomorphism V ’ Q W ’ - +V @
R
R
w
8’ 8 W ’ under this homomorphism is a direct
is injective and the image of
R
summand of the &nodule V @ W. R 1.6. PROPOSITION.
module, M rings.
(Associativity of the tensor product).
an (R,S)-bimodule, and
Then
L @ M
R
N
Let R
a left S-module, where
is a right S-module, M @ N
be a right R-
and
S are
a left R-module, and we have
S
(L@M) @ N r L @ (MQNN) R S R S If, furthermore, L
L
is an ( R ,R) bimodule and 1
above is an isomorphism of ( R ,S )-bimodules. 1
1
N
as &nodules
is an (S,S )-bimodule, then the 1
CHAPTER 1
10
Algebras Let
R
be a commutative r i n g .
By an R-algebra w e u n d e r s t a n d a r i n g
A
which i s
a t t h e same time an R-module such t h a t
rs = ( r * l ) x f o r
I t f o l l o w s d i r e c t l y from t h e d e f i n i t i o n t h a t
r
t h a t t h e map s e n d i n g of
A.
f : R-+
Conversely, i f
R
r-1 i s a homomorphism of
to
Z(A)
r E R, s E A
Z(A)
i n t o the centre
is a homomorphism, t h e n
A
and
can b e r e g a r d -
ed as an R-algebra by s e t t i n g
rx A
I n t h i s way any r i n g
2
morphism Let
v
Z(A), n
-+
=
f(r)z
for a l l
r E R , x f? A.
can be viewed a s a z - a l g e b r a w i t h r e s p e c t t o t h e homo-
!-+ n'l.
be an R-module.
R-+
Then t h e mapping
End(l.')
which sends
r
E R
R t o t h e endomorphism
ul-+
c e n t r e of
Hence t h e r i n g
End(V).
ru
i s a r i n g homomorphism whose image belongs t o t h e
End(V)
R
can b e r e g a r d e d a s an R-algebra by
R
setting
(rf)( u )
rf(u)
=
We s a y t h a t two R-algebras
A
r
for all and
B
E
R , f E End(V1, u E V R
a r e R-isomorphic
a l g e b r a s ) i f t h e r e e x i s t s a r i n g isomorphism
A
-+
E
( o r isomorphic a s R-
which i s a l s o an isomorphism
of R-modules. An R-algebra is s a i d t o be R-free i f it i s f r e e as an R-module.
If
R
is a
f i e l d , then c l e a r l y e v e r y R-algebra i s R - f r e e . Suppose t h a t
A
and
A
1
are R-algebras.
Then one c a n d e f i n e an R-algebra
2
s t r u c t u r e on t h e R-module
A
8 A 2 by t h e formula
1 R
( a 1 8 a 2 )(a' 8 a ' ) = a a' 8 a a' 1
The R-algebra
Al f A 2
A i , i = 1,2.
2
2
i s s a i d t o b e t h e t e n s o r p r o d u c t of a l g e b r a s
I t p o s s e s s e s an i d e n t i t y element e q u a l t o
element of
1 1
2
e
0 e 2 , where
Moreover, t h e mappings
ei
A
and
A
1
i s the identity
.
11
PROJECTIVE, INJECTIVE AND FLAT MODULES
lA1
x c - t xA 18 ?2e
and
a r e homomorphisms of R-algebras such t h a t
f,(a1)f2(a2) = f2(a2’f, (a1) fi are
The homomorphisms
a
for a l l
c a l l e d canonical.
‘A l ,
The t e n s o r p r o d u c t
a2E A2 A
8 A2
is
1 R
c h a r a c t e r i z e d up t o isomorphism by t h e f o l l o w i n g u n i v e r s a l p r o p e r t y : 1 . 7 . PROPOSITION.
i
homomorphism,
all
a
€
A
1
morphism
A
Let =
and
b e a n R-algebra and l e t @
1,2, such t h a t
a
E
A2.
(a ) 1
1
$
and
2
qi (a 1 2
:
Ai+
A
b e an R-algebra
are permutable i n
A
for
Then t h e r e e x i s t s one and o n l y one R-algebra homo-
1
f
:
A
8 A2+
1 R
A
which r e n d e r s commutative t h e diagram
2 . PROJECTIVE, INJECTIVE AND FLAT MODULES Let
R
be a r i n g and l e t
0
4
x-+U
be a s h o r t e x a c t sequence of R-modules.
a(X) i s a d i r e c t summand of ( i ) There e x i s t s a
Y.
y E HomR(Y,X)
Y
B --+
2-
0
We s a y t h a t ( * )
(*)
i s a s p l i t sequence i f
T h i s i s e q u i v a l e n t t o e i t h e r of t h e f o l l o w i n g : such t h a t
ya
=
lX, where
lX d e n o t e s t h e
1L
CHAPTER 1
identity map on
X.
(ii) There exists a The homomorphisms Let
V
y
6 E HomR(Z,Y) such that B6
=
1
z
6 are called splitting homomorphisms.
and
be an R-module.
Then
is called projective, if for any given
diagram with exact bottom row
V
there exists an R-homomorphism y
V
Proof.
=
ny.
is projective
(ii) Any exact sequence (iii) V
6
The following conditions are equivalent
2.1. PROPOSITION.
(i) An R-module
such that
a
0 -+ X - - t
R
Y --+
V
-+
0 of R-modules splits
is a direct summand of a free R-module. (i) =. (ii):
Since
B is surjective and 'I
the following commutative diagram:
is projective, we have
13
PROJECTIVE, INJECTIVE AND FLAT MODULES
y
Thus
i s a s p l i t t i n g homomorphism and (ii)f o l l o w s .
( i i ) * ( i i i ) : Take a s
Y
s p l i t t i n g homomorphism. ( i i i )* (i):
(y.1
E
:
F
V+
Y(V)
Then
Y
and l e t
F.
TI
:
t h e n a t u r a l i n j e c t i o n so t h a t
B
a r e R-modules and t h a t
v-+
:
Y and
s u r j e c t i v e homomorphism, r e s p e c t i v e l y .
F+
V
X-+
be a
F,
and l e t
be t h e n a t u r a l p r o j e c t i o n Now, suppose
TIE = ly.
ci :
Y
Y.
is a d i r e c t summand of
Then we can l e t
v4
:
be a d i r e c t summand of t h e f r e e R-module
be a b a s i s f o r
and
Y
V
Let
a s u i t a b l e f r e e R-module,
Y are a
X
and
homomorphism and a
W e t h e n have a n a t u r a l map
from
F
Y, and we can d e s c r i b e t h e s i t u a t i o n p i c t o r i a l l y a s f o l l o w s :
to
F
Now f o r each g e n e r a t o r
yi
x. E X
@ n ( y.)
exists
A(y.1
a6
=
=
with
6lT(y.) E Y and because
E F, =
a
a(x.). Hence, i f we d e f i n e
is surjective there
a
:
F
X
-+
by
xi, t h e n t h e o u t e r p o r t i o n of t h e diagram i s commutative, t h a t i s ,
an.
Finally,
E
:
V-+
F
so t h a t
6E :
V
-+
X
and
.
Thus t h e lower D o r t i o n of t h e diagram i s commutative, and by d e f i n i t i o n w e
V
conclude t h a t
2.2.
COROLLARY.
i s projective.
t i v e i f and o n l y i f Proof.
Let
V
Let
V
Vl,V2,
be a f i n i t e l y g e n e r a t e d R-module.
Then
V
is projec-
i s a d i r e c t summand of a f r e e R-module of a f i n i t e rank.
...,V n
be a g e n e r a t i n g s e t f o r
V
and l e t
F
be a f r e e
CHAPTER 1
14
module freely generated by
3 1
surjective homomorphism F
summand of
F-+
,x ,...,xn 2
V.
Then the map V
Thus if
by Proposition 2.1.
sition 8.1, the result follows.
.
X
.
e
Ui
determines a
is projective, then V
is a direct
The converse being true by virtue of Propo-
'
If we reverse the direction of all the arrows in the definition of a projective module, we then obtain the definition of an injective module. R-module
V
Thus a given
is i n j e c t i v e , if for any given diagram with exact bottom row
\ P
\ \y
\ a!
Y-0 there exists an R-homomorphism Y 2.3. PROPOSITION.
(i) An R-module
(i)
$ = YU.
The following conditions are equivalent: V
is injective
(ii) Any exact sequence 0 Proof.
such that
\ AX
-+ +P'
a
(ii): Since c1
B X-+ Y -+
0
of R-modules splits.
is injective, we have the following commutative
diagram:
V
15
PROJECTIVE, INJECTIVE AND FLAT MODULES
y
Hence
is a splitting homomorphism and (ii) follows.
(ii) =) (i):
Consider a commutative diagram with exact bottom row
\ P
\ \
0-Y
is the inclusion map, in which case Y
We may harmlessly assume that
X , by hypothesis.
direct summand of by
.
y(y+z) = B f y ) , y E V , z E 2.
required.
2.4. PROPOSITION. @
(i)
Vi
{vili
Let
X
Write
y
Then
X
=
Y @ Z
r
and define
is projective if and only if each
:
X+
V
8 = Ya, as
is an R-homomorphism and
6 11 be a family of R-modules.
is a
Then
Vi
is projective
V.
is injective.
iei
rlVi
(ii)
i€I Proof.
is injective if and only if each
(i) If
V = @ Vi,
if and only if so
ieI is each Vi.
and only if each
Vi
V =
(ii) Let
rlVi
iei
then
V
is a direct summand of a free R-module
CL
:
Vi and
and let TI : V -
exists an R-homomorphism B(Z) =
(B.(z)).
sition 2.3,
V
Then
B
8;
!?-homomorphism.
y
:
X-
:
X-
Vi
:
Vi-+
V
with
TIi
be the natural
Vi
Suppose that each
is injective and
Then, for each =
Bia.
Define
is an R-homomorphism with
=
i
8
E :
I, there
x--t
V
by
ly and so, by Propo-
is injective.
Conversely, assume that
Define
ui
X be an injective R-homomorphism.
V+
is projective if
is projective.
projections and injections respectively. let
V
Hence, by Proposition 2.1,
V
is injective and let a : Vi
Then there exists an R-homomorphism
Vi by y
=
~ ~ 8Then . ya =
TI.(@)
z i -- 1vi
= 7T.p
8
:
-+
X+
X be an injective V
with
ui
=
Ba.
CHAPTER 1
16
and thus, by Proposition 2.3,
@
:
-
Let
R be a ring, V,V'
V
V', $
:
W - + W'
.
Vi is injective.
two right R-modules, W,W'
two R-homomorphisms.
two left R-modules, and
Then the map
defined by
is a homomorphism of additive groups. modules and
@
Moreover, if
is an (S,R)-homomorphism,then
@ @ $
V and V '
are (S,R)-bi-
is an 5'-homomorphism of the
$
left S-modules V @ W and V ' W'. Let R be a ring and let V be a right R R-module. The module V is said to be f l a t if for every injective homomorphism
f
:
W'+
Fi of left R-modules, the homomorphism l v Q D f :
is injective.
{Vili
E
V@W'-+V@W R H
It is an easy consequence of the definition that, given a family
I} of right R-modules, the module
@ .
Vi is flat if and only if each V .
-LEI
is flat. The following elementary result will be useful for our subsequent investigations. 2.5. PROPOSITION.
Proof.
Every projective module is flat.
Since projective modules are isomorphic to direct summands of free
modules (Proposition 2 . 1 ) , flat.
So
modules.
assume that f
it suffices to show that the regular module :
W'-+
RR is hi is an injective homomorphism of left R-
The maps
are R-isomorphisms which render commutative the following diagram
PROJECTIVE, INJECTIVE AND FLAT MODULES
17
0
Hence
18f
is injective and RR
.
is flat.
We close this section by proving
2.6. PROPOSITION.(Dual basis lemma).
An R-module
v
of elements in Hom(V,R) such that every R
{Uili E I }
is projective if and only
{uili E I } of elements in V and a family {fili E I }
if there exists a family
with finitely many
V
0.
distinct from
f.(V)
E
V
can be written in the form
Moreover, if
V
is projective,
can be any generating set.
Proof.
{Uili E I }
Let
module freely generated by
be any generating set of
{ziliE 11, and let f
f(z.1
:
let F be a free R-
V,
F
+
V
be the surjective
Owing to Proposition 2.1, i for all i E I . V is projective if and only if f splits, that is, if and only if there exists a
homomorphism defined by
homomorphism g Assume that with
fg
=
lv
-
:
V
V If
+
= U
F such that fg
is projective.
u E V,
=
lv.
Then there exists a homomorphism g
then write
g(u) =
C iEI
with
ri
E R
depending on 0.
The family
r.z zi
{fili E 11,
where
:
V+
F
CHAPTER 1
18
satisfies
Conversely, let Then
{Ui
Ii E
11 and
{Ui\i E
I} generates V
and a homomorphism
g
V-+
:
{fili t I} be as described in the theorem.
and there is a surjective homomorphism f
:
F
-+
F, where
and
Since fg = l v , V
is projective.
.
We close this section by quoting the following standard facts for the proof
of which we refer to Bourbaki (1959). 2.7.
Let M
PROPOSITION.
(i) If f
:
be a flat right R-module.
Y is a homomorphism of left R-modules, then
X--t
Ker(1, @ f) = M @ (Kerf)
R
and Im(1, 8 f) = M @ ( I m f )
R
(ii) If N ’ , ” ’
are two submodules of a left R-module N ,
M 2 . 8 . PROPOSITION.
Let
Q (N’
n N ‘ ! ) = ( M BN ’ ) n ( M BN “ )
M be
a right R-module.
R
R
then
R
Then the following properties
are equivalent: (i) M
is flat
(ii) For any finitely generated submodule N ’
of a left R-module
ical homomorphism
1,ej
:
M@”-+M@N
R
R
(j being the inclusion map) is injective. (iii) For every exact sequence of left R-modules and homomorphisms
N
the canon-
ARTINIAN AND NOETHERIAN MODULES
19
t h e corresponding sequence
a M@
M 8 N' R
R
N -@?!+M@N" R
i s exact. For e v e r y f i n i t e l y g e n e r a t e d l e f t i d e a l
(iv)
A
of
R,
t h e c a n o n i c a l map
M@A--MA
m 8 a r m a i s an isomorphism.
3. ARTINIAN AND NOETHERIAN MODULES i s s a i d t o be a r t i n i a n ( r e s p e c t i v e l y , noetherian) i f e v e r y descen-
V
An R-module
d i n g ( r e s p e c t i v e l y , ascending) c h a i n of submodules of
is artinian (respectively, noetherian).
n Z RV
V =
i
f o r some
Vl,Vp
The module
,...,U n
as b e i n g f i n i t e l y cogenerated i f f o r e v e r y f a m i l y
V
n Vi
with
=
stops.
0, t h e r e e x i s t s a f i n i t e s u b s e t
in
i s s a i d t o be f i n i t e l y
V.
We s h a l l r e f e r t o
{V,li J
PROPOSITION.
V
Let
be an R-module.
RR
v
I]
E
I
of
V
of submodules of
n V . = 0. jEJ
'
such t h a t
iEI
3.1.
R
The r i n g
a r t h i a n ( r e s p e c t i v e l y , noetherian) i f t h e r e g u l a r R-module
i t s e l f is called
generated i f
V
Then t h e f o l l o w i n g c o n d i t i o n s a r e
equivalent (i) V
is artinian
( i i ) Every nonempty s e t of submodules of (iii) Every factor-module of
( i )* ( i i ) : Denote by
Proof. assume t h a t
x
W'
with
W
2
nonempty.
W'
f o r each
i n f i n i t e descending c h a i n
{Vili
i s f i n i t e l y cogenerated.
X
E
I}
V
a nonempty s u b s e t of submodules of Then, f o r e a c h
W E
and
X, t h e s e t
Hence, by t h e axiom of c h o i c e , t h e r e i s a f u n c t i o n
W E
X.
W 3 W' 2 W" 2
Fixing
...
(iii): It suffices t o verify that i f
(ii)
h a s a minimal e l e m e n t
does n o t have a minimal element.
{W' E XlW' c W} i s WC-L
V
V
i s a f a m i l y of submodules of
W E X,
w e t h e r e f o r e o b t a i n an
of submodules of
W is a
V with
submodule of
n V.
ieI
V.
'
=
W,
then
V
and
W =
n V j w j
20
CHAPTER 1
J
f o r some f i n i t e s u b s e t
of
I.
Put
Y = { n
v Kcr
E d '
Y
By assumption,
W =
Clearly,
is finite}
-
h a s a minimal element, s a y
n Vi
J
with
,jcJ
5I
J
and
finite.
n V ,j€J . i
(iii)* ( i ) : Suppose t h a t
V
h a s a descending c h a i n
v1 3 v 2
so t h e r e e x i s t s
n
n V
W =
of submodules and p u t
61 = V
with
.
n.
1 -
...
By assumption, Hence
V/W
Vn+i = Vn
i s f i n i t e l y cogenerated,
i
for
,..., ,
= 1,2
as
asserted. 3 . 2 . COROLLARY.
V
Let
be a nonzero a r t i n i a n module.
Then
V
has an irredu-
c i b l e submodule. Proof.
Apply P r o p o s i t i o n 3.1. f o r t h e s e t of a l l nonzero submodules of
3 . 3 . PROPOSITION.
V
Let
b e an R-module.
V.
=
Then t h e f o l l o w i n g c o n d i t i o n s a r e
equivalent : (i) V
i s noetherian
(ii) Every nonempty set of submodules of ( iii) Every submodule of
Proof.
V
V
h a s a maximal element
i s f i n i t e l y generated.
.
The proof of t h i s r e s u l t i s d u a l t o t h a t of P r o p o s i t i o n 3 . 1 and t h e r e -
f o r e w i l l be o m i t t e d .
3 . 4 . COROLLARY.
Let
i/
be a nonzero n o e t h e r i a n module.
V
Then
has a
maximal submodule. Proof.
Apply P r o p o s i t i o n 3 . 3 f o r t h e s e t of a l l p r o p e r submodules of
v. =
Note t h a t t h e p r e v i o u s r e s u l t i s a l s o a consequence of t h e f o l l o w i n g observation.
3.5. PROPOSITION.
Let
V
be a f i n i t e l y g e n e r a t e d R-module.
submodule of
V
is c o n t a i n e d i n a maximal submodule.
nonzero t h e n
V
h a s a maximal submodule.
Proof. submodule of
Let
V.
{x ,.. . , x p } be a g e n e r a t i n g s e t f o r Denote by
X
Then e v e r p r o p e r
In particular, i f
V
and l e t
t h e set of a l l p r o p e r submodules of
W
V
is
be a p r o p e r
V
which
21
ARTINIAN AND NOETHERIAN MODULES
W.
contain
wA
then
Vu
Let
x 1
This i s inductive:
i s a submodule.
Wx,
V =
If
,...,xp
W
=
lJ
V,
a contradiction.
W,
.
a s required.
X
Hence
WAi
Wu
Then
R
R
contains
R
1,
v
itself.
As
and t h e submodules a r e
Thus w e o b t a i n t h e
l e f t ideals.
3.6.
i s g e n e r a t e d by t h e s i n g l e element
xi.
f o r some
i s i n d u c t i v e a n d , by
I n p a r t i c u l a r , t h e r e s u l t above i s a p p l i c a b l e t o t h e r i n g l e f t R-module,
x,
i s a chain i n
T h i s i s a maximal submodule of
Z o r n ' s lemma, it h a s a maximal e l e m e n t . containing
E
r
1
and s o
xz.
t h e n each
,...,wA .
W.,,
be t h e l a r g e s t of t h e modules
(w,)
W and i f
it c o n t a i n s
R
Let
COROLLARY.
be an a r b i t r a r y r i n g .
Then any p r o p e r l e f t i d e a l of
I n particular,
R.
i s c o n t a i n e d i n a maxiinal l e f t i d e a l of
R
h a s a maximal
l e f t ideal.
3.7. PROPOSITION.
Assume that
v-
u-
0-
i s an e x a c t sequence of R-modules. i f both
U and
Proof.
U
W
be a r t i n i a n .
is a l s o a r t i n i a n .
V, W
module of
U i s isomorphic t o
Since
U
U and
5V
and
W be a r t i n i a n . W = V/U.
W
u
V,
i s isomorphic t o a f a c t o r -
v
3 -
3 2 -
To prove t h a t
V
i s a r t i n i a n , we
... -3 vn -3 ... V.
Since
V/U
is artinian, there exists
such t h a t
vm + u = vm + i + u Because
a submodule of
Let
be a descending c h a i n of submodules of m
i s a r t i n i a n ( n o e t h e r i a n ) i f and o n l y
S i n c e e v e r y factor-module of
v1 an i n t e g e r
V
Then
is artinian.
Conversely, l e t may assume t h a t
0
are artinian (noetherian).
V
Let
w-
i s a r t i n i a n , t h e r e i s an i n t e g e r
v
n 2 m
n u = vn+i n u
Taking i n t o account modul-arity and t h a t
V
M
3 V n+i'
(i = 1 , 2 ,
..., )
(i =
...,)
such t h a t 1,2,
we have f o r each i = l , Z ,
...,
CHWTER 1
22
Vn = Vn n
vn+i + (vn+i
=
V
Hence
3.8.
V
PROPOSITION.
is
(vn+< + u) = vn+i + (vn n u)
U) = vn+i
@
. .. @ Vn.
Then
is a r t i n i a n (noetherian)
V.
and o n l y i f each
3.9.
v=
Let
n
The proof of t h e n o e t h e r i a n case i s d u a l .
is artinian.
COROLLARY.
(vn+Li) = vn
Assume t h a t
v#
0
.
. v
i s a r t i n i a n (noetherian) i f
i s e i t h e r a r t i n i a n o r noetherian.
Then
a d i r e c t sum of f i n i t e l y many indecomposable submodules.
Proof.
V
For each nonzero module
t h a t does n o t have a f i n i t e indecompos-
a b l e decomposition choose a p r o p e r decomposition
V = where
V'
v'
8 X'
h a s no f i n i t e indecomposable decomposition.
n o t a f i n i t e d i r e c t sum of indecomposable modules.
V
proving t h a t
X'@ X"
C
...
is
Then
Therefore t h e r e e x i s t i n f i n i t e chains
i s a sequence of p r o p e r decompositions.
X'
V # 0
Suppose t h a t
V
and
3
V' 3 V" 3
...
So t h e p r o p o s i t i o n i s
is n e i t h e r a r t i n i a n nor n o e t h e r i a n .
verified.
Let
3.10. PROPOSITION.
,.$ = v1 @
... @ v
with
vi
# 0
i
for
=
1,2, ...,n ,
and w r i t e
l = e Then
1
{e
,. . . ,e 1
< i < n.
potents i n
+
...
+ e
(ei E V . )
i s a s e t p a i r w i s e o r t h o g o n a l idempotents i n
Conversely, i f
R,
1
{e
,. . ., e 1
R
and
Vi
=
Rei,
I s a s e t of p a i r w i s e o r t h o g o n a l idem-
then
n
n
R ( C e i ) = @ Rei i=1 i=l Proof.
Given
r E R,
r = r.1
w e have
=
1-r = r e
particular,
ei = e.e
? - 1
+
... + e.ez n
+
... + re .
Hence, i n
23
ARTINIAN AND NOETHERIAN MODULES
e.e = dijei for all i , j . Furthermore, R e . c V . and z j 2.z @ Ren which implies that V . = Rei for all i.
proving that
...
,,J?= R e l 8
{el,.. .,e 1
Conversely, suppose that
is a set of pairwise orthogonal idem-
n
c ei. Then e 2 = e and eei = eie = e i for all i. i=l Hence Re = c Re on the right by e i' If Eriei = 0 then multiplication n j i=l gives r.e = 0 for all j . Thus Re = 8 Rei, as required. e
potents and put
=
n
I j
i=l
Assume that R
3.11. COROLLARY.
set
1
=
is artinian or noetherian.
Then there exists a
{ e ,,...,e 1 of pairwise orthogonal primitive idempotents of R n 1 ei. Furthermore,
with
i=l
# where each Rei
=
Rel @
... 8 Ren
is an indecomposable module.
Proof. Direct consequence of Propositions 3.10. and 3.9.
.
We next provide a criterion for a completely reducible module to be artinian and noetherian.
3.12. LEMMA.
We need the following preliminary observations
V
Let
be completely reducible. V
(i) Every homomorphic image of V
submodule of
is isomorphic to a submodule of
is isomorphic to a homomorphic image of V
(ii) Every submodule and every homomorphic image of (iii) If
V # 0
then W
Proof. (i) If W'
of
V.
(ii) Let
W
Then
V
V. is completely reducible.
is a submodule of
V/W'
V/W
and
1
ki"
V,
then
U is completely reducible.
V = W 8 W'
for some submodule
as asserted.
So
= V/W.
By (i), it suffices to
assume that
U
V /W
= 1
module of
U.
Then
V = V @ V 1
U
=
U @ (V + W)/W
(iii) Let
=
W 8 W'
W' = W 1 8 W 2
I/
for some submodule 2
of
V
v2
of
is a sub-
1
V.
Hence
is completely reducible.
u # 0 be an element of
submodule W
V
and so
and every
contains an irreducible submodule.
W be a submodule of V and let U
verify that
V
V.
Owing to Zorn's lemma, there exists a
maximal with respect to the property that u
for some submodule W'
of
V.
If W'
f o r some nonzero submodules W I , W p
of
W.
Write
is not irreducible, then V.
Since
CHAPTER 1
24
(wewl)n v 9 W @ Wi
it f o l l o w s t h a t
W.
maximality of 3.13. LEMMA.
v
(i)
.
i
= 1 or
is irreducible.
For a nonzero module
w = 2,
contrary t o the
the following conditions a r e equivalent:
I/
i s completely reducible
(ii) V
i s a d i r e c t sum of i r r e d u c i b l e submodules
(iii) V
i s t h e sum of i r r e d u c i b l e submodules
Proof.
( i i ) : Consider t h e c o l l e c t i o n of sets of i r r e d u c i b l e submodules
(i)
whose sum i s d i r e c t .
'b
of
i
for either
W'
Hence
( w @ w2i =
By Lemma 3 . 1 2 ( i i i ) , it i s nonempty a n d , by Z o r n ' s
{Y.),
lemma, t h e r e i s a maximal e l e m e n t , s a y
V = W@
and l e t
W'.
V'.
i r r e d u c i b l e submodule
{Yi}
i m a l i t y of
.
W' # 0
If
Hence
t h e n by Lemma 3 . 1 2 ( i i ) ,
W
+
V' =
W' = 0 and
Thus
i n t h i s collection.
C" 8
V = W,
W
Let
(iiil,
61'
= @
V
i
c o n t a i n s an
Vi), c o n t r a r y t o t h e max-
(@
as r e q u i r e d .
(ii) * ( i i i ) : Obvious ( i i i ) * ( i ):
W' of
module
W
+ W'
=
W
Let
V
W 8 61'.
V .
3
Vi E
Vi
V
E V
V.
and
d u c i b l e , we i n f e r t h a t
with
v 4 FIB W'.
V . n ( W @ W') # Vi. 3
V . n (W @ W') 3
(W' 8 V . )
W :1
COROLLARY.
3
Let
V # 0
=
=
0.
+
V = V
V, 1 6 i Because
Assume the
< n.
V
j
...
(ii) V
i s irre-
61' @
vj
0, c o n t r a r y t o t h e maximality of
be c o m p l e t e l y r e d u c i b l e .
W'.
.
Then t h e f o l l o w i n g con-
is artinian is n o e t h e r i a n
(iii) V
Proof.
V
Hence
d i t i o n s a r e equivalent: (i) V
+
Thus
v. = W @ 3
Thus
V = W @ W'.
By h y p o t h e s i s ,
i s an i r r e d u c i b l e submodule of
W @ W' +
3.14.
W n W ' = 0.
I t t h e r e f o r e s u f f i c e s t o show t h a t
9 W @ W' f o r some j and so
and t h e r e f o r e
By Z o r n ' s lemma, t h e r e e x i s t s a sub-
maximal w i t h t h e p r o p e r t y t h a t
c o n t r a r y and choose where
Y.
be a submodule of
.
i s t h e d i r e c t sum of a f i n i t e number of i r r e d u c i b l e submodules D i r e c t consequence o f Lemma 3 . 1 3 and C o r o l l a r y 3 . 8 .
.
25
GROUP ACTIONS
3.15.
Proof.
i s completely r e d u c i b l e , t h e n so i s e v e r y R-module.
17 be an R-module and l e t
Let
which sends
RU
fiR
If
COROLLARY.
to
p
by Lemma 3 . 1 2 ( i i ) .
i s a homomorphism.
W
V
Since
C
=
Then t h e map from
Thus
RR
to
i s completely r e d u c i b l e
RV
t h e r e s u l t f o l l o w s by v i r t u e of P r o p o s i t i o n
RV
UEV
=
3.2.
V.
ti E
wetherian.
V
i s s a i d t o b e of f i n i t e length i f
V
An R-module
i s b o t h a r t i n i a n and
We c l o s e t h i s s e c t i o n by q u o t i n g t h e f o l l o w i n g s t a n d a r d f a c t s ( s e e
Bourbaki (1959) ) .
3.16.
R.
PROPOSITION.
V
Then
3.17.
be a f i n i t e l y g e n e r a t e d module o v e r an a r t i n i a n r i n g
i s of f i n i t e l e n g t h .
PROPOSITION.
length.
v
Let
Then
V
(Krull-Schmidt theorem).
can b e w r i t t e n as a d i r e c t sum
ipdecomposable subnicdules.
n = k
t h i s k i n d , then
v=
Moreover, i f
v
Let
be an R-module of f i n i t e
n I.' = 6 V i ,
where V . a r e i=l i s a n o t h e r decomposition of
k @ V'
j=1 j V!) 3
and ( a f t e r p o s s i b l y r e o r d e r i n g t h e
w e have .'L
%
2
V!
Z
i.
f o r each
4. GROUP ACTIONS
Let
G
X
be an a r b i t r a r y group and l e t
X,
symmetric group on By an
action
of
G
be a set.
X we
on
X, g
E
G.
X.
(G,X,f) ( o r simply t o X) f
a s a G-set.
'x
and write
It w i l l
f (g) ( x ) ,
i n s t e a d of
With t h i s c o n v e n t i o n , t h e f o l l o w i n g p r o p e r t i e s h o l d
1x abz Conversely, i f f o r each
=
x
for all
x E X
=
a(bx)
for a l l
x
x
E
X
and
g
E G,
which s a t i s f i e s (1) and (21, t h e n t h e map
f ( g ) !I) = g x , x
f o r the
u n d e r s t a n d a homomorphism
be c o n v e n i e n t t o s u p p r e s s t h e r e f e r e n c e t o E
Sym(X)
i . e . t h e group of a l l p e r m u t a t i o n s of
We s h a l l r e f e r t o t h e t r i p l e
x
We w r i t e
E
X, g E G i s a
E
(1)
X, a , b
E
G
t h e r e i s a unique element
f
:
homomorphism.
G+
Sym(X)
d e f i n e d by
(2)
'x
E
X
26
CHAPTER 1
X
Let
x
By t h e orbit of
be a G-sat.
X, we u n d e r s t a n d t h e set
defined
‘5
by =
‘3:
X
I t is clear that
G(r)
Igxlg
GI
E
i s a d i s j o i n t union of o r b i t s .
f o r t h e stabilizer of
z in
G
z 6 X, w e w r i t e
Given
d e f i n e d by
G(z1 = { g E
GIgx
=
xI
I t i s an immediate consequence of t h e d e f i n i t i o n t h a t
G(gx) = gG(z)g
-1
for a l l
X,g
Z . E
E C.
and t h a t
Now assume t h a t
X is a X.
morphism group of
X
G on
action of
group, module o r an a l g e b r a and l e t
Then c l e a r l y
:
We s h a l l r e f e r t o t h e t r i p l e
(G,X,f)
algebra according t o whether
X
a c t s on a f i e l d
t o be t h e kernel Of
X
i s a subgroup of
X
be t h e a u t o By an
Sym(X).
we u n d e r s t a n d a homomorphism
f
C
Aut
Aut
F,
G
+
AutX
( o r simply t o
X)
a s a G-module o r a G-
i s a module o r an algebra..
we s a y t h a t
the action.
F is a G-field.
Thus i f
G
In particular, i f
f
The k e r n e l of
X
a c t s on
and
is said
Go i s t h e k e r n e l
of the action, then = If
Go = 1,
t h e n we s a y t h a t
X,
f u l l y on
tg
then
Assume t h a t
A
G
E
G
~ l g x=
x
for a l l
a c t s faithfuZly on
z E XI
X.
i s i d e n t i f i a b l e w i t h a subgroup of
i s a G-algebra.
G acts faith-
Thus i f Aut
X.
Then t h e f i x e d s u b a l g e b r a
‘A
of
C is
d e f i n e d by
AG Thus
AG Let
=
{a E ~ l g a= a
i s t h e l a r g e s t s u b a l g e b r a of
F
be a G - f i e l d .
Then
FG
A
for all
g
on which
G
6
GI acts trivially.
i s o b v i o u s l y a s u b f i e l d of
F.
An
GROUP ACTIONS
27
G
i m p o r t a n t p a r t i c u l a r c a s e of G - f i e l d s i s t h e c a s e where t h e a c t i o n of is faithful.
G
I n t h i s case
i s i d e n t i f i a b l e w i t h a subgroup of
G on F i s given by
a c t i o n of
'1
=
g(A)
F.
s i d e r e d a s an e x t e n s i o n of v e c t o r space o v e r written
W e can view
We s a y t h a t
is f i n i t e or infinite.
F
i s a r o o t of a nonzero polynomial o v e r
b r a i c over
F
azgebraic o v e r
i s s a i d t o be A
f
polynomial
F
L
if
over a f i e l d
a l g e b r a i c over tension
F.
f).
f E F[XI
nomial Let
E/F
~1
F.
which h a s a r o o t i n
be a f i e l d e x t e n s i o n .
f i x e d f i e l d of t h e group A field
e v e r y element of Let
E/F
F
i s algebraic o v e r
E
The e l e m e n t s of F
F
alge-
in
E;
E
i f it is a
i s separable o v e r
E/F
F
F
i s separable.
i f it is An ex-
E i s separable over
i f e v e r y element of
E
s p l i t s over
E.
Then t h e group of a l l F-automorphisms of
E
i s a l g e b r a i c over
Gal(E/F).
F
and
F
E
We s a y
is the
Gal(E/F).
i s s a i d t o be p e r f e c t i f e i t h e r
F
and i s
i s s a i d t o b e normal i f any i r r e d u c i b l e poly-
i s a Galois extension i f
E/F
E
of
i s c a l l e d t h e Gulois group of t h e e x t e n s i o n and i s w r i t t e n that
F
a l l of which have o n l y simple r o o t s
An element of
E/F
An a l g e b r a i c e x t e n s i o n
over
= E.
F[Xl,
i s a separable extension
E/F
An element
and i f i t s minimal polynomial o v e r
F
E
i s c a l l e d separable o v e r
F
p r o d u c t of i r r e d u c i b l e polynomials i n ( i n a s p l i t t i n g f i e l d of
and hence a
c a l l e d t h e algebraic closure of
L,
form a s u b f i e l d
con-
i s a f i n i t e or an i n f i n i t e extension accord-
E/F
(E:F)
CY
and t h e
E
f o r the f i e l d
a s an F-algebra
ingly as if
E/F
i t s dimension i s c a l l e d t h e degree o f
F;
(E:F).
E
Aut F
F
g E G, A E F.
for a l l
E be a f i e l d and F a s u b f i e l d ; we w r i t e
Let
on
charF = 0
or
charF = p > 0
and
i s a p - t h power.
be a f i e l d e x t e n s i o n .
g e n e r a t e d by a s i n g l e element o v e r c a l l e d a p r i m i t i v e element o v e r
We s a y t h a t F;
E
i s simple i f
E
such a g e n e r a t i n g e l e m e n t f o r
is E
is
F.
We c l o s e t h i s s e c t i o n by q u o t i n g t h e f o l l o w i n g s t a n d a r d f a c t s ( s e e C o h n ( l 9 7 7 ) ) .
4.1. on
PROPOSITION.
F.
Then
IGl
Let =
F
G
be a G-field where
( F :F I .
G i s f i n i t e and a c t s f a i t h f u l l y
28
CHAPTER 1
The f o l l o w i n g p r o p e r t y i s a n immediate consequence of P r o p o s i t i o n 4 . 1 .
4.2.
F
Let
COROLLARY.
IG/G~I 4.3.
PROPOSITION.
E/F
i f and o n l y i f 4.4.
Let
E/F
Then
)
be a f i n i t e f i e l d e x t e n s i o n .
E/F
A f i n i t e f i e l d extension
4 . 5 . PROPOSITION.
4.6.
be f i n i t e .
Then
E/F
i s Galois
i s normal and s e p a r a b l e .
PROPOSITION.
F
C
(F:F
=
a s p l i t t i n g f i e l d o f a polynomial o v e r
of
G/Go
be a G - f i e l d and l e t
F
A field
B
i s normal i f and o n l y i f
is
F.
i s p e r f e c t i f and o n l y if e v e r y f i n i t e e x t e n s i o n
i s separable.
PROPOSITION.
element
A
E E
Let
E/F
E
such t h a t
L
number of f i e l d s
be a f i n i t e f i e l d extension. =
F(h) F
such t h a t
t h e r e e x i s t s such an element
Then t h e r e e x i s t s an
if and o n l y i f t h e r e e x i s t s o n l y a f i n i t e
5L
E.
If
E
i s s e p a r a b l e over
F,
then
A.
5 . COHOMOLOGY GROUPS AND GROUP EXTENSIONS
N
Throughout t h i s s e c t i o n ,
G
and
G
a r e f i x e d groups.
An extension o f
N
by
i s a s h o r t e x a c t sequence of groups C
E :
Y
B
l--tN--+X---+G---tl
(1)
Assume that
a'
c('
E'
1-
:
i s a n o t h e r such e x t e n s i o n .
a homomorphism
y
:
X+
Y
+G
N-Y
W e say t h a t
E
and
--+
E'
1
are congruent i f t h e r e e x i s t s
which r e n d e r s commutative t h e f o l l o w i n g diagram:
COHOMOLOGY GROUPS AND GROUP EXTENSIONS
29
The following lemma shows that y
is in fact an isomorphism, and therefore the
congruence relation is symmetric.
Because it is obviously reflexive and tran-
N by G; the
sitive, we may speak about the congruence classes of extensions of
E
congruence class of 5.1. LEMMA.
[El.
will be denoted by
Assume that the diagram
p,G31-
a
l-Gl
-G2
is commutative and that both the upper and lower rows are exact. are injective (surjective or bijective), then so is Proof.
and y 3
y
Assume that both
1
y2
is also injective.
Suppose that
y2-
are injective.
y (g)
=
We wish to show that g E GP.
1 for some
2
1 = B
B (g)
from which we get
=
[Y ( g ) I
2
since y 3
1
=
Then we have
Y [B ( g ) 1 3
2
and Y 3
y1
If
1
is injective.
By hypothesis, the upper
1
row is exact.
So we have
= ImCl
KerD 1
such that
g = Nl(gl).
1
.
Hence, there is an element g
GI
Accordingly,
1
=
y (gl
=
y [a (g ) I 2
1
1
which entails, by the commutativity of the diagram, that a by assumption, both
of 1
y1
and
N 2
are injective.
2
So we get
g , = 1 and g = a ( g ) = 1 1
This proves that
Y2
is injective.
Next, assume that both
Y
and 1
show that
h E y2 (C2).
1
Since
y3
y3
are surjective.
Let h E h'
2
.
We must
is surjective, there is an element g 3 of
G3
30
CHAPTER 1
R2(h)
such t h a t
g
that
= y3(g,).
By h y p o t h e s i s , t h e upper row i s e x a c t .
g
R1(g2) f o r some element
=
Bz (h)
=
G
of 2
.
2
h
Put
Because t h e lower row i s e x a c t , w e have
So
o 2 [Yz(9,)1
Y 3 [B1 ( g 2 )1 =
by t h e commutativity of t h e diagram.
=
Kerl?
Y2(g2)k2.
Ima
=
2
h
E
H
1
a (h
such t h a t 1
2
g
element
E GI
)
=
.
h
1
satisfying
y1
As
Yl(g)
=
.
B (h
Then we have
2
)
=1
2
Hence t h e r e i s a n element
i s a l s o s u r j e c t i v e , t h e r e is a n
.
h
We conclude
k2
Thus
=
a2[yl(gl)l= Y 2 [ a ( g l ) l ,
and w e have
.
h = ~ ~ I g ~ @ ~ (Eg Y~ ~l (IG ~ I T h i s proves t h a t
y2
is surjective.
Observe t h a t any e x t e n s i o n (1) i s c o n g r u e n t t o t h e one i n which i n c l u s i o n map.
is t h e
ci
For t h i s r e a s o n , from now on we s h a l l c o n c e n t r a t e on t h e exten-
sions
E :
i
where
l - N ~ X - f t G - - + i
d e n o t e s t h e i n c l u s i o n map.
t
u n d e r s t a n d any map
: G--t
X
I n what f o l l o w s by a section o f
we
such t h a t
t ( l )= 1 and
f't
= 1
G
O f c o u r s e , a p a r t i c u l a r c a s e of a s e c t i o n i s a s p l i t t i n g homomorphism
( i f it e x i s t s ) .
f
Note a l s o t h a t f o r a l l
G
----f
X
x,y E G,
f ( t (2)t ( y ) t ( x y ) - l ) =f ( t ( 2 ))f(t(Y))f(t(?g) 1-1 = x y ( q f ) - l = 1 which shows t h a t
Let
T
=
{T(g)lg
W e say t h a t
n
E N.
E
GI
be a f a m i l y of automorphisms of
( a , T ) i s a f a ct o r s e t of G o v e r
N
N
and l e t
i f for a l l
ci :
G xG-+
r,y,z E G
and
N.
31
COHOMOLOGY GROUPS AND GROUP EXTENSIONS
(The c o n d i t i o n s ( 5 ) are n o t e s s e n t i a l f o r t h e subsequent d i s c u s s i o n s , b u t t h e y
a r e s a i d t o be congruent i f t h e r e e x i s t s a map
(a',T')
(a,T) and
Two f a c t o r sets
h e l p t o s i m p l i f y some of t h e c o m p u t a t i o n s ) .
1 : G+
N
with
h ( l ) = 1 such t h a t
I t i s immediate t h a t t h e congruence of f a c t o r s e t s i s an e q u i v a l e n c e r e l a t i o n .
5.2.
LEMMA.
let
t
:
G-
Let E :
X
2' = { T ( g )ig E G}
A
1 --t N
X
f
be a s e c t i o n of
5 Autu
f , G-
and l e t
a
: GxC-
N
and
(x,y)
=
t (x)t ( y ) t (q)
(z,y
T ( g ) ( n ) = t(g)nt(g)-l
G)
(8)
(g E G,n E N )
(9)
(a,T) i s a f a c t o r s e t ( t o which w e r e f e r as b e i n g a s s o c i a t e d w i t h
t).
corresponding t o t h e s e c t i o n
Furthermore, i f a d i f f e r e n t c h o i c e of
proving ( 5 ) .
Since Given
a(x,l)
t(l) = 1, w e o b v i o u s l y have
n
E
N
x , g E G,
and
E,
or
f
is
(a,T).
made, then t h e corresponding f a c t o r s e t i s congruent t o Proof.
G,
by
be d e f i n e d by N
Then
N
1 b e an e x t e n s i o n of
=
a(l,x)
= 1
and
T(1)
=1.
w e a l s o have
a(x,y)T(xyy) ( n ) a ( x , y ) - l = t (32) t ( y ) t(xy)-lt(qf)nt (q)-lt(q t (y)-lt(x)-l )
proving
(3).
proving ( 4 ) .
Next, f o r a l l
Finally, l e t
Then t h e r e e x i s t s a map
=
t ( 2 )t ( y ) n t( y ) - l t (x)
=
[P(x)T(yIl( n ),
x,y,z E G,
t' 1
: :
G
G
-t
--+
N
w e have
X
b e a n o t h e r s e c t i o n of with
f.
t ' ( g ) = X ( g ) t ( g ) f o r all
g E G.
32
CHAPTER 1
proving ( 6 ) .
x,y E G ,
Finally, for a l l
proving ( 7 ) and hence t h e r e s u l t .
we have
9
I t i s a consequence o f Lemma 4 . 2 t h a t e a c h e x t e n s i o n
E : of
N
over
G
by
N.
N
1-
A
X
L G-
1
-rE
d e t e r m i n e s a unique congruence class o f
o f f a c t o r sets of
G
We a r e now r e a d y t o o b t a i n a s u r v e y of a l l congruence c l a s s e s of
e x t e n s i o n s of a p r e s c r i b e d p a i r of ( n o t n e c e s s a r i l y commutative) g r o u p s . 5.3.
THEOREM.
{El
The assignment
T
&+
E d e t e r m i n e s a b i j e c t i v e correspondence N
between t h e congruence c l a s s e s o f e x t e n s i o n s o f c l a s s e s of f a c t o r s e t s of (i) E
G over N .
(a,T)
a(r,y)
f o r which
= 1
Let
(Ci,T)
E :
define
-+
X
then
N
with
E
i s a s p l i t exten-
A ( 1 ) = 1 such t h a t
T(2)(A(y))-1A(r)-1A(2y)
be a f a c t o r s e t of
1-N
C over N . X f .
G
for all
r , y E G.
be t h e d i r e c t p r o d u c t s e t
N x G.
We c o n s t r u c t a n e x t e n s i o n
1
---L
(a,!!') i s a f a c t o r s e t a s s o c i a t e d w i t h
such t h a t Let
E,
E G.
For t h e sake of c l a r i t y , w e d i v i d e t h e proof i n t o t h r e e s t e p s .
Proof.
S t e p 1.
=
z,y
for a l l
1: G
s i o n i f and o n l y i f t h e r e e x i s t s a map
Cd(r,y)
and t h e congruence
i s t h e congruence c l a s s of a
TE
(a,T) i s a factor set associated with
(ii) I f
G
Furthermore,
i s a s p l i t e x t e n s i o n i f and o n l y i f
factor s e t
by
E.
Given
( n ,g 1
1
)
,f n2 , g p )
E X,
33
COHOMOLOGY GROUPS AND GROUP EXTENSIONS
( n ,g 1 ( n ,g I 1
W e w i l l show t h a t
X
2
1
2
=
(nlT(gl) (n2)a(g1,g2) ,g1g2)
i s a group under t h i s o p e r a t i o n .
(10)
Indeed
and
(by ( 4 ) ) The i d e n t i t y is t h e p a i r (1,l).
proving t h a t t h e given o p e r a t i o n i s a s s o c i a t i v e .
(n,g)
The i n v e r s e of
and t h u s
X
i s a group.
The map
s u r j e c t i v e homomorphism whose k e r n e l
BY
( l o ) , the
fore identify
t(g)
and
=
map
N
(n,l) and
W
K.
(n)-',g-')
(CX(g-',g)-'T(g-')
is
n
f
:
K
X---t G d e f i n e d by
i s an isomorphism of
f.
fl(n,g)l = g (n,l)
c o n s i s t s of a l l p a i r s
Furthermore, t h e map
(l,g) i s a s e c t i o n of
s i n c e by ( 8 )
Since
t
K :
N.
onto
G
-+
X
is a
:
W e may t h e r e -
d e f i n e d by
CHAPTER 1
34
(T,a) i s
we s e e t h a t
S t e p 2. if
H e r e w e show t h a t t h e e x t e n s i o n s
t
Let
corresponding t o
, we
G
+
(f,T)
and we d e n o t e by
and ( 9 )
:
X
E'
and
are congruent i f and o n l y
t
and
g E G.
a and a '
t ' , respectively.
t h e f a c t o r s e t s of
Y
Applying
(f,T,= (f',T'). Thus
see t h a t
and l e t
t'
:
G
Y
t' i s o b v i o u s l y a s e c t i o n
Then
(f',T')
and
6
be a s e c t i o n of
t ' ( g ) = y ( t ( g ) ) for a l l
be d e f i n e d by
O',
E
t.
Assume w e a r e g i v e n a commutative diagram ( 2 ) i n which
TE = T E l .
a r e i n c l u s i o n maps.
of
t h e f a c t o r s e t corresponding t o t h e s e c t i o n
G
over
N
t o t h e f o r m u l a s (8)
TE = TE,.
Conversely, assume t h a t
5X-
N
1-
E :
G
f,
1
and
E' a r e e x t e n s i o n s of of
f
and
l - N - - + iY - + G ~ l
:
N
by
f' and
G
such t h a t
(a,T) and
N
--+
can be uniquely w r i t t e n i n t h e form
:
A ( 1 ) = 1.
with
nt(g),n
Y
one immediately v e r i f i e s t h a t t h e map
y
t
If
= TEl.
and
t' a r e s e c t i o n s
(a',T') a r e t h e c o r r e s p o n d i n g f a c t o r sets, t h e n
A : G
( 6 ) and ( 7 ) hold f o r some
TE
nt (g)
E N, g E
X----t
:
Y
X
N o w each element of
G.
Applying ( 6 ) and ( 7 ) ,
g i v e n by
nX ( g ) - l t 1 ( g )
i s an isomorphism which makes t h e r e q u i r e d diagram commute.
Hence
E
and
E'
are congruent.
S t e p 3.
Completion of t h e p r o o f .
By S t e p s 1 and 2 ,
{ E l c-t
T~
i s a b i j e c t i v e correspondence between t h e
congruence c l a s s e s of e x t e n s i o n s of
G
f a c t o r sets of s i o n of
N
G,
by
corresponding t o
over
let
t.
If
T~
for a l l
G
by
N.
Assume t h a t
t
be a s e c t i o n of
E
a homomorphism i n which c a s e that
N
and t h e congruence c l a s s e s of
E : 1-
E
G.
If
*A
f and l e t
X+
f
G-+
1 i s an e x t e n -
( a , T ) be t h e f a c t o r s e t
is a s p l i t e x t e n s i o n , t h e n we may assume t h a t
a(x,y)
= 1
for a l l
i s t h e congruence c l a s s of a f a c t o r s e t
x,y
N
x , E~ G. (a',T')
t
is
Conversely, assume f o r which
a'(x,y)
= 1
mn
COHONOLOGY GROUPS
E'
N
i s t h e e x t e n s i o n of 2,
E
E'
and
N
1-
:
G
by
Y-
-+
f'
G
E'
Since
1
4
(a',T')
c o n s t r u c t e d from
a r e congruent.
35
GROUP EXTENSIONS
i n S t e p 1, t h e n by S t e p
E,
i s a s p l i t e x t e n s i o n , so i s
.
proving ( i )
By ( i ) , E factor s e t
(a',?")
.
a'(x,y)
such t h a t
=
by a p p l y i n g ( 7 ) and hence t h e r e s u l t . 5.4.
COROLLARY.
E'
*
p?,
G
:
x,y,g E G
N
-+
and
(a,T)
Let
x,y E G .
i s congruent t o
f
f'
'
respectively.
for all
This proves (ii)
l - - + N ~ Y - G - l
:
be s p l i t t i n g homomorphisms.
Proof.
x , y E G.
1 for all
l - n T ~ X - + G - - - t l
be t w o s p l i t t i n g e x t e n s i o n s of
for all
i s congruent t o a
Let
E :
e x i s t s a map
(Cr,T)
i s a s p l i t e x t e n s i o n i f and o n l y i f
N Then
with
G
by
E
and l e t
:
i s congruent t o
G -.+
E'
X
u'
and
G
:
Y
-+
i f and o n l y i f t h e r e
A ( 1 ) = 1 such t h a t
n E N. (a',T'1
and
Since
and
By Theorem 5 . 3 ,
(a?,T').
be t h e f a c t o r sets c o r r e s p o n d i n g t o
i.~'a r e homomorphisms w e have
E
i s congruent t o
E'
u
and
a(x,y)=a'(x,y)= 1
i f and o n l y i f
(a,T)
The d e s i r e d c o n c l u s i o n i s now a consequence o f (6)
and ( 7 ) . We n e x t examine t h e s p e c i a l c a s e where
5.5. LEMMA.
Let
G
(a T ) 2' x
l+ N
i s a c y c l i c group of f i n i t e o r d e r .
n
be a c y c l i c group of o r d e r
E : be an e x t e n s i o n of
G
by
G.
N"-
*
X
f
-+
Then, f o r any
G
g e n e r a t e d by
E
and l e t
1
-+
x
g
X
with
f(x)
= g,
the pair
g i v e n by
(0 G i , j
<M-
1)
36
CHAPTER 1
Tx(gi) ( y )
tx
The map
,
(0 G i
t x ( gi)
G--' X g i v e n by
f.
o b v i o u s l y a s e c t i o n of Then, by ( 8 )
:
ziy3=i
n - 1,y E N )
Q
E.
i s a f a c t o r set corresponding t o Proof.
=
La ,!?
Let
x x
i,
= 3:
0
< i Q n - 1,
is
be t h e corresponding f a c t o r set.
)
we have
i i i+j i + j -1 ax(g ,g ) = z t x ( g )
(0 4 i,j 4 n - 1)
. . If
i+j
then
O G i + j - n < n - 2
.
ax(gi,gj)= xn. follows. 5.6.
THEOREM.
= '"x
tx(g+' )'
then
g
and
and hence
i+j
Tx
The formula f o r
= g
iij-n
.
Hence
= 1.
tx(g
iij
being a consequence of
G be a c y c l i c group of o r d e r
Let
a x ( gi ,g')'
n
i+j 2 n,
If =
)
iij-n
x
and so
(9), t h e r e s u l t
N
and l e t
be a n a r b i t r a r y
group.
(i) I f
E : 1 --+ N
i X f G
-+
@
e x i s t s an automorphism corresponding t o
E
-+
of
+
N,
an element
a
E N
T(g') @ E Aut N
$n(y) = a p - l
and
and
a
=
if
i + j Q n - l
a
if
i + j > n
E N
i
(iil
Given
t o be
i,j
E
T(g).
{O,l,
E
N
(13)
(0G i , j G n-1)
(14)
(0G i G n-1)
(15)
are such t h a t (13) h o l d s , t h e n
(CL,T)
defined
G o v e r N.
( i f I n t h e n o t a t i o n of Lemma 5 . 5 , p u t
@
y
for all
1
$I
by (14) and (15) is a f a c t o r s e t of
and d e f i n e
(a,T)
and a f a c t o r s e t
a(gi,gj)=
Proof.
G, t h e n t h e r e
by
such t h a t
@ ( a )= a
(ii) I f
N
1 i s a n e x t e n s i o n of
(a,!?) = (a ,T
x x
)
and
a
=
xn,
Then ( 1 3 1 , (14) and (15) h o l d by v i r t u e of Lemma 5.5.
...,n-11
and
y E N,
w e have
37
COHOMOLOGY GROUPS AND GROUP EXTENSTONS
Property ( 5 ) being obvious, w e a r e l e f t t o v e r i f y t h a t f o r a l l
proving
(3).
<,j,k
{O,l, ...,n-11,
E
k
a(9i n 9j)a(gi+j ,g 1
i+ j + k
If
Q
a r e equal t o
n-1,
i + j
then
Assume t h a t
1.
n-1
i + j 2 n and
a.
If
(16) a r e e q u a l t o
a.
Finally, i f
i+j
Since f o r
b
j+ k
j+k
i + j
i + j2 n
Q
n-1,
and
(16)
and s o b o t h s i d e s of
n-1 If
Ua(gi+j-n,gk) and
(16) a r e
n
and
i j+k) ,g
j k
( a ( g ,g ) ) a ( g
i + j + k 2 n.
(16) are e q u a l t o
r i g h t hand s i d e of
i
0
=
Q
n-1,
(16)
t h e n b o t h s i d e s of
t h e n a g a i n b o t h s i d e s of
j + k 2 n,
t h e n t h e l e f t and
i jik-n ) , a a ( g ,g
respectively.
j + k 2 n,
and
a(gi+j-n,gk) = cY(gi ,gj+k-n ) t h e r e s u l t follows.
A
Let
be an a b e l i a n group and c o n s i d e r a n e x t e n s i o n
E : A
of
G.
by
f a c t o r set.
T
1 - A A X L G - I p
Let
Aut A
: G - 4
d o e s n o t depend upon t h e c h o i c e of
T'
:
Aut A
G-
Thus t h e congruence c l a s s
2'
T
Then by ( 3 ) ,
e x t e n s i o n and
:
G
4
{El
T
: G-'
Aut A
let
(a,T)
be t h e c o r r e s p o n d i n g
i s a homomorphism. Furthermore, i f
lJ.
It is clear that
E'
i s a congruent
i s t h e c o r r e s p o n d i n g homomorphism, t h e n of
E
T
=
T'.
d e t e r m i n e s a unique homomorphism
and hence a unique a c t i o n of
AutA
homomorphism
f and
be a s e c t i o n of
G
on
A.
From now o n , we
fix
U
and p u t
Then, by ( 3 ) and ( 4 ) , w e have
for a l l
x,y,z
E G.
essential difference: functions
~1 :
GxG-
The assumption t h a t
A
is abelian g i v e s r i s e to another
we can d e f i n e an o p e r a t i o n on t h e s e t
A
s a t i s f y i n g (17) and (18).
Z2(G,A)
Indeed, i f
of a l l
a,B E Z 2 ( G , A ) ,
38
CHAPTER 1
af? d e f i n e d by
then t h e i r product
Z2(G,A).
i s a g a i n an element of
cocycles.
It is clear t h a t
element of
Z2(G,A)
Z 2 ( G , A ) as
W e s h a l l r e f e r t o t h e e l e m e n t s of
Z2(G,A)
i s an a b e l i a n group.
The i d e n t i t y
-1
i s t h e 1-valued c o c y c l e and t h e i n v e r s e
of
i s given
c1
by -1
c1
Then
6t i s e a s i l y seen t o be a c o c y c l e . B2(G,A)
and denote by
for all
(x,y) = a ( z , y ) - l
We s h a l l r e f e r t o
t h e set of a l l coboundaries.
B2(G,A)
i s i n f a c t a subgroup of
Z2(G,A).
H2(G,A)
w i t h r e s p e c t t o t h e g i v e n a c t i o n of
z,y
E
G
6t as a coboundary
I t is e a s y t o v e r i f y t h a t
The second
G on A
cohomology group
i s d e f i n e d t o be t h e
f a c t o r group
2’ (G,A)/B2( G , A ) The elements of
are c a l l e d cohomology c l a s s e s ;
H2(G,A)
t a i n e d i n t h e same cohomology c l a s s a r e s a i d t o be f o r any
c1 6
5.7. THEOREM.
Let
6 for
we write
Z2(G,A),
G
a c t on an a b e l i a n group
u
: G
by
A”.
E : 1-
(i) If
and i f
A
.
X is
4
G X
cohomoZogous.
t h e cohomology c l a s s of
A
ponding second cohomology group and l e t of e x t e n s i o n s of
any two c o c y l e s con-
A,
let
I n what f o l l o w s c1.
H‘(G,A)
b e t h e s e t of a l l congruence c l a s s e s
G
which determine t h e g i v e n a c t i o n of
f -f
G
-+
a s e c t i o n of
1 i s an e x t e n s i o n of
f,
be t h e corres-
A
by
on
G
A.
with
t h e n t h e map
d e f i n e d by
aE ( Z c , Y ) i s an element of
Z’(G,A)
such t h a t
=
-1
uJrIu(y)P(zy)
EE
i s uniquely determined by
{El.
{El
E
h
39
COHOMOLOGY GROUPS AND GROUP EXTENSIONS
(ii) The map
is a bijection (iii) The extension E Let T
Proof.
:
G on A .
action of
aF
splits if and only if
G
-+
Aut A
1
=
be the homomorphism determined by the given
Then, by ( 6 ) and (71, two factor sets
are congruent if and only if the cocycles a
acts on an abelian group A.
A
(CL’,T)
. -
a’ are cohomologous.
and
desired conclusion is therefore a consequence of Theorem 5 . 3 . Again, assume that G
(a,T) and
map
f
:
G
The
A
is
called a crossed homomorphism if f(2.y) =
If f and g
crossed homomorphisms of
fa
:
G
for all x , y E G
are crossed homomorphisms, then their product
is again a crossed homomorphism.
map
f(x)xf(y)
4
A
fg
defined by
Z’(G,A)
It follows easily that the set
G into A
is an abelian group.
Given
of all
a E A , the
defined by =
fa(X)
is obviously a crossed homomorphism.
crossed homomorphism and denote by
x -1 (a ) a
We shall refer to fa
as a principaZ
B’ (G,A) the set of all such homomorphisms.
It is easily verified that B’ (G,A) is a subgroup of
2’(G,A).
The correspon-
ding factor group
is called the f i r s t cohomology group of Assume that and let lJ
:
G
1
4
-+
A -.?+* X
f
+
G
+
G over
1 is a splitting extension of A
X be a splitting homomorphism. a U : G -+X
defined by
A.
For any
by
a E A, the map
G,
CHAPTER 1
40
is also a splitting homomorphism.
We shall refer to
‘U
as being A-conjugate of
It is clear that A-conjugacy is an equivalence relation on the set of all
li.
splitting homomorphisms G Let
5.8. THEOREM.
G, let U
:
G
--+
1
X.
-+
3
.
A
f G X+
G over A
a
E Z’(G,A),
G
with respect to the action of
ga For any
1 be a splitting extension of
X be a splitting homomorphism and let Ii’(G,Al
cohomology group of
(i)
-+
the map
=
be the first on A
u(g)au(g)-’
W
: G
--+
A by
given by
(a E A,g
E G)
X defined by
is a splitting homomorphism.
a
(ii) The map
c1U
is a bijection between
qplitting homomorphisms G hetween H’ ( C , A ) L‘
-+
-+
X.
Z’(G,A) and the set of all
Furthermore, this map induces a bijection
and the set of A-conjugacy classes of splitting homomorphisms
x. (i) Since a ( g )
Proof.
f
A
=
Kerf,
we have for all g E G
C, we also have
Given x,y
proving (i). (ii) It is clear that Let
Y
:
G
----+
c1
u
=
0.
p
implies a
1
= c1
2
for all
X be an arbitrary splitting homomorphism.
to prove the first statement, it suffices to verify that
x,y E G,
we have
c1 1
,a
Since
2
E Z’(G,A).
Y
=
(YU-1 ) P ,
YV-l E Z1(G,A).
Given
41
SOME PROPERTIES OF COHOMOLOGY GROUPS
as required. Let u,B E Z’(G,A). that
uu
UP
and
and
6~
By
To prove the final assertion, it suffices to verify
are A-conjugate if and only if
a-lB
E
B’ (G,A).
are A-conjugate if and only if there exists a E
Since the latter is equivalent to the existence of a E
the result follows.
.
A
A
such that
such that
(9)= a g ( U - l )
(U-lB)
By definition,
for all g E G
6. SOME PROPERTIES OF COHOMOLOGY GROUPS denotes a group acting on an abelian group A.
Throughout this section, G any
n
1,
we put
A~ Then An
is a subgroup of
A
We say that A
{anla
=
E
A}
and the map
is a surjective homomorphism. phism.
Ah1
We denote by
An
is n - d i v i s i b l e if
=
the kernel of this homomor-
A.
It is clear that A
divisible if and only if it is p-divisible for any prime p special case where A
--+
uniquely d i v i s i b l e by istic p > 0 , Let p
For
then
A,a +-t
n.
F*
an
The p-component
P
=
ta
E
A]$
=
1
A
In the is
F is a perfect field of character-
For example, if
n
A
n.
is an automorphism, we say that A
is uniquely divisible by
be a prime.
dividing
is n-
P
p.
of A
is defined by
for some n
>
11
42
CHAPTER 1
A
It is clear t h a t
A
a l l elements of
A.
i s a subgroup of
P
i s a torsion group, i . e . i f
A
If
a r e of f i n i t e o r d e r , t h e n
A = @ A
P P rn < n.
na
i s of f i n i t e exponent
A
W e say t h a t
AG
Finally, we define
m #
An = 1 and
A
t o b e t h e l a r g e s t subgroup of
A
if
1,
1 for
G
on which
acts trivially, i.e.
A PROPOSITION.
6.1.
Let
(i) H2(G,A)n = 1
m-divisible,
A
(iii) I f
A
If
(v)
If
(vi)
G
ta
E
~ l g a= a
A
Proof.
@H2(G,A)
c1 E Z 2 ( G , A )
is n - d i v i s i b l e and
A h ]
is f i n i t e , t h e n
H2(G,A)
p,
H2(G,A)
i s u n i q u e l y d i v i s i b l e by a prime
n,
m, t h e n
i s of f i n i t e exponent
A
then
rn
i s of f i n i t e exponent
then
and i f
A
is
m.
i s cohomologous t o a c o c y c l e of o r d e r
is finite.
P
= 1
H2(G,A) = 1 H 2 ( G , A ) m = 1. and
(m,n)
In p a r t i c u l a r ,
= 1.
H 2 ( G , A l n = 1.
(i) It suffices t o verify that
E Z2(G,A).
n.
pln
i s u n i q u e l y d i v i s i b l e by
A
If
GI.
g E
for all
be a f i n i t e group of o r d e r
H2(G,A) =
and
then
H2(G,A) = 1 if
c1
=
P rn i s t h e o r d e r of t h e cohomology c l a s s of
(ii) I f
(ivl
G
To t h i s e n d , f i x
Because c1 (s,y) a (zy$1
X
=
a( y , Z ) a( X , y z )
we have
an = 6t, p r o v i n g (1).
Thus
-
( i i ) By h y p o t h e s i s , f o r a l l
t
:
G
exists Setting
A
with
P(x) E A
t(l)
= 1.
such t.hat
6 = a(6p),
r,y
E G,
Because
P ( Z ) ~=
we deduce t h a t
am(z,y) = t(z)xt(y)t(xy)-l f o r some A
is m-divisible,
.
-1
t(z)
f o r any
We may assume t h a t
z E G,
there
!J(1) = 1.
43
SOME PROPERTIES OF COHOMOLOGY GROUPS
a s required.
A
(iii) Since
is n-divisible,
and (ii), any
H2(G,A)
Therefore
G x C (iv)
E Z2(G,A)
A[nl.
f
it i s m - d i v i s i b l e
A
A[nl
But, by h y p o t h e s i s ,
i s p-divisible,
A
(v)
(vi)
For any
p
dividing
n.
CY E
Z2(G,A),
w e have
H
be a subgroup of
H
a c t i o n of :
G x G-+
A
on
A,
n,
k
p ,k
0.
Hence, by (ii), w e may assume t h a t
= 1
then
is
for all
A
=
x,y
E
G
1, a s r e q u i r e d .
i s u n i q u e l y d i v i s i b l e by
p
Now a p p l y ( i ) and ( i v ) . Ci(X,y)"
G
= 1
for all
H2(H,A)
and let
induced by t h e a c t i o n of
its restriction t o
H x H
We s h a l l d e n o t e t h i s homomorphism by
r , y E G.
Hence
Res
G
be d e f i n e d w i t h r e s p e c t t o t h e on
A.
Given a c o c y c l e
determines a cocycle
a'
CY-
G,H
a':
H
x H---t
A.
i n d u c e s a homomorphism
( o r simply R e s )
and r e f e r t o i t a s
r e s t r i c t i o n map. H
Assume t h a t
representative
cE
Cora :
G x G -+ A
where
ga
=
G of f i n i t e i n d e x m
i s a subgroup of
s e t of r i g h t c o s e t s of
c
H with
in
G.
c=1
For eacn if
c
Cora
and l e t
C be the
c E C choose, once and f o r a l l , a
= H.
Given
a E Z2(H,A),
let
be d e f i n e d by
ga,a E A , g E G.
I t can be shown ( s e e Weiss ( 1 9 6 9 , p . 8 1 ) ) , t h a t
a
Z2(G,A)
t h e l a t t e r i m p l i e s t h a t CY
One can e a s i l y v e r i f y t h a t t h e assignment
the
CY E
H 2 ( G , A ) m = 1.
Z 2 ( G , A l m = 1 and t h e r e f o r e Let
p,
i s u n i q u e l y d i v i s i b l e by
f o r a l l primes
Ah].
H2(G,A).
i s f i n i t e , hence s o i s
it i s p k - d i v i s i b l e .
i s u n i q u e l y d i v i s i b l e by
A
If
Hence, by ( i )
can be r e g a r d e d as a s u b s e t of t h e s e t o f a l l f u n c t i o n s
k a(x,y)P Since
mln.
i s cohomologous t o a c o c y c l e whose v a l u e s l i e i n
A s s u m e t h a t t h e o r d e r of t h e cohomology class of
Since
a
for a l l
i n d u c e s a homomorphism
Cora E Z 2 ( G , A )
and t h a t
CHAPTER 1
44
We shall denote this homomorphism by as the corestriction map.
The map
Cor HrG Cor H,G
(or simply Cor) and refer to it enjoys the following important
property
Am
CorH,GResG,H(h) =
PROPOSITION.
Let
X
E
H’(G,A) (1)
.
A s an easy application of (1), we now prove
(see Weiss (1969,p.76)1 . 6.2.
for all
be a finite group and let P
G
be a Sylow p-subgroup of
G. 1 ---t H 2 ( G , A )
(i) The sequence
H2(P,A)
P
(ii) The sequence H 2 ( P , A )
H2(G,A)
f
P
is exact
1 is exact
(iii) H * ( P , A ) = Im(Res) x Ker(Cor) Proof.
By
Res restricted to H 2 ( G , A ) p , while Cor is G,P H 2 ( P , A ) into H 2 ( G , A ) because every element of
Res we mean
Cor which indeed maps P,G I f Z ( P , A ) has order a power of Write n = IGl = prm
(p,rn)
=
l,xpr + yrn
1
=
P
p. p
with
1 rn
so that
\PI = pr
for some integers z , y .
and
For any
A
(G:P) = m.
E H2(G,A)
P’
Since we
have
which proves both (i) and (iil. Given p E H ’ ( P , A ) ,
with
we may write
(Res Corpl’ E Im(Res) and
directiness, if
a
=
Resa
and
PfRes Corp)-Y E Ker(Cor1
(see (1)).
As for
Cora = 1, then by (1) 1 = Cor(Resa) = CY
m
r so that
a’”’ = 1 and hence a
6 . 3 . PROPOSITION.
Let
=
aw
=
1.
Thus
@ = 1
and the result follows..
G be a finite group and let P be a Sylow p-subqroup of
SOME PROPERTIES OF COHOMOLOGY GROUPS
G.
If
H2(G,A)
c1 E
o r d e r of
a.
i s such t h a t
I n particular,
Proof.
m
Let
then
i f and o n l y i f
c1 = 1
I; ) m,
6.4.
G
E
Let
G.
Then
:
1
E
--+
A
H2(G,A)
Let
o b t a i n e d from Then
E.
for a l l
am
Let
i --+
X-
f
G
L'.
A
1 b e an e x t e n s i o n of
+
A
by
be d e f i n e d w i t h r e s p e c t t o t h e a c t i o n of c1 E
H'(G,A)
by a
S
of
S splits.
correspond t o
{El
G
on
A
as i n Theorem 5.7.
c o r r e s p o n d s t o congruence c l a s s of t h e e x t e n s i o n of
ResG,S(cl)
o b t a i n e d from
= 1
s p l i t s if and o n l y i f f o r a l l Sylow subgroups
t h e c o r r e s p o n d i n g e x t e n s i o n of Proof.
ResG,S(a)
the a s s e r t i o n follows.
COROLLARY.
f i n i t e group
does n o t d i v i d e t h e
Then, by ( 1 1 , w e have
= (G:P).
1 = Cor p , G R e ~ G , p ( a ) = Since
p
G.
S of
Sylow subgroups
ResCrp(cl) = 1,
45
A
by
S
The d e s i r e d c o n c l u s i o n i s t h e r e f o r e a consequence of Theorem
5 . 7 ( i i i l and P r o p o s i t i o n 6 . 3 .
'
Turning our a t t e n t i o n t o c y c l i c g r o u p s , w e n e x t prove 6.5.
PROPOSITION.
a b e l i a n group
Let
A.
G = < g > be a c y c l i c group of o r d e r
a c t i n g on an
Then
i=0 In particular,
n
N 2 ( C , A ) = 1 i f and o n l y i f
A
n-1
= (
.
C gZ)A. i=0
Proof. induced by map
~1
a
:
Let
a E AG, G =
g.
Then
G x G-
A
is a c o c y c l e .
@(a) = a
and
and l e t = 1
0
be t h e automorphism of
A
and hence, by Theorem 5 . 6 ( i . i ) , t h e
g i v e n by
Hence w e o b t a i n a homomorphism
46
CHAPTER 1
which i s s u r j e c t i v e , by v i r t u e of Lemma 5.5.
a
that
a
i s a coboundary i f and o n l y i f L',
Let
:
X
3A
1
f -+
ding t o
aa.
Setting
j = 0, we o b t a i n
now p u t
z=
i=o G
-+
1 be a n e x t e n s i o n o f
U
Then t h e r e i s a s e c t i o n
!J ( g ) ,
i
!J(g
i
of
for a l l
i
) =
n-1 Pi(g)!J(g) = 1
s p l i t s i f and o n l y i f
. ..
( b ~= )bgb ~
n-i E
Hence
a E
s p l i t s i f and o n l y i f
(
1 g
E
f
by
G
correspon-
such t h a t
{ o , l , ...,a-
E
x
I f we
1).
n
f o r some
'
.
s p l i t s , t h e r e s u l t follows.
--
n- 1 bzn
)A.
Z=0
and only i f
-*
A
t h e n w e have
n- 1
E
x
: G
U(g)
=
a = aa(g,g The e x t e n s i o n
W e are t h e r e f o r e l e f t t o v e r i f y n- 1 E ( C gi)A.
B u t , by (2).
...
= b'b
Since
b E A.
i s a coboundary i f
0:
7 . MATRIX RINGS AND RELATED RESULTS
Let
R
be an a r b i t r a r y r i n g and, f o r any
R.
of a l l n r n - m a t r i c e s o v e r
Mn(R)
n 2 1, l e t R
We shall identify
w i t h i t s image i n
c o n s i s t i n g of a l l s c a l a r m a t r i c e s d i a g ( r , r , . - . , P ) , r E R.
i , j G n,
(i,j),1 elsewhere.
e
let
ij
e
The elements
denote t h e r i n g
For e v e r y o r d e r e d p a i r
b e t h e matrix with ( i , j ) - t h e n t r y
ij'
called the
mtriz units,
Mn(R)
1 and
0
s a t i s f y t h e following
properties (i) e . . e = 0 1.3 k s 1= e
(ill
11
+
if
...
j# k + e
and (iii) determine
7.1. PROPOSITION. {vij(l G i , j 4 n } set in
S.
e i j e k s = eis
j
if
=
k
nn teij)
( iii) The c e n t r a l i z e r of
Note a l s o t h a t
and
R
e
11
M (I?) Let
S
in
M (R)ell. n
M (R)
is
R.
Our f i r s t o b s e r v a t i o n shows t h a t ( i ),( i i ),
up t o isomorphism. be a r i n g t h a t c o n t a i n s a s e t o f elements
s a t i s f y i n g (i) and ( i i ) and l e t
Then t h e map
R
b e t h e c e n t r a l i z e r of t h i s
MATRIX RINGS AND RELATED RESULTS
47
Furthermore, R
is an isomorphism of rings and R-modules.
Sv
21 11
11
.
We first note that Mn(R) is a free 8-module freely generated by the
Proof. matrix units
ei j .
eij
Hence the map
$.
modules, which clearly coincides with
extends to a homomorphism of R-
Vij
t+
In view of (i), (ii) and (iii), $
is
easily seen to be a ring homomorphism.
c a . .U
Suppose that
t
=
i , j 23 i j
0 and fix k , s E 11 ,...,n l .
Then, for all
t {l, ...,n l ,
c
0 = Vtk(
s t - aksVtt
U..V..)V
i ,j
ZJ 1-3
and therefore
u t ks tt - a k s (tC u t t ) = a ks
Q=Ca
(a..)
= 0 and so ii, is injective. z3 Finally, fix s E S and for each i , j
Thus
Pij
Then, for all
Vrt,
put
= FkiSUjk
we have
vrtrij
=
yrtvkisvjk
=
vpisujt
and
r Hence
rij
u = L u s v v - u s v ij rt k i j k rt - p i j t and so r
commutes with all Vrt
proving that $
is surjective.
.
Given a (left) R-module direct power of
V.
V
E R.
Moreover, by the foregoing,
Taking into account that
$ ( e l l M n ( R ) e l l )= u the result follows.
ij
Su 11
and 11
e
11
M (R)ell n
and a positive integer n ,
2
R,
we write
v"
for n-th
The following result illustrates that matrix rings arise as
48
CHAPTER 1
endomorphism r i n g s of d i r e c t powers of modules.
7.2.
PROPOSITION. proof.
End(v")
Mn(End(V)) R i , j E {1,2 n}, d e f i n e
R
Given
2
,...,
E..(U
7-3
Then o b v i o u s l y
$ ( u l ,...,un) $
= $ 1
= 2
able w i t h
=
EijEks
1
,...,v
6jk~is
)
and
=
i' 0,...,0)
ZE..
7-7-
... = Qn.
End(?) R centralizes a l l E..
v"
in
t3
is identifi-
End(?)
R
v"
v"
a s an R-module.
But one a l s o
M (R)-module i n t h e f o l l o w i n g n a t u r a l way.
as an
A E M (R)
as column v e c t o r s and, f o r each
(i) I f
W
We v i s u a l i z e
x
and
E
v",
We a r e now r e a d y t o prove
as t h e m a t r i x m u l t i p l i c a t i o n .
v,
i s a submodule of
V
t h e n t h e map
wk-+ w"
o n t o t h e l a t t i c e of
M
i s an
(R)-sub-
v".
(ii) End([/?)
9
M (R)
End(V)
( i i i ) TEe map
R V-
i s m c l a s s e s of
R
W w e
Proof.
I€..}
'
isomorphism from t h e l a t t i c e of submodules of
g i v e n by
i f and o n l y i f
The d e s i r e d c o n c l u s i o n i s t h e r e f o r e a consequence of Propo-
1 . 3 . PROPOSITION.
modules of
and
E
$3
Hence t h e c e n t r a l i z e r of
End(V). R
t h e elements of
Ax
$
If
= 1.
I n t h e preceding d i s c u s s i o n w e regarded
define
by
1
s i t i o n 7.1.
can r e g a r d
End(?) R
E
ii
(0I . . . , u
(JI ( u l ) ,...,$ , ( u n ) ) , t h e n $
=
E
11
v" and
i n d u c e s a b i j e c t i v e correspondence between t h e isomorph-
M
(R)-modules.
The i n v e r s e of t h i s correspondence i s
W.
( i ) The correspondence
f
W
---f
w" i s
obviously order-preserving.
It
t h e r e f o r e s u f f i c e s t o show t h a t i t h a s an i n v e r s e which i s a l s o o r d e r - p r e s e r v i n g . Consider t h e p r o j e c t i o n on t h e f i r s t f a c t o r
:
'TT 1
module
X
of
v",
put
order preserving.
v", V
w e see t h a t of
V.
g(X)
=
71
(X).
It is clear that
X
Thus
proving t h e a s s e r t i o n .
V
and f o r any sub-
Then t h e correspondence
(gf)( W ) = W.
can be w r i t t e n i n t h e form
1
v"+
X
X r t g(X)
From t h e a c t i o n of =
v"
is
M (R)
on
f o r a s u i t a b l e submodule
49
MATRIX RINGS AND RELATED RESULTS
S
(ii) Put
=
M (R)
f
and f i x
E End(?).
S
Then, by t h e n a t u r e of a c t i o n of
S
J", f
on
h a s t h e same p r o j e c t i o n s , s a y
$ E End(!/)
d e t e r m i n e s an element of
End(p) S
R $.
a r e equal t o
e
(iii) W e need o n l y show t h a t
R
and
M
f+-+Xf
Thus t h e map
11
?
whose p r o j e c t i o n s on a l l f a c t o r s
(ellWIn
and
W,
V
where
e
By t h e d e f i n i t i o n of
F = V X...O X O~ P
11
Conversely, any
p r o v i d e s t h e d e s i r e d isomorphism.
V
(R)-modules, r e s p e c t i v e l y .
e
on a l l f a c t o r s .
f'
11
and
W
are
'
V
Observe a l s o t h a t
W = e Because
eklw
=
e
0 i f and o n l y i f w
e 11
1
+
11
. . _i
We ...
11
w
enl F/
@J
0, 1 Q k
=
f
e w t-ni n
LJ
(2 11
1
Q
a s R-modules
n , w E W,
,...,e
i s w e l l d e f i n e d and i s a t l e a s t an R-isomorphism of
W
11
t h e map
w 1 n
onto
(ellki)".
I t there-
fore suffices t o verify that
f (eijeklw)
=
e . . f ( e k w1 ZJ
(w E W, i , j , k E {1,2
1
.
The l a t t e r b e i n g a consequence of t h e a c t i o n of follows.
I
Given an i d e a l
7.4.
PROPOSITION.
of
R.
I+
Mn(I) i s a b i j e c t i o n between t h e s e t s of
R i s simple i f and o n l y i f so i s
In particular,
n
(ellW)n, t h e r e s u l t
on
we write
( i ) The map
R and M (R).
i d e a l s of
M,(R)
,...,n } )
M, (R). (ii)
Mn(R)/Mn(I) p M n ( R / I )
(ii) I f
R = I
@J
... @I,
i s a two-sided decomposition of
Mn(R) = M n ( I l ) 0 i s a two-sided decomposition of p o s a b l e i f and o n l y i f so i s Proof.
(il
Mn(R).
R,
then
... 63 Mn(Is)
Furthermore, t h e i d e a l
Ii
i s indecom-
Mn(Ii).
It is clear t h a t
M (I1
i s an i d e a l of
Mn(R)
and t h a t t h e
CHAPTER 1
50
given map is injective.
J be an ideal of M ( R ) and let I C R consist
Let
J.
of all entries of elements in
R
(ii) The natural map
-+
I is an ideal of R such that J = M (I).
Then
induces a surjective homomorphism M ( R )
R/I
Mn(R/T)
--f
whose kernel is M n ( I )
.
Z ( M ( R ) ) = Z(R)
(iii) This is a direct consequence of the fact that
r e R , Mn(R)r
f o r any
=
Mn(Rr).
and that,
n
To present our next result, we require the following preliminary observations. V
V =
Vi
7.5. LEMMA.
Let
ible and let
W be an irreducible submodule of V . i for which W
ranges over those Proof. V .
be an R-module, let
Choose nonzero
W
A
Rw
=
Rvi.
V. =
to
V
into rvi
Since
W,Vi V
completely reducible R-module
Then
V
V = 8 V
Let
+I i'
such that
V j , j E S,
I.
For each
Vi,i
E
58 'Lj
is irreduc-
j
where
for r E R
vi
E
Vi.
If
is a nonzero homomorphism
are irreducible, W
2
Vi
as required..
is said to be homogeneous if it can be The sum of all irreducible V
which are isomorphic to a given irreducible submodule of
called a homogeneous component of 7.6. LEMMA.
W
with
= cVi
written as a sum of isomorphic irreducible modules. submodules of
V.
where each
i'
and write w
w E W
# 0 then the map sending rw
from
2
@
is
V.
where each
Vi
is irreducible and let J
5I
be
are the representatives of all isomorphism classes of
j
J,
let
Xj
denote the sum of all
Vi
with
V.
z
2
V
j-
Then (1)
The X j E J, are all homogeneous components of
V
j'
(ii) V = G3
x
i€J j Proof. (i) Apply Lemma 7.5. (ii) Direct consequence of the definition of Let
V
be any R-module.
invariant in
V
7.7. LEMMA.
Let
V
V.
j*
Then a submodule W
V
of
if it admits all R-endomorphisms of
is said to be fuZZy
V.
V # 0 be a completely reducible R-module.
is fully invariant if and only if
ents of
X
W is
A submodule
W of
a sum of certain homogeneous compon-
MATRIX RINGS AND RELATED RESULTS
Proof.
By Lemma 3.13,
v
51
can be written as in Lemma 7.6.
If
f E End(V)
R
V.
and
V
then either f(V,l
j' Hence
z
ducible.
f(xj) 5 X.,
=
V. z
0 or f(V.)
X
proving that each
j'
is fully invariant.
F/
P+Q
so
Let f
:
P, then Q
Q
=
5 W.
Q
If
P @ Q and therefore V
=
It suffices to show that If
P, this is clear; otherwise P n Q
Q
= P @
U for some submodule U of
@
p--+ Q be the given isomorphism and define an endomorphism $ l J J ( z + y+ a )
COROLLARY.
7.8.
is irre-
W and Q is an irreducible submodule of V
is an irreducible submodule of
such that
j
Vi
is fully invariant.
Conversely, assume that P
since
and hence the sum of
J.'
3
X
certain
V
Let
v
=
+
f ( z ) + f-l(y,
V by
Q,%
E
# 0 be a completely reducible R-module and express V
the sum of its homogeneous components
6X
V =
:
0,
V.
of
(z E P,y E
z
=
U)
as
Then
j € J j' End(V) R Proof. V
of
(f.1
Any family
3
2
of endomorphisms of
and it is clear that the correspondence
morphism
rl End(X.1 3
jEJ
-
X
(f.1 3
j I-+
defines an endomorphism f
f is an injective homo-
End(V) R
is fully invariant (Lemma 7 . 7 ) , any endomorphism f of V maps j into itself. Setting fj = flXj,j E J , it follows that (fi) W f
Since each each
n End(X.) R
j€J
X
j
X
.
as required. 7.9. LEMMA.
Let
V be an R-module and let e be an idempotent of R. Hom(Re,V) R
Similarly, if
V
eV
Then
as additive groups
is a right R-module, then Hom(eR,V)
Ve
R Proof. map V E
If f E Hom(Re,Vl, then ef(e) = f(e2) = f(e) E eV.
R
fF+ f(el is a homomorphism from Hom(Re,V) to eV. R eV
then the mapping g,
:
xe-
33u
Hence the
Conversely, if
is an R-homomorphism from
Re
to
V.
52
CHAPTER 1
Because t h e map
V L-+ gv
f kf ( e ) ,
i s an i n v e r s e of
t h e f i r s t isomorphism
The second isomorphism can b e e s t a b l i s h e d by a similar argument.
follows.
Given a r i n g
R
groups of
R, R"
and
Ro
we denote by
i.e.
t h e opposite r i n g ,
Re
c o i n c i d e and m u l t i p l i c a t i o n i n
the additive
aob
i s g i v e n by
ba
=
a,b E R. 7.10.
LEMMA.
Let
e
be an idempotent of a r i n g
R.
Then
End(eR) EZ eRe
End(Re) 1 ( e R e ) " and
R
R
In particular, End( R )
R R
= R"
End(R ) h' R
V = Re,
Invoking Lemma 7 . 9 f o r
Proof.
and
an isomorphism of t h e a d d i t i v e group of
R
2
f W f(e)
w e see t h a t t h e map
EndtRe)
is
eRe.
o n t o t h e a d d i t i v e group of
R
f,g
Given
E EndfRe),
R
f(e)
write
=
er e
g ( e ) = er e
and
for some
,r E R.
P 1
2
Then
($9)(el = f ( e r e ) = er e r e 2
=
1
(er e ) (er e )
proving t h a t t h e given map r e v e r s e s t h e m u l t i p l i c a t i o n . t i t y element of t h e r i n g
eRe,
= g(e)f(e)
Since
e
i s t h e iden-
t h e above map p r e s e r v e s i d e n t i t y e l e m e n t s .
This
e s t a b l i s h e s t h e f i r s t isomorphism and t h e second f o l l o w s by a s i m i l a r argument.
7.11.
LEMMA.
( S c h u r ' s Lemma).
is a division ring Proof.
z e r o submodule of and
Kerf
7.12.
V
=
LEMMA.
R/X
V
b e a nonzero R-homomorphism.
and K e r f
# V
i s a p r o p e r submodule o f
0, p r o v i n g t h a t f i s an isomorphism. Let
V
be an R-module.
Clearly
R/X
Conversely, assume t h a t
RV
V
f o r some maximal l e f t i d e a l
Proof.
Then
V
Then
X
Then
f(V)
V.
End(V)
R
V
R-homomorphism.
Hence
R/X
V = RV. 2
=
v
a
i s i r r e d u c i b l e i f and o n l y i f
R.
of
and s o
f(V)
So t h e lemma i s t r u e .
is i r r e d u c i b l e and choose a nonzero
V
i s a non-
Thus
X
i s i r r e d u c i b l e f o r any maximal l e f t i d e a l
i s a nonzero submodule of
surjective
be an i r r e d u c i b l e module.
.
$ : V+
Let
V
Let
V
The map
R--t
f o r some l e f t i d e a l
v
of
V.
in
R. Then
V,r
++
rV i s a
X
of
R.
53
MATRIX RINGS AND RELATED RESULTS
Since
is irreducible, X
V
is maximal, as required
We have now accumulated all the information necessary to prove the following result. 7.13. PROPOSITION.
fl
Assume that
artinian and there exist primitive idempotents e ,e 1
,...,n,
integers n
Then R
is completely reducible. 2
,...,e
of
is
R and positive
such that are all nonisomorphic irreducible R-modules
(i) Rel,Re2,...,Rer
nMni P
(ii) R
(eiRei) and each
eiRei
is a division ring
i=l
Proof.
#,
1 lies in the sum of finitely many irreducible submodules of
Since
Hence R
the same is true for RR.
is artinian and so by Corollary 3.11,
there exist primitive orthogonal idempotents el,e2, ...,en ,@ = Rel @
... @ Ren
and each
Rei
of
R such that
is indecomposable (and hence irreducible).
.
We may assume that Re ,Re2,.. ,Rer
are all nonisomorphic among the Rei,1 Q i G n .
Then, by Lemmas 7.12 and 7.6, Re ,Re ,...,Rer 1
are all nonisomorphic irreducible
2
R-modules. (ii) Let
Xi
be the homogeneous component of
Then, by Lemma 7.6, RR = X @ integer n
i' 1
G
i
4
r.
... @ X,
fl
and X.
corresponding to Re
i'
(Re.)
"i
lGiQr.
for some positive
Invoking Lemma 7.10, Corollary 7.8 and Proposition 7.2,
we derive
and therefore
The fact that e.Rei is a division ring being a consequence of Lemmas 7.10 and 7.11, the result follows. The Jacobson rudical maximal left ideals of Let by
V
J(R)
R.
be an R-module.
of
R is defined to be the intersection of a11
The ring
R is said to be semisimpZe if J(R)
The annihilator of
V,
written
= 0.
ann(V), is defined
54
CHAPTER 1
{r E R ( r V
ann(V) = I t is clear that
module.
I
An i d e a l
of
i r r e d u c i b l e module.
is faithfuZ i f
V
W e say t h a t
V may be viewed a s an R/ann(V)-
i s an i d e a l and t h a t
ann(V)
01
=
ann(V) = 0.
i s s a i d t o be primitive i f t h e r i n g
R
has a f a i t h f u l
I is
i s p r i m i t i v e i f and o n l y i f
I
Thus
R/I
the annihilator
of an i r r e d u c i b l e R-module. 7.14.
Every maximal l e f t i d e a l of
LEMMA.
R
c o n t a i n s a p r i m i t i v e i d e a l and e v e r y
p r i m i t i v e i d e a l i s t h e i n t e r s e c t i o n of t h e maximal l e f t i d e a l s c o n t a i n i n g it. Proof.
X
Let
d e n o t e a maximal l e f t i d e a l of
an i r r e d u c i b l e R-module.
whose a n n i h i l a t o r i s
Mv
Clearly
Mv
M
0 # v E V,
J(R)
5I
J(R)
R/X
is a primitive ideal
R/X
is
R.
R
in
V
b e an i r r e d u c i b l e R-module
V,
put
MV
is i r r e d u c i b l e ,
.
I
Since
=
{r
E
Rlrv
V = RV
=
i s t h e i n t e r s e c t i o n of a l l
be a r i n g .
i s an i d e a l of
I
In
R.
i s t h e a n n i h i l a t o r of a c o m p l e t e l y r e d u c i b l e R-module. if
01.
R/Mv
i s t h e i n t e r s e c t i o n of t h e a n n i h i l a t o r s of i r r e d u c i b l e R-modules.
particular, J ( R ) (ii)
Since
the r e s u l t follows. Let
v
Given a nonzero
R.
V
and l e t
i s a maximal l e f t i d e a l i n
PROPOSITION.
(i) J(R1
I.
is a l e f t ideal i n
V
with
7.15.
R
I be a p r i m i t i v e i d e a l i n
Let
By Lemma 7.12,
X.
containing
and so
Hence t h e a n n i h i l a t o r of
R.
Furthermore,
i s t h e a n n i h i l a t o r of a completely r e d u c i b l e R-module.
( i ) D i r e c t consequence of Lemmas 7.12 and 7.14.
Proof.
Vi,
= n ann(Vi’i).
S e t t i n g {Viji E I) t o b e t h e i€I s e t of r e p r e s e n t a t i v e s of t h e isomorphism classes of i r r e d u c i b l e R-modules, i t (ii) I f
V = 69
then
ann(V1
iEI
follows t h a t
J ( R ) = ann(V).
The second s t a t e m e n t i s a consequence of t h e f a c t
.
t h a t any completelv r e d u c i b l e R-module i s e i t h e r of i r r e d u c i b l e submodules.
7.16.
PROPOSITION.
Then
f(JUi)l
5J ( S )
Let
f
:
R
4
S
with e q u a l i t y i f
0
o r a d i r e c t sum of a f a m i l y
be a s u r j e c t i v e homomorphism of r i n g s . Kerf
J(R).
55
MATRIX RINGS AND RELATED RESULTS
Proof. taining
{Mil<
Let
Kerf.
E I}
be the set of all. maximal left ideals of
{f(Mi) li
Then
€
R
con-
s
I } is the set of all maximal left ideals of
and
f(J(R) 1 5f
c
n Mi’
n f(Mi) = J ( S ) {€I
=
iEI
If Kerf Hence
5J ( R ) ,
nM i€I i
J(R) =
2
(J(R)
If
is the set of all maximal left ideals of
be an ideal of a ring R.
+ I)/I
(ii) If I C_ J ( R ) , then J ( R / I ) = J ( R ) / I . (iii) If J(R/I) = 0, then J ( R ) (iv) I = J ( R )
if and only if
In particular, J(R/J(R))= 0.
5I
I C_ J ( R 1
and J ( R / I ) = 0.
(i), (ii): Apply Proposition 7.16 for the case where f : R -
Proof.
R.
and, by the above f(J(R)) = J ( S ) .
Let I
7.17. COROLLARY. (i) J ( R / I )
{Mi/<
then
R/I
is the natural homomorphism. (iii) Direct consequence of (i) (iv) Direct consequence of (ii) and (iii). An element x exists y
€
element x
is called a left Irespectively, r i g h t ) unit if there
R
of
such that yx = 1 (respectively,
R
of R
=
By a Unit we mean an
1).
which is both a left and a right unit. -1
x
and only if there exists y E R(denoted by 7.18. PROPOSITION.
)
= y x = 1.
such that
x be an element of a ring R.
Let
only if for all P E R , 1-rx is a left unit.
x is a unit if
Thus
Then X E J ( R )
In particular, J ( R )
if and
contains
no nonzero idempotents. Proof. Assume
that
Note that
x
6 J(R)
R(1-rx) c - I where
x
E
I, hence rx
1-rx but
is a left unit if and only if
R(1-rx) # R
for some r E R .
I is a maximal left ideal of R .
6 I
and
1 = (1-PX)
+ rx
Conversely, assume that R(1-rx) = R ducible R-module. for some r E R ,
x
E ann(V)
Given i.e.
V E
V,
if
I,
Since I E J ( R )
r E R.
# 0, then R m =
x
Rv
By Proposition 3.5, we have
a contradiction.
for all
(1-rx)V = 0. Hence
and, by Proposition 7.15(i),
E
R ( 1 - m ) = R.
Let
v,
V
be an irre-
hence
V =
= 0 and therefore V = 0 .
E J(R).
Pm Thus
CHAPTER 1
56
e
Assume t h a t some
x E R.
7.19.
PROPOSITION.
R
1 of
y
Hence
be a r i n g .
2(1- e) = 1
By t h e above,
e
yields
Then
J(R)
1 - r z is a unit for a l l r
E R,
e
0.
=
(1-x) i s a l e f t u n i t ,
Since
is in
i s t h e unique l a r g e s t i d e a l
x E I.
R
Let
y(1-2)
l=y'(l-z)
and s o
1 - z = y-l
i s a u n i t and s o
1 . 2 0 . COROLLARY.
J(R)
for
8
Owing t o P r o p o s i t i o n 7.18, it s u f f i c e s t o show t h a t
1 - y = -yx
2 =
R
Let
x E J(R).
for a l l
J(R).
M u l t i p l y i n g on t h e r i g h t by
such t.hat
Proof.
Then
i s an idempotent i n
=
1
=
y'y
is a unit
1-2
f o r some
y E I?.
y' E R .
f o r some
is a l s o a unit.
be a r i n g .
J ( R " ) = J(R1
(i)
(ii) J ( R )
Proof.
R
i s t h e i n t e r s e c t i o n of a l l maximal r i g h t i d e a l s of ( i ) D i r e c t consequence of P r o p o s i t i o n 7.19
(ii) Follows from ( i ) . 8
7.21.
COROLLARY.
Proof.
for any
I is
If
I
Assume t h a t
i s a l e f t n i l i d e a l of
I
and l e t
is a unit.
Hence, by P r o p o s i t i o n 7.18,
.
7.22. PROPOSITION.
Let
E I}
{Rili
X E
J(R).
I.
Then,
Since
1.
R"
i s a r i g h t n i l i d e a l , t h e n t h e same argument a p p l i e d t o
result.
I
then
i=o
i=o
1-rx
R
( r z ) n= 0 f o r some n
r E R , rz E I and so
it f o l l o w s t h a t
R,
a l e f t o r r i g h t n i l i d e a l of
be a f a m i l y of r i n g s .
I &J(R).
If
yields the
Then
J(nRi) = nJ(Ri) i€I
iEI
Proof.
An element
(ril E n R i
i s a l e f t u n i t i f and o n l y i f
ri
is a
,LEI
l e f t u n i t of 7.23.
R i
for a l l
PROPOSITION.
Proof. s i t i o n 7.3.
Let Then
i
6
I,
For any r i n g
V
Now a p p l y P r o p o s i t i o n 7.18.
R
and any p o s i t i v e i n t e g e r
be an R-module and l e t ann(J?) =
M
v"
b e an
M
n
8
n,
(R)-module a s i n Propo-
( a n n ( V ) ) , by t h e d e f i n i t i o n of
J?.
Hence t h e
57
MATRIX RINGS AND RELATED RESULTS
required assertion is a consequence of Propositions 7.3 and 7.15(i). Let R
7.24. PROPOSITION.
be a ring.
Then the following conditions are
equivalent (i) R
is semisimple artinian
(ii) RR
is completely reducible
#
(iii)
is a direct sum of finitely many irreducible modules
(iv) Every R-module is completely reducible (v) R
is a finite direct product of full matrix rings over division rings
Proof.
(i)
=)
(ii): Since R
is artinian, RR
is finitely cogenerated.
n n li = 0 for some maximal left ideals i=l n 11,1 2,...,1n of R . Hence RR is isomorphic to a submodule of (R/li) and i=1 so is completely reducible, by Lemma 3.12.
Since d ( R ) = 0, this implies that
n
(ii) * (iii): By Proposition 7.13, R
is artinian and so the required assertion
follows by virtue of Corollary 3.14. (ii
)
* (iv): App1.y Corollary 3.15 and Lemma 3.13
(iv * (v):
Direct consequence of Proposition 7.13(ii).
(v) * (i): By hypothesis, there is a direct sum decomposition
R =
li is a two-sided ideal of R
where each
Di.
and some division ring
Di
r o ... O
and I 'M
(D.) i- ni z
By Propositions 7.22 and 7.23,
is artinian, so is li by Proposition 7.3.
artinian.
I ~ for some
n 2- 1
i
J ( R ) = 0.
Since
Hence, by Corollary 3.8,
R
is
'
7.25. COROLLARY.
Let R
be a ring.
Then the following conditions are equiv-
alent: (i) R
is a simple artinian ring
(ii) R
*M
Proof.
(D) for some n
2- 1
and some division ring
(i) * (ii): By hypothesis, R
Proposition 7.24, division rings. division ring.
R
D.
is artinian and
J ( R ) = 0.
Hence, by
is a finite direct product of full matrix rings over
But R
is simple, hence
R
is a full matrix ring over a
58
(ii)
CHAPTER 1
*
(i): By Proposition 7.24,
is also simple.
R
is artinian and, by Proposition 7.4(i), R
59
2 Group-graded algebras and crossed products: General theory
been a tremendous s u r g e of a c t i v i t y i n t h e t h e o r y
I n t h e past ten years th ere has
of graded r i n g s and t h e i r i m p o r t a n t s p e c i a l case, namely c r o s s e d p r o d u c t s .
The
aim of t h i s c h a p t e r i s t o p r o v i d e t h e f o u n d a t i o n s of t h e t h e o r y of c r o s s e d p r o d u c t s w i t h emphasis on t h e r e f i n e m e n t s and e x t e n s i o n s i n scope a c h i e v e d i n r e c e n t years.
The bulk of t h e c h a p t e r c o n s i s t s of a number of r i n g - t h e o r e t i c
r e s u l t s r e q u i r e d f o r subsequent i n v e s t i g a t i o n s .
These i n c l u d e MascNce's theorem,
t h e c e n t r e of c r o s s e d p r o d u c t s , t h e Jacobson r a d i c a l , induced modules, H-project i v e and i n j e c t i v e modules, and p r o j e c t i v e c r o s s e d r e p r e s e n t a t i o n s .
The c h a p t e r
c u l m i n a t e s i n p r o v i n g a b e a u t i f u l r e s u l t due t o Montgomery (1976) which asserts that i f
IG1-l E R ,
then
G = J ( R ) n 'R
J(R
=
J(R)
G
1. DEFINITIONS AND ELEMENTARY PROPERTIES
Throughout t h i s s e c t i o n , Given R-submodules
X
A
and
d e n o t e s an a l g e b r a o v e r a commutative r i n g
Y
of
A,
we w r i t e
XY
R.
f o r t h e R-submodule of
A
c o n s i s t i n g of a l l f i n i t e sums
CXiYi Let
G
be a m u l t i p l i c a t i v e group.
W e say t h a t
with
xi
E
X,yi E Y
i s a G-graded algebra i f
A
t h e r e e x i s t s a family
{ A g l g E GI of R-submodules c o n d i t i o n s hold:
of
A
indexed by t h e e l e m e n t s of
G
such t h a t t h e following
CHAPTER 2
60
( d i r e c t sum o f R-modules)
r,y E G
for all By (1) each
a
A
in
ag # 0.
w i t h f i n i t e l y many
The
support of
a,
written
&G
a 9'
E A
9'
i s d e f i n e d by
Suppa,
a
i s a f i n i t e s e t t h a t i s empty i f and o n l y if
i s d e f i n e d as
c
=
Gla # 01 9
s u p w = Ig E Thus Suppa
a
can be w r i t t e n u n i q u e l y i n t h e form
=
0.
l e n g t h of a
The
/Suppa/.
W e s h a l l r e f e r t o (11 as a G-grading of
A
and t o
A
9
as t h e g-component o f
A. When ( 2 ) i s r e p l a c e d by t h e s t r o n g e r c o n d i t i o n .
A A
(3)
3 y
A
z,y E G
Of c o u r s e , i f
R =
k,
then w e
i s a G-graded r i n g ( r e s p e c t i v e l y , s t r o n g l y G-graded ring) i n s t e a d of
A
say t h a t
for a l l
w
i s a s t r o n g l y G-graded a l g e b r a .
A
we say t h a t
= A
i s a G-graded a l g e b r a ( r e s p e c t i v e l y , s t r o n g l y G-graded a l g e b r a )
Let
1.1. LEMMA.
Al
(i)
A
be a G-graded a l g e b r a .
A
i s a s u b a l q e b r a of
( i i ) For each
g E G, A
1E Al
and
i s an
g
( A ,A )-bimodule under l e f t and r i g h t m u l t i p l i 1
1
.
A
c a t i o n by t h e e l e m e n t s of
Then
1
(iii) A
1E A A
i s s t r o n g l y G-graded i f and o n l y i f
Proof.
( i ) By d e f i n i t i o n ,
multiplicatively closed.
Al
9 g-1
i s an R-submodule of
g
for all
A
E G.
and, by ( 2 ) ,
I t therefore suffices t o verify that
1E A 1
.
A]
By (1)
t h e r e i s an expansion
1 =
c a gEC;
where
a
E A
9 F i x some
(4) By ( 2 )
9 t E G A =
for a l l
g E G,
g
and a l l b u t a f i n i t e number of
a
a r e zero.
9 and
a;
E
At.
Because
G
i s a g r o u p , (1) i m p l i e s t h a t
@ A
g€G gt t h e p r o d u c t a a' l i e s i n A g t gt
i s p r e c i s e l y t h e expansion of
for a l l
g E G.
Therefore
a' = 1.a' = z a a t t g€G g t ' ui i n t h e decomposition (4).
But
a;
is
already
DEFINITIONS AND ELEMENTARY PROPERTIES
l i e s i n t h e summand
aiai
z e r o and
t
all
I n view of
A.
al
Thus
=
a
(1) t h i s f o r c e s
a
E Al
imply t h a t
A
Hence a l l t h e
Therefore
(ii) Direct consequence of (iii) Suppose t h a t
i n (4).
At
a;.
must be
E G.
algebra
A l.t
g 9-1
for
9 t
g # 1 must be At
for
t o be a l e f t i d e n t i t y f o r a l l t h e
1
i s t h e i d e n t i t y element of
A,
proving ( i ) .
(2).
Then ( 3 ) and t h e f a c t t h a t
1E A
g E C.
for a l l
Conversely, assume t h a t
a a'
a c t s as a l e f t i d e n t i t y on
i s s t r o n g l y G-graded.
1E A A
61
1 E A A -1 g g
for all
g E G.
Then, a p p l y i n g ( 2 ) , we
derive
for all
x , y E G.
T h u s (31 h o l d s and t h e lemma i s proved.
For t h e r e s t of t h i s s e c t i o n ,
denotes the unit group o f
UiAl
u
We s a y t h a t a u n i t We shall refer t o
g
E
as the
denotes a G-graded algebra over
A
1.2.
LEMMA.
(ii) C r U ( A )
(i) If
U ( A ) i s graded i f it l i e s i n
degree of
u
deg:
A 9
f o r some
and
u E GrU(A)
g E G.
and w r i t e =
deg ( u !
A
d e n o t e s t h e s e t of a l l graded u n i t s of
i s of d e g r e e
g,
then
u-l
U(A). g-'.
i s of d e g r e e
U(A1
i s a subgroup of
(iii) The map
(iv)
GrU(A)
R
A.
g = deg(u) I n what f o l l o w s
a
C r U ( A ) -+ G
i s a group homomorphism w i t h k e r n e l
U(A1)
The map
1
GrU(A)
-f
Aut A
u-iU
where
i i s a homomorphism.
(2) =
uxu
-1
for a l l
x
E
Al
1
62
CHAPTER 2
u
(v) Right multiplication by any
A
E
g
A - A u 1
n U(A)
9
is an isomorphism
= A
9
19
of (left) A -modules.
u-l
(i) Write
proof.
number of them are zero.
C ax with 35G
=
a
Since uax E A A
E Ax (I
A
gx 1 uax is a unique expansion for uu-l
that uu-l =
gx-
such that all but a finite
x
for all
E
G , it follows
in the decomposition
&G
A =
@ A
uu
But
-1
= 1
@ Ax
=
&GgX
zSG -1
by Lemma l-l(i). Hence the a for x # g X 39-l lies in A -1 Therefore L4-l = a 1 must equal Ua -1 . -1
lies in A
must be zero, and
=
A
g
g
as
(ii) Direct consequence of (i) and the fact that A A
x v -C
(iii) Since A A
C A
X Y -
A
g
required.
1
U(A).
1
for all z,.!f
?Y
E G.
the given map is a homomorphism with kernel
w'
BY (i), A
A
n U ( A ) C_
and since the opposite containment is
U(A1)
trivial, the required assertion follows. (iv) By hypothesis, u E A
Since G r U ( A )
for some g E G.
Hence, by (i), we have
is a group, this implies (iv). is a unit of A ,
(v) Since u g phism of A l
g
onto
A u 1 g'
right multiplication by
Because u
9
E A
g'
is an A -isomor-
u
9
1
we have A u C A . 1 g - 9
On the other
hand, by (i),
proving that A
g
=
A u
9'
1
S o the lemma is true.
Owing to Lemma 1.2(iii), the sequence of group homomorphisms (5)
1
+
U ( A 1-
GrU(A)
G-
1
is always exact except possibly at C. We say that a G-graded algebra A written A = A * G ,
is a crossed product of
provided the sequence (5) is exact.
G
over A ,
Expressed otherwise,
1
a G-graded algebra A
is a crossed product of
G over A
1
if and only if
63
DEFINITIONS AND ELEMENTARY PROPERTIES
A
# 0
$7
I n case ( 5 ) i s an exact s p l i t t i n g sequence, w e s h a l l r e f e r t o
G over A .
ring of
g E G,
i f f o r each
g E C,
p a r t i c u l a r c a s e , where
over
G
skew group r i n g ,
A
Let
9 E Ag
there exists
algebra of
Al
Al
=
-
A
*G
Z(A)
C 1 -
.
n U(A) such t h a t
Finally, i f
A *G
in
G
be a c r o s s e d p r o d u c t of
A
Let
over
A
Al.
*G
i f for In the
as a twisted group
i s a t w i s t e d group a l g e b r a and a
Al
.
G
A
over
.
A map
)
i f , f o r all
g E G,
t h e r e is a u n i t
9
1 = 1 such t h a t
with
-1
0 ( g ) (x) = B
Al
A
-
9 (6)
over
i s n o t h i n g e l s e b u t t h e group a l g e b r a of
i s c a l l e d an automorphism system f o r
A
G
centralizes
we shall refer t o
5 : G+Aut(A
of
i f and o n l y
xy
i s s a i d t o be a twisted group ring of
A
G-graded a l g e b r a
all
Al
over
G ,
E
-xy A
G
i s a s k e w group r i n g of
as a skew group
A
there e x i s t s
x,y
such t h a t f o r a l l
A
Thus
g E G
for a l l
jxg
for a l l
be an R-algebra.
zE A
1
Given maps
and
we say t h a t
(G,R,u,a)
x , y , z € G , and
b E B,
i s a crossed system f o r
over
B
i f , for a l l
t h e following p r o p e r t i e s hold:
-1
(7)
x(yb)
(8)
a ( x , y )a (xy, z )
(9)
a(x,1) = a ( l , r ) = 1
=
G
LY(z,y)% =
CI(x.,y) X
a ( y J ) (x.y2)
where for all
g E G, b E B
64
CHAPTER 2
1 . 3 . THEOREM.
g E C,
For each :
0
G
+
AutA
A
L e t a G-graded R-algebra
let
9
G
be a c r o s s e d p r o d u c t of
A
be a f i x e d u n i t of
A
in
with
g
=
over
A l
let
1,
be t h e corresponding automorphism system g i v e n by ( 6 ) and l e t
1
G x G “U(Al)
:
be d e f i n e d by
---
-1
Cc(x,y) = x y z y
(10)
Then t h e f o l l o w i n g p r o p e r t i e s hold: (i) A (ii)
(G,Al,U,a) i s a c r o s s e d system f o r
-
A g = gA 1
A
over
1
( t o which we r e f e r a s
1
1
For a l l
1
2
z,y E G
and
P ,P E A 1
(rlcC)(r2y)= r l z p 2 a ( x , y ) G
(11)
B
Conversely, f o r any R-algebra t h e f r e e B-module
and any c r o s s e d system
r ,r 1
g E G)
(G,B,U,Cc)
C f r e e l y g e n e r a t e d b y t h e elements
p l i c a t i o n given by (11) ( w i t h for a l l
=
i s a f r e e ( l e f t and r i g h t ) A -module f r e e l y g e n e r a t e d by t h e elements
(iii) A
B,
G
-
g
A)
corresponding t o
(iv)
A
i s a s t r o n g l y G-graded a l g e b r a w i t h
E
B) is
-
g,g
E
a G-graded R-algebra
for
G
over
C
=
B
over
G with multi(with
2
which i s a c r o s s e d p r o d u c t of
G
C g
=
Bg
and having
(G,R,o,n) a s a corresponding c r o s s e d system Proof.
Assume t h a t
A
71 E
1 = uu-l E A A
g
n U(A) f o r some g 6 G.
Then, by Lemma 1 . 2 ( i ) ,
S i n c e t h e above i s t r u e f o r all g g-l s t r o n g l y G-graded by v i r t u e of Leinma l . l ( i i i ) . The e q u a l i t y
and so
.
consequence of Lemma 1 . 2 ( v ) . p l i c a t i o n a l s o shows t h a t Fix
b
E
Al
and
A
proving ( 7 ) .
=
,(yby Given
z,y,z
g
= /I g
1
is is a
-
= gAl,
g
x,y E G.
,r
A
-
The argument of t h i s lemma a p p l i e d t o l e f t m u l t i -
=
E
proving ( i ) .
Then
- --1
-1 -1
“Pb)
g 6 G, A
-1
Cc(z,y)zybxy a ( x , y ) G,
we have
=
a ( x , y ) q b a(z,y)-l
65
DEFINITIONS AND ELEMENTARY PROPERTIES
proving ( 8 ) .
i
Since
=
i s a l s o t r u e and (ii)i s e s t a b l i s h e d .
1, ( 9 )
P r o p e r t y (iii) i s a d i r e c t consequence o f ( i ) . Given
x,y E G ,
E A
r ,r 1
2
and
I
w e have
(r Z) (r I
proving ( i v )
2
y)
( 22Z
r
=
I
-1
(2;)
)
=
-
X
r 1
r a(x,y)xy 2
.
To prove t h e c o n v e r s e , i t s u f f i c e s t o show t h a t t h e m u l t i p l i c a t i o n g i v e n by
(11) i s a s s o c i a t i v e .
So f i x
r
,P 1
,r
2
E
B
x , y , z 6 G.
and
3
Then
a s required. Let
RG
G
d e n o t e s t h e group a l g e b r a of
G.
a commutative r i n g .
G o v e r R. G
f r e e l y g e n e r a t e d by t h e elements of by t h a t i n
R
be an a r b i t r a r y group and
Thus
RG
I n what f o l l o w s
i s a f r e e R-module
and t h e m u l t i p l i c a t i o n i n
RG
i s induced
W e n e x t p r e s e n t some examples of c r o s s e d p r o d u c t s a r i s i n g from
RG. Let
N
G.
be a normal subgroup of
Then t h e group a l g e b r a
RG
c a n be
r e g a r d e d as a (G/N)-graded R-algebra by s e t t i n g
(RG)gN = g ( R N ) Since f o r a l l of
RG,
g E G, g
RG
i s a u n i t of
which l i e s i n t h e gN-component
g(RN)
we conclude t h a t
RG
=
RN*(G/N)
The f o l l o w i n g p a r t i c u l a r c a s e s d e s e r v e s p e c i a l a t t e n t i o n .
Case I. N
Assume that
and a subgroup
B
G of
s p l i t s over
G.
Then
RG
N,
i.e.
G
i s a s e m i d i r e c t p r o d u c t of
i s a skew group r i n g of
B
over
RN
CHAPTER 2
66
Assume that N
Case 2 .
N.
Then R G
is a twisted group ring of
Assume that N
Case 3.
group algebra of Case 4 .
G/N
T in G
has a transversal
G/N
such that
RN.
over
is a central subgroup of
T centralizes
G.
Then R G
is a twisted
over R N .
Assume that N
G,
is a direct factor of
say G = N x B .
Then R G
B over R N .
is a group ring of
We next present circumstances under which a crossed product is a twisted group ring.
Then A
inner.
* G be such that every automorphism of A l is
Let A = A
1.4. PROPOSITION.
is a twisted group ring of
G over
Al
.
choose u E A n U ( A ) . Since conjugation by u g g g -1 is an automorphism of A l , there exists V E UIA ) such that u 2, central9 g g -1 Setting = u v for all g E G, it follows that E A n U(A) izes A l . g g g For each g E G,
Proof.
9
and
-
,
centralizes A
g
as asserted.
For future use, we next record the following observation. Let S
1.5. LEMMA.
of
=
e
1
+e + 2
... + e
be a decomposition
1 into a sum of orthogonal idempotents, and let K = e Se 1
subgroup of the unit group of
Proof.
Because
suitable g E G , 1 Q
i
=
-1
gi elgj.
11
which permutes the set
M (K)
+
.
G be a
Let
{ e l , . . . , e n } transi-
as rings and K-modules.
-1
i
n, with g
=
1.
Given
Our aim is to verify that the
+ ...
1
G acts transitively, we may write ei = gi e l g i
(a) vijvks = 0 if j # k (b) 1 =
S
Then S
tively by conjugation.
Vij
1
be a ring, let
and
V z3 ..V
ks
=
vis
vij
i , j E {l, ...,nl,
set
satisfy conditions
if j = k
'nn
Once this is accomplished, the result will follow from Proposition 1.7.1. this end, fix
k # j
for a
and observe that g-1 . e g. # g -1 k elgk 3 1 3 -1 -1 ( g . e g.) (gk e g ) = 0 3 1 3 i k
so that
To
67
DEFINITIONS AND ELEMENTARY PROPERTIES
-1
-1 -1 (9.e g .) ( g k e l g k ) 3 13
= (g. g.)
2 . 3
= G and
vijvjt
=
-1
-1
(gi e 1g 3.) (gj e ~t g
)
-1
=
9; elgt
=
Vit'
proving ( a ) . By d e f i n i t i o n ,
V
.
= e
ii
(b) and hence t h e r e s u l t .
i
1 = vI1 t.
and t h e r e f o r e
A
A s s u m e t h a t a G-graded R-algebra
vZ2 +
... + vnn, G
i s a skew group r i n g of
proving
over
A 1
.
Then, by d e f i n i t i o n ,
i s a s p l i t e x a c t sequence.
such t h a t f o r a l l
x,y
Hence w e may i d e n t i f y identification,
E
G
x
9
E
G,
there exists
G
{slg E G}
w i t h t h e subgroup
each element of
with f i n i t e l y many
9
T h i s means t h a t f o r each
# 0.
A
of
U(A).
With t h i s
can b e u n i q u e l y w r i t t e n i n t h e form
Furthermore
( g E G,h E H ) where
1 . 6 . PROPOSITION. p r o d u c t of
n
Let
c o p i e s of
S
S,
be a r i n g , l e t and l e t
A
= S x
G =
S
x
... x
S
be a d i r e c t
be a c y c l i c g r o u p o f o r d e r
n
68
CHAPTER 2
A
on
G
with t h e a c t i o n of
g i v e n by
8(S1,S2,.
If
A *G
.. ,sn)
Proof.
< i < n.
1= e
Then
+ e2 +
1
... + e
e
particular,
h e
1
S.
1
12
then
as r i n g s and S-modules
1 is i n the i - t h position,
i s a decomposition of
i he.
and f o r a l l
G
The a c t i o n of
h
on
1
S
1 i n t o a sum of
qua r i n g s .
ensures t h a t
.
{el, e 2 , . . ,e
U(A) , permutes t h e s e t
a subgroup of
A,
over
M, 6 )
ei = (0, ...,l,o,...,0) where
Let
o r t h o g o n a l c e n t r a l idempotents of
1
1.7. PROPOSITION.
G.
A
Then
A
Let
N
be a normal subgroup
a l g e b r a by means of
Ax
@
AgN =
.
t r a n s i t i v e l y by conju-
b e a G-graded a l g e b r a and l e t
can be regarded as an G/N-graded
In
G, r e g a r d e d a s
The d e s i r e d c o n c l u s i o n i s t h e r e f o r e a consequence of Lemma 1 . 5 .
gation.
of
G
i s t h e corresponding skew group r i n g of
h *G
1
,. .. ,Sn-l)
= (S,,S
figN A
Moreover, i f
G
is a c r o s s e d p r o d u c t of
regarded a s a c r o s s e d p r o d u c t of
G/N A")
over
Al,
then
A
can a l s o be
over
8 Ax
=
&N Proof.
7!'
Let
A = Given
t ,t 1
Assume t h a t
E 2
A
T,
A,
written
I t i s obvious t h a t
@ A tET tN
G
i s a c r o s s e d p r o d u c t of
-
g
For any R-algebra in
G.
we have
there exists a unit
X
in
N
be a t r a n s v e r s a l f o r
of
A
CACX),
A
in
A
g
and a s u b s e t by
over
.
A
i n which c a s e
X
of
A,
Then, f o r a l l
-g E
AgN,
w e define the
g E G,
a s required.
centralizer of
69
DEFINITIONS AND ELEMENTARY PROPERTIES
cA (X) = la A
If
Alas
E
g
i s a s t r o n g l y G-graded a l g e b r a and
=
xu1 t h e n by Lemma 1.1, 1 E A A
G,
n Hence
1
c
=
99 aibi
ai E A g' bi
n > 0 and some
f o r some i n t e g e r
i=1 Using t h i s f a c t , w e now prove t h e f o l l o w i n g r e s u l t .
1.8. PROPOSITION.
A
Let
1=
g
be a s t r o n g l y G-graded a l g e b r a , l e t
n 1 aibi f o r some i n t e g e r
n > 0
E
A -l,lG
G
E
a z. E A g' bi E A -1' 1
and some
i
n
4
and w r i t e
G
n.
g
i=l
y
For any
E CA(A1),
y'
define
by n
Then t h e f o l l o w i n g p r o p e r t i e s h o l d :
gy
(i)
y'
Furthermore,
G
g E G
w i t h any
(iii) Z ( A ) =
{c
CA(A1)'
algebra
Proof.
y
s e n d i n g any
y
C ( A ) c'/
= c
E
A
I
g
i
E
A
E {1,2
E CA(Al)
i
E Z(Al),
g E GI,
then
9'
b.a
z9
,..., n } . C a.yb.a
i=1 2
2
E
A
I
respectively, i n t o
Z(A)
i.e.
(Al
and
gy .
i s t h e f i x e d sub-
A
y-l 9
=
A
b.a 29
and so 1
commutes
I t follows t h a t
!7
=
n C a.b.a y i=l z -L g
x be any s o l u t i o n t o
Let
E {1,2 ,...,n } ,
y
and
for a l l
n gyag = proving (12).
z( A ) a l g e b r a s cA( A l )
Z ( A l ) tgy
E
G.
If
for a l l
y
a c t s a s automorphisms of t h e
of
(i)
satisfying
and, p r o v i d e d
E CA ( A l )
( i i ) The group
with
A
i s a unique element of
=
(12).
1-a y g
Then
=
a y,
a.y
g
=
xui
for a l l
and so
n
aiybi i=1 Since
If
y E Al
a
g
E
implies t h a t
A
g
and
z E Al,
y'
E
9
Therefore
Al,
('yz -zgy)A
we must have
9
'yz
57
=
=
za
9
) = (za
and s o
=
x
=
'y
we are l e f t t o verify t h a t E
A A
)y
9
1 g
=
=
A
9'
z ( a y) 9
(gyz-Zgy)A A
0 and hence
zgy
C xaibi
i=1
Al,
then
(9yz)a = Sy(za
=
=
E CA(A1).
E CA(Al).
so (12) i m p l i e s t h a t
zPya 1 9 =
0.
g g-l
y'
'y
= ( 29y ) a
9
Since
1
lies in
CHAPTER 2
70
( i ), i t s u f f i c e s t o t r e a t t h e a l g e b r a
( i i ) I n view of
A
1 of
identity
,
A
lies in
g , s E G, a
A
g
a E A
and
g
x
a ’Jya
=
“
9
s(gy) (g,yl
proving t h a t Now f i x <norphism of
g
+ Jy
G.
E
y,t
E
CAMl),
a (yt) g ’(yt)
By ( i f ,t h i s i m p l i e s t h a t
CAM1),
of t h e a l g e b r a ( i i i ) Since
SfG g’ = n c (A = g€G A g c
So t h e p r o p o s i t i o n i s t r u e .
1.9. COROLLARY.
A
g Z(Al),
and
.
Proof.
g E G,
g E G
y’
I---t
i s an R-auto-
t h e n (12) g i v e s
gya t 9
’ygt.
=
gygta g
f o r all
y +--+
Therefore
Y’
a
g
=
E
A
i s an automorphism
E
c
{ c E C ~ ( A ~ E) ~ C(A 1
CA(A1)
y ’
y ’
by
G -1
A
over
1
,
= syi
.
where
y
E CAM )
y
and
E
c ’
= c.
For each
CA(A1)
Z(A 1,
G CA(A1) .
n
Apply P r o p o s i t i o n 1.8 w i t h Let
A
g
E G,
a
b e a s t r o n g l y G-graded
t h e l e f t and r i g h t A -module Given
= 1,
A
9
=
-
g
A
i s any u n i t of
a c t s as automorphisms of t h e a l g e b r a s
sending any
GI
E
i f and o n l y i f
$7
’
define
G
A
centralizes
g
for a l l
A g
and
algebra.
b
= 1
and
respectively,
Furthermore, t h e above a c t i o n does n o t depend upon t h e c h o i c e of
1.10. PROPOSITION.
Proof.
y
t h e map
be a c r o s s e d p r o d u c t of
Then t h e group
Z(A)
g E G,
A
Let
w i t h any
’y
into
and
.
’Y,
CAW1).
a s required.
Owing t o ( 1 2 ) , a n element
in
and so by (i),
x9
A = CB A Z(A)
y E CAM
=
=
x(gy)a a x g
G on t h e s e t
i s a n a c t i o n of
If
y E CA(Al)
for all
1
“qY,
=
By t h e d e f i n i t i o n of
CA(Al).
=
A
g e n e r a t e t h e submodule
“ g
Y
.=
lY.l
t h e n (12) g i v e s
sr
x g
a a
1.y =
=
(a a ) y The p r o d u c t s
Since t h e
e q u a t i o n (12) y i e l d s
y If
CA(Al).
-- 1-
g
.
g
8
Then, f o r a l l
i s f i n i t e l y g e n e r a t e d and p r o j e c t i v e .
we may w r i t e , by v i r t u e of Lemma l . l ( i i i ) ,
g
71
EQUIVALENT CROSSED PRODUCTS
1= xy 1
f o r some
n
c a t i o n by
xi
1 and some
xi
E A
-l,yi 9
d e f i n e s a homomorphism
x
t h a t f o r each
E A
...
+
1
E A
pi
A
----+
9
n.
Q
A
The r i g h t m u l t i p l i -
of l e f t A -modules such 1
9
x
=
v 1 ( x ) y1 +
... + lJn(x)yn
Hence, by P r o p o s i t i o n 1.2.6, t h e l e f t A -module projective.
i
1
g'
:
"nYn
+
A
9
i s f i n i t e l y g e n e r a t e d and
A
A s i m i l a r argument e s t a b l i s h e s t h e c a s e when
9
i s r e g a r d e d as a
r i g h t A -module. 1
1.11. PROPOSITION. is local.
A
Then
Proof.
A
Let
Since
be a s t r o n g l y G-graded a l g e b r a whose s u b a l g e b r a
A A
xY
= A
XY
for a l l
J(Al)
a
So t h e r e e x i s t
9
E A
A
- J(A
)
=
U(A
is a r i g h t inverse t o
A
A C A g-l g - 1 -
C
since
a
in
9
g E G
for a l l
such t h a t
9
=aa' Al
is a local ring.
A.
Hence
Now t h e idempotent
a''
-lag 9
lies in
I t c a n n o t be z e r o s i n c e
Thus it is t h e unique nonzero idempotent a two-sided i n v e r s e t o
a
g
in
A.
A
1 of t h e l o c a l r i n g
Al,
and
a''
a
Hence
A proving t h a t
Al*
we have
Al = A A
A -1
9
)
x,y E G,
9 g-l
and
9
u But
G over
i s a c r o s s e d p r o d u c t of
Al
9
UIA) #
i s a c r o s s e d p r o d u c t of
P
G over
.
.
for a l l
-1
is
g E G,
2. EQUIVALENT CROSSED PRODUCTS Throughout t h i s s e c t i o n , Assume that
A
A
d e n o t e s a G-graded a l g e b r a o v e r a commutative r i n g
is a c r o s s e d product of
G
over
A
.
By Theorem 1.3,
any
R
12
CHAPTER 2
choice of a unit
of
A
A
in
for each g E
9’ (G,A , o, a)
a corresponding crossed system
for
G, with
1=
1, determines
G over A , where
U
a
and
are given by
{gig
Suppose that
€
GI
(G,A , U f , a ’ )
is another such set of units, and let
the corresponding crossed system.
-
Since g E A g ,
be
it follows from Theorem 1.3(i)
that there is a function
with
u(1) = 1
such that for all g E G
It is then immediate that the following properties hold: CJ’(g) =
(3)
i
for all g E G
u(g)
u(g)
where
i (2)= LL(q1z u ( g ) - l u( g )
for all
I% E
A ,g E G
and (41
3:
for all x , y
a‘(z,y) = u ( 2 ) u(y)a(Z,y)u(Zy)-l
We s h a l l refer to two crossed systems
(G,Al,U,a) and
c q u i v a h z t if there is a map
U(A
)
with
u(l)
such that ( 3 ) and
= 1
G over
determines a unique equivalence class of corresponding crossed systems for
over A 1
G
.
Let A
B
: G4
as being
It follows from Theorem 1.3 that any given crossed product of
( 4 ) hold.
.’Il
?A
(G,Al,U’,af)
G
and
B
be two crossed products of
G over A , .
are equivaLent if there is a graded isomorphism
isomorphism of A -modules. that f ( A
g
) =
2.1. LEMMA.
B
g
f
:
A-
We say that
B
The latter means that f is a ring isomorphism such
for all g E G
and f ( a ) = a
Two crossed products of
Let A
and
B
and
which is also an
for all a E A
G over A l
.
are equivalent if and only if
they determine the same equivalence class of crossed systems for G Proof.
A
be two crossed products of
over
G over A
and let 1
EQUIVALENT CROSSED PRODUCTS
(G,A ,a',a')
(G,A ,o,a) and
73
be t h e r e p r e s e n t a t i v e s of t h e c o r r e s p o n d i n g e q u i -
1
G
v a l e n c e c l a s s e s of c r o s s e d systems f o r
.
over Al
{Gig E GI
GI and
{ilg E
above c r o s s e d s y s t e m s .
f(g)
Since
E
A
9
XrY
E
and
Suppose t h a t
g E G,
for a l l
u ( 1 ) = 1 such t h a t
A
in
f(s)
= U(g)g
B,
Choose systems of u n i t s
r e s p e c t i v e l y , which g i v e r i s e t o t h e
B
and
A
a r e equivalent v i a
u
there is a function
g E
for all
G.
:
f
:
B-A.
G-+ U ( A l ) w i t h
It follows t h a t f o r a l l
G U(xylG
proving (4).
Given
g E G
-1- -
f(a'(x,y) x y )
=
f(&
=
a'(z,y) f ( . 7 ) f ( y )
=
a'(x,y)
=
a' ( x , y ) -lu( 5 ) ( y ) Z
=
a' ( x , y )-lu ( x ) z u ( y ) a ( x , y ) ~ ,
=
-1
-1
( U W 5 )(u(y)Y)
y
a E A , w e also have
and
1
-1 O'Q)
(a)
=
gag
-1 =
f(,4ag
)
-1
proving ( 3 ) .
Hence
(G,A
,U,a)
=
f(G)f(a)f(g
and
(G,A ,a',a')
i t i s immediate t h a t t h e
p r o v i d e s an e q u i v a l e n c e of
are equivalent.
1
C o n v e r s e l y , assume t h a t t h e r e is a f u n c t i o n such t h a t ( 3 ) and ( 4 ) h o l d .
)
u :
G-+ U(A,) w i t h
u(1) = 1
Then by a p p l y i n g
A - l i n e a r e x t e n s i o n o f t h e map
A
and
R.
Our n e x t aim i s t o s u r v e y a l l c r o s s e d p r o d u c t s w i t h a f i x e d automorphism system. 2.2.
LEMMA.
Let
A
be a c r o s s e d p r o d u c t of
G
over
A
1
.
For e a c h
g E G,
74
CHAPTER 2
-
fix a unit
g
g3c =
9.9
(5)
of -1
A
A
in
Then t h e formula
g'
zE U ( Z ( A l ) )
for a l l
G
p r o v i d e s an a c t i o n of
U(Z(A
on
) )
1
-
which i s independent of t h e c h o i c e of
g,g E G. I t is c l e a r t h a t
proof.
3c,y E
for a l l
U(Z(A
'z = z
for a l l
Because f o r a l l
)).
g
u E I l ( f l l ), we have
-1 -1
proving t h e f i r s t s t a t e m e n t . t Ag
n U(A)
of
G
on
E
G,
-
--1
1 4 2
i s a n o t h e r such system of u n i t s .
Z2(G,U(Z(A U(Z(A ) )
E U ( Z ( A 1 ) ) and t h a t '(q) =g$y
-
glg2 = u g l g 2
f o r some
-]
{SIg
To prove independence, assume t h a t
and hence b o t h a c t i o n s c o i n c i d e . Let
3c
Then, f o r a l l
E
GI w i t h
g E G,
.
be t h e group of a l l 2-cocycles w i t h r e s p e c t t o t h e a c t i o n
) )
given by ( 5 ) .
Recall t h a t
Z2(G,U(Z(A
) )
c o n s i s t s of a l l
1
functions
satisfying (6)
a(z,y)n(qj,z) =
(7)
n(r,l)
=
2
n(y,z)a(z:,yz)
a(1,x) = 1
For any such f u n c t i o n s
6,
a and
R e c a l l also t h a t f o r any
t
:
G
-f
t h e i r product
U(Z(A
))
with
n4
i s d e f i n e d by
t(l) = 1, t h e 2-coboundary
1
6t is d e f i n e d by (6t)(z,y) AS u s u a l , w e w r i t e
B 2 ( G , U ( Z ( A 1 ) ) f o r t h e subgroup of
of a l l 2-coboundaries and p u t
(r,y
= t(3c)3ct(y)t(zy)-1
Z2(G,U(Z(A
) )
E
G)
consisting
75
EQUIVALENT CROSSED PRODUCTS
W e a r e now ready t o prove t h e f o l l o w i n g r e s u l t .
THEOREM.
2.3.
fl
Let
c
be a c r o s s e d p r o d u c t of
over
E = ( G , A ,O,a) be a corresponding c r o s s e d system f o r
G
1
6E
put
Z2(G,U(Z(A1)),
EB
(i)
ED
=
(G,A
6,)’ E
( i i ) For any
T of
B2(G,U(Z(A1)) G
a r e a l l n o n e q u i v a l e n t c r o s s e d systems for
D
H2(G,U(Z(A
E
Y
Eg,O E T ,
I--+
Z2(G,U(Z(A1)),
in
over
Al
with f i x e d
=
z,y,z E G
Fix
(i)
z(yb)
Since
the
ED , B E T ,
U.
c
over
A
as an automorphism system.
5
Proof. (8)
6
i n d u c e s a b i j e c t i v e correspondence between
1
having
For any
i f and o n l y i f
and t h e e q u i v a l e n c e c l a s s e s of c r o s s e d p r o d u c t s of
))
.
A
over
Z 2 ( G , U ( Z ( A 1 ) ) , ED i s e q u i v a l e n t t o
(iii) For any t r a n s v e r s a l
The map
A
y.
i s cohomologous t o
(iv)
over
and l e t
Then t h e f o l l o w i n g p r o p e r t i e s h o l d :
,O,Ra).
G
i s a c r o s s e d system f o r
Al,
and
b
E
A
1
.
Then
a(s,y)~ba(x,y)-l
b’(z,y) E Z ( A l ) ,
w e have
s(yb)
=
B ( ~ , y ) ~ ( ~R(z,y)-’ b)
=
D (z,y ) N (z, y 1 qba (z ,y ) - l R (z, y)
= (Ra) ( ~ , y ) ~ b ( B(z,y)-’ a)
Ra.
proving ( 6 ) f o r (ii) Assume t h a t
u
:
all
G
-+
U(Z(A
g E C.
) ) 1
y
Property ( 7 ) being obvious, t h e a s s e r t i o n follows. i s cohomologous t o
with
u ( 1 ) = 1.
6,
Since
On t h e o t h e r hand, f o r all
z,y
say
y
= B(8u)
f o r some
u ( g ) E Z ( A l ) , i u ( g ) U ( g ) = U(g) E G
for
CHAPTER 2
76
EB
proving the equivalence of
and
EY' u
Conversely, assume that there is a map that ( 3 ) and ( 4 ) hold with tively.
EB and
Because
:
U(A1) with u ( l ) = 1 such
G--t
a'
and
E
have the same automorphism system, it follows from
Y
being replaced by
c1
and
Yc1
@a, respec-
( 3 ) that
By ( 4 )
,
we also have
and so y = B ( 6 u ) . as required. (iii) Assume that E' = (G,A
1
G over A
is a crossed system €or
,O,ct')
Owing to (ii), it suffices to verify that
~ 1 '=
Ra
B
for some
.
E Z2(G,U(Z(A1)).
By the definition of a crossed system, we have
a'(z,y)%a'(X,y)-l which shows that
a'(T,y)R(z,y)
-1
B
:
for all ~ , Ey G,b E A
a(x,y)%X(X,y)-l
=
E U(Z(A
G x G
))
-
for all U(Z(A
X,Y
E G.
Define
))
by
-1
for all ~ , Ey G
B ( S , ? I ) = a'(X,y)a(X,y)
Then
a'
= BY ..
(iv) Direct consequence of (iii) and Lemma 2.1. Assume that A
is a crossed product of
is a corresponding crossed system for belong to A
.
and a straightforward calculation shows that
Z(fll).
Then
CT
:
G
--+
as a group of automorphisms.
G over A l ,
B E Z z (G,U(Z( A l ))
.
that E = ( G , A , o , d
G over A , and that the values of a 1
AutM
) 1
is a homomorphism and so G
Furthermore, by the definition of
acts on
a, a
is an
1
element of
Z2(G,U(Z(A
) )
,
where the action of
G on U ( Z ( A
) ) 1
is induced by U.
EQUIVALENT CROSSED PRODUCTS
Let A
2.4. COROLLARY. (G,Al,o,B)
and A '
77
over A l
be crossed products of
such that
(G,A ,O,y) are their corresponding crossed systems, where the
and
1
B
values of (i) A
'i
and
is equivalent to A'
(ii) If B (iii) If
Z(A
belong to
).
is a coboundary, then A
(i) Since the values of
homomorphism.
(C,A
Hence
apply Theorem 2.3(ii)
B
(ii) Assume that
skew group ring of
a
1
.
By (l),
commutative.
Let A
skew group ring of Proof.
Y
over A
1
.
Now
is equivalent to A '
A
is a skew
and so A
is a
' and let A
G over A
if and only if
G over A l .
Since A l
by appealing to Corollary 2.4(i) for A '
=
is a
Assume that A
is commutative,
G over A
another corresponding crossed system for
be
is a coboundary.
N
The "if" part is a consequence of Corollary 2.4(ii).
is a skew group ring of
is a
)
is a corresponding crossed system, then A
over A l
G
Aut(A
--+
so that A '
1
=
be a crossed product of
(C,Al,O,a)
If
.
G over A
.
G over A
(iii) Direct consequence of (ii). 2.5. COROLLARY.
C
.
1 and Lemma 2.1.
=
is a coboundary and take
G over A
group ring of
:
0
, a r 1) is a crossed system for G
1
for
Z(Al),
lie in
over A
G
is a skew group ring of
B
Y
is cohomologous to
is a skew group ring of
H 2 ( G , U ( Z ( A 1 ) ) = 1, then A
proof.
B
if and only if
1
.
Thus
(G,A
1
,o,l)
is
is a coboundary
'
A.
For future use, we next record the following observation. LEMMA.
2.6.
Let A
and
automorphism system. CB(A
)
5A
1
of
B.
B
be crossed products of
Assume that (!
%$(z)=
and that
J:
:
A
having the same
,
where
8
is an automorphism
1
Then A
and
B
are equivalent.
It suffices to show that
Choose units
{ g i g E G}
and
8$
{ g i g E G}
determine the given automorphism system. E A
over A i
is a ring isomorphism, that
for all z E A
1
Proof.
a
B
-----f
G
and so
:
A
-f
in A Then
B is a graded isomorphism.
and
B , respectively, which -1 -1 = Gas for all g 6 G ,
70
CHAPTER 2
-1 (8$)
Hence for a l l
g
(9)
E G,
A
centralizes
a s required.
and t h e r e f o r e lies i n
A
.
(e$) ( Ag )
Thus
= Bg
8
3 . SOME RING-THEORETIC RESULTS
R
Throughout,
d e n o t e s a commutative r i n g and a l l c r o s s e d p r o d u c t s a r e assumed
t o be R-algebras.
As u s u a l ,
c r o s s e d p r o d u c t of
G
H
If
d e n o t e s a m u l t i p l i c a t i v e group.
,
t h e n by abuse of language, w e w r i t e
A
over
i s a subgroup of
G
G,
A
H
All s u b r i n g s of a r i n g
(').
element as
S.
is a
A = A *G 1
then
is o b v i o u s l y a c r o s s e d p r o d u c t of of
A
If
over
and we o f t e n w r i t e
A *H
instead
are assumed t o have t h e same i d e n t i t y
S
S
An automorphism of
Al
i s s a i d t o be an outer automorphism i f i t
i s n o t an i n n e r automorphism. 3.1.
PROPOSITION.
(G,A
,U,a)
6
:
A
--+
Let
A
b e a c r o s s e d p r o d u c t of
be a corresponding c r o s s e d system.
S
a homomorphism o f R-algebras and
G
Let
$
S
over
A)
and l e t
b e an R-algebra,
a mapping from
G
to
U(S)
1
s,y E G
such t h a t f o r a l l
a
and
E Al
UJ(z)$(y) = $ ( a ( x , y ) )uJ(Xy) =
$
i ( X I 0 (a) i (2)
where
Then t h e map
h
:
A
-t
S
g i v e n by
i s a homomorphism of R-algebras.
Proof.
The map
h
t o verify that for a l l
i s o b v i o u s l y an R-homomorphism.
a,b
E A
and a l l
h ( (a:) ( b y )) = Takinq i n t o account t h a t
z,y E G
A ( ~ 2X ()b y )
It therefore suffices
SOME RING-THEORETIC RESULTS
.
t h e r e s u l t follows. 3.2. of
PROPOSITION.
G
(noetherian)
Let
Al.
over
79
G
Al
If
A
be a f i n i t e group and l e t
be a c r o s s e d p r o d u c t
A
is a r t i n i a n ( n o e t h e r i a n ) , then
is artinian
.
Proof.
A
If
i s a r t i n i a n ( n o e t h e r i a n ) , t h e n so i s t h e A -module
1
.
v i r t u e of Theorem 1 3.
as r e q u i r e d .
Hence, by C o r o l l a r y 1.3.8,
A
A
by
i s a r t i n i a n (noetherian)
The n e x t r e s u l t is a g e n e r a l i z a t i o n of a well-known
p r o p e r t y of o r d i n a r y group
algebras. 3.3.
PROPOSITION.
If
T is
i s a u n i t of
A
be a c r o s s e d p r o d u c t of
G
over
Al
H
and l e t
be
G.
a subgroup of
li)
Let
A
g e n e r a t e d by
H
a l e f t (right) transversal for in
At, E
A
then
in
G
and, f o r each
is a f r e e r i g h t ( l e f t )
t
E
T, t
A(H)-module f r e e l y
2'1
( i i ) The n a t u r a l p r o j e c t i o n
71
:
A
3
A(H)
d e f i n e d by
n ( C a ) = Zah hEH
SfG
i s a homomorphism o f l e f t and r i g h t A(H)-modules. (iii) U ( A ) n A ( H ) = U ( A ( H ) )
and, i n p a r t i c u l a r ,
U(A) n A l Proof. 1.3(i), A
(il
Assume t h a t
T
=
U(A1)
i s a l e f t transversal for
H
in
G.
By Theorem
i s a s t r o n g l y G-graded a l g e b r a w i t h
A
t = ? A
1
for a l l
t E T
CHAPTER 2
80
Therefore
A a s required.
The c a s e where
2‘
( i i ) Choose
i
+
7, t
$ 1,
such t h a t
@ A A(*) ET
=
T
1E
Q
=
tfT
i s a r i g h t t r a n s v e r s a l i s handled s i m i l a r l y .
-
T and
1
=
t c--t 0 ,
Owing t o ( i ) ,t h e map
1.
e x t e n d s t o a homomorphism TI‘ :
of ( l e f t and r i g h t ) A(H)-modules.
A-
A
(H)
n’
Bearing i n mind t h a t
= TI,
t h e required
assertion follows. u
(iii) I t s u f f i c e s t o show t h a t i f
u n i t of
A(H’.
i s a u n i t of
A
uu = Uu = 1 f o r some
So assume t h a t
in
A (H),
U E
A,u E A ( H ) .
then
u
is a Then,
by ( i i ) ,w e have
as required. Let
L
. If
V
i s a n S-module, we w r i t e
An S-module
V
i s s a i d t o be L-projective i f
S.
be a s u b r i n g o f a r i n g
t h e r e s t r i c t i o n of
V
to
L.
e v e r y e x a c t sequence of S-modules
0
3
u
-
w-
VL
for
v-0
f o r which t h e a s s o c i a t e d sequence of L-modules
G--tU--tW-V-O L L
L
s p l i t s i s a l s o a s p l i t e x a c t sequence of S-modules. An 5’-module
!/
i s c a l l e d L-injective i f e v e r y e x a c t sequence of S-modules
f o r which t h e a s s o c i a t e d sequence of L-modules
0 -+ V L
-
WL
-+
UL
-
G
s p l i t s i s a l s o a s p l i t e x a c t sequence of S-modules. Let
H
be a subgroup of
i s a s u b r i n g of
A
G
and l e t
A
b e a G-graded a l g e b r a .
Then
A (H)
and we r e f e r t o A C H ) - p r o j e c t i v e and A ( H ) - i n j e c t i v e modules
SOME RING-THEORETIC RESULTS
as H-projective and H - i n j e c t h e modules, respectively.
completely reducible. Let A
3 . 4 . THEOREM.
containing 1. V
and
the map
$*
:
V
G over A , ,
be a crossed product of
For each
(i) If
is
is also
We are now ready to prove the following important result.
G of finite index n
group of
Thus, if A ( H )
is H-projective, then A
V
completely reducible and each A-module
81
g
and let G,
be a sub-
T be a left transversal for H
-
g of A
choose a unit
W are A-modules and $
let H
:
V
in A
9
with
in G
1=
1.
W is an A(H)-homomorphism, then
--f
W given by
--+
-1
$*(U)
=
z t@(ZU ) tfT
is an A-homomorphism which is independent of the choice of (ii) If (iii) If
n
is a unit in A, is a unit in A
n
T.
then any A-module is H-projective is semisimple artinian, then A
and A(’)
is semi-
simple artinian. Proof. (i) Assume that (G,A
,O,a)
Zt =
tht
t E T with
for each
ht E H .
Let
be the crossed system corresponding to our choice of the i , g E G.
Then, f o r all
U E
V.
-1
=
c t$(t
U),
ET $*
proving that If a E A
is independent of the choice of and
1
V E I/,
then
T.
CHAPTER 2
82
so it s u f f i c e s t o v e r i f y t h a t @*(&7,
for all
= ic$*(U,
g
E G,V
Since
(i)
follows
(ii) Assume t h a t
0-u-w-v-0
f
is an e x a c t sequence of A-modules f o r which t h e a s s o c i a t e d sequence of A
(H)-
modules
o - + u - wH splits.
H
-+vH-o
Then t h e r e e x i s t s a n A(H)-homomorphism c$ : V
-.--f
Thus
fc$*(u, =
c ?(f@)(? E T
-1 u) =
-1
c ET
Z? v
=
nu
W
with
f.4
=
lv*
V
SOME RING-THEORETIC RESULTS
83
and therefore the A-homomorphism
is a right inverse to f. (iii) Direct consequence of (ii) and Proposition 1.7.24. Let S
3.5. COROLLARY.IMaschke's theorem). finite qrou? of order n artinian, then S*G Proof.
n
such that
.
be an R-algebra and let
S.
is a unit of
If
S
G be a
is semisimple
is semisimple artinian.
Apply Theorem 3.4(iii) for the case
H = 1.
We next provide some elementary results pertaining to the Jacobson radical of
s
crossed products.
In what follows,
3.6. PROPOSITION.
For any subgroup H
always denotes an R-algebra. of
G,
Assume that x E J ( S * G ) n S*H.
Proof.
.
S*H for all r
is a unit of as asserted.
.
for all r E S*H
S*G
is a unit of
E S*H.
1-rx
Then, by Proposition 1.7.19,
Hence, by Proposition 3.3(iii), Thus, by Proposition 1.7.18,
1-rx
2 E J(S*H)
For future use, we next record 3.7. LEMMA. (Nakayama's lemma). J(S)V = V ,
Proof.
v
then
since
V
be a finitely generated S-module.
v=
SU
and so v
An E J ( S ) , 1 - An
+
1
=
... + sun
for some U 1
. .. + Xnvn
Xlvl +
is a unit.
with
,...,U in v. Then xi E J(S). In particular,
This yields n-1
(1- X n ) V n =
c xpi
i=1
so that
n-1
vn
c
=
(1- h n ) - l X i V i
i=1 Thus
is generated by
generators of
V
If
V = 0.
Write
E V = J(S)V
Let
Ul,...,Vn-l.
.
By continuing this process until all
have been eliminated, we derive
V = 0.
CHAPTER 2
84
Assume that L
{l
s.
is a subring of
Following Passman ( 1 9 7 7 ) , we say that
x 1 , x 2 , . . . , z n } is a normalizing basis for S over L
=
if the following
two properties holds (a) Every element s
5’ can be written uniquely
of
s = R x i R x +...+ 1 1
ki
with
2
as
Rnxn
2
L
E
(b) There exist automorphisms a ,a 1
2
,...,0n
of
L
such that
x.R = a.(R)z
z
for all 3.8.
R
E
LEMMA.
{l = x ,x 1
2
L,
iE
Let
I1,2
N
,...,xnl
i
,... ,nl. N
is a transversal for
normalizing basis for
S*G
of finite index n .
G
be a normal subgroup of
{T,x
G, then
in
If
*,...,zn}
is a
S*N.
over
.
Property (a) is a consequence of Proposition 3.3 (i) Let 0 -1 the automorphism of S*N defined by a . ( Q ) = 7 R for all R E S*N Proof.
z
i
{1 =
x ,z ,...,X n j
.
be
.
Then -1
for all
k E S*N,
3 . 9 . LEMMA.
Let L
proving (bl.
be a subring of
izing basis for S over L ,
Proof. have
If
1
2
is a normal-
then
and a nonzero
V
in
V.
Then we
Owing to (a),
v VL
S.
Fix an irreducible 5’-module V
V = Sv.
so that
.
= Lx v 1
iLx
1
v +
... .+
is a finitely generated L-module.
we have by (b)
Lx v
Given
i
E
{1,2,
...,n l ,
R E L,
SOME RING-THEORETIC RESULTS
J ( L ) V i s an S-submodule of
Hence, by ( a ) ,
J ( L ) V = V. J(L)
J(L)V
and s o e i t h e r
=
0
The l a t t e r e q u a l i t y i s i m p o s s i b l e , by v i r t u e of Lemma 3 . 7 .
V.
annihilates
J(L)
conclude t h a t 3.10.
V
85
LEMMA.
5J ( S ) .
H
Let
.
V
Since
or Thus
i s an a r b i t r a r y i r r e d u c i b l e S-module, w e
G
be a subgroup of
{Hiji
and l e t
E
I}
be a f a m i l y of
subgroups such t h a t
H. 3 H
(i)
i
for a l l
2.-
I
€
X
( i i ) For any f i n i t e s u b s e t
x
Let
E S*H.
G there exists
If
i f and o n l y i f
Conversely assume t h a t = s u p p ~ ~ ~I. f
X # 0
that
.
X
A
of
I E
=
8,
z E
X
with
5H..
x
E
J(S*Hi
i
for a l l
)
E I
X
1-yr
n S * H ~5 J ( s * H ~ )
JIS*Hi) f o r
then
By ( i i l ,
Proposition 1.7.19, Thus,
I
x E J ( S * G ) , t h e n by (i) and P r o p o s i t i o n 3.6, w e have E J(S*G)
X
i
Then
zE J(S*G) Proof.
of
y
5 Hi
=
i E I.
all
0 and c l e a r l y f o r some
i s a u n i t of
i
S*Hi
t I
y E S*G
Fix
1-yr
and l e t
is a u n i t .
and so
y
E
S*Hi.
So assume
Hence, by
and t h e r e f o r e i s a u n i t o f
S*G.
J (S*G) , by v i r t u e of P r o p o s i t i o n 1.7.18.
group
G i s s a i d t o be locally f i n i t e if e v e r y f i n i t e l y g e n e r a t e d subgroup
G is f i n i t e
3.11.
PROPOSITION.
finite.
Let
N
b e a normal subgroup of
G
such t h a t
G/N
is locally
Then
J ( S * N ) = J ( S * G ) n S*N Proof.
Let
Owing t o P r o p o s i t i o n 3 . 6 , we need o n l y v e r i f y t h a t
{ H i l i E I} b e a f a m i l y o f a l l t h o s e subgroups
finite.
Then because
G/N
Hi
with
i s l o c a l l y f i n i t e , it follows t h a t
s a t i s f i e s t h e assumptions ( i )and (ii)of t h e p r e v i o u s lemma. 3.9, we have
H. 2 N z -
and
Hi/N
{ H i l i E I} By Lemmas 3.8 and
86
CHAPTER 2
J(S*N) c - J(S*G)
Hence, by Lemma 3 . 1 0 , 3.12.
G
Assume t h a t
COROLLARY.
.
a s required.
i s a l o c a l l y f i n i t e group.
Then
J ( S ) = J(S*G) n S Proof.
Given a nonzero
x i n S*G , t h e supporting subgroup of
be t h e subgroup o f
G
3.13.
0 # x E S*G
LEMMA.
Let
Let
{ H i l i E I}
ing
[I.
Then
{Hi}
x
i s defined t o
g e n e r a t e d by S u p p .
H
and l e t
E J(S*G)
J:
z € J(S*Hi
i f and o n l y i f
H, we
x.
b e t h e s u p p o r t i n g subgroup of
G
be t h e f a m i l y of a l l f i n i t e l y g e n e r a t e d subgroups of
By t h e d e f i n i t i o n of
Proof.
.
N = 1.
Apply P r o p o s i t i o n 3 . 1 1 f o r t h e c a s e where
iE I
for all
x
have
for a l l
)
S*H.
i
€
contain-
I.
Moreover, t h e f a m i l y The r e q u i r e d
s a t i s f i e s t h e assumptions ( i )and (ii)of Lemma 3.10.
a s s u r t i o n now f o l l o w s by v i r t u e of Lemma 3.10. 3.14.
JCS*H) = 0 .
G,
J ( S * G ) = 0.
Then
'
D i r e c t consequence of Lemma 3.13.
Proof.
H of
Assume t h a t f o r any f i n i t e l y g e n e r a t e d subgroup
COROLLARY.
The proof of t h e f o l l o w i n g r e s u l t i s e x t r a c t e d from Passman ( 1 9 7 7 ) . 3.15.
PROPOSITION.
Let
S
normalizing b a s i s f o r
L
b e a s u b r i n g of
S
and l e t
{l
=
x ,x 1
over
2
,.. . ,x 1
be a
L.
J ~ s c_) J ~( L ) S c_ J ( S )
(i)
( i i ) I f e v e r y S-module i s L - p r o j e c t i v e ,
Proof. (a),
S = W
i s a f r e e l e f t L-module of rank
t h e s e a c t i o n s commute. homomorphism of
5 , s = cs 7:jxj E L,
J(L)S5J(S)
( i ) The i n c l u s i o n
then
S
into
= U.(R)z
z
S
i
i s a consequence of Lemma 3.9.
n
and a l s o a r i g h t S-module,
a c t s f a i t h f u l l y on
M ( L ) g i v e n by
sij E L
and t h e
x.11
Since
J(L)S= J(S).
then
s k--+
(s
ij
W, E
il
and
t h e r e i s an i n j e c t i v e
Mn(L)
a r e u n i q u e l y determined by ( a ) .
so t h a t
By
where Note t h a t i f
maps t o t h e d i a g o n a l m a t r i x
87
SOME RING-THEORETIC RESULTS
V
Given a ( r i g h t ) L-module
-
V
d e f i n e t h e r i g h t S-module
-
t o be t h e s e t of a l l n - t u p l e s
-
V
a c t s on
position.
V
L
Now
a c t s d i a g o n a l l y on
-
... @ vn.
VL = V @ V 6
V is
if
2
L. V1
f o r e t h e map
n
sum of
V, s o
V
--+
i’
V
F-+
1
vlxi is
V,
S
and
U
x
V
-
of
VL
= 1,
V
Vi,
since
Oi
= 1,
xlxi
laxi and t h e r e -
V.
N o t e also t h a t
=
i s an auto-
-
VS
By t h e f o r e g o i n g ,
Therefore
-
an L-submodule
a bijection.
an i r r e d u c i b l e L-module.
which i m p l i e s t h a t
Vi i s
each
then s o i s each
i r r e d u c i b l e L-modules.
n,
length
-
F i n a l l y , o b s e r v e t h a t because
V is
Assume t h a t
We t a k e
,. . . , V n )
U S = (V
Moreover, s i n c e
an i r r e d u c i b l e L-module,
morphism of
as f o l l o w s .
( s . .). Denote by ‘i 23 c o n s i s t i n g of t h o s e n - t u p l e s t h a t a r e z e r o e x c e p t i n t h e i - t h
-
1
V
with e n t r i e s i n
-
by way o f t h e m a t r i x p r o d u c t
t h e s u b s e t of
and
,...,Vn)
V = (V
-
-
Vs
is a d i r e c t
h a s a composition series of
h a s a composition series of l e n g t h
Q
n.
J(S) a c t s t r i v i a l l y on each f a c t o r of t h i s series, it f o l l o w s t h a t J ( S ) n n a c t s t r i v i a l l y on 7. F i x x E J ( S ) n and w r i t e x = z kixi, e . E L. Then i=1 Vx = 0 and so f o r a l l V E V l , n Since
Since V
i
k
=
i
Villi
E V1
0.
But
and
V1
V
I-+
i s L-isomorphic t o
V
and t h u s
S i n c e t h i s i s t r u e f o r a l l i r r e d u c i b l e L-modules
x
and hence t h a t
E
J(L)S.
Thus
JIS)n
i n (i).
Since
-VL
V,
5J ( L ) S ,
Assume t h a t e v e r y S-module i s L - p r o j e c t i v e ,
(ill
VI
Ulzi i s a b i j e c t i o n
of
V
implies
x
-
V J ( S ) = 0.
E J(L)S,
Vk.
= 0
f o r i E {1,2
ti E J ( L )
we infer that
V
and l e t
-
V is
Then, by t h e argument of
and t h u s
J(S)
.
5J ( L ) S .
consequence of ( i ) ,t h e r e s u l t f o l l o w s .
and
v
b e as
Hence e v e r y S-submodule completely reducible.
-
( i ) , Vx = 0
f o r a l l such
Our f i r s t a p p l i -
c a t i o n i s an e x t e n s i o n of V i l l a m a y o r ’ s theorem (1958) t o c r o s s e d p r o d u c t s .
Let
N
It
The r e v e r s e containment b e i n g a
The r e s u l t above h a s a number of i m p o r t a n t a p p l i c a t i o n s .
3.16. THEOREM.
,...,n }
i s a d i r e c t sum of i r r e d u c i b l e L-modules i t i s c o m p l e t e l y
i s a l s o a d i r e c t summand and t h e r e f o r e
follows t h a t
we d e r i v e
a s required.
r e d u c i b l e , and e v e r y submodule i s a d i r e c t summand.
-
Vi,
--+
be a normal subgroup o f
G
of f i n i t e i n d e x
n.
V
CHAPTER 2
88
5
J ( s * G ) ~ (s*G)J(s*N)
(i)
(ii) If n
is a unit in
Proof. (i) Let x
{GI ,i2,.. . ,G 1
3.8,
5 J(S*G)
S, then J ( S * G ) = ( S * G ) J ( S * I V )
,...,3:
= 1,5
be a transversal for N
is a normalizing basis for S * G (S*G) J ( S * N ) = J ( S * N )
in
S*N.
over
G.
By Lemma
Note also that
(S*G)
since ~ J ( s * N )= C & J C S * N I ~ - ' ) = S J ( s * N ) ~for all g
E
:3)
G.
NOW
apply Proposition
3.15.
(ii) Direct consequence of Proposition 3.15 and Theorem 3.4(ii). G
Let of
be an arbitrary group.
H
If
G,
is a subgroup of
then the Core
H in G is the largest normal subgroup of G contained in 14. Let H
3.17. LEMMA.
G
be a subgroup of
and let
H*
be the core of
H
in G.
Then
n,$ SEG ( G : H * ) < n! H*=
(G:H) = n,
Moreover, if Proof.
zE G ,
Let
we have
N
then
G
be a normal subgroup of
N =
rs" 5 3
and so N
5
n
contained in H.
9.
(G:H) = n <
a.
Then G
permutes the n
right multiplication and thus we have a homomorphism G symmetric group of degree n.
HN
and
= H.
is clearly
the first assertion is proved.
Assume that
N 4G
K"
&G
Z5G
normal in G ,
n
Since
Then, for all
(G:N)
Hence
Q
ISn[ = n!
5H
N
If N
--+
right cosets of
s,
Sn
H
by
being the
is the kernel of this homomorphism, then
Moreover, since N
fixes the coset H ,
we have
and the result follows.
As a second application of Proposition 3.15, we now prove 3.18. PROPOSITION.
Then
J(S*G)
Proof. J(S*H:
Let
Suppose first that Hence, if
assume that J ( S * H ) S*G
be a group and let H
is nilpotent if and only if
5 J(S*G).
ideal of
C
.
H O G.
J(S*G)
is nilpotent.
Thus J ( S * G )
J(S;H)
be a subgroup of finite index. is nilpotent.
Then, by Proposition 3.11,
is nilpotent then so is J ( S * H ) . Then, by ( 3 )
( S * G ) J ( S* H )
Conversely,
is a nilpotent
is nilpotent, by virtue of Theorem 3.16(i).
SOME RING-THEORETIC RESULTS
H*
Turning t o t h e g e n e r a l c a s e , l e t
H*
by Lemma 3 . 1 7 ,
i s of f i n i t e i n d e x i n
and h a s f i n i t e i n d e x i n
H
G.
so t h a t
J(S*H)
H.
H* 4 G
Then
H*
Moreover,
and,
J(S*G)
Hence, by t h e f o r e g o i n g ,
J(S*H*) i s n i l p o t e n t .
is n i l p o t e n t i f and o n l y i f
is nilpotent.
be t h e c o r e of
89
is normal J(S*H*)
i s n i l p o t e n t i f and o n l y i f
So t h e p r o p o s i t i o n i s t r u e .
The n e x t r e s u l t i s a sharpened v e r s i o n of Theorem 3 . 1 6 ( i i ) . 3.19.
PROPOSITION.
0#
Let
by P r o p o s i t i o n 3.6.
G/N
N
be a normal subgroup of
X E
J(S*G)
Since
H/N
Thus
J(S*G)
i s a u n i t of
5 (S*G)J(S*!f)
and l e t
s.
H/N
G with
S, t h e n
a r e u n i t s of
H = <Suppx>N
.
a locally
G/N
Then
i s f i n i t e l y g e n e r a t e d , so i s
<Suppx>
is l o c a l l y f i n i t e , hence
o r d e r of
G/N
If t h e o r d e r s of elements of
f i n i t e group.
Proof.
Let
is finite.
H/N.
But
Furthermore, by h y p o t h e s i s , t h e
Invoking Theorem 3 . 1 6 ( i i ) , w e conclude t h a t
and t h e o p p o s i t e i n c l u s i o n b e i n g a consequence of
P r o p o s i t i o n 3.11, t h e r e s u l t i s e s t a b l i s h e d . W e n e x t p r o v i d e c i r c u m s t a n c e s under which
J(S*C)
is a n i l i d e a l .
As a
p r e p a r a t i o n , we prove t h e f o l l o w i n g r e s u l t of Amitsur (1956).
3.20.
LEMMA.
Let
A
F
be an a l g e b r a o v e r a f i e l d
and suppose t h a t
dimA < IF1
F a s an i n e q u a l i t y of p o s s i b l y i n f i n i t e c a r d i n a l numbers.
a
(if L e t
a
have t h e p r o p e r t y t h a t
i s a l g e b r a i c over
F,
is a unit f o r a l l
h E F.
Then
F.
IF1 >
dim A e l e m e n t s ( 1 - h a ) F and so t h e s e must be F - l i n e a r l y dependent. I t follows t h a t
Proof. E
1 - ha
is a n i l ideal.
(ii) J ( A )
h
E A
(i) By h y p o t h e s i s , there e x i s t
-1
with
CHAPTER 2
90
,...,1
for suitable distinct t h a t a l l t h e elements
E'
E 3'
with
1-XU
i i l , l i p ,...,!An E
and nonzero
F.
Note n e x t
and t h e i r i n v e r s e s commute.
Therefore,
n
n (1- Aiu),
i f w e m u l t i p l y t h e g i v e n e q u a t i o n by a l g e b r a i c e q u a t i o n s a t i s f i e d by ( i i ) Given
a l g e b r a i c over
F.
U
is a unit for a l l
U E j ( A ) , 1 - XU
n,r
l+b
= IJ
is a unit.
PROPOSITION.
3.21.
H
Now
2 E
F.
E
+
a + p a2 2
is
Hence, by ( i ) , U
= 0
However,
...
= 0
+ li ar
€
J(A)
and t h e r e s u l t f o l l o w s . S
b e a nondenumerable f i e l d , l e t
G
J(S*G)
be a group.
H
and l e t
is f i n i t e l y g e n e r a t e d , so
IHl
of
Let
Vi
Thus 'U
F
Let
1
countable F-dimension and l e t Proof.
2
and some
b
... + ll,uri
a + p u2 +
1
so t h a t
E 3'.
We can c l e a r l y w r i t e t h i s polynomial e q u a t i o n a s an(l+li
f o r some i n t e g e r s
w e c l e a r l y o b t a i n a nonzero
i=l over F .
H
Then
J(S*G)
S,
be 3'-algebra of is a n i l ideal.
Then, by P r o p o s i t i o n 3.6,
= <SuppZ>.
i s countable.
c o p i e s of t h e v e c t o r s p a c e
m
S*H i s
Since
w e conclude t h a t
a d i r e c t sum
i s countable.
dimS*H
F IF1
On t h e o t h e r hand, by assumption,
s o , by Lemma 3 . 2 0 ( i i ) ,
J(S*H1
i s uncountable.
is a n i l ideal.
Thus
Hence 3:
dimS*H <
F
IF/
and
i s a n i l p o t e n t element,
a s required. I n what f o l l o w s t h e a c t i o n s of
G
on
C,,,(Sl
and
Z(S)
a r e d e f i n e d as i n
C o r o l l a r y 1.9.
3.22.
PROPOSITION.
S*G
Let
be a c r o s s e d p r o d u c t of
e i t h e r a domain o r a simple r i n g .
G
CSC , S)
on
z(S)
In
if
S*Go Proof.
S
S
i n case
Z
=
length.
We c l a i m t h a t
S,
where
s g, s
Suppa
S
E S,
i s a nonzero i d e a l of
S*G,
b e a nonzero element i n
I
G on
then
of minimal
g
ZG
g
is
Go t o b e t h e k e r n e l of t h e a c t i o n of
S*Go.
-
Sfcg
I
If
i s a nonzero i d e a l of
a
over
i s a domain, and t h e k e r n e l of t h e a c t i o n of
i s a simple r i n g .
Let
Define
G
f o r some
g E G;
i f s u s t a i n e d , it w i l l
SOME RING-THEORETIC RESULTS
91
-1
follow that a s
1 0 S*Go.
is a nonzero element of
Assume, by way of contradiction, that X - l y
‘ Z(S) there exists s 2
Then, by the definition of Go, -1 that s.’ # s in which case
(zs -’s)s field.
Y
# 0.
Suppb
Hence x
- ’ s # 0.
X
s -as.
5 Suppa.
(“s
s 0#
- ’s)s
Y
or
S E
in Suppa.
Cs*G(S)
such
is a domain, then clearly “S-’s
E Z(S)
S is a
and
# 0.
Then
However, since
X
s
E C,,,(S),
we have
X
2
ssX - sX s
0.
=
9 Suppb and, on the other hand, y E Suppb since
.
by virtue of the preceding paragraph. the proposition is true. 3.23. COROLLARY.
This contradicts our choice of a
Let S*G be a crossed product of
either a domain or a simple ring. (i) If
If
is a simple ring, then
Hence in this case we also have
Put b =
and
If S
9 Go for some x , y
Define Go
G over S, where
and so
S
is
to be as in Proposition 3.22.
S*Go is simple, then so is S*G.
(ii) If J(S*G 1 = 0, then J(S*C) = 0.
The converse is true if
G/Go
is
locally finite. Proof. (i) Direct consequence of Proposition 3.22. (ii) Assume that J(S*G) # 0.
Then, by Proposition 3.22,
J(S*G)
S*Go # 0.
Applying Proposition 3.6, we deduce that J(S*Go) # 0. Assume that
G/Go
is locally finite and that J(S*Go) # 0
.
Then, by
Proposition 3.11,
as required.
0 # J(S*Go)
.
3.24. COROLLARY.
Let
5 J(S*G)
S*G be a crossed product of
of characteristic p 2 0, and let G
G over a division ring
be the kernel of the action of
C
on
S
CHAPTER 2
92
cS*G (Sl . If Go
(il G
if
p
is a finite group and
C,,,(Sl,
acts faithfully on
(ii) If G
1,
IG
then J(S*G) = 0.
In particular
then J(S*G) = 0.
is finite and acts faithfully on
CS*G(S), then S*G
is simple
.
artinian
Proof. (i) By Corollary 3.5, J(S*Go) = 0.
(ii) Direct consequence of Proposition 3.2 and Corollary 3.23(i). 3.25. LEMMA.
S*G be a crossed product of G over S.
Let
Proof.
Given
S E
sx
S,
c
xs
=
.
follows.
(2 g ) s =
sx
g
=
c
2
9s;
@Gg
SfC
Since the latter is equivalent to
Then
if and only if
)S = z
(SZ
SfG
.
Now apply Corollary 3.23(ii).
'S
g
for all g E Suppz, the result
4. THE CENTRE OF CROSSED PRODUCTS OVER SIMPLE RINGS Throughout this section, H*G group G
(,?*GI
9
denotes a crossed product of a (possibly infinite)
over a simple ring R .
with
-
1
=
1
For each g
Go
which conjugation by g
g
E Go,
Then
ij
let
1 g g
assume that
G consisting of all those g R.
-
Go does not depend upon a choice of units g,g 4
E
in
U(R)
induces an inner automorphism of
E
G for
Of course, the E G.
For each
U(R) be such that
-
=
G x G
for the normal subgroup of
definition of
G, we fix a unit 9 of R*G
and define ~1 :
We write
-
E
is clearly in
CRkG(R). Thus we may, and from now on we shall,
93
THE CENTRE OF CROSSED PRODUCTS OVER SIMPLE RINGS
-
g
CR*G(R)
for all
g
G
E
(1)
-1
'r
The formula
C
on
Z(R)
jrg
=
,r
E
CR*G(R).
and
Z(R)
r
or
G
Since
E
CR,,(R) ,g
E G,
Z(R)
and
a c t s on
a G - f i e l d and t h e f i x e d f i e l d
r e s u l t i n t h i s s e c t i o n provides a d i s t i n g u i s h e d b a s i s f o r
Z(R)G
R
Z(R)
i s simple,
Z(R*G).
i s contained i n
G
of
p r o v i d e s an a c t i o n of
Z(R*G)
is
Our main
over t h e f i e l d
.
W e s t a r t by p r o v i n g some g e n e r a l r e s u l t s of independent i n t e r e s t
4.1.
PROPOSITION.
With t h e above n o t a t i o n , t h e f o l l o w i n g p r o p e r t i e s h o l d :
( i ) C R k G ( R ) = Z(R)*Go
i s a t v i s t e d group a l g e b r a of
( i i ) Z(R*G! = ( Z ( R )* G o )
Proof.
Go
over t h e f i e l d
Z ( R ).
C
( i ) I t f o l l o w s from (1) t h a t
x
Conversely, l e t
c
=
5
9E
cR,,(R)
g E Supp.
and l e t
Then, by Lemma 3 . 2 5 ,
5-
rx
g
Hence
Rx
9
=
follows t h a t
x R 9 x
i s a nonzero i d e a l of i s a u n i t of
g
R
g E G
x gr g
R
and t h e r e f o r e
R
=
z
which shows t h a t
=
x-lr g
3:
g
E
$7
Z(R).
gr
= z-lp
9
9
T h i s proves t h a t
C,,,(R)
PROPOSITION.
Assume t h a t
Go
=
1
.
Then
i s simple
(ii) CR*G(R) = Z ( R ) ( i i i ) Z(R*G) =
Proof.
9
R. 9
It
= 2
for a l l
r
for a l l
P E R
5 Z (R)*Go
( i i ) D i r e c t consequence of ( i ) and C o r o l l a r y 1.9.
( i ) R*G
RX
E
R
and, by (l), p =
4.2.
r E R
for a l l
such t h a t
gr Thus
=
Z ( R )G
(i) Let
I
be a nonzero i d e a l o f
O # x =
R*G
C X j E I g€fc g
and choose
as r e q u i r e d .
94
CHAPTER 2
-
-1 Multiplying by an appropriate g ,
of minimal length.
x #
assume that
is an (R,R)-bimodule and R
Since I
0.
1
also assume that
x
=
1.
for some
is simple, we may
Thus we may write
1
x
If x = 1, then
if necessary, we may
I=
+ z
zgz g+l R*G, so we may assume that x # 1 in which case xh # 0
1 # h E G.
Now for any
= 1.i
r E R
c
rx-n=
(rx - x gr)
E
I
&G
and
Supp(rx-xr)
5
Suppx.
Moreover, since x = 1, 1 9 supp(rz-xr).
x
is of minimal length, we conclude that
E CR*G(R).
since x
h E Suppx, so h = 1
But
by Proposition 4.1(i), a contradiction (ii) Apply Proposition 4.l(il. (iii) Apply Proposition 4.1(ii). 4.3.
COROLLARY.
(i) R*G
Assume that G
. acts faithfully on
Z(R).
Then
is simple
(ii) CR*G(R) = Z ( R )
G
(iii) Z ( R * C ) = Z ( R ) Proof.
E Go.
Assume that
.
by hypothesis, g = 1. Proposition 4.2.
Then g
acts trivially on
field
Z(R)
Let
V
Z(R*G) over the
.
be a vector space over a field
A semilinear transformation of
V
that there exists an automorphism
Note that the automorphism formation f
and hence,
Thus Go = 1 and the result follows by virtue of
We now head towards providing a distinguished basis for
G
Z(R)
of
9
F
(possibly of infinite dimension).
is any additive homomorphism f : V-+
9
of
F
such
with
is uniquely determined by
is said to be nonsinguzar if f
V
V
f.
A semilinear trans-
is a bijection.
It is
clear that under composition of mappings the set of all nonsingular semilinear transformations of
V
constitutes a group; we denote this group by
GS(V)
and
95
THE CENTRE OF CROSSED PRODUCTS OVER SIMPLE R I N G S
V.
refer to it as the general semizinear group of 4.4.
LEMMA.
For each
f E GS(V), let @
F.
be the associated automorphism of
f
Then the map
is a homomorphism whose kernel is cL(V).
Proof.
Given f , g E GS(V),U E V
In particular,
h E F,
and
we have
@ @
= @fg, proving that the given map is a homomorphism. f-9 if and only if f E GL(V), the result follows.
Thus
Since
@
’
By a monomial space over a field F we understand a triple where
V indexed by X
~y a semilinear
monomiaL representation of a group G on
1
(V,X,(Vx)),
V is a vector space over F , X is a (possibly infinite) set and
is a family of one-dimensional subspaces of
s=
(Vx)
such that
(V,X,(Vx)) we mean a
homomorphism
g E G , T(g)
such that for all
a G-field and we shall write to
h
Vx,x 6
permutes the
x.
By Lemma 4.4, F becomes
for the automorphism of F
+-+
corresponding
r ( g )Note also that
group of the set
r
determines a homomorphism y
X, where for all g
E G
y(g)x Thus
and =
from
to the permutation
x,y E X
y
C acts on the set X and we denote by
G
if and only if
G(x)
that is G(z) = { g E Gly(g)z =
XI
r(g)Vx
the stabilizer of
3: E
=
X,
V
Y
CHAPTER 2
96
x
We s a y t h a t an element
Vx
X
of
ox
i s r-regular i f t h e r e e x i s t s a nonzero
in
with
nglvx
g E G(x)
for all
a s b e i n g r-regular i f each element of t h i s
X
W e s h a l l r e f e r t o a G-orbit of
ux
=
o r b i t i s T-regular.
r
By t h e fixed-point space of
we understand t h e s e t of those
v E V
for
which
I t i s c l e a r t h a t t h e fixed-point
i s a v e c t o r space over
PROPOSITION.
(v,x,(vI )
Let
r
:
G
be t h e sum
G on
(V,X,(Vx)).
of one-dimensional s u b s p a c e s of
of t h e o r b i t c o n t a i n i n g for a l l
g E G(z).
(i) If
x E
x
x
V
Let
Z
be a s e t
and, f o r each
r,
t h e f i x e d - p o i n t space of
(a)
h.’ = 0 (h.’ n L,)
(b)
W n LZ
Z E
E
Vz
with
r(g)V,=
x
then
ZEZ
C gXT(g)uz\xE FC(”>
sf T z
T,
where
= {
i s a l e f t transversal for
(c)
If
{Aili
(d)
If
dimV F
E
G
I > i s an P - b a s i s of
m
and
G
FG(”
,
G
containing
1.
then
dimW <
m
and
FC FG
( i ) Let
in
i s a f i n i t e group, then dimW =
Proof.
G(z)
C (dimFG ( z ) ) zEZ
FG
x E X be r - r e g u l a r and l e t
y E X
2,
indexed by t h e e l e m e n t s
0 # Vz
e E Z, f i x
F o r each
2.
and l e t
is i’-regular, t h e n so a r e a l l t h e elements i n t h e G-orbit of
W is
(ii) I f
the
GSIV)
-+
of a l l r e p r e s e n t a t i v e s f o r t h e f i n i t e r - r e g u l a r o r b i t s of
LZ
F
be a monomial space o v e r a f i e l d
be a s e m i l i n e a r monomial r e p r e s e n t a t i o n of
let
F’,
G.
f i x e d f i e l d of 4.5.
”
r
space of
be any element i n t h e
V,
97
THE CENTRE OF CROSSED PRODUCTS OVER SIMPLE RINGS
x.
G-orbit of
G(y1
gG(r)g-'.
=
x
Since
T ( h ) with
by a l l
g E G
Then t h e r e e x i s t s
u
f o r some nonzero
in
Y
I/
Y
Since
.
,V
T(g-l)V
=
E G(y),
say
Y
t
Let
=
X
t h e r e i s a nonzero
is r-regular,
G(x).
h
r(g)V
such t h a t
V
and
Y
u
=
ghg-l
fixed
-1
u
we may w r i t e
t
Vx
in
X
X
=r(g )uy
h E G(x).
with
Then we have
a s required.
(iil
Let
each
y
Y
Y,
X
d e n o t e a s e t of a l l r e p r e s e n t a t i v e s f o r t h e o r b i t s of
L
let
V
b e t h e sum of one-dimensional subspaces of
Y
y.
t h e elements of t h e o r b i t c o n t a i n i n g
and, f o r indexed by
Then
v=
@ L
*y V
i s a decomposition of
i n t o d i r e c t sum of G - i n v a r i a n t s u b s p a c e s .
w=
Hence
@ ( w ~ L ) *fy
c
U =
Let
U
fixx'
Then, f o r a l l
U
IC
E
V,,
g E G, T ( g ) U t E V y ( g , t
In particular, i f
g
E
and h e n c e , by ( i ) ,t h a t
0 # Ux E VX
number of Hence
Iv' =
Given
G(t),
and suppose t h a t
t
is finite,
# 0 f o r some t r(g)u=
t
E
U ,
X.
that
is r-regular
Furthermore,
since the
belongs t o a f i n i t e r - r e g u l a r o r b i t .
@ (hi n L z ) , p r o v i n g ( a ) . ZEZ V E Lz, we may w r i t e u n i q u e l y
{T(g)Vzlg E
=
c x r(g)u,
TZl i s an F - b a s i s f o r =
c
E C
and
g
€ 4'z,
let
(A
gfTz
U,.
hxqr(hg)-JZ =
u E W
Hence
c x
r(g)uz
L,
9
E F)
i f and o n l y i f f o r a l l hEG ( 2 )
gfTz
gfTZ
h
t
proving t h a t
Vt,
t belongs t o a r - r e g u l a r o r b i t .
r(h)v Given
U
which i m p l i e s , i n view of
r(g)Ut =
then
v since
W
belong t o
gh E Tz
be d e f i n e d by
h g = g t h h
hg t g h G ( z ) . f o r some
Then
th E G ( z )
CHAPTER 2
98
and s o
Hence ( 2 ) i s e q u i v a l e n t t o
i g h l g E T ~ =I T ~ ,( 3 ) i s e q u i v a l e n t t o
Since
hA
= A
h t G,g
for a l l
(4)
E Tz
gh h = A
Set
and assume t h a t ( 4 ) h o l d s .
1,h
Then, t a k i n g
g
h E G,g E imz
and w r i t e
=
T
E
and
h
E Gtz),
we o b t a i n
Conversely, assume t h a t ( 5 ) h o l d s .
t
some
E
G(z).
Fix
hg
=
g t for h
Then we have
proving (4), and hence Ib) i s e s t a b l i s h e d . Given P 1 , . ..,p,
A
E
in
FG(z), w e may w r i t e
FG.
t h e n by ( 6 ) we have proving ( c ) .
A =
!
pixi i=l
f o r a unique
n
1 and unique
Then
C
gAr(g)Vz = 0.
But t h e n
A
=
0
and hence each
vi
0,
=
6T..6
F i n a l l y assume t h a t
dimz' <
and t h a t
G
i s a f i n i t e group.
Then
X
is a
4 f i n i t e s e t , hence s o i s
Z
and, by C o r o l l a r y 1 . 4 . 2 ,
dimFG(Z) <
m
for a l l
z E
FG T h i s proves ( d ) by a p p e a l i n g t o ( a ) and ( c ) .
W e have now accumulated a l l t h e i n f o r m a t i o n n e c e s s a r y t o p r o v i d e a d i s t i n g uished
2 ( R )'-basis
for
Z (R*G) .
2.
THE CENTRE
We say that g
99
OF CROSSED PRODUCTS OVER SIMPLE RINGS
G Is a-regular, provided g satisfies the following two
E
conditions: (a) g E Go (b) There exists a nonzero V
in
(Z(R)*G ) 0 9
&
such that
for all
= VG
zE cG(g). Since each nonzero
0# A
E
Z(R)
g
and
(Z(R)*G
in
V
0 9
E Go,
exists 0 # A E Z(R1
is of the form V =
we see that g 6 G
is a-regular if and only if there
such that
"Ja(r,g)= h a ( g , z ) Thus, if
for all
G acts trivially on Z ( R ) , then g E Go
R
(1.e. if
a(z,y)= 1 for all x , €~ G I ,
.Z
CG(g)
E
(7)
is a-regular if and only if
a(s,g)= a(g,z) for all z E C G ( g ) , while if R*G over
for some
G
is a skew group ring of is a-
g 6 Go
then each
regu1as. The following observation will enable us to take full advantage of Proposition 4.5.
4.6. LEMMA. ii)
F
Let
iV,Go,(V
g
=
Z ( R ) , v = F*Go
and, for each g E G o ,
(ii) For each g E G ,
the map
r(g)
:
r(g)( v )
v =
-f-
ZV.;
is a nonsingular semilinear transformation of
(V,Go, (V
g
A
=
6
F}.
r
: G
-+
GS(V)
V
defined by
-1 (V E V )
V
is a semilinear monomial representation of
G on
such that
) )
(a) For each
3:
E
Go, G(z)
(b) An element g E Go
=
CG(z)
is r-regular if and only if g
ticular, by Proposition 4 . 5 ( i ) , if gate of
g
is a monomial space over F
))
(iii) The map
put V
g
E Go
is a-regular.
In par-
is regular then so is any G-conju-
g.
(c) Z(R*G) Proof.
is equal to the fixed-point space of
V.
(i) Direct consequence of the fact that
{ilg E G
1 is an F-basis of
F*Go (ii) The map
T(g) is obviously additive and is a bijection.
Since for all
CHAPTER 2
100
the assertion follows.
r(g)
(iii) By (ii), each
E
GS(W
G
is a semilinear monomial representation of
V
X
(V,Go,(V 1 ) .
on
r
,zE Go,
Let f g to the permutation group of the set
G
corresponding homomorphism from
r ( g ) permutes the
and since
be the GO.
Then, for each
Y(g)x
g E G , 2 6 Go,
=
gzg-'
G(X) = C G ( z ) .
and hence
This
proves (a) and also (b), by applying (a) and the definitions of ®ularity r-regularity.
Property (c) being a consequence of Proposition 4.1(ii), the
result follows.
'
It is now an easy matter to obtain our main result.
G
C of
class
and
Go
contained in
(hence for all) g
is a-regular if
is a-regular for some
C.
in
4.7. THEOREM. (Karpilovsky (1986)).
C over a simple ring R 'r-regular classes of
g
We say that a conjugacy
G.
R*G be a crossed product of a group
Let
and let 2 be a full set of representatives for finite For each
z E
Z, let 0 # rz
E
Z(R)
be such that for all g t C (z)
c
let
{Ai,,li E I z l
sal f o r
G be a
Cr(B)
Z ( R ) -basis of
Z(R)
,
Tz be a left transver-
let
C (z) in G containing 1, and p u t G
(ii) If
G is finite, then dimZ(R*G) is also finite and is given by the Z(R)G
following fornula
c
(z)
dim Z(K*G) = 1 (dim Z(R1 G Z(R)G z€Z Z ( K ) G (iii) Z ( R * G ) = Z ( H ) G
if and only if
{l\
,
is the only finite a-regular class of
C. Proof. choice of
(i) Keeping the notation of Lemma 4.6, put
r.. 0
ensures that
v
# 0 is in
Vz
and that
V
- rzz. -
T(g)Vz = Vz
Then our for all
THE CENTRE OF CROSSED PRODUCTS OVER SIMPLE RINGS
101
The d e s i r e d c o n c l u s i o n i s t h e r e f o r e a consequence of P r o p o s i t i o n 4.5
( a ) , ( c ) and
Lemma 4 . 6 ( a ) , ( c ) .
(iil
G
If
i s f i n i t e , then
dimV
I
=lG
F
is also finite.
Hence t h e r a q u i r e a
O
a s s e r t i o n i s a consequence of P r o p o s i t i o n 4 . 5 ( i i ) and Lemma 4 . 6 ( a ) , ( c ) . ( i i i ) D i r e c t consequence of
(i).
9
We c l o s e t h i s s e c t i o n by p r o v i d i n g a number of consequences of Theorem 4 . 7 . 4.8.
R
ring
R*G
Let
COROLLARY.
G over
Tz
Z
Let
z
For each
is the twisted
2, p u t
E
C
G
G
in
(2)
Z(R*G).
Z ( R ) - b a s i s of
is a
R*G
(e.g.
be a f u l l s e t of r e p r e s e n t a t i v e s f a r f i n i t e
i s a l e f t t r a n s v e r s a l of
( i ) b Z l z E 21
Z(R)
a c t s t r i v i a l l y on
R).
G.
a - r e g u l a r c l a s s e s of
where
G
and assume t h a t
group r i n g of
G o v e r a simple
be a c r o s s e d p r o d u c t of a group
1.
containing
is f i n i t e ,
Go
In particular, i f
t h e n dim Z(R*G) i s a l s o f i n i t e and i s e q u a l t o t h e number o f a - r e g u l a r
classesofG.
2( R ) (ii) If
G
b a s i s of
Z (H*G).
Proof.
( i ) Keep t h e n o t a t i o n of Theorem 4 . 7 .
r
Z ( R ) , we can choose
Z(HG
=
z
for a l l
1
.
Z(R) = Z(R)
=
G
i s a subgroup of
i s a b e l i a n , then
/I I
Hence
E
2.
and
Since
G
a c t s t r i v i a l y on
Furthermore, we a l s o have
A
and we can choose
= 1
is a Z(R)-
21
E
i ,z
= 1.
Now
.
a p p l y Theorem 4.7 ( i )
G
( i i ) Suppose t h a t
G
of
Z(R*G).
af.2 ,z 1
z
and, f o r each
proves t h a t of
is abelian.
2
)
E
vz
=
i
Assume t h a t
Z(R)
1
since
is a-regular. 2
Z,CG(z)
z
for a l l
Z ( R ) _C Z ( R * G ) , ,l = 1 , 2 ,
z z
E
z ,z 1
=
2
E Z.
z l , z p E Go
TZ
G, s o a s
2.
E
2 c o n s i s t s of a l l a - r e g u l a r elements
Then,
we can choose
Hence, by (i),
-z
Then
(see (1)).
we conclude t h a t
F i n a l l y , assume t h a t
z z 1
2
2 E
z
1
2
=
This
i / z E Zl i s a Z ( R ) - b a s i s
-
a ( ~ ~ ,).z?z
Since
2
z
i
Z(R*G).
z.
11).
Since
E
1
2
Z(R*G)
and and
Thus, by d e f i n i t i o n ,
CHAPTER 2
102
-
-1 z
we see t h a t 4.9.
R,
Z
Z(R*GI.
E
Let
COROLLARY.
let
z 2-l =
a(2,Z-l)
z
Hence
-1
-i E
Z(R) c Z(R*GI,
E Z
and t h e r e s u l t f o l l o w s .
G
be a skew g r o u p r i n g of a g r o u p
R*G
o v e r a sirnple r i n g
be a f u l l s e t o f r e p r e s e n t a t i v e s of f i n i t e conjugacy c l a s s e s of
contained i n
Go
Z E Z,
and, f o r e a c h
G
G
{xi,ZliE I 1 be a Z ( R ) - b a s i s of
let
CI(Z)
Z(R)
,
(;'
T,
let
be a l e f t t r a n s v e r s a l f o r
G
in
CG(Z)
containing
1,
and
let
u
u
Then
li
{U
E
i ,Z
1 gx.
=
1 ,z
(;zg
G @T,
I Z j is a
Z ( R ) - b a s i s of
-1
(iE IZ)
)
Z(R*G).
,€Z
Proof.
Since
H*G
i s a skew group r i n g , e a c h
Furthermore, i n t h e n o t a t i o n of Theorem 4 . 7 ,
g E Go
w e can p u t
Y
is a-regular.
Now a p p l y
= 1.
.
Theorem 4 . 7 ( i ) 4.10.
simple r i n g charR of
Let
COROLLARY.
k,
G
R
/GI.
R*G
be a c r o s s e d p r o d u c t of a f i n i t e g r o u p
R
and assume t h a t Let
Z (R)'
over a and t h a t
2 be a f u l l s e t of r e p r e s e n t a t i v e s f o r t h e a - r e g u l a r c l a s s e s
n(R*G)
and l e t
is finite-dimensional over
G
b e t h e number of nonisomorphic i r r e d u c i b l e R*G-modules.
Then
CG ( 2 1
Z
n(R*G)
(dim
ZEZ
with equality i f Proof.
Z(R)G.
R*G
Furthermore, s i n c e
.
R*G.
i s a finite-dimensional a l g e b r a over t h e f i e l d
charR
k,
/GI,R*G
i s semisimple by Maschke's theorem
Hence
n(R*G) with e q u a l i t y i f
)
Z(R)G
G G Z ( R ) is a s p l i t t i n g f i e l d f o r t h e Z f R ) -algebra
By h y p o t h e s i s ,
(Corollary 3.5)
Z(R)
Q
::R*C)
dim Z Z (R)
2 (RIG i s a s p l i t t i n g f i e l d f o r
R*G.
N o w a p p l y Theorem
4 . 7 (ii). 4.11.
COROLLARY.
simple r i n g (i) G
R.
Let
R*G
be a c r o s s e d p r o d u c t of a f i n i t e g r o u p
G
over a
Assume t h a t
a c t s t r i v i a l l y on
Z(R)
(e.9.
R*G
i s a t w i s t e d g r o u p r i n g of
G over
PROJECTIVE CROSSED REPRESENTATIONS
103
R) (ii)
R
(iii)
charR
Z(R)
is finite-dimensional over
1
/GI
Then the number of nonisomorphic irreducible R*G-modules does not exceed the
G.
number of a-regular classes of field for the Z(R)-algebra
The equality holds if
Z(R)
is a splitting
R*G.
By hypothesis, Z ( R ) G = Z(R) and so the result follows by virtue of
Proof.
Corollary 4.10.
m
5 . PROJECTIVE CROSSED REPRESENTATIONS Throughout this section, G vector space over a field over F*
and
Z2(G,F*)
v
Thus F"G
we write
F"G
a finite-dimensional
the group of all 2-cocycles of
defined with respect to a specified action of
a E ZZ(G,F*), F.
F
denotes a finite group,
G
on F .
Given
for the corresponding crossed oroduct of
is a free left F-module with basis
{ z \ g E GI
G
G
over
and with multi-
plication defined distributively by using the identities L -
2
where
h
denotes the image of
In what follows we write
2.
= a(z,y)zj
for all
Ich
="A;
for all 5 E G,h E F
1 under the automorphism of F
Gs(V)
corresponding to
for the general semilinear group of
that is the group of all nonsingular semilinear transformations of A
z,y E G
xy
V,
V.
mapping p :
G
-+
GS(V)
is called a p r o j e c t i v e crossed representatior, of
G
over F
if there exists a
mapping :
G x G-
F"
for all x,y E G
CHAPTER 2
104
To stress t h e dependence of
G on
a-representation o f p
d e t e r m i n e s an a c t i o n of
ponding group 5.1.
LEMMA.
g E G,
each
on
F
a
p :
Let
@
let
G-
9
and
CX.,
P as an
we shall often refer t o
O u r p o i n t of d e p a r t u r e i s t h e o b s e r v a t i o n t h a t
on
F
and t h a t
a
i s an e l e m e n t of t h e c o r r e s -
G S ( V ) b e an a - r e p r e s e n t a t i o n of
be t h e automorphism of
= @
E Z 2 (G,F*)
9
( A ) ,g E G,X E F,
, where
F
G
V
on
and, f o r
P(g).
determined by
p r o v i d e s an a c t i o n of
G
on
Then
F
Z 2 (G,F*) i s d e f i n e d w i t h r e s p e c t t o t h e a c t i o n of
given i n ( i ) .
Proof. all
V. G
V
on
Z2(C,F*).
(i) The formula (ii)
P
E
F
(i) Given and
u E V.
and
which shows t h a t
Since t h e e q u a l i t y
X,y E
G, i t s u f f i c e s t o v e r i f y t h a t
w e have
'$ @
X Y
=
QXy.
For
G
PROJECTIVE CROSSED REPRESENTATIONS
p(1) = lv,
i s a consequence of t h e assumption t h a t
105
.
t h e r e s u l t follows.
W e now proceed t o develop o u r v o c a b u l a r y . Assume t h a t
i s an a - r e p r e s e n t a t i o n
p
say t h a t
is a
p(g) E CL(V),
G
Finally, i f
GS(V).
--+
I n c a s e each
G over
F.
By
Let
p : G
written
degp,
----3
i s b o t h c r o s s e d and p r o j e c t i v e
P
i s s a i d t o be a line ar representation.
p
G over
w i s e , a l i n e a r r e p r e s e n t a t i o n of GS(V)
F
be an a - r e p r e s e n t a t i o n of
V.
invariant i f
W
is irreducible, i f
P
G
V.
on
The
degree
A subspace
w
of
GL(V). of
V
P,
is said
W
We r e f e r t o
0
and
V
ar e t he only invari-
i s s a i d t o be completely reducibze
P
The r e p r e s e n t a t i o n
i f f o r any i n v a r i a n t subspace
V = W 8 W'.
P :
i s s e n t i n t o i t s e l f by a l l s e m i l i n e a r t r a n s f o r m a t i o n s
We s a y t h a t
V.
G
-
Expressed o t h e r -
i s a homomorphism
i s d e f i n e d a s t h e dimension of
a n t subspaces of
that
P : G
then w e
Thus a c r o s s e d r e p r e -
as a project7:ve re presentction of
p
F.
on
representation, then
p(g),g E G.
C over F .
x,y E G,
i s a p r o j e c t i v e r e p r e s e n t a t i o n i f and o n l y i f i t d e t e r m i n e s t h e
t r i v i a l a c t i o n of
t o be
a(z,y) = 1 for all
If
i s j u s t a homomorphism
we r e f e r t o
P
V.
on
crossed representation of
G over F
s e n t a t i o n of
Lemma 5 . 1 ,
G
of
t h e r e e x i s t s a n o t h e r such subspace
p
as being indecomposable i f
V
W'
such
c a n n o t be
w r i t t e n a s a n o n t r i v i a l d i r e c t sum of i n v a r i a n t subspaces. Let
pi
:
same a c t i o n of
GG
GS(V.),i on
F.
=
b e two a - r e p r e s e n t a t i o n s which d e t e r m i n e t h e
1,2,
Then t h e map P1
+ P2
:
C
--t
GS(V1 C B Vz)
d e f i n e d by (PI
+ P2) ( g ) ( V l + v z )
= P
1
(g)v + P2(g)v2 1
(vi
Vi'
E
i s e a s i l y s e e n t o be a g a i n an a - r e p r e s e n t a t i o n which d e t e r m i n e s t h e same a c t i o n of
G on 8'.
W e warn t h e r e a d e r t h a t u n l i k e t h e s i t u a t i o n f o r l i n e a r r e p r e s e n -
t a t i o n s , t h e r e i s no n a t u r a l way t o d e f i n e a sum of an C ( - r e p r e s e n t a t i o n and r e p r e s e n t a t i o n when
a # 6.
R-
CHAPTER 2
106
Two projective crossed representations: pi
:
G
(i =
GS(Vi)
1,2)
are said to be p r o j e c t i v e l y equivalent if there exists a mapping
v
G
:
F*
-+
with
P(l) = 1
and a vector space isomorphism
f : v - v such that for all g p(g)
If
for all
1
=
g E G,
then
P1
are said to be l i n e a r l y equi-
P2
and
G
E
valent. 5.2. LEMMA. If
(i) Let
pi,i
=
1
action of P1
G
on
F and
,
p
is projectively equivalent to
C
PI
then
is cohomologous to a:
CI
vi.
1,2, be an n.-representation on the space
1
2
and
.
P2
determine the same
Furthermore,
=
c(
if
CL
1
2
is linearly equivalent to p2.
p1
(ii) Let
be an u -representation on the space
V.
Then, for any cocycle
1
al,
that is cohomologous to
c1 2
there exists an
which is projectively equivalent to
P 1
then
P1
.
CL 2
-representation p 2
In particular, if
V
on
is a coboundary,
U 1
is projectively equivalent to a crossed representation.
Proof.
(i) Assume that
is projectively equivalent to
PI
P .
Then there
2
is a map
P
f
v
:
V1 -.+
:
G
+
F*
with
p(l) = 1
such that P2
Fix g E G
and a vector space isomorphism
(9)=
1-1 ( 9 ) f P 1
(g)f-l
and let $i be the automorphism of
Then, for all
V E
V
and
h E
for a l l
F determined by P i ( g ) , i
F , we have
2
P (9)( h v )
( h ) P z( g ) u =
az (h)!J(g)J?
=
$2
=
u'g)fP1 (g)f%0)
= P ( g q (A) P1
(g)f-lv
(g)J'--lv
g E G =
1,2.
107
PROJECTIVE CROSSED REPRESENTATIONS
0,)
which shows that
x,y
Given
= @
(1) and hence that
G, we have
E
a
(Sc.Y)
= P 2 (x)P2 ( y ) Pz (xy) = (U(+)fPl = (Ll ( S ) ” U
proving that a
and
E
G, then
(ii) Let
2
Define
U ( 1 ) = 1.
1
-5 (f-h(Sty) -l)
(P(5)zU(y)P(Scy)-1) (fP1(z)P1(Y)Pl(q)-lf-l)
=
(6U) ( x , y ) a l(z,y) ,
ai,
2
with
1
( W L f ) f P l (xg)f-l)-l
=
be cohornologous to
0
( y ) f ) (P1 (Sc) P ( y ) P (zy
)
Furthermore, if
p(x) = 1 for all
.
= 0 1
(x)S-l)( i J ( Y ) f P 1 ( y ) f -
are cohomologous.
0 2
1
x
@1 = Q 2 .
2
P2
:
say
=
0
(6ii)a
f o r some
G
-+
Then
p2
ii :
F*:
2
G --.-+ GS(V) by
P2 ( g ) =
U(g) P 1 (gl .
.
a projective crossed representation which is projectively equivalent to
. 2
whose corresponding cocycle coincides with
c1
is
and
PI
The following result shows that the study of &-representationswith a fixed
G on F
action of
5.3. THEOREM.
is equivalent to the study of F‘G-modules.
Let F be a G-field and let c1 E Z2(G,F*), where
F.
Z2(G,F*) is
defined with respect to the given action of
G
correspondence between 0-representations of
G which determine the given action
of
G on F and FaG-modules.
on
Then there is a bijective
This correspondence preserves sums and maps
bijectively linearly equivalent(irreducible, completely reducible, indecomposable) a-representations into isomorphic (irreducible,completely reducible, indecomposable) FaG-modules. Proof.
Let
p
be an a-representation of
rise to the given action of zl
G.
V,g
G on F.
:
FOG
f(Cx st g
V
P(g)
(10) =
’XP(g)V
for all
1 E F,
Due to Proposition 3 . 1 , we can therefore define a homomorphism
f
Hence
Then
G on the space V which gives
-+
=
End(V)
zx p ( g )
becomes an FNG-module by setting
g
CHAPTER 2
108
(ZZ
;)u =
9
V,
Conversely, g i v e n an F'G-module
P ( g f E End(V)
gv
p(g)U =
by
lies i n
p(g)
Thus each
for a l l
p ( 1 ) = lv
p(x)P(y) =
and
a-representation
of
G
V
E V.
=
=
g.
on
V
E
9
P(g)
Then
'hP(g)V
F
F
Ti)
and w e d e f i n e
i s i n v e r t i b l e and
1 E F,g
for all
G S ( V ) and t h e automorphism of
c o i n c i d e s w i t h t h a t determined by have
(x E F,u
9
i s a v e c t o r s p a c e over
V
P ( g )( X U ) = ; ( X U )
Ex p ( g ) v
p(g)
determined by
P,
Furthermore, by t h e d e f i n i t i o n of
@(;c,y)P(xy) f o r a l l x , y E G.
P
Thus
G
which d e t e r m i n e s t h e g i v e n a c t i o n of
V
E G,v E
we
i s an
on
F.
W
of
T h i s s e t s up t h e d e s i r e d b i j e c t i v e correspondence. Let
p
be an a - r e p r e s e n t a t i o n
i s i n v a r i a n t under a l l
G
of
p(g),g E G
V.
on t h e s p a c e
W is
i f and o n l y i f
A subspace
V
a n FaG-submodule.
Hence t h e correspondence p r e s e r v e s sums and maps b i j e c t i v e l y i r r e d u c i b l e (comp l e t e l y r e d u c i b l e , indecomposable) a - r e p r e s e n t a t i o n s i n t o i r r e d u c i b l e ( c o m p l e t e l y r e d u c i b l e , i n d e c o m p s a b l e ) FaG-modules. We n e x t observe t h a t a n F-isomorphism FaG-isomorphism
suppose now t h a t
f
Pi
:
G-
i s l i n e a r l y equivalent t o : I/
V
-t
1
:
-f
V
2
of F'G-modules
=
fcsu)
GS(V.),i = 1,2, p2
for a l l
i f and o n l y i f t h e r e i s an F-isomorphism
such t h a t
V
1
.
p2(g)f(U) = fpi (g)U
a r e isomorphic.
A s an a p p l i c a t i o n of Theorem 5 . 3 , THEOREM.
for all
or t o
SfCV)
= f(gV),
g
E G
for a l l
.
Thus two a - r e p r e s e n t a t i o n s are l i n e a r l y e q u i v a l e n t i f and o n l y i f
t h e corresponding F'G-modules
5.4.
Then
2
The l a t t e r i s e q u i v a l e n t t o G,V E
u t Vl,g E G
a r e two a - r e p r e s e n t a t i o n s .
P 2 ( g ) P = f P 1 (9)
g
i s an
i f and o n l y i f
gfw p1
f
Let
F
be a G - f i e l d ,
we now prove let
a E Z2(G,F*),
of r e p r e s e n t a t i v e s f o r t h e a - r e g u l a r c l a s s e s of
G
let
and l e t
X
be a f u l l s e t
charF
k
IGI.
by n t h e number of l i n e a r l y nonequivalent i r r e d u c i b l e @ - r e p r e s e n t a t i o n s of
Denote
G
109
PROJECTIVE CROSSED REPRESENTATIONS
G on F.
which determine the given action of
Then each &representation
G
of
is completely reducible and
n
Proof. Hence
G
G is finite, the field extension F/FG is also finite.
Since
F"G
CG (x) C dimF 6 X FG
is a splitting field for the F -algebra FaG.
FG
with equality if
Q
is a finite-dimensional algebra over the field
FG.
Therefore FaG
is semisimple artinian by virtue of Maschke's theorem (Corollary 3.5).
It
follows, from Proposition 1.7.24, that each F"G-module is completely reducible, Thus each a-representation of
G is completely reducible, by Theorem 5.3.
The
second assertion being a consequence of Theorem 5.3 and Corollary 4.10, the result follows. We next discuss the matrix form of projective crossed representations. Let
{?Il , ? I p , .
V
mation of
. . ,V,}
V
be an F-basis of
and let f
which determines the automorphism @
be a semilinear transfor-
of F.
We may write
n
in F.
for uniquely determined h i j
We shall refer to the matrix
as the matrix of t h e semiZinear transformation f basis).
If F
For any
(a..)
=
1-3
F
E
M,(F)
and any
@ E Aut(F),
g E G, then we write ' A
is a G-field and
automorphism of so is
A
(with respect to the given
corresponding to
g.
for ' A ,
Of course, if A
' A E M,(F)
define
@
where
by
is the
is nonsingular, then
'A.
5.5. LEMMA.
Let f and g
be the automorphism of
Proof.
Write
A
f
be two semilinear transformations of
F determined by f.
=
(A.
.)
1-3
and A
g
Then
= (Ll. . ) . 1-3
Then
V
and let
@
CHAPTER 2
110
n
n
.
proving the required assertion.
It is now an easy matter to provide the matrix form of projective crossed representations. P :
Let
G
+
GS(U
be an &representation P
We know, from Lemma 5.1, that becomes a C-field.
P
By Lemma 5.5,
P
(g) E GL(n,F).
corresponding to
G,
Since
let
r P ( g ) be the
P(1) = l V
a ( z , y ) r (q)
, we have
for all z , y
P
E
C
G - - + GL(n,F) as the matrix a-representation
:
It is clear that
p.
and let n = dimv. F on F so that F
Furthermore, by Lemma 5.5, we have
r P m 2 r P (y’) = rP
V
G
g
and, for each
(1) = 1, the identity nxn-matrix.
We shall refer to the map
G on
determines an action of
Fix a basis of
P(g).
matrix of
of
is determined uniquely by
p,
up to
conjuyacy. Conversely, let F
r
be a G-field and let
G
:
---f
GL(n,F) be any map that
satisfies the following conditions: 5
r)I(
r (Y) =
wx,y)
r(l) f o r some
F.
over
o(g) on
a
E
V
:
CXG
--+
F*.
Let
V
For any g E G, define
GS(V)
and the map
P
:
G-
p,
we have
r
=
rP
(S,Y E G)
r
=
be the vector space of all nxl-matrices P(g)
:
V--
GS(V)
which determines the given action of
definition of
r (VY)
V
by
P(g) (0)= r(g)’U.
Then
is obviously an a-representation of
G on F .
G
Furthermore, by the
with respect to the natural basis of
.‘L
Thus we may treat the terms a-representation and matrix a-representation as interchangeab1e. For future use, we now record the following result. 5.6. PROPOSITION.
Let F
be a G-field and let a E Z * ( G , F * ) ,
is defined with respect to the given action of (i) Let
p
be an a-representation of
G
where
Z2(G,F*)
G on F .
over F
of degree n
and let
P
PROJECTIVE CROSSED REPRESENTATIONS
G
determine t h e g i v e n a c t i o n of
J
(ii) L e t
F.
on
be a l e f t i d e a l of
FaG
an
Then
111
i s a coboundary.
of F-dimension
n.
an
Then
i s a co-
boundary. Proof.
(i) L e t
P
P.
be t h e m a t r i x a - r e p r e s e n t a t i o n c o r r e s p o n d i n g t o
Then
Taking t h e d e t e r m i n a n t s of b o t h s i d e s and s e t t i n g
P(2)xll(y)
an
Thus
6P,
=
(ii) S i n c e
=
P(g)
we obtain
= detrp(g),
an(5,y)P(q)
x,y E G
for a l l
a s required.
J i s a n FaG-module of F-dimension
J
12,
d e t e r m i n e s a n a-represen-
n which d e t e r m i n e s t h e g i v e n a c t i o n of
t a t i o n of d e g r e e
G
F (Theorem 5 . 3 ) .
on
Now a p p l y ( i ) . W e c l o s e t h i s s e c t i o n by g i v i n g a n i n s t a n c e of how p r o j e c t i v e c r o s s e d r e p r e -
s e n t a t i o n s can a r i s e i n t h e s t u d y of l i n e a r r e p r e s e n t a t i o n s .
G
Let
b e a f i n i t e group and l e t
F.
=
X(q-lng).
F
and each
G.
'x
N
Then
'X
i s o b v i o u s l y an i r r e d u c i b l e c h a r a c t e r of
over
g E G,
i s a f f o r d e d by a r e p r e s e n t a t i o n
the representation
d e f i n e d by
'~(Yz) =
r,
r(g-'ng).
define
'X
then
We s a y t h a t =
g E G.
'r
The l a t t e r i s e q u i v a l e n t t o t h e r e q u i r e m e n t t h a t
Irr(N)
G a c t s on
G(X)
=
Let
G(X)
over
r)
X
(or
x
for all
is
are e q u i -
Irr(N) F
w i t h each
t h e s t a b i l i z e r of
g
x.
E G
X
sending each
Thus
X
E Irr(N)
F
over
F.
to
'X.
We
i s i n v a r i a n t i f and o n l y i f
G.
E
be t h e a l g e b r a i c c l o s u r e of
'$ E G a l ( E / F ) , each
N
d e n o t e t h e s e t of a l l i r r e d u c i b l e c h a r a c t e r s of
F
denote by
N
g E G.
valent for a l l
Then
and
by
is a f f o r d e d by
G-invariant ( o r simply inuari'ant i f no c o n f u s i o n can a r i s e ) i f X'
Let
F o r each
of
x
In fact, i f
be a normal subgroup of
x
irreducible character
'X(n)
N
define
'$ E Gal(E/F)
'$1 by
F.
E Irr(N)
F o r each
E
$1 = '$01. Then Gal(E/F)
sending each
1E I r r ( N ) E
to
@A.
and
Irr(N) w i t h E Following I s a a c s (1981), a c t s on
CHAPTER 2
112
A
we say that given
is F-semi-invariant
if its Galois orbit is G-invariant, i.e.
and g E G ,
@ E Gal(E/F)
g(%) =
@A
for some $ E Gal(E/F).
Example.
Let
X
Irr(N). F
E
Then
A. E Irr(N) are distinct and constitute an orbit under Gal(E/F). ' E The positive integer m is called the Schur index of each of the Ai over
where the
and rn = 1 if
X.
charF > 0.
The characters hi
X.
They are uniquely determined by
permuted by
G
hi
and hence each
are called the E-constituents of
Thus if
i(
is invariant, the
Assume that
X'
(i) The formula
=
$ex
on the Gal(E/F)-orbit of (ii) For each g E G ,
(iii) The map
G
-f
xi
are
is F-semi-invariant,
h E Irr(N) is F-semi-invariant and let F ( A ) E the field generated over F by the values of 1.
5.7. LEMMA.
F
provides a regular transitive action of
denote
Gal(F(A)/F)
1.
there exists a unique
Qg
Gal(F(A)/F),g !--+
$
g
E Gal(F(A)/F)
such that
is a homomorphism with kernel equal to
G(X).
(iv) G ( A ) 4 G Proof.
and
Hence Gal(F(A)/F)
F,
every
B E
Gal(E/F)
restricts
Since every automorphism of every subfield of the extends to
E,
this restriction map is onto.
acts transitively on the Gal(E/F)-orbit of
the identity in Gal(F(A)/F) A
is normal over
Gal(F(A)/F).
algebraically closed field E
(ii) Since
is abelian.
Since F ( A )
(i)
to an element of
G/C(A)
A,
can fix
is F-semi-invariant,
apply (i). (iii) Given g , h E G , we have
A'
A.
Since only
the above action is also regular.
is in the
Gal(E/F)-orbit of
1.
NOW
PROJECTIVE CROSSED REPRESENTATIONS
113
i Also, if g E G ( A ) , then A = 1 ' and hence $J = 1. = Qgh. g h 9 Conversely, if $J = 1, then = A and g G(1). 9 (iv) Direct consequence of (iii) and the fact that Gal(F(A)/F) is abelian. and thus $ $J
Let F be a G-field with the trivial action of corresponding group
Z2(G,F*).
we may form the group
Then
Z2(G/N,F*).
N
on
F
'
and form the
F can also be regarded as a G/N-field, so
Given
Z2(G/N.F*),
E
define infaEZ2(G,F*)
by
Then the map
is a homomorphism which carries coboundaries to coboundaries. morphism
H 2 (G/IY,F*)
Assume that
The induced homo-
H 2 (G,F*) is called the inflation map.
--+
A E Irr(N) is F-semi-invariant. E
By Lemma 5.7(iii),
F(A) is a
G-field with the trivial action of N
on F(A).
shall, regard F ( A )
We now form the corresponding groups
as a G/N-field.
Z2(G/N,F(A)*) and Z2(G,F(A)* ) .
Hence we may, and from now on
The following result in somewhat different form
was established by Isaacs (1981). 5.8. THEOREM.
be a normal subgroup of G,
N
Let
A
let E
be an algebraically
Irr(N) be F-semi-invariant for some subfield F of E. E Assume that the Schur index of A over F is equal to 1 so that A is affordclosed field and let
E
ed by some irreducible representation (i) There exists a cocycle w
r
=
D
=
inf ( w
(iii)
r
of
F(X). such that for
cr
=
inf(w),
G. E Z2(G/N,F(A) * )
1
1
to a &representation
over
G
for some w
),
??
of
w ( A ) E Z2(G/N,F(h)*)
extends to an a-representation of
(ii) If
r
G, then w
and
,
is s u c h that
r
extends
are cohomologous.
0
1
extends to a crossed representation of
G if and only if w
is a
coboundary. Proof.
(i) Let
T
be a transversal for
N in G containing 1.
For each
CHAPTER 2
114
tE T,
n
the map
I-+
tr(t-lnt) is an irreducible F-representation of N
affords the character 't(*X) = 1 is equivalent to
r
(see Lemma 5.7(ii)).
Hence this representation
and so we can choose a nonsingular matrix
M(t)-lr(n)M(t)= tr(t-l,t) We can (and do) take
M ( 1 ) = I, the identity matrix. P w ) =
and note that
p
M(t)
such that
n E N (1)
for all p
Now define
rmM(t)
(tE T,n
E
on
G
by
N)
(2)
I- on N .
agrees with
Our aim is to show that
which
p
is a projective crossed representation of
G such
that
P(ng)
=
P(n)P(g)
p(g)gp(n)p(g)-l =
and prove that (i) follows from the above.
g,h
E
G.
If N E Z ~ ( G , E A ) * )
PP~)
G,n
N
(3)
for all g E ~ , En N
(4)
for all g
E
E
To see the latter, fix m,n E N
corresponds to
p,
and
we need only show that
To prove (5), we apply (3 , (4) and the fact that N
acts trivially on
F(l),
to
deduce that ci (ng,rnh)
as required. Next it will be shown that ( 3 ) and (4) hold.
Write
g = ms
with r n E N , s E T .
Then
p ( n g ) = p ( ( m ) s ) = r(n)r(m)d!(s) = P ( n ) P ( g ) , proving ( 3 ) . we have
Using (1) and the fact that the action of
N
in F ( X )
is trivial,
115
PROJECTIVE CROSSZD REPRESENTATIONS
proving ( 4 ) . W e are l e f t t o verify that
P
is a p r o j e c t i v e crossed representation with
r e s p e c t t o t h e g i v e n a c t i o n of
C
on
F(X).
-1 To t h i s end, we f i r s t p u t n = g mg
i n (4) t o obtain p ( g ) -lr ( m ) P ( g ) = gr ( g - l m g )
Since
r
affords
X
for a l l
m
E
N
(6)
it i s a b s o l u t e l y l r r e d u c l b l e and so, by S c h u r ' s
IrrrN),
E p(x)"p(y) = a(z,y)p(xy)
lemma,
(ii) Assume t h a t
f o r some
w
r
f o r some
a(x,y) E F ( X ) * , pl
extends t o a B-representation
of
as required.
G,
where
B= inf(o
)
1
E Z2(G/N,F(X)*).
N
Since
a c t s t r i v i a l l y on
F(X)
and s i n c e
1
O(g,n)
=
B(n,g) = 1 for a l l
n E N,g
E
G,
=
gr ( g l n g )
we have
I t f o l l o w s from ( 8 ) and ( 7 ) t h a t
p1 ( g ) - l r
( n )p ( g )
Comparing (9) and 161, w e o b t a i n
p
:
G-
G(X)*
since
i n Lemma 5 . 2 . we have
r
p1 ( g ) = ~ ( gp () g )
for all
n E N,g
E G
(9)
f o r some f u n c t i o n
is absolutely irreducible.
Hence, by t h e c a l c u l a t i o n s
CHAPTER 2
116
D
We are therefore left to verify that Given n E N , g E G,
Hence
U ( n g ) = D(g1
r
(iii) If
for all g
G and n
proving (ii).
E N,
N
is a coboundary.
k!
Then there is a function p : G
G, then we can take
W
=
Conversely, assume that
W
is
--+
F(X)*
1
which is constant on
and such that V ( 1 ) = 1 and
p1 (g) = iJ(g)p(g) for all g E G.
Define p1
E
extends to a crossed representation of
a coboundary.
N.
it follows from (7) and ( 3 ) that
in (ii) and hence, by (ii),
cosets of
is constant on cosets of
r
agrees with
on
N.
U(n) = 1 for all n
Then
Then, by (lo), P I
E
N
and so
is a crossed representation of
G
and the proof is complete. In view of Theorem 5.8 the extedibility of U(A)
G
For this reason we shall refer to
r
is governed by
E z'(G/N,F(x)*) W(xl
G
as the obskruction
COCyCZe COPFeSpOnding
to X. The result above can be applied to investigate the extendibility of representations from normal subgroups in the case where the base field is not necessarily algebraically closed. Let N of
N
be a normal subgroup of
over
F and let E
r
Yo.
character of (i) V
'?
and put
YL
and let
V'
It is no loss to assume that
ro.
L =F(X).
Let
when viewed as an L-representation of
irreducible constituent of affords
F.
be the algebraic closure of
constituent of the character of representation
G, let T be an irreducible representation
N
Denote by
X an E-
TL denote the let
To be an
be an irreducible LN-module which h
is an E-constituent of the
The following two facts are standard (see Isaacs (1981)).
is irreducible as an FN-module and as such it affords
r.
PROJECTIVE CROSSED REPRESENTATIONS
( i i ) The c e n t e r of
117
i s t h e s e t of s c a l a r m u l t i p l i c a t i o n s by e l e m e n t s
End(V1 FN
L.
of
Using ( i ) and ( i i ) ,w e now prove t h e f o l l o w i n g r e s u l t . 5.9.
THEOREM. ( I s a a c s (1981)).
G
r e p r e s e n t a t i o n of
I n t h e above s i t u a t i o n ,
r
ro
i f and o n l y i f t h e L - r e p r e s e n t a t i o n
G
a c r o s s e d r e p r e s e n t a t i o n of
L
over
e x t e n d s t o an F-
N
of
extends t o
w i t h r e s p e c t t o t h e a c t i o n of
G
L
on
g i v e n by Lemma 5.7 ( i i i ) . Proof.
V
Let
be t h e r-dimensional
column space o v e r
L,
where
P = degr 0
V
Then
i s an LN-module v i a t h e a c t i o n
n*v V
and
affords
representation
.
rO
r.
For each
g E G
N
and
V E V,
g
w e define
F,
L
and
d e f i n i t i o n s of
n*v f o r n
r
G
extends t o
I---+
Put
g-u
D
G
to
E V
t h e map
P(n) = N
E
r
v
agree.
G
over
L
Now d e f i n e
V.
( g E G,u E V )
all
i s 6'-linear.
n
E N,
Also,
FN
since
N
w e see t h a t t h e t w o
g , h E G, w e have
r
extends t o
G
so t h a t we have an F - l i n e a r a c t i o n
such t h a t
= End(V).
r .
which e x t e n d s
by a p p l y i n g t h e f i e l d automor-
hg'V
Given
it a f f o r d s t h e
amounts t o an F - l i n e a r
P(g)gv
(n) for
(11)
G.
to all
'V
=
V,n E N )
a s required.
Conversely, assume t h a t
v
r
t o each e n t r y i n t h e column v e c t o r
i s t r i v i a l on
acts t r i v i a l l y on
Thus
(V E
i s viewed a s an FN-module,
V
on
g'v Qg
(n)v
i s a c r o s s e d r e p r e s e n t a t i o n of
Qg induced by
Since
r
Thus an e x t e n s i o n of
p
Assume t h a t
=
V
By ( i ) , when
e x t e n s i o n of t h e a c t i o n (111 of
phism
.
For each
g E C
and
9 E D,
o b s e r v e t h a t t h e map
CHAPTER 2
118
N.
i s F - l i n e a r and commutes w i t h t h e a c t i o n of
'*$
we denote
D
lies i n
Owing t o ( i i ) ,t h e c e n t r e of on
V.
on
L
@ ++ '*$l
and w e see t h a t
G
W e have t h e r e f o r e d e f i n e d an a c t i o n of
D is
I t f o l l o w s t h a t t h i s map, which
D
on
t h e s e t of s c a l a r m u l t i p l i c a t i o n s of
g*)
Thus w e have an induced a c t i o n (which we c o n t i n u e t o d e n o t e
G
G
I t remains t o show t h e * - a c t i o n of
F,
=
then Jl*'
)-1
s i n c e t h e a c t i o n of
E
Since
on
N
A l s o s i n c e t h e a c t i o n of
n
V.
a c t s a group of s e m i l i n e a r t r a n s f o r m a t i o n s of
corresponding ( m a t r i x ) c r o s s e d r e p r e s e n t a t i o n of
all
v i a automorphisms.
L
of
G
such t h a t
Hence
'd E
D.
i s an automorphism of
V
G,
Thus, i f p
t h e n by ( 1 2 )
is the original action. is F-linear.
G
is L-linear,
we s e e t h a t
is the
extends
rO-
Note t h a t i f
g* E G a l ( L / F ) .
Thus
n*
p
is t r i v i a l for
N.
$g
i s t h e unique
a
E Gal(L/F)
such t h a t
u(gA)
=
A,
t h e proof w i l l
be complete when w e show t h a t
g*A(g-lng)
Since
n*
i s t r i v i a l and
=
h(nf
for a l l
g E G,n E N
P I N = To, w e o b t a i n
-'r0
p(g)
( n )P (g)
=
g*ro (g-lng)
and t a k i n g t r a c e s y i e l d s
trr ( n ) But
trr
=
mh.
.
w e can c a n c e l theorem.
where
rn
Let
N
trP*ro( g - l n g ) )
i s t h e Schur i n d e x , and s i n c e
rn and o b t a i n
5.10. COROLLARY.
=
cG
( n ) = *' A (g-lng).
and l e t
r
m
= 1
if
charL > 0,
T h i s completes t h e p r o o f o f t h e
be an i r r e d u c i b l e F - r e p r e s e n t a t i o n of N
119
GRADED AND G-INVARIANT IDEALS
G.
which is invariant in
A
E
Let
E
be the algebraic closure of
r.
Irr(N) be an E-constituent of the character of
E
index of
over
F
is equal to
To.
and only if
r
Then
TF(”
A
is a coboundary.
G, A
is F-semi-invariant.
such that
By Theorem 5.9,
To
Let
be
is an E-constituent of the
extends to an F-representation of
G if
To extends to an F(h)-crossed representation of G with respect to G
the action of over F
charF > 0).
G if and only if the obstruction cocycle
w ( x ) E Z2(G/N,F(X)*) corresponding to G Proof. Since r is invariant in
character of
Assume that the Schur
1 (e.9. assume that
extends to an F-representation of
an irreducible constituent of
F and let
F(h)
on
1,
is equal to
given by Lemma 5.7(iii).
ro
A.
affords
.
Since the Schur index of
Now apply Theorem 5.0.
5.11. COROLLARY.
In the situation of Corollary 5.10, assume that for each Sylow
subgroup P/N
G/N,
of
T extends
carries
From the definition of
00
to of
Hence, by Corollary 5.10,
W(x)
=
1 for all Sylow
P By Proposition 1.6.3, we deduce that w(x,
G/N.
Corollary 5.10 again,
T
it is clear that the restriction map
W(x),
P
G subgroups P
w(x).
Then
G.
extends to an F-representation of Proof.
P.
to an F-representation of
r
=
1.
By
G extends to an F-representation of
G.
6. GRADED AND G-INVARIANT IDEALS Throughout this section, A An ideal I of A
denotes a G-graded algebra over a commutative ring R .
is called a graded i d e o l if
I = @ ( I ~ A ) Let B
be another G-graded algebra.
f
SfG A homomorphism :
A-B
of R-algebras is said to be graded if f ( A 1
g
5B
g
for all g
E
G.
The following
observation ties together graded ideals and graded homomorphisms.
6.1. LEMMA.
(i) If
I is a graded ideal of A ,
then A/I
is a G-graded
CHAPTER 2
120
-
(A/I)
a l g e b r a by s e t t i n g (i i ) I f
f
:
A
B
i s a graded i d e a l . isomorphism
I
homomorphism A-
(i)
proof.
= (A
9
+I)/I
3
B.
of
A
B
G.
f
Kerf
i s s u r j e c t i v e , then t h e r e i s a graded
I
i s graded i f and o n l y i f
i s t h e k e r n e l o f graded
of G-graded a l g e b r a s .
Since
A = @A
I = @ (I n A SEG
and
SEG
Furthermore, f o r a l l
g
for a l l
i s a g r a d e d homomorphism of G-graded a l g e b r a s , t h e n Furthermore, i f
A/Kerf
(iii) An i d e a l
9
s,y
),
w e have
G,
a s required. (ii) F i r s t of a l l ,
Since
B = @B &G $7'
Kerf
A.
i s an i d e a l of
x
w e deduce t h a t each
9
E
Let
Kerf
I E Kerf
and w r i t e
and hence t h a t
Kerf
is a
A/I
is a
graded i d e a l . Assume t h a t
f
i s s u r j e c t i v e and l e t
f*
: A/Kerf
be t h e induced isomorphism of R-algebras.
proving t h a t (iii) I f
I
f*
--+
B
Then, f o r all
g E G,
i s i n f a c t a graded isomorphism.
A,
i s a graded i d e a l of
I.
graded homomorphism w i t h k e r n e l
A A
i s t o d e s c r i b e a l l graded i d e a l s of
X
of
A
A
--+
The c o n v e r s e is a consequence of ( i i ) .
From now it w i l l be a s s m e d t h a t
d e f i n e d t o be t h o s e i d e a l s
t h e n t h e n a t u r a l map
1
i s a stron g l y G-graded algebra.
i n terms of G-inuarinat i d e a l s of f o r which
Our aim
A 1
GRADED AND G-INVARIANT
121
IDEALS
A X A = X g g-l
then
A
i s a c r o s s e d p r o d u c t of G -1 Thus i n t h i s c a s e X A -l = ; X i
of c o u r s e , i f
over
.
A
g
9
;xi
-1
for all
x
Al
and i f
g E U(A)
g E G
n Ag,
i s G - i n v a r i a n t i f and o n l y i f
=x
for a l l
g E G
To m o t i v a t e t h e d e f i n i t i o n above, w e make t h e f o l l o w i n g o b s e r v a t i o n .
6.2.
LEMMA.
Let
Al.
be t h e s e t of a l l i d e a l s of
Id(A )
Then t h e group
G
1
a c t s on
Id(Al)
according t o t h e following r u l e
'X=A Furthermore, f o r any g i v e n
9
XA
for a l l
g-l
g E G,
i n t e r s e c t i o n s and p r o d u c t s of i d e a l s i n Proof.
If
X
E
Id(A )
A gXA 1
proving t h a t
'X
and
1
E Id(Al).
g E G,
a,b
E G,
g
E
A X A -1 = X ' g 9 w e have
=
1E Al,
1
XA
= X 1
then
and t h u s we have an a c t i o n of For a given
p r e s e r v e s i n c l u s i o n s , sums,
then
= A A X A -lA l g g Since
k-+ X '
and a l l g e
Id(All.
'X=A Finally, i f
X
t h e map
X E Id(A1)
G,
G on
t h e map
X
Id(Al).
'X
L--P
c l e a r l y preserves inclusion.
f o r e i t p r e s e r v e s sums and i n t e r s e c t i o n s .
If
(gX) (gY) = (A X A
X,Y € I d ( A
), 1
There-
then
1) ( A Y A
g 9 = A X A 1 Y A -1 = A X Y A 9 g 9-l = S(XY)
9 9 -
Hence t h e g i v e n map a l s o p r e s e r v e s p r o d u c t s and t h e r e f o r e t h e proof i s complete. W e now d e s c r i b e a l l graded i d e a l s of
6.3. then
THEOREM.
X
(Dade ( 1 9 7 0 a ) ) . ( i ) I f
i s a G - i n v a r i a n t i d e a l of
Al
A.
I is
a g r a d e d i d e a l of
such t h a t
A
and
X = I
nA
G
CHAPTER 2
122
I=AX=XA ( i i ) For any G-invariant
I n A
and
X
ideal
A
of
9
,
1
= A X = X A
9
I = XA = AX
g E G
for all
9
A
i s a graded i d e a l of
such t h a t
I (iii) J ( A
=A A X = X A g g
9
A
is a G - i n v a r i a n t i d e a l of
)
1
for all
A*J(A
and hence
1
) =
J(A )A
E G
is a
1
1
g
A.
graded i d e a l of Proof.
~
I = @ I
( i ) By h y p o t h e s i s ,
X = I
in particular,
1
.
x,y E
If
A I X Y
=
where
I
SEG G, t h e n
A ~ (n I A
Y
)
=
g
I nA
g
for a l l
g
E G;
c A ~ n I A A -
" Y
C - I n AXY
= I x?4 A
so
I C I x-l xi? - Y
and t h u s
'xy
C
A 1I x y = Ax ( A z - l I q )
C A I - " Y C
I
- XY Hence
I A
lg
=
A I = I and, i n p a r t i c u l a r , A I = I for a l l " Y gl g I f o r a l l g E G. Given g E G , w e a l s o have
g E G.
Similarly
g
A I A g
Il
provinq t h a t
1
g-l
= I A = I , g 9-1 I
is G-invariant.
Finally,
I = @ I
=
@Gg
@ A I
=
@Ggl
@ I A
@Gig
=A.I
=
Il.A
as required. ( i i ) W e have
A
A X A -1 = X g
yields
n
A X = XA 9 9
g
for a l l
g
for a l l
g
E
E G,
G.
Since
so m u l t i p l i c a t i o n on t h e r i g h t by
A = @ A
SfG I = A X = @ A X = @XA
g
E G.
I
i s a graded i d e a l of
= X A ,
SfG
sfGg proving t h a t
it f o l l o w s t h a t
g'
A
such t h a t
I nA
g
=
A X = XA g
g
for a l l
GRADED
AND G-INVARIANT IDEALS
(iii) Owing to (ii), it suffices to show that J ( A Al.
Let
each
g
E
V
M
be an irreducible A -module, and 1
'M
G, we shall show that
1
123
is a G-invariant ideal of
)
the annihilator of
V.
For
defined by
is the annihilator of some irreducible A -module.
Because
J(A1)
is the inter-
section of all such annihilators, the required assertion will follow. We may choose a maximal left ideal I A A
Z Y
=
A
for all
Z,IJ
E G
the set of all left ideals X
Y
with
-+
A - 1Y
of A l
g
1
g'
V.
Since
the map
X
--+
9'
1
as the inverse map.
9 so A I is an A -submodule of A
so that A /I
Al
A X sends bijectively g onto the set of left A -submodules of A
1E Al,
and
of
These maps obviously preserve inclusion. An elementary calculation shows that the
annihilator of the irreducible A -module A / A I is precisely A
9 9
noted above, this completes the proof of (iii). 6.4.
PROPOSITION.
(i) A / A ' J ( A
)
Let A
be a G-graded algebra over a commutative ring R.
is a G-graded algebra by setting for all g E G
(A/ASJ(A1)Ig= (A +A*J(A1))/A.J(A1) g
and (A/A'J(A1))
(ii) If A
A1/J(Al)
is a crossed product (skew group ring, twisted group ring, twisted
group algebra) of
G over A l ,
then A / A ' J ( A
)
is a crossed product (skew group
ring, twisted group ring, twisted group algebra) of (iii) If A
is a crossed product of
G over A l
sum of additive groups), then A / A * J ( A 1 )
G over A l / J ( A 1 ) . and A
J(A1)
(direct
is a twisted group algebra of
G over
= R @
R.
Proof.
(i) That A / A * J ( A )
is a G-graded algebra with the given g-component 1
is a consequence of Lemma 6.l(i) and Theorem 6.3(iii).
and so
By definition,
124
CHAPTER 2
Now
and t h e r e f o r e
A ~ A - J U ) = A -J(A 1 1
,
= J(A1)
1
as required.
A
( i i ) Assume t h a t
g E G,
there e x i s t s a unit
A/A'J(A1)
(A/A.J(A1))
in
G
c r o s s e d p r o d u c t of
A
Assume t h a t units
-
9
of
g
A
for a l l
9'
over
of
A1/J(Al),
in
A
g
G.
E
9
-
A
over
-
Then, f o r e a c h
+ A.J(Al)
g
Hence
g'
i s a u n i t of
A/A.J(A1)
This proves t h a t
is a
by a p p l y i n g ( i ) .
i s a skew g r o u p r i n g of
in A
A
G
is a c r o s s e d p r o d u c t o f
G
over
A
1
.
Then we may choose
such t h a t
-
-XY
z,y
for a l l
=ccy
E
G.
But t h e n
( ~ ~ + A . J I ( A ~( )~) + A . J ( A) ) =
;c,y
for a l l
E
G,
proving t h a t
(Zj +
A/A*JIA1)
)),y
A . J ( A ~ ) )E ( A / A . J ( A
is a skew g r o u p r i n g of
G
over
A1/J(A1).
A
Assume t h a t
g E G,
i s a t w i s t e d group r i n g o f
we may choose
f o r each
A/A.J(A1)
g E G,
-
g E U(A) n A g
(i+A.J(A1))
A
A
C Z(A)
G
1 -
G
a t w i s t e d group a l g e b r a of
centralnes
over
G
5 Z(A/A*J(A1)).
Al/J(Al),
over
Al
.
B u t then
proving t h a t
A1/J(A1 I .
i s a t w i s t e d g r o u p a l g e b r a of
(Al + A . J( A 1) ) /A *J( A l)
and hence
g
Then, f o r each
(Al + A . J ( A 1 ) ) / A * J ( A l ) ,
centralizes
is a t w i s t e d group r i n g o f
F i n a l l y , assume t h a t
such t h a t
Al.
G over -
over Thus
Al
.
Then
A/A*J(A1)
is
as r e q u i r e d .
(iii) Our assumption g u a r a n t e e s t h a t
R and hence
A /J(A
A/A*J(A1)
1
) 1
(A:
A*J(A1))/A*J(A1)
is a t w i s t e d group a l g e b r a of
5 Z(A/A.J(A1)) G
over
R.
'
For f u t u r e u s e , we n e x t r e c o r d t h e f o l l o w i n g o b s e r v a t i o n .
6 . 5 . PROPOSITION.
Let
A
be a c r o s s e d p r o d u c t of
G
over
A
1
,
let
N
be a
INDUCED MODULES
normal subgroup of
G
and l e t
( i ) For any i d e a l
X
of
T
tX
where
(iil
=
Z X x
A") =
@
t X-t
( d i r e c t sum of A('"-modules)
. {Xi(i
E
I} of
X,Y of
(iii) For any i d e a l s
G/N
A ( ' ) ,X
c Y
A
1.7,
By P r o p o s i t i o n
A(N),
i d e a l s of
=
t
-1
'x
, w e have
=
A-X =
(
c AY
can b e r e g a r d e d a s a c r o s s e d p r o d u c t of
-
A ( ~ , ) by s e t t i n g AtN = @ A = ?.A"' and -1 gEtN X? = tX and so we may assume t h a t N = 1.
( i ) Since
AX
i f and o n l y i f
tN
over
tNX
G.
in
lST
For any f a m i l y
Proof.
N
be a t r a n s v e r s a l f o r
A-X -1
125
gxj
=
iX
=
g E G.
for all
Z.
Then
Hence,
@ A )X = 8 (;A ) X = @ G X = 8 gXi s f G g SfG SfG sfC
a s required.
A.X.
( i i ) By ( i ) ,w e have
'
n A*Xi
8 gx.g
=
SfG
8 ( f? g X . ) g S ~ CE I
=
'
iEI
and t h e r e f o r e
'
=
Og(
&G
X.).;
iEI
'
=
A * ( n Xi) iEI
as asserted. (iii) By ( i ) ,we have
A-X Hence, i f
AX
5 AY.
AX
C
AY
AX If
C
AY,
@ tX.; E T
X c Y.
then
AX = A Y ,
=
and
A'Y = 8
Conversely, if
X
t h e n by t h e above,
=
Y,
X
tY*t c Y,
then c e r t a i n l y
a contradiction.
Thus
m
a s asserted.
7. INDUCED MODULES
Throughout t h i s s e c t i o n ,
A
d e n o t e s a G-graded a l g e b r a o v e r a commutative r i n g
S t a r t i n g from Lemma 7 . 8 , i t i s assumed t h a t Al.
For any subgroup
H
of
G,
we write
A
i s a c r o s s e d p r o d u c t of
G
over
R.
126
CHAPTER 2
so t h a t
A(H)
A(')
i s an H-graded a l g e b r a .
i s a s u b a l g e b r a of
product
A 63 V
A,
V
Let
be a ( l e f t ) A(H)-module.
Since
we can d e f i n e a n A-module s t r u c t u r e on t h e tensor
by
A(H) y(z Q
0) =
yz Q v
T h i s i s t h e induced module and we denote i t by module, then we w r i t e 7.1.
H
LEMMA.
5 L,
WH
t o indicate that
VG
W
or
@.
z,y E A If
W
and
2,
E V
i s a n A-
i s r e g a r d e d a s an A(H)-module.
( T r a n s i t i v i t y of t h e i n d u c t i o n ) .
t h e n f o r any A(H)-module
for a l l
If
6,H
are subgroups of
G with
V,
= VG
(VL)G
The d e s i r e d isomorphism f o l l o w s from
Proof.
(+)G
A @
=
8 V)
(A(L)
A(H)
8 A(L)) @ V
(A
(by P r o p o s i t i o n 1 . 1 . 6 )
P (L) G
V = V .
= A @
(by P r o p o s i t i o n 1.1.1)
A(H) 7.2.
LEMMA.
Proof.
For any A(H)-modules
V 1 and
W e have
=A
as required. 7.3.
LEMMA.
. For any subgroup
A = A ( H ) @ ( 8 A ) &H Proof. module. all
=
H
of
8 Vl @ A '8 V2 A(H) A(H)
vG1 8 vG2
G ( d i r e c t sum of l e f t and r i g h t A(')-modules)
I t suffices to verify that
M =
@A
&H
i s a l e f t and r i g h t
The l a t t e r w i l l f o l l o w provided we show t h a t
h E H.
Since
(by P r o p o s i t i o n 1.1.5)
A M C M h -
and
MA
A(H)C
h -
M
for
INDUCED MODULES
A M c
-&H
9
and
hg
MAh
5 M.
7.4.
H
for
LEMMA.
9 H,
g
Let
V
w e have
C A A
AhM
be an A(H)-module.
i
C
g-&H
v-
M.
in
VG
i s a d i r e c t summand of
Proof.
1.1.5,
By Lemma 7 . 3 ,
A(H)8 V
hg
A s i m i l a r argument shows t h a t
Then t h e map
vG
V r - - t l @ V
i s an i n j e c t i v e homomorphism o f A(H)-modules.
V
Z A
127
Furthermore, t h e image
1@ V
of
G
( V )H.
A ( H ) i s a d i r e c t summand of
and so, by P r o p o s i t i o n
A
@.
i s a d i r e c t summand of t h e A(’)-module
S i n c e t h e map
i s an isomorphism of A ( H ) - m o d u l e s , t h e r e s u l t f o l l o w s . W e s h a l l r e f e r t o t h e i n j e c t i v e A(H)-homornorphism
canonicai? injection.
v
G
-+
V
,V +-+
1 8
V,
as the
With t h e a i d of t h i s i n j e c t i o n , we now p r o v i d e a u n i v e r s a l n
c h a r a c t e r i z a t i o n of
v“.
7.5.
H
f
:
THEOREM.
V
+
fl
Let
b e a subgroup of
be t h e c a n o n i c a l i n j e c t i o n .
G,
let
(H)
w
be an A(H)-rnodule, and l e t
Then, f o r any A-module
@ E Horn(Ti,WH), t h e r e e x i s t s one and o n l y one A . commutative t h e f o l l o w i n g diagram:
V
G
hi
and f o r any
$ 6 Hom(V ,W) which r e n d e r s A
CHAPTER 2
128
Since a 8
proof.
(@ V l)
V = ~
,a
E
A,# E V,
1 @ V.
is uniquely determined by its restriction to one such $.
A
Consider the map
X
V
w,
+
E
V.
Thus there exists at most
(a,#)-+
balanced and hence determines a %homomorphism
v
any A-homomorphism VG --+ W
W, a c-3
VGL
Then this map is
U$(U).
a$(#) ,U E A
V
,
Since
Q(a ( a 8
a,a
for all
A
E
and
=
Vl)
V E
V,
$ ( a a c-3
Q
u
V ) =
U$fU)
=
a Q(u @
is an A-homomorphism. $(l Q v )
V)
Bearing in mind that
u E V,
for all
= @(U)
the result follows. Let S R E L
and
9
E Hom(L,M), define
h$
-
and let M
be a subring of a ring L
:
L
S
( a @ )(3c)
be a (left) S-module.
M by the rule
$(d)
=
Given
for all
It is straightforward to verify that the additive group
3:
E L
Hom(L,M) becomes a (left) S
L-module.
H be a subgroup of G and let V be an A(H)-module.
Let
A,
a subring of
Since A ( H )
we may apply the foregoing construction so that
is
(A,V)
Hom A (H)
becomes a (left) A-module.
We shall denote this module by
‘V
and refer to
GV
v
be
as a coinduced module. 1.6. THEOREM. (Frobenius reciprocity).
an A(H)-module, and let W (i) Hom(@,W)
2
A G
(ii) Hom(kr, V ) A Proof.
Let
be an A-module.
H be a subgroup of
let
Then
Hom (v,WHl A(H) Hom
G,
as R-modules a s R-modules
(vH,V)
A (H)
(i) Given
E
Hom (V,WH ) A(H)
we may, by Theorem 1 . 5 , define
A * E Hom(I/G,W) by
A X*(a 8 v ) The map
X
F-+
A*
is obviously R-linear.
X* (1 8 u )
=
u E V,u E A
n(X(u)) Furthermore, if
= X(U) = 0
A* = 0 ,
then
for all u E V
INDUCED MODULES
1 i-
Then clearly
-*@ ( A )
is an R-linear map.
If
129
c1 E
Hom
(W,,V),
then the
A (H) map
$ ( a ) defined by [ $ ( a ) (w)1
G
Hom(W, V ) .
belongs to
Furthermore,
(a)
w
= a(m)
($0) ( A )
=
A
and
(@$) ( a ) =
a.
E
W,a E A
Hence
A the map
h
@(A)
+-+
Assume that R
is the required isomorphism.
U and V be any A-modules which are
is a field and let
R.
finite-dimensional over
Then the intertuining number for
U and V
is
defined to be
i ( U , V ) = dimHom(U,V)
R A Assume that
U
is completely reducible and
nonnegative integer m
of
is the multipZicity of
is irreducible.
We say that a
V as an irredueible constituent
U if a decomposition of U into irreducible components contains exactly rn
V.
submodules isomorphic to
be a field, let d i d < m and let R Suppose further that V is an irreducible A(H)-module Let R
7.7. THEOREM. (Nakayama reciprocityl.
H be a subgroup of G. and
V
W an irreducible A-module.
ducible constituent of
Let
m
be the multiplicity of
VG/JifiF, and let n
irreducible constituent of
Soc(WH).
W as an irre-
be t h e multiplicity of
V
as an
Then
In particular, is a splitting field for A
(i) If
R
(ii) V
is isomorphic to a submodule of
a factor module of
and A
(H), then rn = n
WH if and only if W is isomorphic to
p.
(iii) For any irreducible factor module
X of
e,
V
is isomorphic to a sub-
CHAPTER 2
130
module of
xH*
Proof.
Hom(U,W) = 0 for all irreducible A-modules
We first note that
u
A
U ? W.
with
Therefore Hom(fi,W) I H o m ( # / J ( v
G ) ,W)
A
A
Hom(hr,W) @
... @ Hom(W,W)
i(p,W)= r n - i ( W , W ) .
Hence
On the other hand,
(V,WH)
Hom
Horn
(V,Soc(WH)) A (H)
A (HI
(V,V)@
P Horn
...@Horn
A (H) and therefore c ( l J , W
G i ( V ,W) = i ( V , W , ) ,
) =
H
n.i(V,V).
the result follows.
g E G,
For each
Al.
( n times)
(It,v)
A (HI
Taking into account that, by Theorem 7.6(i),
’
For the rest of this section, we assume that A over
( m times)
A
A
is a crossed product of
we may therefore choose a unit
-
g
A
of
in
A
G
g
in
which case
- -
A =Ag=gA 9
Let
H be a subgroup A
Proposition 3.3, I n particular,
V,
A(H’-module
A
of
for all 9 E G
1
G and let T be a left transversal for H in G.
{;It
is a free right A(H1-module freely generated by is a flat A(H)-module.
and let j
:
U-
V
Suppose that
E
By
TI.
U is a submodule of an
be the inclusion map.
Since
A
is f l a t ,
the canonical homomorphism
is injective. image in
fi.
For this reason, from now on we shall identify UG
The following lemma collects together some elementary properties
of induced modules. 7.8. LEMMA. (i)
Ir c 1 -
(ii) VG = 1
(iii) (iv)
Let
8
Vl
and
V2
if and only if
2
$.2
with its
if and only if
(vl + y Z l G = VC1 + VC2 ( v n~ v21G = VC1 n VC2
be submodules of an A(H)-module
5V2
V1
V
= V 1
2
V.
INDUCED MODULES
Proof.
131
Since A
We first observe that (ii) is a consequence of (i).
is a
flat A(H)-module, it follows from Proposition 1.2.7 (ii) that
To prove (i), note that
proving (iv).
Vc c 8. 1 -
Let T
2
V
cV
1 -
be a transversal for H
vG=
obviously implies that
2
G containing 1.
in
Since
@t@V tfT
fi
each element of Therefore, if
E
V
5V2,
implies V1
V1
is such that
i 8 ul
E
@ Ut
with
tfT
f,
then v 1 E V2.
Thus
Vt
E
V.
fi c Vc 1 -
2
proving (i).
To prove (iii), note that the containment
(i).
t
E
can be uniquely written in the form
8+
c ( V l + V 2 ) G follows from
1
2 -
The opposite containment being a consequence of the equality 'ESCV
1
+ u ) = Z Q u 2
+ ? Q v
(v
1
6
v1
,u
2
E
v*,tE T )
the result follows. The following theorem contains some important formal properties of induced modules. 7.9. THEOREM.
Let H
G and let
be a subgroup of
quence of homomorphisms of A ( H )-modules.
U-
x
V
-!!+ W be a se-
Then the following properties held:
(i) The sequence
0-u-
A
V
A W - 0
(1)
is exact if and only if the corresponding sequence of A-modules
0 -f
#2!3-+ WG
---+
0
is exact. (ii) Suppose that the sequence (1) is exact.
Then (1) splits if and only if
( 2 ) splits.
(iii) If
U is a submodule of V,
then
fi/UG
(V/U) G
.
Proof. (i) That exactness of (1) implies that of ( 2 ) is a consequence of the fact that
A
is a flat A(H)-module (see Proposition 1.2.8(iii)).
assume that ( 2 ) is exact.
By Proposition 1.2.7(i),
Conversely,
CHAPTER 2
132
Since Ker(1 8 ll) = Im(1 @ A ) , A similar argument shows that
Im(1 @ 1J) =
8
:
W
+
V
0
U
-+
x
--+
Kerp = I d .
Ker(1 8 1) = 0 implies Kerh = 0 and that
implies I m U = W.
(ii) Assume that
y
it follows from Lemma 7.8(ii) that
Thus the sequence (1) must also be exact.
W
V
-----3
0 is a split exact sequence and let
be a splitting homomorphism, 1 8 y :
Then
c
wc-v
is obviously a splitting homomorphism for the induced sequence. Conversely, suppose that the exact sequence
O d 8 % f l % # + O Let T
of A-modules splits.
be a transversal for H
in
G containing 1.
Consider the mappings
I Then
0
and
T
w
we
are well defined A(H)-linear maps.
Q(i8w)
W E
=
w. ~
fET
v
Ct8vt?+v 16T Let $ : WG
W t 4 1 @ W
splitting homomorphism €or ( 3 ) and let
-
VG
-f
fi
be a
Then, on the one hand,
2
8
~
t
8 ut)
for some t E V
and, on the other hand,
Hence
W = 1J(V )
and therefore (UTW
m) =
(UTU
(i@ u ) =
(UT)(
c ET
= U(U ) =
T h i s proves that
T$O
:
W
3
v
u
is a splitting homomorphism for (1) and the
required assertion follows. (iii) Direct consequence of (i) applied to the natural exact sequence
0-u-v-v/u-o. 7.10. COROLLARY.
Let
V
be an A(H)-module.
If
8
is completely reducible,
133
INDUCED MODULES
V.
then so is Proof.
Let
W be a submodule of
V
and let
0-w-v-v/w-0 be the natural exact sequence. 0---t
splits by hypothesis.
wG
-
lF-
(v/mG
-
0
Hence, by Theorem 7.9, the sequence
V
splits, proving that
The corresponding sequence
is completely reducible.
We next establish the following result which will allow us to identify, under certain circumstances, induced and coinduced modules. 7.11. THEOREM. A (H)-module.
Proof.
4l
E
G
v,
H be a subgroup of G of finite index and let V be an
Let Then
Let
{ g l , g *,...,g 1
For every
put
Then an elementary calculation shows that 8
To prove that
G, there is an h .
For every
g
index set
{1,2
E
,...,n1
E
E
:
GV-+
fl
is an A -homomorphism
U(Al).
=
ge ($1
-1 -1 such that g . g = h '.Lg T(i)
It follows that
and
--
-
for all g E G , $ E
H and a permutation i
--1- --1 g . g = a hz. gT ( < )
a
0
is an A-homomorphism, it therefore suffices to verify that
ec&
for some
H in G .
be a left transversal for
i
g g T ( i )=
hence
gi
-
-
a 9i h i
T ( i )
of the
G
V
CHAPTER 2
134
8 is an A-homomorphism. -1 -1 ) gi 8 @(gi ) = 0, then
proving that
n If
@(gi
z ,. . . ,gn-13
=
0 for all i E {1,2
,...,n].
Since
li=l
{gil,gi g
=
a
-1
hgi
for some h
U(All
H in G, given g
is a right transversal for 11
i
and
E
-
{1,2 ,... , n } .
G, we have
E
- --1
Hence g = ah gi
for some
and therefore
-1 =
@(;)
0
=
)
9)
= cz = 0 , proving that e is injective. -g -1g for some ai E_tj_lAl) and since A is a free left gi -1 - aigi -
But then
@(CZ
Since
-1 -1
A(H)-
-1
as a basis, we deduce that module with the elements g ,g , . . . , g n 1 2 -1 -1 --1 I u\ g l ,g2 , . . . ‘ S n is another A“”-basis for A. It follows that for every set -1 V l 1 G i G n} there is a @ E GV such that ) = Ui,l < i < n . Hence ~
@(z.
{Vi
0
is surjective and the result follows. Let H
7.12. COROLLARY.
A(H)-module and let ki
be a subgroup of
be an A-module.
Then
Horn(W,fl)
Hom (kiH,V) A(H)
A
Apply Theorems 7.61ii) and 7.11.
Proof. Let
G of finite index, let I/ be an
S be a ring and let
ideal I of
s,
let
IV
V
as R-modules
’
be a left S-module.
Given a (left or right)
denote the set of all finite sums of the form
Caivi Then
IV
is the additive subgroup of
is in fact a submodule of
7.13. PROPOSITION. (L)
(Y/X)‘
(ii) Let
Let X
V
with
I
and if
ai E I , v i
is a left ideal, then
E V
IV
V.
5Y
be left ideals of
A(H)’
A*Y/A*X I
be a right ideal of
A
such that
I C_ A.X.
If
V
is an A (H)
module then there is a canonical injective map
i
:
-
r(8)
(xv)G
of additive groups.
135
INDUCED MODULES
Proof.
All tensor products below are to be taken over
A(')
(i) Consider the diagram
where all maps are canonical. a flat right A(')-module,
both rows are exact, and
1 denote the natural homomorphism
f,q are isomorphisms (see
Let x E I , y E A ,
and U E V .
v/XV.
V-
1QA
O-A@XV-A@V-A@ is exact.
is
This proves the required assertion.
Proposition 1 . 2 . 8 ) . (ii) Let
Then the diagram is commutative and since A
IV/XV)
Then the sequence
--to
Then 3 ~ l l E I F A.X.
so
XI{
can be
written in the form xy = y a 1
1
+
...
ynan
(yi
f
A,ai E X)
Hence
natural projection. (i) For any left ideal I of
A , I C_ A . ~ r ( l )
(ii) For any right ideal I of Proof. argument.
A,
1 5 n ( 1 )* A
We shall only establish (i) since (ii) is proved by a similar Let
T
be a left transversal for H
in
G.
By Proposition 3.3(i),
CHAPTER 2
136
a E I can b e u n i q u e l y w r i t t e n i n t h e form
each
a
A (H),
t a t , at
=
with
tfT
at.
f i n i t e l y many nonzero
t
s i d e d i d e a l s of
A
A
contained i n
Let
S
a t
a E I and
X
of
contained i n
X.
=
a) E T(lj
71(?
M,
1 . 1 5 . LEMMA. 71 :
A
written
Let
--+
(ii) Id(A-X)
A,
let
Then
d e n o t e t h e sum of a l l two-
Id(X)
i s t h e unique l a r g e s t i d e a l
Id(X)
X.
be a r i n g and l e t
h i l a t e r of
IT(C
-1
-1
For any a d d i t i v e subgroup
let
we have
a E A * n ( I ) and t h e r e s u l t f o l l o w s .
Hence
of
t E T,
For each
H
M
b e a ( l e f t ) S-module.
R e c a l l t h a t t h e anni-
i s d e f i n e d by
ann(M),
G,
be a subgroup of
X be an i d e a l o f
let
A(H)
and
A ( H ) be t h e n a t u r a l p r o j e c t i o n
I
i s t h e unique l a r g e s t two-sided i d e a l
of
A
satisfying
5x
(iii) Id(X'A) = Id(A'X)
H
(iv) I f Proof.
G, t h e n
i s a normal subqroup . . of ( i ) Since
ann(A/A'X)
i s an i d e a l of
i s a consequence of t h e d e f i n i t i o n o f
Id(A.X)
i s an i d e a l of
A,
Id(A'X) = A (
Id(A*X).
A,
Since
fl 'X) SfG
the equality
Id(A'X)
5A'X
we have
Theref o r e
- n AgX Id(A'X) C But
n A aX
SfG
SEG i s o b v i o u s l y an i d e a l of
A
contained i n
A.X.
Hence
and
INDUCED MODULES
137
a s required.
I
(ii) L e t
and s o
AX.
I s Id(A.X).
Then
X
and
A
b e an i d e a l of
W ( I )
n ( I ) 5 X.
such t h a t
5 lT(A.X) 5 X,
i s a l e f t i d e a l of
A
since
(H)
A
contained i n
i s a homomorphism of r i g h t A(H)-modules
T
. I of
i s a l s o t h e unique l a r g e s t two-sided i d e a l
A
satisfying
T ( I )
Id(X.A)
X.
Thus
a s required.
Direct consequence of
(iv)
be any i d e a l of
and t h e argument of ( i i ) ,w e deduce t h a t
(iii) Applying Lemma 7 . 1 4 ( i i )
Id(X'A) = Id(A*X)
I
On t h e o t h e r hand, l e t
Then, by Lemma 7 . 1 4 ( i ) ,
( i )and P r o p o s i t i o n 6 . 5 ( i i ) .
We are now r e a d y t o d e s c r i b e t h e a n n i h i l a t o r s of induced modules. 7.16. PROPOSITION. and l e t (i)
H
Let
arm(#)
= Id(A*X) =
H Q G,
Proof. Let
I : AX
that
G I(v )
n AgX
sEG G
ann(V ) = A (
then
gX). SEG a n n ( f l ) = Id(A'X).
By Lemma 7 . 1 5 ( i ) , ( i v ) , it s u f f i c e s t o prove t h a t be a n i d e a l of
0.
=
Hence
A.
Since
XV = 0, P r o p o s i t i o n 7 . 1 3 ( i i )
t e l l s us
I : a m ( @ ) and t h e r e f o r e
5 ann(fl)
Id(A*X)
T
be a l e f t A(H)-module
X = ann(V).
(ii) I f
Let
G, l e t
be a subgroup of
be a l e f t t r a n s v e r s a l f o r
H
in
G.
Then each element
a
E
A
can be
u n i q u e l y w r i t t e n i n t h e form
a = w i t h f i n i t e l y many
1 tat tfT
at # 0 ( P r o p o s i t i o n 3 . 3 ( i ) ) so t h a t
vG=
otev tET
If
c
a E ann(V ) ,
then f o r a l l
o
=
u
a(i Q v )
€
= (
V,
c tat)O v S T
Hence each thus
a
ann(p)
t
E
X
and t h e r e f o r e
5 Id(A'X).
a E A'X.
=
c
t
8atv
ET It f o l l o w s t h a t
So t h e p r o p o s i t i o n i s t r u e .
ann(8)
A*X
and
138
CHAPTER 2
7.17. COROLLARY. only if J ( A ) Proof.
Let
H be a subgroup of G.
Then
J ( A ) E A ’ J ( A ( H ) ) if and
5J ( A ( ~ ) ) A
Direct consequence of Lemma 7.15(iii).
To dispel any notion that
Id(X)
is an exotic type of ideal, we now offer the
following applications due to Knorr (1981) for group algebras.
artinian.
be a subgroup of
Let ff
7.18. THEOREM.
G of finite index and let A ( H ) be
Then the following conditions are equivalent:
(i) For any irreducible A(H)-module
W,(#),
is completely reducible
5
(ii) ~ ( ~ ( ~ ’ 1 . A4 - J ( A ( ~ ) ) (iii) J ( A ( ~ ) ) =A A - J ( A ( ’ ) ) Moreover, if these conditions are satisfied, then
J(A(H))nA = ( J ( A ( H ) ) A ) n for a l l
n E N.
Proof.
(i) =* (ii):
sition 3 . 3 ) , artinain, W
A
Since
A
is a free A(H)-module of finite rank (Propo-
is also artinian.
Put
w
=
A‘”/J(A‘”).
is completely reducible, so by assumption
reducible; hence J ( A f H ) )C - ann(WG)
.
Since A ( ~ ) is G
( W )H is completely
Now
ann(I$;) = Id(A.J(A(’))) by Proposition 7.16(i), so
J(A(H))A (ii) * (iii):
It follows from (ii) that
whence
by Lemma 7.15(iii) assertion follows.
Consequently, A . J ( A ( H ) )
5 J(A(H))A
and the required
INDUCED MODULES
139
(iii) * (i): It is clear from (iii) that A * J ( A ( ' ) ) be an irreducible A(*)-module and let X
=
is an ideal of Then
ann(V).
J(A(')
)
A.
5X
W
Let and so
5A - X
A
Invoking Proposition 7.16(i), we infer that
J(A(')) so
J(A('))WG
=
0 and thus
5 A - J ( A ( ' ) ) 5 Id(A'X)
artinian.
Let
arm(#),
(fi)' is completely reducible.
The final assertion follows by induction on 7.19. THEOREM.
=
n , using (iii).
.
H be a subgroup of G of finite index and let A ( H ) be
Then the following conditions are equivalent:
(i) For any irreducible A(H)-module
W,
the A-module
8
is completely redu-
cible.
5A * J ( A ( ' ) ) J ( A ) 5 J(A('))A
(ii) J ( A ) (iii)
That (ii) is equivalent to (iii) is a consequence of Corollary 7.17.
Proof.
w
Put
=
A(')/J(A(')).
requirement that
8
since A")
is artinian, (i) is equivalent to the
is completely reducible.
Owing to Proposition 7.13(i), we
have
vG 2 A / A . J ( A ( ~ ' ) Now A
is artinian, so
8
is completely reducible if and only if J ( A ) #
=
0.
Hence (i) is equivalent to (ii), as required. 7.20. LEMMA. artinian.
Let
H be a subgroup of
G
of finite index and let A ( H ) be
Then the following conditions are equivalent:
(i) For any irreducible A-module
V,
the A(H)-rnodule
VH
is completely reduc-
ible. (ii) J ( A ( ' ) ) Proof. artinian.
c J(A) Since
H is of finite index and A ( * )
is artinian, A
Hence any irreducible A-nodule is a direct summand of
(i) is equivalent to the requirement that the A(')-module reducible.
Since A ( H )
A/J(A)
is also
A/J(A).
Thus
is completely
is artinian, the latter is equivalent to the condition
140
CHAPTER 2
that J(A(H)) annihilates A/J(A). only if J ( A
)
Let H
7.21. THEOREM. artinian.
5 J(A),
But
J(A(H'
annihilates A/J(A)
if and
hence the result.
G of finite index and let A(H) be
be a subgroup of
Then the following conditions are equivalent:
M,( M H ) G
(i) For any irreducible A-module
V , VG
(ii) For any irreducible A(H)-module any irreducible A-module
M,
MH
is completely reducible
is completely reducible, and for
is completely reducible
(iii) J ( A ) = A.J(A ( H ) ) (iv) J(A) = J(A(~))-A Owing to Theorem 7.19 and Lemma 7 . 2 0 ,
Proof.
(iv) are equivalent. implies (ii), let M
conditions (ii), (iii), and
It is evident that (ii) implies (i). be an irreducible A-module.
To prove that (i)
Then, by assumption
(MHIG is
MH is completely reducible by Corollary 7.10. be any irreducible A(*)-module and let M be an irreducible submodule
completely reducible and therefore Let
I/
Then, by Corollary 7.12,
of .'/I
and therefore there is an exact sequence
v
MH-
-
0
By Theorem 7.9, the latter gives rise to the exact sequence
(MH)GSince
(MH)'
.
is completely reducible,
theorem is true.
VG
8
-
+
0
is also completely reducible.
So
the
We next provide a characterization of H-projective A-modules in terms of induced modules.
The following two preliminary observations will clear our path.
7.22. LEMMA.
Let H be a subgroup of
versal for H
in G
f
:
CVH)'
-t
I/
by
containing
1
G of finite index, let T be a trans-
and let
V
be an A-module.
Define
INDUCED MODULES
141
G
Then f is a surjective A-homomorphism and It is clear that f
Proof. and so f
is surjective.
Kerf
is a direct summand of
preserves addition.
If u E A ,
For any
U E
“dH .
V,f(i@U)
=U
then
1
-1
-1
= =
TO prove that
that
f(i( 7 8 U,))
=
gt
if(1 t @ U , )
and note that for each
-- - -
c tu ST t
g E G.
for all
t E T, there exists t‘
Since g t = t’ht‘A
t’ht.
=
a
teT
tET
such that
=
UtU,)
@ U,)
f;ET is a surjective A-homomorphism, we are therefore left to verify
f
Fix g E G
c t(Z E T af( z t
for some
c Z’X,.X
=
E
T and h, E H
A E U(A1), we have
8 ut
E T
Therefore
=
--
zgtw,
ST =
Sf( c
t 8 vtl
*T as required. Let
E :
VH+
( V H ) G be the canonical injection.
Obviously
E ( V i n Kerf = 0
H
Since C t @ vt tET
- i8
(
c tv,)
E Kerf
%T
it follows that
So
the lemma is true.
7.23. LEMMA.
Let
H be a subgroup of G of finite index, let T be a trans-
1 42
CHAPTER 2
versa1 for H
G
in
containing
1 and let
V be an A-module.
Define
--1 t
v
Then f is an injective A-homomorphism which is independent of the choice of the transversal Proof.
T
and
Assume that
there exist ht E H some
t
E
( f ( V ) )H
U(Ai).
and
T'
is a direct summand of
The map
t' E T '
such that
t
=
t'ht.
)H. Then, for each t E T ,
t
Hence
=
t'zt.At
for
Accordingly
t'et'
C
t 'ET'
-1
v ,
is independent of the choice of the transversal.
f obviously preserves addition.
T,
we may write
g
t
so
proving that f is an A-homomorphism. Because
f is an injection.
Now
V E
V
uf ( V )
-
--IE
If a E A,, then for all
ET
E T =
Given g E G , t
G
(V,)
is another such transversal.
=
proving that f
(
=
g-lt A
g,t
for some
X
gJ
E lJ(Al) and
INDUCED MODULES
W'
-
6
=
t @
143
VH
LET - 11) is obviously an AfH)-submodule of ((V,) If
f(v)= C 7
@
t
G
CCV,)
G
)H
=
and we show that
jH
W ' 6 (f(V)),
-1
v
E
f(V) n W',
then v = 0.
E T
- z t @ ;
C Z @ v ET
On the other hand
,
-1 VtEW'
ET
'
thus completing the proof.
The next theorem establishes the equivalence of several important characterizations of H-projective and H-injective modules.
It was originally proved by
Higman (195413) for group alqebras. 7.24. THEOREM.
Let
module and let T
H be a subgroup of
be a transversal for
G
of finite index, let
H in G containing
1.
V
be an A-
Then the
following conditions are equivalent. (i) V
is H-projective
(ii) V
is isomorphic to a direct summand of
(iii) V
is isomorphic to a direct summand of
(iv) There exists $ E End A
(V
(H)
H)
(V,)
G
8,where
W is an A(H)-module
such that
tfT (v) 'L
is H-injective
Proof. (i) =. (ii): Owing to Lemma 7.22,
0-u-
there exists an exact sequence
(V)G_f,V'O
H
of A-modules such that the associated sequence of A(H)-modules splits. thesis, V
is H-projective, so Kerf
is a direct summand of
(VHjG = Kerf@ where
V'
(V,) G
By hypo-
.
Thus
8'
V.
(ii) * (iii): Obvious (iii) * (iv):
We first show that any A-module
W is an A(H)-module, satisfies (iv).
V
of the form
Define $ : V
.-+
V
by
V = W
G
,
where
144
CHAPTER 2
q is an A(H)-homomorphism.
By Lemma 7.4,
t'
E
Furthermore, for all w E W
and
T, we have
tfT
It'@#
=
,
ET proving that
UJ
satisfies (iv).
Turning to the general case, we may harmlessly assume that
fi = Let
be the projection of
71
8.
to
Then
To$
4
V ( B V'
v
onto
(direct sum of A-modules)
and let $
satisfy (iv) with respect
V-
induces an A(H)-homomorphism
Y and,
f o r all
U E
v,
we have
tET
E T since
T
is an A-homomorphism.
. Assume that (iv) * (v).
V
The desired implication follows.
is an A-submodule of the A-module
U such that
uH = vH @ w for some A(H)-module
@
:
U-
V
W.
Let
71
be the projection of
UH onto VH
and let
be defined by
-1
1
$(u) =
t($.Ti)t u
(u E
u)
ET $ E End (VH) satisfies (iv). A (H) so Ker6 is an A-submodule of U. If -1 -1 TT(% 0) = U. Thus
where
proving that
V
By Theorem 3.4, 0 E
V,
6
is an A-homomorphism
then @ ( U ) = v
since
is H-injective.
(v) * (i): Assume that
V
is H-injective.
morphic to a direct summand of
(V,)
G.
infer that there exists $ E End (V,) A(H)
Then, by Lemma 7.23,
Applying implication (iii) such that
V
is iso-
=)
(iv), we
145
INDUCED MODULES
z 7IJJ;
(
-1
for all
) U = 2,
~1 6
V
tET Assume that f
U-
:
V
is a surjective A-homomorphism for which
U The restriction f
@
and define
=
U
:
-+
fjW
Kerf @ W
=
(direct sum of A(H)-modules)
is an A(H)-i~omorphismof
IJ onto VH.
Put cp =
f;'
U by -1 @ ( u )=
z t(@$f,t u
E T
Then
8 is an A-homomorphism, by Theorem 3.4(i), and
This proves that
U
=
O(u1 + Kerf. e(u)
Finally, if
8(U) 6
-1
-1
c LpjJf7
=
U
=
8 ( U ) 8 Kerf
Kerf,
then
c Z@$Z fu = a
=
tET
tET Thus
u
@(u)n
and the result follows.
'
We next investigate how the process of restriction and induction influences the projectivity of A-modules.
All subgroups of
G in the theorem below are
assumed to be of finite index. Let S
7.25. THEOREM.
5H
be subgroups of
G
and let
V
ba an A-module
.
n V =
(i) Let
@
Vi, where Vi
is an A-module, 1 4
i
n.
Then
V
is H-
i=1 projective if and only if each V
(ii) If (iii) If
is S-projective, then V
is H-projective and
(iv) For any A(H)-module (v) I f
IJ
summand of Proof.
Vi is H-projective.
W,
V
VH is S-projective, then
the A-module
is an S-projective A(')-module,
bf
is also H-projective
8
V
is S-projective.
is H-projective
then any A-module which is a direct
is S-projective.
(i) Assume that
V
is H-projective.
Then there exists
146
(V,)
End
E
)I
CHAPTER 2
which satisfies Theorem 7.24(iv).
Let
Tri,1
4 i 4 n,
be the
A (HI V
projection of
Vi
onto
defined by the decomposition
the proof of implication (iii) homomorphism of
Vi
9i E
Conversely, if each
(Vi)li
End
Let 0
(ii)
U
-+
IJ
----3
V
---+
+
the associated sequence 0 - - - + UH
0
quence
sequence
US
-+
0
-+
-+
U
IJS -+
--+
W
P
V
and
V"
-+
0 be an exact sequence of A-modules such that
-----+
WH
VH
-~-+
0
--+
splits. V
Since
0 must a l s o split.
( V H ) G = V' 8 X
Then the se-
is S-projective, the
Thus
?'
is H-projective
( V S ) H = V" 8 Y, where
and
Hence (v,)~
v
then
obviously satisfies the same conditions.
S
--+
VH.
satisfies Theorem 7.24(iv),
V --.+ 0 splits.
(iii) Owing to Theorem 7.24,
=
CV,)
G
G
v f 8 x 8 YG ,
8Y =
is isomorphic to a direct summand of
CV,)
G
.
Now apply Theorem 7.24.
This was established in the course of the proof of Theorem 7.24.
Civ)
(W,jH =
Owing to Theorem 7.24,
W' 8 X where W'
(WS,G
By (iv), (IJS)'
A
8
W so that
8 XG
8
is S-projective.
is S-projective, again by appealing to (i).
further link between induction, restriction and projectivity is provided by
7.26. THEOREM. V
(i) I f
Let
H be a subgroup of
is a projective
The converse is true if (ii) If V
= WG
2
is S-projective and therefore, by (i),
Hence any direct summand of
if
Hence
A (2)
$1
(v)
induces an A(H)-
which also satisfies Theorem 7.24tiv).
End (VH) defined by Vi = Q i , A (H) Hence I i is H-projective, a s required. $
so
Then, as in
VZ: is H-projective.
each
Vf
Tie$
(iv) of Theorem 7.24,
=)
Vi,
onto
V = 8 Vi.
V
n
A-module, then
=
G.
VH is a projective A(H)-module.
(G:U) is finite and n
is an A(H)-module, the induced module
is a unit of
fi
A.
is projective if and only
is projective.
Proof.
(i) Since A
module to A ( H )
is a free A(H)-module, the restriction of any free A-
is a free AIH)-module.
modules are projective A(')-modules.
Thus the restrictions of projective A-
MONTGOMERY ' S THEOREM
V
Conversely, suppose that
n
A.
is a unit of
147
is an A-module such that
is projective and
Let
0-u+w-v-0 be an exact sequence of A-modules.
VH is projective.
splits, since
Then the associated sequence
By Theorem 3.4(ii),
v
is H-projective;
hence the sequence
also splits. (ii) Let taining
V
V
Thus
is a projective A-module.
be an A(H)-module and let T
be a transversal for
H in
G
con-
Then
1.
(@)H
=
(i8 V ) o
TBV)
I:
(
(direct sum of A(H)-modules)
tfT-IlI where, of course, projective.
7
@ V
V.
is identifiable with
(fiIH
Then, by (i),
Suppose now that
is also projective.
Hence
V
VG is
8V
is
projective. Conversely, assume that
V
is projective.
If
I, is a free A(H)-module,
then the A-isomorphism
A
@
A(")
P
A
A (HI implies that
p
case we have
W
Then
is a free A-module.
n V @
8 fi 8 UG
Thus
$.
is projective.
U for some free A(H)-module
and hence
VG
In the general
W and some A(H)-module
U.
is projective.
8. MONTGOMERY'S THEOREM
Let G of
G.
be a finite group acting on a ring
and let RG be the fixed subring
R
Our aim is to prove an elegant result due to Montgomery (1976) which
asserts that if
1GI-l E R ,
then
G
J(R
)
=
G
J(R) n R
=
G
J(R)
Tile proof we offer is extracted from a work of Passman (1983a).
It is based on
CHAPTER 2
148
the following special case of Theorem 3.16(ii). 8.1. LEMMA.
If
E
R, then J(R*G) = J(R)*G
Apply Theorem 3.16(ii) for the case N = 1.
Proof.
The following standard fact will be required for subsequent investigations. Let e # 0 be an idempotent of a ring R.
8.2. LEMMA.
Proof.
Let
I be a primitive ideal in R and let V be a faithful irre-
ducible R/I-module.
Then
eV
is an eRe-module.
J(eRe) 5 eRe Assume that V = RW
eV # 0. V
since
Let
w
0c
=
is irreducible.
eV
=
W.
5
If
I
ejJ f er/ where
eV = 0, then
. I%' is an eRe-module.
Then
Hence eV = eReW
and so
Then
It follows that f?V
5W
is an irreducible eRe-module. whence
S ( e R e ) V = J(eRe)eV = 0 Thus J(eRe) 5 I. d(eRe)
5 J(R)
and
Since I was an arbitrary primitive ideal in R, so
Assume that a E J ( H ) .
Then
yields eae + ebe = ebeae. unit.
eae
E
K, in which case beae
for some b
Since
this yields
Hence
J(R) and
=
s o , by Proposition 1.7.18,
eae+b.
Multiplying by
(e-ebe) ( e - eael
=
e
so that
e
on both sides
e - ecle
is a left
eJ(R)e is an ideal of eRe, Proposition 1.7.18 implies that
eJ(R)e 5 J ( e R e ) . Let G
be a finite group acting on a ring R.
Define the trace map on R
by
MONTGOMERY'S THEOREM
Given t h e a c t i o n o f
G
R.
on
149
R*G.
w e form t h e skew group r i n g
Since
G
is
c o n t a i n e d i s o m o r p h i c a l l y i n t h i s r i n g , we do n o t u s e t h e o v e r b a r n o t a t i o n . Define
g E G
and o b s e r v e t h a t f o r a l l
we have
G
G+g =
8.3. LEMMA. (i) I f
I
(ii) I f
G
L e t a f i n i t e group
G+rG+
then
=
=
gG
R
tr(r)G+
+
R.
a c t on
i s a G - i n v a r i a n t i d e a l of
r E R,
+
.
hand, i f
(i)
r
E
( i i ) Given
I
Since
G
I ,
r
then
r E R,
=
tr(lG1- P )
E
=
tr(1).
5 RG*Gf
G t r ( 1 )5 I n R
i s G-invariant, 1
then
Hence
G+(R*G)G+ = t r ( R ) . G + Proof.
E R,
lG1-l
and i f
tr(1).
=
IG .
On t h e o t h e r
IG C - t r ( I ) , as asserted.
Hence
we have
C grG+
G+PG+ =
sfc
=
grg-lG+
C
=
tr (r)G+
SEG
a s required. The f o l l o w i n g o b s e r v a t i o n w i l l e n a b l e us t o t a k e f u l l advantage of Lemma 8.1. 8.4. LEMMA.
L e t a f i n i t e group
ICl-lG+
i s an idempotent of
e
=
(i) e(I*G)e =
G I e
Proof.
I
=
tr(1).
R
/GI-'
and l e t
E
R.
Then
with
I of
R
RG where t h e l a t t e r i s a r i n g isomorphism.
In
G G J ( R el = J(R ) e . Since
is an idempotent.
G
R*G
a c t on
f o r any G - i n v a r i a n t i d e a l
(ii) e ( R * G ) e = eRG = RGe
particular,
G
gG+ Let
= G+
(G+)
it f o l l o w s t h a t
I be a G - i n v a r i a n t i d e a l of
=
I GI .G+ R.
and hence t h a t
e
Then, by Lemma 8 . 3 ( i ) ,
Invoking Lemma 8 . 3 ( i i ) , we d e r i v e
G
I C I e U * G ) lGle = tr(I)./Gle = I l G l e which i m p l i e s ( i f . By ( i ),
G
e ( R * G ) e = R e.
o u s l y commute wPth
e,
-
On t h e o t h e r hand, s i n c e t h e e l e m e n t s of
we have
R Ge = eRG
.
The map
RG-
R Ge,x
RG
obvi-
re, i s
CHAPTER 2
150
Furthermore since each element of
obviously an isomorphism of additive groups.
RG clearly commutes with e ,
R
the isomorphism
G
2
G
R e
does indeed preserve the
ring structure. We have now come to the demonstration for which this section has been developed. 8.5.
THEORXivl. !Montgomery (1976)). Let a finite group
let
1GI-l
E
R.
Then
JW G
=
C
Proof. The equality d(R) 0 R
G
J(R
)
=
G
J(R) ,
G act on a ring R and
form R*G
J(R) n R
= J(R)
G
=
G
J(R)
is obvious.
and let e = /GI-lG+.
G
Then we have
J(R) e = e(J(R)*G)e
(by Lemma 8.4(i))
=
eJ(R*G)e
(by Lemma 8.1)
=
J ( e (R*G)e )
(by Lemma 8.2)
=
G J(R e )
(by Lemma 8.4(ii))
G = J(R )e G
To prove that
Thus we see that J(R)G = J(R I , as required.
(by Lemma 8.4(ii))
151
3 The classical theory of crossed products
Let
E/F
Z2(G,E*)
G = Gal(E/F).
be a f i n i t e Galois extension and l e t
G on E*.
with r e s p e c t t o the n a t u r a l action of
we denote by
01
E G
crossed product
t h e corresponding crossed product of
E'G
Consider t h e group Given
G
over
U E
E.
Z2(G,E*),
This
was introduced by Noether and played a s i g n i f i c a n t r o l e i n Our aim i n t h i s chapter i s t o
the c l a s s i c a l theory of c e n t r a l simple algebras.
provide t h e foundattons of t h e c l a s s i c a l theory of crossed products and c e n t r a l simple algebras. the
U
E G
That the crossed products
occur i n an unavoidable way i n
study of c e n t r a l simple algebras i s i l l u s t r a t e d by Lemma 3 . 4 which a s s e r t s
A
t h a t any finite-dimensional c e n t r a l simple F-algebra
A
c e n t r a l i z i n g s u b f i e l d of
i s isomorphic t o
U
E G
main r e s u l t of the chapter a s s e r t s t h a t t h e map i s an isomorphism.
such t h a t
-
f o r some
H 2(G,E*)
U
E
is a self-
E Z 2 (G,E*).
Br(E/F),
The [E'G]
In p a r t i c u l a r , every Brauer c l a s s contains a crossed product.
1. CENTRAL SIMPLE ALGEBRAS
A
Throughout t h i s s e c t i o n ,
contains an isomorphic copy identify
F
and
F-1.
central F-algebra. A
F.1 = { h * l l h E
I f t h e c e n t r e of
If
the subalgebra of
denotes an algebra over a f i e l d
A
and
A2
generated by
Al
A
F} of
F.
and
A2.
Thus
Z(A)
For convenience, we s h a l l
F,A i s s a i d t o be a
is p r e c i s e l y
a r e subalgebras of
F.
A,
we w r i t e
AIAz
for
Our aim i n t h i s s e c t i o n i s t o
examine c e r t a i n p r o p e r t i e s of c e n t r a l simple F-algebras.
We s t a r t by providing
some background information, Suppose t h a t s t r u c t u r e on
the
A
and F-space
B
a r e F-algebras.
A OB F
Then one can define t h e F-algebra
by t h e formula
152
As
CHAPTER 3
is customary, we shall often identify the algebras A
images A 8 1 and
1 8 B
A 8B. F
in
and
B
with their
With these identifications,
become elementwise commuting subalgebras which generate A @ B F
A
and
B
and satisfy
dim(A 8 B ) = ( d i d )(dimB) F F F F The converse is also true, by virtue of the following observation.
1.1. LEMMA. of
C be a finite-dimensional F-algebra.
Let
Given subalgebras A , B
C, if (il dime
(iii) A
and B
Proof.
= (did](did?], (iif C = AB and F F F commute elementwise, then C A 8 B .
F
The mapping
(s,y)
from A x F
P?
A 8B
induces a linear mapping from
to
C.
to
C
js
bilinear and so
morphism by (iii), and both algebras have the same F-dimension by (i). Let V,
U,V
be vector spaces over F .
.
This is surjective by (ii), a homo-
F Given subspaces U'
of
U and V'
of
we have an injective homomorphism
U ' 8 V'-
F induced by the inclusions image in space of
U 8 V. F U '8 V . F
1.2. LEMMA.
Let
5U,V' 5 V.
v
We shall identify
With this identification, U' @ V'
U' 8 V' F
will be regarded as a sub-
The following elementary result is often useful. U,V
be vector spaces over F
and let
U',V'
Then
We can write
U for some subspaces
with its
F
U and V, respectively.
Proof.
U'
U 8 F
U",V"
=
of
U' @ U" and
V = V' @ V"
U and V,
respectively.
Now the left-hand side of the required equality is
Hence
be subspaces of
CENTRAL SIMPLE ALGEBRAS
153
U' '8 V'.
and i s c l e a r l y
F Let
V
be an R-module,
R
where
i s an a r b i t r a r y r i n g , and l e t
S = End(V).
R V
Then
i s a l s o an S-module,
@v
r
Each
E R
S
t h e o p e r a t i o n of
b e i n g g i v e n by
fr(v)
:
=
V
@ E S,v E V
for a l l
= @(V)
f
i n d u c e s an S-homomorphism
V
on
+
V
g i v e n by
ru
v
for a l l
V
Thus we g e t a r i n g homomorphism
family
acts d e n s e l y on
R
We s a y t h a t
V1,V2,
...,V
V
in
V
6 E End(V)
i f f o r each
r E R
there exists
and each f i n i t e
S such t h a t
e w .t1. = fr ( vz. ) 1 . 3 . LEMMA.
Let
T I be an R-module,
s
let
n)
(1 G i
= End(V)
and l e t
v
be f i n i t e l y
R g e n e r a t e d a s S-module.
Then
R
acts d e n s e l y on
V
i f and o n l y i f t h e homo-
morphism
is surjective.
R
P r o o f . I f t h e g i v e n homomorphism i s s u r j e c t i v e , t h e n o b v i o u s l y on
V.
Conversely, assume t h a t
g e n e r a t i n g s e t f o r t h e S-module exists
r E R
8 (U .1
with
R
a c t s d e n s e l y on
V.
1.4.
THEOREM.
submodules. Proof.
f
and l e t
V
be a
,...,Z,
1
8 E EndCV),
If
= fr(Vi).
S Hence f o r any
n
proving t h a t
V
a c t s densely
t h e n by h y p o t h e s i s t h e r e
X1,X2 ,...,An
in
s,
n
= 6.
Assume t h a t an R-module Then
R
a c t s d e n s e l y on
v
i s a f i n i t e d i r e c t sum of i r r e d u c i b l e
V.
For t h e sake of c l a r i t y , we d i v i d e t h e proof i n t o two s t e p s .
154
CHAPTER 3
S t e p 1.
S = End(V)
Put
and f i x
v E V
f E End(V).
and
R
rv f o r some
r E R. W.
submodule
T h i s shows t h a t
S t e p 2. Let
Because
V
: V
-+
Let
ii
f(V)
We c l a i m t h a t
S =
f(v) V = Ru @ W
i s completely r e d u c i b l e ,
Ru
be t h e p r o j e c t i o n .
f o r some R-
TT E S
Then
and hence
a s claimed,
E RV,
Completion of t h e p r o o f .
0 E End(v) S
and l e t
,...,V n
2,
e,(s
s'
and p u t
=
E
v.
0,
Define
,...J,) =
(o(X )
f+ J"
by
,...,o(z,)) S'
Then, by P r o p o s i t i o n 1 . 7 . 2 ,
End(P).
:
i s a f u l l matrix r i n g
R over
S.
lies i n
End(ljz). S
which i s what we wanted t o prove.
A
For any F-algebra
A"
one immediately v e r i f i e s t h a t
Hence, by S t e p 1, t h e r e e x i s t s an element
( r v l ,...,rv 1
where
V,
Using t h i s and t h e assumption on
=
[O(v 1
A
E
R such t h a t
,...,O ( v , ) )
' A @A",
w e c a n form t h e s o - c a l l e d enveloping a l g e b r a
A.
denotes t h e F-algebra opposite t o
l e f t m u l t i p l i c a t i o n by
r
A,
on
U
3
A
c
R -
On
and
End(A)
a
Given
a
E A,
let
right multiplication.
F aK. d e n o t e
Then t h e maps
,
F
F
where
A,
=
iaRla E A }
and
A p = {aria E A )
a r e i n j e c t i v e homomorphisms of F - a l g e b r a s .
aebr
=
braa. f o r a l l
a,b E A.
Therefore
By t h e a s s o c i a t i v i t y of
A
becomes a l e f t
( a 8 b ) r = (aRbr)(x) = uzb 1.5. LEMMA.
Let
A
( i ) End(A) = Z ( A ) R
R
be an F-algebra and l e t
Z(A)
R
=
A @Ao F
w e have
(A @ Ao)-module
F upon d e f i n i n g
A,
CENTRAL SIMPLE ALGEBRAS
(ii) A
A
(iii) I f
i s an i r r e d u c i b l e R-module
Z(A)
i s simple, then
is a field.
( i ) The i n c l u s i o n one way i s c l e a r .
Proof.
f
A
i s simple i f and o n l y i f
Then, by an e a s y computation,
E EndiA).
155
f
Conversely, assume t h a t = fcl),,
and one t h e n checks t h a t
R f(1)E Z ( A )
since
f E End(A). R
A
( i i ) The two-sided i d e a l s i n
A
(iii) I f
i s s i m p l e , t h e n f o r any
a
.
a s required. 1.6. THEOREM. (i) A
a r e p r e c i s e l y t h e R-submodules of
@A" F
Let
A
t
0
P a
U(A) n Z ( A )
E
Z(A)
Aa
=
Aa
=
A.
A. Hence
= U(Z(A
b e a c e n t r a l simple F-algebra.
A
a c t s d e n s e l y on
d i d = n C m, t h e n A 8 A " f M,(F) F F Proof. ( i ) P u t R = A 8 A' and observe t h a t A i s a n i r r e d u c i b l e R-module F Hence, by Theorem 1.4, R a c t s d e n s e l y on A . by L e m m a 1 . 5 ( i i ) . (ii) I f
( i i ) By Lemma 1 . 5 ( i ) ,
End(A)
2
F
R
and so by Lemma 1 . 3 , t h e map
----+
EndA,
F
R
r C-L f,
i s a s u r j e c t i v e homomorphism of F - a l g e b r a s .
Since
EndA P M n ( F )
we
F have d i d = n 2 = dim EndA F F F and t h e r e s u l t f o l l o w s . 1.7.
Let
THEOREM.
Then t h e map
A
9
be a c e n t r a l simple F-algebra
I uA @I
and l e t
B
b e any F-algebra
i s a l a t t i c e isomorphism between t h e i d e a l s of
B
and
F A @ B.
t h o s e of
The i n v e r s e of t h i s isomorphism i s g i v e n by
F J - J n B where
B
1 - 8B.
i s i d e n t i f i e d with
Proof.
We c l a i m t h a t t h e g i v e n mappings a r e i n v e r s e t o e a c h o t h e r .
This
w i l l prove t h e r e s u l t , f o r t h e y e s t a b l i s h a b i j e c t i o n which i s c l e a r l y o r d e r -
p r e s e r v i n g and hence a l a t t i c e isomorphism. Let
I be an i d e a l i n
B.
Then
A 8I F
i s o b v i o u s l y an i d e a l i n
A 8B F
and
156
CHAPTER 3
(~€9') n B = i F by v i r t u e of Lemma 1 . 2 .
A €9 B and p u t F A €9 I. J , so w e need o n l y v e r i f y t h a t J ly A @ I c F F each nonzero c E J can be w r i t t e n a s c = a €9 b +
J
Conversely, l e t
be a n i d e a l i n
I
5
1
0 # bi
B
E
and w i t h
{u
,...,a
1
Hence
x3: j j
.
J C- A B I . F
Hence t h e r e e x i s t
id r.
i j
bi
b E 1 and s i m i l a r l y
€9 by
U
I
E
for
with
A.
b e i n g a s u b s e t of a n F - b a s i s of
Theorem 1.6 ( i ), A €9 A " a c t s d e n s e l y on A . F j E {l,Z s} such t h a t u y = Ail, 1
,...,
Then obvious-
To t h i s e n d , n o t e t h a t
... +
1
B.
= J
By
Zj,yjE
A
for
But t h e n
i E {2,...,P I ,
proving t h a t
The n e x t r e s u l t d e s c r i b e s c e n t r a l i z e r s i n a t e n s o r p r o d u c t .
1 . 8 . PROPOSITION.
Let
A
and
A
1
of
Ai, i
= 1,2.
be F - a l g e b r a s and let
A2
Al
element of
b E B 1
(ZU
,
2
%
E CABA
u E CA ( B 1 ) and so
2
Invoking Lemma 1 . 2 , we deduce t h a t
.
( b €9 1) ( X U i €9 Lli)
( B 1 €9 B z ) , F
1
as required.
A
Then e v e r y
Xui €9 vi w i t h a
Al.
w e have
. €9 Vi) ( b €9 1) -
La. €9 b .
{oil of
can be u n i q u e l y w r i t t e n i n t h e form
iF
fore
.
Then
To prove t h e o p p o s i t e i n c l u s i o n , f i x an F - b a s i s
Hence, i f
be a s u b a l g e b r a
It i s clear that
Proof.
Given any
Bi
2
then
u.b
= =
z (Uib - bail €9 V i
bai for a l l b
E
B
.
There-
157
CENTRAL SIMPLE ALGEBRAS
1.9.
Proof.
A
Let
COROLLARY.
A
and
be F - a l g e b r a s .
2
B
Apply P r o p o s i t i o n 1.8 f o r t h e c a s e
Then
1
=
A
1
B
and
2
=
A
2
.
.
We n e x t p r o v i d e a number of i m p o r t a n t consequences of Theorem 1.7.
1.10. PROPOSITION.
Let
A
be a c e n t r a l simple F-algebra and l e t
B
b e any F-
algebra. (i) A @ F
B is
B is
simple i f and o n l y i f
( i i ) Z ( A @ Bl
simple
Z(B)
F I n p a r t i c u l a r , t h e t e n s o r p r o d u c t of c e n t r a l s i m p l e F - a l g e b r a s i s a g a i n c e n t r a l simple. Apply Theorem 1 . 7 and C o r o l l a r y 1.9.
Proof.
1.11. PROPOSITION.
B
algebra
A
A,
of
H zy
cA (Bl
and
F BCA(B).
whose image i s
tive. 1.12.
Thus
=
PROPOSITION.
A
B 8 c,(B),
which by Theorem F i s t h e k e r n e l of t h e r e s t r i c t i o n of
0 and t h e r e s u l t f o l l o w s . Let
c e n t r a l simple s u b a l g e b r a
Proof.
-
B 63 I, where I F But t h i s i s t h e i n c l u s i o n mapping, which i s i n j e c t o CA(B).
A
I
CA(B)
I t s k e r n e l i s an i d e a l i n
1.7 i s of t h e form --+
commute elementwise, s o t h e mapping
g i v e s r i s e t o a homomorphism
B8
B 8 CA(B) F
Then, f o r any c e n t r a l s i m p l e sub-
be an F-algebra.
By d e f i n i t i o n , B
Proof.
(z,y)
Let
9
A
be an F-algebra.
B
of
Then, f o r any f i n i t e - d i m e n s i o n a l
A,
By P r o p o s i t i o n 1.11, we need o n l y v e r i f y t h a t
A
=
BCA(B).
W e
a s l e f t (B@B")-rnodule, i . e . by Theorem 1 . 6 ( i i ) , a s l e f t M (F)-module, F where n = dimB. The module i s completely r e d u c i b l e , s o A = @ Ai, where Ai F i s i r r e d u c i b l e , isomorphic t o A . L e t u E Ai correspond t o 1 i n t h i s i s o regard
A
i
CHAPTER 3
158
morphism,
A
=
Then
u.a = au
i
for all
a E
A, so ui E CA ( B ) and thus
BCA(B). We next prove the following basic result known as the Skolem-Noether theorem. Let A
1.13. THEOREM. B
be a finite-dimensional central simple F-algebra and let
be a simple F-subalgebra of
A.
Then every isomorphism of F-algebras
@ : B-B'CA can be extended to an inner automorphism of in A
A, that is, there exists a unit
a
such that
@ ( b )= aba-l
€or all b E B
A
In particular, isomorphic simple subalgebras of
are conjugate and hence have
conjugate centralizers. Proof.
Let
V
be an irreducible (left) A-module, and let D = End(V). A
Then
D is a division algebra with centre F'lV
F-algebra by Proposition 1.10.
We can regard
and V
B 8 D is a simple F
as an B 8 D-module according to
F the rule
we can define a second left (B 8 D)-module V ' , whose F underlying vector space over F is V , and with the action of B @ D given by $,
Using the automorphism
(b 8 dlV' Then
V
and
V'
=
@(bltdv'1
are of the same F-dimension.
(b
Since B 8 D
F B,d
E
E
D,vr E V r )
is simple artinian,
F we deduce that there exists an isomorphism of ( B 8 D)-modules 0 : V * V ' . Since F 8 E End(V), it follows from Lemma 1.3 and Theorem 1.4 that 0 1 s a left multi-
D plication at
for some
Finally, because
8(=a )
L
a E A.
Moreover, a E U(A)
since
0
is a ( B 8 D)-isomorphism, we have F
a ; b ( d v ) ) = @ ( b ) d ( a v ) for all b Taking that
d = 1, and applying the fact that A
ab
=
@(bfa,
so
is an isomorphism
$ ( b l = aba-l
E
acts faithfully on
for all b E B.
~ , 6d D,V V,
E I/
it follows
CENTRAL SIMPLE ALGEBRAS
1.14. COROLLARY.
159
Every automorphism of a finite-dimensional semisimple F-algebra
which leaves the centre elementwise fixed, is inner. Let A
Proof.
=
A @
identity element of Ai.
E
A.
zt
@
Let
Since e
elementwise fixed.
a
... 8 Ar i
E
where each Ai
is simple and let ei
be an automorphism of Z(A), we have
@ ( e.)
=
A e
be the
which leaves Z(A) 1G
i'
i
G n.
If
then
@ ( a )= @ ( a e i ) = @ ( a lie and thus
induces an endomorphism of Ai.
@
induces an automorphism of Ai,
@
and hence
bi,
fixed by
we may regard
algebra Ai.
Qi
E
Ai
By a similar argument, so does
bi
say
.
Since
is left
as an automorphism of the central simple Z(A.1-
By Theorem 1.13, there exists a unit ui in Ai
@i(") for all z 6 A . .
Z(Ai)
@-'
such that
-1
= u .xu
z i
Setting u = Cui, it follows that
@ ( z )= uxu
-1
for all
TEA. O u r next result is due to R. Brauer.
1.15. THEOREM.
B
Let A
be a finite-dimensional central simple F-algebra and let
be a simple subalgebra with centre E.
(i) CA(B) is simple with centre E (ii) CA(CACB)) = B
(iv) dimA = (dimB)(dimC (B11 F F F A Proof. Put n = dimA, k = dimB and s = dimC ( B ) . P F F A with the algebra of all F-linear transformations of B . subalgebras BP. of left multiplications and Clearly B
P.
Y
B and A
B"
We may identify Mr(F) such it contains the
As
Bp of right multiplications.
and, by Proposition 1.10, A @Mk(F)
is central
F
B 8 F and F 8 EL are isomorphic simple subalgebras of F F Hence, by Theorem 1.13 and Proposition 1.8,
simple.
Now
C ( B ) 0 MkV) A
P
2
A 8 B,
F
A
@ B"
F
A @ Mk(F). F
CHAPTER 3
160
Since C A ( B ) €4 Mk(F) obtain nk =
A '8 B o
Now
Mk(CA(B)),
F k2s or n
=
ks,
this proves (v), and comparing dimensions, we
i.e. (iv).
is simple, hence by (v), so is
cA(B) (using Proposition 1.7.4
F (ill.
B by C A ( B l
Replacing
in (iv) we find
whence
(c
dimC F A Since we clearly have
CA(CA(B)l
( B ) ) = dimB
A
2 B,
F
it follows that
C A ( C A ( B ) ) = B , proving
.
(ii)
K
Finally, if
E
we also have
2 K,
E
i.e.
.
1.16. COROLLARY. let E
Let A
=
K.
CA(L'.l.
bras, both subalgebras of 1.12.
CA(B),
is the centre of
then K
2 E,
and since
Hence, B,C (B) are central simple E-algeA This proves (iii) by appealing to Proposition
be a finite-dimensional central simple F-algebra and
be a subfield of A
containing
F.
Then
C,(El €4 Mk(F) , where k = dimE F F F (ii) dimA = (die) (dim C,(E) ) F F E In particular, (dimE) divides did.
A €4 E
(i)
*
F
F Proof.
(i) Put B = E
B
(ii) Setting
as required,
.
1.17. COROLLARY. let E
=
and apply Theorem 1.15(v).
E, it follows from Theorem 1.15(ivl that d i d = (dimE)(dimCA(E)) = (dimE)'dimCA (E) F F F F E
Let A
be a subfield of A
be a finite-dimensional central simple F-algebra and containing
F.
Then the following conditions are
equivalent: (il
CA(E) = E
(ii) E
is a maximal commutative subring of
A
THE BRAUER GROUP
161
(iii) d i d = (dimE) F F (iv) d i d = (did)2
E
F Proof.
It is clear that (I) and (ii) are equivalent.
(i) and (iii) follows from Corollary 1.16(ii).
The equivalence of
Finally, the equivalence of (iii)
and (iv) is a consequence of the equality
So the corollary is true,
Let D
1.18. COROLLARY.
.
d i d = (did)(dimE)
E
F
F
be a division ring with centre F, let
did <
and
F let E be a maximal subfield of (il
d i d = (didl2 = (dimD1'
(ii)
D 8E F
F Proof.
x
E
M (E) where
k
k = did
F
(i) Obviously C D ( E )
CD(EI,B(x)
Then
E
F 4
D.
is a subfield of
2E
and equality must hold, since for all
D containing E.
Now apply Corollary 1.17 for
A = D . (ii) Apply Corollary 1.16(i) for the case Let A
1.19. COROLLARY. A
=
A
=
D, CA (E) = E.
.
be a finite-dimensional central simple F-algebra, say
Mn(D) where D is a division ring with centre F, and let E be a maximal
subfield of
D.
Then A '8 E
Mnk(El where k
=
dimE
.
In particular,
F
F d i d = n 2 k 2 is a perfect square.
F Proof.
Applying Corollary l.l8(iil, we have
as required.
2. THE BRAUER GROUP
Throughout this section, F denotes a field and all F-algebras are assumed to be
finite-dimensional. Assume that division algebra
A
is a central simple F-algebra.
Then there is a central
D and a natural number n such that A
M
(D)
D 8 Mn(F)
F
162
CHAPTER 3
The algebra refer to
D is unique up to F-isomorphism, while n is unique.
D as the s k e w f i e l d c o m p o n e n t of A .
are called s i m i l a r , written A
-
B,
TWO central simple F-algebras A , B
if their skewfield comp0nent.s are isomorphic.
[A]
This is clearly an equivalence relation and we denote by class of
We shall
the equivalence
A.
2.1. LEMMA. The similarity classes of (finite-dimensional) central simple F-algebras form an abelian group with respect to the multiplication induced by the tensor product. and B
Then A €3 B B 8A F F is again a central simple F-algebra by virtue of Proposition 1.10. Assume that Let A
Proof.
be central simple F-algebras.
Then
D2.
and so A 8 B -. D1
Hence, if
- A',B - B',
A
F
then
A ' B B I -
F proving that the multiplication
D l ; D 2 - A 8 B
[ A ] [BI
=
IA 8 Bl
F is well defined.
F The multiplication of similarity classes is associative by the corresponding law for tensor products.
Bearing in mind that A 8 F
"=
A
and that by Theorem
F 1.6(ii), A €3 A"
Mn(E') I
we also have
F [A1 IF1 = IAI This shows that
[Fl
,
[A1 LA"] = [ F l
is the identity element and that
[A"]
is the inverse of
A. The group of similarity classes of central simple F-algebras is called the
Brauer group of P
and is denoted by
Br(F).
The elements of
Br(F)
are
called the Brauer cl asse s of central simple F-algebras.
2.2. LEMMA. Proof.
The Brauer group of an algebraically closed field is trivial. Let
Given d E D , F ( d )
D
be a division algebra over an algebraically closed field F .
is a finite extension of F.
Since F
is algebraically
163
THE BRAUER GROUP
F(dl = F
c l o s e d , w e have
and s o
d E F.
T h i s proves t h a t
D
= F
and hence t h e
result. Let
A
be a c e n t r a l simple F - a l g e b r a . d i d
We know, from C o r o l l a r y 1 . 1 9 , t h a t =
r2
F
r
f o r some i n t e g e r
and t h a t t h e r e i s a f i n i t e e x t e n s i o n
A 8E F
r
The i n t e g e r
&
=
2
E
F
of
such t h a t
Mr(E)
i s c a l l e d t h e degree o f
A
and i s denoted by degA
F Note t h a t i f
then degA = (degD)n By t h e index of
A.
component of
w r i t t e n indA, w e u n d e r s t a n d t h e d e g r e e of t h e s k e w f i e l d
A,
Thus indA i s a measure of how f a r
Br(F).
i d e n t i t y element o f determines
A
E/F
f o r some p o s i t i v e i n t e g e r
i s s a i d t o be a s p l i t t i n g f i e l d f o r
r.
4.
w h i l e degA
F.
A
if
Mr(E) h a s always a s p l i t t i n g f i e l d
A
By t h e f o r e g o i n g ,
which i s a f i n i t e e x t e n s i o n of
E
[AI,
Note t h a t indA i s an i n v a r i a n t of
A @ E F
that
d e v i a t e s from b e i n g t h e
up t o isomorphism w i t h i n i t s Brauer c l a s s .
A f i e l d extension
E
[A1
[A1
W e say t h a t
is a splitting f i e l d f o r the class
T h a t t h e n o t i o n of a s p l i t t i n g f i e l d f o r
[A],
if
E
is s p l i t by
E, o r
is a splitting field for
[A]
i s w e l l d e f i n e d i s a conse-
F.
Then t h e map
quence o f t h e f o l l o w i n g simple o b s e r v a t i o n . 2.3.
LEMMA.
E
Let
be a f i e l d e x t e n s i o n of
i s a homomorphism whose k e r n e l c o n s i s t s of t h o s e proof.
Let
A
[A]
which a r e s p l i t by
be a c e n t r a l simple F-algebra and w r i t e
A
2
E.
D 8 Mn(F). F
Then
CHAPTER 3
164
A and hence
D 8 Mn(Fl 8 F F
8E F
[A 8 El = ID 8 E l .
F
D 8 Mn(El
E
F
Thus t h e g i v e n map i s w e l l d e f i n e d .
I t is a l s o a
F
homomorphism, s i n c e
(A @ B ) 8 E F F
(A 8 E) 8 ( B 8 E) F F
2
The f i n a l a s s e r t i o n i s a consequence of t h e d e f i n i t i o n of t h e s p l i t t i n g f i e l d f o r
[A].
8 E) where
W e n e x t p r o v i d e a r e l a t i o n between i n & and ind(A
E/F
is a f i n i t e
A
be a c e n t r a l
F f i e l d extension
2 . 4 . PROPOSITION.
Let
simple F-algebra.
Then
E/F
be a f i n i t e f i e l d e x t e n s i o n and l e t
I
1 ( d i d ) i n d (A 8 El
i n d ( A 8 El i n &
F
by
D'
F
t h e s k e w f i e l d component of
F
D is
A =
W e may h a r m l e s s l y assume t h a t
Proof.
D 8 E.
a d i v i s i o n algebra.
Denote
Then we have
F D8E F f o r some
k 2 1.
s
D' 8 Mk(E) E
Comparing F-dimensions, (in&)
we deduce t h a t
(dimEl = Iind(A 8 F F
E)I 'k'
(dimE)
F
and so ind(A
8 E) I indA F
Put
r
= dimE
F homomorphism of
D 8E F
E
and observe t h a t t h e r e g u l a r r e p r e s e n t a t i o n of
E
into
i s embedded i n
D 8 M,(F). F Because
E
Hence
i s embedded i n
M (F) r
and t h e r e f o r e
Invoking (l), we a l s o have an embedding of
2 F,
c o n t a i n s M (F) a s c e n t r a l k F r simple subalgebra. L e t B d e n o t e t h e c e n t r a l i z e r of Mk(F) i n D 8 M r ( F ) . F Denoting by D' t h e Then B i s a c e n t r a l simple F - a l g e b r a ( P r o p o s i t i o n 1 . 1 2 ) .
Mk(E) i n
D 8 Mr(F).
Mr(F).
i s an i n j e c t i v e
E
D 8 M ($7
F
s k e w f i e l d component of
f o r some
s
>
1.
B,
it f o l l o w s from P r o p o s i t i o n 1 . 1 2 t h a t
By uniqueness, we have
D
s
D'
and comparing F-dimensions,
we
THE BRAUER GROUP
r = ks.
find 2.5.
165
Thus
klr
and so by ( 2 ) i n d A / i n d ( A 8 E ) r , F
Let
E/F
be a f i n i t e f i e l d e x t e n s i o n and l e t
COROLLARY.
.
a s required.
A
be a c e n t r a l
In dimE i s prime t o indA, t h e n indA = i n d ( A '8 E). F F p a r t i c u l a r , i f A i s a d i v i s i o n a l g e b r a such t h a t d i d i s prime t o degA, t h e n F A 8 E i s a l s o a divis i o n algebra. F Proof. The f i r s t s t a t e m e n t i s a d i r e c t consequence of P r o p o s i t i o n 2 . 4 . simple F-algebra.
Assume that
A
If
=
D
is a d i v i s i o n algebra such t h a t
dimE F
i s prime t o
indD = i n d ( D 8 E l .
Then, by t h e f i r s t s t a t e m e n t , we have
.
degA
Write
F
D8E
P
F
D'.
f o r a d i v i s i o n E-algebra
D ' 8 Mn(E) E
indD = i n d ( D '8 E ) , F
Since
w e have d i d F
=
did'.
E
Thus
.
dimD = dimD F E
n
and s o 2.6.
=
1,
a s required.
COROLLARY.
Let
A
= dimLh
F
be a c e n t r a l simple F-algebra
f i e l d e x t e n s i o n which i s a s p l i t t i n g f i e l d f o r
A.
and l e t
be a f i n i t e
E/F
Then
I
indA d i d F Proof.
A 8E
By h y p o t h e s i s ,
Mn(E1
f o r some
n
1 and so
F
ind(A 0 F
E)
=
1.
Now a p p l y P r o p o s i t i o n 2.4. W e next p r e s e n t a c r i t e r i o n f o r a given f i n i t e f i e l d extension
PROPOSITION.
Let
A
f i n i t e f i e l d extension. exists
B E
This algebra Proof.
[A]
B Let
be a c e n t r a l s i m p l e F-algebra Then
such t h a t
V
Now any
B
E
[ A ] E Br(F)
splits
contains
E
and l e t
E/F
be a
i f and o n l y i f t h e r e
as a s e l f - c e n t r a l i z i n g s u b f i e l d .
n e c e s s a r i l y ha5 degree
D
dimE. F be t h e d i v i s i o n a l g e b r a i n t h e c l a s s
t h e l e a s t i n t e g e r such t h a t that
to split
P.
a g i v e n Brauer class o v e r 2.7.
E/F
i s an E-module.
E
5Mn(D).
Put
By t h e m i n i m a l i t y of
[ A ] , and l e t
B = Mn(D) , V = On
n, V
n
be
and o b s e r v e
i s an i r r e d u c i b l e E-module
0 # C E C (E) d e t e r m i n e s a nonzero endomorphism of B
V,
hence
c
is
166
CHAPTER 3
invertible. ( i ),
E
C (E)
Thus
B
is a d i v i s i o n a l g e b r a . CB(F).
i s t h e c e n t r e of
Invoking C o r o l l a r y 1 . 1 6 ( i ) , we a l s o have
CB(El
-B @E -D0E F
E
T h i s shows t h a t
splits
D
Furthermore, by Theorem 1.15
F
CB (E) = E,
i f and o n l y i f
i s a s e l f - c e n t r a l i s i n g s u b f i e l d of
B.
E
i f and o n l y if
i.e.
The l a s t a s s e r t i o n i s a consequence of
Corollary 1.17. I t w i l l n e x t b e demonstrated t h a t t h e s p l i t t i n g f i e l d o f a c e n t r a l s i m p l e
a l g e b r a can always be t a k e n t o b e a s e p a r a b l e e x t e n s i o n of t h e ground f i e l d . T h i s w i l l be achieved w i t h t h e a i d of t h e f o l l o w i n g g e n e r a l r e s u l t . 2.8.
PROPOSITION.
F:
maximal s u b f i e l d Proof. show t h a t
D
Let
E/F
such t h a t
LJ
Then
c h a r F = p > 0.
A
such t h a t
contains a
Our f i r s t t a s k i s t o
F.
c o n t a i n s a p r o p e r s e p a r a b l e f i e l d e x t e n s i o n of
F.
i s algebraic over
D
D
i s a separable extension.
We may h a r m l e s s l y assume t h a t
n o t e t h a t each element of
h $2 F
be a c e n t r a l d i v i s i o n F-algebra.
F,
i s separable over
then
To t h i s end,
Hence, i f t h e r e e x i s t s
FCX)
i s t h e required extension.
F.
Assume, by way o f c o n t r a d i c t i o n , t h a t t h e r e a r e no s e p a r a b l e e x t e n s i o n s o v e r Then, by t h e f o r e g o i n g , each element
2'
E F
Let
f o r some
r
xw
On t h e o t h e r hand, w e have such t h a t
i-1
c = f
Setting
f(b)# 0
( b ) h a s t h e form c -1 v = X c,
-Ax.
fp kc)
and l e t
i =
we s e e t h a t
-1
w = uz>
X Thus
hi = 1
+
Xwh-l,
,
xpx
f # 0 since =
0 f o r all fi-'(b)
be such t h a t
uX-
hu and s a t i s f i e s
v
commutes w i t h
Xu
=
X $2 F
A
1, g F
x
# 0
E
but
f ( c ) = 0, and
uX-Xu
f o r some
q
=
pn.
But t h e n
E
F.
Z(D1 = F .
and
D.
- Au)v-l = u v - 5 , - xuv -1 = w x - XW
wq E F
Xp
such t h a t
we d e r i v e
= (UX
but
Then
= 3:-hP -
c = Hence, p u t t i n g
purely inseparable, i . e .
We may t h e r e f o r e choose
= p(x).
f be t h e mapping
D is
of
3:
Choose
b
E
fi(b)
=
i.e.
c h = Xc.
0.
D
Then
THE BRALIER GROUP
q,
1 = 0,
and t h e r e f o r e
=
1
+ (hwX-l)q
=
1
+
E/F
if
But
CD(E) # E ,
then
D E
T h i s would imply t h a t Thus
2.9.
COROLLARY.
Let
2.10.
D
contains a proper
D of maximal d e g r e e .
F
D.
in
By Theorem l . l 5 ( i ) ,
i s separable over
i.e.
D
E
F,
D
such t h a t
i s simple Therefore,
by t h e f i r s t p a r t .
D.
i s a maximal s u b f i e l d of
in
cD (E)
is a
c o n t r a r y t o t h e maximality of
be a c e n t r a l d i v i s i o n F-algebra.
E/F
E
Then
CD(E).
E',
has a proper separable extension
separable f i e l d extension Proof.
E F)
U'
i s a d i v i s i o n a l g e b r a , hence so i s
E'
C (E) = E, D
E.
Awqh-l
Consequently,
be a s e p a r a b l e e x t e n s i o n i n
E.
+
F.
maximal s e p a r a b l e e x t e n s i o n of with c e n t r e
(since
W'
a contradiction.
s e p a r a b l e f i e l d e x t e n s i o n of Let
1
=
167
E
splits
Then t h e r e i s a
D.
Apply P r o p o s i t i o n 2 . 8 and C o r o l l a r y 1.19.
f i n i t e G a l o i s e x t e n s i o n of Proof.
F
Every Brauer c l a s s of
COROLLARY.
h a s a s p l i t t i n g f i e l d which i s a
F.
D be a d i v i s i o n a l g e b r a i n a g i v e n Brauer c l a s s of
Let
D
P r o p o s i t i o n 2.8, r a b l e extension.
c o n t a i n s a maximal s u b f i e l d
E
Then
E
such t h a t
E/F
F.
By
i s a sepa-
i s c o n t a i n e d i n a f i n i t e G a l o i s e x t e n s i o n of
F, and
D.
this w i l l also s p l i t
W e c l o s e t h i s s e c t i o n by p r o v i d i n g c i r c u m s t a n c e s under which t h e t e n s o r
This w i l l
p r o d u c t of two c e n t r a l d i v i s i o n a l g e b r a s i s a g a i n a d i v i s i o n a l g e b r a be achieved w i t h t h e a i d o f t h e f o l l o w i n g g e n e r a l r e s u l t .
2.11. THEOREM.
D
Let
and 1
t h e i r centres such t h a t
Do
F,E
II
be d i v i s i o n a l g e b r a s of d e g r e e s
E
r e s p e c t i v e l y , where
i s embedded i n
1
Do, we s e e t h a t such
q
P,S
over
2
M (D ) 4
>_
F.
Denote by
q d r2).
exists, in fact
Then
1
M ( D )= D o @ D 4
the l e a s t integer
( t a k i n g a r e g u l a r m a t r i x r e p r e s e n t a t i o n of
2
f o r some c e n t r a l d i v i s i o n E-algebra
q
' F
2 D
of d e g r e e 3
rt
=
sq
t,
and
CHAPTER 3
168
D1
Moreover, Proof.
:
t over E.
D 2 h a s index
Since
Do C M ( D ) 1 - 4 2
it f o l l o w s from P r o p o s i t i o n 1 . 1 2 t h a t
M ( D )g D " 8 D 4 2 ' F 3 where
U
Do
i s t h e c e n t r a l i z e r of
3
E,
with c e n t r e
4
8D
h a s index
t.
D3
Write
E
t d e n o t e t h e d e g r e e of
q 2 S 2 = t 2 r 2 , proving t h a t
We a r e t h e r e f o r e l e f t t o v e r i f y t h a t U
D 3 i s simple
The a l g e b r a
2
Let
i n (3) we find
Comparing F-dimensions
M (D 1.
in
by P r o p o s i t i o n 1.10.
(3)
.
D
rt
= Sq.
i s a d i v i s i o n a l g e b r a and t h a t
D3
Mk(D),
D
where
i s a s k e w f i e l d component of
' F D 3
.
Then
M (D )
Do 8 D ' F
4 2
>
R
(4)
' F
' F
Dr 8 D
Denote t h e s k e w f i e l d component of some
Do 8 M k ( D ) p M ( D o @ D)
.s
by
Db.
Then
D
Df
ML(Dlt)
for
F
1
M (D )
and s o
9 2
Mka(D
P
4 in
S u b s t i t u t i n g t h i s value f o r
D
By u n i q u e n e s s ,
).
(41, w e f i n d
Do 8 D
2 2
D
4
4 = kR.
and
Ma(D-).
I t follows
1 F
Do
that
c Mk(Dz)
1 -
ensures t h a t
k
Finally,
D
=
and s o , by t h e m i n i m a l i t y of 1 and t h e r e f o r e
z M (D ) d 5
€3 D
' F
M
D
E D
q, we have
d 2 1 and some s k e w f i e l d
f o r some
Mr2(F) 8
This
i s a skewfield.
( D ) p D 8 M ( D ) E D 8 5 I F 4 2 1 F
dq
q 4 R.
D5-
Hence
D; 8 D3 F
o3
F 2 M,2
dq
Thus
=
r 2 and
D
5
D
(D3)
h a s degree
t.
It follows t h a t
3
D1
:
D 2 h a s index
t, a s r e q u i r e d . From t h e proof above, w e deduce
2.12.
where
COROLLARY.
Let
D 1 , D 2 , F , E , q , r be a s
D i s a d i v i s i o n a l g e b r a and dq
d i v i s i o n a l g e b r a i f and o n l y i f
q
=
r2.
=
r2.
COROLLARY.
Then
Dl
Let
U ,D 1
2
Then
I n particular,
'
A s a f u r t h e r consequence of Theorem 2 . 1 1 ,
2.13.
i n Theorem 2.11.
w e prove
be c e n t r a l d i v i s i o n F - a l g e b r a s of coprime d e g r e e s .
D 2 i s a g a i n a c e n t r a l d i v i s i o n F-algebra.
169
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
Proof.
2
rt
t, where q
=
-
1.
r e s p e c t i v e l y and choose
=
rst
rst
=
qs2
Then
=
r 2 p ; but
m
and so
D1
;
D2
minimal
D2
has index
( r , s ) = 1,
t
r2s2m, i . e .
=
tlrs.
D 2 , so by t h e d e f i n i t i o n of index,
D1
p,q
D1
Owing t o Theorem 2 . 1 1 ,
q
Thus
r 2 m f o r some rn
rs
M (D2).
rp.
= sq, s t =
t h e i n d e x of = 1,t =
C
1 -
1
p
r,s
be of d e g r e e s
D o c M (D ) , D o
subject t o
i.e.
D,,D2
Let
so
rsm.
Now
r2/q,
t is
I t follows t h a t
has index equal t o i t s degree.
D1
Hence
;
D 2 is
a d i v i s i o n F - a l g e b r a which i s a l s o c e n t r a l , by P r o p o s i t i o n 1.10.
3. CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
E/F
Throughout t h i s s e c t i o n ,
Br(E/F)
W e d e n o t e by
is a f i n i t e G a l o i s e x t e n s i o n and
G = Gal(E/F).
t h e k e r n e l of t h e homomorphism
Br(F) 1/11
-
Br(B)
ct 1.4 El El
F Hence
BrIE/F)
s p l i t by of
G
E.
on
E*.
Given G
GI
c1 E
over
Z2(G,E*), E.
c o n s i s t i n g of t h o s e
ZZ(G,E*)
Consider t h e group
crossed product of E
Br(F)
i s t h e subgroup of
EaG
which a r e
with respect t o the n a t u r a l action
w e d e n o t e by
Thus
LA1
EaG
t h e corresponding
i s a f r e e l e f t E-module w i t h b a s i s
such t h a t
-
-1 =
and
=
Ct(x,y)xy
where
gh By Theorem 2.2.3,
n a t u r a l a c t i o n of
any c r o s s e d p r o d u c t of
G
on
I n what f o l l o w s we w r i t e
F)
-
CL
=
g(X)
G over
is equivalent t o
E%
E
for a suitable
f o r t h e cohomology c l a s s of
CI.
F o r any
c1 E
Z2(G,E*),
a
E
Z2(G,E*).
O u r main r e s u l t of
i s an isomorphism. LEMMA.
E E,g E G
(with r e s p e c t t o t h e given
t h i s s e c t i o n a s s e r t s t h a t t h e map
3.1.
X
for a l l
t h e f o l l o w i n g p r o p e r t i e s hold:
CHAPTER 3
170
(E)=
C
i s a c e n t r a l simple F-algebra and
( i ) E"C
E
E"G
E
(ii)
dimfG = \ G I 2 F ( i ) T h i s i s a p a r t i c u l a r c a s e o f C o r o l l a r y 2.4.3.
EaG
i s a s p l i t t i n g f i e l d f o r t h e F-algebra
Proof.
and
The f i r s t a s s e r t i o n f o l l o w s from ( i )and C o r o l l a r y 1 . 1 6 ( i ) .
(ii)
d i d = F
second, n o t e t h a t by P r o p o s i t i o n 1 . 4 . 1 ,
IGI
.
To prove t h e
Hence
dimEaG = (GIdimE = / G I 2 F F a s required. 3.2.
0
For a l l
LEMMA.
a
E G
(i)
6a G
Proof.
So
(1)
as F - a l g e b r a s
8
and
(iii)
the following conditions a r e equivalent:
E'G
i s equivalent t o
( i i ) EaG
Since
Z2(G,E*),
cl,a E
a r e cohomologous.
The e q u i v a l e n c e of
( i l and ( i i r l f o l l o w s from C o r o l l a r y 2.2.4.
o b v i o u s l y i m p l i e s ( i l l , w e a r e l e f t t o v e r i f y t h a t (ii) i m p l i e s ( i ) .
$ : ERG
assume t h a t
-t
ERG i s an isomorphism of F - a l g e b r a s .
i?'
Lemma 3 . 1 and Theorem 1 . 1 3 , t h e isomorphism
E B G.
i n n e r automorphism of
$(i= )
T-
kx
f o r all
y E E DG,
for a l l 3.3.
1
(ii)
E"G
(iiil
N
E.
Hence, if
8
E
for a l l
rr. E Z 2 ( G , E * ) ,
M (PI, where
( i ) G'B
4 (E)
Thus t h e r e e x i s t s a u n i t
( 8 4 ) (1)= k
then
For any
LEMMA.
f E
+
a Aut(E G)
E.
.Q. E
Owing t o
c a n be extended t o an
B
E G
of
2
such t h a t
i s d e f i n e d by
N o w a p p l y Lemma 2.2.6.
t h e following conditions a r e equivalent:
n = /GI
i s a skew group r i n g of
G
E
over
i s a coboundary
Proof.
By C o r o l l a r y 2.2.5,
( i i ) and (iii)a r e e q u i v a l e n t .
Invoking Lemma
3 . 2 , we a r e l e f t t o v e r i f y t h a t ( i i ) i m p l i e s ( i ) . Assume t h a t
E"G
i s a skew group r i n g of
$(x) E End(E) be d e f i n e d by F
@(x)( y )
=
$ : G by
b(g)
=
g.
xy -+
G
over
for a l l
E.
y E E.
Given
z E E,
let
Define
Aut(E) F
Then c o n d i t i o n s (1) and ( 2 ) of P r o p o s i t i o n 2.3.1
are s a t i s f i e d
171
CLASSICAL CROSSED PRODUCTS AND THE BRUAER GROUP
and so the map
is a homomorphism of F-algebras.
End(E1
EiiG
Since
is simple and both
E0"G
and
are of the same P-dimension, we conclude that
F EaC
End(E)
P
M,(F)
F as required.
ExarnpZe.
Then
G
9
Let
=
E
{l,g}
=
Q ( < ), i 2=-1,
and let g E Aut(E) be defined by
is a group of automorphisms of
G
Thus E / a ) is a Galois extension and
F whose fixed field is 0.
Gal(E/'Q)
=
Let
a
:
G x G
--f
E*
be
defined by
a ( l , l ) = Ci(g,l) = Ci(l,g) = 1, Cx(g,g) = -1 Then obviously where
1
c1 E
Z2(G,E*1 and EaG
is a free E-module with basis
E"C
is the identity element of
{i,;},
and
-1
jxz By identifying
o@(& with
E
and
=
g(x:) ,
j'
j
with
i,
=
-1
(x E E)
we see that
EaG
is isomorphic
to the quaternion algebra
over Q.
On the other hand, if
is a coboundary, then
by virtue of Lemma 3.3.
3.4. LEMMA.
Let A
be a (finite-dimensional) central simple F-algebra such that
E is a self-centralizing subfield of A. Proof.
For each g E G,
subalgebras of A.
-
the map
g
-
Then A :
E
F
2
ERG
for some
c1
E Z2(G,E*).
is an F-isomorphism of simple
Hence, by the Skolem-Noether theorem (Theorem 1 . 1 3 ) , there
exists g E U(A) such that
CHAPTER 3
172
-1 ;A;
=
d i d = (did)
By C o r o l l a r y 1 . 1 7 , we have
*
g(X)
A E E,g E G
for a l l
(1)
I n o r d e r t o prove t h a t
= ]GI2.
F
F
BE;
A =
(2)
SfG it s u f f i c e s t o show t h a t t h e r i g h t hand e x p r e s s i o n i s a d i r e c t sum, f o r t h e n b o t h
’.
s i d e s w i l l have F-dimension
]GI
I f t h e sum i s n o t d i r e c t , l e t
-
s = A g
i
1 1
k.
be a r e l a t i o n w i t h minimal Choosing
E*
A
such t h a t
... f h k g k
I t is clear t h a t
g (A) # g ( A ) ,
(Ai E E*)
= 0
k > 1 s i n c e each
-
g E U(A.1.
one e a s i l y v e r i f i e s t h a t
g1(Xls-Sh = 0
-
g i v e s a s h o r t e r n o n t r i v i a l r e l a t i o n connecting
-
This i s a contra-
g2, ...,gk.
d i c t i o n and t h u s ( 2 ) i s e s t a b l i s h e d .
-1
x,y
Suppose now t h a t
E
G.
h E E.
commutes with each
Since
E
A
Let
LEMMA.
A
D and some n 2 1.
v e c t o r space over
LJ
on which
an i n n e r automorphism of
where
I,
A.
A,
is the i d e n t i t y
A
A
P
M (Dl n
c1 : c1
G
X
G
we o b t a i n
----f
E Z2(G,t’*).
E*
and
in
Br(F).
[A1
=
[El
is a
B
c e n t r a l simple F-algebra,
[ A ] = [eAel
Then
B y Wedderburn’s theorem,
s i o n algebra
f o r some
be a ( f i n i t e - d i m e n s i o n a l )
e be a nonzero idempotent of Proof.
implies t h a t
EG ‘
A
cocycle and t h u s we have shown t h a t
A,
i s i t s own c e n t r a l i z e r i n
The a s s o c i a t i v i t y of m u l t i p l i c a t i o n i n
3.5.
x y q
Then t h e e q u a t i o n (1) shows t h a t
and l e t
f o r some d i v i -
By changing t h e b a s i s of an n-dimensional a c t s i r r e d u c i b l y and f a i t h f u l l y , i . e .
w e may assume t h a t
rxr-matrix.
Then
applying
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
eAe
and so
M,(D).
[eAel = [ D l = [ A ]
Thus
a s required.
173
.
The n e x t p r o p e r t y i s c r u c i a l f o r t h e proof of t h e main r e s u l t . t h e r e i s a s i m i l a r i t y of F - a l g e b r a s .
- Ea’G
E ~ G E’G F Proof.
Let
C
=
A €9 B ,
where
F
A = E% Then
‘L
=
B
and
8 E; SfG
=
E’G
=
i s a c e n t r a l simple F-algebra by Lemma 3 . 1 and P r o p o s i t i o n 1.10.
e
5
E E @ E
eCe
C such t h a t F y i e l d t h e d e s i r e d r e s u l t by a p p e a l i n g t o Lemma 3 . 5 . aim i s t o f i n d an idempotent
W e f i r s t note t h a t t h e s u b f i e l d s
mentwise.
Since
E
E 63 1 and
i s s e p a r a b l e over
F,
E = F(A) ( P r o p o s i t i o n s 1 . 4 . 3 and 1.4.6). nomial of
8 E :
SfG
A.
degf(X) = n ,
Then
n
E @ E
F E E
there exists Let
where
of
1 8 E
F aaG.
Our
This w i l l
commute e l e such t h a t
f C X ) E F [ X ] be t h e minimal poly-
= dimE.
Now d e f i n e
F
g E G - {l}.
where t h e s e p r o d u c t s a r e t a k e n o v e r a l l
# g(x)
a t e r i s not zero since d i s t i n c t from z e r o i n
E
@
E,
Observe t h a t t h e denomin-
g E G - {l}.
f o r each
s i n c e t h e elements
The numerator i s a l s o
{ A z @ 110 4
i
n - 11
are
F l i n e a r l y independent o v e r
1 @ E.
Hence
e # 0 in
E @ E.
F Next we observe t h a t
where
g
ranges over
which shows t h a t
G - {l}.
Consequently, w e must have
(1 8 Ale = (1@ l ) e
in
E @ E.
Thus, by i n d u c t i o n ,
F (1’ Q 1 ) e = (1
Az)e
for all
n-1 Because
E = @ FAi
i=o
and m u l t i p l i c a t i o n i n
E@ E F
i s commutative, we d e r i v e
i >0
174
CHAPTER 3
Thus
proving t h a t
e
E 8 E.
i s an idempotent i n
We are l e f t t o v e r i f y t h a t
eCe
BG.
eCe =
c
To t h i s end, n o t e t h a t
e ( E Q E) (Lt. 8 i ) e
x,$G =
c
e(E Q l ) e - e ( l Q ~ ~ e - eQ( y’)e x
XI YEC
But
e ( l @ E)e = e ( E @ l ) e by ( 3 1 , and
t o E.
W e now compute
e ( E Q l)e = E ‘
i s a f i e l d F-isomorphic
e ( ; 8 g ) e , by u s i n g t h e formulas
Then
where
z
ranges over a l l e l e m e n t s of
On t h e o t h e r hand, when
x
=
y
A s i m i l a r argument proves t h a t
The foregoing shows t h a t
where
G - {I}.
we o b t a i n
From ( 3 ) , we o b t a i n
175
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
E'
=
e ( E @ 1 ) e and
3
=
(g E GI
e ( i @ $)e
and a l s o shows t h a t
Clearly
G and, f o r
Gal(E'/F)
3
Hence, c o n j u g a t i o n by
8's
Thus t h e
all
g
a c t s as
!J E
E,g
on
E'.
E G,
Furthermore, f o r a l l
multiply according t o t h e cocycle
eCe
9
E
x,y
E G
T h i s shows t h a t
C$.
aBG
and t h e r e s u l t f o l l o w s . We have now come t o t h e d e m o n s t r a t i o n f o r which t h i s s e c t i o n h a s been developed. Let
E/F
be a f i n i t e G a l o i s e x t e n s i o n , l e t
Z2(G,E*),
let
E'G
3.7.
THEOREM.
any
a
E
t o t h e n a t u r a l a c t i o n of
be t h e c r o s s e d p r o d u c t of
G on
E.
G = Gal(E/F) and f o r
G over E with r e s p e c t
Then t h e map
i s an isomorphism. Proof.
Consider t h e map
Then, by Lemma 3 . 6 ,
f
i s a homomorphism. Kerf
and t h u s
?(a)
f
= [E'"G].
=
Furthermore, by Lemma 3 . 3 ,
B' ( G , E * )
i n d u c e s an i n j e c t i v e homomorphism Now assume t h a t
[A] E Br(E/F).
7
:
H2(G,E*)
--+
BF(E/F) g i v e n by
Then, by P r o p o s i t i o n 2 . 7 ,
there
CHAPTER 3
176
exists
B
E
[A]
B
such t h a t
E
contains
Invoking Lemma 3.4, w e deduce t h a t
B
E"G
as a self-centralizing subfield. f o r some
c(
€
Z2(G,E*).
Thus
and t h e r e s u l t f o l l o w s . 3.8. COROLLARY.
A
Let
be a ( f i n i t e - d i m e n s i o n a l ) c e n t r a l simple F-algebra.
Then t h e r e e x i s t s a f i n i t e G a l o i s e x t e n s i o n
G
= Gal(E/F),
Proof.
such t h a t
and
a E Z2( G , E * ) ,
i s s i m i l a r t o t h e crossed product
By C o r o l l a r y 2.10,
[A] E B r ( E / F ) .
that
A
E/F
where
b"G.
t h e r e e x i s t s a f i n i t e G a l o i s extension
E/F
such
Now a p p l y Theorem 3.7.
The above r e s u l t t e l l s u s t h a t e v e r y Brauer c l a s s c o n t a i n s a c r o s s e d p r o d u c t . I n p a r t i c u l a r , a c e n t r a l d i v i s i o n a l g e b r a i s always similar t o a c r o s s e d p r o d u c t . A n a t u r a l q u e s t i o n i s t o ask whether a c e n t r a l d i v i s i o n a l g e b r a i s always a
crossed product.
T h i s q u e s t i o n was i n v e s t i g a t e d i n d e t a i l i n t h e 1930s
(by
Hasse, A l b e r t and o t h e r s ) and i t was d i s c o v e r e d t h a t e v e r y c e n t r a l d i v i s i o n algebra i s a crossed product.
e-
However, i t was o n l y i n 1 9 7 2 t h a t Amitsur gave
examples of c e n t r a l d i v i s i o n a l g e b r a s t h a t a r e n o t c r o s s e d p r o d u c t s . To examine t h e Brauer group i n more d e t a i l , w e need t h e f o l l o w i n g s i m p l e observation 3.9.
LEMMA.
[A] E R r ( P ) Proof. !e may w r i t e
The Brauer group h a s index
P,
Br(F)
then
[Alp
i s a t o r s i o n group. = 1.
Keeping t h e n o t a t i o n of C o r o l l a r y 3.8,
A = I
+)
... @ In
More p r e c i s e l y i f
where each
w e may assume t h a t
A
=
PG.
li i s a minimal l e f t i d e a l of
A.
Then, by C o r o l l a r y 1 . 1 7 and Lemma 3 . 1 ( i ) ,
r
= indA = (degA)/n = ( d i d ) / E = (ndiml
E Invoking P r o p o s i t i o n 2.5.6, Theorem 3.7,
[A]'
[A].
as a s s e r t e d .
[A] E Br ( F )
The o r d e r of simple a l g e b r a
= 1
A,
.
w e deduce t h a t
)/n
= dim1
E l
E l
a'
i s a coboundary.
i s c a l l e d t h e exponent of
[A].
Hence, by
For any c e n t r a l
i t s exponent i s d e f i n e d t o be t h e exponent of i t s Brauer c l a s s
Thus Lemma 3.9 may be e x p r e s s e d by s a y i n g t h a t f o r any Brauer c l a s s , t h e
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
177
Although i t i s n o t t r u e i n g e n e r a l t h a t t h e exponent
exponent d i v i d e s t h e index.
i s e q u a l t o t h e i n d e x , w e do have t h e f o l l o w i n g g e n e r a l r e s u l t . 3.10.
[ A ] E B r ( P ) , t h e i n d e x and exponent of
For any
THEOREM.
[A1
have t h e
same prime f a c t o r s . Proof.
LA]
Let
r
have index
p, p / r
I t t h e r e f o r e s u f f i c e s t o v e r i f y t h a t , f o r any prime
t h e n o t a t i o n o f C o r o l l a r y 3.8,
G and
p-subgroup of
let
Then
3.9.
In
S b e a Sylow
Let
S.
b e t h e f i x e d f i e l d of
p(n.
implies
A = E'G.
w e may assume t h a t
Eo
n ( r , by Lemma
n so t h a t
and exponent
dimE
=
(G:S)= W
F o i s prime t o
A @ E,. F
p,
while
pk
IS] =
k
f o r some
1.
d e n o t e t h e index of
1J
Let
Then, by P r o p o s i t i o n 2.4,
lilrlliv Furthermore, s i n c e
E is a splitting field f o r A,
we have
(A@Eo) @ E = A @ E - E
F d i d = p
Taking i n t o account t h a t
Thus
If
11
= 1,
then
PIV,
p.
is a p o s i t i v e power of
p
,
F
w e deduce t h a t
plp
k
.
Hence
I-r
is a
EO
p.
power of
F k
'A C3 Zo F and t h e r e f o r e t h e exponent of
A
but
PIP
and
p k
W,
a contradiction.
Now by Lemma 3.9,
(A @ Eo)p F
i s d i v i s i b l e by
-1 p,
a s we wished t o show.
We now d e r i v e t w o consequences of i n t e r e s t . 3.11.
where t h e
where
Di
pi
D
be a c e n t r a l d i v i s i o n F-algebra of d e g r e e
a r e d i s t i n c t primes.
Then
i s a c e n t r a l d i v i s i o n F-algebra of d e g r e e
Proof.
where
Let
COROLLARY.
0 < mi
By Theorem 3.10 and Lemma 3 . 9 ,
Q
n
i'
1
< i < s.
the class
p
i '
[D]
h a s exponent
By t h e b a s i s theorem for a b e l i a n g r o u p s ,
[DI
9
CHAPTER 3
178
[Dl
can be written as a product of classes which are powers of Yower exponent.
mi exponent pi
,
Di be a division algebra similar to a power of
Let
1
with prime
Q
s.
with
Then
D
- D 1 ‘8 D2 63 ... 8 Ds F
Owing to Theorem 3.10, the
[Dl
F
U . have coprime degrees.
Hence, by Corollary 2.13,
we have a division algebra on the right, and therefore the two sides are isomorphic. A central simple F-algebra is called primary if it is # F
and contains no
proper central simple F-subalgebra. 3.12. COROLLARY.
(i) Every central simple F-algebra is a tensor product of
primary ones (ii) A primary P-algebra is either a division algebra of prime power degree or of the form M (F), where
P
Proof.
p
is a prime.
(i) Direct consequence of Proposition 1.12.
(ii) Apply Corollary 3.11.
= G over E
An important special case of crossed products of E/F
is a cyclic extension, i.e. the group
G
=
Gal(E/F)
is that where
is cyclic.
This is
due to the fact that the formulae become simpler, and secondly this case is important in applications.
cyclic o f order
For the rest of this section we assume that
generated by
n
g.
We say that an F-algebra A Fix
E
F*
and define
ax
G is
is cyczic if A Z2(G,E*)
E
EuG
for some
N E.
Z2(G,E*).
by
-
if
i +j9n
if
i + j > n
a
Z2(G,E*) is cohomologous to
1
(04 i,j G n - 1)
X We know, from Proposition 1.6.5, that each for a suitable
A
E
F.
E
In what follows, we write
h
E G
for E
G.
T h u s , from
what we have said above, it follows that each cyclic algebra is isomorphic to
E h i; for some h E E’. E-module with basis
Moreover, by the definition of {;‘I0
i
n
-
11
such that
BA G,EX G is a free left
CLASSICAL CROSSED PRODUCTS AND THE BRAUER GROUP
C o n v e r s e l y , t h e above d e s c r i p t i o n d e t e r m i n e s a u n i q u e asso-
gll = g ( u ) .
where
179
c i a t i v e F-algebra.
u
F o r any
NEIF(p)
It is clear that
3.13.
THEOREM.
h
( i i ) E’G
E G
(iii)
(iv)
-
F
E
and
p
of
i s d e f i n e d by
NEIF(E*) i s a subgroup o f
and t h a t
F*.
F*.
belong t o
1-I
F
over
h h E NEIF(E*)
E~~J.G
M (F)
%
NE,F(p)
i f and o n l y i f
Q EPG
h F
h
Let
E’G
(i) E C
t h e norm
E E,
i s a norm
i f and o n l y i f
The map
i s an isomorphism. Proof.
h
F G
( i ) By Lemma 3 . 2 ,
By P r o p o s i t i o n 1 . 6 . 5 ,
homologous.
E’G
i f and o n l y i f
the l a t t e r is equivalent t o
( i i ) T h i s i s a d i r e c t consequence of t h e f a c t t h a t
c1
X1-I
a
and
CLA
= clrclu
1-I
a r e co-
h/v and Lemma 3.6.
( i i i ) Apply ( i ) and Lemma 3.3.
(iv)
Apply P r o p o s i t i o n 1.6.5 and Theorem 3.7.
3.14.
THEOREM.
Let
( i ) The exponent o f
A
=
x
E G
f o r some
[A] E B r ( F )
[A]
(ii) I f t h e exponent o f
h E P.
t
i s the least positive integer
n
is equal t o
= dim??,
A
then
such t h a t
is a division
F algebra. Proof.
[AIt =
1
( i ) W e have
kt
i f and o n l y i f
( i i ) Since
n
= dirnE,
in
Br(E).
Thus, by Theorem 3.13,
i s a norm.
w e have dimA = n 2 .
A
If
M,(D),
where
D
in
Br(F),
is a
F
F d i v i s i o n a l g e b r a of i n d e x v i r t u e of Lemma 3.9.
[A]* = [EAt G]
m, t h e n
n
=
mr.
Hence t h e exponent of
But
[A]
[Aim
= 1
divides
m.
by
Hence, i f t h e
exponent of
A
is equal to
algebra, as asserted.
n,
then
m = n
and
r = 1.
Thus
A
is a division
181
4
Clifford theory for graded algebras
A
Let
b e a G-graded a l g e b r a .
An i m p o r t a n t method f o r c o n s t r u c t i n g i r r e d u c i b l e
A-modules c o n s i s t s i n t h e a p p l i c a t i o n ( p e r h a p s r e p e a t e d ) of t h r e e b a s i s o p e r a t i o n s : 1.
Restriction t o
A
2.
E x t e n s i o n from
A
3.
I n d u c t i o n from
A
1
1 1
T h i s i s t h e c o n t e n t of t h e s o - c a l l e d C l i f f o r d t h e o r y f o r g r a d e d a l g e b r a s o r i g i n a l l y developed by C l i f f o r d f o r t h e c a s e
A
= FN,
A = FG
where
FG
i s regarded
1
as a G//N-graded a l g e b r a o v e r a f i e l d
F
whose gN-component i s g i v e n by
(FG)gN =
Q
Fs
sfsm
The a i m of t h i s c h a p t e r i s t o p r e s e n t t h e g e n e r a l t h e o r y developed by Dade.
A
number of a p p l i c a t i o n s i s a l s o provided.
1. GRADED MODULES
Throughout t h i s s e c t i o n ,
A
d e n o t e s a G-graded a l g e b r a .
While w e s h a l l have
o c c a s i o n t o use r i g h t modules, any u n s p e c i f i e d module w i l l always b e understood t o be l e f t and unitary. W e s a y t h a t an A-module
of A -submodules of
M
M
i s graded p r o v i d e d t h e r e e x i s t s a f a m i l y
indexed by
G such t h a t t h e f o l l o w i n g two c o n d i t i o n s hold.
1
M = @M SfG AzMy c_ M
w
( d i r e c t sum of A -modules)
(1)
1
for all
x,y
The above d e f i n i t i o n c e r t a i n l y i m p l i e s t h a t t h e r e g u l a r module
E
AA
G
(2)
i s graded
182
CHAPTER 4
(with M = A for all g E GI. We refer to M g g g the g-component of m E M we understand unique m
as the g-component of E M
m E N
M
of a graded module
and all g E G, m
g
E N.
m =
By
C m
S f C g' is said t o be a graded submodule if for all g
A submodule N
defined by
M.
g
Expressed otherwise, N
is graded if
C B ( N ~ M )
N =
Sfc Thus if
N
is a graded submodule of N
M,
$7
=
then N
~
g
N
is a graded module with
M
A
A graded left A-submodule I of
for all g E G.
is called a graded l e f t
ideal of A. We call (11 the G-grading of
M
and refer to
as the y-component of
M
g
M.
When ( 2 ) can be replaced by the stronger condition
A M = M Y ZY M
we say that Let M
for all r , y E G
(3)
is a strongly graded A-module. N
and
be two graded A-modules.
An A-homomorphism
f:M---tN is said to be graded if
f ( M ) C_N g
In case
f
A-moduZes
.
is an isomorphism, we say that
1.1. LEMMA.
Let
M and N .
Then
(i) Kerf (ii) f ( M )
f
:
M
d N
M
and
(4)
are isomorphic a s graded
N
be a graded homomorphism of two graded A-modules
is a graded submodule of
is a graded submodule of
(iii) M/Kerf
for all g E G
g
M N
f(M) as graded A-modules, where the g-component of
M/Kerf
is
defined by (g E
(M/Kerf) = ( M +Kerf)/Kerf g g Proof.
Given m E M, we have
f(m)
=
f( C m ) sfGg
=
C f(m
&G
)
G)
where each f(m
g
)
GRADED MODULES
N
is in
T h i s shows t h a t
g'
f(m) g
rn
If
183
E Kerf,
f(m1
then
=
f(m
=
9
0 and hence, by ( 5 ) , f ( m
-
9
) =
p r o v i n g ( i ) . P r o p e r t y ( i i ) i s a d i r e c t consequence of To prove ( i i i ) ,l e t
f*
: M/Kerf
f(M)
g E G,m E M
for all
)
0.
Thus
m g
(5)
E Kerf,
(5).
f.
be t h e isomorphism induced by
Then
f*(m + Kerf) 9 which shows t h a t If
V
f*
=
f(m
) =
g
i s a graded isomorphism.
i s an A-module,
.
f(m)
t h e n we s h a l l d e n o t e by
by t h e r e s t r i c t i o n of a l g e b r a ;
V
thus
=
V
E f(M) g
g
m
for all
M
E
t h e A -module o b t a i n e d
VA
as a d d i t i v e g r o u p s b u t o n l y a c t i o n
Al
of
A
i s d e f i n e d on
T h i s p r o c e s s w i l l b e c a l l e d r e s t r i c t i o n and i t
VA-.
1
Y
p e r m i t s u s t o g o from any A-module
t o a u n i q u e l y determined A -module
A'
As i n t h e c a s e of modules o v e r group a l g e b r a s , t h e r e i s a d u a l p r o c e s s of induction.
T h i s p r o c e s s p e r m i t s us t o go from any A -module
determined graded A-module
@
d e f i n e d i n Chapter 2.
V
t o a uniquely
For convenience, l e t u s
r e c a l l t h e f o l l o w i n g p i e c e of i n f o r m a t i o n . Let
V
be any A -module.
Then t h e t e n s o r p r o d u c t
1
$ = A @ V A 1
i s n a t u r a l l y an A-module w i t h
y(x@v) W e s h a l l r e f e r to
@
=
yx '8 v
as t h e induced module.
f o r any
g E G.
and
v
E
V
In view of
and Lemma 2 . 1 . 1 ( i i ) , w e may i d e n t i f y t h e A -module
@,
f o r a l l x,y E A
A
9
@ V
w i t h i t s image i n
A,
With t h i s i d e n t i f i c a t i o n , w e have
8=
@ (A
'8 V)
( d i r e c t sum of A -modules)
Setting
(fl)g = A g
Q V A,
for a l l
g
E G
(6)
CHAPTER 4
184
i t i s immediate from (6) and t h e containment
@
that
i s a g r a d e d A-module.
fl
we s h a l l p u r s u e o u r s t u d y o f
i n o u r sub-
Here w e w i l l o n l y b e c o n t e n t t o p r o v i d e a u s e f u l
sequent i n v e s t i g a t i o n s .
c h a r a c t e r i z a t i o n of s t r o n g l y g r a d e d a l g e b r a s which i n v o l v e s
@
(see Theorem 1 . 2
below). Let
M
be a g r a d e d A-module.
g E G,
For e a c h
put
4 = M and d e f i n e
(@Ix Then
=
Mxg
for all
I& i s o b v i o u s l y a g r a d e d A-module which d i f f e r s from
fl
We s h a l l r e f e r t o
M.
a s a conjugate of
g E G,
we s a y t h a t
M
G
(7)
o n l y by q r a d i n g .
In c a s e
M=@ for a l l
M
zE
( a s g r a d e d A-modules)
i s G-invariant.
We a r e now r e a d y t o prove t h e f o l l o w i n g r e s u l t .
1.2.
THEOREM. (Dade(19801).
The f o l l o w i n g p r o p e r t i e s of a G-graded a l g e b r a
A
are e q u i v a l e n t t o each o t h e r : (i) A
i s s t r o n g l y graded
( i i ) Every g r a d e d A-module i s s t r o n g l y g r a d e d
M,
( iii) For any g r a d e d A-module
t h e n a t u r a l map
i s a graded isomorphism
M,
( i v ) For any graded A-module
t h e r e e x i s t s an A -module 1
Me (i)
proof. graded,
A A
zY
= A
( i i ) : Let
zY
for a l l
M
fl
be a g r a d e d A-module.
z,y E G,
V
such t h a t
( a s g r a d e d A-modules) Since
s o by Lemma Z . l . l ( i ) ,
A
is strongly
w e have
185
GRADED MODULES
Mq = AIMxy = AxAX-1M zy C - AzMy C Mq Hence A M
"Y
where we have used (2) twice. (ii) =. (iii):
so
di
Given g E G,
=
The kernel N
(Lemma l.l(i)) whose g-component N
9
A l @ MI A,
0 and thus $
=
A N
91 onto MI, so N g
Put 'I = M
(iii) =) (iv):
is a graded A-module
is given by
is the kernel of the natural isomorphism of =
$
of
is strongly graded so that for a11 g E G, N
By hypothesis, N
N
w'
i
as required.
we have
is a graded epimorphism.
Therefore
M
.
N1
Now =
0.
is a graded isomorphism.
and apply (iii)
(iv) * (i): Owing to Lemma 2.1.1(i) ,tii), we have
A A =A Hence, for any A -module 1
for all g
9
91
E
G
V,
Applying the hypothesis, we therefore conclude that
A M = M 91
for any graded A-module M. g-component A some
x
E
9
A
The regular A-module
for all g E G.
Letting
M
is a graded A-module with
to be the conjugate module
A"
for
G, we obtain
A A 9" by (7).
9
Hence
1.3. COROLLARY.
A
=
A (A")l
=
g
(AX) 9
=
A
92
is strongly G-graded, as asserted. Let A
be a strongly C-graded algebra, let M
N be a graded homomorphism.
graded A-modules and let f : M--+
and
N be two
Then f
monomorphism, epimorphism, or isomorphism if and only if its restriction
f
1
: M - N 1
1
is, respectively, a monomorphism, epimorphism, or isomorphism of
A -modules.
is a
CHAPTER 4
186
M
By Theorem 1 . 2 ,
Proof.
N
and
a r e s t r o n g l y graded so t h a t
A M = M g l g
A N
and
91
= N
gE G
for a l l
g
Hence
A
Because
f (M
91
)
=
1
A N
91
if and o n l y i f
1
epimorphism i f and o n l y i f s o i s
f
If
fl m
A
f(m ) 9
Then
Y'
since
)
= All,
m cM
g-l g -
=
f
w e infer that
i s an
1
fl.
i s a monomorphism, t h e n c l e a r l y so i s
i s a monomorphism and l e t
E M
f (M
f(m)
=
0
f,. m
f o r some
C o n v e r s e l y , assume t h a t
E
M.
Write
f ( A -lmg) = 0
0 f o r a l l g E G, hence
m
=
C m SfG
and t h u s
9
But t h e n
1'
m
which i m p l i e s t h a t
g
=
.
0, as r e q u i r e d .
2 . RESTRICTION TO A
Throughout t h i s s e c t i o n ,
A
d e n o t e s a G-graded a l g e b r a .
i n v e s t i g a t e t h e behaviour of t h e module module.
'A '
where
V
Our a i m i s t o
i s an i r r e d u c i b l e A-
The main r e s u l t i s a g e n e r a l i z a t i o n o f a c l a s s i c a l C l i f f o r d theorem
which d e a l s w i t h t h e c a s e
A = FG,A
=
FN
where
FG
i s r e g a r d e d as a G/N-
1
F
graded a l g e b r a o v e r a f i e l d
For any A -module
V
and any
whose gN-component i s g i v e n by
g E G,
put
1
Then, by Lemma 2 . 1 . 1 ( i i ) ,
conjugate of
V.
2.1.
Let
LEMMA.
module and l e t
A
'V
i s an A -mod:le.
W e shall refer to
be a s t r o n g l y G-graded a l g e b r a , l e t
U be any A-module.
M
-Y'
as a
be any g r a d e d A-
with
RESTRICTION TO A
187
G on the set of isomorphism
(i) The formula (1) provides an action of the group
classes of A -modules, i.e. for all x,y E G
'V
"=
V
and any A -module
and
x(yV)
M
gM
g
2
V,
qV
(2)
as A -modules
(3)
1
1
I& A B g M 1
(as graded A-modules)
(4)
A1 (iii) For any irreducible A -submodule W module
A W
of
and all
9
v
(i) Let
be any A -module and let x,Y
E
G.
By definition,
and z(yV)
€3 V )
= x(A
=
A
X
A,
is an isomorphism of A -modules. (ii) The isomorphism morphism ( 3 ) . 2
@.
$
Because
€3
(A
(@)
Thus x(YV)
€3
V)
4,
A1
Next by Theorem 1.2(iii), applied to M =
A €3 M
G, the A 1 -
(as A -modules)
'WEAW
'v v
E
is also irreducible and
g
Proof.
g
the map
xyV,
proving (2).
of Theorem l.2(iii) is graded and hence induces an iso= Mg,
(@I1
it follows from Theorem 1.2(iii) that
The isomorphism
A1
induced by ( 2 ) is obviously graded, proving ( 4 ) (iii] Assume that submodule of
X is a nonzero submodule of A W.
W and hence A
- 1X =
W.
?i
by
Ag,
we derive
X
=
A W. g
Then A -lX is a nonzero ?i n Y Multiplying both sides of this equality
Thus
A W is irreducible.
is obviously a surjective homomorphism. A
€3 W Al
is irreducible.
The natural map
g
So assume that
Therefore it suffices to show that
X is a nonzero submodule of A
g
€3 Al
W.
CHAPTER 4
188
get
5A
A -lX
Then
9
X
8
= A
t,
W,
so
X
A
-1
=
A
8 W.
A
Again, m u l t i p l y i n g by
we
9’
Al
as r e q u i r e d . g.
Al
A
Let
8 W
A,
V
be a s t r o n g l y G-graded a l g e b r a and l e t
be an A -module.
Consider
1
the s e t
H H
Then, by Lemma 2 . 1 ( i ) ,
i n e r t i a group
V;
of
E GIV
= {g
i s a subgroup of
H
i n case
=
G
2
’V}
G.
W e shall refer t o
we s h a l l s a y t h a t
V
H
as the
i s G-invariant.
We
have now come t o t h e d e m o n s t r a t i o n f o r which t h i s s e c t i o n h a s been developed. 2.2.
V
and l e t
VA
(i)
G
Let
THEOREM.
A
be a f i n i t e q r o u p , l e t
be an i r r e d u c i b l e A-module.
be a s t r o n g l y G-graded a l g e b r a
Then
U
c o n t a i n s a n i r r e d u c i b l e submodule, s a y 1
(iil
H
If
U
i s t h e i n e r t i a g r o u p of
and
W
t h e sum of a l l submodules of
VA 1
isomorphic t o
VA
(a)
U,
then there e x i s t s a p o s i t i v e integer
e(g1Ue
e
such t h a t
... o gn u)
1
where
{l
=
gl,g2,...,gn}
H
i s a l e f t transversal for
a r e p a i r w i s e nonisomorphic i r r e d u c i b l e A -modules.
G
in
and
In p a r t i c u l a r ,
9 1
gn
U,..., U
VA
is 1
c o m p l e t e l y r e d u c i b l e of f i n i t e l e n g t h . (b)
hi
i s an i r r e d u c i b l e A(H1-module such t h a t WA
2
eU
and
8
V
1
cU
(Here, of c o u r s e , Proof.
d e n o t e s a d i r e c t sum o f
(i) Since
V
sfGg V
in
g e n e r a t e d A -module.
V.
Because 1
We now s e t
M
of
u).
C A V
V.
G
(5)
SfG’
By P r o p o s i t i o n 2 . 1 . 1 0 ,
f i n i t e l y g e n e r a t e d A -module. p r o p e r A -submodule
copies of
is irreducible,
V = A v = ( C A ) v = f o r a l l nonzero
e
each
A
9
is a finitely
i s f i n i t e , i t f o l l o w s from ( 5 ) t h a t
Hence, by P r o p o s i t i o n 1 . 3 . 5 ,
V
is a
t h e r e i s a maximal
RESTRTCTION TO
Since A M so
C
Mo
for all g E G,
M
A
189
is an A-submodule of
90= 0 and thus there is an injective homomorphism
M
V
V.
8 [V/A M
--f
Mo
But
5M #
Y,
of A -modules
SfG We are therefore left to verify that each A -module V / A M is irreducible or, 9
equivalently, that each A M
is a maximal A -submodule of
9
V.
1
5 M' c I.' for some A -submodule M' of 9 plication by A -1 on the left yields g M 5 A -lM' C A -1V 9 9 so M = A M' and hence M' = A M, as required. g 9-1 c A U is a nonzero 4-submodule of V , we have (ii] Because assume that
So
A M
sfc
Thus, by Lemma 1.3.13,
9,U , ...,gn U
VA
Then multi-
v= c
A U.
SfG
is an irreducible A -module isomorphic to 'U.
A U
Owing to Lemma Z.l(iii), each
V.
9
is completely reducible.
Moreover, by Lemma Z.l(i),
1
U.
are all nonisomorphic conjugates of VA
all nonisomorphic irreducible submodules of submodules of
VA
isomorphic to A
i
.
U,. ..,A
U are gn be the sum of all
Hence A
g1
Wi
Let
U, 1
Q
n.
Then, by Lemma 1.7.6,
W1 Q
... Q Wn
(Wl = W)
gi
1
VA = 1
Thus, to prove (a) it suffices to verify that each of irreducible direct summands.
Now
Wi contains the
for each g E G, A V = V
same number
and the modules
g
ASWi,
A
9
W
i'
i # j, have no composition factor in common. W.=A 7,
Hence
W 9<
(2
i Q n)
which proves (a) and also that
92
1
Because
A hi =
h
W for all h A(H) @
A = A'H) @ A 92 SO
by ( 6 ) the natural map
E
W
[email protected]
= W 8 A
VA
gn
H, W is an A(H)-module.
... 8 A
A (HI gn
Now
(direct sum of right A (I'
-modules)
CHAPTER 4
190
S i n c e i t i s o b v i o u s l y a homomorphism
i s a t l e a s t an isomorphism of A -modules. 1
of A-modules,
8=A
t h e A-modules
W
@
and
V
a r e isomorphic.
A (H) Y
F i n a l l y , assume t h a t
9%
92 V.
W
V = V'
Hence
Then
Y
Y O ...@A
V ' = Y @ A i s an A-submodule of
W.
i s an A'H)-submodule of
Y
a n d , by ( 6 ) ,
=
W.
This proves t h a t
i s i r r e d u c i b l e and t h e r e s u l t f o l l o w s .
2.3.
algebra.
.
VA
a s required.
PROPOSITION.
5S(A).
J(.AI)
A
Let
Proof.
$;
V
Let
be an i r r e d u c i b l e
S(A
is completely r e d u c i b l e , s o
=
0.
Thus
1
G
be a c r o s s e d p r o d u c t of
A
f i n i t e group whose o r d e r i s a u n i t of Then t h e A-module
be a s t r o n g l y G-graded
5d ( A ) .
1
By Theorem 2 . 2 ,
J(A1) 5 J ( A )
A
be a f i n i t e group and l e t
I t s u f f i c e s t o show t h a t
A-module.
2.4.
A*JU
Then
Proof.
G
Let
COROLLARY.
and l e t
V
over
Al,
where
G is a
be an i r r e d u c i b l e A -module.
is completely r e d u c i b l e of f i n i t e l e n g t h .
#,
By t h e d e f i n i t i o n o f
w e have
vc= SEG
@ ( A @ V)
where, by Lemma Z . l ( i i i 1 ,
each
@ V
A
gAl i s a n i r r e d u c i b l e A -module.
A1
A -module
VG
i s completely r e d u c i b l e of f i n i t e l e n g t h .
module of
fi
and l e t
0
- - vc - -
be t h e n a t u r a l e x a c t sequence. s p l i t s , since
8
vc/w
w
Let
W
be an A-sub-
0
Then t h e c o r r e s p o n d i n g sequence of
i s a completely r e d u c i b l e A -module.
now f o l l o w s by v i r t u e of Theorem 2 . 3 . 4 C i i ) ,
Hence t h e
applied f o r
A
-modules
The d e s i r e d a s s e r t i o n
H = 1.
m
The f o l l o w i n g r e s u l t p r o v i d e s some i n f o r m a t i o n on r e s t r i c t i o n t o
A
of
indecomposable A-modules. 2.5.
THEOREM.
Let
over an a r t i n i a n r i n g
be a f i n i t e group, l e t
G
Al,
indecomposable A-module.
and l e t Then
V
A
b e a c r o s s e d p r o d u c t of
G
be an A - p r o j e c t i v e f i n i t e l y g e n e r a t e d
i s a d i r e c t sum of f i n i t e l y many G-conjugate
GRADED HOMOMORPHISM MODULES
191
indecomposable A -modules, each isomorphism type occurring with the same multi1
plicity in the sense of the Krull-Schmidt theorem. Since A l
Proof. sition 2 . 3 . 2 1 .
is artinian and
Since
1
VA
tion 1.3.16,
is a finitely generated A -module.
is of finite length.
generated A-moduie.
is artinian (Propo-
is a finitely generated A-module and A
V
generated A -module, V
G is finite, A
Hence, by Proposi-
(VA )G
Note also that
is a finitely
is a finitely
(VA )G 1s of finite length.
Hence, by Proposition 1 . 3 . 1 6 ,
I
We may therefore apply the Krull-Schmidt theorem (Proposition 1.3.17) to both
VA
and
(VA
G.
)
1
1
There exist indecomposable A -modules V1,V 2 , . 1
VA
=
..,Vk
such that
v1 0 v2 0 ... CD Vk
(7)
1
and hence
V is A -projective and therefore, by Theorem
By hypothesis,
G
(VA ) ,
morphic to a direct summand of V
may assume that
is isomorphic :t
is isomorphic to a direct summand of
2.7.20,
v
is iso-
Thanks to the Krull-Schmidt theorem, we
a direct summand of
CvG1
.
Vy.
Consequently,
vA
Now
Al
(direct sum of indecomposable A -modules) since
@ V
=
theorem, V .
V'
Thus, by the Krull-Schmidt
is obviously indecomposable.
'V
for some g
is also a decomposition of
E
G.
But
SV
V'
and
into indecomposable A -modules.
VA
Hence the
1
multiplicity of if and only if
V1
in (7).
Vi V.
z
in ( 7 ) is equal to that of
SV
Vj' the multiplicity of SV
Thus all the
Vi
in ( 8 ) .
Because
SV i
' 5
in ( 8 ) is equal to that of
in ( 7 ) occur with the same multiplicity and the
result follows.
3 . GRADED HOMOMORPHISM MODULES
Throughout this section,
A denotes a G-graded algebra
over a commutative ring
CHAPTER 4
192
R
and
M,N
two graded A-modules.
g E G,
For any
we d e f i n e t h e R-submodule
Hom(M,N)
A Hom(M,N1
A
if
=
Hom(M,N)
E
g
fi
i s d e f i n e d by
(fl)x
for a l l
2
62
by
G}
(1)
fi =
N
and
= N
for a l l
xcg
x
E
G
(2)
can be i n t e r p r e t e d as t h e R-submodule
Hom(M,N)
A
g GrHom (M,
fi)
H O ~ ( M , N ) c o n s i s t i n g of a l l graded homomorphisms from A
of
Hom(M,N) A
A
R e c a l l t h a t t h e graded A-module
By (1) and ( 2 ) ,
5 Nxg
jf(Mx)
of
'
M
fl.
to
In fact,
we have a s t r o n g e r p r o p e r t y , namely Hom(M,N)
A L
If
= GrHom(18,flg)
g
for a l l
x,g
E
G
(3)
A
i s a l s o a graded A-module,
w e e a s i l y compute t h a t :
where t h e p r o d u c t of homomorphisms i s t h e u s u a l composition.
3.1.
The R-submodule
LEMMA.
Hom(M,N)G of HomfM,N)
g e n e r a t e d by t h e
A
A Hom(M,N) , g E
A
G,
g
i s t h e i r d i r e c t sum
Hom(M,NIG = @ Hom(M,N1
A Proof.
A
( a s R-modules)
Assume t h a t we have homomorphisms
a l l b u t a f i n i t e number of t h e
f9
f
f E Hom(M,N)g,g E G , ' A
such t h a t
are zero and
Cf SfG W e must prove t h a t t h e
(5)
g
= o
( i n Hom(M,N))
A
a r e a l l zero.
T o t h i s end,
fix any
x
E C
g
mx
E
M,.
Then, f o r a l l
g
E
G,f
(m ) N =
@ N
@Gg the equality
lies in
9 2 =
@ N
@Gxg
N
x9
by (1).
Because
and
193
GRADED HOMOMORPHISM MODULES
x
f (rn ) = 0 f o r a l l g x
ensures t h a t
f
.
G.
Thus, f o r a l l
x
C,f
E
= 0, So t h e lemma i s t r u e . g W e a r e now i n a p o s i t i o n t o p r o v i d e a d i r e c t d e s c r i p t i o n of
hence
g
(M ) x
= 0, and
Hom(M,N)G.
A 3.2.
(Dade (1980)).
THEOREM.
Hom(M,N)G
The R-module
c o n s i s t s Of a l l
A f E Hom(M,N)
A f o r which t h e r e e x i s t s a f i n i t e s u b s e t
I n p a r t i c u l a r , by Lemma 3 . 1 ,
Proof.
f
of
T
If
c
G
if
!I' of
such t h a t
i s f i n i t e , then
i s any f i n i t e s u b s e t of
Hom(M,W)t
G
s a t i s f i e s (6).
G , t h e n (1) i m p l i e s t h a t any element
Therefore a l l
f
E Hom(M,N)G
f o r some such
s a t i s f y (6)
A
i3T A T
by ( 5 ) .
Conversely, assume t h a t
f
E
Hom(M,N)
s a t i s f i e s ( 6 ) f o r some f i n i t e s u b s e t
T
A of
G.
Since
G
i s a group, t h e decomposition
N = 8N
i s equivalent t o
SEG
f o r any f i x e d
g,x E G ,
g 6 G.
T h e r e f o r e , t h e r e e x i s t unique
f
gtgx
m
I t f o l l o w s from ( 6 ) t h a t a l l t h e
for a l l
Ho%
)
for
=
@ M S f G g'
fx
= 0,
g gx
such t h a t
€ M a r e z e r o f o r a l l b u t a f i n i t e number of g g t h i s i s w e l l - d e f i n e d and s a t i s f i e s
whenever
(A4 ,N
E
z
g.
Since
fg,gx,g E G , a r e z e r o , and hence t h a t
o u t s i d e t h e f i n i t e set
2'.
Moreover, we have
M
CHAPTER 4
194
So
each
f
will lie in HomiM,N)
X
that each f
is an A-homomorphism.
X
Fix y , g , x E G , m
z,y
E
and the theorem will be proved, once we show
A 9
and a
E M
E A
Y
9'
Then (7) and
Y'
A M C M C N x y - xyrAx#y - xy'
G, yield:
and
f (a m ) x Y 9
E
f (A M 1 C_fx(M
Y9
x Y 9
)
Nygx
Since f is an A-homomorphism, we therefore have two expansions:
a f ( m ) = Z a f ( r n ) Y' 9 & G Y X c7 and
f ( a m ] = C f (am)
Y g
for the same element a f ( m
Y
g
f(a m 1
=
)
in the decomposition
Y g
N =
@ N 3cEG
which is equivalent to
N = 8N
because
ygx G
is a group.
We may therefore
SfG
conclude that
a f (m 1 Y X g Since fX
Y g
;LEGx
=
fx(am 1 Y g
for all
y E C,a E A Y Y
is A -linear it follows from this equality and the expansion
A
=
@ A SfG
that
afz(mg) for any a E A, g E G
and m
9
E
M
=
fx(m ) 9
Since fx
9'
this equality and the expansion M = 8 M
is A -linear it follows from
that
SfG afx(m) f o r any
m
E
M.
Thus each
f,
=
fx((am)
is an A-homomorphism and the result follows.
M, we denote by E n d W the R-algebra Hom(M,M) of all A A A-endomorphisms of M, and by lM the identity element of that algebra, i.e. For any A-module
the identity map of M naturally turns
M
onto itself.
into an
Application of A-endomorphisms on the right
( A , End (MIo)-bimodule,where
A
End
A
(M)"
is the
GRADED HOMOMORPHISM MODULES
opposite algebra t o
End
195
(M).
A
M
Let
and
CJ
be two graded A-modules.
k
i f there exists a positive integer
f f(M1
such t h a t
:
weakly d i v i d e s
M
N
and a g r a d e d monomorphism
... @
N 6
M -
W e say t h a t
N = N (k)
N(k).
i s a d i r e c t summand of
( k copies)
We s h a l l s a y t h a t
weakly isomorphic i f each weakly d i v i d e s t h e o t h e r .
N
and
are
Of c o u r s e , t h i s i s a n e q u i -
M
Finally, we say t h a t
v a l e n c e r e l a t i o n among graded A-modules.
M
i s weakly G-
@.
invariant i f i t i s weakly isomorphic t o a l l i t s G-conjugates
For a r b i t r a r y A -modules w e d e f i n e t h e n o t i o n s of weak d i v i s i b i l i t y and weak I
isomorphism i n a s i m i l a r f a s h i o n , by r e p l a c i n g “graded monomorphism” w i t h “monoWe s h a l l r e f e r t o an A -module
morphism”.
V
i s weakly isomorphic t o a l l i t s G-conjugates
3 . 3 . THEOREM (Dade ( 1 9 8 0 ) ) . a l g e b r a and l e t
E
define
weakly G-invariant i f it
’V. A
be a f i n i t e group, l e t
be a graded A-module.
E = End(M1 A
Put
i s a G-graded R-algebra w i t h
g E G,
and, f o r each
E -1
a s i t s g-component,
g
(ii) E
b e a G-graded
by
9
E
(1)
M
G
Let
a s being
i s t h e R-submodule
g (iii) E l
GrHom($,$’)
E
of
x,g
for all
f o r any
g
E
G.
E G
A i s t h e subalgebra
of
GrEnd(91
E
f o r any
5 E
G
A
M
(ivl
i s a g r a d e d (A,E’)-bimodule
i n the sense t h a t
A M E
for a l l
C M
Y g x - Ygx
E
(v)
i s a s t r o n g l y G-graded a l g e b r a i f and o n l y i f t h e graded A-module
weakly (vi)
x,y,g E G M
is
G-invariant If
A
i s a s t r o n g l y G-graded a l g e b r a , t h e n
o n l y i f t h e 4 -module 1
Proof.
M
1
i s s t r o n g l y G-graded
i f and
i s weakly G - i n v a r i a n t .
(i) By Theorem 3.2,
E = @ E
SfG
proving ( i )
E
’
( d i r e c t sum of R-modules)
and, by ( 4 ) ,
196
CHAPTER 4
(iii) D i r e c t consequence of
(ivl
x,y
Ex
A M z 5'
and t h e f a c t t h a t
C Mx5'
for a l l
G.
(v)
By (i) and Lemma 2 . l . l . ( i i i ) ,
1 E
E E f o r a l l g E G. 9 g-l
M
= N.
( i ) ,(ii) and Lemma Z . l . l ( i ) .
Follows from t h e d e f i n i t i o n of E
M
( 3 ) applied f o r
( i i ) Direct consequence of
E
i s s t r o n g l y G-graded i f and o n l y i f lM i s a l s o
Since
x E G, we know
for a l l
1
J
from ( i i ) t h a t t h i s i s e q u i v a l e n t t o : 1
J
E GrHom(~,Jg)GrHom(Jg,~c)
for a l l
x,g
M"
weakly
By t h e d e f i n i t i o n of weak d i v i s i b i l i t y , t h i s happens i f and o n l y i f
2'
divides
M
says that (vi)
x,g
for a l l
E
G.
Because
is weakly G - i n v a r i a n t ,
A
Suppose t h a t
w e nay i d e n t i f y
@
G
G i s a group, t h i s l a s t c o n d i t i o n j u s t
as r e q u i r e d .
i s a s t r o n g l y G-graded a l g e b r a . with
E
A
A @M '
g E G.
for a l l
Then, by Lemma 2 . 1 ( i i ) , Hence, f o r all
g,s 6 G , 2
A1
?@
weakly d i v i d e s
M
that
i f and o n l y i f
'MI.
weakly d i v i d e s
IcM1
MI.
i s weakly G - i n v a r i a n t i f and o n l y i f so i s
The l a t t e r i m p l i e s The d e s i r e d a s s e r t i o n
now f o l l o w s by a p p l y i n g ( v ) .
4.
A
EXTENSION FROM
1
G
Throughout t h i s s e c t i o n ,
a commutative r i n g and
A
a G-
All c o n v e n t i o n s and n o t a t i o n i n t r o d u c e d i n t h e p r e v i o u s
R.
graded a l g e b r a o v e r
R
d e n o t e s a group,
s e c t i o n s remain i n f o r c e .
M
Let
let
be a graded A-module,
E = End(M)
and, f o r each
g E G,
put
A E
{ f E End(M)
=
9
A
we know, from Theorem 3 . 3 , t h a t i f with
E -1
a s i t s g-component,
f(Mx) C_ M xg G
for all
E
is f i n i t e , then
f o r any
2 E
GI
i s a G-graded R-algebra
g E G.
9 4.1.
LEMMA.
(i) If
A
Let
V
and
W
be a s t r o n g l y G-graded a l g e b r a . a r e A -modules,
(ii) I f
V
Hom(@,d) o n t o Hom (gY,gxFI) for all A A1 i s an A -module, t h e n r e s t r i c t i o n t o Y'
R-module
B
= End(#)x
t h e R-module
x
A
onto
Hom (gV,gzVl Al
gV
then r e s t r i c t i n g t o
for a l l
x,g
i s an isomorphism of E
G.
is an isomorphism of t h e
x,g
E
G.
In particular,
EXTENSION FROM A
is an isomorphism of the algebra E
'V
restriction to
197
1
onto
End ('Vl
for all
g 6 G. Proof.
'V.
(i) Let
f
be the restriction of
f to
9
By definition,
g so
and let f
Hom(@,iflIx A
E
=
Ckp)
'V,
9
=
'W
- gxV
and f ( ' V )
C
that
fg
E Hom
(gV,gxW)
Al
Assume that
f(gv) =
fg
f++
It is clear that the map
is an R-homomorphism
1
0.
Then
f ( A -1' V ) =
f(Al
9
@
V) =
€3 V i ) ) = f(#) = 0,
f(A(Al
Al
Al
so that the given map is injective.
Finally, let $
Horn (gV,gxW). Al
Then
'V
induces a homomorphism
A @ gV -!@+ A @ gxW
*,
A, NOW
$
is the 1-component o f
(@)g,
while
gxW
is the 1-component of
(bb)'x.
Let
be the isomorphisms defined as in Theorem l.Z(iii).
.
is an element of
whose restriction to A (ii) Direct consequence of (i).
4.2. LEMMA.
Assume that
Then
'v
coincides with
$.
G is a finite group, M a graded A-module and let
End(M). Let E be defined by (l), for all g G. A g (i) For any g E G, the elements of Eg n U ( E ) are precisely the graded iso-
E
=
morphisms of (ii) M
M onto
fl.
is G-invariant if and only if
E
is a crossed product of
G over El
(iiil
If A
is strongly G-graded and
M
=
J'
for some A -module
V, then
V
CHAPTER 4
198
E is a crossed product
is G-invariant if and only if
Proof. (i) Assume that f E E
End (V). A, is an automorphism of M
Then f
U(E).
9
G over E
of
such that
Since f
is an automorphism, we must have M
isomorphism of
M
automorphism of
=
M xg
and so f is a graded
#.
onto
Conversely, let f
f(Mz)
M onto
be a graded isomorphism of
f(Mx)
such that
=
M
29
for all
X E
I@.
G.
Then f
is an
Thus f E E
U(E),
g
as asserted. (ii) Direct consequence of (i) and the definition of crossed products. (iii) By Lemma 2.1(ii), we may identify
4
'M
A €3 ' V . Hence, by A1 is G-invariant. The fact
A Q
with
=
A1
V
Corollary 1.3,
is G-invariant if and only if
M
'
that E
e End (V) being a consequence of Lemma 4.1(ii), the result follows. A, Let S be an arbitrary ring and let V be an S-module. We say that
v
V is hoth artinian and noetherian.
of finite l e n g t h if
V
The module
is
is
called s t r o n g l y indecomposable if its endomorphism ring End(V/) is a local ring. S
4.3. LEMMA.
V
If
is strongly indecomposable, then
is indecomposable.
Suppose that
a direct decomposition of
V
is strongly indecomposable and that
V.
If
71 :
V-
V'
V
Let
let C
u E End(V),
un.
Suppose that
L
=
L
and 2
u ( z ) = u 2(yl ,
X
= 1
X 2
.
1.
V ' @ V"
is 71
Hence either
It suffices to
un,n > 1, and let X
Then, for any x E V,
in which case
=
form an ideal.
We first show that for a sufficiently large
1
such that
be the image of
n
S kernel of
TT =
is indecomposable and of finite length.
verify that the set of all nonunits in End(V1
V
is the projection map, then
is an idempotent of the local ring End(V), so 71 = 0 or S V ' = 0 or V' = V , proving that V is indecomposable. Assume that
The
V is of finite length.
converse is true if Proof.
V
be the
'2,
there exists y
E
V
EXTENSION FROM A l
For any x E L 1 n X I , we have
x
u(y).
=
V
Because
Xn
y
Then
X2
t
=
Xl
u(x)
=
whence
199
0 and there exists y 3c
= u(y) =
V
such that
0 and therefore
is of finite length, there exists n 2 1 such that L
n
v= =
L
L
X
@
1
1
.
and
zn
whence by the above (2) holds.
= X2n,
Now assume that u
End(V). Because V is indecomposable, S is nilpotent. Moreover, the endomorphisms uV and
is a nonunit of
it follows from ( 2 ) that
u
vu
V E
are nonunits for all
End(V).
Finally, assume by way of contradiction,
S
that v
and
W
are nonunits of
End(V)
such that
S
u = v + w is a unit.
Then
1
=
v
'
+ w',
hence nilpotent endomorphisms.
where
v
'
= u - ~ V , ~ '= u-lW
However, V '
.
( ~ ' =1 (~ ~ ' =1 0 ~ for a sufficiently large a contradiciton.
and
W'
are nonunits and commute, so that
= 1-U'
1
implies that
YI
= (V'+W')
2n-1
0,
As a final preliminary observation, we prove 4.4.
LEMMA.
Let
G be a finite group, let A
be a G-graded algebra and let
M be a graded A-module such that the ring GrEnd(M) is local.
Y
Denote by
d
A
strongly indecomposable A -module. 1
(i) M
is weakly G-invariant if and only if it is Ginvariant
(ii) If A
is strongly G-graded, then
is weakly G-invariant if and only if
it is G-invariant. Proof.
It suffices to verify that "weakly G-invariant" implies "G-invariant".
(i) Suppose that
3.3(v),
M is weakly G-invariant and put E
E is strongly G-graded.
Because
End(M). By Theorem A = GrEnd(M) (Theorem 3.3(iii)),
E 1
is a crossed product of
G
over E
=
E
A
(Proposition 2.1.11).
Hence, by Lemma
1
4.2(ii),
M
is C-invariant.
(ii) Assume that so,
V
is weakly G-invariant and set M =
by Theorem 3.3(vi),
B
is strongly G-graded.
ring, Proposition 2.1.11 implies that E Thus
V
#.
Then
M
V
and
Since E
End(V) is a local ' A is a crossed product of Z over
is G-invariant by applying Lemma 4.2(iii).
El-
CHAPTER 4
200
From now on we assume t h a t
G
i s a f i n i t e group
A
i s a s t r o n g l y G-graded a l g e b r a
V
i s an a r b i t r a r y A -module
V
Our aim i s t o d i s c o v e r c o n d i t i o n s under which i.e.
V
t h e r e e x i s t s an A-module
V*,
can be extended t o an A-module,
whose a d d i t i v e group c o i n c i d e s with t h a t of
while t h e m u l t i p l i c a t i o n :
t
A x V*-
V*
V*
in
satisfies
a*v=au I n t h e n o t a t i o n of
(1) we p u t
M
fl
=
v E V,a
for a l l
A
(3)
i s a G-graded a l g e b r a
E = End(#)
so t h a t
E
A g
E a s i t s g-component, f o r any -1 R w i t h i t s image i n A l 8 V so t h a t with
E
G.
For convenience, we i d e n t i f y
v
Al
Owing t o Lemma 2 . 1 . 2 ( i i i ) ,
t h e sequence of homomorphisms
G.
i s always e x a c t e x c e p t p o s s i b l y a t
A
s p Z i t t i n g homomorphism Y f o r t h e Y
sequence (5) i s , as usual, a homomorphism
of t h e group
G
into
GrU(E)
such
that d e g ( Y ( g )1 = g S i n c e t h e g-component of
E
is
E -1 , (6) i s
for a l l
g
E
G
(6)
g
E
G
(6')
equivalent to
57
y ( g ) E U ( E ) n E -1
for a l l
9 E v i d e n t l y a s p l i t t i n g homomorphism
y
e x i s t s i f and o n l y i f t h e sequence ( 5 ) i s
both e x a c t and s p l i t ( r e c a l l t h a t i n c a s e ( 5 ) i s an e x a c t s p l i t t i n g sequence, we refer to
E
a s a skew group r i n g of
G
El).
over
We have now accumulated a l l t h e i n f o r m a t i o n n e c e s s a r y t o prove t h e main r e s u l t of t h i s s e c t i o n .
4.5.
THEOREM (Dade (1981)).
The
A
-module
V
can be extended t o an A-module i f
EXTENSION FROM A
201
and only if the G-graded algebra B = End('/) is a skew group ring of G over A E End ( V ) . Indeed, there is a bijective correspondence between all extensions
4
V*
to A-modules and all splitting homomorphisms Y
V
of
corresponds to y
V*
for ( 5 ) , in which
if and only if
V
In particular, if
Y(g) (a
=
Ug*V
9
for all
@ V )
Ag
V , g E G,ag
V
(7)
V
can be extended to an A-module, then by Lemma 4.2(iii),
is G-invariant. Proof.
Y* is an extension of V
Assume that
implies that there is a unique A -homomorphism of
ag&,
into
for all g E
G,V E V ,
and a
g
deduce that there exists a unique Y(g) l E E El
( 3 ) and ( 4 ) , the element
a
*V =
Hence the unique element y ( l ) For any
a
z
E
A b
E
x,y
E
A
and
Z'Y
Y
$,
morphism of
a v 1
E
V,
E
into
V
sending a
g Applying Lemma 4.1(ii),
9'
8
V
we
Owing to
E -1 such that ( 6 ) holds. 9
E
satisfies =
1 (a 8 v ) E l
must be
for all
d E
and a
V
1
Al
lE.
G, the product yCx)y(y) v
A
Then ( 3 )
to an A-module.
lies in
E -lE -1 C- E " Y (Xy)-l
If
-
then by ( 7 ) and the fact that y ( ~ : ) is an A-endo-
we have
y ( x ) y ( y ) (a b @
"Y
V) =
y,(y
(a ( b 8 v ) ) )
Y " Y
"( aXyY ( bY 8 v ) )
= y
= y,(a =
(b *v)1
" Y
yx(ax8 ( b * u ) )
Y
= az*(b * v )
(8)
Y
This last expression is just
A A X
So
Y
=
A
XY
f o r all
x,y
E
(a b ) * v
"Y
since
V*
is an A-module.
G, the above products axby
Because
generate the R-module.
these equations force y(z)y(y) to be the unique element y ( z y ) .
strates that y
This demon-
is a splitting homomorphism for ( 5 ) , the only such homomorphism
satisfying ( 7 ) . Now suppose that y
v E
and
a
E
g
A
the element y(g) g
For any g E G,
is any splitting homomorphism for (5). of
E g-1
sends a
g
@ V E
8 V
A
g
A,
=
'V
CHAPTER 4
202
'V
into an element of
=
V by (1) applied to M
=
#.
Thus we may use ( 7 ) to
define a bilinear product * : A
xV-V
9
for any
g
E
since A = @ A
G.
there is a unique extension of these
9'
SfC
products to a bilinear product
Because the homomorphism
Y sends
lC into the identity
If v E V,aZ E AX
follows from ( 4 ) and ( 7 ) that ( 3 ) holds. any
Z,Y E
G , then
and
GrU(E), it b
Y
E
A
for
Y'
$.
y(x) is an A-endomorphism of
( 8 ) holds by ( 7 ) since
Taking into account that Y
lE of
is a homomorphism, this implies that
(a b
XY
)*U =
Y ( q )(a b
X Y
@ V)
=
Y ( Z 1 Y ( Y ) (ab @ u )
=
ax*(b* V )
X Y
Thus the product module
V
*
is associative and therefore defines an extension of the A
.
to an A-module
theorem is true. The group
U(E 1
Y
V*,
the only such module satisfying ( 7 ) .
So
-
the
acts naturally on the set of all splitting homomorphisms for
1
( 5 ) , with any
u E U(E ) 1
ting homomorphism
sending any such homomorphism into the conjugate split-
for ( 5 ) defined by
uY(g) = u y ( g ) u
E EIE
g
-1E 1 C-
E -1
for all g
E
G
(9)
g
This action is related to isomorphisms of extension modules by the following result. 4.6. THEOREM (Dade (1981)).
Two extensions of the A -module
V
to A-modules are
1
isomorphic as A -modules if and only if the splitting homomorphisms for ( 5 ) cor1
responding to them in Theorem 4.5 are
U(E )-conjugate.
Thus the correspondence
1
of Theorem 4.5 induces a bijective correspondence between a l l A-isomorphism classes of extensions of
V to A-modules and a l l U ( E )-conjugacy classes of 1
splitting homomorphisms for ( 5 ) .
EXTENSION FROM
proof. and
y'
V*
Let
V*'
203
V
be two extensions of
V*'
By Lemma 4.1(ii), the A -automorphisms of
(L').
End
is also an A -automorphism of
V*
onto
Al
restrictions of elements u E U(E1).
y
to A-modules, and let
be, respectively, corresponding splitting homomorphisms for ( 5 ) .
A-isomorphism of of
and
Al
V,
Any
that is a unit
L' are precisely the
Since 0 A
A =
-?c u
the restriction of any such
V*'
is an A-isomorphism of
onto
V*
and
a
if and only
if
u(a * ' U ) g
=
a
g
for all
*U(U)
* ' and *
lpplying ( 7 ) for
li E V , g E
together with the fact that
G U
g
E
A
(10)
g
is an A-endomorphism,
we derive
and
a
*U(V) =
$7
Because the products a '8 U
y(g) ( a 9
@JU(V)) =
@,
generate
9
ay'(g1 which holds if and only if
Y'
=
Y'
=
[ - f ( g ) u l ( a' 8 V ) g
we deduce that (10) is equivalent to for all g
Y(g1U
by ( 9 ) .
E
G
This completes the proof of the
theorem. 4.7. COROLLARY.
Assume that ( 5 ) splits and that
ting homomorphism.
Assume further that
U(E )
u
:
G
--+
GrU(E) is a split-
is abelian and denote by
1
H'(G,U(E 1 ) the first cohomology group of
G
over
U(B )
G
with respect to the action of
1
on
U(E )
given by
1
gu = ll(gIull(gY1
(u E U ( E ),g E
Then there is a bijective correspondence between the elements of the A-isomorphism classes of extensions of Proof.
V
to A-modules.
Direct consequence of Theorems 4.6 and 1.5.8(ii).
B
H' ( G , U ( E 1
)
G). and
CHAPTER 4
204
4.8. COROLLARY. A
only if
The regular
A -module A l
over A
G
is a skew group ring of
can be extended to an A-module if and
correspondence between all extensions A : homomorphisms
Y for the sequence
in which A*
corresponds to
a *a g
y
Indeed, there is a bijective
A l to A-modules and all splitting
of
if and only if
Y(g)aa gl
=
1
.
for all a E A l ,g t G
and a
g
E
A
(11)
g
This correspondence induces a bijective correspondence between all A-isomorphism classes of such extensions A* morphisms
sends
A
‘A
into the regular A-module A ,
= A @ A
A1 A ,a E A.
1
1
=
9
‘8 A
sending
a 8a
into aa
Evidently, this isomorphism is grade-preserving, i.e.
1
A
there is a natural isomorphism of the induced A-
1 E Al,
Since
a t
for all
U(Al)-conjugacy classes of such homo-
Y.
Proof. module A
and all
onto A 1
g
=
A
for all g E G .
A g I’
Because the A-endomor-
A1
phisms of the regular A-module are just right multiplications by elements of A, this isomorphism induces an isomorphism
A
+
End(A1) A A
sending any
a
E
A
into the endomorphism
a’ 8 lA Since A A
x Y
=
A
for all x , y E G,
XY morphism maps A into E: 9
where
g’
grade-preserving isomorphism of A
F-+
(a’a)c3 lA
for all a‘ E A
it follows from (1) that this latter iso-
A
E = End(A ) and g E G . Hence it is a A ’ onto E . The corollary now follows directly
from Theorems 4.5 and 4.6 once we translate their statements about E
to statements about A
4.9. COROLLARY.
Let A
fl = A:
and
via the above isomorphisms. be a skew group ring of
is commutative and denote by
H’(G,U(A
) )
G over
Assume that A
the first cohomoloqy group of
G
1
over
1
U(A
)
with respect to the action of
G on A
given by 1
1
-1
ga
=
;as
(aE A l ,g
E
G)
INDUCTION FROM A
205
H1 (G,U(AI
Then t h e r e i s a b i j e c t i v e correspondence between t h e e l e m e n t s of t h e A-isomorphism Proof. 4.10.
Al
c l a s s e s of e x t e n s i o n s of
and
j
t o A-modules.
Apply C o r o l l a r y 4.8 and Theorem 1 . 5 . 8 ( i i ) . m
v
Let
COROLLARY.
A
be a G - i n v a r i a n t
-module such t h a t
i s an
Aut(V)
1
/GI.
a b e l i a n group u n i q u e l y d i v i s i b l e by
can be extended t o an A-
Then
module. Proof.
V
E
Put
= End(#)
E
i s G-invariant,
e
so t h a t
I End(V)
u(E
and
AutV
1
A i s a c r o s s e d p r o d u c t of
A1
G
E
over
.
Since
(Lemma 4 . 2 ( i i i ) ) . 1
Hence t h e sequence ( 5 ) i s e x a c t . v i r t u e of P r o p o s i t i o n 1.6.1(v)
5 . INDUCTION FROM A
Furthermore,
( 5 ) i s a s p l i t e x a c t sequence by
Now apply Theorem 4.5.
and Theorem 1 . 5 . 7 ( i i i ) .
1
G
Throughout t h i s section, algebra over a f i e l d
F
A
d e n o t e s a f i n i t e group, dim4 <
with
a s t r o n g l y G-graded
V an indecomposable ( f i n i t e - d i m e n -
and
F s i o n a l over that
A
F) A -module. 1
I n o r d e r t o avoid t e c h n i c a l i t i e s , w e s h a l l a l s o assume
i s a c r o s s e d p r o d u c t of
-
g
exists a unit
A
of
in
A
$7'
G
g
A l , i . e . t h a t f o r each
over
E G,
there
Note, however, t h a t some o f our r e s u l t s (e.g.
A
s e e P r o p o s i t i o n 5.1 below) a r e v a l i d under t h e more g e n e r a l assumption t h a t
is
a s t r o n g l y G-graded a l g e b r a . F o r any subgroup
H
of
G , A'')
H
i s a c r o s s e d p r o d u c t of
over
Al
defined
bY
A(H)
=
Ah
@
E H To s i m p l i f y t h e n o t a t i o n , w e s h a l l w r i t e pectively.
fl
For t h e same r e a s o n , we w r i t e
and
End(fi)
f
@
for and
and
End(V)
for
$(H),
res-
End(#)
A and
End ( V ) ,
respectively.
F i n a l l y , all modules c o n s i d e r e d below a r e assumed t o
'4
be f i n i t e dimensional o v e r t h e ground f i e l d
F.
Our aim i s t o i n v e s t i g a t e t h e decomposition of t h e induced module s c r u t i n i z i n g t h e s t r u c t u r e of t h e endomorphism r i n g of
fl.
fi
by
For convenience of
r e f e r e n c e , we r e c o r d t h e f o l l o w i n g consequence of some p r e v i o u s r e s u l t s . 5.1.
PROPOSITION.
The r i n g
C End(V 1
i s a G-graded F - a l g e b r a whose g-component i s
206
CHAPTER 4
given by {$ E End(VC)
Furthermore, V
-
If(,
8 V) A,
5 q-' 8
for all cc E GI
V
A1
G End(V )
is G-invariant if and only if
is a crossed product of
G over End(V).
=
Apply Theorem 3.3 and Lemma 4.2(iii).
Proof.
The following result will enable us to concentrate on the case where
V
is C-
invariant. 5.2.
Let ff
THEOREM.
v
be the inertia group of
and write
vl'= v1 8 ... 0 v, V. are indecomposable A
where the
f 8 ... 8 f ,
(i) VG =
<
V! 3
V.
implies that
(ii) If
V
where the
2
-modules.
Then
I$ are indecomposable A-modules such that
V
j.
is irreducible for some
i
(1)
...,s } ,
i E {1,2,
then
G and V are
V .
irreducible. Proof.
(i) The equality V
G
=
G
V 8 1
... 8 :V
follows from the fact that
induction is transitive and preserves direct sums (Lemmas 1.7.1 and 1.7.2). prove that each
V!
is indecomposable, assume that
By Lemma 1.7.4,
is a direct decomposition. summand of
({IH
,
a direct summand of
V. is identifiable with a direct
so we may assume, by the Krull-Schmidt theorem, that
Vi
is
XH, say Xff =
Next we note that
To
vi 63 X'
(2)
is the direct sum of isomorphic A -modules of the form
(?) *I
&
@ V.
Invoking the Krull-Schmidt theorem, we obtain from (1)
A1
(ViIA for some positive integer m
i'
2
rn .V
(1 s
i s s)
(3)
Hence, for any g E G , we have an A -isomorphism
INDUCTION FROM A
201
which in turn yields
T is a transversal for H in G
where
Restricting ( 2 ) to A l ,
containing 1.
we see by ( 3 ) that XA
CF
8 Xi
m.V
1
X is an A-module,
But
proving that for each
so
t E T,; @ V appears as direct summand of
@ V
Bearing in mind that for distinct t l , t PE T,;
are nonisomorphic A -modules, we deduce from ( 4 ) that 1
8
possible only when
Hence
Assume by way of contradiction that
({I, V;
V! 3
and
=
Vie Vi
fi
S dimX,
P Z
A1
is
Vf
V? 3
but
which is
F
Vi
$ Vj,
( VGi ) H ,
we may write
and
(V.) = V . 8 V! 3 H 3 3
Vi
Because
V'
.
1
1
Consequently,
j'
(V!) A, isomorphic to an A -direct summand of ( V ! )
morphic to an A -direct summand of
is
G
Thanks to the Krull-Schmidt theorem,
are A(H)-modules.
is isomorphic to an A(H'-direct summand of
(V.)
'
i'
is isoA1
It now follows from ( 3 ) that
V is
and that
A l
V
However, by (41, the multiplicity of is exactly m
(V!)
dim
xA 1 is indecomposable.
identifiable with a direct summand of
V1
@
G (V.) A1
.
A1
isomorphic to a direct summand of
where
with
A1
multiplicity at least mi. and
1
Al
i'
This contradiction proves the desired assertion
(ii) Assume that there exists an Then, by Theorem 2.2,
as an indecomposable direct summand of
(Vi)A
i
...,s }
E {1,2,
such that
is completely reducible.
Vi
.
is irreducible.
Hence, by ( 3 1 ,
V
is
1
irreducible. To prove that
V:.
fi
is irreducible, let
Then, by Theorem 2.7.7(iii),
Therefore
(Vi)A 1
Vi
W be an irreducible factor module of
is isomorphic to a submodule of
is isomorphic to a submodule of
wH-
and hence, by ( 3 ) ,
CHAPTER 4
208
contains a submodule X
A -submodule SX 1
Let T
be
m .(;
m.V. @ V)
It follows that for all
g
E
G, WA
.
contains the 1
" A, a transversal for
H in G.
Then the
-
A -modules t 63 V,t A,
t
T,
are irreducible and nonisomorphic and they all appear as direct summands of
wA 1 with multiplicity at least mi.
Invoking ( 4 1 , we deduce that dimW 2 d i m E
F G
V.
Thus
F"
W is irreducible and the result follows.
=
a
It is now easy to provide a sufficient condition for complete reducibility of induced modules. 5.3.
V be an irreducible A -module and let H be the inertia 1
V.
group of VG
Let
COROLLARY.
If the characteristic of
P does not divide the order of H, then
is completely reducible. Owing to Theorem 5.2, we may harmlessly assume that
Proof. the order of
G
is a unit of
A.
H
G.
=
Then
The desired conclusion is therefore a conse-
quence of Proposition 2.4. a Owing to Theorem 5.2, the decomposition of of
$,
where
H is the inertia group of V.
lessly assume that Let
fd
VG
is entirely reflected by that
For this reason, we may harm-
V is G-invariant.
be an A-module.
W is
Following Huppert (1982), we say that
a b s o h t e l y indecomposable if
which means that the F-algebra End(W)/J(End(W)) is isomorphic to
A
A
absolutely indecomposable, then by Lemma 4 . 3 ,
W is indecomposable.
verse is true if
F is algebraically closed.
a local ring.
End(W)/J(End(M)) A A is algebraically closed, it follows that
over
F.
As
End(W)/J(End(W))
A and
W
If
W is
The con-
Indeed, by Lemma 4.3,
End(W) is A is a division algebra of finite dimension
Hence
F
F.
2
F
A
is absolutely indecomposable.
The special case of the following result in which A
is assumed to be a
INDUCTION FROM A
G over F and A
twisted group algebra of over A")
G/N
5.4. THEOREM.
E
=
End(fi)
N
where
E
and let
E
=
G, is due to Conlon (1964).
be the 1-component of
{f E End(i/;)
If(;
52
63 V ) Al
(i) E
=
i.e. for all
z E GI
2
End(V)
E
such that
E/E'J(EI)
is a
G over F.
(i) Direct consequence of Proposition 5.1.
I
J ( E )'E
is an ideal of
(ii) The fact that
E'J(E
Theorem 2.6.3(ii).
By Theorem 2.2, J(E )
module and so J(E1)
5 J(E),
=
is a consequence of
annihilates each irreducible F-
E*J(E
Thus we must have
",J(E) (and hence E * J ( E I ))
dimE
@ V
J ( E 1 ) * E is a nilpotent ideal of
twisted group algebra of Proof,
E,
Al
G over E
is a crossed product of
(ii) E . J ( E
is regarded as a crossed product of
be absolutely indecomposable G-invariant A -module, let
V
Let
-4
209
1
)
c J(E).
Since
-
is nilpotent.
F Now
V
is absolutely indecomposable, so
P.
over
=
F @ J ( E 1 ) as F-spaces. Apply-
we deduce that E/E.J(EI)
ing Pronosition 2.6.4(iii), of G
E
is a twisted group algebra
9
The discussion has now reached a point where, in order to make further progress, we need to bring in certain ring-theoretic results.
R, and idem-
we first exhibit a relationship between idempotents in a ring
potents in a factor ring R/N, denotes the image of
5.5. LEMMA.
r
=
R
in
=
-
N is a nil ideal.
In what follows P
R/N.
N be a nil ideal of a ring R.
Let
(il Each idemDotent
-e
E
where
E
E
can be lifted to an idempotent e
E
R,
that is,
E.
(ii) If
e
and f
,E z T E
and only if (iii) Let
-
1 =
E
R.
1
idempotents in
+
R, then eR
are idempotents of
4
fR as right R-modules if
as right E-modules. E 2
+
... +
be a decomposition of
E
-
1 into orthogonal
Then there exist orthogonal idempotents e , e 1
such that l = @ 1
+ e
2
+...+
e
ei = 'i
2
,...,e
E
R
CHAF'TER 4
2 10
Furthermore, Proof.
(i) Fix an idempotent
N and so
u-u2
ei
is primitive if and only if
Ei
( U - U ? ) ~=
is also.
and choose u
E E
0 for some rn > 1.
while the first rn of the first m
Now
terms are devisible by
terms, then
u(1-u) E N,
1
=
u
.
R with 1*
e + (1-u) g ,
rn
(1-u)
is divisible by
e
Hence, if
rn
Then
= E.
We have
Note that on the right each term after the first rn
rn
E
9
where
,
denotes the sum
is a polynomial in
U.
so
uZrn + 2mu
e
2m- 1
- -
Because
e,f
be idempotents in R.
that is, e = u = t.
(1-u) +
e(1-e)
...
= c(1-u)
rn
5
g
u(moM),
0, e
=
is a required idem-
potent. (ii) Let
modules if and only if there exist a , b E R
a
0 : eR
Assume first that to
f.
fb
Then
eaf
similarly
=
=
then x
* bx
b
=
eaf,b = fbe,ab
---3
and
a.
fh'
:
be,
B(e)e = O k ) ,
( 0-18) (el
=
a,b
and
a
=
E
R
eR E
If
Now put a
= 2
f
(5)
e
so
to
b
and
=
b,
and
fbe
a
0-'(f)b
=
ab
a and b
satisfy ( 5 ) ,
fR with inverse y
satisfy ( 5 ) .
so Z E S
E,E
Then
75.
I-+
ay. satsify the
Next suppose that
are such that
eaf,b
fbe,ab
Then a b
has an inverse 1 - z
= e z 'e, we have
R
to
, E SFE,
1
z
=
Conversely, if
fR and let a,b
caf,bl = fbe.
nilpotent, 1 - z
=
be = b,
0-'(fb)
proving ( 5 ) .
a and put
and ba
hence
O-'(b)
Same conditions for the isomorphism
.sE
e,
is an isomorphism that maps
is a homomorphism from
Assume that eH
eR
=
such that
Hence
e and similarly f
=
fR as right R-
We first show that eR
z
+
z"
=
zz"
a l , b 2= b ( e - z " ) , 1
=
e,ba =
1
'
z"z
then
e-z,
f (modN), where
z E eNe.
and so z + z ' = z z
'=
Because z
Z'Z.
Setting
and it follows that a b ( e - z " ) = e. 1
1
is
INDUCTION FROM A
a
a b2a2 = f - y r
Next write
b
E
211 1
a2b2 = e
b(mocW1
then y E fNf.
Since
( b 2 a 2 ) ' = b ea 2
=
2
b a 2
2
we have
f-y
y2
Hence
=
y
a b
Because
2
= (f-y)2
and since y =
2
and b a
e
2
=
+
f2-fy-yf
y2 =
e,f
2
=
f
imply a
ea2f b2
=
that there exists an idempotent g
h2
=
h,he
such that
is nilpotent and so
h Then
--
such that e f
be idempotents in R
end, note that fe
E
0, and
=
y2
b a =f.
is nilpotent, we flnd that y = 0 and thus =
fb2e,
=
fe
ertion follows. (iii) Let
+
f-2y
=
=
7.
=
7
-_
2
0.
=
We first show
eg = g e = 0.
and
1 - f e is a unit.
2
the required ass-
To this
Put
(~-fe)-'f(~-fe) g = (1-elh.
Now put
Then
proving the auxiliary assertion.
e1,e2,...,e
We now prove that if
i# j,
then there exist idempotents e !
< For
n
=
1
5 and
.
Put
e!e! = 0
for all
1-3
i j
2
+
... + en'
then e
--
--
1
n > 1 and use
0 for i # j ,
=
=
0.
1
Applying the auxiliary assertion, there exists an idempotent e r
-
e' = e 1
and 1
ee'
=
1
gonal idempotents.
e r e = 0.
0 for
is an idempotent such that
e e = e e
-
=
i# j
In other words, we may assume that e e
e = e
e.e 1-j
such that
there is nothing to prove, so we may assume that
induction on n
i,j > 1.
=
--
are idempotents such that
It follows that
1
e',e 1
2
,...,e
such that
are pairwise ortho-
It is now an easy matter to prove the required assertion.
Indeed, by (.i)and the above, there exist pairwise orthogonal idempotents
e ,e 1
2
,...,e
-
such that e .= 1-
E
if
1G
i
G n.
Thus for
u
=
e
1
+ e + 2
... + en
we
212
CHAPTER 4
-
-
u = 1,
have
e E R
To prove t h e f i n a l a s s e r t i o n , assume t h a t
e # 0,
Since
R,
x,y
e
0.
- -
e = e
e
+
1
2
e
0 or
=
e
.
xf y
0.
i s a p r i m i t i v e idempotent.
f o r some o r t h o g o n a l idempotents
e 1 , e 2E R
le + e ) R ,
2
1
y
0 or
=
1 = xte2
=
which i s p o s s i b l e o n l y
2
cc
Hence e i t h e r
-
with
0 and t h e r e f o r e
=
is primitive.
+
= u
where
V,
u = 0
is primitive.
5.6.
LEMMA.
(i) R
e
is primitive.
-
or
2, =
0.
Thus e i t h e r
R
Let
I
0 and so
-
e =
then
0 or
U =
U
+
2, =
T h i s completes t h e proof of t h e lemma.
e # 0.
If
- - - -V,uV=
V U
=
0
0, p r o v i n g t h a t
8
be an a r t i n i a n r i n g .
0 and
i s l o c a l i f and o n l y i f
(ii) I f
e#
Then
u,V a r e o r t h o g o n a l idempotents,
c
and so e i t h e r
e
=
By ( i i ) , eR
=
Conversely, assume t h a t
e
e
If
t h e n t h e r e e x i s t o r t h o g o n a l idempotents
i n t h e case
-
e#
w e have
and hence w i t h
u = 1.
1 - u i s a n i l p o t e n t idempotent, and hence
so
i s a n i l p o t e n t i d e a l of
1 a r e t h e o n l y idempotents of
R,
i s l o c a l i f and o n l y i f s o i s
R
then
R
R/I. Proof.
R
(i) If
p o t e n t s of
R.
R.
Since
J(R)
of
H/J(R1,
i s l o c a l , then obviously
i s a n i l p o t e n t i d e a l of
by v i r t u e of Lemma 5 . 5 ( i ) .
matrix r i n g s over d i v i s i o n r i n g s . therefore
R
LEMMA.
G
Let
C
a l g e b r a of
Hence
and
and
1 a r e t h e o n l y idem-
1 a r e t h e o n l y idempotents of
R,
0 and
1
But
R/J(Rl
i s a d i r e c t p r o d u c t of f u l l
R/J(R)
a r e t h e o n l y idempotents
must be a d i v i s i o n r i n g and
is local
( i i ) D i r e c t consequence of
5.7.
0
Conversely, assume t h a t
0
Proof.
E:
Let
homomorphism A
be a p-group,
F.
over
( i l and Lemma 5 . 5 1 i ) .
A
Then
8
l e t charF = p
and l e t
A
b e a t w i s t e d group
i s a local ring.
be t h e a l g e b r a i c c l o s u r e of
F.
Then t h e r e i s an i n j e c t i v e
E @ A i s a t w i s t e d group a l g e b r a of S o v e r E. F i s l o c a l i f and o n l y i f 0 and 1 are t h e o n l y idempotents of
--i
E @ A and
F
A
By Lemma 5.6,
A.
Hence we may assume t h a t
1.6(iv), an F - b a s i s
A
p
FC.
{g
Let
- 11 1 #
I(G)
g E
GI
F
i s a l g e b r a i c a l l y closed.
be t h e augmentation i d e a l of and each
g
-1
Then, by P r o p o s i t i o n
FG.
i s nilpotent since
Then
I(G)
has
LNDUCTION FROM A
Hence
.
I(G) is a nilpotent ideal of FG.
is a local ring.
Let F
5.8. LEMMA. and let A
G over F.
be a twisted group algebra of
Proof.
FG/I(G) e F and therefore FG
G
=
charF = p
1 or
Then A
0 and G
Assume that f G
G has an element g of order n
local ring and assume that
H =,
H'F
we have
e
=
is a local ring
is a p-group.
The "if" part is a consequence of Lemma 5.7.
Then, for
G be a finite group
be an algebraically closed field, let
if and only if either
F.
But
213
3
1 where
is a
# 0 in
y1
FH (Proposition 1.6.5) and n-1 i (Vnl C g i=0
is a nontrivial idempotent of contrary to Lemma 5.6(i).
FH.
Thus
Hence G'F
charF = p
3
.
has a nontrivial idempotent,
0 and G
is a p-group.
The following result generalizes a classical theorem of Green(1959). 5.9. THEOREM.
be an absolutely indecomposable A -module and let H
V
Let
be
1
V.
the inertia group of
If
charF = p > 0 and
H is a p-group, then VG is
indecomposable. Proof. V
By Theorem 5.2(i), we may harmlessly assume that H = G, i.e. that
is G-invariant.
Put E = En<(e)
and note that, by Theorem 5.4(ii),
E / E * J (El) is a twisted group algebra of group, E/E.J(E
G over F.
Since
charF = p > 0 and
is a local ring, by virtue of Lemma 5.7.
)
But
E-J(E )
nilpotent ideal of Thus
is a pis a
1
1
(ii)).
G
fi
E
(Theorem 5.4(ii)), hence
.
E is a local ring (Lemma 5.6
is indecomposable, by virtue of Lemma 4.3.
The result above can be strengthened under the assumption that F
is algebra-
ically closed. 5.10. THEOREM.
Let F
be an algebraically closed field, let
posable A -module and let H 1
conditions are equivalent: (i)
fi
is indecomposable
be the inertia group of
V.
V
be an indecom-
Then the following
CHAPTER 4
214
?
(ii)
is indecomposable
H
(iii) Either
E
=
H is a p-group
and
That (i) is equivalent to (ii) is a consequence of Theorem 5.2(i).
Proof. Let
charF = p > 0
1 or
=
so that by Lemma 4 . 3 ,
End(#)
(ii) is equivalent to
E being local.
By Lemma 5.6(ii), the latter is equivalent to E/B*J(B1) being local.
is H-invariant, so by Theorem 5.4(iii,
H over F.
But
E/E*J(E1) is a twisted group algebra of
The desired conclusion is therefore a consequence of Lemma 5.8.
fl
To examine the decomposition of
in the general case, we shall need the
following ring-theoretic results. 5.11. LEMMA.
N be a nil ideal of a ring R and let the ring R'
Let
=
R/N
admit a decomposition
ii = L
@ L @...@Ln
1
2
Then there exists a decompo-
into finitely many indecomposable right ideals.
R,
sition of
R
=
I @ I2@ 1
... @ I n
into indecomposable right ideals such that (i) Li
is the image of
(ii) L .
z
=L
Proof.
-
with
1=
Ti
under the natural map
as right ??-modules if and only if
j
1
E
1
+
+ e +
E
9
+
...
t. E
and
E.E. z
L. =
z
2
... + e .
1
I. I z
j
R. as right R-modules.
z
1
into indecomposable right ideals of R.
z
2
E
i
E
R'
Owing to Lemma 5.5(iii), there 2
Setting I.= e . R ,
R = I @I
(i) holds.
c
-t.
By hypothesis, there exist orthogonal primitive idempotents
exist orthogonal primitive idempotents e , e 1 = e
R
,...,e
€
R
with
-
ei =
E.
z
and
we have the decomposition
0 . . . @In Since
72.
=
e. 2
= E.R =
L
i'
property
Property (ii) being a consequence of Lemma 5.5(ii), the result
follows. We next prove a general fact which provides the link between a module and its ring of endomorphisms.
Let R
any subset S of
,
EndfJ)
R
be a ring and let
let SV
V
be a left R-module,
denote the set of all finite sums
For
INDUCTION FROM A
215
ISi ( V i )
Then, clearly, SV
cipal right ideal of
End(V1 R
Let R
5.12. LEMMA.
V.
is an R-submodule of
V
I is
Note also that if
4,
generated by
be a ring, let
S,Vi E V )
(Si E
a prin-
@(W.
then I V =
be a left R-module, and let End(v)
R admit a decomposition
... @ I n
End(v) = I @
R
into finitely many indecomposable right ideals. (i)
=
... 0 I,V
I V @ I V 0 1
2
= 1.V
(ii) 1 . V a modules.
Then
is a decomposition into indecomposable submodules.
I.E I
as left R-modules if and only if
3
z
(i) Put E = End(v1
Proof.
as right End(Vi)-
R
j
and write
R 1
4;
Then
=
# j,Ii
= 0,;
$i,$.Q.
=
Q1 +
Q2
= QcE,
... Qn
+
(Qi E 1;)
Ii is indecomposable, each of
and since
Z J
the
idempotents Qi
If v
is primitive.
V,
E
then
and so
v= Suppose that
0 1
,v
2
,...,v
v+
I
I
2
... + InV
v+
are such that
1
0.
Thus
Q
Now put
=
Q .,I
=
.
Y
+ X..
tents such that for each
v
E
Then
X
=
X 0X
1
v
also have
=
or
1
2
Then
z
@(v) for each
must be
IT
for some R-modules X
V and we denote by
and
1
IV,v = L, E
n
(u)
TI
IV.
+
n
(v).
TI
0.
But
2
X. z
=
IT
i
E
2
IV
=
@ ( V ) , we must
2
Hence
= T
I#I
TI.(V), so
z
X
=
1
+ n
and therefore either 2
0 or X
=
0, proving that
2
IV is indecomposable. (ii) Assume that
I. 2 I
z
i
E
are orthogonal idempoBecause
1 1
= 0.
2
are direct summands of
I
T
2
1
the canonical projection
2
2
I., and assume that I V
X
X
... Q n ( V n )
1 + Q (v 1 +
V = 69 I.V.
and
and
1
@i, we can deduce that each Q . ( v . ) z z
Then, taking the images of both sides under is
(v
Q
as right E-modules.
We may regard
V
a s a left E-
2 16
CHAPTER 4
module via
,@
@V = $ ( V )
V.
E E,V
E '8 V becomes a left E-module via E
Then
$ ( $ '8 0) = $+ '8
v
E,v
($,+ E
E
V)
Now the canonical map
Since Ii
is an isomorphism of E-modules.
is a direct summand of
identify I.'8 V with its image in E @ V. " E E is I.V, we therefore derive
I. @ " E which in turn implies that 1.v z
8
Now suppose that E Ii, we have
I$
1.V
Im$C IiV
(8$)( V ) = @ . ( v
that
:
3
1
Lz
Ii 8 V
Because the image of
in
V
E
I.V
I.V. 3
I.V is an R-isomorphism and fix
-f
3
so Im(e@)
for some
)
v
E, we may
V
5I j V .
E V.
Because
I.V 3
= @
V E
v.
.(V),
3
Given
we deduce
Thus
@ .(84J)(v) = @ . ( V ) = (8@,( V ) 3 3 1 and so
8@ = @.(6$)
c2
3
If we now define A : Ti + I
j
morphism of right E-modules.
by
A(@)
=
I j
8$, then
is obviously a homo-
A similar argument shows that the mapping
)l:I-Ii j defined by A,
p($) = 8
-1
q)
the result follows.
5.13. LEMMA.
Let R
.
is also an E-homomorphism.
be a local ring.
Since 1.1
is the inverse of
Then all finitely generated projective
R-modules are free. Proof.
v
1
,...,v
Let
V
be a finitely generated projective R-module, and let
be a minimal generating set for
V.
Suppose that we are given a
relation
r 1v 1 + rz vz + . . . +
r n vn = O
(Pi E R )
INDUCTION FROM A l
We claim that all such r
i
E
217
Indeed, if pi 4 J ( R ) , then it is a unit in
3(R).
R, and we can solve for the corresponding
vi
as an R-linear sum of the
This, however, is impossible since otherwise the generator
V
j
can be deleted.
vi
r.E J(R).
Thus all
W be a free R-module freely generated by
Let
u 's.
by sending w i
vi.
to
contained in J ( R ) E / .
W/Vr
..., n
and map
W
By the foregoing the kernel of this map, say
By hypothesis, V
W Since S
W1,W2,
=
S 8 V'
S, is
is projective so that we have V'
with
it must be finitely generated.
V
Furthermore,
and hence the inclusion S 5 J ( R ) V forces S = J(RIS. lemma (Lemma 2.3.7) we conclude that
W onto
Invoking Nakayama's
S = 0 and hence that V
W.
So the
=
End(fi).
lemma is true.
Then
be an indecomposable G-invariant A -module and let
V
Let
E
g
End(V) is a local ring (Lemma 4 . 3 ) ,
F
so any finitely generated projec-
1
tive E -module is free (Lemma 5.13).
In particular, if a right ideal
I of E
1
is a direct summand of
then I is a projective E -module since
E,
free E -module (Proposition 5.1).
Thus
1
tifying
I is in fact a free E -module.
Iden-
1
V
and
5.14. THEOREM.
ideals.
7 63 V , Let
V
we now prove be an indecomposable G-invariant A -module, let
and let E = @ Ii be a decomposition into indecomposable right
E = End(@)
(i)
B is a
1
Then the following properties hold:
9 = @ 1.V
(ii) I.e! I z
i
is a decomposition into indecomposable A-modules as
E-modules if and only if
1.V 2
2
I,V 3
as A-modules
(iii) dimI,V = ( E -rankla)(dimV) F Z
(iv) If
V
1
Z
is irreducible and
F
charF
[ \GI,
A- module. Proof.
(i) Owing to Lemma 5.12(i),
vc
=
8 IiVG
then each
I .V
is an irreducible
CHAPTER 4
218
so we need only verify that
ly have IiV C_ Ii#.
IiVG for all i.
=
Iifl =
Because
-
spaces of the form g €3 V,
9
71 E
V,
V
5 VG,
we obvious-
be defined by
and since
fl
for
is a sum of sub-
we need only show that
CPc; Let x E E
Since
To prove the reverse inclusion, observe that I.= @/?
I$ E E .
some idempotent
IV.
8 V)
x ( v ) = $,v 9
E
5 IiV
.
V
g
for all
E
G
and for all
we have
G
and thus I.V = I.V
,
proving (i)
.
(ii) Direct consequence of Lemma 5.12(ii). (iii) We first claim that the map
is an F-isomorphism. IGi,
Indeed, by Proposition 5.1,
6' is a free E -module of rank
so
1
$ be a unit in E Then 0 (i8 V ) = g -1' 9 g the above map is surjective. This substantiates our claim.
For each g E G,
let
8
v
and so
li is a direct summand of E,I. €3 V is identifiable with its
Because
Z
image in E €3 V
E 1
and so the induced map
El
G
I . 8 v - V El
is injective.
Taking into account that dim(I 8 V) = ( E -rankI.) (dimV) F iEl Z F
and that
T.V
is the image of
I. 8 V ,
(iv) Direct consequence of (i) and Corollary 5.3. Assume that
E
=
End(l/;).
V
is an absolutely indecomposable G-invariant A -module and put
Then, by Theorem 5.4,
where the group
.
the required assertion follows.
1
E/E'J(E
)
e F"G
for some
a E Z2(G,F*),
Z2(G,F*) is defined with respect to the trivial action of
G
on
INDUCTION FROM A
F.
E.J(El)
Since, by Theorem 5.4,
219 1
is a nilpotent ideal of
E, it follows from
Lemma 5.11 that any decomposition F'G
= L
=
Q L
2
F"G
into indecomposable right ideals of
E
1
Q
into indecomposable right ideals of
L
~
can be lifted to a decomposition
...
I 0I Q 1
... Q
2
E.
In
We are at last in a position to prove
the following generalization of a theorem of Conlon (1964). V
Let
5.15. THEOREM.
be an absolutely indecomposable G-invariant A -module.
Then, in the notation above, the following properties hold: (i)
fi
=
0 IiV is a decomposition into indecomposable A-modules
L
(ii) L .
z
as FaG-modules if and only if
j
1.V
z
I.V as A-modules 3
(iii) dimI .V = (dimL.) (dimV) F " F" F Proof. Properties (i) and (ii) follow from Theorem 5.14(i), (ii) and L e m a 5.11(iil.
Thanks to Theorem 5.14(iii), to prove (iii) we need only show that for
1 = 1
i
E -rankI
=
dimL
1
Because
F"G
is identifiable with
F i
E/E*J(EI), the latter will follow provided we
prove that for n = E -rankI 1
dim I F Choose an E -basis
{dl,d
summands of
,dn}
{dl,d2,..
,...,dn,dn+l,...,dm 1
r+
of
E
m
and so
= 1
of
JCE
) 1
@
F and since
(6)
~1
I and extend it to an E -basis
(this is always possible since all direct
E are free E -modules). n
Since F
E - S ( E ~ ) ) / E * J ( E ~ )=I
Then
rn
CHAPTER 4
220
(I + E*J(E1)l/E*J(E1)is a unique F-linear
it follows that each element in
combination of the images of d l ,d2,.
..,d,
in
(I + E*J (El 1 1 / E * J (El> .
This
proves ( 6 ) and hence the result. The following simple observation will enable us to take f u l l advantage of some of our previous results. Let V
5.16. LEMMA.
an indecomposable direct summand of of
W, respectively.
V and
w
be an indecomposable G-invariantA -module and let
fi.
Then n
multiplicity of the isomorphism type of
n
Denote by
and m/n
divides m V
and m
be
the F-dimensions
is equal to the into
in the decomposition of
wA 1 In particular, V
indecomposable A -modules. 1
an indecomposable direct summand L Proof.
By hypothesis,
fl
=
of
W@X
4
is extendible to A
if there is
such that d i d = dimv.
F
for some A-module
F X and therefore
Note also that
where all the summands are isomorphic to
V
since
V
is G-invariant.
It
follows, from the Krull-Schmidt theorem, that all indecomposable components of hl A1
are isomorphic to
V.
t denotes the number of such components, then rn = n t
If
and the required assertion follows. 5.17. THEOREM.
Let
I: be an algebraically closed field and let
invariant indecomposable A -module. 1
charF
)
lG/,
Proof.
summand of
Then
'
V
Suppose that
is extendible to
A.
Owing to Lemma 5.16, it suffices to exhibit an indecomposable direct VG having the same F-dimension as
ideal of F-dimension 1.
V.
Now the assumption that
FG, while the assumption that charF
Since F
be a G-
G is cyclic and that
By Theorem 5.15(iii), the
latter holds provided the twisted group algebra F'G
FO"C
v
./'
IGI
has an indecomposable right G
is cyclic guarantees that
ensures that FG
is semisimple
is algebraically closed, all irreducible FG-modules (equivalently, all
INDUCTION FROM A
indecomposable right ideals of able right ideals of
F'G
FGI
221 1
are one-dimensional.
Thus all indecompos-
are one-dimensional and the result follows.
This Page Intentionally Left Blank
223
5 Primitive and prime ideals of crossed products
P r i m e and p r i m i t i v e i d e a l s a r e i m p o r t a n t b u i l d i n g b l o c k s i n noncommutative r i n g
theory.
I t i s t h e r e f o r e a p p r o p r i a t e t o i n v e s t i g a t e t h e r e l a t i o n s between prime
R
and p r i m i t i v e i d e a l s of
G
R.
over a r i n g
R*G,
and
R*G
where
i s a c r o s s e d p r o d u c t of a group
f i n i t e , s i n c e v i r t u a l l y n o t h i n g i s known i n t h e g e n e r a l case.
P
0
over a f i e l d .
P
of
R*G
R = 0 and G-prtme i d e a l s of a c e r t a i n f i n i t e - d i m e n s i o n a l a l g e b r a We t h e n a p p l y t h i s correspondence t o examine t h e r e l a t i o n s h i p
R*G
between t h e prime i d e a l s of
and t h e G-prime i d e a l s of
R.
The i n f o r m a t i o n
o b t a i n e d a l l o w s us t o prove t h e I n c o m p a r a b i l i t y and Going Down Theorems. u s i n g a d i f f e r e n t approach, w e a l s o e s t a b l i s h a Going Up Theorem. s e c t i o n is devoted t o some a p p l i c a t i o n s . t h a t t h e prime (or p r i m i t i v e ) r a n k of
1GI-I E R,
is
One of o u r a i m s
i s t o e s t a b l i s h a b i j e c t i v e correspondence between t h e prime i d e a l s satsifying
G
Throughout we d e a l e x c l u s i v e l y w i t h t h e case where
By
The f i n a l
These i n c l u d e t h e proof of t h e f a c t
R*G
i s e q u a l t o t h a t of
t h e n t h e prime ( o r p r i m i t i v e ) r a n k of
R
R
and t h a t i f
i s e q u a l t o t h a t of
RG,
1. PRIMITIVE, PRIME AND SEMIPRIME IDEALS Throughout t h i s s e c t i o n ,
R
denotes a ring.
Our aim is t o p r o v i d e r i n g - t h e o r e -
t i c i n f o r m a t i o n r e q u i r e d f o r subsequent i n v e s t i g a t i o n s . W e s t a r t w i t h a few well-known
i f f o r a l l nonzero i d e a l s c a l l e d a prime
idea2
if
A , B of R/P
definitions.
R
w e have
is a prime r i n g .
The r i n g
AB#O.
R
An i d e a l
i s s a i d t o be prime P
of
R is
Prime r i n g s and prime i d e a l s are
i m p o r t a n t b u i l d i n g b l o c k s i n noncommutative r i n g t h e o r y . W e say t h a t
R
is
semiprime i f i t i s a s u b d i r e c t p r o d u c t of prime r i n g s o r
e q u i v a l e n t l y i f t h e i n t e r s e c t i o n of a l l prime i d e a l s of
R
i s zero.
An i d e a l
I
224
CHAPTER 5
i s s a i d t o be s e m i p r i m e i d e a l i f
R
of
i s a semiprime r i n g .
i s s a i d t o be a p r i m i t i v e ring i f it h a s a f a i t h f u l i r r e d u c i b l e
R
The r i n g
R/I
Such a r i n g should r e a l l y be c a l l e d left
module.
a r e l e f t modules.
p r i m i t i v e f o r a l l modules u s e d
r i g h t p r i m i t i v e rings.
We could s i m i l a r l y d e f i n e
In f a c t
t h e s e c o n c e p t s are d i s t i n c t , a s examples show, b u t w e s h a l l o n l y b e d e a l i n g w i t h l e f t p r i m i t i v e r i n g s and i d e a l s and t h e r e f o r e omit t h e q u a l i f y i n g a d j e c t i v e .
I
ideal
i s s a i d t o be p r i m i t i v e i f
R
of
R/I
i s t h e i n t e r s e c t i o n of a l l p r i m i t i v e i d e a l s of
R
Assume t h a t
A
R
c o n t a i n s no nonzero
A
of
c o n t a i n s a nonzero i d e a l
1s a n a r b i t r a r y prime i d e a l In Thus
R.
0.
i d e a l of square Proof.
Thus J ( R )
is a primitive ring.
R i s semiprime i f and o n l y i f
The r i n g
1.1. LEMMA.
An
R,
A'A = 0 c -P
then
0.
square
P
If
A C_ P.
and t h e r e f o r e
i s c o n t a i n e d i n t h e i n t e r s e c t i o n of a l l such prime i d e a l s and so
R
is
n o t semiprime. Conversely, assume t h a t nonzero
R
R,
ci E
X
=
Bn
E
R
X n {@I= @.
so
CY = c1
A
and
B
properly contain If
1.2.
# 0,
ananan # 0, and we p u t
n n n.
Owing t o Z o r n ' s lemma, t h e r e e x i s t s an i d e a l
rn
be i d e a l s of
P,
We now c l a i m t h a t
ci
m+l
= max(i,j),
LEMMA.
Proof.
P,
Since
R
A
with
then c l e a r l y
=
we have
Let
R
so by t h e maximality of
'm+l Since
ax
P
i s prime.
P
of
Because
9 P, t h e r e s u l t w i l l f o l l o w .
Let
i,j.
of nonzero e l e m e n t s of
Second, g i v e n
we have
P n X = 0.
maximal w i t h r e s p e c t t o
..I
2
Given a
= a B a
n+1
0 @ X,
,a , ...,an,.
= c1.
R,
CY NGW
1
F i r s t put
Thus, f o r some
0.
c o n t a i n s no nonzero i d e a l of s q u a r e
d e f i n e a sequence
i n d u c t i v e l y a s follows.
(RanR) # 0.
R
CYm'marn AB
E
c1
m
P.
E P + A , CY
rn
5P
P which p r o v e s t h a t
is primitive,
R
Then
R
E P+B
+
P+A
Then
P, a , € P + A ,
(P+A)(P+B)
be a p r i m i t i v e r i n g .
R
P,B
CY
j
E P+B
P+B
f o r some
and t h e r e f o r e
AB
P
and
i s prime.
.
i s prime and semisimple.
h a s a f a i t h f u l i r r e d u c i b l e l e f t module
V.
PRIMITIVE, PRIME AND SEMIPRLME IDEALS
J ( R ) V = 0,
Since
A
suppose t h a t
we conclude t h a t
B
and
V
a c t s f a i t h f u l l y on
. V,
module of
i s prime. 1.3.
R.
proof.
RyR
B # 0
V = BV.
But t h e n
i s semisimple.
AB
and
BV
i s an i d e a l , w e s e e t h a t
x
E
Z(R) and l e t
i s a nonzero i d e a l of
x = 0,
a s required.
R
Let
LEMMA.
AB = 0,
with
R
that
0.
=
Since
R
i s a nonzero R-sub-
so t h a t
A V = ABV = 0
Now,
R
A = 0
and
R
i s not a zero
I n p a r t i c u l a r , t h e c e n t r e of a prime r i n g i s an i n t e g r a l domain.
Let
therefore 1.4.
B
i.e.
A nonzero element i n t h e c e n t r e of a prime r i n g
LEMMA.
divisor in
and
R
a r e i d e a l s of and
and s o
J ( R ) = 0,
225
. R.
=
0
0# y
f o r some
E
R.
Then
R, RZ = 0 and
Hence, b y t h e primeness of
A,B
be a semiprime r i n g and l e t
BA = 0.
then
q
R.
b e i d e a l s of
If
I n p a r t i c u l a r , l e f t and r i g h t a n n i h i l a t o r s of i d e a l s a r e
equal. proof.
If
AB = 0,
then
(BAl2
B(AB)B = 0
=
and s o
BA = 0.
2. PRIMITIVE IDEALS I N CROSSED PRODUCTS
Throughout t h i s s e c t i o n , Given a c r o s s e d p r o d u c t for a l l
g
E
-1
gr=Ljrg Let
X
G d e n o t e s a f i n i t e group and R R*G
G, and p u t
of
G o v e r R,
= 1.
an arbitrary ring.
r
A s usual, given
E
R
lG1-l
R
E
or
X
i s i n h e r i t e d from
G is solvable.
Let
H
be a subgroup of
i r r e d u c i b l e R*G-module, t h e n
VH
G
X is
Then
X
not i nher i t ed
The main r e s u l t of t h i s s e c t i o n R*G
to
R
and t h a t t h e converse
As a by-product of t h e p r o o f , w e
o b t a i n some i n f o r m a t i o n on t h e p r i m i t i v i t y of
Proof.
g E G,
(R*G) g’ we put
d e n o t e t h e c l a s s of r i n g s whose p r i m i t i v e i d e a l s a r e maximal.
a s s e r t s t h a t t h e property
LEMMA.
in
.
by s u b r i n g s and o v e r r i n g s as e a s y examples show.
2.1.
g
and
i s o b v i o u s l y s t a b l e under homomorphisms a l t h o u g h t h e p r o p e r t y
is true i f
-
w e choose a u n i t
R
and
such t h a t
R*G.
1Hl-l E R.
If
V
i s an
i s completely r e d u c i b l e o f f i n i t e l e n g t h .
Owing t o Theorem 4.2.2,
VR
i s completely r e d u c i b l e of f i n i t e l e n g t h .
226
CHAPTER 5
(VR)H
Therefore
s i t i o n 4.2.4.
i s completely r e d u c i b l e of f i n i t e l e n g t h , by v i r t u e of PropoNow t h e map
d e f i n e d by
i s a s u r j e c t i v e R*H-homomorphism.
Hence
VH
i s c o m p l e t e l y r e d u c i b l e of f i n i t e
length. 2.2.
THEOREM (Lorenz ( 1 9 7 8 ) ) .
R
t h e p r i m i t i v e i d e a l s of
1GI-l E R
or
Proof.
G
is solvable.
P
R
i s a p r i m i t i v e i d e a l of
-1
Note t h a t
P
hence
M.
-P C 3
a r e maximal, t h e n
The converse i s t r u e i f e i t h e r
a r e maximal.
Suppose t h a t
maximal i d e a l
R*G
If a l l p r i m i t i v e i d e a l s of
g E G,
Given
put
'P
=
;Pi , F
s i n c e e q u a l i t y would imply t h a t
would b e maximal.
P
=
3
properly contained i n a
&
and
E
=
-1 § M g f o r some
@ g E G
. and
T h e r e f o r e , by P r o p o s i t i o n 2 . 6 . 5 ( i f ) ,
(R*G)P c ( R m Z . Let
hi
Then
V
P
be an i r r e d u c i b l e R-module such t h a t
= ann
(v), and
let
W =
8.
i s o b v i o u s l y of f i n i t e l e n g t h and hence we may f i x a f i n i t e composition
series
Set
Q
i
=
ann(Wi-l/Wi),
1
< i Q n.
Then, by P r o p o s i t i o n 2 . 7 . 1 6 ( i i ) ,
ann(w) =
(R*G)F
and so
...
Q ~ Q ~Q c ann(w) nSince
=
(R*G)P c (R*G)E
(R*G)E i s a p r o p e r two-sided i d e a l of
N of R*G c o n t a i n i n g
i t follows t h a t
Wi-l/Wi
(R*G)&
and o b t a i n
R*G,
&z. C -N
c o n t a i n s a copy of
we can choose a maximal i d e a l
f o r some
@ V
i.
f o r some
Because
g E G
and hence
PRIMITIVE IDEALS I N CROSSED PRODUCTS
-
g 8 'J
b e i n g G - i n v a r i a n t c o n t a i n s a copy of
and hence
Q. C
N.
g
for a l l
IG/ = 1 i s t r i v i a l , w e may assmue t h a t f o r groups of s m a l l e r o r d e r .
If
N
P
To t h i s end, l e t
V
W.
and
3Q
and
L.
Let
H
t r e a t t h e cases
H # G
and
H
Assume t h a t
H # G.
V
module
1
G
R*H
=
Q
R*G.
= ann(W)
W e have t o show t h a t
.
P
Set
= ann(V ) . 1
VR
c o n t a i n s an i r r e -
L.
I n what f o l l o w s w e
separately.
I(
V
WH
The r e s t r i c t i o n
f o r some i r r e d u c i b l e of
W
E*H-
i s completely
1
I n case
1GI-l E R
t h i s follows from Lemma 2 . 1 and
i s s o l v a b l e it f o l l o w s from t h e f a c t t h a t i n t h i s c a s e
R.
P=Q.
f a r some i r r e d u c i b l e R*G-modules
b e t h e i n e r t i a group of = G
G, t h e n
G i s simple.
By Theorem 4 . 2 . 2 ( i i ) ,
r e d u c i b l e of f i n i t e l e n g t h . when
lC\ 3 1 and t h a t t h e a s s e r t i o n is t r u e
are p r i m i t i v e i d e a l s of
= annCV1
A.
Since the case
/GI.
By Theorem 4 . 2 . 2 ( i ) , t h e r e s t r i c t e d module
d u c i b l e submodule
It follows t h a t
i s a p r o p e r normal subgroup of
Hence, we may h a r m l e s s l y assume t h a t
P
G.
E
Qi i s a nonmaximal p r i m i t i v e i d e a l of
Therefore
To prove t h e second a s s e r t i o n , we use i n d u c t i o n on
Assume t h a t
227
H = 1 and
Write
f o r some i r r e d u c i b l e R*H-modules
W.
and p u t
Q. =
ann(W.).
Then
n
Therefore
Qi
=
P
.
Qi C_ P 1 ,
s a y , and by i n d u c t i o n
(IHI < /GI), w e
conclude t h a t
Consequently, i t may b e deduced t h a t
G
a n n ( 8 ) = ann(V ) = ann(V) = The embedding of
Wi i n t o
W
P
e x t e n d s t o an R*G-homomorphism
WE
-+
W
which i s
CHAPTER 5
228
W
surjective since
is irreductble.
P
Hence
ann(8) c ann(W) = Q
=
z--
proving the case H # G . Now assume that
VR is isomorphic to a finite direct sum of
Then
L (Theorem 4.2.2(ii)).
copies of
R.
H = G.
Moreover, by Theorem 4.2.2
P n R = ann(L)
Hence
is a maximal ideal of
R
and by assumption on Q
~
R
R,
we obtain
= ngM
SfG for some maximal ideal M (R*G) ( P
The left ideal
flR )
It follows immediately that Q
R.
of
is actually a graded ideal of
5
=
-
R.
automorphism of
E
Let
E
Go
Since G
1.
consist of all those
5.
and hence
Z(E)
E
of
Otherwise, by Corollary
Then Go
g E G
9
such that
is a field since
E
is an inner
for which conjugation
zg =
is simple.
G = G
.
Hence, for each
-
rgi centralizes R. Since fi*G =
ZCE)
c Z(E*G) and we may view E*G c1 : G X
G+
by
we g'
as an Z(E)-algebra.
~
Define
The
EZ
SfG have
G
is a normal subgroup of
is simple, we deduce that
G, there exists a unit r E 9
centre
n R).
G,g.# 1, such that conjugation by
is an inner automorphism of
distinct from
g
P = Q = (R*G) ( P
is simple, we have
2.3.23(i), there exists g
9'
P n R.
R / P n R.
If E*G
by
=
E*G
R*G/(R*G)(P n R ) where
R*C
nR
- - - -1 Cl(~,y) = " Y x ~ .
Then, for all
r,y E
G,
we
have
follows that Y
algebra of
E*G
and obviously
simple subalgebra R*G
K
8- A ,
Z (R) ideal of A .
E,
E*G
-
=
E
the algebra E*G
A.
Because
A
A
is a Z(6-sub-
centralizes the central
h a s the structure of a tensor product,
E
'8 I, where I is an Z(R) (See Proposition 3.1.10 and Theorem 3 . 1 . 7 ) . Obviously I has to and the ideals of E*G
are of the form
229
PRIMITIVE IDEALS LN CROSSED PRODUCTS
R'
@ I
i s prime. (R) deduce t h a t a l l prime i d e a l s i n be prime i f
A
Since
E*G
is a finite-dimensional
a r e maximal,
algebra, we
In particular,
=
Q
thus
8
has
completing t h e p r o o f .
2.3.
THEOREM (Loren2 ( 1 9 7 8 ) ) .
E*G
(i) I f
R
i s prime and
R
(ii) I f
i s prime and
Proof.
(i) L e t
'L
The f o l l o w i n g p r o p e r t i e s h o l d :
is p r i m i t i v e , t h e n
R*G
i s p r i m i t i v e , then
R*G
R
is primitive
is primitive.
b e a f a i t h f u l i r r e d u c i b l e R-module.
Then
w=
f i n i t e l e n g t h and, by P r o p o s i t i o n 2 . 7 . 1 6 ( i i i ) ann(u = A (
a n n ( v ) ) =0
g-=
Fi
F i x a composition s e r i e s
= W
3
W
0
Q. = Then
Q1Q2
... Qn 5 ann(W) =
i, p r o v i n g t h a t
f o r some
v
(ii) L e t
0.
R*G
...
3
W
=
and p u t
Q.
i s prime it f o l l o w s t h a t
R*G
=
0
is primitive.
be a f a i t h f u l i r r e d u c i b l e R*G-module.
n >1
W
0 of
ann(FJi-l/Wi).
Since
VR = f o r some
3
1
w1
@
w2
@
Owing t o Theorem 4.2.2,
... @ wn
and some i r r e d u c i b l e R-modules
wi.
If
Qi d e n o t e s t h e a n n i -
Wi, t h e n
h i l a t o r of
n
Q1Q2
Hence
therefore
Remarks. (e.9.
... Qn 5
R
n Qi i=1
0.
Since
R
i s prime,
R
does n o t i n g e n e r a l imply
c o n s i d e r t h e group a l g e b r a of a f i n i t e group
R = F x F , where
R
'(?I,)J)
f o r some
i and
R*G
F is a
= (p,?I),
G
over a f i e l d
does n o t imply t h e primeness of
field, let
1 , E~ F.
R*G
t o b e prime
F
with
G)
dividing t h e order of
( i i ) The p r i m i t i v i t y of
and p u t
Q. = 0
is primitive.
( i ) The p r i m i t i v i t y of
charp = p > 0
over
=
G =
R.
Indeed, l e t
Then t h e c o r r e s p o n d i n g skew group r i n g o f
i s s i m p l e , hence p r i m i t i v e , b u t
R
2
b e t h e c y c l i c group of o r d e r
i s n o t prime.
(iii) I t w i l l be shown ( s e e Theorem 6.4), t h a t t h e second s t a t e m e n t of Theorem
G
CHAPTER 5
230
G.
i s t r u e f o r any f i n i t e group
2.2
However, t h e proof o f t h i s r e q u i r e s a form-
i d a b l e t e c h n i q u e t o be developed i n subsequent i n v e s t i g a t i o n s .
3. PRIME COEFFICIENT RINGS
G
Throughout t h i s s e c t i o n ,
R
i s a prime r i n g .
G-prime i f B
or
C 2 -
ideals
B1-B
A.
P
C 2 -
R
Let
R.
i n v a r i a n t i d e a l of
A
d e n o t e s a f i n i t e group and, u n l e s s s t a t e d o t h e r w i s e
A c - ii*G
be an a r b i t r a r y r i n g and l e t
be a G -
A
Following Lorenz and Passman ( 1 9 7 9 ) , we s a y t h a t for
Bi
G-invariant i d e a l s
of
R
implies t h a t
B1
is
5A
Our a i m i s t o e s t a b l i s h a b i j e c t i v e correspondence between t h e prime of
R*G
?
satisfying
f i n i t e dimensional a l g e b r a
E
n R
0
=
and t h e G-prime i d e a l s of a c e r t a i n The r e s u l t s o b t a i n e d w i l l be
over a f i e l d .
a p p l i e d i n t h e n e x t s e c t i o n i n which w e examine t h e r e l a t i o n s h i p between t h e prime
R*G
i d e a l s of
R.
and t h e G-prime i d e a l s of
We s t a r t by r e c o r d i n g two g e n e r a l o b s e r v a t i o n s , namely Lemmas 3 . 1 and 3.2 below, which a r e v a l i d f o r an a r b i t r a r y r i n g 3.1. of
R.
ideal
p
R*G
of
A
I
P
=
A =
A
R,
R
A*G
and
i s a G-prime i d e a l
t h e n t h e r e e x i s t s a prime
R*G,
i s a two-sided
R*G.
B
I
then since
i s a G-invariant i d e a l of
i s a prime i d e a l of
Suppose
Pn
then
A.
i s an i d e a l of
A. (R*G) = (R*G) - A
l e a s t G-invariant.
c ? n R.
p n R
Conversely, i f
Assume t h a t
R*G,
i s a G-prime i d e a l of
such t h a t
Note t h a t i f
I n R.
2.6.3(ii),
AB
i s a prime i d e a l of
Conversely, i f
Proof.
so i s
p
If
LEMMA.
R.
i deal of
R,
i s G-invariant t h e n by Theorem
R*G
Then, by t h e above,
a r e G - i n v a r i a n t i d e a l s of
with
P R
n R
is a t
with
Then (A*G) (B*G) = (R*G)AB(R*G)
c
(R*G) (P n R ) (R*G)
C P so t h e primeness of
B 5 P n R,
P
proving t h a t
implies t h a t
P n R
A*G
5P
i s G-prime.
or
B*G
5 P.
Thus
A EP n R
or
PRIME COEFFICIENT RINGS
A
Conversely, l e t
R
i d e a l of
A x G
and s o
R.
be a G-prime i d e a l of i s an i d e a l of
R*G
P nR
to
A.
=
J n R
and
I,J be i d e a l s of
Let
R
a r e G - i n v a r i a n t i d e a l s of
(I n R ) ( J
G-prime t h i s y i e l d s
R*G
n R)
A
A
Then
i s a G-invariant
n R
(A*G)
with
P
Z o r n ' s l e m m a , we may t h e r e f o r e choose an i d e a l
231
of
fi*G
A.
=
maximal w i t h r e s p e c t
P.
properly containing
A.
properly containing
P.
IJ
and hence
Applying
Then I
A
Since
P is
Hence
R
is
prime
as r e q u i r e d .
T
If
LEMMA.
I
Let
*,...,gn}
T = {gl,g
i
E {1,2,
G
be a s u b s e t of
g
with
R*G
wlth
=
and l e t
Assume t h a t
= 1. 1
...,n } ,
In
and
for a l l
T'
C
T
E I with
r
=
r .}
(R*T') = 0
Bi by
define
n B.
R*G
be a nonzero l e f t and r i g h t R-submodule of
I n (R*T) # 0 For each
f o r t h e s e t of e l e m e n t s of
T.
support i n 3.2.
R*T
G, w e w r i t e
i s a subset of
{r E R l t h e r e e x i s t s
fi
a nonzero i d e a l o f
R
=
C r
.;
j=1 3 3
z
Then
Bi i s
( i ) Each
f 7,.
( i i ) There e x i s t s a n a t u r a l b i j e c t i o n
(a) f i b a t ) = r f . ( a ) g i ,
for all
:
B1-
Bi
r,t E R
such t h a t
a
and
E
B 1
i s t h e i d e n t i t y map
(b) f 1
In
( i i i ) The elements of
(R*Tl
a r e p r e c i s e l y t h e e l e m e n t s of t h e form n
Proof.
Because
B
t h a t each each
i
Bi i s
I
i s a l e f t and r i g h t R-submodule
R.
i s a n i d e a l of
nonzero.
Note a l s o t h a t f o r each
and
(3'
s m a l l e r s u p p o r t and so
-
fi
- 0'
=
By t h e f o r e g o i n g , f o r each
fi
b
= a. 1
7,
i'
ensures t h a t
t h e m i n i m a l i t y of
T
=
0.
a
E
B
-
n 1 b.;. E I i=l 2 Bi by s e t t i n g f . (a)= b ..
t h e r e e x i s t s a unique 1
with
b. E B
it is c l e a r
T h i s i s so C r j g j E I w i t h ri = b.. 2 j=1 a r e two such e l e m e n t s , t h e n 6 - B ' E I i s an element of
i m p l i e s t h a t t h e r e e i x s t s a unique
fi
T
Moreover, t h e m i n i m a l i t y of
n
s i n c e if
R*G,
of
W e may t h e r e f o r e d e f i n e
f
:
B1
a
=
232
CHaPTER 5
fi,f,
By the definition of
is the identity map and
n
Conversley, it is obvious that each element of
I n (R*T) is of this form.
Moreover, each fi is clearly an additive bijection. Finally, let a €5 B
and
be as above and let r,t
CY
E
R.
Then
-
and since g, = 1, this implies that
So the lemma is true.
.
For the r e s t o f t h i s section,
R
w i l l be a prime r i n g .
In order to make
further progress, we briefly discuss a certain ring of quotients
S =
which
QoL(R)
is defined in Martindale (1969).
X of all left R-module homomorphisms
Consider the set
where
f
I?
:
A
A
ranges over all nonzero ideals of
-3
with
R
g
and
C EA n B
:
B--+
of
and with
R, A n B
We call two such homomorphisms
R equivalent if there exists a nonzero ideal C of f(c1 = g ( c )
obviously reflexive and symmetric. ideals A , B
R.
c
for all
E
C.
It is a l s o transitive, since for all nonzero
is a nonzero ideal of
an equivalence relation on the set X.
R.
For each f E
We have therefore defined
X, let [ f , A j
no confusion can arise) denote the class of the function f define an arithmetic on the set Qo(R)
I f 4 where
fg
This relation is
IgrBI
above.
(or
=
called the Martindale ring of q u o t i e n t s of
We can
[fg,BA] In this way we obtain a ring,
R.
The following five lemmas provide some basic properties of the ring For any f : A
--+
if
of these classes by
is the composition first f then g.
3.3. LEMMA.
7
R in X,
7=
0 if and only if f
=
0.
Qo(R).
In
PRIME COEFFICIENT RINGS
particular if g
B-
:
R
233
x,
is another element of
-
-
f=
then
g
if and only if
on A n B .
f = g
Then there exists a nonzero ideal B
F=O.
7=
It is clear that f = 0 implies
Proof.
that f ( B ) = 0.
5B
Now BA
R
of
F=
.
For every a E R ,
let a
E HomW,R)
‘ 3.4. LEMMA.
contained in A
The second
if and only if
a
be defined by
(2)=
za.
R
4
defined by
a
is an element of
X, then
aTf
Qo(R1
R
(i) The map
and such
and hence that f = 0.
assertion follows from the first and the fact that
f-g=o.
Conversely, assume that
and so
is prime, it follows that f ( A ) = 0
Since R
0.
C--+
ar
is an injective ring
homomorphism (ii) If f
:
A
R
-+
(if If a,b E R ,
proof.
then
7r
=
1, a b
Hence the given map is a ring homomorphism. 3.3.
Hence
a
x
every
m,
(ab)
and
.r
ar = 0 ,
If
then
a
R
E A,
C A, a-
(a,f) proving that
hence
f
is defined on R
U
“,f
(5) =
cf -
=
=
f(a),.
.
f(xa)
=
xf(a)
In view of Lemma 3.4(i), we may identify R
=
a? 3.5. LEMMA.
Let
sE
S
Asi
2
=
S = Qo(R)
As
and
s ,s 1
each
(a+b),
0, by Lemma
.
Therefore for
f(e),(x)
with its image in
this identification, Lemma 3.4(1i) tells us that for f : A
(ii) If
=
U
=
R,
E
(i) If
ar+br
0 and (i) follows.
=
(ii) For every
=
r r
=
,...
,S
=
f(a1
for all
Qo(R).
R
in
a E A,
7~
-f
With
X, Q~(R)
(1)
be as above
0 for some nonzero ideal A
of
R,
E 5, then there exists a nonzero ideal
then s = 0 A
of
R
with
in R.
(lii) S is a prime ring (iv). If ism of
S I
S.
i s
an automorphism of R ,
then a
extends uniquely to an automorph-
CHAPTER 5
234
Z(S) = C S ( R )
(v)
ti)
Proof.
f(a)
=
0
s
fore
(ii) L e t a l l the
s
Let
a
for a l l =
Z(S)
and
E
7 = 0. sl,s ,...,s
fi
is a field.
A.
E S
a
-
afi
-
f,
=
t h e n (1) shows t h a t
Then we may h a r m l e s s l y assume t h a t
1 b e a nonzero i d e a l of
S.
i
A.
Then, by (11,
fi(a)
=
E
R
A.
S meets R Let
si.
=
=
0 # As E I n R
(iv)
si
with
a r e d e f i n e d on t h e common domain
€
(iii) L e t
s
If
f v a n i s h e s on an i d e a l i n i t s domain and t h e r e -
Hence
as for a l l
A s = 0.
with
E S
nontrivially.
R
Since
'f
0#
of
R.
I,
S E
and l e t
t h e n by (ii) and (i),
Thus e v e r y nonzero i d e a l of
S.
i s prime, so i s
R
be an automorphism of
0
A
f o r some nonzero i d e a l
If
f
A-
:
R.
Then t h e map
: A'-R
d e f i n e d by
f'(a9
=
i s s u r e l y a l e f t R-module homomorphism.
From t h i s it f o l l o w s e a s i l y t h a t t h e
-
map
f'
?I--+
f(aP
g i v e s r i s e t o a n automorphism of
S
extending
8.
To prove uniqueness of e x t e n s i o n , i t s u f f i c e s t o show t h a t i f S
morphism of and l e t
A
fixing
R
elementwise, t h e n
be a nonzero i d e a l of
R
i s an a u t o -
s
To t h i s end, l e t
T = 1.
A s C_R.
with
T
Then, f o r a l l
€
S
a E A,
we
R,
it i s
have T as = ( a s ) =
Hence (v)
A(s-s')
=
s E CS(R)
Assume t h a t
clear that
T
=
T = 0 by ( i ) . Then t h e map
f
l e f t R-modules.
0 and ( i ) i m p l i e s t h a t and t h a t
a Ts T = asT .sT = s .
s # 0.
Since
s
centralizes
0} i s a two-sided i d e a l of
R
and so w e must have
By (ii), w e may choose a nonzero i d e a l
A
of
{t
:
€
AIf
Rlts
R
=
d e f i n e d by
B = f(4).,
then
f(a)
=
as
B = As = sA
R
with
As
5 R.
i s a n i n j e c t i v e homomorphism of since
s
E CS(RI.
Now t h e r e
235
PRIME COEFFICIENT RINGS
exists an inverse map g 8,
R,
E
S
and
:
R so that fg
B-
is an inverse of -1
s
we must have
E
C3(R)
7=
s.
Because
lA.
B is an ideal of s centralizes
Furthermore, since
C (R) is at least a division ring.
and therefore
S
s induces an automorphism of S which is trivial on R.
Finally, conjugation by
S.
Hence by (iv) the automorphism must also be trivial on
Consequently s
Z(S)
E
and the result follows. Qo(R)
By the foregoing, the structure of next result will indicate that Q ( R )
is very close to t.hat of
R.
Our
i s large enough to contain certain needed
additional units. 3.6. LEMMA.
Let
elements.
u be an automorphism of R and let a , b E R be fixed nonzero
If for all
r
E R,
arb then there exists a unit
=
s E S = Qo(R)
CTu br a such that
-1
s rs=T Put A = RaR, B = RbR
Proof.
TO prove that
f
b
=
as and such that
U
for all
and define the maps
f
:
is well-defined, it suffices to show that
A
-f
r
E
R
B and
Cxsayi
=
0 implies
i that Zxiby:
i
=
0.
To this end, assume that
the formula arb = braaa
and so
Tx.ayi
=
0.
Then for all
r
E
R
27'
Cxiby:
then for all
=
yields
0 since a
U
#
0 and R
is prime.
Similarly, if
Czibyi i
=
r E R we have
since they are clearly homomorphisms of left R-modules, we have
7=
sE
S
and
0,
CHAPTER 5
236
9 E S.
gp
Moreover, fg = lA and
Note that arf
is defined on R (a f)(;c) = 21
.$
Thus
=
gr,
,
lB
=
-
g
so
-1
and
= S
S
and for all
z f R we have
f(m) = xb
br(x)
=
is a unit in
S.
or equivalently as = b .
Finally, let c
E
R.
Then gc f
I3
is defined on
and for all xby tE B
we
and let A , B
be
have
-
c 7 = 'e
Thus
3.7. LEMMA.
u xbye
=
and so S
Let
-1
s cs = ,'c
Qo(R),
=
let
as required.
. R
be an automorphism of
U
Assume that f
nonzero ideals of R.
a cp(xbyl
=
:
B
A+
is an additive bijection which
satisfies
for all r , t E R
r
E
and f ( a l
R
Proof.
and a E A .
for all a
as
=
are elements of
-1
S.
q
Hence gq,f
7
E R.
=
qz
we have
Let u
and
b
f
A
is also an
B,
g are left R-module homomorphisms so that
Furthermore, since g
is an inverse of
Then g q f is defined on
(gq,S) (b)
=
ra for all
S =
f
and
f, we have
.
Let
s
=
A.
Moreover, for all P,t E R
Note also that f and
-
S, s-lrs
is a unit in
Since f is a bilection, its inverse g : B-+
additive bijection.
g = s
e
7
s =
Then
0
=
f(g(b)ql
and this yields
f(a)
= as
for all
be an automorphism of
=
-1
s qs a
R.
B and for all b E B ,
(gf) (bq') =
=
bq'
=
we have
qz(b)
qa for all q
R.
Finally, because
A. Following Kharchenko (19751, we say that
is X-inner if it is induced by conjugation by a unit of
words, these automorphisms arise from those units s
E S
S = Q,(R).
with
s-lRs = R.
In other If
231
PRTME COEFFICIENT RINGS
s
s
and
a r e t w o s u c h u n i t s , t h e n c l e a r l y so i s
2
s1s2.
Invoking L e m m a 3.5
R
( i v l , we s e e immediately t h a t t h e s e t of a l l X-inner automorphisms of Aut ( R )
normal subgroup of
R*G
Now l e t Then
G*
R
G*/U(R)
Bearing i n mind t h a t
{ g E Glr
=
G* be t h e g r o u p o f g r a d e d u n i t s o f
i s a normal subgroup of
t--t gi-
G
w e deduce t h a t
G.
r
k-+
'r
{gig
be t h e f r e e S-module f r e e l y g e n e r a t e d by
E GI
R
of
which w e d e n o t e by t h e same symbol.
S
R1
i s an X-inner automorphism o f
Owing t o Lemma 3.5 ( i v ) , t h e automorphism automorphism of
P
R*G.
U ( R ) s u r e l y g e t a s X-
by c o n j u g a t i o n and t h e e l e m e n t s of
i n n e r automorphisms.
Ginn
.
b e g i v e n and l e t
a c t s on
is a
e x t e n d s t o a unique
We now d e f i n e
S*G
to
and w i t h m u l t i p l i c a t i o n g i v e n
by
(a5( b y ) = a"bEa(33,ylrcy for
a,b
E
S
x,y
and
H e r e o f course
E G.
c1 : G
--
a(r,y) = "y
G
3.8.
S
over Let
LEMMA.
E
=
Cs*,(S),
t w i s t e d group a l g e b r a of
H
(iii) I f
S*G
S*G.
L
If
(S*G)Ln
s
Proof.
F
5 S*Ginn Ginn
,
S*G. =
,
S*G i n n = S €3 F E
and
some
F.
E,
and
E
then
f3
t (S*H) = F H
L ( S * G ) = (S*G)L
( S * G ) L c o n s i d e r e d as a l e f t S-submodule of Furthermore,
t
E = F Ginn,
then
(E n (S*H))
i s a G - i n v a r i a n t i d e a l of
Moreover,
summand o f
= S €3
i s i n f a c t a unique c r o s s e d
S*G e x t e n d i n g R*G
over the f i e l d
'inn
i s a subgroup of
S*H (iv)
E
then
i s d e f i n e d by
F = Z(S).
and l e t
(i) There e x i s t s a unique c r o s s e d p r o d u c t (ii) I f
U(R)
R*G.
extending
S = Q (R)
-+
q
The f i r s t p a r t of t h e lemma below shows t h a t p r o d u c t of
x G
--1
(S*G)L n (S*Ginn) = SL
and i f
S*G
i s a n i d e a l of is a direct
L # E,
then
0. ( i ) B y Lemma 3 . 5 ( i v ) , t h e automorphisms of
R
extend uniquely t o
2 38
CIIAPTER 5
S.
automorphisms of R*G.
extension of
R
S*G is the only possible
It therefore suffices to verify the associativity of the
Let G’
multiplication. acts on
Hence, the above definition for
Note that G*
R*G.
be the group of graded units of
and therefore the uniqueness of extension implies that we obtain a
G* on S .
group action of
This fact will be used implicitely in the computa-
Given a , b , c
tions below.
E
S and x,y,z E G, we have [
(&I ( b y )I (cz)
=
-
s 1 xyz
and
s l , s z E S.
f o r some
An
Ccyz
[ ( b y )( G ) I= s 2
c&)
easy computation shows that
and s
where
r
and r
and
case when
a
Hence
s
=
b
=
c
and 2
(ii) Let Y E E (i),
(ii),
=
a,b
and c.
StG
Since
a(x,y)xyca ( x , y )-1 But in the special
the above products belong to R*G
1
and so r
= 1
r
2
.
is associative.
-
and let x E Suppy, say y = sx
there exists a E R
commutes with
independent of
c terms in the two expressions are equal.
= S 1
=
axb x P c ) r 2
R
are elements of
z(yc?l the a , b ,
=
R*G
ccy E
with
+
... .
and with
3:
Owing to Lemma 3.5
E Suppay.
Since y
r a , we have aray
=
ayra
for all
P E
R
We conclude therefore that
-
arasx Since b
=
as
=
-
asxra
is a nonzero element of
R,
=
as xrxax
we see that the identity
arb = bxyxa holds f o r all r E R.
Invoking Lemma 3.6, we see that
r
x~
is an X-inner
239
PRIME COEFFICIENT RINGS
R and thus
automorphism of
g
For each on R
E
and put
3:
E
This proves that E
Ginn.
Ginn choose a unit s =
for all g
s-ls.
Then the elements g The elements g, g E Ginn
basis for S*Ginn.
2
Applying Lemma 3.5(iv), we see that E
E
5 S*Ginn.
is clear that each S n
%E
E
=
E.
Yg?
centralizes S.
of
$
Since
2
E
.
E
with Y E S and it g g is a unit of E, we have
CY
form an F-basis for
E.
More-
E is an associative F-algebra with
Moreover, for z , y E Ginn
, %;€
E
and
2;
=
s
6
for
t is isomorphic to F Ginn, some twisted group
It follows that E
S.
E
=
s, i.e.
C_ Z ( E ) we must have S*Ginn = S @F E .
{ZIg E Ginn}.
s
some
Y
Then we can write
To prove the last assertion, note that basis
To this end,notethat each
must centralize all of
F and we deduce that the elements
since F
over,
Ginn form an S-
S*G which acts by conjugation on S centralizing all of R .
is a unit in
Assume that y
E
are obviously F-lineary indepen-
dent and we claim that they form an F-basis for
z
'r
S inducing the automorphism r
E
g
5 S*Ginn.
over the field F .
Ginn
(iii) Direct consequence of the way the algebra FtGinn is constructed. (iv) Let LS = S L
L
be a G-invariant ideal of
and the G-invariance yields
follows that
L
=
5 ,;
Then, by definition of for all g
SL
€
in E.
Since
S*Ginn = S @ E
=
G.
E, we have
It therefore
S*G.
is an ideal of
Y be a transversal for Ginn in G and let L'
Let for
L ( S * G ) = (S*G)L
E.
SL 0 S L ' ,
be the F-complement
we have
F S*G =
c
c
SLY @
YEY
SL'y
YEY
Furthermore, since
L
which implies that
Z (S*Ginn)LY = C SLY fiy yf y (S*G)L is an S-direct summand of S*G.
is G-invariant, (S*GIL =
Z SLY is
Moreover,
L@y
clearly a direct sum and thus
If
L # E,
But then
then we can choose L '
1 SL'y fi7
contains S
above to contain the identity element
and therefore
(S*G)L n S
=
1.
0, as required,
'
240
CHWTER 5
From now, w e p u t S
E = CS*,LS)
= Q0(R),
F = Z(S)
and
O u r aim i s t o a p p l y t h e f o r e g o i n g r e s u l t s i n o r d e r t o e s t a b l i s h a b i j e c t i v e c o r r e -
P
spondence between t h e prime i d e a l s
E.
prime i d e a l s of
R*G
of
satisfying
P n R = 0 and t h e G-
G a c t s on E by t h e r u l e
R e c a l l t h a t t h e group
-1
gx=gxg
, r E E , g € G
A
and t h a t , by d e f i n i t i o n , a G - i n v a r i a n t i d e a l f o r G-invariant
L
If
ideals
Ai
E
of
of
E
A
C 1 -
implies t h a t
i s a G - i n v a r i a n t i d e a l of
A A
i s G-prime i f
A
or
A
C
2 -
1
2
A
C -
A.
then w e set
E,
L~ = L(S*G) n R*G L'
so t h a t
R*G
i s an i d e a l of
by Lemma 3 . 8 ( i v ) .
I
For any i d e a l
R*G,
of
we s e t
P P
Observe t h a t
A y
with
C 1 1 -
zero ideal
=
2
of
5I
f o r some nonzero i d e a l
E.
is a G - i n v a r i a n t i d e a l of
I, A y
B
{y E E l A y
C
I
and l e t
t E E.
A
of
Rj.
Indeed, assume t h a t
€Id YlIYL
By Lemma 3 . 5 ( i i I , t h e r e e x i s t s a non-
2 -
R with
B t &R*G.
and
Because
t centralize R,
YlrY2
we have
and
B A (ty 1 1
y1
Hence
+ yl,ylt,tylE
#
so
(Bt)( A y
=
1
9
)
1 1
i s an i d e a l of
5I E
which i s c l e a r l y G-invar-
iant.
Let that of
h*G
I R , Ax
I be an i d e a l of
R+G.
Following Lorenz and Passman ( 1 9 7 9 a ) , w e s a y
i s R-cancelable i f f o r any
EI
satisfying
implies t h a t
P nR = 0
x E I.
x
E
R*G
and any nonzero G - i n v a r i a n t i d e a l A
I t is obvious t h a t any prime i d e a l
i s R-cancelable.
P
of
241
PRIME COEFFICIENT RINGS
3.9. LEMMA.
Let
L,L
and L 1
(i) L'
E.
be G-invariant ideals of
Then
2
is R-cancelable
(ii) L'*L" 1
2
c (L~L~)'
-
(i) Assume that AX c - L"
Proof.
variant ideal A
of
R.
(S*G)L
Lemma 3.8(iv),
for some x E R*G
Lu,
By definition of
and some nonzero G-in-
we have Ax
is a left S-module direct summand of
for some S-submodule K
of
S*G.
5 (S*G)L. S*G and so
x = x 1 + x 2 with x
Writing
Owing to
E
(S*G)L,x2E Ic
we have
proving (i). (ii)
(S*G)Li = L i ( S * G )
By Lemma 3.8(iv), we have
as required.
and hence
.
The next lemma is crucial. 3.10. LEMMA.
With the notation above, we have
E, L
(i) For any G-invariant ideal L
of
(ii) If I is an ideal of
then
Proof.
(i) Given
some nonzero ideal R therefore L
A of R.
5L . Ud
Hence
y of
R*G,
E L,
R.
=
I s
L
lid
flu.
we have by Lemma 3.5(ii) that By Then B y C L(S*G) n R*G = L",
Conversely, if
y E LZAd,
then Ay C_ L'
so
5 R*G y E LUd
for and
for a nonzero ideal
242
CHAPTER 5
by Lemma 3 . 8 ( i v ) .
y
=
L'
Denote by
y, + y 2 w i t h
y1 t L
Y
and
E L'.
A(y-Y
(ii) Fix
3:
E
i n d u c t i o n on
I. The c a s e Suppl.
YE I
elements
if
y E Y',
Now 2'y
n SL'
E SL
ISuppz = 0
Choose
I n R*T # 0.
= {g
=
1
l,g2,...,g A
There e x i s t nonzero i d e a l s :
Bi
A-
=
R
,...,Rn
,B2
1
.
1 E Ti1
Siii
E
R
f.(a)
and
= as i
-
=
x;
S i n c e it s u f f i c e s t o
T
and
R
of
Ti'.
by
and a d d i t i v e b i j e c t i o n s
R.
t r i v i a l l y on
0 E B.
Lemma 3 . 2 ( i i i )
R.
s
z.ii
that for a l l
a
such t h a t
- --1 = girgi
=
.s'
=
a
Id.
E
asi
for a l l
-
S
Sigi
a
E A,
E
A.
E
=
ax
f,
gi
r
is
x
-
aOr
and l e t
a
acting
F:
= CS*G(S)
i t f o l l o w s from
cf;(a)giE I
be t h e i d e n t i t y c o e f f i c i e n t of
Y
S-lrs=
r E R, we see t h a t
for a l l
E A
z i
A,r,t E R
1.
=
fi(a)
Moreover, s i n c e
E
Furthermore, s i n c e
Invoking Lemma 3 . 5 ( i v ) , w e conclude t h a t
T h e r e f o r e , by d e f i n i t i o n , w e have
r
r
E A.
S
of
t h e n y i e l d s an automorphism of
aa = Cas
Let
7
=
S
Since
But
a
a
for a l l
s1 ,s2,...,s
for a l l
g;
1sI,. gz. E S*G.
centralizes
and t h u s
Indeed,
-1
r f i ( a ) "t
=
there exist units
t h e i n d e n t i t y f u n c t i o n , we have
B
T.
1 E
satisfying
Hence, by Lemma 3 . 7 ,
Let
by
minimal w i t h
g.
r
@
t h e n Lemma 3 . 2 a p p l i e s and we employ i t s n o t a t i o n .
f(rat)
for a l l
E
then
p,
fi
7 ' C _ Suppx
W e may assume t h a t
a l s o h a s t h i s minimal p r o p e r t y and -1 show t h a t xy E w e can r e p l a c e x by
T
x
# 0 and t h a t t h e r e s u l t i s t r u e f o r all
3:
-1
If
0
=
b e i n g t r i v i a l , w e show t h a t
of s m a l l e r s u p p o r t s i z e .
respect t o the property t h a t
and w r i t e
L.
E
Assume t h a t
B
in
Then
Ay
=
)
y = y
and Lemma 3 . 5 ( ) i m p l i e s t h a t
L
a n F-complement f o r
Put
PRIME COEFFICIENT RINGS
Because
Suppa
y
Furthermore,
6E
I?
if
D =
=
I since
E
aRr
and
T C- Suppz, 1
n gA,
E
R*G
uar
D
s
and
w e see t h a t
= 1,
ax
and t h u s
6
i s a nonzero G - i n v a r i a n t
SfG
R
i d e a l of
.
r e s u l t now f o l l o w s by i n d u c t i o n .
y
p.
E
for a l l
E
i s R-cancelable and hence
But, by Lemma 3 . 9 ( i ) ,
)SuppyI*)Supp).
I, so by i n d u c t i o n we have
E
Ul?
so
then
!7'
243
x
E
u E A.
Cx
with
Idu.
Note t h a t Hence,
5 Id".
The d e s i r e d
We have now come t o t h e d e m o n s t r a t i o n f o r which t h i s s e c t i o n h a s been d e v e l oped. 3.11.
THEOREM (Lorenz and Passman ( 1 9 7 9 a ) ) .
G
t h e f i n i t e group i z e r of
S
=
R
o v e r t h e prime r i n g
Qo(R) i n
S*G
where
R*G
Let
and l e t
F = Z(S).
Let
P
G-prime i d e a l of
E.
Proof.
(i)
Pa i s
( i i )'L
P
E
Pa'.
E*G
with
E
=
E ii*G
and
R*G
D =
n
8".By Lemma and
x
Lu n R 3.1O(iilI
(S*G)Fd
E
L
* L'.
=
0, and l e t
and
L
=
L be a
=
0
P
we have
=
I: 0.6 i
6.D
t h e r e e x i s t s a nonzero i d e a l
5P B
for all of
R
ud
5 Pdu.
D
i.
Assume t h a t
Di of
R
with
S*G and 6
i
E
Moreover, by Lemma 3 . 5 ( i i ) ,
B
BB. C R*G z-
for a l l
i s G-invariant.
Pa
0.6 . = & . D S P. a z zz-
is a nonzero G - i n v a r i a n t i d e a l of
such t h a t
t h e f o r e g o i n g argument we may a l s o assume t h a t
ai E
with
i=l
it follows t h a t
L
so
Pd, t h e r e e x i s t nonzero i d e a l s n n gDi,
&G i=1 t h e prime r i n g R w i t h
sion is that
P n R = 0
with
P = Pdu
with
x
Setting
yields a
P nR
n
By d e f i n i t i o n of
P +-+ Pa
W e must prove t h a t
P
x
R*G
be a prime i d e a l o f
i s a prime i d e a l o f
Then
of
be t h e c e n t r a l -
The i n v e r s e of t h i s map i s g i v e n by
a G-prime i d e a l of
W e f i r s t show t h a t
x
E.
t
E = F Ginn
Then t h e map
b i j e c t i v e correspondence between t h e prime i d e a l s and t h e G-prime i d e a l s of
be a c r o s s e d p r o d u c t of
i
and by
The conclu-
244
CHRPTER 5
P
But
x
Thus
P
i s prime and E P,
Pdu
so
T o prove t h a t
pd
M, ,M2
E.
ideals
of
does n o t c o n t a i n e i t h e r
5P
i s G-prime,
Mi
E
Since
Pd i s
Hence
C
# L
( M1 2M ) ~ c _ ~ Z ~ = P cP
M?
L'
L
maximality of
C
R*G.
I
f o r some i d e a l
I
nR
Lu
P
=
E
F,
.
0,
r e s u l t follows.
L
we see t h a t
w e conclude t h a t
ud
Then
I
. 8 - Lud
2&=
Lu.
L
=
3
and
Hence t h e
G-
I n R # O.
and s o
c 11,L' c I
L'
i s prime, assume t h a t
Then, by t h e above,
=
L = L
R*G.
of
II n R # 0
R # 0,
I2
and
f o r some i d e a l s whence
5 1112n R
0 # (Il n 81 (I2n R ) L'
a n d , by Lemma 3 . l O ( i ) ,
G-prime and ( i ) i s e s t a b l i s h e d .
and by Lemma 3 . 1 0 ( i ) , we have
implies t h a t
To prove t h a t
Because
i
f o r some
2 -
s i n c e o t h e r w i s e Lemma 3 . 1 0 ( i i ) would y i e l d
11,12 of
f o r some G - i n v a r i a n t
1 2 -
Furthermore, it i s a n immediate consequence of Lemma 3.8
nR = 0
'L
Assume t h a t
d I
c Pd
M M
i s a finite-dimensional algebra over t h e f i e l d
i s i n f a c t G-maximal. (iv) that
P n R = 0.
since
Then, by Lemma 3 . 9 ( i i ) ,
P i s prime, t h i s y i e l d s
Cd - Pd.
=
D*G
p.
=
assume t h a t
MU#C 1 2 -
Because
P
and t h e r e f o r e
or
B*G
I I 9 Lu.
Thus
1 2 -
Lu
i s prime and t h e
We c l o s e t h i s s e c t i o n by p r o v i d i n g two consequences of Theorem 3.11. 3.12.
Let
COROLLARY.
prime r i n g an i d e a l of Proof.
R.
If
R*G
R*G
P
i s a prime i d e a l of
properly containing
By Theorem 3 . 1 2 ,
P
=
L'
P,
PROPOSITION.
t h e prime r i n g
Let
R*G
R*G
with
P nR = 0
if
I
and i f
is
I n 8 # 0.
then
d
L = P
f o r t h e G-prime i d e a l
t h e argument i n t h e proof of Theorem 3.12, 3.13.
G over t h e
be a c r o s s e d p r o d u c t of t h e f i n i t e group
I
3
P
= 'L
then
I
.
f?
Hence, by
R # 0.
be a c r o s s e d p r o d u c t of t h e f i n i t e g r o u p
G
over
R.
( i ) A prime i d e a l
P o€ R*G
i s minimal i f and o n l y i f
( i i ) There a r e f i n i t e l y many s u c h minimal p r i m e s , s a y
P
f-
Pl,P2,
R = 0.
...,P
and i n f a c t
245
PRIME COEFFICIENT RINGS
(iii). S = P n P 1
2
n
... n Pn
is the unique largest nilpotent ideal of R*G.
In
fact
J where
=
E.
J(E) is the Jacobson radical of Since E
Proof.
I 'inn I = o ,
and J
J(E)U
is a finite-dimensional algebra over a field, there are
only finitely many, say
...,Ln
,
Ll,L2, n
of G-prime ideals of
dimgtGinn = IGinn/ =
E.
Then clearly
/GI
and J(E) =
,
If Pi = L:
R*G
P
then by Theorem 3.11,
n nL
i=1 i
,...,Pn
are the unique prime ideals of
Jd 5 L':
=
which are disjoint from R - (01. Put J
=
P.. 2
i=1
Then
J
5 Lui
so
,
n
C
Invoking Lemma 3.lO(ii), we see that J
Jdu 5 J ( E ) ' .
pi
Since
J ( E ) U and
we conclude, from Lemma 3.9(ii), that both other hand, each
Li by Theorem 3.11, and thus
J
J(E)
is nilpotent
are nilpotent.
surely contains all nilpotent ideals of R*G.
On the
Thus
n J = np. i=l 2 and therefore J = J(E)' IG i n n Furthermore, since S(E)
contains J(E)' of
R*G.
1 'inn 1 J
=
is clearly the largest nilpotent ideal
I
0
=
(see Theorem 2.3.16)'
we have
0.
Finally, let P be any prime ideal of
Since J
R*G.
is nilpotent,
n
and hence
P 3
Pi for some i.
ma1 members of the set
{Pl,
Hence the minimal primes of R*G
...,Pn}.
However, P . 3 P 2 -
L . = Pci'3 2
and hence, since L PI,
...,P
i
2 -
pciJ
=
implies that
Lj
is G-maximal, we must have
are precisely the minimal primes of
j
are the mini-
i
R*G
=
j.
This proves that
and the result follows.
.
CHAPTER 5
246
4 . INCOMPARABILITY AND GOING DOWN
R*G
Let
R
We s a y t h a t
Thus
R
R.
over an a r b i t r a r y r i n g
A,B
i s G-prime i f f o r a l l nonzero G - i n v a r i a n t i d e a l s
AB # 0.
have
G
be a c r o s s e d p r o d u c t o f a f i n i t e group
0
i s G-prime i f and o n l y i f
is a
R,
of
we
R.
G-prime i d e a l of
The problems t h a t m o t i v a t e t h i s s e c t i o n are t h e f o l l o w i n g IncornparahiZity.
If
P cP 1
a r e prime i d e a l s of
2
P
Down.
Going
P
n R
=
A
? nR = A ?
R
of
d o e s t h e r e e x i s t a prime
2'
does it f o l l o w t h a t
n R c p 2 n R ?
Al c A2
Given G-primes
R*G,
PI
and of
a prime
R*G
P2
of
satisfying
R*G
? C 1
P
with 2
o u r aim i s t o p r o v i d e a p o s i t i v e answer t o b o t h problems.
and Even
more, it w i l l be shown t h a t f o r I n c o m p a r a b i l i t y t h e primeness of t h e l a r g e r i d e a l
P2
i s unnecessary.
P
and i f
PcI
Indeed, it w i l l be proved t h a t i f
a r e i d e a l s of
R*G
i s prime, t h e n
The f o l l o w i n g two simple o b s e r v a t i o n s w i l l j u s t i f y our r e s t r i c t i o n t o G-prime coe f f i c i e n t rings. 4.1.
The f o l l o w i n g c o n d i t i o n s a r e e q u i v a l e n t .
LEMMA.
P C I a r e i d e a l s of
(i) I f (ii) I f
p C
I are ideals
R*G R*G
of
P prime, t h e n
with
n R# InR
R a G p r i m e and
P prime,
with
?
n R
= 0,
I n R # 0.
then
p flR = 0 i m p l i e s t h a t
(Note t h a t , by Lemma 3.1, Proof. 2.6.3(ii),
Note t h a t
p
f'
R
i s &prime).
R i s a G i n v a r i a n t i d e a l of
( p n R ) * G i s a graded i d e a l of
R.
Hence, by Theorem
R*G so t h a t
(P n R ) * G
I t f o l l o w s t h a t i n d e a l i n g w i t h ( i )w e may f a c t o r o u t t h e i d e a l
R*G
and t h u s reduce t o t h e c a s e
4 . 2 . LEMMA.
If
P
R
PnR
=
0
of
which i s t h e c o n t e n t of ( i i ) .
The Going Down problem i s e q u i v a l e n t t o t h e f o l l o w i n g s t a t e m e n t :
i s a G-prime r i n g and
P
a prime i d e a l of
R*G,
then
P
c o n t a i n s a prime
INCOMPARABILITY AND GOING DOWN
Q
ideal
of
R*G
Proof. that
Q nR
with
=
241
0. At
D i r e c t consequence of t h e assumption
i s G - p r i m e and t h e f a c t
(R/R1)*G.
(R*G)/(A1*G)
F o r t h e r e s t of t h i s s e c t i o n , w e s h a l l assume t h a t R
i s a G-prime r i n g
Note t h a t t h e above c o n d i t i o n i s s a t i s f i e d i f t h e r e e x i s t s a prime i d e a l
n gQ
such t h a t
0.
=
A A 1
=
2
A.,
For i f
A ,A
A A
0, then
1
we i n f e r t h a t
2
A.
cQ
so
C
fl gQ
' --&G
2
A.C Q
LEMMA. ( i ) R
i.
f o r some
-L-
and hence t h a t
Ai
c o n t a i n s a prime i d e a l
Q
Using t h e G-invariance of
0.
=
R
i n g o b s e r v a t i o n shows t h a t i n any G-prime r i n g 4.3.
R with
a r e G - i n v a r i a n t i d e a l s of
1
6 G
R
of
Q
P a r t ( i l of t h e follow-
one can f i n d such a prime
n gQ
with
=
0.
Q.
In particular,
&G
R is
semiprime.
(111
Any prime i d e a l of
R
contains a conpgate
a r e p r e c i s e l y t h e minimal primes of (iii) L e t
H
be the s t a b i l i z e r of
h i l a t o r of
Q
in
R
(by Lemma 1 . 4 ,
gQ
of
Q
{gQlg
and s o
E G)
8. Q
in
G
and l e t
N = ann(Q) R
be t h e anni-
l e f t and r i g h t a n n i h i l a t o r s of
are equal).
Q
Then
and
f o r all (iv)
g
E G-H.
A
If
i s any nonzero i d e a l of
R
with
A C_N,
then
annA = Q.
R
(il
Proof.
Q of
e x i s t s an i d e a l Assume t h a t
G
Because
A A
C Q
B
and
B
f o r some i d e a l s
A ,A 1
of
R
B.
and p u t
2
a r e G - i n v a r i a n t i d e a l s of
=
a
R
and s i n c e
B B
R
i s G-prime,
so
Bi
=
=
0.
qEG
C
1 2 -
But
n gQ
R maximal w i t h r e s p e c t t o t h e p r o p e r t y t h a t
1 2 -
Then
is f i n i t e , it f o l l o w s from Z o r n ' s lemma t h a t t h e r e
n gA.,i 6 G
A"A 1
2
,
= 1,2.
we have
0 f o r some i E {1,2} and t h e n t h e maximality of
Q
CHAPTER 5
248
A.
implies t h a t
=
Q.
Q
Hence
R
(ii) Any prime i d e a l of
R.
0 = n 'Q and hence, s i n c e G SfG Moreover, t h e r e are no i n c l u s i o n r e l a t i o n s between
certainly contains
I
'Q.
i s f i n i t e , c o n t a i n s some
1 s a prime i d e a l of
n E GI.
t h e primes
n
taking
Q
For, i f
C
'Q,
then
NQ = 0 ,
(iii) Since
t h e n by (ii),
w e have
N C_ gQ
NQ C - 'Q
for a l l
N
C_ n ' Q , . &H
and t h e r e f o r e
for all
>
n
1,
so by
Q c Q.
/GI w e o b t a i n t h e c o n t r a d i c t i o n
=
Q
Q C
g E
G.
g E G-H,
Hence, i f
Conversely, because
w e have
N = annQ
2
n
'Q ,
&H N =
proving t h a t
n 'Q.
Q
I n any c a s e , by ( ii)we have
N
g
9 H,
then
N
C
'Q
and so
N
P 0.
n $'Q = R .
N =
&H
Furthermore,
~ n g n= g g = o SEG
-1
and i f
H = G, t h e n by d e f i n i t i o n
Note t h a t i f
@=
so
'N
5 Q.
Hence f o r
g 9 H
we have
and hence
(iv)
If
A
5N
If
A
# 0
implies t h a t
3
5 Q.
A R = 0.
AQ = 0, so 4 5 a n d .
then c l e a r l y
A
then
Thus
Q,
Q
since
= annA
Conversely, assume t h a t
N n Q = 0 by ( i i i ) . Hence
AB = 0
and t h e r e s u l t f o l l o w s .
For t h e rest of t h i s s e c t i o n , t h e f o l l o w i n g n o t a t i o n w i l l be used
Q
i s a minimal prime of t h e G-prime r i n g
N
= annlQ)
H
i s t h e s t a b i l i z e r of
R
R M
=
4.4.
C 'N
(so t h a t
LEMMA.
( i ) Let
V
M
Q
in
G
i s a nonzero G - i n v a r i a n t i d e a l of
R)
With t h e n o t a t i o n above, t h e f o l l o w i n g p r o p e r t i e s hold: be a nonzero r i g h t R-submodule
of
NZ
and l e t
T = tg, ,g2,.
..,gnl
5Q
249
INCOMPARABILITY AND GOING DOWN
be a subset of
G with g
=
Assume that
1.
1
for all 2"
c T.
(ii) Let I
T
Then
V
n R*T
# 0
V
but
n R*T' = 0
5 H.
be an ideal of R*G.
Then there exists a nonzero G-invariant ideal
E of R with EI c- %J(I n R + H ) E Proof.
!I
C_ M?
o#
(i) By hypothesis, there exists
r
we have
i
E N
for all
T ensures that all r . # 0.
i
E
Given
{1,2,.
n 3: =
..,nl
z r z. g- i
E V
i=l
fi
R*T.
Because
and the minimality condition on Er; Q, we have xq = ri qgi E V and
z
q
i=lg. rlq
E
NQ = 0.
Thus the minimality condition on -1
and we have r g i Q = 0, 1
i
< i S n.
Thus
gir.Q
T implies-lrizq =
0, so 9i r i
0 for all i
=
E N = ann(Q)
and
therefore, since r . E N, we have
i is in H, as required. Since I is G-invariant, we have
By Lemma 4.3(iii), we conclude that each (ii) First assume that NI = 0.
MI and so we may take Observe that right ideal of
V
R*G.
E
=
M
= (
V
Let
N.
V
o = ZV(I n R*H)Z We may therefore assume that V = N I # 0. is a
X be the set of all subsets i" of G such that
is a right ideal of
empty collection of subsets of
Since
~M)I =
satisfies the hypothesis of (i) and, furthermore, V
V n R*T # 0, V Since
c
gfG in this case.
g
R * T 1 = 0 for a11 !?'C T R*G,
G.
is an R-subbimodule of
it is easy to see that
X is a finite non-
Moreover, by (i), ! ? & H
NF, AT
1E T
and
for all
is a nonzero ideal of
R
T E X.
contained in
Put
D= n A EX Because X is finite and A C N for all T TD # O.
E
X, we have by Lemma 4.3(iii) that
CHAPTER 5
250
I t w i l l now be shown t h a t
gm+lz c - E N ( I n R*H)E
m
The c a s e
Y
E
V
E V
0 being t r i v i a l , we u s e i n d u c t i o n on m.
=
ISupplcI = m
given with
x
for
m
and
5 Suppx
Choose
T
1E
Tg-',
Of c o u r s e ,
r i g h t i d e a l of
R*G.
If
g
then
Tg-l
E
T,
Suppxi
-1
= (Suppx)g
a l s o has t h i s minimal p r o p e r t y s i n c e
is
3 5''g-l
V =
and
NI i s a
Because w e need o n l y show t h a t
$r+l --l
xg
we can r e p l a c e
E I/
minimal w i t h r e s p e c t t o t h e
-1 R*T # 0.
V
x
So assume
(1)
0 and suppose t h e a s s e r t i o n h o l d s f o r a l l e l e m e n t s
of s m a l l e r s u p p o r t s i z e .
property t h a t
= /SuppxI
--1
x by xg
T
and
c - ZN(r n R*H)E
by
Tg
-1
.
Thus w e c a n assume t h a t 1 E Suppx
T E X.
and
c
Let
trx
=
d e f i n i t i o n of
and, s i n c e
AT,
SuppU
x and l e t d
be t h e i d e n t i t y c o e f f i c i e n t of
6E V n
w e may choose
5 suppx
and
R*T
with
Y
=
&-Be E V
try
=
0
trR
I SuppY/
we have
=
E
d.
<
m.
D LAT. Hence
Thus, by induc-
tion,
dnY C_ E N and hence
Now
T
11 C_
N,
5H
and s o
Be
E
V
n R*H C_ I n R*H.
c_
~ " O C
N U n R*H)
Therefore, because
c_ F N ( I
n R*H):
The c o n c l u s i o n is t h a t
which i m p l i e s t h a t
om+ 1 x and proves (1) by i n d u c t i o n .
c_ GNU
n
R*H)G
By
m B 1 and
INCOMPARABILITY AND GOFNG DOWN
k
Applying (1) f o r
=
IGl, we deduce t h a t Dk+l v
Note t h a t
DktlN
Hence, i f
E
# 0 since
c EN (In R*H) 5 -
D~+'NI
=
D C - N , D # 0 and
n
Q
N = 0,
by Lemma 4 . 3 ( i i i ) .
i s d e f i n e d by
E E # 0
then
251
E
since
R*G, E
a r e i d e a l s of
{r E R l r I
=
2 Dki'N
# 0.
C -
?%"I n R*H)G?
On t h e o t h e r hand, s i n c e
R.
i s a G - i n v a r i a n t i d e a l of
I
and
aN(1n R*H)z
Lf
and t h e r e -
Since
G,
EI c - EN (In
R*H)
t h e r e s u l t follows. For any i d e a l
I;
R*H,
of
we s e t
n
LG =
SfG If
I
i s a n i d e a l of
and hence, Lemma 2.7.15
LG
i s t h e unique l a r g e s t i d e a l o f
LG
i s t h e unique l a r g e s t i d e a l
where
R*G-
:
R*H L
4.5.
LEMMA.
(i)
L ~ cL ( L~~ L ~ ) '
If
SEG
R*G, t h e n w e s e t
L ( R * G ) = LE
Note t h a t
~ L C
n
g(1;0 =
R*G
I
implies t h a t
contained i n
R*G
of
LE
satisfying
V ( l )
5 L,
is natural projection.
L2
and
a r e i d e a l s of
R*H,
then
1 2 -
Proof.
( i ) Because
Lc
R*G, w e have
i s a n i d e a l of
ELG
c
2 -
fore
L
~
c LL ~EL^ c L
1 2 -
LGLG
Since
(iil
1
2
i s an i d e a l o f
I t i s obvious t h a t
1
1
IG
2
C -
~(LY
n
we have
-
c L L
L~
1 2 -
1
2
2
it f o l l o w s from ( 2 ) t h a t
R*G,
(L 0L
2
LG 1
c 2
-
n LG. 2
n L~
I ; ~
G G
L L
C
1 2 -
Conversely, s i n c e
(L L
1 2
)
G
.
CHAPTER 5
252
L'
n
1
c (L n
~~j~
1
2 -
by virtue of 13). 4.6. LEMMA. ideal of
Let
L
be an ideal of
R*H
with
LnR 3 - Q and let I be an
R*G.
(i) CNLC C_ LG C_ L z
(ii) M U H )
G
C_
I.
and
G
L C_ ( L I H
Moreover, there exists a nonzero G-invariant ideal E
R
of
with
Proof. g @
(i) If g E H,
then ZNLZ
since L
H*H.
is an ideal of
If
H, then
GNLG because of
5 Lg
g N C_ Q E L
for g
!$
g f ~ Zc_ Q(R*G) 5
=
H.
Hence
LZ
ZNLz g L z ,
and since
GNLZ
is an ideal
R*C, we derive CNLG
In particular
NL
E LG and
c_ 'L
LE
s o , by definition, we have
L
C
( LG )H.
(ii) We have
N ( I I I I G & NIHz C I z = I , where the second inclusion is a consequence of definition of and
( I H ) Gare
proving the first assertion.
c_ Z N U n II*H)F
for some nonzero G-invariant ideal E
as required.
B
Because
Owing to Lemma 4.4(ii),
EI
is an ideal of
IH'
G-invariant, we deduce that
R*H
containing
&.
of
R.
Moreover, I n R*H
5 IH
and
IH
Invoking (i), we therefore conclude that
INCOMPAWLBILITY AND GOING WWN
4.7. LEMMA. (i) If
With the notation above, the following properties hold:
I is an ideal of R*G
(ii) If L
with
is an ideal of R*G
then I f- R = Q and 0 = Q*H. H H L n R = Q, then L G n R = 0. In
I n R = 0,
with
(Q*H)G = 0.
addition, Proof.
(i) Because
by Lemma 4.3(iv). (iii
253
InR
=
0, we have
Similarly, because
and R
=
Q, we have 0
H
=
Q*H.
Bearing in mind that
we have L
G nR
=
n 'J(r;ij) n R SEG
=
( Q * H ~=~ n
"QF
n ~ ( L n F R) = n gg = 0 SEG SfG
Similarly
.
and (ii) follows.
= (
SfG
n gQlz = 0 SEG
We have now accumulated all the information necessary to prove the main result of this section. 4.8. THEOREM
(Loren2 and Passman (1980bl).
the finite group
P
* PH
with
R
with H
the stabliizer of
PnR
L n R
Proof.
=
0 and the prime ideals L
=
(ii) LG
in G.
Then the map
of
L t-+ L G ,
R*H
with
L
n R
=
Q.
The inverse
for any prime ideal L
of
R*H
Q.
Let P be a prime ideal of R*G
prime ideal of R*H (i) PH
Q
is G-prime and let Q be
yields a bijective correspondence between the prime ideals P of R*G
of this correspondence is given by with
be a crossed product of
Assume that R
G over the ring R.
a minimal prime of
Let R*G
with
L n R = &.
is a prime ideal of R*H is a prime ideal of
R*G
with
PnR
=
0 and let L
be a
We must show that
with with
PH n R = Q and P L G n R = 0 and
=
(pH)G
L = (LG)
It will be convenient to start with the following observation.
H
Let I be an
254
CHAPTER 5
and let E be a nonzero G-invariant ideal of
R*G
ideal of
R.
We wish to show
that
ET C - LG Assume that
EI
5 LG.
implies I C - LG
Then
and, by applying the natural projection TI : R*G+
NOW
is G-invariant and L n R = Q
E
Hence, since L G
that I C - L
,
(4)
satisfies
is prime, we infer that
TI(I1
5
R*H,
we have
n gQ = o and s o E*H L. SfG L. Applying ( 3 ) , it follows
proving (4).
(i) Setting
L = PHI it follows from Lemma 4.7(i) that L ~ R = F~
H
R
=
Q
Note a l s o that, by Lemma 4.6(ii),
Since
p
is prime and
M*G
L
It will next be shown that of
Pli
containing
L
we conclude that P
P,
with
is prime.
2
L
L L 1
and so
P
2
LE
for some
i
€
2
{1,2,},
.
2
(PHIG = LG.
Indeed, let
L
and 1
L
be ideals 2
Then, by Lemma 4.5(i),
since
F
is prime.
Since
Lemma 4.6(i) yields
L = PH Hence L
is a prime ideal of
R*H
2
with
ailH G 2 Li L n R = Q.
By Lemma 4.6(ii), there exists a nonzero G-invariant ideal E EP C_ ( P H ) G = LG
Invoking (41,we deduce that P
5 LG
and hence
P = LG by ( 5 ) .
of R
with
255
INCOMPARABILITY AND GOING DOWN
P
(ii) S e t t i n g
L
=
G
,
i t f o l l o w s from Lemma 4 . 7 ( i i l
Owing t o Lemmas 4.6Li)
I
and
H
R = Q.
f-
:
R*H
be t h e n a t u r a l homomorphism.
-I - -
Hence
3
H -
Invoking C o r o l l a r y 3.12,
Q*H.
3
-
0.
Assume
I nR
and
=
0.
G
> PH = ( L ) H z L
H -
Let
t h i s map.
=
and 4 . 7 ( i ) , we have
I
L
I? P
I i s any i d e a l ( p o s s i b l y P i t s e l f ) s a t i s f y i n g
that
PnR
that
L, L
--).
(R*H)/ ( Q * H ) = ( R / Q ) * H
-
IH and
Then b o t h
L
R*H
i s a prime i d e a l of -
of
I n H
IH = L
and hence t h a t
Q*H
-
and o b v i o u s l y
*
IH = L
w e deduce t h a t
contain t h e kernel
=
0.
since
G
I = P = (15 I H = L . H H
I n p a r t i c u l a r , it f o l l o w s t h a t
Moreover, by Lemma 4 . 6 ( i i ) , t h e r e i s a nonzero G - i n v a r i a n t i d e a l
B
R
of
with
EI
5 (qG = LG = P r 5 LG =
Applying (41, w e t h e r e f o r e conclude t h a t
P is
now be shown t h a t
I
Let
I.
above,
n R # 0, <
= 1,2.
o so
I I 7 P and 1
4.9.
2
-
G
a f i n i t e group
P nR
=
0
and i f
Proof. i d e a l of ideal
L
Let
R*G of
#
R*G
(rIn R )
P.
It w i l l
Then, by t h e
is G-prime,
P i s prime.
o v e r a G-prime r i n g
I is P
an i d e a l of
R*G
be a prime i d e a l of
with
L
nR
=
Q.
R.
P.
Let
G
R*G
b e t h e c r o s s e d p r o d u c t of
P is a
If
prime i d e a l o f
properly containing
R*G
P nR
with
=
P
By Theorem 4.8,
P,
then I
0 and l e t
=
LG
R*G fl
with
R # 0.
I be an
f o r some prime
Moreover, t h e argument of t h e l a s t p a r a g r a p h
COROLLARY (Lorenz and Passman ( 1 9 7 9 a ) ) .
of a f i n i t e group
=
n R) c - I 1I 2 n R
i n t h e proof of Theorem 4.8 now a p p l i e s t o y i e l d 4.10.
R
T h e r e f o r e , because
properly containing R*H
P.
properly containing
(Lorenz and Passman L1980b)).
COROLLARY
I
and t h u s
prime.
I2 be i d e a l s of
and
P
o v e r an a r b i t r a r y r i n g
I
fl R #
Let
R.
R*G If
P
0, a s a s s e r t e d . b e t h e crossed product
C
I are
i d e a l s of
i?*C
CHAPTER 5
256
P
where
P n R # I n R.
i s prime, t h e n
Apply Lemma 4 . 1 and C o r o l l a r y 4 . 9 .
Proof.
i s g i v e n by
A f u r t h e r a p p l i c a t i o n of Theorem 4.8
THEOREM (Lorenz
4.11.
and Passman ( 1 9 7 9 a l ) .
G over a G-prime r i n g
a f i n i t e group
'I
( i ) A prime i d e a l
of
R*G
R*G
Let
be t h e c r o s s e d p r o d u c t of
R.
\GI
(iii) J =
::
h*G and
P i s t h e unique l a r g e s t n i l p o t e n t i d e a l of i= 1 i
proof.
Let
Q
t h e s t a b i l i z e r of
R,
be a minimal prime of
Q
G.
in
-
:
R = 0
P 1 , P L,...,P
(ii) There a r e f i n i t e l y many such minimal p r i m e s , say
n
P
i s minimal i f and o n l y i f
and i n f a c t
J I G ' = 0.
H
a s i n Lemma 4 . 3 ( i ) , and l e t
be
Consider t h e n a t u r a l map
R*H
(R*If)/LQ*H) = (R/Q)*H
---+
c
By P r o p o s i t i o n 3.13,
L
primes
R*H
of
with
by P r o p o s i t i o n 3.13,
L n R = 0.
R*H.
.
Then, by Theorem 4 . 8 ,
-T
Furthermore,
Q
Li
let
P. =
L1,L 2,...,Ln,
n Li
=
i s t h e unique
a g a i n by P r o p o s i t i o n 3.13.
be t h e complete i n v e r s e image o f
G
Li,
then
Y
i=1
T I H I = 0,
and
i E 11,. , , n } ,
t h e r e a r e f i n i t e l y many
Indeed, i f t h e s e a r e
-R*H \GI. -
n 4 \HI
l a r g e s t n i l p o t e n t i d e a l of For each
R*H,
applied t o t h e crossed product
1
i
n,
-Li
in
a r e p r e c i s e l y t h e prime i d e a l s
n of
R*G having t r i v i a l i n t e r s e c t i o n w i t h
complete i n v e r s e image of
T
in
R*H. TIH'
Setting
J
=
C
T ,
R.
-
Put
Because
T
=
Li
so t h a t
T
z=1 T I H ' =O , we t h e n have
5 Q*H.
it f o l l o w s from Lemmas 4.5 and 4 . 7 ( i i )
that
and
Hence
S i n c e each
Pi,
being prime, c o n t a i n s all n i l p o t e n t i d e a l s of t h e c r o s s e d
is the
251
INCOMPARABILITY AND GOING DOWN
J
product t h e latter implies t h a t
n n Pi
ideal
and
SO
3 Pi
P
i.
f o r some
i=1
=
P. n R 3
0,
=
{PI , P 2 , .
of t h e s e t
t h e minimal members
P. n R
R*G.
P be any prime i d e a l of
Finally, l e t
P
Then
contains the nilpotent
Hence t h e minimal primes of
..,P 1 .
If
i
C o r o l l a r y 4.9 e n s u r e s t h a t
2 pj,
Pi =
j.
R*G,
4.12.
R*G
G o v e r an a r b i t r a r y r i n g P
and t h a t prime
P1
k*G
of
.
P
R
Q nR
=
show.
.
4.13.
R*G
P
1
P
{l,
P
...,n } .
Let
S
R
.
=
Then t h e r e e x i s t s a
By Lemma 4.2,
it s u f f i c e s t o
c o n t a i n s a prime i d e a l
Q
of
R*G
Then, i n t h e n o t a t i o n
n n Pi and hence
P
3 Pi
P. n R
R.
Let
R*G
Then R*G
=
0 a s w e wished t o
be a c r o s s e d p r o d u c t of a h a s a unique l a r g e s t n i l -
J I G ' = 0. be t h e sum o f a l l n i l p o t e n t i d e a l s of
and from t h i s it w i l l f o l l o w t h a t
J
R*G.
W e s h a l l show
i s n i l p o t e n t (hence a unique
R*G.
l a r g e s t n i l p o t e n t ) i d e a l of
Since
A
But, by Theorem 4.11(11,
and
Q
.
R
i=l
J
If
P2 n R = A
surely contains the nilpotent ideal
potent ideal
JIGi= 0
a r e C-prime i d e a l s o f
i s a G-prime r i n g .
G o v e r a semiprime r i n g
that
cA2
R*G.
THEOREM (Lorenz and Passman ( 1 9 7 9 a ) ) .
Proof.
1
P nR
and
2
be any prime i d e a l of
f i n i t e group
R
A
wfth
c P
so assume t h a t R
0.
E
be a c r o s s e d p r o d u c t of a f i n i t e group
i s a G-prime r i n g , t h e n
of Theorem 4 . 1 1 ,
i
t h u s completing t h e p r o o f .
Assume t h a t
satisfying
Let
prove t h a t i f
for some
R.
i s a prime i d e a l of
Proof.
with
Let
are
Therefore
.
a r e p r e c i s e l y t h e minimal primes of
R*G
then since
P1,P2,. .'Pn
COROLLARY (Going Down).
R*G.
f s t h e unfque l a r g e s t n i l p o t e n t i d e a l of
R,
i s a prime i d e a l of
n gQ gfc
then
i s c l e a r l y a G-prime i d e a l of
i s semiprime, i t f o l l o w s t h a t t h e i n t e r s e c t i o n of a l l C-prime i d e a l s of
i s zero.
Assume t h a t
P
i s a G-prime
-
i d e a l of
: R*G---t
be t h e n a t u r a l homomorphism,
Then
R
&*G)/(P*G)
J
and l e t
(R/P)*G
i s a sum of n i l p o t e n t i d e a l s of
(R/P)*G
R
CHAPTER 5
258
and
R/P
i s a G-prime r i n g .
-
JIG(= 0.
n i l p o t e n t with
X
T h e r e f o r e if
JIG(
Hence
5 P*G
f o r a l l G-prime i d e a l s
R,
i s t h e s e t of a l l G-prime i d e a l s o f
JIG1
c - n (p*G) E X n P
B u t , by t h e second p a r a g r a p h ,
P
R.
of
then
P)*c
E X JIG' = 0
0, so
=
n
= (
is
3
rnvokfng Theorem 4 . l l ( i i i ) , we deduce t h a t
a s required.
E X The f o l l o w i n g s i m p l e o b s e r v a t i o n , due t o Montgomery and Passman (19781, w i l l be needed t o p r o v i d e a f u r t h e r a p p l i c a t i o n of Theorem 4.11. 4.14.
F
field
F"G
Let
LEMMA.
p > 0.
of c h a r a c t e r i s t i c
a r a b l e f i e l d e x t e n s i o n of
F.
Then
i s a f i n i t e p u r e l y insep-
FCIG/J(F"G)
In particular,
over a
J(F'G)
1 s a unique prime l d e a l of
.
F'G
Proof.
E
If
J(EG) = I ( G ) ,
E"C
EG.
I
E.
F-subalgebra o f
I
fore
= J(F'G).
Thus
then F"G/I
I
E.
BG.
charF =
p,
In particular,
Hence t h e same i s t r u e of
FaG
i s a f i n i t e f i e l d e x t e n s i o n of
with
F
EnG.
FaG/I
an
and t h e r e -
Furthermore, t h e f i e l d e x t e n s i o n i s p u r e l y i n s e p a r a b l e s i n c e
n f o r some
i s a p-group and
i s a n i l p o t e n t i d e a l of
.
i t i s g e n e r a t e d by t h e images o f t h e e l e m e n t s E F
then
I(C) i s t h e augmentation i d e a l of
where
J ( E " G ) n F"G,
=
G
Since
J(EG) i s a n i l p o t e n t i d e a l w i t h EG/J(EGI Thus if
F,
i s t h e a l g e b r a i c c l o s u r e of
and, by P r o p o s i t i o n 1.6.1,
$'
G
be a t w i s t e d group a l g e b r a of a f i n i t e p-group
n.
-
g
g E P
with
and t h e s e s a t i s f y
We a r e now r e a d y t o prove 4.15.
THEOREM (Lorenz and Passman ( 1 9 7 9 a ) ) .
G
be a c r o s s e d p r o d u c t of a f i n i t e p-group istic
p.
Then
Proof. and o n l y i f Theorem 3 . 1 1 ,
R*G
Cl
R = 0.
o v e r a G-prime r i n g
P
E =
R
of c h a r a c t e r -
i s a minimal prime i d e a l o f
L e t u s f i r s t assume t h a t
t h e number of minimal primes of
G-prime i d e a l s of
p > 0 be a prime and l e t R*G
h a s a unique minimal prime i d e a l .
Owing t o Theorem 4 . 1 1 ( i ) ,
P
Let
F tGinn.
But
F
R*G
R
R*G
i s a prime r i n g .
if
Then, by
i s t h e same a s t h e number o f
i s a f i e l d of c h a r a c t e r i s t i c
p
and
259
A GOING UP THEOREM
i s a f i n i t e p-group,
Ginn Thus
E
E
so
Bas a unique prime i d e a l by v i r t u e of Lemma 4.14.
h a s a unique G-prime i d e a l and t h e r e f o r e
Turning t o t h e g e n e r a l c a s e where
number of minimal primes of p
i s G-prime,
Then t h e number of minimal primes of
i t s notation.
teristic
R
R*C
H
and
(R/Q)*H.
R/Q
Because
h a s a unique minimal prime.
w e a p p l y Theorem 4.8 and i s t h e same a s t h e
R*G
i s a prime r i n g of charac-
i s a f i n i t e p-group, t h e r e s u l t f o l l o w s from t h e prime c a s e
proved above.
5 . A GOING UP THEOREM
R*G be a c r o s s e d p r o d u c t of a f i n i t e group
Let
G
R.
o v e r an a r b i t r a r y r i n g
The main q u e s t i o n we a r e i n t e r e s t e d i n i s t h e f o l l o w i n g : Going Up. with
A
Let
P
R
C
A.
be a G p r i m e i d e a l of
R
P
and l e t
Does t h e r e e x i s t a prime i d e a l
Q
b e a prime i d e a l of R*G w i t h
of
P
C
Q
R*G
and
Q ~ R = A ? The main t a s k of t h i s s e c t i o n i s t o p r o v i d e a p o s i t i v e answer t o t h e above The r e s u l t w i l l emerge a s a consequence of more g e n e r a l c o n s i d e r a t i o n s
problem.
p e r t a i n i n g t o i n t e g r a l i t y and n o r m a l i z i n g e x t e n s i o n s of r i n g s . Let
S
R
be a r i n g and l e t
be a s u b r i n g w i t h t h e same
Lorenz and Passman (197913) w e s a y t h a t t h e e x t e n s i o n
{zili E I} of e l e m e n t s of
t h e r e e x i s t s a family (a)
The elements
(b)
x.R
=
Rxi
x.(<
E
I)
generate
S
S
R
5S
1.
Following
i s normaZizing i f
such t h a t
a s a l e f t R-module
iE I
for all
The normalizing e x t e n s i o n
R
5S
i s c a l l e d finite i f
S,
r e g a r d e d a s a l e f t R-
module, i s f i n i t e l y g e n e r a t e d o r , e q u i v a l e n t l y , i f t h e above s e t
bili E I}
can
be chosen t o be f i n i t e . Of c o u r s e , i f
R c - R*C
i s a crossed product of
R*G
i s normalizing.
s a t i s f y ( a ) and ( b ) .
G
over
T h i s i s so s i n c e t h e e l e m e n t s
Furthermore, i f
G
i s f i n i t e , then
R,
then t h e extension
{s/gE G I
R
5 R*G
clearly
is a f i n i t e
normalizing e x t e n s i o n . For l a t e r a p p l i c a t i o n s , i t i s n e c e s s a r y to d e a l w i t h r i n g s w i t h o u t i d e n t i t y , i.e.
w i t h r i n g s which do n o t n e c e s s a r i l y have a
1.
Given such a r i n g
R,
we
CHAPTER 5
260
for the ring of all n
Mn(A)
write
subring of M ( A )
consisting of diagonal matrices.
... x A
isomorphic to A x A x be a transversal if B
B
Let B
(n times).
Dn(A)
Then
A subring
A x A x
...
X
A.
Dn(A)
and
for the
is naturally
B of D n ( A )
projects onto each direct factor A
is a subdirect product of
bedded i n
n matrices over A
X
is said to
or, equivalently, if
For example, A
naturally em-
M ( A ) as scalar matrices is a transversal. be a transversal in M n ( A ) .
monomial in al,a2,...,ak
If
I
,a
,...,ak
Z
Mn(A),
then by a B-
we understand a product in some order of the
ai's,
each occurring finitely often, and of elements of
B with at least one element of
B
then a 3 b a2b a
Thus, for example, if bl,b2 E B ,
occurring.
and a 2 , while Q2a 0
mial in a
is a B-mono-
1 1 2 2 1
By the degree of any such monomial
is not.
1 2 1
we will mean the total degree in the a i r s . The following theorem is a generalization of a result due to Pare and Schelter (1978).
5.1. THEOREM (Loren2 and Passman (1979b)). B
n
such that for any
CL E
=
of degree less than
.., n } ,
k E {l,Z,.
For each integer
t.
M ( A ) be the subring of
let
Mn(A) consisting of those matrices of the form k.
1 depending
@(a)
@ ( a ) is a sum of B-monomials in a
Proof.
1 and let
Mn(A),
at where
be a ring without
Then there exists an integer t
be a transversal in M,(A).
only upon
Let A
[i
Ey
with
*
a block of size
Note that
Our aim is to show, by induction on such that if
E
Mk(A),
then
as where
that there exists an integer s = s ( k )
k,
=
@(a)
@ ( a ) is a sum of B-monomials in a of degree less than
accomplished, the result will follow by taking call the relation as
- $(a)
satisfied by
a,
t
=
s(n).
S.
Once this is
In what follows we
a monic polynomial over
B
of
A
degree
261
GOING UP THEOREM
S.
k
Jf
= 1,
then
b E B
there exists
E]
ci =
k
= 1
E
A.
:]
b =
such t h a t
This settles t h e case
a
with
by t a k i n g
Then, because
and so
satisfies
ci
where
fir
B
Mk+l(A)
( k + l ) s t row and column.
Indeed, because
B,y,
B is
6,
R,Y E M k + L ( A )
0.
k < n
and c o n s i d e r
.
Mk+l (A)
Y1
E
we p u t
6
we w r i t e
yp,
there exists
then c l e a r l y
b
E
B
~ - b =
y to indicate that
-Byl -BYZ.
b
E
B
y" 0
O]
and t h e n
a s claimed. Applying i n d u c t i o n on
j 2 0, i t now f o l l o w s t h a t i f
6" = Y".
Moreover, we
such t h a t
a transversal, there exists
*
then
=
and t h e remaining nonzero e n t r i e s come from t h e
Given such
We n e x t o b s e r v e t h a t i f
c l a i m t h a t given
a2-ba
can be p a r t i t i o n e d a s
i s i t s i n i t i a l k-block
Furthermore, f o r any
a transversal,
s(1) = 2.
Assume t h a t t h e r e s u l t i s t r u e f o r a l l i n t e g e r s Any m a t r i x
B is
such t h a t
262
CHAPTER 5
... -B Y
gjEj-l where
i
j + 1.
... Blr + @
RjBj-l
i s a sum of nonconstant B-monomials i n t h e
j
Indeed, t h e c a s e
=
0 i s t r i v i a l and i f
... B I Y Eor some s u i t a b l e
for a suitable
Ql
Q
and
=
B .Q
3 1
- Bjb
i s an a p p r o p r i a t e sum of B-monomials
j + 1.
of d e g r e e l e s s t h a n c1 E
monic polynomial
+
t h e n by t h e above
b E B
W e now f i x
of d e g r e e smaller t h a n
Bi'S
-S c1
Mk+l(A)
-
and n o t e t h a t , by i n d u c t i o n ,
-@(a)
B
over
of d e g r e e
s = s
(2)
c1
.
s a t i s f i e s a suitable
Then w e c e r t a i n l y have
and we a p p l y t h e o b s e r v a t i o n of t h e p r e c e d i n g p a r a g r a p h t o each of t h e monomials i n t h i s expression.
Bearing i n mind t h a t
BMk+l ( A )
5 Mk+l
(A)
and
Mk+l ( A ) B
we s e e t h a t each nonconstant B-polynomial i n longing t o
Mk+l(A).
5 Mk+l
(A)
a i s a p r o d u c t of f a c t o r s a l l be-
W e conclude e a s i l y from t h e above t h a t
( a S - @ ( a ) ) +a $(a)= ( E S - . @ ( E ) ) O
0,
with
@(a) being a s u i t a b l e sum of n o n c o n s t a n t B-monomials i n
than
s +l.
I t follows t h a t t h ere e x i s t s
a monic polynomial o v e r We now observe t h a t
B
over
B
0,
of d e g r e e less
B
-b
in
c1
of d e g r e e
D
0. -0
E M (A) k
a l s o s a t i s f i e s a monic polynomial
and t h a t t h e c o n s t a n t term i n we conclude t h a t
s + l , with
(p)
€I
m u s t belong t o
B
n
-
B"-@(U)
Mk(A).
= 0
Because
A
Y
=
263
GOING UP THEOREM
P-em
,"
=
3
0 0 0 for some a E A.
with
b
But the latter expression is a monic polynomial in
a
It follows that if
b
E B
=
diag(*,a,*),
then
y2-by = 0
a monic polynomial in cx
2s(s+l).
of degree
hence it is
.
Therefore the induction step
s ( k + l ) = 2 s ( k ) ( s ( k ) + 11, as required.
follows with
2s;
of degree
Our next aim is to apply the above result to finite normalizing ring extensions n
An ideal A
First, however, we must develop our vocabulary.
norma2 if 2.A
of
R
is called
,...,
Ax for all i E 11'2 n}. Thus, if A is a normal ideal i of R, then the right ideal AS of S is in fact a two-sided ideal. In particular, AS
=
is a subring of
previous terminology, if
S
s E AS
degree of such monomials.
(without 1) containing
A.
In analogy with our
we will speak of A-monomials in
s
and of the
We are now ready to prove the following result.
n 5.2. THEOREM (Lorenz and Passman (1979b)).
c Rzi
Let R c S=
be a finite
i=l
normalizing extension of rings and let A exists an integer
t
1
be a normal ideal of
depending only upon
R.
n such that for any s
Then there 6
AS,
st = @(s) where
s of degree less than t.
@ ( s ) is a sum of A-monomials in
proof.
V = R x R x
Put
module of rank
n.
... x R
( n times1 so that V
Define the left R-module homomorphism
n(rl,r
*,...,r
is a free left R7 :
v-
n )
=
z1rixi i=
Then 7' = {@E EndR(V)\@(Kern) c - Kern} is a subring of
EndR(v)
=
Mn(R) and there is a natural homomorphism
s
by
264
CHAPTER 5
A defined as follows.
If
:
T
T and s
@ E
EndRIS)
+
=
n C r . x . E S, then A($) i= 1 2 2
E
EndR(S)
is
given by
1 ($) ( s ) =
A(@)
That
..,r 1
Ti@(rl ,r2,.
A
is well defined and that the map
is a ring homomorphism follows by
straightforward verifications. Denote by
z i =
with
U.X
B the set of all diagonal matrices of the form
for some a E A .
2.0
z
B
follows easily that
is a transversal in M n ( A ) .
B(v)
(r1a 1 ,r 2 a 2
=
)
and
n
n
6
E
T
Moreover, if
,...,rn) E V ,
,...,rnan
The latter implies that
is normal, it
(rl,r2
=
2)
then
Using the assumption that A
and that
h(B1
is a right multiplication by
a € A. We now fix s E AS
x.s
E
AS.
and note that since AS
La. .x . jZJ3
=
2
and we set u = ( a
) E Mn(A).
ij
1
(TU) (0)=
plication by
s
'T(v)s.
2
,...,r
It follows that
U E
$(a)
for some a
ij
E A
)
E V,
T and that
A(U)
is right multi-
E S.
B is a transversal, Theorem 5.1 may be employed to infer that there
is an integer t 2 1, depending only upon n ,
where
we have
Then one easily verifies that if
v = (r ,r
Because
S,
Hence
x .s
then
is an ideal of
is a sum of B-monomials in
0
such that
of degree less than
to this relation and letting the resulting R-homomorphism of
S
t.
Applying
act on
l E
A
s, we
265
A GOING UP THEOREM
obtain
@(s) i s a sum of monomials o b t a i n e d from t h o s e i n
where by
s
L? E B
and each
by an a p p r o p r i a t e
a
E
A.
Hence
$!G)
by r e p l a c i n g
@(S)
i s of t h e
D
r e q u i r e d form and t h e r e s u l t f o l l o w s .
I be a ( l e f t o r r i g h t ) i d e a l of a r i n g
Let
bounded d e g r e e i f t h e r e e x i s t s an i n t e g e r 5.3.
I
Let
LEMMA.
bounded degree. Proof.
R.
zn
n 2 1 such t h a t R
be a nonzero r i g h t i d e a l o f a r i n g
=
0
>
t
Given an i n t e g e r
X(t)
1 and a s u b s e t =
X
of
x
for all
E
R,
R.
we s e t
x}.
Ixt[xE
n 2 1 and so I ( n ) I = 0.
Then, by h y p o t h e s i s ,
I ( n ) = 0 f o r some i n t e g e r
may c e r t a i n l y choose
t 2 1 minimal such t h a t t h e r e e x i s t s a nonzero r i q h t i d e a l
J C -I
J(t)J
with
t
The c a s e
a
E
J,a2J = 0
for a l l
x
E
= 1
0.
To complete t h e p r o o f , i t s u f f i c e s t o show t h a t
J,
w e have
=
t
0, ( a z ) J
=
0.
Hence
( a x )t-1aJ
J C_ I,
f o r some hence t h a t
n
>
1.
J2
=
so
J
=
=
= 0
We
J 2 = 0.
W e f i r s t show t h a t f o r
=
t , we deduce t h a t aJ
By t h e m i n i m a l i t y of
1.
0. Indeed, l e t a E J w i t h a 2 J = 0. t ( m i a) J = 0. However, a 2 J = 0 i m p l i e s t h a t
aJ
( a J ) (t-l)( a J )
Now
t >
b e i n g obvious, we may assume
implies t h a t
J(t’J
and, s i n c e
=
Then,
and t h e r e f o r e
0
0.
is a n i l right ideal.
Hence, f o r any
b E J , bnJ bJ
Therefore t h e preceding paragraph implies t h a t
=
=
0
0 and
0, a s r e q u i r e d .
We can now o b t a i n a Lying Over Theorem f o r f i n i t e n o r m a l i z i n g e x t e n s i o n s . 5.4.
THEOREM (Lorenz and Passman
(197913)).
Let
R
5S
=
n 1 Rxi
be a f i n i t e
i=1 normalizing e x t e n s i o n and l e t then
A
L
which i s n i l of
I c o n t a i n s a nonzero n i l p o t e n t r i g h t i d e a l of
Then
i s n i l of
I
W e say t h a t
b e a normal i d e a l of
R.
If
A
i s semiprime,
266
CHAPTER 5
A
Furthermore, i f
i s prime, t h e n t h e r e e x i s t s a prime i d e a l
s
P of
with
~ n ~ = . 4 Proof.
R
of
I
Put
n R.
AS
A.
containing
r E I.
=
r E AS,
Since
t
Let
i s a normal i d e a l of
r
r
belongs t o
A
A
=
Thus
A = I = AS
r
Now assume t h a t
X
Let
A
i=
the ideal
is
and where
I/A
i s n i l of bounded
as a s s e r t e d .
R.
i s a prime i d e a l of
P n R = A. and
A
r
$(r) E A
P
lemma e n s u r e s t h e e x i s t e n c e of an i d e a l property t h a t
o r belong t o
is a
i s semiprime and hence Lemma 5.3 i m p l i e s t h a t
R
n R,
=
R/A,
-
But, by h y p o t h e s i s ,
0.
=
b(r)
where
=
and l e t
and t h e r e f o r e
T h i s shows t h a t i n t h e r i n g
7
t
I t f o l l o w s t h a t each such monomial i n
must occur.
rt
degree.
i s an i d e a l
R e c a l l t h a t , by d e f i n i t i o n , each A-monomial i n
a f i n i t e product whose f a c t o r s a r e e i t h e r e q u a l t o a t l e a s t one f a c t o r from
R, I
2 1 be t h e i n t e g e r g i v e n by Theorem 5.2
Theorem 5.2 t e l l s u s t h a t
r.
sum of A-monomials i n
A
Since
Because
s
of
AS n R = A ,
maximal w i t h r e s p e c t t o t h e
W e are therefore l e f t t o verify t h a t
X 2 b e i d e a l s of
S
Zorn's
properly containing
P.
P
i s prime.
xi n R
Then
3 A -
and t h e r e f o r e
A
since
i s prime.
Hence
(XIn R ) ( X 2 n R )
A
X I X 2 n R (r A ,
X X
so
1
2
eP -
and
P is
a prime i d e a l
s.
of
We a r e now r e a d y t o prove a Going Up Theorem f o r c r o s s e d p r o d u c t s . 5.5.
THEOREM (Lorenz and Passman ( 1 9 7 9 b ) ) .
f i n i t e group if Q
P
G o v e r an a r b i t r a r y r i n g R . R*G
i s a prime i d e a l of
R*C
with
Proof.
Let
of
Let
P C Q S = R*G
and
with
PnR
Q n R = A.
and l e t
$ : S
If C
-
A,
S/P
R*G A
be a c r o s s e d p r o d u c t of t h e
i s a G-prime i d e a l of
R
and
t h e n t h e r e e x i s t s a prime i d e a l
be t h e n a t u r a l homomorphism.
CHAINS OF PRIME AND PRIMITIVE IDEALS
n(S) i s a f i n i t e normalizing e x t e n s i o n o f
Then
R c - R*G
T h i s i s so s i n c e
s,g E G. g
E
G.
A
Because
P
n
R C A
A
i s G-prime,
i s G - i n v a r i a n t and so
T(A)
R/A
;A
=
Ag
for all
n(R).
i s a normal i d e a l o f
Moreover,
w e have
R/A
n(R)/n(A) Now
n(g),gEG.
with generators
i s a f i n i t e normalizing extension with g e n e r a t o r s
I t t h e r e f o r e follows t h a t
since
K(R)
261
R/A
i s a G-prime r i n g , so
i s a semiprime i d e a l .
W e may t h e r e f o r e a p p l y Theorem 5.4
Tr(R)
izing extension
By Theorem 5.4,
i s semiprime by Lemma 4 . 3 ( i ) .
Hence
II(A)
t o t h e f i n i t e normal-
5 n(S).
K(A)Tr(S)
n Tr(R) =
nu)
and t h u s , by t a k i n g complete i n v e r s e
images, w e d e r i v e
(A*G
+ P) n R I
I t f o l l o w s t h a t t h e r e e x i s t s an i d e a l
Owing t o Z o r n ' s lemma, w e may choose respect t o t h i s property.
Q
=
A.
R*G
of
I 3 P
such t h a t
R*G
t o b e a n i d e a l of
I t w i l l n e x t be shown t h a t
Q
I n R=A.
and
maximal w i t h
i s prime and t h i s w i l l
clearly yield the result. Let
X2
X1 and
Xi n R i s
be i d e a l s of
a G - i n v a r i a n t i d e a l of
R*G
R
properly containing
Q.
A
so
properly containing
(X1 n RI ( X 2 n R ) since
A
i s G-prime.
T h i s shows t h a t
Q
I t follows t h a t
i s a prime i d e a l of
gA
X X nR E A 1
Then each
and hence t h a t
2
X X
1 2 -
Q.
R*G, a s r e q u i r e d .
6. CHAINS OF PRIME AND PRIMITIVE IDEALS T h i s s e c t i o n i s devoted t o some a p p l i c a t i o n s of p r e c e d i n g r e s u l t s t o r i n g s w i t h f i n i t e group a c t i o n s .
R,
G
i s a f i n i t e group a c t i n g as automorphisms on a r i n g
t h e n we can form t h e skew group r i n g
given action. ring
If
R~
of
W e use
R*G
R*G
of
G
over
R
t o s t u d y t h e r e l a t i o n s h i p between
with r e s p e c t t o t h e
R
and t h e f i x e d
G.
R e c a l l t h a t f o r any r i n g
R,
the
prime rank of
R
i s d e f i n e d t o be t h e
268
CHAPTER 5
n
largest
such t h a t
R
h a s a c h a i n of prime i d e a l s
primitive rank of prime i d e a l s .
R
i d e a l s of
R
Thus t h e p r i m i t i v e rank of
S,
P
Let
i f and o n l y i f a l l p r i m i t i v e
A
containing
p/A
such t h a t
A
is
i s a minimal
s/A.
o v e r an a r b i t r a r y r i n g
6.1. LEMMA.
0
t h e n a minimal covering prime of
S
of
From now on, up t o Theorem 6.4,
G
is
n.
i s an i d e a l of any r i n g
d e f i n e d t o be a prime i d e a l
group
R
The length of t h e c h a i n
a r e maximal.
prime of t h e r i n g
The
i s d e f i n e d a n a l o g o u s l y u s i n g p r i m i t i v e i d e a l s i n s t e a d of
i s d e f i n e d t o be t h e number
If A
i s s a i d t o have infinite prime rank.
R
I f no such maximum e x i s t s , t h e n
A
R*G
d e n o t e s a c r o s s e d p r o d u c t of a f i n i t e
R.
be an i d e a l of
R
and s e t
A ' = (R*G)( n gA) = ( n
'A)*G SEG
SEG
A'
Then
i s an i d e a l of
i s prime, t h e n
A
(i) I f
R*G
A'
with
A'
R =
'A
.
SEG h a s f i n i t e l y many, s a y
Furthermore
P1,PP,
...,P n ,
minimal
c o v e r i n g primes and
pi n R = n SA n 5n i s n i l p o t e n t modulo (b)
(1 G i G n )
(a)
i=1 (ii) I f A
A'
i s p r i m i t i v e , t h e n t h e minimal c o v e r i n g p r i m e s of
A'
a r e a l s o prim-
itive
A
(iii) I f
Proof. (1)
If
A
i s maximal, t h e n s o a r e t h e minimal c o v e r i n g primes of
The f i r s t s t a t e m e n t being o b v i o u s , w e need o n l y v e r i f y ( i )- ( i i i ) .
-
R = R/
i s prime, t h e n by t h e i n t r o d u c t o r y remark t o Lemma 4.3,
i s a G-prime r i n g .
Moreover, t h e minimal c o v e r i n g primes of
t h e minimal primes of t h e r i n g 4.11.
A'.
(R*G)/A'
-R*G
A'
n 'A
SEG
correspond t o
which a r e d e s c r i b e d i n Theorem
Hence, t h e y a r e f i n i t e i n number, a l l of them s a t i s f y
t h e i r i n t e r s e c t i o n i s t h e unique l a r g e s t n i l p o t e n t i d e a l of
7 nB R*G.
= 0,
and
This clearly
CHAINS OF PRIME AND PRIMITIVE IDEALS
269
proves (i). (ii) Assume that A
is primitive and let
v
be an irreducible left R-module
Consider the induced R*G-module VG = R*G €3
with annihilator A .
R
V.
Then we
have (direct sum of R-modules)
v
€3
where each
is irreducible since
v
is irreducible.
a finite composition series, and therefore VG
ann(VG)
=
Now let Pi = ann(Xi-L/Xi), 1 G
i
'
i, P. z - A'
also have ideals
=
ann(V
G
PIPZ.. .Pt5A'.
Pi.
subset of
(8)R
has
has a finite composition series,
By Proposition 2.7.16(ii), we have
as an R*G-module.
for each
Hence
)
.
(R*G)( n '(annv)) SEG
t.
=
A'
Then the ideals Pi
are primitive and
Since PIP2.. .Pt clearly annihilates
Thus any prime containing
8,we
A ' contains one of the
The conclusion is that the minimal covering primes of A'
form a
..
{PI,P2,. ,Pt? and therefore are primitive
(iii) Suppose that A ing prime of
A'.
is a maximal ideal of
R and let p be a minimal cover-
Assume by way of contradiction that P
=I
for some ideal I
Then, by Corollary 4.10, we have
of R*G.
p n ~ > p n ~n =g
~
SfG where the latter equality holds by virtue of (i).
R containing I n R. Because A
Choose a maximal ideal
B
of
X
n 'A c B and therefore A f B for some X G. SfG is maximal, we have B = " A . Taking into account that I R is a
G-invariant ideal of
Then
R
and that
~
I
R c - B,
we deduce that
n n g ~g = nc g ~ = p n ~ , -&G SEG
a contradiction. 6.2. LEMMA.
of
R.
Let I be an ideal of R*G.
Furthermore,
Then
I n R is a G-invariant ideal
270
CHAPTER 5
I
(ii) I f
P
i s semiprime, t h e n so i s
I
(i) If
R
of
InR
i s prime, t h e n t h e r e e x i s t s a u n i q u e , up t o G-conjugacy,
prime i d e a l
such t h a t
I nR
n gp
=
SEG
I
(iii) I f
p
i s p r i m i t i v e (hence p r i m e ) , t h e n t h e i d e a l
i n (ii) i s a l s o
primitive. The f i r s t a s s e r t i o n i s obvious
Proof. (i) L e t
I
be a semiprime i d e a l of
n i l p o t e n t modulo
I n R.
B =
ideal
51
A
We must prove t h a t
observe t h a t t h e G-invariant
A
and l e t
R*G
c 'A
R
b e an i d e a l of
R,
which i s
To t h i s e n d , we f i r s t
i s a l s o n i l p o t e n t modu1.o
I
cI R ,
SEG G
since
is f i n i t e .
B*G
Now
R*G
(1n R ) *G
i s contained i n
B*G
Hence some power of
i s a n i d e a l of
such t h a t
and hence i n
I.
Since
I
B*G c- I and hence t h a t
i s semiprime, t h e l a t t e r i m p l i e s t h a t
as r e q u i r e d .
InR
( i i ) Owing t o Lemma 3 . 1 ,
g u a r a n t e e s t h e e x i s t e n c e of s u c h
i s a G-prime i d e a l of
P.
R.
The u n i q u e n e s s of
p,
Hence Lemma 4 . 3 ( i ) up t o
G-conjugacy,
i s a consequence of Lemma 4 . 3 ( i i ) (iii) Assume t h a t
with annihilator
I
v
i s p r i m i t i v e and l e t
I.
vR
By Theorem 4 . 2 . 2 ,
be a n i r r e d u c i b l e l e f t R*G-module c o n t a i n s a n i r r e d u c i b l e submodule
W and
Let Then
Q
denote t h e a n n i h i l a t o r of
gQ i s
W
Q
so t h a t
t h e a n n i h i l a t o r of t h e R-submodule
r
R The uniqueness of
P
.
now e n s u r e s t h a t
primitive, a s asserted.
SfG P
R =
gw
(SW)
n R = annv = n ann X
Q
8.
is a primitive ideal of
=
of
vR
and so
nQ'
SEG f o r some
X
G.
Thus
P
is
271
CHAINS OF PRIME AND PRIMITIVE IDEALS
W e now combine Lemmas 6 . 1 and 6.2 i n o r d e r t o e x t e n d Theorem 2.2.
The
following a u x i l i a r y a s s e r t i o n w i l l c l e a r our path. 6.3. of
Let
LEMMA.
e,
a unit in
e
E
R*G
R.
If
be an idempotent w i t h
tre,
the identity coefficient
h a s prime ( o r p r i m i t i v e ) rank 2 n ,
R
then t h e r e
e x i s t s a chain Po
c
R*G
of prime ( o r p r i m i t i v e ) i d e a l s of Proof.
PI
... c P,
c
such t h a t
e 7 P,
.
By h y p o t h e s i s , t h e r e e x i s t s a c h a i n
Qoc of prime (or primitive) i d e a l s of
...
Q1
R.
Qn
N o t e t h a t f o r each
i, n gQ, C n gQi+l @G
'-&G
and we c l a i m t h a t
Q . II n gQi+l
Indeed, i f e q u a l i t y o c c u r r e d , t h e n
' -&G
Qi >_ xQi+l f o r some
x
E
G.
s t r i c t inclusion
and t h e r e f o r e
3 X Qi
But, as has been observed i n t h e proof of Lemma 4 . 3 ( i i ) , a Qi 3
X
Qi i s i m p o s s i b l e s i n c e
G
is finite.
This s u b s t a n t i -
a t e s o u r claim.
e B QA
W e next note t h i t
tained i n
nQ ' ,. SfG
p o t e n t modulo
Q'.
ma1 c o v e r i n g prime
= ( f?
'Qn)*G
.sEG Moreover, because
Pn
of
QA
such t h a t
have
2
Pi,
Pi
e 9 P,
t h e l a t t e r being a consequence of
b e i n g a u n i t , i s n o t con-
e
cannot be n i l -
it f o l l o w s t h a t t h e r e e x i s t s a mini-
e ? P,.
W e now s u c c e s s i v e l y a p p l y
Pn-l,Pn-2,...,P
i s a minimal c o v e r i n g prime of and
tre
i s an idempotent,
Invoking Lemma 6 . 1 ( i ) ,
C o r o l l a r y 4.12 and f i n d prime i d e a l s
Pi+l
e
since
Q!
of and
P.
R*G
such t h a t
R = :GgQi.
Then w e
272
CHAPTER 5
This establishes the prime case.
Qi are primitive, then each Pi
In case all
is also primitive, by virtue of Lemma 6.1(ii).
So
the lemma is true.
We are now ready to prove the following important result.
6.4. THEOREM (Lorenz and Passman C1979a)).
G over an arbitrary ring R.
finite group
equal to the prime rank of
be a crossed product of the
Then the prime rank of
R*G
is
R and the primitive rank of R*G is equal to the
R.
primitive rank of Proof.
Let R*G
e = 1
By taking R
primitive rank of
in Lemma 6.3, we see that the prime rank and the
do not exceed the corresponding ranks of R*G.
To prove
the reverse inequalities, assume that
Fo
C
PI C
...
is a chain of prime (or primitive) ideals of
P71
R*G,
Then, by Corollary 4.10, the
intersections P . n R
are all distinct and, by Lemma 6.2(ii), we may write
for a prime ideal Qi
of
Q,. Qi+l
If
uniquely determined up to G-conjugacy.
so that
SEG
Hence by successively choosing
Qi C Qi+l
the two ranks must be equal. 6.2 (iii), each
Qnll-l,Qn,...,QO we obtain the chain
since
This shows that the prime rank of R*G
Qi
does not exceed that of
Finally, if each
is also primitive.
So
Pi
R.
the theorem is true.
Since
R*G
as formal sums
G
is a finite group acting
Using this action, we form the skew group ring
R.
R and therefore
is primitive, then by Lemma
For the remainder of the section we assume that G on a ring
We now fix
we may choose Qi
is given, then because Qi+l 2 - n 'Qi,
Qi+l
3 Qi.
where
R
li*G
of
G
over
is isomorphically embedded in R*G, we may treat the elements of C r g g , rs E SfG
R; addition in R*G
is defined componentwise
and multiplication is defined distributively by means of the rule
2 13
CHAINS OF PRIME AND PRIMITIVE IDEALS
(rxx)( P y ) Y
X
rz r xy
=
x,y E G, r
for a l l
Y
r Y
2'
[GI-'
For t h e remainder of t h i s s e c t i o n , w e a l s o assume t h a t
E R.
R
Then, by
Lemma 2.8.4,
e
]GI-
=
1
c
g
SfG
R*G
i s an idempotent of
such t h a t G
e ( R * G ) e = eRG = R e
RG
R c - R*G,
Our aim i s t o a p p l y our p r e v i o u s knowledge on t h e e x t e n s i o n w i t h some c l a s s i c a l r e s u l t s on t h e e x t e n s i o n of t h e t y p e
RG
i n f o r m a t i o n about t h e p a i r
eR*Ge
- R*G,
together t o derive
5R.
Our p o i n t of d e p a r t u r e i s t h e f o l l o w i n g g e n e r a l r e s u l t .
6.5.
Let
LEMMA.
f
-
@
P
:
fPf
=
and
P
f
are i n
X,
i d e a l of
2
fsf.
P,P1
Moreover, i f
P2
and
@(p) i s
fSf, t h e n
SAS
i s an
i f and o n l y i f
P
1 -
A
i s an i d e a l of
S such t h a t
p r o p e r i d e a l of 1
S not
of prime i d e a l s of
i s primitive.
W e f i r s t note t h a t i f
Now assume t h a t P E
A A
5@ ( P 2 )
P
X
of a l l prime i d e a l s of
@(Pl)
then
p r i m i t i v e i f and o n l y i f Proof.
Y
and t h e s e t
Then t h e map
P n fSf
s e t s up a b i j e c t i v e correspondence between t h e s e t containing
s.
be a nonzero idempotent o f t h e r i n g
c fPf. -
X.
fSf.
Since
f?P
Suppose t h a t
we have
Al
and
f 9 $(P) = fpf A2
a r e i d e a l s of
and so
@(p) i s a
fSf w i t h
Then
(SAIS)( S A z S ) = SA1 ( f S f I A z S C -
SAIAzS
C -P and so t h e p r i m e n e s s of over, f o r t h i s
P
i, w e have
implies t h a t
z c - P f o r some
SA.S
<E
[1,21
More-
274
CHAPTER 5
P1,P2 E X
Suppose t h a t
Since
P cP 1 -
2
5
@
be inferred t h a t
P
c P2.
$
@
SqS
P n fSf
&.
=
P
P
P E
x.
By o u r c h o i c e o f
is primitive.
f '? P and l e t
Then
fV
V
So assume t h a t
Indeed, l e t
U
5 fV SU
V
=
Moreover, t h e a n n i h i l a t o r of
@Pl= fPf.
Thus
Conversely, l e t i b l e l e f t fSf-module
maximal and
i s prime.
@ ( P )= & ,
we have
a p r i m i t i v e i d e a l of
P
f $? P ,
=
S
ann(V).
fV
we claim t h a t
of
fV.
Then,
and t h e r e f o r e
fV
Q be a p r i m i t i v e i d e a l of Q.
with a n n i h i l a t o r
fSf/I.
Il
P is
primitive i f
in
fSf i s e a s i l y s e e n t o be
@(PI i s p r i m i t i v e .
Then
I l = SI
I n fSf and t h e r e f o r e
P,
be a nonzero fSf-submodule
a s claimed.
W
P
X, $(P) i s
P E
which i s nonzero, s i n c e
is i r r e d u c i b l e , w e have
fSf w i t h
S which i s maxi-
be a n i r r e d u c i b l e l e f t S-module s u c h t h a t
i s a l e f t fSf-module
is i r r e d u c i b l e . V
&.
=
is surjective.
with
since
of
b e a prime
SQS n fSf
S satisfies
Then i t f o l l o w s e a s i l y t h a t
f $? P, we have
Q
To t h i s end, l e t
of
We a r e t h e r e f o r e l e f t t o v e r i f y t h a t f o r a l l and o n l y i f
We have t h e r e -
I n p a r t i c u l a r , i t may
1 -
is s u r j e c t i v e .
Then, by (11, t h e i d e a l
m a l with respect t o
proving t h a t
Con-
1 -
f = $ ( P 1.
Applying Z o r n ' s lemma, w e may t h e r e f o r e choose a n i d e a l
Moreover, s i n c e
P c Pz.
and so
i s injective.
I t w i l l now be shown t h a t
/;Sf.
5 fP
@ ( P 1 )= fP,f
then obviously
X, @(P) E Y.
E
fP,f 5 fP,f
Then
therefore implies t h a t
@ ( P 1 ) $(P2) i f and o n l y i f
f o r e shown t h a t
i d e a l of
5 @(P2).
P
P
Thus f o r a l l
@(PI)
and t h a t
f $? P2 , t h e primeness of
versely, i f
fSf.
@tP) i s a prime i d e a l o f
proving t h a t
=
Let
@ S(1-f)
SI
f-
1, n
i s a l e f t i d e a l of
fSf = fSI
fsf =
I.
be an irreduc-
I be a maximal l e f t i d e a l of
=
Hence
S with
I
i s c o n t a i n e d i n a maximal l e f t i d e a l
f 9 I,, we have
W
fSf and l e t
I2
of
S.
Since
I
is
275
CHAINS OF PRIME AND PRIMITIVE IDEALS
v = s/r as fSf-modules. S
(fsf+r2)/r2
3
2 -
P
Therefore, if
=
annS(U,
= fsf/r p w P is a primitive ideal of
then
such that P n fSf =
fPf
5 ann
W
=
Q
fSf In particular, we must have
f 9 P.
It therefore suffices to verify that
To this end, observe that
and
Hence &S =
and therefore SQS
Qsf + S
is an ideal of
QS(1- f) C_ I 2 contained in
5.
It follows that
SQS 5 ann(S/12) = P and thus
Q
5 fpf =
6.6. COROLLARY.
P n fsf.
If
f
This proves ( 2 ) and hence the result.
is a nonzero idempotent of a ring
S,
.
then the prime
rank and the primitive rank of fSf are bounded above by the corresponding ranks of
s. Proof.
fSf.
Let
Q
C Q2 C
---
4,
C
Then by applying the map
be a chain of prime (or primitive) ideals of as given in Lemma 6.5, we obtain a chain
$I-'
.
PI c P 2 c of prime (or primitive) ideals of
S.
... c P,
We are now ready to prove the main result of this section. 6.7. THEOREM (Lorenz and Passman (1979a)). a ring R rank of
and let
RG
1Gl-I
E R.
Let
G be a finite group acting on
Then the prime rank of R
and the primitive rank of R
is equal to the prime
is equal to the primitive rank of
RG.
2 76
CHAPTER 5
Proof.
Let r a n k s
whichever i s r e l e v a n t .
d e n o t e e i t h e r t h e prime o r p r i m i t i v e rank of a r i n g
R*G
Denote by
G
r r e s o n d i n g t o t h e g i v e n a c t i o n of
e
t h e skew group r i n g of
R.
on
(GI-
=
1
G
over
R
S,
co-
Then, by Lemma 2 . 8 . 4 ,
cg
SfG i s an idempotent of
R*G
such t h a t
e (R*G)e T h e r e f o r e we may i d e n t i f y
eRG
=
RG w i t h
RG
Rce
=
e(R*G)e
5 R*G.
BY C o r o l l a r y 6.6 and Theorem
6 . 4 , w e have rank
tre
Note t h a t II,
=
RG
rank R*C = r a n k R Hence, i f rank R
R.
1GI-l i s a u n i t of
n,
f o r some i n t e g e r
t h e n Lemma 6.3 i m p l i e s t h a t t h e r e e x i s t s a c h a i n
Po c PI of prime ( o r p r i m i t i v e ) i d e a l s of S = R*G
and
f = e,
R*G
= ... c Pn with
e 9 P,.
Invoking Lemma 6.5, w i t h
we i n f e r t h a t
$(Po)= $ ( P I ) c
... c $(Pn)
i s a chain of prime ( o r p r i m i t i v e ) i d e a l s of
rank R
e(R*G)e
rank R
R
G
.
Thus
G
and t h e r e f o r e rank RG
=
rank R*G = rank R
a s required.
6.8.
COROLLARY (Lorenz ( 1 9 7 8 ) ) .
1GId1 E R.
Then a l l p r i m i t i v e i d e a l s of
i t i v e i d e a l s of Proof. i d e a l s of
L e t a f i n i t e group
RG
R
G
a c t on a r i n g
R
and l e t
a r e maximal i f and o n l y i f t h e prim-
a r e maximal.
The p r i m i t i v e r a n k of a r i n g
S a r e maximal.
S is
0 i f and o n l y i f a l l p r i m i t i v e
Now a p p l y Theorem 6 . 7 .
277
6 Semiprime and prime crossed products
Let
G
b e a c r o s e d p r o d u c t of a group
R*G
.
over a r i ng R
In t h i s chapter,
w e are concerned w i t h f i n d i n g n e c e s s a r y and s u f f i c i e n t c o n d i t i o n s f o r
R*H
depend upon t h e analogous c o n d i t i o n s f o r t h e c r o s s e d p r o d u c t
H
of
be
The main r e s u l t shows t h a t t h e s e c o n d i t i o n s u l t i m a t e l y
prime or semiprime.
subgroup
R*G t o
G
of t h e f i n i t e
and upon t h e i n t e r r e l a t i o n s h i p between t h e n o r m a l i z e r s of
t h e s e subgroups and t h e i d e a l s t r u c t u r e of
R.
The b u l k of t h e c h a p t e r h a s a
unique s o u r c e , namely a fundamental p a p e r of Passman (1984).
S i n c e Passman's
method a p p l i e s e q u a l l y w e l l t o s t r o n g l y G-graded r i n g s , w e have d e c i d e d t o work i n t h i s more g e n e r a l c o n t e x t .
1. COSET CALCULUS
G
Throughout t h i s s e c t i o n ,
R.
ring
d e n o t e s a group a c t i n g on t h e set of i d e a l s of a
Thus, f o r any i d e a l
R
mined i d e a l of
I of
R
and any
g 6 G,'1
i s a uniquely deter-
and
z(Yr) = W I 'I FO
=
g E G
and i d e a l s
(i) i f
I 5J
then
(ii) ' U J )
=
'1
T,J
of
G strongly p e m t e s R,
G
i d e a l s of
t h e i d e Is of
R
5' J
'FJ G
comes from a homomorph-
i n t o t h e group of i n c l u s i o n p r e s e r v i n g p e r m u t a t i o n s of t h e set o f
R.
if
we have
Observe t h a t c o n d i t i o n ( i l i m p l i e s t h a t t h e a c t i o n of
i s m of
r,y E G
I
w i n g Passman (1984), w e s a y t h a t
for a l l
for all
I n p a r t i c u l a r , t h e y also p r e s e r v e t h e l a t t i c e o p e r a t i o n s o f
278
CHAPTER 6
a r b i t r a r y sums and i n t e r s e c t i o n s ,
Thus we must a l s o have
0
=
Ax
and
A
Suppose
1.1. LEMMA.
B
and
z
If
E
'R
Then, f o r a l l
Ax n By,
=
R G
and t h a t t h e r i g h t c o s e t s
Ax
2 E
A x = A z , By = B z
then
g E G
for a l l
a r e subgroups of
By a r e n o t d i s j o i n t .
Proof.
and
By,
and t h e r e f o r e
as r e q u i r e d .
n 1.2.
G
Assume t h a t
LEMMA.
s t r o n g l y permutes t h e i d e a l s of
R
u
and l e t
Gigi
i=1
that
J
R
i s a nonzero i d e a l of
G.
Gi of
be a f i x e d union of r i g h t c o s e t s of t h e subgroups
Assume f u r t h e r
such t h a t
gJJ = 0
n U G g i=l i i
g E G-
for a l l
nY . p
Then t h e r e e x i s t s a subgroup
H
G
of
and a nonzero p r o d u c t
K =
'J,
with
i=1 some
yi
= 1,
such t h a t
~ K =K o
for all
g E G-H
Moreover,
IH Proof. the
Gls
.
Let
Then
X
:
X
i s f i n i t e , c l o s e d under i n t e r s e c t i o n s and
conclusion hoth a s s e r t t h a t
/XI
>
1
1x1.
gJJ = 0
Let
A
X' 11'1 < (XI and
of r i g h t c o s e t s
i E { l , ~ ,...,n l
9X'
1x1
If
for a l l
g
E
=
{GI ,GZ,.
..,Gn} 51.
0 t h e n t h e h y p o t h e s i s and
G.
and t h a t t h e r e s u l t i s t r u e f o r any p r o p e r s u b s e t of
c l o s e d under i n t e r s e c t i o n s .
Then
f o r some
be t h e s e t of a l l p r o p e r ( t h a t i s , nonempty) i n t e r s e c t i o n s of
We prove t h e r e s u l t by i n d u c t i o n on
Assume t h a t
IH n G ~ )
be a maximal member of =
X
X
and s e t
X - {A}.
i s c l o s e d under i n t e r s e c t i o n s .
Given a f i n i t e union
2 79
COSET CALCULUS
S = U A g
iij
Ai E X ,
with
S,
by t h e s u p p o r t of
written
SuppS
,
A!s
w e mean t h o s e
which
occur i n t h i s r e p r e s e n t a t i o n . K
Assume that
g
G-S.
E
9
A
If
=
y 1 ~ 2 J . . . y pf J0, some
SuppS,
e x i s t s a f i n i t e product
g E G- H,
for a l l some
I=
Now
u1
H
where
A . E SuppS.
SuppS c -
then
I H : ( H n Gi)I I i s a l s o a s u i t a b l e product
Since
G
i,
f o r some
=
and i n d u c t i o n a p p l i e s .
i s a subgroup of
-L-
’KK
and t h a t
1,
=
# 0, w i t h some
K 2K...USK
c G?,.
A.
X’
yi
Uj = 1,
H n Ai
so
Hence t h e r e and
I H : ( H n Ail
with
0 for a l l
‘II
= 0
O3
for
1
5H
n Gi
J,
the r e s u l t follows
and hence
< m
of t h e c o n j u g a t e s of
i n t h i s case. Thus we can assume, for a l l such p a i r s
K
A E SuppS,
as above, t h a t
S
and
Of c o u r s e , t h e r e i s a t l e a s t one such p a i r by h y p o t h e s i s ( t a k e
r
=
y,
1 and
= 1,
n k
so t h a t
= J,
and
S = i=l U G g
t h e smallest number, s a y
rn
W e now choose
).
2 1,
of c o s e t s of
A
K
S
and
so t h a t
S
has
occuring i n i t s represention.
Then
S = Az T
where
U 1
Az
... U A z m U 1
U
2
m
Then
H
i s o b v i o u s l y a subgroup o f
Assume t h a t
5
KK # 0
G-conjugates of 2
KK
c_ 2K
X’.
i s a f i n i t e union of c o s e t s of groups i n
r l
J
f o r some
x
E
G
Define
H
5-G
by
m
and we show t h a t
G.
Then
“KK
i s a nonzero f i n i t e p r o d u c t of
w i t h some c o n j u g a t i n g element e q u a l t o
1.
Moreover, s i n c e
K, YKK
=
o
for a l l
y
E G-S
for all
y
E
implies
’(“KK) (“KK)
5 ’KK
= 0
G-S
CHAPTER 6
2 80
and
Thus
g(*KK) (*KK) for a l l
g with g E (G - S ) U (G -S)z-'
Because
0
=
x
-
= G
(S
i s c l o s e d under i n t e r s e c t i o n s , Lemma 1.1 t e l l s u s t h a t
x.
a l s o a f i n i t e union of r i g h t c o s e t s of members of
s
SuppSX-l = Sup$, with
n Sx-1)
A
i s equivalent t o
B =
c
= A.
=
B
~
i s a maximal member of
A
Sz-'
is
Indeed, s i n c e c l e a r l y
i s a union of c o s e t s of groups of t h e form Observe a l s o t h a t s i n c e
B , C E SuppS.
S
B n
c
X,
C
By o u r c h o i c e of
m, S
Sx-'
contains
m' 2 m
A.
c o s e t s of
Because
S n SX-'
t h e A-cosets of
come from
rn u AziX-5 i=1
m "hi n
(
(
i=1 A
But cosets of
rn terms.
rn'
a r e e i t h e r d i s j o i n t or i d e n t i c a l , so t h i s i n t e r s e c t i o n h a s
m' = m,
I t follows t h a t
rn U
m Azi
= ( U
i=1 and t h u s Now
H
z E H,
G
Az E Y,
Azi)z-l
i=1
proving (1).
permutes t h e s e t of r i g h t c o s e t s of
i s t h e s t a b i l i z e r of t h e f i n i t e s e t
c1 =
so
then
and it f o l l o w s t h a t
Y
= {Az
1
A
by r i g h t m u l t i p l i c a t i o n and
,Az , 2
..., A z r n } .
Hence, i f
281
COSET CALCULUS
H n Azi = @
Assume t h a t
on
K,
i
for a l l
E {1,2,
...,ml.
By (1) and t h e assumption
w e have
H n
where, by t h e above h y p o t h e s i s ,
H n T is
S =
a f i n i t e union of c o s e t s of t h e
groups i n
Y IYI
Since
lXrl <
/XI
=
and s i n c e
{H n Y
AIA
E
X'I
i s c l e a r l y c l o s e d under i n t e r s e c t i o n s , U
induction a p p l i e s here.
u
with some
with
Hence, a s b e f o r e , t h e r e e x i s t s
I
U
U
'K # 0
'K 'K...
=
= 1 such t h a t
j
~I= I
- -
IH : ( H n B )
1
< a
o
B E Y.
f o r some
for a l l
I
Since
-
g E G-H
h a s a n a p p r o p r i a t e form, t h e
r e s u l t follows i n t h i s case. Finally, i f
H n Az. # @
i,
f o r some
z = z
w e may assume t h a t
i
E H.
Since,
by ( Z ) ,
IH : H n ( z - l ~ z ) (
.
c o n j u g a t i n g t h i s e x p r e s s i o n by proved.
Remark.
2
E
H,
gJJ
The lemma above h o l d s w i t h
merely r e p l a c e a l l o c c u r r e n c e s o f
lH
yields
gKK
r e p l a c e d by
#K.
by
P-lJ
=
0
g
and s i n c e r e p l a c i n g
H n A( <
$J.
and t h e lemma i s
Indeed, i n t h e proof
Moreover, b o t h of t h e s e h o l d T h i s i s so s i n c e
i f r i g h t c o s e t s a r e r e p l a c e d by l e f t c o s e t s .
valent t o
:
g-
by
gJJ = 0
i s equi-
e f f e c t i v e l y interchanges
l e f t and r i g h t c o s e t s .
1.3.
LEMMA.
Assume t h a t
G
s t r o n g l y permutes t h e i d e a l s o f
R,
that
H
is a
n subgroup of
G
of f i n i t e index and t h a t
r i g h t c o s e t s of t h e subgroups
Hi
of
u Hihi i=1
H.
Let
i s a f i x e d f i n i t e union of
J
b e a nonzero i d e a l of
R
that
~ J = J
o
n
such
CHAPTER 6
282
L
Then t h e r e e x i s t s a subgroup
yi
w i t h some
Proof.
and a nonzero p r o d u c t
such t h a t
= 1,
( L n Hi) I <
1L:
Moreover,
G
of
Because
IG
:
i.
f o r some
HI <
and
J
# 0 , w e c a n choose t h e sequence
t o be of maximal s i z e s u b j e c t t o : !i) 1 E X
(ii)
Let
K
X E
=
and t h e
'~JSZJ...'SJ
G
gi
a r e i n d i s t i n c t l e f t c o s e t s of
l e f t c o s e t of
H
X
XXH =
XH.
f o r some
KK # 0.
Xgi E
H
G
f o r some
Then, by c o n s i d e r i n g t h e l e f t c o s e t s of
Indeed, i f
i
{1,2,.
X.
c o n t r a r a r y t o t h e maximality of that
in
# 0
and assume t h a t
H, we must have
H
i.
xXH # XI],
then
Xgsi
would be i n a new
.
T h i s would t h e n y i e l d
Because
1E X
.,S}.
and
XXH =
XH, w e
deduce
Moreover,
W i
JJ 3 - "KK # 0
s o , by h y p o t h e s i s s i n c e
xgi
H,
w e must have
n
o
~ K K=
We can now a p p l y Lemma 1 . 2 t o t h i s s i t u a t i o n t o o b t a i n a nonzero p r o d u c t
I w i t h some
i
yj
= 1
Y = Yl:2K...
and a subgroup
L
of
tK # 0
G with
and
~ I =I
o
L
:
(L n
Hi) <
f o r some
COSET CALCULUS
1 E X, I
Because
283
is a desired product of G-conjugates of
J,
as we wished to
show. 1.4. LEMMA.
Let
G
strongly permute the ideals of R
and let I
be an ideal
R
of
(i) GI =
C ' 1 is the smallest G-invariant ideal of R containing I SEG (ii) If I is G-invariant, then so are r ( l ) and L ( I ) , where P(l) and I, respectively.
are right and left annihilators of Proof.
(i)
J
then ' J = J 2 -'I
G preserves
2I
Conversely, if
J
for all g
and hence
E
G
and
G I.
(iii) Since I r ( 1 ) = k ( I ) I = 0
Hence g r ( I )C_ r ( I ) and ' E ( 1 ) follows by acting by Let I if
Since the action of
G-invariant ideal of R.
is a G-invariant ideal of R ,
J >
2 I.
It is clear that
arbitrary sums, GI is a
k(I)
and
'1
5 e(l)
=
for all g
I
for all g
E
G.
E
G, we have
The opposite containment
g-1 on both sides of the above containments.
be a G-invariant ideal of
R.
Then I
is said to be G-nilpotent-free
I contains no nonzero G-invariant nilpotent ideal of R.
G-semiprime if R
Thus R
is G-nilpotent-free.
We say that R
is
is G-semiprime if and only if
0 is the only G-invariant nilpotent ideal of R. Let
I
be a G-invariant ideal of R.
Then I
f r e e if for all nonzero G-invariant ideals A , B if
I
we have AB # 0 .
is G-annihilator-free, then it is G-nilpotent-free.
G-prime if and only if R 1.5.
5I ,
is said to be G-annihilator-
LEMMA.
ant ideal of (i) If (ii) If
Let G
Obviously,
Note also that R
is
is G-annihilator-free.
strongly permute the ideals of
R
and let I
be a G-invari-
R.
I is G-nilpotent-free, then r ( l ) = r ( 1 2 )
H is a subgroup of
C
of finite index, then I
is G-nilpotent-free if
and only if it is H-nilpotent-free. Proof.
(i) It is clear that r ( I ) f r ( 1 2 ) . Furthermore, by Lemma 1.4(ii),
I r ( 1 2 ) is a G-invariant ideal of
R.
Since
1 2 r ( 1 2 )= 0 ,
we see that
CHAPTER 6
284
I is G-nilpotent-free,
But
(ii) If
P(12)
5? ' ( I ) ,
as required.
I is H-nilpotent-free, then it is clearly G-nilpotent-free. I is G-nilpotent-free and let J
ely, assume that ideal of R sum
I r ( 1 2 )= 0 and
so
contained in I.
be an H-invariant nilpotent
H is of finite index, GJ is a finite
Because
of the nilpotent ideals ' J
Convers-
GJ is nilpotent.
and so
Thus
GJ = 0 and
therefore J = 0, as required. 1.6. LEMMA.
Let
G
R,
strongly permute the ideals of
G and let I be an ideal of R.
Suppose that
~ I =I o (i) HI is an H-invariant ideal of
R
If
for all g E G - H
with
g(Hr, )I'( (ii) Let I C -J
let H be a subgroup of
=
o
for all
g E G- H
I an H-invariant ideal and with J a G-invariant ideal.
with
is G-nilpotent-free (or G-annihilator-free), then I is H-nilpotent-free
,7
(or H-annihilator-free, respectively). (i) Let a,b
Proof.
E
H and g
E
Then b-lga
G-H.
9 H and therefore
-1
ga91 = b ( b g a I I ) proving that
H
' ( I)(
H7)
=
0 for all g
=
.
G - II
E
0 The fact that I '
is H-invar-
iant being obvious, (i) follows. (ii) Assume that
AB
=
0.
If g
E
A
and
B
H, then 'AB
are H-invariant ideals of = AB = 0.
~ Ac B ~ I =I Thus ' A B = 0 for all g E G. invariant ideals of R free then either Hence
GA
=
I is H-annihilator-free.
E
G-H,
contained in
=
with
o
'AGB
=
O.
GA
and
If J
0 which implies that either A Finally by taking A = B
gous result for nilpotent-free ideals. m
I with
then by hypothesis
It therefore follows that
contained in J
0 or GB
g
If
R
GB
are
G-
is G-annihilator=
0 or B
=
0.
we obtain the analo-
285
A - METHODS
2.
8 - METHODS
A
Throughout t h i s s e c t i o n ,
d e n o t e s a s t r o n g l y G-graded r i n g .
A
=
Thus
@ A gEG
( d i r e c t sum of 2-modules)
= A
for all
and
A A
Z Y
w
a c t s on
.
A
be t h e s e t of a l l i d e a l s of
Let
Owing t o Lemma 2.6.2,
Z,Y
E G
C
t h e group
by t h e r u l e
Id(A ) 1
g X = A XA
9
G
Furthermore, If
H
g-l
s t r o n g l y permutes t h e i d e a l s of
G,
i s a subgroup of
s t r o n g l y H-graded r i n g .
we w r i t e
A(H)
and a l l
g E G
Al.
@Ah hYi
for
R e c a l l from Lemma 2.7.3
A =A(H) 0 ( @ A )
X 6 Id(Al)
for a l l
so t h a t
A(H)
is a
that
( d i r e c t sum of l e f t and r i g h t
A(H) -modules)
9@ Hence t h e n a t u r a l p r o j e c t i o n
nH ( A )('
i s an
,A
) -himodule
:
(H)
A -A
homomorphism.
We s t a r t by r e c o r d i n g t h e f o l l o w i n g elementary o b s e r v a t i o n .
2.1.
LEMMA.
Let
H
be a subgroup of
9XX = 0 for a l l
with
( i ) XA X = 0
9
( i i ) XAX
for all
5X A ( ~)c A
g E G-H.
=
0.
i s a n i d e a l of
Then
then
g-l E G - H ,
-1 XX = 0
so
Hence
9
=
X(AoXA
a s required.
A = @ A S f G g'
and ( i i ) i s proved.
X
( ~ )
XA X
(ii) Since
and suppose t h a t
g E G-H
Proof,(i) If g E G - H ,
flX
G
m
w e have
"
)A
= flXA
g-l g
g
=O
and t h e r e f o r e
A,
CHAPTER 6
286
The following notation and terminology is extracted from Passman (1984).
H
be a subgroup of
The almost c e n t r a l i z e r of H
G.
in G ,
Let
written D G ( H )
is
defined by
D ( H ) = { g E G / I H : C,(g)I G It is clear that D (H) is a subgroup of G
G
<mj
H.
normalized by
W e put
A(H) = DH(H) so that H n DG(H) = A(H)
For t h e r e s t of t h i s s e c t i o n , we put that 0
(i.e.
and a s s m e t h a t
is the only G-invariant nilpotent ideal of
1 and
assume t h a t
R = A
are nonzero i d e a l s o f
J
IJ
=
A
R
i s G-semiprime Furthermore, we
R).
with
0 is a form for 1,J
(H,D,X,B)
Following Passman (1984), we say that the 4-triple if: (i) H (ii) X (iii)
G
is a subgroup of
and
D
=
is an H-invariant ideal of R
6E
J , XR # 0, and
I,J #
0
1 E R,
and
with
'XX
( G , A ( G ) ,R,R)
is a form for a l l
where
$ =
Z6
SuppB
with
.
G
-H
H
=
G, D
-
{Ol
= A(G)
and
X = R.
E
0# B
E
E
J
J.
t h e s i z e of t h e form, to be the number of right D-
Here
fig
E
we have
We define n = ( H , D , X , B ) , cosets meeting
0 for all g
Indeed, take
XI # 0 , XB # 0 for all B Thus
=
XI # 0
Observe that such forms always exist. Since
D (H) G
SuppB
is defined by
Ag
SEG 2.2.
LEMMA.
A = A W .
Assume that
(H,D,X,Bl
is a form whose size
n
is minimal and let
A - METHODS
287
is H-nilpotent-free and r ( X ) = r ( X 2 ) A A (ii) XITA(I) # 0 and there exist g G and a (i) X
we may assume that X n A (iii) If y E A ( D )
Proof. and
'XX
free.
=
(@)
-1
In particular, upon replacing B
(and hence that v D(6) # 0
# 0
(H,D,X,Ba)
such
is a
g
with X f i ~ ( B a ) # 0.
form of size n
A
since D
b,
by
>_ A )
,
v D ( B ) # 0, then X y B = 0 if and only if Xyv (6) = 0 . D (i) By hypothesis, f? is G-semiprime. Because X is H-invariant and
0 for all g
Hence, by Lemma 1.5(i),
ed in 2 1 ( X 2 ) .
Lemma 1.6(ii) implies that X
E G-H,
r(X)
Assume that y = Iy
= 21(X2).
R
R
is H-nilpotentis contain-
g
Then
A x2yg
=
for all g E G
0
and therefore
xyg
This yields X y g A -1 = 0, so
=
0 and hence y
Thus
E 21(X).
A
g
(x2)C
21
r(X)
- A
A
and the opposite inclusion being trivial, (i) follows. (ii) Because that
X I # 0 we have
XIA
g
# 0 for all g E G . X 7 i A ( I ) # 0.
I A C I, it follows immediately that g -
Because x6 # 0, we have xf3 # 0 for some g
g
E
G.
Taking into account
6
CB g' Therefore X B g A -1 # 0 and Now write
=
g
with X% a # 0. It is now clear that ( H , D , X , B a ) is -1 g g also a form with the additional property that X v D ( B a ) # 0. Moreover, we have
we can choose a E A
(H,D,X,Ba)#G ( H , D , X , O ) # so
(H,D,X,Lh)
(iii) If
XyB
also has minimal size =
0, then applying
?iD
n. yields X y n D ( 6 ) = 0.
Xyn. (61 = 0. Then, for any x E X , u less than n right cosets of D since that
By definition of
n,
we deduce that
we have xyB E J
(H,D,X,OB)
XXyB = 0
and therefore X2ye = 0.
Conversely, assume and
SuppxyB meets
cannot be a form.
Hence
for all x E X
Invoking (i), we conclude that xYB = 0, as required.
288
CHAPTER 6
Owing to Lemma 2.2, we may, and from now on we shall, choose a form whose size n
I,X, and g satisfy:
is minimal and
xnAUl #
A
where
o
Assume that
XnA(I)-XB # 0.
of finite index, a = La E I n A(H)
and
g
(iil
and
XnA(B) # 0
= A(H).
2.3. LEMMA.
(i) W
(H,D,X,B)
centralizes S u p p ~ ~ ( a )and
for some
Then there exists a subgroup
d
E Suppn
D
SuppnD(B)
and
W of
H
(a) such that
X
is W-nilpotent-free
w
u E
XnD(a)B >U(XadA-l)~D(a)B # 0 d (iii) for all y E F/,
Therefore, since D n ,4 = We now choose
A,
so that c1 E
Let
n (d =
D
ISuppnD(a) 1
IT
A
(a) and
XnD(ct)B # 0.
is minimal subject to
I n A ( H ) and XnD(a)
# 0
W be the intersection of the centralizers in H of the elements of
Supp?rD(Co and of Of
we have
D
=
SuppnD(B).
Since
SuppnD(a)
U
SuppnD(B)
D (H), it is clear that W has finite index in H . G
is a finite subset By Lemma 2.2(i),
is H-nilpotent-free, and hence it is also W-nilpotent-free by Lemma 1.5. (i) is established. (ii) Put
Y
=
n (a10 and write
n
c1=
cc1
SEG
, b =
c 6 SEG
and
y= c y SEG
Thus
X
-
L?
n (a)
Then
E ud &D
=
Y
and we d e n o t e by
Given
d
E
D
y
and
E
G,
g i v e n by
)
d-’
CZED XYT
R
t h e W-invariant i d e a l of
Z W( R n A
Y = I t w i l l n e x t be shown t h a t
289
METHODS
# 0.
D
w e have
= Y which i m p l i e s t h a t
YgAg-l XYy = 0.
Now assume t h a t
CY
XYY A
Then
9 g-l
0 so
=
g
for a l l
-
E
r,(XY).
E G
(1)
I t follows
from (1) t h a t
xyga -1 9
5 XY n r R ( X Y )
=
0
X
s i n c e t h e l a t t e r i s a W-invariant n i l p o t e n t i d e a l c o n t a i n e d i n
i s W-nilpotent-free, t o o u r c h o i c e of Since
XY
Hence
a.
=O
Thus w e have
X Y n D ( a ) a # 0,
d E SuppnD(n) and
9
u
and t h e r e f o r e
i t f o l l o w s from t h e d e f i n i t i o n of
E W
uE W C H
and
Y
A -1
Y
,
b
E
d y
X is
Note t h a t , s i n c e
y
-1
in
gy,
D
y
respectively. since
y E H
A
d-ly
Y
that there exist
with
nD ( a )6 # 0
H-invariant,
(ii)f o l l o w s .
(iii) T h i s p a r t u s e s t h e minimal n a t u r e of
a -1
XY = 0, c o n t r a r y
X Y T D ( a ) B # 0.
x“ (Rad AdBecause
XTID(C1)a =
and, by ( i ) ,X
.
Suppn
D
y E W
(a). F i x
and choose
Consider t h e f o l l o w i n g element
centralizes
d,
t h e summands on
In particular, i f normalizes
D.
g 8: D
I n case
O(g)
have g r a d e s
g
and
t h e n n e i t h e r of t h e s e g r a d e s i s
g E D, w e have
g E SuppTID(a)
so
CHAPTER 6
290
y
g
commutes w i t h
71
D
and t h e r e f o r e both t h e s e summands have g r a d e
g E D.
Hence
(y) = C u ( g ) and SED
I SUPPITD(y) 1 In f a c t , since
U(d) =
0, t h e above i n e q u a l i t y i s s t r i c t .
I SuppTD(a)I
W e now u s e t h e m i n i m a l i t y of hence, by a p p l y i n g
1 SuppTrD(a)I
that
ITo,
xTD(Y)TD(B)=
t o infer that
0.
XTD(y)6 = 0
As w e observed above,
a.
comes p r e c i s e l y from t h e D-homogeneous components of
and
TD(Y)
Therefore
and t h u s
for a l l
a -1
€
A -1
.
But
XA -1
= A
w
u
1/
w
f o r e f o l l o w s t h a t w e can c a n c e l t h e
-1X
since
X
i s II-invariant.
I t there-
f a c t o r and o b t a i n
A -1
Y
for a l l
b
E A
d-ly
d-ly 2.4.
.
So t h e lemma i s t r u e .
With t h e assumptions and n o t a t i o n of Lemma 2 . 3 ,
LEMMA.
-
t
t
H.g Here U Hisi i s a f i x e d f i n i t e union of r i g h t c o s e t s i = 1 2 i' i=l of t h e subgroups Hi and each Hi i s t h e c e n t r a l i z e r i n W of some element of for a l l
y
E
W
U
SuppB Proof. for a l l Let
R e c a l l t h a t , by h y p o t h e s i s ,
h E H.
-
D
X
i s H - i n v a r i a n t and hence
A X = XA h h
The above f a c t w i l l be f r e q u e n t l y used i n o u r d i s c u s s i o n below.
y E W and assume t h a t
Then w e have
and t h e r e f o r e , by Lemma 2 . 2 ( i i i ) ,
291
It follows that
and therefore finally
Write
a
D (a)+ 5
= IT
and
+
B = n,(B)
g.
Then, since
we have
We now examine in detail the four summands obtained from the above expression. Note that yu E W
so
y
normalizes D
and centralizes d E SuppnD(a).
In
particular, we have
a C - A -1 Y
Ad-l-1 d
Y
and from this follows easily that the sets
have supports disjoint from
D.
On the other hand, by ( 2 ) , we have
Thus, by ( 3 ) , this expression must be cancelled by terms from the fourth summand
In particular, the latter two summands must have a su-oport element in common.
.-.
Hence there exist f E SuppE,g
E
with -1
SuppB, a E Supp.irD(a) and
b E SuppnD(@)
CHAPTER 6
292
y
Because
E
W centralizes b
E Supp71D(61,
this yields
y-lgy = f - l a b Thus y
E
CW(g)x, some fixed right coset of C W ( g ) depending only on the
finitely many parameters f,g,a,b.
Because -,
g E SuppB = SUPPI?
the result follows.
-D
'
We have now accumulated all the information necessary to prove the main result of this section.
Let A
2.5. THEOREM (Passman (1984)). that the base ring R = A ideals of
A
with
IJ
H-invariant ideal X
X (i) ~ X =
of
is G-semiprime.
0. R
)
Suppose that
I and J are nonzero
Then there exists a subgroup H (3 E J
and an element
of
G, a nonzero
such that
for all g E G - H
X T ~ ( I )# 0, XnA(I?) # 0 where b
(ii (ii
o
=
be a strongly G-graded ring and assume
=
A(H)
XnA(I)*XO = 0
Proof.
We use the notation of the preceding three lemmas.
2.2, we may choose H,X
and
6 which satisfy (i) and (ii).
satisfy (iii), then the result is proved.
Owing to Lemma
If in addition they
Thus we will assume that x'l~(1) O X 6 # 0
D
and we derive a contradiction. Set y =
ir
D
(a)a= C y 6 G g'
Then, by Lemma 2.3(ii),
there exists
with
Since X
is H-invariant and
is a nonzero ideal of
we have
u E
W C_ H, we deduce that
R contained in X.
Moreover, since
3:
E
Suppy
n
- mTnons
293
-1
Y CXoldA
a-l
and t h e r e f o r e
w e deduce t h a t
Applying Lemma 2 . 4 ,
o r equivalently t h a t
y
for a l l since
u E W.
and hence
'YyX
In particular,
'YY
=
0.
IH
Because
L
t h a t t h e r e e x i s t s a subgroup
:
=
0
WI
.<
H
of
y
for these
E
t W - u H.g.u-' i=1 z. 'l
so
Lemma 1.3 may be employed t o i n f e r
m,
K =
and a nonzero p r o d u c t
Ol
O.5
Y 'Y...
Y
such t h a t
hKK = 0 Moreover,
Oi = 1
i
f o r some
so
K
5Y 5X
\L
and
:
-L
h E H
for a l l
( L n Hi) 1 <
f o r some
i. L
(L,DG(L) , K , 7 r D ( c l ) 6)
I t w i l l now be shown t h a t we f i r s t n o t e t h a t s i n c e
'KK
=
Hence, by Lemma 1 . 6 ( i )
,
LK
h E H-L
0 for a l l
gKK
=
i s a l s o a form.
0
R
i s a nonzero L - i n v a r i a n t i d e a l of
i s G-semiprime,
f r e e and i n p a r t i c u l a r
( L K ) 2 # 0.
But
a s claimed. L
W # 0.
for a l l
We c l a i m t h a t
LKY =
L
LK
g E G-L with
g E G-L i s &nilpotent-
KTD(a)B# 0.
LK(RYxA -1) = 0
so
w e have
and hence
Indeed,
LKY = 0.
X
rR( K)
f o r e , since
R
Lemma 1 . 6 ( i i ) i m p l i e s t h a t
LKyx = 0
i f t h e c o n t r a r y is t r u e , t h e n
L
2 X,
for a l l
g ( L ~ ) =L o~ Moreover, s i n c e
K
and
To t h i s end,
is L-invariant,
K f Y,
w e have
by Lemma 1 . 4 ( i i ) , so t h i s y i e l d s (
L
K)
= 0,
Moreover, t h i s i m p l i e s t h a t
a contradiction.
Ll?Y Hence
LKTI ( U ) # 0 so s i n c e
F i n a l l y , t a k i n g i n t o account t h a t
D
TI
(ol)6
E J,
0.
=
L
There-
XTID(a).B# 0 ,
13E
I w e have
we deduce t h a t
CHAPTER 6
294
L (L,DG(L),K,nD(a)O) i s a form. (H,D,X,B)
R e c a l l t h a t , by o u r c h o i c e ,
H ?L,
w e have
DG ( L ) Hence, s i n c e
lTD(a)E
A(D),
IL and
ffi = C V ( g I
g E DG(L)
Dg
and
f o r some
n
:
which meet elements i n
r i g h t c o s e t s of
- D.
5 DG(L). Suppa
D
DG(L).
right
But
m
IL
Accordingly,
Because
TID(B)
# 0,
:
CL(g)
n
I
<
so
D
t h e two D-cosets
DG(L).
merge t o t h e s i n g l e c o s e t
SuppnD(N)B meets less t h a n
conclusion i s t h a t
n
SuppITD(a)B meets a t most
(L n H ~ I ) <
g E Suppa
Dg
and, i n f a c t ,
1 DB(H) =
we see t h a t
0 and t h e r e f o r e a t most
c o s e t s of
W e now
( L , D G ( i ) ,LK,TID(a)B).
d e r i v e a c o n t r a d i c t i o n by computing t h e s i z e of t h e new form Because
n.
i s a form of minimal s i z e
r i g h t c o s e t s of
The
DG(L)
and
thus
(H,D,X,B),
S i n c e t h i s c o n t r a d i c t s t h e minimal n a t u r e of
t h e r e s u l t follows.
3 . THE MAIN THEOREM AND ITS APPLICATIONS
Throughout t h i s s e c t i o n , If
H
A
d e n o t e s a s t r o n g l y G-graded r i n g w i t h b a s e r i n g R = A
G, we w r i t e A ( H )
i s a subgroup of
s t r o n g l y H-graded r i n g w i t h b a s e r i n g Let For
h
E
H,N H
be subgroups of and
I
G
an i d e a l of
so t h a t
A (H)
is a
N
H).
R.
such t h a t
A('),
@Ah E H
for
H
normalizes
N
(e.g.
we d e f i n e
hI = A h I A
k-l Then, by t h e argument of Lemma 2.6.2, on t h e i d e a l s of (i) I f (ii) h
1 C_ J , ~
Note t h a t i f
=
A(N) then
h h
~I J ~
A(H)
t h e above formula d e f i n e s an a c t i o n of
such t h a t
hJ C - hJ
)
i s a c r o s s e d p r o d u c t of
H
over
R,
then
H
.
THE MAIN THEOREM AND ITS APPLICATIONS
295
-1
hI where
%
=
ZIE
A C H ) contained in
is a unit of
An ideal I of A")
for all
h
E
H
Ah'
is said to be H-invariant if
hI = I for all
h E H.
Since
IA
this clearly occurs if and only if if
N = 1, then A")
=
R
h
=
for all h E H.
AhI
and we obtain the action of
Observe also that
G on the ideals of R
exhibited in Sec. 2 . In what follows W
H denote subgroups of G and N is a normal sub-
Suppose I is a nonzero ideal of A ( ki)
K.
group of
and
the span of all elements a # 0 cosets of
of
d
Then we denote by min
I whose support meets the minimal number of
N.
Assume that N
3.1. LEMMA.
H-invariant ideal of A ( ' ) , (i) m i n d
and
W
are normalized by
(i) By definition, minNI# 0 .
If
W E
If
I is a nonzero
A(')
is a nonzero H-invariant ideal of
Proof.
H.
then
(ii) nN(I) is a nonzero H-invariant ideal of
mind.
.
A (N)
Let
E I be any generator of
W, then SUPPAW
c_
(suppol)w
and supPAwa c_ w (Suppol) Since N 4 CJ, Given
it follows that m i n d
h E H.
is an ideal of A ( ' ) .
we have
Since I is H-invariant and
N
is normalized by
H, we deduce that min I is A
H-invariant, as required. (ii) Because is an ideal of
rN
:
A
A(N)
.
--+ A(')
is an
Moreover, if
(A")
aE I
,A") and
1-bimodule homomorphism,
TI
N
(I)
h E H then, since H normalizes
296
N,
CHAPTER 6
w e have
‘“1)
and hence
g E Sup@,
i s H-invariant. w.A -1
then
5I
and c l e a r l y
X
f i n i t e subsets
Y
and
G,
of
G
If
sf”
is a right-orderedgroupif
<
respect t o the relation
g E G
t h e r e e x i s t s a t l e a s t one element
g = q
h a s a unique r e p r e s e n t a t i o n i n t h e form that
+ 0.
4
a E I.
C
=
is s a i d t o be a unique product group i f , g i v e n any two nonempty
G
A group
-1)
ITN(&
4
.
0# a
F i n a l l y , choose
xc X
with
G
t h e e l e m e n t s of
E
Y.
W e say
are l i n e a r l y o r d e r e d w i t h
z,y,z E G, x < y
and i f , f o r a l l
y
and
that
z z < yz.
implies
The f o l l o w i n g f a c t s are s t a n d a r d (see Passman ( 1 9 7 7 ) ) :
G
(i) I f
h a s a f i n i t e subnormal s e r i e s
1 = G~ Gi+l/Gi
with q u o t i e n t s
a
G
Q
... a cn
= G
G i s a right-
which are t o r s i o n - f r e e a b e l i a n , t h e n
o r d e r e d group.
G
(ii) I f
G
i s a r i g h t - o r d e r e d group, t h e n
3 . 2 . LEMMA.
W/N
Assume t h a t
ordered group).
If
I
T~
Proof.
i s a unique p r o d u c t group ( f o r example, r i g h t J
and
are nonzero i d e a l s of
A(’)
with
IJ
=
0,
then
( m i n d ) nN (minNJ) = 0
A(‘)
By P r o p o s i t i o n 2 . 1 . 7 ,
s t r o n g l y ) W/N-graded
i s a unique p r o d u c t group.
can be regarded as a ( n e c e s s a r i l y
AwN = €B Ax.
r i n g by p u t t i n g
Thus each
a E A(’)
can be
6wN w r i t t e n uniquely i n t h e form
a
=
-1
a-
&W/N Let
a = Ca-
and
X
W/N
Because
@ =
Z6Y
be g e n e r a t o r s of
x mind
i s a unique p r o d u c t group, we c a n l e t
and
--
x y
0 0
min-J, 1V
respectively.
be a unique p r o d u c t
element i n
a- D- = 0. I t f o l l o w s t h a t Cia- E I xo y o y0 s u p p o r t of t h i s element meets less c o s e t s of N t h a n does SuppcC.
Since
a6
=
0,
w e must have
and t h e
297
THE MAIN THEOREM AND ITS APPLICATIONS
Thus t h e minimal n a t u r e of
a6-
dictates that
c1
= 0
a-a- = 0 yo
and so
YO
3: E W/N.
A
s i m i l a r argument shows t h a t
N
meets l e s s c o s e t s o f
2,i
for a l l
E
W/N.
than does
a-6 X
Supp6.
J
E
for a l l
and t h e s u p p o r t o f t h i s element
Therefore
a R
E-
=
0
a-6-
and so
=
0
X Y
In particular,
a s asserted.
3.3.
LEMMA.
H ,H
Let
1
n
H
nH
=
,...,H
2
G,
has f i n i t e index i n
i
Ie
HI
:
G
be subgroups o f
of f i n i t e index.
Then
and i n f a c t
Ic
: H1/Ie :
H ~ I ... IG : Hni n
Proof.
x
If
G, t h e n c l e a r l y HX = n H.x.
E
.
IG
t h e r e are a t most t h e r e s u l t follows.
3.4.
G',
Then
[z,yl
Let
1
,x
2
=
H21
,...,x
1
IG :
Z
for
-1 -1
x y xy
G'
it f o l l o w s t h a t
By h y p o t h e s i s , {X
:
x
=
uz y
i'
=
HnI
z/ <
in
j
~ , Ey G.
u,V
gij =
G.
Indeed, i f
in
2
i'
=
J
s,y E G
5Z(G),
= IX
i' XJ. 1
of f i n i t e i n d e x .
G.
Since
[Z,yI -1
=
n
g
=
3: 6
n , we have
E
[+,yIn E 2.
[ r , y I n-1 y
-1 - 2 2 -1 n-1 y "11' y [ X r Y I y
Ix,y21 ly-lXY,yln-l
=
Zzi,y
Zx
j'
ij
= X -1y -1z L z , y l n y
b e i n g a n automorphism of
y-lIz,Yln-ly
and
and t h i s c l e a r l y i m p l i e s t h a t
i s of o r d e r
= z-ly-lzyIs,yl
= 3:
y
,...,H X ,
and o b s e r v e t h a t t h e s e a r e ,
[X X.1
= X-ly-lx(X-ly-lXy)
s i n c e c o n j u g a t i o n by
2
G, i s f i n i t e .
Put
G/Z
Because
iX,yIn+l
z
1
and s o w e may choose a f i n i t e t r a n s v e r s a l
G.
with
2 1 z
H x,H x
choices f o r
d e n o t e a commutator i n
Iz,yI We now f i x
:
is t h e s e t o f a l l f i n i t e p r o d u c t s o f commutators.
i n f a c t , a l l t h e commutators o f then
... IG
t h e commutator subgroup o f
Proof.
Iy,zl,
I
H I IG
G be a g r o u p w i t h a c e n t r a l subgroup
Let
LEMMA.
:
Taking i n t o a c c o u n t t h a t
i=1
'
G
implies t h a t
Iy-lzy,y-1y y ] n-1 = [y-lXy,y]n-l
Hence
CHAPTER 6
298
We now show t h a t e v e r y element o f
G'
can be w r i t t e n a s a p r o d u c t of a t most
n3
commutators, and t h i s w i l l c l e a r l y y i e l d t h e r e s u l t .
G' and u
u E
Assume t h a t
=
c c 1
rn > n3 t h e n s i n c e t h e r e a r e a t most qiij,
g = [z,yl
say,
2
...ern
n2
i s a product of
distinct
n+l
occurs a t l e a s t
gij,
times.
rn
commutators.
If
i t f o l l o w s t h a t some
We shift
n + l
of t h e s e
successively t o t h e l e f t using
and o b t a i n
i s a p o s s i b l y new commutator.
c!
where each
u
we c a n t h e n w r i t e
G' 3.5.
as a p r o d u c t of
H
Let
H
group of Proof.
and t h e r e f o r e
be a f i n i t e l y g e n e r a t e d subgroup of
H
t h e s e t of e l e m e n t s o f
of f i n i t e o r d e r .
and
H/Ho
Let
H =< h ,h
Hence e v e r y e l e m e n t of
commutators.
n3 o f t h e gij
i s a p r o d u c t o f a t most LEMMA.
rn-1
Applying
Then
Ho
G'
A(G)
is f i n i t e . . and l e t
Ho
be
i s a f i n i t e normal sub-
i s a f i n i t e l y g e n e r a t e d f r e e a b e l i a n group. 1
2'
...,hn>.
H
Then
I G : CG(hi)I
5 A(G)
implies t h a t for a l l
< 0 3
iE
{1,2,
...,nl
Moreover, it i s c l e a r t h a t
s o t h a t Lemma 3.3 y i e l d s
/ G : CG(H) 1 <
m.
Since
Z(H) = H n CG(H),
w e deduce
that
H'
Hence, by Lemma 3 . 4 , Now
H/H'
is finite.
is f i n i t e .
i s a f i n i t e l y g e n e r a t e d a b e l i a n group so i t s t o r s i o n subgroup
Since
H'
i s f i n i t e , w e have
W = H /H'
and t h u s
Ho
is a f i n i t e
THE MAIN THEOREM AND ITS APPLICATIONS
H.
normal subgroup of
H/Ho
I t follows t h a t
299
i s a f i n i t e l y generated t o r s i o n -
=
f r e e (and hence f r e e ) a b e l i a n group.
W e come now t o t h e main r e s u l t of t h i s s e c t i o n .
THEOREM (Passman (1984)).
3.6.
.
R = A
ring
A
Then
A
Let
be a s t r o n g l y G-graded r i n g w i t h b a s e
c o n t a i n s nonzero i d e a l s
I,J
IJ = 0
with
i f and o n l y
i f there exist:
H C G
N
( i ) subgroups
N
with
X
(ii) an H - i n v a r i a n t i d e a l
finite
R
of
?,2
(iii) nonzero H - i n v a r i a n t i d e a l s
I
Furthermore,
=
J
perties.
0
g E G-H
for all
with
y , ? c- X A ( N )
J
and
e x i s t and s a t i s f y t h e a p p r o p r i a t e pro-
A?A
=
J = AYA
and
so t h a t t h e s e a r e nonzero i d e a l s of t h e s t r o n g l y G-graded r i n g
IJ
shown t h a t
=
g E H,
i s H-invariant,
-I A -J g
g E G-H,
On t h e o t h e r hand, i f
=
--
A
Ng
and
5G -
(1)
A IJ = 0 g
-I,J- C_ X A ( N ) ,
we have
c X A ( ~ ) XA A ( ~ )
9 -
C
g E G
we have
then since
?A J =
I t w i l l be
for all
g
-I
t h e n because
A(N)A
A.
0 or equivalently t h a t
-I A -J = O
Now
-IJ = O
Set
I
If
and
T = J.
H,N,X,I
Assume t h a t
=
A(')
of
- -
i f and o n l y i f c
Proof.
'XX
with
9
H
so Lemma 2 , l ( i ) y i e l d s
$Ng XA(N)A X = 0
9
proving (1). implies t h a t
We have t h e r e f o r e shown t h a t
I
ideal
I
of
Moreover,
-I -J =
clearly
= J.
Conversely, assume t h a t Suppose t h a t
IJ = 0.
I
and
J
R
i s n o t G-semiprime.
R
with
t h e r e s u l t follows.
-I 2
=
0.
a r e nonzero i d e a l s of
A
with I J
=
0.
Then t h e r e e x i s t s a nonzero G - i n v a r i a n t
Hence, t a k i n g
H = G, X = R , N = 1 and
-J -I, =
CHAPTER 6
300
By t h e f o r e g o i n g , w e may h a r m l e s s l y assume t h a t a l l o w u s t o employ Theorem 2.5.
R,
r i a n t i d e a l of
ti) ~ X = X (ii)
B
and
o
E
X
a nonzero H-inva-
such t h a t
g
for a l l
E
G-H
A
where
=
A (H)
Xr (II.X@ = 0
(iii)
-
A
H
We have t h e r e f o r e found an a p p r o p r i a t e and
i s G-semiprime which w i l l
H C- G ,
Thus t h e r e e x i s t
J
X r A (I)# 0 , XnA (6) # 0
R
-J .
X,
and
and we are l e f t t o f i n d
N,I
To t h i s end, p u t
I
=
XnA(II
II
By Lemma 3 . 1 and ( i i ) above,
J
and
= X*H[A(A)nA(fi)A(A)]
A
i s a nonzero H - i n v a r i a n t i d e a l of
It
follows from (iii) t h a t
XTTA(I).XnA(a, and s o , by ( i i ) a g a i n , w e s e e t h a t
I J
with
0.
=
J
=
0
i s a nonzero H - i n v a r i a n t i d e a l of
A")
Observe t h a t
1 1
X r A ( B ) C_ X n A ( J ) , w e have
and, s i n c e
I
Because b o t h
and
J
1
f i n i t e l y many H-conjugates,
t h e r e e x i s t s a normal subgroup
(bf
W
(c)
both
Obviously
2
W
H
of
W & A
with
i s f i n i t e l y generated
I
I n A(')
= 2
2
has only
it i s c l e a r t h a t :
(a)
I J
Q(H)
are nonzero and e v e r y element of 1
and
1
I
and
J
J
n A(')
= J 2
a r e nonzero
1
A(')
are I l - i n v a r i a n t i d e a l s of
XA(')
contained i n
with
=o.
By ( b ) ,
W
i s a f i n i t e l y g e n e r a t e d subgroup of
N
h a s a f i n i t e c h a r a c t e r i s t i c subgroup
I
= mindz 3
,I
= IT 4
N
with
(I ) , J 3
= 3
W/N
min
A(H)
torsion-free abelian.
and NJZ
so by Lemma 3 . 5 ,
J
= TI ( J 4
N
3
1
W Set
301
THE M a I N THEOREM AND ITS APPLICATIONS
N 4 H,
Decause
I
Lemma 3.1 i m p l i e s t h a t
J
and
a r e nonzero H - i n v a r i a n t 3
A(')
i d e a l s of
A(')
i n v a r i a n t i d e a l s of
W/N
XA(')
contained i n
I
and t h e n t h a t
XA(N)
contained i n
.
IJ 4
=
0.
N
I
F i n a l l y , assume t h a t
J J'
w e have
=
0.
J.
=
C
1 -
-
I = S
t h u s we can t a k e
=
J
4
.
2
= 0
and
I
and
For g e n e r a l 5
I = I
as above,
J = J
and
4
.
Then s i n c e
Xn ( J ) A
=
XnA(l)
=
I
1
J? = 0 for a l l
I t t h e r e f o r e f o l l o w s a s above t h a t 5
2
4
*
t h e r e s u l t now f o l l o w s by t a k i n g
IJ
Moreover, s i n c e
i s an o r d e r e d group, Lemma 3 . 2 i m p l i e s t h a t
J,
are nonzero H-
J4
and
.
i
T h i s completes t h e proof of t h e theorem.
and
The rest of t h i s s e c t i o n w i l l b e devoted t o some a p p l i c a t i o n s of t h e preceeding r e s u l t .
R
I n what f o l l o w s t h e assumption t h a t
i s G-semiprime
i s made i n
o r d e r t o avoid t r i v i a l i t i e s .
G
Let
R
s t r o n g l y permute t h e i d e a l s of
X b e an i d e a l of
and l e t
R.
Then w e d e n o t e by
G ( X ) = {g E G I g X t h e s t a b i l i z e r of
X
=
X}
X i n G. Following Passman (1984), w e s a y t h a t t h e nonzero i d e a l
R i s a t r i v i a 2 i n t e r s e c t i o n idea2 i f f o r a l l
of
'X n X
Note t h a t
=
0 implies
=
0 since
gXX
a
E
5 gX
G,
f!
X.
The f o l l o w i n g
simple o b s e r v a t i o n w i l l clear o u r p a t h . 3.7.
Let
LEMMA.
R
z e r o H - i n v a r i a n t i d e a l of
G(X)
(i)
(ii)
H
be a G-semiprime r i n g ,
R.
If
a subgroup o f
gXX = 0 f o r a l l
g E G-H,
G
and
X
a non-
then
= H
X is a t r i v i a l intersection ideal
T is
(iii) I f
a l e f t transversal for
H
in
G,
then
t
GX = @ X
eT Proof.
Because
R
n i l p o t e n t - f r e e and hence
g E G-H
gX
=
X
,
then since
gX # X
is G-semiprime, it f o l l o w s from Lemma 1.6 t h a t
X
f-
i l R ( X ) = 0.
and so
g 4 G(X).
X is &invariant.
Thus
In particular,
X 2 # 0.
On t h e o t h e r hand, i f
g E G(X)
and t h e r e f o r e
X i s H-
Hence, i f
g
E
G(X)
H, =
then
H,
CHAPTER 6
302
proving ( i ) . Next observe t h a t
and t h i s c l e a r l y y i e l d s t h e d i r e c t sum i n ( i i i ) . P r o p e r t y (ii)b e i n g a consequence of (iii), t h e r e s u l t f o l l o w s .
8
The f o l l o w i n g r e s u l t i s a r e f o r m u l a t i o n and s l i g h t e x t e n s i o n of Theorem 3.6. 3.8. ring
Let
A
be a s t r o n g l y G-graded r i n g whose b a s e
Then
A
c o n t a i n s nonzero i d e a l s
THEOREM (Passman ( 1 9 8 4 ) ) .
R = A
is G-serniprime.
I,J
with
1
id = 0
i f and o n l y i f t h e r e e x i s t :
X
(i) a t r i v i a l intersection i d e a l
N
( i i ) a f i n i t e group
G(X)
with
R
of
equal t o i t s normalizer i n
-I,J-
(iii) nonzero G ( X ) - i n v a r i a n t i d e a l s
X A") Furthermore, Proof.
I = J If
Conversely, assume t h a t use i t s notation.
and
>
I
and
with
-IJ-
and
=
0
-I -J .
i f and o n l y i f
X,N,?
A(N)
of
G
=
I
e x i s t , t h e n so do
J
N
J
by Theorem 3.6.
Then we may a p p l y Theorem 1 . 3 and we
exist.
In particular,
and
i s a f i n i t e group and
N C_ H
5 NG( N )
=
z.
Because
gxx
=
0
for all
G-H
g E
5G - H
f
i t f o l l o w s from Lemma 1 . 6 ( i ) t h a t
HX
=
gy? Owing t o Lemma 3.7,
?
i s an ??-invariant i d e a l of for a l l
= Q
R
with
g E G-8
i s a t r i v i a l intersection i d e a l with
G(X) L e t u s c o n s i d e r t h e a c t i o n of
=
3H
=
on
N (N)
G
A(').
-
?=
H-
I
;=
and
so t h a t t h e s e are k i n v a r i a n t i d e a l s of
gxx
A") =
0
Put
z-J contained i n
?A'".
for a l l
Because
g
E
G-K
303
THE MAIN THEOREM AND ITS APPLICATIONS
5 H,
N
and s i n c e
i t follows e a s i l y t h a t
.
g(XA"))
XA(N) = 0
for all
g E
ii-H
for all
g E
ii-H
and t h e r e f o r e t h a t
-I
However,
gI":
--
H I is
s i n c e t h e r i g h t a n n i h i l a t o r of
77
f i n i t e normal subgroups of
G.
z-iI J=O
=
A+(G)
W e d e n o t e by
A+(G)
=
R,
of
A
i s G-prime.
1
of
R
N
and
A
Let
A
t o b e prime o r semiprime.
be a s t r o n g l y G-graded r i n g whose b a s e
A
Then
W e use Theorem 3.8.
Proof.
h a s no non-
Assume t h a t f o r e v e r y t r i v i a l i n t e r s e c t i o n i d e a l
A + ( G ( X ) ) = 1.
we have
G
We now u s e Theorem 3.8 i n o r d e r t o o b t a i n
COROLLARY (Passman ( 1 9 8 4 ) ) .
R
t h e j o i n of a l l t h e
= 1 i f and o n l y i f
s u f f i c i e n t c o n d i t i o n s f o r t h e s t r o n g l y G-graded r i n g
ring
Therefore,
5 - i n v a r i a n t , w e conclude t h a t
Thus
t r i v i a l f i n i t e normal subgroups.
3.9.
-
I J = 0.
so t h i s y i e l d s
8
G be an a r b i t r a r y group.
Let
0
j+
I J = (3,
i s H - i n v a r i a n t and
as required.
=
i s prime.
X
Assume that
G
a f i n i t e subgroup of
X
with
is a t r i v i a l intersection ideal
NG(N) = G ( X ) .
Then
Nc_A+(G(X)) = 1 and so
N = 1, A ( N ) = R A(N) = R
follows t h a t
and
G ( X ) = N ( N ) = G. G
Because
R
i s G-prime,
c o n t a i n s no nonzero G - i n v a r i a n t i d e a l s
-I
and
J
it with
*-
IJ
=
0.
Hence, by Theorem 3.8,
3.10.
COROLLARY (Passman ( 1 9 8 4 ) ) .
ring
R = A Proof.
i s (;-prime. Let
.
is torsion-free. C o r o l l a r y 3.9.
X
If
G
be an i d e a l of Hence
i s prime. Let
A
be a s t r o n g l y G-graded r i n g whose b a s e
i s torsion-free,
R.
Since
then
G(X)
A
i s prime.
i s a subgroup of
G, G ( X )
b + ( G ( X ) ) = 1 and t h e r e s u l t f o l l o w s by a p p e a l i n g t o
3.11.
COROLLARY (Passman (1984)).
ring
R = A
i s prime.
A
Then
A
Let
A
be a s t r o n g l y G-graded r i n g whose b a s e
i s prime ( o r semiprime) i f and o n l y i f , f o r
CIIAPTER 6
304
N
e v e r y f i n i t e normal subgroup
G,
of
i s G-prime
( o r G-semiprime,
respec-
tively). Proof.
Since
i s prime, it i s c e r t a i n l y G-semiprime.
R
R
t r i v i a l i n t e r s e c t i o n i d e a l of
g E G - H , we i n f e r t h a t H = G
G(X) =
with
and hence
X
H.
'XX
From
X be any
Let = 0
for all
i s a G - i n v a r i a n t i d e a l of
now f o l l o w s t h a t i f ( i ) , (ii)and (iii) of Theorem 3.8 are s a t i s f i e d f o r
X
they a r e s a t i s f i e d f o r
=
R.
R.
X,
It
then
The d e s i r e d c o n c l u s i o n i s t h e r e f o r e a consequence
of Theorem 3.8.
The problem o f p r o v i d i n g n e c e s s a r y and s u f f i c i e n t c o n d i t i o n s under which t h e crossed product open.
R*G,
i s f i n i t e , i s prime and semiprime i s s t i l l wide
G
where
Using Theorems 5.3.11
and 5.4.11,
it i s e a s y t o show t h a t t h e primeness of
a f i n i t e c r o s s e d p r o d u c t e v e n t u a l l y depends on t h e G-primeness of c e r t a i n s u i t a b l y c o n s t r u c t e d t w i s t e d group a l g e b r a .
But even t h e q u e s t i o n of primeness of a
f i n i t e t w i s t e d group a l g e b r a i s e x t r e m e l y complicated.
Some p a r t i a l r e s u l t s i n
t h i s d i r e c t i o n a r e e x h i b i t e d i n Sec. 5.
R
Let
H
be G-prime,
be t h e s t a b i l i z e r of
Sylow p-subgroup of
G
let
Q
Q in
R
be a prime i d e a l of
G.
H
= 0
SfG
G P
let
and l e t
be a
and d e f i n e
Then t h e r e i s a n a t u r a l l y determined c r o s s e d p r o d u c t ring.
p,
F o r e a c h prime number
e x t e n d i n g one of
n Q'
with
R *G
P
and
P
R
P
i s a G-prime
We c l o s e t h i s s e c t i o n by q u o t i n g some r e s u l t s which a r e r e l a t e d t o t h e
work of t h i s s e c t i o n , b u t which w i l l n o t be used i n t h e s e q u e l . 3.12.
THEOREM (Lorenz and Passman
a f i n i t e group notation, 3.13.
R*G
over
R
where
G over
so i s
R*H.
R
R
and l e t
H
Let
R *G
P
R*G
P
Let
R*G
Then, w i t h t h e above
i s prime for a l l
p.
b e a c r o s s e d p r o d u c t of t h e f i n i t e
be a subgroup of
THEOREM (Passman ( 1 9 8 5 a ) ) .
be t h e c r o s s e d p r o d u c t of
i s a G-prime r i n g .
i s prime i f and o n l y i f
THEOREM (Passman ( 1 9 8 5 a ) ) .
group
3.14.
G
R*G
(1980~)). Let
G.
If
R*G
i s semiprime, t h e n
be a c r o s s e d p r o d u c t of t h e f i n i t e
SUFFICIENT CONDITIONS FOR SEMIPRIMENESS
G
group
f o r each prime
p
G.
G
HI-torsion,
:
H
and l e t
Let
be a subgroup of
then
R*G
G.
i s semiprime.
R*G
R*G
Then
has p-torsion,
THEOREM (Passman ( 1 9 8 5 a ) ) .
group
IG
.
R
such t h a t
a Sylow p-subgroup of
3.15.
R.
o v e r t h e semiprime r i n g
305
i s semiprime i f and o n l y i f
R*G
we have
P
G
semiprime f o r
P
be a c r o s s e d p r o d u c t of t h e f i n i t e
.
If
R*H
R
i s semiprime and
h a s no
4. SUFFICIENT CONDITIONS FOR SEMIPRIMENESS
A
Throughout t h i s s e c t i o n ,
R = Al.
Our aim i s t o o b t a i n some s u f f i c i e n t c o n d i t i o n s f o r
V
Let
torsion i f say t h a t
be an a d d i t i v e group.
lGlv V
V
of
=
0
for
v
E
If
I?
implies
S
Let
i s called essextial i n
V
V
I n what f o l l o w s , i f
4.1.
A
t o be semiprime.
v = 0.
If
G
V
H
X
then
VR
=
no IGI-
G, V
of
b e a l e f t S-module.
implies
has
i s a r b i t r a r y , t h e n we h a s no
A submodule
X
i n c a s e f o r e v e r y submodule
i s an A-module,
V
i s f i n i t e , w e say t h a t
be any r i n g and l e t
W n X = 0
to
G
h a s n o IGI-torsion i f , f o r a l l f i n i t e subgroups
IHI-torsion.
W
d e n o t e s a s t r o n g l y G-graded r i n g w i t h b a s e r i n g
V,
of
0 V
d e n o t e s t h e r e s t r i c t i o n of
R. Let
LEMMA.
V
be an A-module and l e t
(i) A W
i s a submodule of
(ii) I f
W
9
(iii)
is essential i n
nA W sfG
Proof.
VR,
for a l l
VR,
then
i s an A-submodule o f
B
W
be a submodule of
Then
vR-
g E G
A W
is essential i n g contained i n W
( i ) T h i s i s a d i r e c t consequence of t h e e q u a l i t y
VR,
RA
for all
=
g
A
9'
g
E G
for a l l
g E G. (ii) Assume t h a t
X
f o l l o w s from ( i ) t h a t essential in
VR,
i s a nonzero submodule of
A
-1X
9
w e have
i s a nonzero submodule of
- lX #
W nA
9
proving t h a t
A W
9
VR.
is essential i n
vR-
0
so
Because
vR*
A
A
9-1 9
Since
= R,
W
is
it
306
CHAPTER 6
nA W
v E
(iii) I f
x
and
E G,
then
@C g
A X v C_ AxAgW
=
A
xg
W
g
for all
E G
I t follows t h a t
and t h e lemma i s proved. 4.2.
gi
E A
g
w
W
5V
i s f i n i t e and t h a t
VR.
Write
b -1 E A -1, and l e t g i 9
and
~
@Cg
-sfCxg
G
Assume t h a t
LEMMA.
i s a d i r e c t summand of
WR
a
.
w= n
n~
A V C
1 :
TT
=
V
Cagz.b -1
4 W
g
are A-modules such t h a t
,
a f i n i t e sum w i t h
i
b e t h e p r o j e c t i o n map.
Then
t h e map
i s an A-homomorphism such t h a t
Ibl Proof.
If
w E W,
then
b -1 IJ
E
=
W
/G/w and so
W
A
CB 1
= X
i s an A-submodule of
V, A
n ( b -1
W) =
g i
g i
Since
for a l l
w
b -1
E
W Hence
W.
9 ;
i s c l e a r l y an a d d i t i v e map.
Because
we a r e l e f t t o v e r i f y t h a t
a
A(a
V ) 2
To t h i s end, observe t h a t f o r each
!loreover, s i n c e
=
g E G,
E R , TT
axX(vi
for a l l
ax
E
Ax, v
we have
i s an R-homomorphism and :apib
-1
9 i w e have
E V
=
1,
SUFFICIENT CONDITIONS FOR SEMIPRIMENESS
307
as required. As an application, we now establish a version of Maschke's theorem for strongly G-graded rings.
4.3. PROPOSITION (Nastacescu (1983)).
1 GI -torsion and there exists an A-submodule U of v are A-modules with no
Furthermore, if Proof,
U
=
V
U
Then
i ( V ) = lGlV =
Now let
0.
V E
vR
with
5v
We ll
such that
is essential in
Then
vR
-
W be the A-homomorphism defined in Lemma 4.2 and let
v.
If
V has no IGI-torsion,
Since V
hf
is finite, that
WR is a direct summand of VR.
is an A-submodule of
and set
and so X ( l G l V - w ) = 0. module of
that
G
that
W is a direct summand of V .
/GIV, then
h : V-+
Let
Keri.
=
Assume
W = x(V) E
x
=
= 0
V
W n U , then by Lemma and hence
WnU
=
4.2,
0.
Then
l G l l r C _ W @ U.
Thus
( W @ U) n
W.
V E
In particular, if
X is a
sub-
0, then
lGiXZ(W@U) n X = O Since
v
tial in
has no IGI-torsion, we conclude that VR.
Finally, if
v=
( G ( V , then V
x =
=
W
0.
@
U.
Hence
W @ U is essen-
m
The following result is known as the "essential version" of Maschke's theorem for strongly G-graded rings. 4.4. PROPOSITION (Passman (1984)).
Assume that
are A-modules with no /GI-torsion.
Then
W is essential in Proof.
If
G
is finite and that
WC -V
W is essential in V if and only if
vR'R.
W is essential in VR,
Conversely, assume that
W
maximal with respect to
IJ n L = 0.
then certainly IJ
is essential in Then
I/
and let
We L
is essential in
L be a submodule of
is essential in
VR
V. vR
and so, by
CHAPTER 6
308
Lemma 4.l(ii), for all g E G
A (W @ L) is essential in vR g is the finite intersection E =
E
Furthermore, if essential in
VR
and E
submodule of
V,
it follows that
Hence
WR
exists an A-submodule U
WnU
=
W
We u
0,
so
ER
We U vz.
essential in
But
G
E
ER.
is
is an
W 6 (E n L)
=
A-
.
Thus there
The latter
.
be a finite normal subgroup of
If X
E
W is essential in V
W is essential in vR.
U = 0 and
variant ideal of R .
and let X
and
be a G-in-
is G-nilpotent-free with no lNl-torsion, then XA")
A W"
is a G-nilpotent-free ideal of Proof.
and thus
and Proposition 4 . 3 applies.
E with
of
is essential in
Let N
4.5. LEMMA.
5E 5W @ L
then
Since W
is an A-module by Lemma 4.ltiii).
is a direct summand of
implies that
n A (W@ L ) , &G
We first claim that XAw)
has no INI-torsion.
Indeed, assume that
c X. Because X l G l y g A -1 = 0 and Y A g g-1 9 'ygAg-l = 0 and hence that y = 0 for all
Then, for all g E G, we have no IG/-torsion,we deduce that
g
has
g
E
Hence y = 0 as claimed. Now assume that I with 1.'
=
0.
If L
is a G-invariant ideal of A(N) is the right annihilator of
two-sided ideal of 4").
Indeed, assume that K
XA'").
If
IK # 0, then 1 '
is a nonzero left ideal of =
Since L
X
=
L
then L
is a
is essential in
A(N)
contained in
+ IK S L n K
0, then 0 # K c - L n K.
is essential in
from Proposition 4.4 that L n
=
L
XA(N)
0 implies that
o IK
I in X A " ) ,
Moreover, as left A(')-modules,
XA").
On the other hand, if
contained in
XAcm)
and
X A ( m ) has no IN/-torsion,we deduce
is essential in the R-module
r X U ) is essential, as a left R-module, in X L' = r X U )
XA").
5 XA(N).
Hence Now put
G.
309
SUFFICIENT CONDITIONS FOR SEMIPRIMENESS
and note that L'
is a G-invariant ideal of R
X we have
aL' = 0
X
X and that X is
Hence L' n R (5') = 0 and since L'
G-nilpotent-free, by hypothesis. essential in
contained in
is
X
Finally, if
( L ' ) = 0.
c1 =
ICYE I 5 XA"),
then
9
implies that
A
5 X,
Hence, since A
for all g
u L ' = O
9-1 Y
G
we have
9
A -lag C_ i l X ( L ' )
= 0
9 and therefore
Thus I = 0 and hence
0.
CY =
CORnLLARY (Passman (1984)).
4.6.
ring
R = A
is G-semiprime.
X of R,X
has no
Proof.
Let X
Let A
XA(N)
is G-nilpotent-free.
be a strongly G-graded ring whose base
Suppose that, for every trivial intersection ideal
)A+(G(X))I-torsion.
Then A
is semiprime.
be any trivial intersection ideal of R
and set H = G ( X ) .
Then
gxx c SI n I = so Lemma 1.6(ii) implies that
finite normal subgroup of
o
for all g E G - H
X is H-nilpotent-free.
H.
Then N
5 A'(G(X))
so,
IN/-torsion.
Invoking Lemma 4.5, we conclude that
ideal of A(')
and Theorem 3.8 yields the result.
of
G
over a semiprime ring R
is a
by hypothesis, X
XA")
The special case of the following result, in which skew group ring
Assume that h'
has no
is an H-nilpotent-free
G is finite and A
is a
is due to Montgomery and Fisher
(1978). 4.7. COROLLARY (Passman (1984) 1. ring R
=
A
Proof. ideal
is G-semiprime. Since R
X of R.
Let A If R
be a strongly G-graded ring whose base
has no /GI-torsion,then A
has no /GI-torsion, X has no
is semiprime.
)A+(G(X)))-torsion for any
The desired conclusion is therefore a consequence of Corollary
4.6.
We close this section by providing an important application of Corollary 4.7. Let G
be a finite group acting on the ring R.
relationship between R
and the fixed subring
We are concerned with the
CHAPTER 6
3 10
for all
R
R e c a l l t h a t t h e t r a c e map on
GI
g
i s d e f i n e d by
i
R %RG
r
Given an a c t i o n of
R.
G
Since
notation.
G
R,
on
w
Igp
SfG
we now form t h e skew group r i n g
R*G
of
G
over
i s c o n t a i n e d i s o m o r p h i c a l l y i n t h i s r i n g , w e do n o t u s e t h e o v e r b a r
G'
A s u s u a l , we d e f i n e
by
G'
C g E R*G
=
SEG and observe t h a t for a l l
g E G,
we have Gg '
4.8.
Gi = gG+
=
THEOREM (Bergman and I s a a c s ( 1 9 7 3 ) ) .
R
the ring
,-d
I # 0
(i) If
suppose t h a t
R
G
Let
be a f i n i t e group a c t i n g on
i s semiprime w i t h no IGI-torsion.
R,
i s a G - i n v a r i a n t r i g h t or l e f t i d e a l of
IG# 0 and
then
trl # 0. R~
i s semiprime.
Proof.
(i) Let
(ii
I
is s u r e l y an (R,R)-bimodule. and s i n c e
RG'I
G'
R.
be a nonzero G - i n v a r i a n t r i g h t i d e a l of
R
Furthermore, s i n c e b o t h
G,
a b s o r b s f a c t o r s from
i s i n f a c t a nonzero i d e a l .
RG'I
and
i s a n i d e a l of
I
RG'I
Then
are G - i n v a r i a n t
R*G.
I # 0,
Since
We c l a i m t h a t
Indeed, by i n d u c t i o n
But
G'IRG'
=
G'IG'
=
By C o r o l l a r y 4.7, T h u s , by ( * ) ,
where
I
(ii) L e t
G'trI, R*G
i s semiprime so t h e i d e a l
t r l i s not n il p o ten t.
i s a l e f t i d e a l of
A
by Lemma 2 . 8 . 3 , so t h e r e q u i r e d e q u a l i t y h o l d s .
R
In particular
RG'I
is not nilpotent.
0 # trI
2 1'.
The c a s e
is handled s i m i l a r l y .
be a nonzero r i g h t i d e a l of
nonzero G - i n v a r i a n t r i g h t i d e a l of
R
RG
and s e t
and so, by ( i ) ,
I = AR.
trI
Then
I
is a
is not nilpotent.
TWISTED GROUP ALGEBRAS
trI
But
5A
RG
so
311
is surely semiprime.
5. TWISTED GROUP ALGEBRAS
Throughout this section, G
denotes a finite group, F
the group of all F*-valued 2-cocycles of
G on F*.
of
G over F .
group algebra of
F'G
Thus
Z2(G,F*)
G with respect to the trivial action
E Z 2 ( G , F * ) , we write
Given
a field and
F'G
for the corresponding twisted { i l g E GI
is an F-algebra with F-basis
and with multiplication determined by
-_
z y =
-
'(x,y)xy
for all
z,y E G
1
dimF'G = GI m, F'G is semiprime (respectively, prime) if and only if F is semisimple (respectively, simple). The aim of this section is to examine
Since F'G
the following problems:
Problem A . and
G
Given F'G,
for F'G
Problem B. and
and
G for F f G /xJ V aGI
G
Given
for F'G
Problem D.
ci
to be a separable F-algebra?
F'G,
what are necessary and sufficient conditions on
u E ZZ(G,F*)
H x H
c1 :
such that F'G
+F*
GxG
is also a cocycle.
G and F
and its restriction to
Z2(H,F*).
with the subalgebra of
F'C
{Xjh E
and a subgroup H
of
G, the restriction of
To prevent our expressions from becoming too
Z2(C,F*)
With this convention, we may identify H'F
consisting of all F-linear combinations of the ele-
HI.
We start by providing some useful general observations. extension.
for the
is a central simple F-algebra?
cumbersome, from now on we shall use the same symbol f o r an element of
ments
F, a
to be simple or central simple?
Given a cocycle to
what are necessary and sufficient conditions on F ,
What are necessary and sufficient conditions on
existence of
ci
c(
to be semisimple?
Given F'G,
Problem C.
F,
what are necessary and sufficient conditions on
Then the embedding
F-
E
Let
E/F
be a field
induces an injective homomorphism
Thus Z2(G,F*) can be identified with the subgroup of
z z (G,E*)
consisting of all
312
CHAPTER 6
For this reason, from now on we shall use the same symbol
F*-valued cocycles.
Z2(G,F*)
for an element of
Z2 (G,E*).
and its image in
will always be clear from the context.
The precise situation
The foregoing, however does not mean that
we can identify H2(G,F*) with its image in morphism
c1
-
TP(G,F*)
H2(G,E*) since the natural homo-
H~ (G,E*)
need not be injective. Let c1 E Z2(G,F*) and let E
5.1. LEMMA.
F.
be a field extension of
Then
the map
is an isomorphism of E-algebras.
z,y
B
E
GI be an F-basis of FzG
g
E
G, put
Let
G.
For each
E
@ G'F
F
{gig
proof.
=
1 8 g.
with
Then
.;
=
Gig
5 '
a(Z,y)G
for all
is an E-basis of
and, for all z,y E G,
-sy
=
(1 8 5) (1 8 y,
= a ( z , y ) (1 €3
= 1 8
;.y'
= 1 8
a(z,y)G
.y,
.
= a(s,g)6
So the lemma is true.
5.2. LEMMA.
Let
G be a cyclic group of order m
G'F In particular, F'G nomial in FIX].
Conversely, for any
The map
G.
x"- A
g,
as
F [ x I / ( ~ " -A )
is a field if and only if
twisted group algebra of Proof.
3
generated by
let
F-algebras
is an irreducible poly-
A E F*, F[XI/(X"
- 1)
is isomorphic to a
313
TWISTED GROUP ALGEBRAS
is obviously a surjective homomorphism of F-algebras.
grn = A * i ,
Because
we
have
>_ Cx"- A )
Kerf
f induces a surjective homomorphism
and thus
/!x"- A)
FIX1
F"G
--+
Since both algebras have the same F-dimension m,
of F-algebras.
the required
isomorphism follows.
A E F*,
The final assertion is a consequence of the fact that for any
the
elements
CP-A),x + CP-A), ..., P-l + CP-A,
+
1
F"X]/(P-A)which is multiplicatively closed up to
constitute an F-basis of
multiplication by the elements of
is said to be (von Neuman) regular if for any r E R
Recall that a ring R there exists an 5.3. LEMMA.
A
r'
F*.
R with rr 'r
E
ring R
r.
=
is regular if and only if every finitely generated right
ideal is generated by an idempotent. ian, then R Proof. idempotent.
=
rr'
R is semisimple artin-
is regular. Assume that every finitely generated right ideal is generated by an
r
If
E
R,
then r R
there exists an idempotent e
e
In particular, if
for some
r'
E
is a finitely generated right ideal, and hence
R
of
with
eR = rR.
and since r E e R
R,
we have
Since
r
=
er
e
E =
rR,
we have
rr'r;
hence R
is regular. is regular and let r R
Now assume that R there exists r '
so that
e
E
R
with
rr'r
is an idempotent and
=
e
so that r E eR.
Hence
rR
=
eR
Set e = rr'.
r.
E rR.
er
be a right ideal of
=
rr'r
R.
Then
Then
Note also that =
r
and therefore every singly generated right ideal
314
of
CHAPTER 6
R
is generated by an idempotent.
NOW consider the right ideal x R + YR. idempotent e .
By the foregoing, XR = e R
for some
Since
it follows easily that
+
xR Let
yR = e R
+
(l-e)yR
f be an idempotent that generates (1- e ) y R . Setting g = f ( l - e ) ,
ef = 0.
Then certainly
f2
=
f and
we therefore derive
g$ = f ( l - e ) f =
f2
f
=
g 2 = g f ( 1 - e ) = f(1-e) = g
ef
fR and f
Observe that g E
= gfE
=
gR so that f R
;cR + y R and g
Finally, since e
0 = ge
=
=
gR
and
e R + gR
are orthogonal idempotents, e + g
eR
+ gR
=
is an idempotent and
(e+g)R
.
The desired conclusion now follows by induction on the number of generators of the given right ideal.
H be a subgroup of G.
5.4. LEMMA.
Let
(i) If
is semisimple, then so is F"H
F"G
/ G : HI # 0 in F and if F"H
(ii) If (iii) If
char?' = p > 0 and P
semisimple if and only if Proof.
for
=
z
x. =
is a Sylow p-subgroup of
Hence, if
is semisimple.
x
E
by Lemma 5.3,
E
FaH.
J(F"H)
is semisimple
G, then F"G
Then, by Lemma 5.3,
F"H, then there exists y E F"G
Thus by applying the projection map
nH(y)
F"G
is
is semisimple.
(i) Assume that G'F
von Neumann regular.
xyx
F"P
is semisimple, then
It follows that F"H
H '* F"G
71
--+
F"G
is
with
F"H we have x z x
is a l s o von Neumann regular.
=
Hence,
is generated by an idempotent and therefore J(F"H) = 0.
(ii) Direct consequence of Theorem 2.3.4(iii).
x
315
TWISTED GROUP ALGEBRAS
(iii) Direct consequence of (i) and (ii). The following result is essentially due to Passman (1970a). 5.5.
Let 124E Z2(G,F*) and let the values of
THEOREM.
subfield of Proof.
F.
Then
If
charF
(Corollary 2.3.5).
F"G
1:
is semisimple if and only if
belong to a perfect charP
) /GI.
is semisimple by Maschke's theorem
(GI, then G'F
Conversely assume that charF = p
G and let P be a Sylow p-subgroup of G. show that J ( F n P ) # 0.
Ci
7
0
divides the order of
By Lemma 5.2(iii), it suffices to
By hypothesis, there exists a perfect subfield L,
of
F
such that
F"P C -
Thus J(L'P)
J(F"P).
But Lap
where
I(P)
J(F'P)
# 0, as required.
F C3 L"P L
LP by Proposition 1.6.1 and J(LP) = I(P) # 0
.
is the augmentation ideal of
5 . 6 . LEMMA.
F"G
=
Let
Proof.
Hence J ( L " P ) # 0 and therefore
G be a p-group and let charF
is semisimple if and only if G'F
extension of
LP.
=
p.
Then, for all
is a finite purely inseparable field
F. Direct consequence of Lemma 5.4.14.
5.7. COROLLARY. charF = p ,
Let
G be a cyclic group of order pn
let a E Z'(G,F*I
and let
1
=
8cx(g,gi).
generated by
g,
Then the following
i=1 conditions are equivalent: (i) G'F
is semisimple
(ii) f G (iii)
is a field
A
is not a p-th power of an element in
Direct consequence of Lemmas 5.2 and 5.6.
Proof. 5.8.
F.
LEMMA.
Let P be a p-group and let charF = p .
(i) P'F
is commutative if and only if P
(ii) If
P
5 G',
is abelian
the commutator subgroup of the finite group
G,
and if
let
316
P'F
CHAPTER 6
C -
F'G,
then ?P
(il Let E
Proof.
P'E
FP.
In particular, by Lemma 5.4(i), if FC"G
is semi-
IG'/.
simple, then p
E '8P'F F
2
be the algebraic closure of F.
and, by proposition 1.6.1, Since EP
if and only if so is EP.
fP
%
EP.
Then, by Lemma 5.1,
Hence P'F
is commutative
is commutative if and only if P
is
abelian, (i) follows. (ii) Put S finite.
=
{iglx
E
F*,g
Then F*
E G}.
S'
Hence, by Lemma 3.4,
is central in S and S/F*
is finite.
Moreover, since S/F*
I
IG
is
G we
have
S'/V Now
P 5 G'
QV/V'
P.
=
F'P,
where
V = F* n S'
and S' is finite, so there exists a finite p-subgroup Q c - s' But
Q fl V = 1
since F*
and we conclude that for each
- fxz x E
G'
2
Q.
Because
has no elements of order p.
Thus
5 F*
x e P. there exists a unique fx E V
-xy xy
and the elements
=
{Glx f2 PI
with
Q
P
with
form an F-basis of
we deduce that FaP % FP. In view of Maschke's theorem and Lemma 5.4(iii), the following result (essen-
tially due to Passman ( 1 9 7 8 ) ) reduces problem A to the case where
charF = p
and
G is an elementary abelian p-group. Let P be a p-group and let
5.9. THEOREM.
charF = p.
Then the following
conditions are equivalent: (i) P'F
is semisimple
(ii) For any elementary abelian central subgroup P, (iii) P
is abelian and F ' P o (i) * (ii):
Proof.
FaPo
If
for all subgroups Po
simple. in
is semisimple, where
simple. mutative.
subgroup of
of
P.
ig
E
Plgp = 1)
Conversely, assume that P'F
2
Because J(F"P)
so that
so is
is not semi-
to be a subgroup of order p
FPo and the latter is not semi-
P is abelian so that by Lemma 5.8(i), P'F
Put P = { g E P i g p = 1)
P.
=
is semisimple
F P is semisimple, then by Lemma 5.4(i),
Then, by Lemma 5.8(ii), , P ' F
Now assume that
P
P, ~ P ' F
0.
If P is nonabelian, then we can take Po
Z(P) n P'.
of
is com-
Po is a central elementary abelian
is a nonzero nilpotent ideal, we can choose
317
TWISTED GROUP ALGEBRAS
J(FaP),Z # 0 with
2
multiply Z
by some
g
with
preserving the fact that
Moreover, since FaP is commutative, we can
= 0.
Zp
'2
E
P to guarantee that trz
0.
=
Thus
Zz$
2 =
with
z
g
+0
while still
F and
E
z
=
trz #o.
Then, again using the commutativity of the ring, we have
Thus, if y = a nontrivial
z
(2) =
71
g p : nilpotent ideal FO
, in
then y # 0 and FaPo
FaPo
and
FaP
is semisimple. If S
semisimple. S
is abelian, by Lemma 5.8(i).
-F c , .
P is abelian and that FaP
is any elementary abelian central subgroup of FaS
FaF is a
Then, by Lemma 5.6,
Furthermore, by Lemma 5.4(i),
Conversely, assume that
Hence, by Lemma 5.4(i),
generates
is not semisimple.
(i) * (iii): Assume that FaP is semisimple. field and so P
Hence y
yp = 0.
P,
Thus 5%
is semisimple.
is
then
is semisimple
by virtue of the implication (ii) * (i). It is not true that F"P of
is semisimple if and only if for all subgroups
P of order p, FaPo is semisimple.
The following example is extracted from
Montgomery and Passman (1978). 5.10. EXAMPLE.
Let
K be a field of characteristic p
the rational function field over
K
twisted group algebra of the group
If
c
is a nonidentity element of
0 < i,j < p
in the variable x.
P
=
P,
FaP
Let
be
be the
< a > x < b > of order p 2 given by
i j
then c = a b
for some
i,j
with
and
is not a p-th power in
is semisimple.
F"P
and let F = K(T)
F.
Hence, by Corollary 5.7,
On the other hand,
! l =
1
+
Z-X
F'
satisfies
is a field and thus
vp
=
0 so that
is not semisimple. To present a partial solution of Problem B , we first provide the following
background information.
In what follows A
denotes a finite-dimensional F-
algebra and all A-modules are assumed to be finitely generated.
CHAPTER 6
318
V
Let
be a v e c t o r space o v e r
VE = E
form t h e tensor p r o d u c t
,...,U
{vl,u2
}
E.
VE over
form a b a s i s of
9 V, V
is a b a s i s of
F,
E
let
b e a f i e l d e x t e n s i o n of
E.
which is a v e c t o r s p a c e o v e r
over
F , t h e n t h e elements
h
F,
@.vl
and
If
,...,1 @ v V n }
With t h e a i d o f t h e i n j e c t i v e F-homomorphism
[:2?Jv we s h a l l o f t e n i d e n t i f y
VE
each element of
v
E
n 1 xiVi
with
i=1
xi
1 AiVi E V if and o n l y i f a l l E F. Therefore, i n passing i=1 VE w e a r e e x t e n d i n g t h e f i e l d of o p e r a t o r s from F t o E.
and
V
from
With t h i s i n d e n t i f i c a t i o n ,
h a s a unique r e p r e s e n t a t i o n of t h e form
n
.E
VE.
w i t h i t s image i n
to
A
Suppose now t h a t
is an F-algebra and l e t
AE = E @ A .
AE
Then
i s an
F E-algebra and t h e map
i s an i n j e c t i v e homomorphism of F-algebras.
AE
image i n element of
and d e n o t i n g by
AE
ta
Again, i d e n t i f y i n g
,...,arn1
a n F - b a s i s of
C
can be u n i q u e l y w r i t t e n i n t h e form
I/
w e see t h a t each
1. E E ;
Xiai w i t h
more-
i=l
1 xiai i=1
If
with i t s
m
rn over,
A,
A
E
A
i f and o n l y i f a l l
Xi
E F
i s an A-module, t h e n t h e v e c t o r s p a c e
VE
becomes an A -module under
E
a module a c t i o n
V
By i d e n t i f y i n g
A
and
VE
w i t h t h e i r images i n
AE,
and
respectively, the
a c t i o n above can be w r i t t e n as
Thus t h e a c t i o n of
A
on
VE
i s j u s t t h e e x t e n s i o n of t h e a c t i o n of
A
on
by E - l i n e a r i t y . Suppose now t h a t homomorphism We w r i t e
BE
B
i s an F-subspace of
--+ E '8 A F F f o r t h e image of B
8
:
E '8 B
d e f i n e d by under
0.
A.
Then t h e r e i s an i n j e c t i v e E-
e(L @ b ) If
B
=
L @ b, R
E
E, b
E B.
i s a subring ( i d e a l ) , then
V
319
TWISTED GROUP ALGEBRAS
i s a s u b r i n g ( i d e a l ) of
BE
5.11.
LEMMA.
E
Let
AE'
be a f i e l d e x t e n s i o n of
F
A
and l e t
b e an a l g e b r a o v e r
F. I
(i) If
i s an i d e a l of
(ii) J ( A I E
8a
( i ) The map
(A/IIE
onto
(A/I)E
then
%
AE/IE
a s E-algebras
5J(AE)
Proof.
AE
A,
-+
J(A)
IE.
The d e s i r e d c o n c l u s i o n now f o l l o w s
(A/I)E
AE/IE
and
J(AIE
i s n i l p o t e n t , so i s
V
An A-module
i s a n E-algebra homomorphism of
whose k e r n e l c o n t a i n s
from t h e f a c t t h a t t h e a l g e b r a s (ii) Since
'8 ( a + I )
and hence
i s s a i d t o be separable i f
E
module f o r e v e r y f i e l d e x t e n s i o n
of
F.
VE
.
a r e of t h e same E-dimension.
5
c T ( A ) ~ J(AE).
i s a completely r e d u c i b l e A
E-
Thus a s e p a r a b l e module i s complete-
l y r e d u c i b l e ; t h e c o n v e r s e need n o t be t r u e as w e s h a l l s e e below. For convenience of r e f e r e n c e , w e now quote t h e f o l l o w i n g s t a n d a r d f a c t .
5.12.
PROPOSITION.
be a completely r e d u c i b l e A-module.
V
Let
s e p a r a b l e i f and o n l y i f f o r e v e r y i r r e d u c i b l e submodule
W
of
v
Then
V
t h e c e n t r e of
End(W) i s a f i n i t e s e p a r a b l e f i e l d e x t e n s i o n of A See Bourbaki (1959).
F.
i s s a i d t o be separable i f t h e l e f t r e g u l a r module
AA
the d i v i s i o n algebra Proof.
A
The a l g e b r a separable. of
AE
F,
5.13.
Expressed o t h e r w i s e ,
PROPOSITION. If
The F-algebra
charF
)
IGl
,
E 8 F"G
By Lemma 5.1,
semisimple. If
Lemma 5 . 4 ( i ) ,
E'P
+ EP
5.14.
Thus
F"G
charF = p
E"P
and so
LEMMA.
valent:
E
i s separable i f f o r every f i e l d extension
then
E"G
F"G
i s s e p a r a b l e i f and o n l y i f
charE
)
IGI
and so, by Maschke's theorem,
is separable. divides
J(E%
Let
A
# 0,
Conversely, assume t h a t
E
E
8 C'F
F FaG
\GI and P i s a Sylow p-subgroup of
i s semisimple, where
charF
) /GI. E
f o r every f i e l d extension
F
rable.
is
i s a semisimple E-algebra.
Proof.
F.
A
is
of
is
i s sepa-
G,
i s t h e a l g e b r a i c c l o s u r e of
t h e n by
F.
But
a contradiction.
be an F-algebra.
Then t h e f o l l o w i n g c o n d i t i o n s are e q u i -
CHAPTER 6
320
(i) A / J ( A )
is separable
B of
(ii) For every f i e l d e x t e n s i o n
E
(iii) For e v e r y f i e l d e x t e n s i o n
In particular,
E
extension
A
for a l l
f o r any f i e l d
F.
F
be an a l g e b r a o v e r a f i e l d
A
say t h a t
AE/J(AE)
D i r e c t consequence of Lemma 5.11.
Proof. Let
of
F , ( A / c J ( A ) ) ~ AdJ(AE)
of
i s s e p a r a b l e , t h e n so i s
A/J(A)
if
F , J(AE) = J(AIE
1s
1c ,1c 1
2
definable over
E X, z
Expressed o t h e r w i s e ,
1c 1
2
A
L
and l e t
i f t h e r e e x i s t s an F - b a s i s
L
F.
be a s u b f i e l d of
X
of
A
such t h a t
X.
i s an L - l i n e a r combination of t h e elements of i s d e f i n a b l e over
L
A
if
F
8B
We
f o r some L-algebra
L
B. 5.15.
PROPOSITION.
Let
A
be an a l g e b r a o v e r a f i e l d
F.
d e f i n a b l e over a p e r f e c t s u b f i e l d of Proof.
By h y p o t h e s i s ,
A
F 8B
Then
A/J(A)
F
and assume t h a t
is separable
B
f o r some L-algebra
A
is
.
and some p e r f e c t
L subfield
L
of
F.
By Lemma 5.14, we need o n l y prove t h a t
and
t h e r e s u l t follows.
We a r e now r e a d y t o p r o v i d e a p a r t i a l s o l u t i o n of Problem
B.
E Z2(G,F*)
c1
5.16.
is a
The latter being a consequence of P r o p o s i t i o n s 1.4.5
s e p a r a b l e L-algebra.
5.12,
B/J(B)
PROPOSITION.
pe rfe c t subfield of
Let
c1
F.
Then
and l e t t h e v a l u e s of
FOIG/J(FaGI
belong t o a
i s a s e p a r a b l e F-algebra.
In partic-
u l a r , by Lemma 5.14,
f o r any f i e l d e x t e n s i o n Proof.
E
of
S i n c e t h e v a l u e s of
F. CY
belong t o a p e r f e c t s u b f i e l d of
d e f i n a b l e over a p e r f e c t s u b f i e l d of
i s a s e p a r a b l e F-algebra.
F.
Hence, by P r o p o s i t i o n 5.15,
F , FaG
is
FaG/J(FolG)
m
To prove our n e x t r e s u l t , it w i l l b e c o n v e n i e n t t o r e c a l l t h e f o l l o w i n g p i e c e
of i n f o r m a t i o n . if
Given
a E 2' ( G , F * ) ,
an element
g E G
i s s a i d t o be
a-regular
TWISTED GROUP ALGEBRAS
321
"(g,x) = a(x,g). g
By Lemma 2.4.6,
if
conjugacy class
C of
is ®ular.
x
If
E G
- - --1
5.17. COROLWY.
1 gzg
, where
'p,
CG(z) in
FaG.
a E Z2(G,Ff), then the following conditions are equivalent:
If
is the only a-regular element of
(iii) For any
1 # g E G,
G
there exists It: E CG(g) such that a(g,It:) # n(z,g)
Conditions (ii) and (iiil are equivalent by the definition of aSince (i).and (iil are equivalent by virtue of the quoted fact above,
regularity.
the result follows.
'
We are now ready to attack Problem 5.18. THEOREM.
C.
Let c1 E Z2(G,F*) and assame that the values of
F.
perfect subfield of (i) G'F
is a left transversal for
G and,
F
=
Proof.
C
9'-T,
is an F-basis of the centre of
(i) Z(F'C)
The
is called a-regular if one (hence, all) element of
G
is a set of all representatives for the a-regular classes of
G, then
a belong to a
Then the following conditions are equivalent:
is central simple
(ii) charF Cl(g,z)
G.
is a-regular, then so is any conjugate of
The following fact is a consequence of Theorem 2.4.7:
for each s E X, kX =
(ii) 1
z E C,(g)
for all
[ IG/
and, for any
1 # g E G, there exists
3:
E CG(g)
such that
# a(z,g).
Proof. Theorem 5.5,
(i) * (ii): Since charF
4
lG/.
FaG
is simple, it is semisimple and so, by
By hypothesis, Z(F'G)
=
F
and so the second asser-
tion follows by Corollary 5.17. (ii) =) (i):
Since
charF
) IGl, F"G
Fdrthermore, by Corollary 5.17,
is semisimple by Maschke's theorem.
Z(F"G) = F
Hence FaG
is central simple, as
required. In what follows, A
denotes a finite-dimensional algebra over
F.
CHAPTER 6
322
5.19. LEMMA.
Let
E be a field extension of F.
V,W be A-modules and let
Then
IV ,W ) AE E E
Hom If
V
=
(HomA(v,W))E
as E-spaces
W, the two sides of this are isomorphic E-algebras.
-
1 @ f E HomA(VE,WE).
Given f E HomA(V,W), we have
Proof.
lA
(Hom(v,W))
x
E
AE @ ).
It therefore suffices to verify that eachelement
(V W ) is of the form $ = Z X i ( l @ fi) with A. AF: E’ E i Let {Xili E I} be an F-basis of E. Then each element
of Hom
uniquely written in the form
C i€I
Define fi
0.
distinct from
Hence the map
(vE,WE)
Hom
8 f w X(1
is an injective E-homomorphism. $
1
:
xi
8vi
with
V
E B,f. Z
of
E
vE
many ‘i i E V and with finitely
z xi
(v E V )
8fi(U)
i€I Then obviously each f. is an /“i-homomorphismsuch that $ = E X ,(1 @ f .)
i z
V
of F , VE
sion E
5.20. LEMMA.
Let
A
can be
W by
V--+
$(l @ v ) =
An A-module
Hom(v.W).
.
8
is said to be a b s o l u t e l y i r r e d u c i b l e if for every field extenis an irreducible A E-module.
V be an irreducible A-module.
Then the following statements
are equivalent: (i) V
is absolutely irreducible
(ii) If
E is the algebraic closure of F, then VE is an irreducible A E -
module. (iii) End(V) = F ,
i.e. each A-endomorphism of
V
is a left multiplication by an
A element of
F.
Proof.
(i) =) (ii): Obvious
(ii) * (iii): By Schur’s lemma and Lemma 5.2,
E
=
End
(V
AEE
E ‘8 End (V) A
This shows that dirn$ndA(V)
=
dim End
(IfE) = 1,
AE as required. (iii)
=)
(i): We may assume that
V
is faithful in which case A
B
M (F) for
TWI5TED GROUP ALGEBRAS
n 2 1 and hence A
E
P
Mn (El for any field extension E of F.
with a minimal left ideal of M n ( F ) , dimEVE = n.
with
A
for
VE
Hence
we see that
VE
Identifying V
is a left ideal of
is irreducible and the result follows.
be an algebra over a field F.
Let A
323
Then F
Mn(E)
=
is called a splitting f i e l d
if every irreducible A-module is absolutely irreducible.
F is a splitting field for A
5.21. COROLLARY.
A/J(A) for some positive integers n 1
field for A , Proof.
then A
Let
=
if and only if
r
nMn
(F)
i=1 i
,...,nr.
In particular, if
is simple if and only if A
Vl,...,Vr
F is a splitting
is central simple.
be all nonisomorphic irreducible A-modules and let
r
D.
=
End ( V ) , 1 G i 4 r .
A n ,...,n 17.= D9 z
i
7
,
for some positive integers i=1 i is a splitting field for A if and only if
F
By Lemma 5.20
r'
1
TIMn (D;)
Then A / J ( A )
...,PI.
F for all i E fl,
So
the corollary is true.
We are now ready to prove 5.22.
THEOREM.
Let
perfect subfield of
E Z2(G,F*) and assume that the values of
F.
If F
is a splitting field for F'G,
CL
belong to a
then the follow-
ing conditions are equivalent: (i) G'F
is simple
(ii) F"G
is central simple
(iii) charF "(g,z)
1
IGI
and, for any
1# g
E
G, there exists z E CG(g) such that
# '(x,g).
Proof.
The equivalence of (ii) and (iii) follows by Theorem 5.18, while the
-
equivalence of (i) and (ii) is a consequence of Corollary 5.21.
Remark.
'
Iwahori and Matsumoto (1964) conjectured that if af'G
simple for some a E Z2(G,C*), then
G is solvable.
is central
Using the classification of
finite simple groups Howlett and Isaacs (1982) verified this conjecture. over, DeMeyer and Janusz (1969) proved that ffG
Po p l
is central simple for any Sylow subgroup P Let a E Z2(G,F*) and let V
a mapping
p :
G
--+
GL(V)
More-
.
is central simple if and only if of
G.
be a (finite-dimensional)vector space over F
is called an a-representation of
G
or (projective
324
CHAPTER 6
representation, if a
F if
is not pertinent to the discussion) over for all z , y E G
(i) P ( z ) P ( ~ ) = a(z,y)~(zy) (ii) p ( 1 ) = 1 V is faithful if
P
We say that
P(g)
=
1.1
1 E F*, implies g
v'
c1
Suppose that a E Z2(G,F*) is such that F G
5.23. LEMMA.
all irreducible a-representations
G
over
=
1
is simple.
Then
F are linearly equivalent and faith-
ful. Since FaG
Proof.
ducible F'G-modules.
is simple, there is only one isomorphism class of irre-
Taking into account that the isomorphism classes of irre-
ducible F"G-modules correspond bi jectively to the linear equivalence classes of irreducible a-representations of Assume that
p
G,
the first assertion follows.
FaG
V C D
=
It is obvious that
P
if and only if any linear equivalent of P
Write
... 69 V,
as a direct sum of irreducible F'G-modules.
assume that
G.
is an irreducible a-representation of
is also faithful
.
p
is faithful
Hence we may
V 1 in which case
is afforded by
sv
=
for all g E G, v E V
P(g)v
n
By the foregoing, there exist F G-isomorphisms
$i Assume that g E G
is such that
-
U E V
, gv
= XV,
Vi E
vi .
1.i.
5.24. LEMMA.
(1GiG.) for some
h E F*.
Then for all
and thus
which implies that =
v
p ( g ) = h*l
-
= g$;(Vi)
= A$J.(V.I $
2
= $.(All.) $
7
,
Hence, for all V i E Vi,
;vi
obtain
v2-
v,
+jVi)
for all
:
=
Ax
for all
=
xui
'
z E F G.
Thus g = 1 and the result follows.
Let a E Z2(G,F*) and let F
.
In particular, taking
z =
be a splitting field for F'G.
i
we
If
325
TWISTED GROUP ALGEBRAS
G admits a faithful irreducible %representation over F, then 1 is the only central element of
Assume by way of contradiction that g E Z(G)
Proof.
g # 1.
G which is a-regular.
P
Let
:
G
--+
GL(V)
is a-regular and
G.
be a faithful irreducible a-representation of
Then the map : G'F
P*
.-t
End ( V )
F defined by P* (CZ );
=
9
E Z(F"G).
regular, we have for some is true. 5.25.
.
E
F,
be such that
1
9
Because g
E
p
G.
is the only a-regular element of
1
and
is faithful.
G be an abelian group of exponent m
primitive m-th root of
Z(G)
(in particular, charF
k
and let
Then
E
G)
is a-
b
Hence, by Schur's lemma, P ( g ) = P*(;)
contrary to the assumption that
Let
LEMMA.
9
FaG.
is an irreducible representation of
(z E F , g
Ex P ( g )
=
1.1
V
So the lemma
U E
Z2(G,F*)
F contains a
IGI) and PUG is a central
simple F-algebra. Owing to Corollary 5 . 1 7 ,
Proof. J: E
Z(FaG)
=
F.
Fix
g
E
G and, for each
G, put $ (z)= a(g,z)n-l(z,g)
g
Then a routine calculation shows that
I.-
G
is a homomorphism. therefore Hence
F
g
=
1.
$ (2)= 1
g Because
g
E Hom(G,F*)
@g for all
/Hom(G,E'*)
LEMMA.
z E G, then g is a-regular and
/GI
,
Let
it follows that G
Hom(G,F*).
1, which in turn implies that
Thus, by Maschke's theorem, F"G
it follows that F"G
and that the mdp
Hom(G,F*)
contains a primitive m-th root of
charF ) /GI.
5.26.
If
----+
@
is in fact central simple, as required.
G be an abelian group, let
.
is semisimple.
E
Since Z(FaG) = F ,
Z2(G,F*) and let F
be a
CHAPTER 6
326
splitting field for FaG.
Then the following conditions are equivalent: such that G
E Z2(G,F*)
(i) There exists
admits a faithful irreducible a-
representation (ii). FaG
Mn(F) for a suitable positive integer n
2
(i) * (ii): By Lemma 5.24,
Proof.
is the only a-regular element of
1
G.
Now apply Lemma 5.25 and Corollary 5.21.
(ii) =, (i): Direct consequence of Lemma 5.23. Let G of
and G 2
G and G 1
be groups.
into F*
2
A map
:
G x G 1
-3
2
is called a pairing
F*
if
f(zy,z)
=
f(z,z)f(y,z)
(z,y
f(z,yz)
=
f(x,y)f(z,z)
(X E G
The set P(G ,G ;F*) of all pairings of 1
f
G1
and
G2
into
2
F*
E
GI, z E G 1 r
~
E G i P )~
constitutes an
abelian group under the multiplication defined by the rule
(f,f,)( z r y ) 5.27. LEMMA.
Let G
(i) Any pairing (ii) Let
41
f
Define the pairing
Then a E Z 2 t G , F * )
@
x G
--f
a@
F*
: G
--+
F*
-
G
=
>x
G
... x
,y
E G2)
is a cocycle
be a pairing such that @(z,z)= 1 for all
x G
zE
G.
F* by the formula
and, for all 2,y E G,
@(x,y) = proof.
(x E f, (4f2 (2,y)
be an abelian group, say
: G
: G x C
=
(i) Fix z,y,z
E G.
-1
CY
3
(Z,Y)CY@
(y,z)
Then we have
For any g E G, f ( g , l ) = f ( g , l ) f ( g , l ) , which shows that f(g,l) = 1. argument shows that for all g E G, f ( 1 , g ) = 1, as required.
A similar
327
TWISTED GROUP ALGEBRAS
(ii) That c1
@
ZZ(G,F*)
is a consequence of (i).
@(s,y) =
-1
a
(X,Y)cg
@
The equality
(x,y)
can be established i n a straightforward manner. The next observation is crucial. Let
5.28. LEMMA. n,
and let F
G =
>
>X
where
gI
contain a primitive n-th root of
and
g2 are of the same order
1.
Define a pairing
+F*
@ : G x G
by the rule
@ ( g l , g l )= @ ( g 2 , g 2=) 1 @ ( g 1 , g 2 )=
where
E
is a primitive n-th root of
E
1
, @ ( g z l g l =) in
F.
E
-1
Then for
a = a@t
FaG and FaC is central simple ij that g1g2, 0 < i,j < n - 1 , is a-regular.
is the
only a-regular element of Proof.
Assume
ij w 119192)
=
Then
ij
a(glg2,g1)
and
ij
ij
w g 2 ' g l g p ) = a(!31gz'g*) which implies that
ij
ij
@(g1,g1g2) = @ ( g 2 q g 2= ) 1 It therefore follows that
E3
the only a-regular element of
= ECz =
G.
1 and hence that
i
=
j
= 0.
Thus
1
is
The desired conclusion is now a consequence of
Lemma. 5.25. Following Yamazaki (1964b1, we say that a group
G is of symmetric type if
G S H x H for a suitable group
H.
We are now ready to solve Problem D for abelian groups.
5.29. THEOREM (Yamazaki, 1964b).
rn and let F
be a field.
Let
G
be a finite abelian group of exponent
Then the following conditions are equivalent:
328
CHaPTER 6
( i ) There e x i s t s
N
( i i ) There e x i s t s
E Z2(G,F*) N E
F"G
such t h a t
Z2(G,F*l such t h a t
i s a c e n t r a l simple F-algebra i s t h e o n l y a - r e g u l a r element of
1
G. (iii) G
i s of symmetric t y p e and
c o n t a i n s a p r i m i t i v e m-th r o o t of
F
( i ) * ( i i ) : D i r e c t consequence o f C o r o l l a r y 5.17.
Proof.
,
g E G
(ii)a f i i i ) : For a f i x e d
4
let
:
9
G
3
F*
be d e f i n e d by
-1 $ ( x ) = a(g,r)a (r,g)
for a l l
9
Owing t o Lemma 5.25,
F
$g
i s a l s o of o r d e r
some p r i m i t i v e rn-th r o o t
Hence
i
i s d i v i s i b l e by
G
onto
m,
so
E
n
of
Hom(G,F*).
@ (t)= 9
1 in
F.
Let
g
f o r some
t
2 E
E
G
t
E
be o f o r d e r
G
of o r d e r
rn. rn and
We t h e r e f o r e deduce t h a t
and t h e r e f o r e
tZ = gJ = 1.
T h i s shows t h a t
i s of symmetric t y p e ,
Now we i n t r o d u c e a n o t h e r subgroup of
tigj
Assume t h a t
Thus
i Let
E
and
j
A
{@zIx
=
GL.
E
GI}, l e t C
I t is clear t h a t
G d e f i n e d by
Then we have
are d i v i s i b l e by
n.
Hence
B = Hom(G,F*)
tigj
= 1
and l e t
= {x E Hom(B,F*)IX(A) =
11
G
1 and t h e map gl-+
c o n t a i n s a p r i m i t i v e m-th r o o t of
p r o v i d e s an isomorphism of Then
1.
and so
GI n G 2
= 1.
$ g
329
TWISTED GROUP ALGEBRAS
C ICI
and so
=
Then the map
G~ onto C.
a
Let
IG X
:
I-+
I.
G
For a fixed z 5 G
$
is an isomorphism of
X
IG,~
Hence
denote the restriction of
a-regular.
Then
I$
Y
= 1
(Z)
CY
Y
G1
cause
=
1
1 E B,
to
so that the pair
Qx(X)
put
=
A(+).
G onto Hom(B,F*) which carries
G
for all z E G
tp) = I$ Y It follows that
and
and therefore G = G
/ G : G1j
=
B/A
Hom(B/A,F*)
2
X 2
G
2
1
x
G
2
.
and assume that y E G 2
is
and, by definition of GZc
(2) =
for all z E G
1
1
B,Gz satisfies condition (ii).
Be-
is of symmetric type, the desired conclusion follows by induction on
IGI.
(iii) =) (i): be such that
Let 1
GI and G 2 be abelian groups and let a .
is the only Ci.-reqular element of
x a
CY 1
E
Z2(G
x
Gi.
E
Z2(Gi,F*),i= l,2,
Define
G2,F*)
2
by (a1 x
gl,gz
E
GI, h , , h 2
regular element of
cl2)
Gz. G
x I
( ( g l , h l,)( g 2 , h 2 ) )=
CY1
(g1,h1)CY2(g2,h2)
Then it is easy to see that
G 2
.
(1,l) is the only ( a x cx 1
2
)-
On the other hand, any finite abelian group of
symmetric type is decomposed into a direct product of subgroups Gi, 1 Q
i
such that Gi
Hence the
is a direct product of two isomorphic cyclic subgroups.
desired assertion follows by applying Lemmas 5.28 and 5.25.
G k,
This Page Intentionally Left Blank
331
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339
Notation
Number Systems
N
t h e n a t u r a l numbers
z
the rational integers
Q
t h e r a t i o n a l numbers
R
t h e r e a l numbers
c
t h e complex numbers
z/d
t h e i n t e g e r s mod
m
S e t Theory
C
proper i n c l u s i o n
1x1
inclusion
X-Y
t h e complement of
C
t h e c a r d i n a l i t y of t h e s e t
Y
in
X
X
Number Theory
alb
a divides b
a@
a
(a,b)
g r e a t e s t common d i v i s o r of
n
t h e p - p a r t of
P
nP '
does n o t d i v i d e
t h e p'-part
of
n n
b a
and
b
Notation
34 0
Group Theory
the multiplicative group of a field the subgroup generated by
X
the cyclic group of order
n
G1
direct product of
F
G2
and
H is a normal subgroup of G the centralizer of
X in G
the normalizer of -1 -1
= z =
y
X in G
xy
< [xJ] jx E x , y
E
Y> G
the commutator subgroup of
nG~
direct product of
@ Gi
restricted direct product (or direct sum) of
Gi,i
E
I
iEI
Gi,i
E I
iEI
G
2 (G)
the centre of
GL ( V )
the group of all nonsingular transformations of the vector space V
x under the action of g
the image of
x
the G-orbit of
€
the stabilizer of symmetric group on
x
E
X
X
the automorphism group of
RG
the fixed subring of
An
=
tan/aE A 1
AM1
=
ta
A
the p-component of A
P GS ( V )
X
G
Alan = 11
general semilinear group of = t3 E =
G
X
AutX
E
E
G / \ H: cH(3)I
V
<mI
DHCH)
= {3 E
GlgX
=
XI
the join of all finite normal subgroups of
G
Notation
341
Rings and Modules
R0
opposite ring
Z (R)
the centre of
A@B
tensor product
R
the characteristic of
ji
direct sum direct product of rings
X over R
polynomial ring on
the Jacobson radical of a ring
R
V
the radical of an R-module
V
the algebra of 8-endomorphisms of the principal ideal generated by
nxn
r
matrices with entries in R
induced module
=L@V F =L@A F
M
the annihilator of
X
the left annihilator of the right annihilator of
X
the sum of all two-sided ideals of contained in X G-conjugate of
V
the intertwinning number inf ( V )
inflated module
ext ( V )
extension of
e..
the matrix unit
v"
n-th direct power of
SUPP
support of
A
the g-component of
9
R*G
V
v
2
A
the crossed product of
G
over R
R
342
Notation
the unit group of A =
19 E G ~ P + g r is an X-inner automorphism of R ) u
the degree of
graded units of A the inner automorphism induced by
u
8 Ax x%N crossed system
=
=
@A
5supporting subgroup of
x
the natural projection projective crossed representation irreducible characters of
N
over F
X
the stabilizer of coinduced module restriction of =
c
'1
&G
V
R*H
to
the smallest G-invariant ideal of R contajlning I
trace map equivalence class of A
degree of
Brauer group of index of
A
F
A
subgroup of are split by
consisting of those
BY(?')
F ~ G
twisted group algebra of
FG
group algebra of
Hom(M,N) A g Grnom (M,N )
A Qo ( R )
g
1.41
which
E'
G
G over F
over F
R-submodule of Hom(M,N) consisting of all graded homomorphisms fromAM to fl Martindale ring of quotients of
R
343
Notation
Cohomology Theory
t h e group of a l l A-valued 2-cocycles of
G
a coboundary
t h e subgroup of =
Z2(G,A)
Z2 ( G , A ) / B 2 ( G , A )
t h e cohomology class of =
c o n s i s t i n g of a l l coboundaries
f E Z2(G,A)
2’ ( G , A ) /B’ (G,A)
t h e c o r e s t r i c t i o n map t h e r e s t r i c t i o n map o b s t r u c t i o n c o c y c l e corresponding t o
F i e l d Theory
f i e l d extension
(F
:
K)
F
d e g r e e of
K
over
G a l o i s group of
F
over
t h e smallest s u b f i e l d c o n t a i n i n g
S
and
t h e smallest s u b f i e l d c o n t a i n i n g
K
and
t h e a l g e b r a i c c l o s u r e of
N~/~(?”)
K
norm of
1-1
over
P
F
K 1
,...,a
This Page Intentionally Left Blank
345
Index
Action, 25 faithful, 26 Algebra, 10 central, 1 5 1 definable over a subfield, 3 2 0 enveloping, 1 5 4 isomorphism of, l o R-free, 10 similar, 162 skewfield component of, 162 splitting field of, 163 tensor product of, 10 Algebraic closure, 27 algebraic element, 27 almost centralizer, 286 Arnitsur, 8 9 annihilator, 53 of induced modules, 137 automorphism system, 63 Bergman, 3 1 0 bimodule, 7 Bourbaki, 18, 25 Brauer, 159 Brauer classes, 1 6 2 Brauer group, 162 canonical homomorphism, 11 canonical injection, 1 2 7 centralizer, 3 characterization of, artinian modules, 19 flat modules, 1 8 induced modules, 1 2 7 injective modules, 1 4 Jacobson radical, 5 4 noetherian modules, 2 0 projective modules, 12, 13 semisimple artinian rings, 57 simple artinian rings, 57 strongly graded algebras, 184 Clifford theorem for strongly graded algebras, 188
coboundary, 3 8 Cohn, 27 cohomology class, 3 8 cohomologous cocycles, 3 8 commutator, 3 commutator subgroup, 3 conjugate of a module, 1 8 6 core, 88 corestriction map, 4 4 criteria for algebra to be crossed product, 7 1 algebra to be skew group ring, 77 A / J I A ) to be separable, 3 2 0 complete reducibility of induced modules, 1 3 9 crossed product to be artinian, 7 9 crossed product to be noetherian, 7 9 crossed product to be simple, 9 1 direct sum to be artinian, 22 direct sum to be noetherian, 22 element to be in J ( R ) , 5 5 equality of prime ranks of B and
RG,
275
equivalence of crossed products, 7 2 existence of faithful projective representation, 3 2 6 existence of maximal submodules, 2 0 existence of irreducible submodules, 2 0 extendibility of modules, 2 0 0 extendibility of regular module, 2 0 4 extension to be Galois, 28 extension to be normal, 2 8 extension to be simple, 28 extension to be split, 45 field to be perfect, 28 finite decomposition, 2 2 finite irreducible decomposition, 24 @G to be central simple, 321, 323 to be semisimple, 314, 315 F"G to be separable, 3 1 9 @G to be simple, 323 P P to be commutative, 315 induced module to be indecomposable, 213
346
Index
[criteria for] isomorphism of extensions, 202 G-grading, 60 isomorphism of induced and G-module, 26 coinduced modules, 133 Cset, 25 module to be completely reducible, 24 g-component, 60 module to be flat, 16 general semilinear group, 95 module to be irreducible, 52 Going down, 246 module to be projective, 17 Going up, 259 module to be of finite length, 25 Going up theorem, 266 prime ideal to be maximal, 226, 276 graded homomorphism, 119 prime ideal to be minimal, 244, 256 graded ideal, 119 ring to he matrix ring, 46 graded unit, 61 ring to he regular, 313 Group, 2 R*E to he primitive, 229 cyclic, 3 R*G to have unique prime ideal, 258 direct product of, 4 RG to be semisimple, 310 direct sum of, 4 submodule to be fully invariant, 50 exponent of, 3 ZCFGl = F, 321 extension, 28 Galnis 27 Dade, 121, 184, 193, 195, 200, 202 locally finite, 85 degree of extension, 27 multiplicative, 2 degree of unit, 61 n-divisible, 41 DeMeyer, 323 p-component of, 41 diagram, 2 of finite exponent, 42 commutative, 2 of symmetric type, 327 domain, 5 restricted direct product of, 4 dual basis lemma, 17 splitting of, 3 torsion, 3, 42 endomorphism ring, 6 torsion-free, 3 equivalent crossed products, 72 uniquely divisible by n, 41 essential submodule, 305 essential version of Maschke's Higman, 143 theorem, 307 homomorphism, 5 extension, 27 factoring, 5 algebraic, 27 Howlett, 323 congruence class of, 29 Huppert, 208 congruent, 28 factor set of, 30 Ideal, 54 finite, 27 G-annihilator-free, 283 finite normalizing, 259 G-nilpotent-free, 283 Galois, 27 nil of bounded degree, 265 infinite, 27 prime, 223 normal, 27 primitive, 54, 224 normalizing, 259 R-cancelable, 240 primitive element o f , 27 semiprime, 224 separable, 27 idempotent, 5 simple, 27 orthogonal, 5 primitive, 5 Field, 27 incomparability, 246 perfect, 27 inertia group, 188 first cohomology group, 39 inflation map, 113 fixed-point space, 96 intertwinning number, 129 fixed subalgebra, 26 irreducible constituent, 129 Frobenius reciprocity, 128 Isaacs, 117, 310, 323 fully invariant submodule, 50 Iwahori, 323 C-algebra, 26 G-field, 26 G-graded algebra, 59 G-graded ring, 60
Jacobson radical, 53 Janusz, 323 Karpilovsky, 100
347
Index
kernel of the action, 26 Kharchenko, 236 Kngrr, 1 3 8 Krull-Schmidt theorem, 25 left transversal, 3 left unit, 55 length, 60 linear representation, 1 0 5 L-injective module, 80 Lorenz, 2 2 6 , 229, 2 3 0 , 2 4 0 , 243, 2 5 3 , 255, 256, 2 5 7 , 258, 259, 260, 2 6 3 , 265, 266, 272, 215, 276, 304 L-projective module, 80
Map, 1 balanced, 8 composite, 1 identity, 1 inclusion, 1 restriction, 1 Martindale, 232 Martindale ring of quotients, 232 Maschke's theorpm, R 3 for strongly graded rings, 3 0 1 matrix of semilinear transformation,lo9 matrix a-representation, 110 matrix unit, 46 module, 6 absolutely indecomposable, 208 absolutely irreducible, 322 artinian, 19 basis of, 6 coinduced, 1 2 8 completely reducible, 6 composition factor of, I composition series of, 7 decomposable, 7 direct sum of, 6 faithful, 54 finitely cogenerated, 1 9 finitely generated, 1 9 flat, 16 free, 6 homogeneous, 50 homogeneous component of, 5 0 indecomposable, I induced, 1 2 6 injective, 1 4 irreducible, 6 noetherian, 1 9 of finite length, 25 projective, 1 2 regular, 6 separable, 319 strongly indecomposable, 1 9 8 weakly G-invariant, 1 9 5 unital, 6 Monomial representation, 9 5 monomial space, 9 5
Montgomery, 5 9 , 1 4 7 , 150, 258, 317 Montgomery's theorem, 1 5 0 multiplicity, 1 2 9 Nakayama's lemma, 8 3 Nakayama reciprocity, 1 2 9 Nastacescu, 3 0 1 nilpotent element, 5 nil ideal, 5 nilpotent ideal, 5 norm, 1 7 9 normal closure, 3 normalizer, 3 normalizing basis, 8 4 normal subgroup, 3 obstruction cocycle, 116 orbit, 26 outer automorphism, 78 Par;, 259 Passman, 8 6 , 1 4 7 , 2 3 0 , 240, 243, 253, 255, 265, 296, 307,
256, 266, 299, 309,
257, 2 5 8 , 259, 260, 263, 272, 2 7 5 , 277, 286, 292, 3 0 1 , 3 0 2 , 3 0 3 , 304, 305, 316, 317 prime rank, 267 primitive rank, 267 principal crossed homomorphism, 39 projective crossed representation, 1 0 3 projective representation, 105
regular class, 100 regular element, 96 regular orbit, 96 restriction map, 4 3 representation, 105 completely reducible, 1 0 5 degree of, 1 0 5 indecomposable, 1 0 5 irreducible, 1 0 5 linear, 105 linearly equivalent, 1 0 6 projectively equivalent, 106 right transversal, 3 right unit, 55 ring, 5 artinian, 19 associative, 5 centre of, 5 characteristic of, 5 direct product of, 6 G-prime, 2 3 0 G-semiprime, 283 homomorphism, 5 left primitive, 224 matrix 46 noetherian, 1 9 opposite, 52 primitive, 224
348
[ring 51 regular, 313 right primitive, 224 semiprime, 223 semisimple, 53 Schelter, 259 Schur's lemma, 5 2 sequence, 4 exact, 4 natural exact, 4 short exact, 4 split, 11 second cohomology group, 38 section, 3 semilinear transformation, 94 group of, 94 nonsingular, 94 separable element, 27 separable polynomial, 27 set 2, cardinality of, 2 size of the form, 286 skew group ring, 63 strongly G-graded algebra, 6 0 strongly G-graded ring, 60 strong permutation of ideals, 277 splitting field, 323 splitting homomorphism, 12 support, 60 supporting subgroup, 86
Index
Tensor product, 7 associativity of, 9 characterization o f , 11 of modules, 7 Transitivity of induction, 126 Trivial intersection ideal, 301 Twisted group algebra, 63 Twisted group ring, 63 Unique product group, 296 Unit, 55 Villamayor, 87 Villamayor's theorem, 87 Weak divisibility, 195 Weak isomorphism, 195 Weiss, 43, 44 X-inner automorphism, 236 Yamazaki, 327
