503 79 10MB
English Pages 239 Year 1983
Table of contents :
Preface
Contents
Chapter 1. Prerequisites
1. Notation and Terminology
2. Background Sketch
Chapter 2. Preliminary Results
1. Definitions
2. Elementary Properties
Chapter 3. RingTheoretic Properties
1. Chain Conditions. Regular and Completely Reducible Group Algebras
2. Characters, Direct Decompositions, and Irreducible Modules
3. Idempotents
4. Nil and Jacobson Radicals
5. Krull Dimension
Chapter 4. Integral Domains
1. GCD Domains
2. Unique Factorization Domains
3. Krull Domains
Chapter 5. Study of Units
1. Units of Finite Order
2. The Unit Group of Group Algebras of TorsionFree Groups
3. Splitting Unit Groups
4. The Isomorphism Class of U(ZG)
5. The Unit Group of Z[ε_n]
6. Effective Construction of Units of ZG
Chapter 6. Isomorphism Questions
1. Preliminary Results
2. Finite Groups
3. May's Conjecture
4. Algebraically Closed Fields
5. Countable pGroups
6. Real Group Algebras
Chapter 7. Research Problems
References
Notation
Index
COMMUTATIVE GROUP ALGEBRAS
PURE AND APPLIED MATHEMATICS A Program of Monographs, Textbooks, and Lecture Notes
EXECUTIVE EDITORS .
Earl J. Taft
Zuhair Nashed
Rutgers University New Brunswick, New Jersey
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Edwin Hewitt
University of California, Berkeley Berkeley, California
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EDITORIAL BOARD
Glen E. Bredon Rutgers University Sigurdur Helgason Massachusetts Institute of Technology Marvin Marcus University of California, Santa Barbara
W. S. Massey Yale University Leopoldo Nachbin Universidade Federal do Rio de Janeiro and University of Rochester Donald Passman University of Wisconsin
Irving Reiner University of Illinois at UrbanaChampaign Fred S. Roberts Rutgers University Paul J. Sally, Jr. University of Chicago Jane Cronin Scanlon Rutgers University Martin Schechter Yeshiva University Julius L. Shaneson Rutgers University
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Other Volumes in Preparation
COMMUTATIVE GROUP ALGEBRAS GREGORY KARPILOVSKY La Trobe University Bundoora/ Victoria Australia
MARCEL DEKKER, INC.
New York and Basel
Library of Congress Cataloging in Publication Data Karpilovsky, Gregory, [date] Commutative group algebras. Includes bibliographical references and index. 1. Commutative algebra. 2. Group algebras. 3. Abelian groups. I. Title. QA251.3.K37 1983 512'.24 837173 ISBN 0824719182
.COPYRIGHT© 1983 by MARCEL DEKKER, INC.
ALL RIGHTS RESERVED
Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. MARCEL DEKKER, INC. 270 Madison Avenue, New York, New York 10016 Current printing (last digit): 10 9 8 7 6 5 4 3 2 1 PRINTED IN THE UNITED STATES OF AMERICA
FOR HELEN, SUZANNE
&ELLIOTT
Preface
The commutative group algebra RG is an Ralgebra which reflects properties of the abelian group G and the ring of coefficients R. These algebras may be studied for their own sake or as a tool for tackling other mathematical problems.
There exist·s a long history
of interaction between commutative group algebras and other mathematical concepts.
One of the first applications was a theorem of
Krull [l] which states that any abelian ordered group is the value group of a valuation.
Since then a number of results of Iwasawa
[1,2] in algebraic number theory have led to purely algebraic theorems concerned with the group algebras of Galois groups of cyclotomic fields.
In quite a different area, some important results in
algebraic coding theory have involved ideals of group algebras of cyclic groups over Galois fields.
A further instance is furnished
by algebraic and differential topology.
Here the reader may profit
ably consult the excellent article of Milnor [l]. The theory of commutative group algebras as an independent area of study has developed only in relatively recent times, following the fundamental work of Higman [1,2].
During the last two dec
ades the subject has been pursued by a number of researchers and many interesting results have been obtained.
The present book is
intended to give a systematic account of this work, but in no way is the coverage intended to be encyclopedic.
Our main goal is to
tie together various threads of the development in an effort to convey a comprehensive picture of the current state of the subject.
V
vi
Preface
We consider here only commutative group algebras; noncommutative group aigebras are extensively presented in the books of Passman [1,3], Passi [2], and Sehgal [4]. This book is written on the assumption that the reader has had the equivalent of a standard firstyear graduate algebra course. Thus we assume a familiarity with basic ringtheoretic and grouptheoretic concepts and an understanding of elementary properties of modules, tensor products, and fields.
Apart from a few specific
results and the general knowledge we have presupposed, the book is entirely selfcontained.
We have included, for the convenience of
the reader, a chapter on algebraic preliminaries.
This chapter
provides a brief survey of topics needed later in the book.
A sys
tematic description of the material is supplied by the introductions to individual chapters. The book can be roughly divided into two parts, which will not preclude, however, some strong interrelations between these.
The
first part is devoted to the investigation of various ringtheoretic properties of the group algebra RG.
Among the properties con
sidered are those of being reduced, indecomposable, semisimple, local, and semilocal.
We also describe the nil and Jacobson radi
cals and investigate the Krull dimension. devoted to integral domains.
A separate chapter is
Here we determine necessary and suf
ficient conditions on Rand Gin order that RG should be a GCD domain, UFD, PID, Dedekind domain, euclidean domain, and Krull domain.
Using Pontryagin's result on torsionfree groups of rank 1,
we construct examples of nonnoetherian unique factorization domains of arbitrary characteristic and arbitrary Krull dimension~ 2. In the second part, comprising Chapters 5 and 6, we are concerned with the unit group U(RG) of RG and the isomorphism problem. After introducing a class of rings R for which RG has only trivial units of finite order, we discuss the case where G is torsion free. We then concentrate on splitting properties of U(RG) and, in particular, give circumstances under which U(RG) is free modulo torsion.
Special attention is drawn to the case where R
=
Z.
vii
Preface
Extending a classical result of Higman, we determine the isomorphism class of U(ZG) for an arbitrary group G.
A separate section is de
voted to the problem of effective construction of units of ZG.
The
last topic is an investigation of the relations between groups G and Hin the case that the group algebras RG and RH are Risomorphic.
Finally we list and discuss a number of open problems in the
field. I would like to express my gratitude to my wife for the tremendous help and encouragement she has given me in the preparation of this book.
I am also indebted to Kenneth Pearson, who read a por
tion of the manuscript and made many helpful suggestions, and to Warren May, who provided the result concerning the isomorphism class of U(ZG).
Finally, my thanks go to Helen Cook and Judy Westwood for
their excellent typing. Gregory Karpilovsky
Contents
PREFACE CHAPTER 1.
V
PREREQUISITES 1. 2.
CHAPTER 2.
PRELIMINARY RESULTS
RINGTHEORETIC PROPERTIES
44
2. 3. 4. 5.
Chain Conditions. Regular and Completely Reducible Group Algebras Characters, Direct Decompositions, and Irreducible Modules Idempotents Nil and Jacobson Radicals Krull Dimension
INTEGRAL DOMAINS 1.
2. 3. CHAPTER 5.
27 29
1.
CHAPTER 4.
1 5
Definitions Elementary Properties
1. 2.
CHAPTER 3.
1
Notation and Terminology Background Sketch
3. 4. 5. 6.
44 48 59 68 75 83
GCD Domains Unique Factorization Domains Krull Domains
STUDY OF UNITS 1. 2.
27
83
90 108
115
Units of Finite Order The Unit Group of Group Algebras of TorsionFree Groups Splitting Unit Groups The Isomorphism Class of U(lG) The Unit Group of Z1En] Effective Construction of Units of lG
ix
115 120 127 136 141 151
Contents
X
CHAPTER 6.
CHAPTER 7.
ISOMORPHISM QUESTIONS
160
1. 2. 3. 4. 5. 6.
161 166 174 182 185 190
Preliminary Results Finite Groups May's Conjecture Algebraically Closed Fields Count ab le pGroups Real Group Algebras
RESEARCH PROBLEMS
198
REFERENCES
203
NOTATION
215
INDEX
219
1 Prerequisites
The aim of this chapter is to give the reader some idea of the mathematical background we are assuming, as well as to fix conventions and notation for the rest of the book.
Because we presuppose a
familiarity with various elementary ringtheoretic terms, only a brief description of them is presented. 1.
NOTATION AND TERMINOLOGY
All rings and algebras in this book are commutative with 1 I 0, and subrings of a ring Rare assumed to have the same identity element as R.
Each ring homomorphism will be assumed to preserve identity
elements.
All modules are unital.
The word "group," with the exception of Galois group and automorphism group, will mean, throughout, an abelian group.
Unless
explicitly stated otherwise, all groups are assumed to be multipliWe use pointed brackets to denote "subgroup generated by."
cative.
The cyclic group of order n is denoted by Zn. We shall write A  B for the complement of the subset B in the set A, while Ac B will mean that A is a proper subset of B. The symbol for a map will be written before the element affected, and consequently, when the composition fog of maps is indicated, g is the first to be carried out. If f: X + Y is any mapping, then £IX' denotes the restriction off to a subset X' of X. Let R be a ring. 1
+ · ··
+
The mapping Z
+
R defined by n
+
n•l
=
1
+
1 (n summands) is a ring homomorphism whose image is called 1
Chap. 1.
2
Prerequisites
the prime subring of R; its kernel is an ideal mZ for a unique m ~ 0, called the characteristic of Rand denoted by char R. We shall in future identify Z/mZ with its image in Rand write n instead of n•l. uv
=
An element u of a ring R is said to be a unit if
1 for a suitable v ER.
The set U(R) of all units in R con
stitutes a group, called the unit group of R.
Note that if
w:
S is a ring homomorphism, then w induces a group homomorphi_sm
R +
U(R)
+
U(S) whose kernel is
U(R)
n
(1 + Ker W)
Let R be a ring.
An element x ER is nilpotent if xn
some positive integer n.
=
0 for
An ideal Jin R is nil tf every element
of J is nilpotent, while J is nilpotent i f there is a positive integer n such that Jn 0, where Jn is the product of J with itself n times.
An element x of a ring R is idempotent if x
idempotent is trivial if it is O or 1. orthogonal if uv = 0.
2
=
x.
An
Two idempotents u,v are
A nonzero idempotent is primitive if it can
not be written as a sum of two nonzero orthogonal idempotents. A ring R is reduced if O is the only nilpotent element of R, while R is indecomposable if it has no nontrivial idempotents. A ring R is local (respectively, semilocal) if it has precisely one maximal ideal (respectively, if it has only finitely many maximal ideals). xy
=
An element x of a ring Risa zero divisor if
0 for some nonzero y ER; in case x f O and xis a zero divisor,
we say that xis a proper zero divisor. An integral domain is a ring without proper zero divisors.
An
ideal P of a ring R is prime if R/P is an integral domain, while P is primary if every zero divisor in R/P is nilpotent. The Jacobson radical J(R) of a ring R is the intersection of the maximal ideals of R; equivalently, J(R) consists of all x ER such that for ally ER, 1  xy is a unit.
A ring R is semisimple
i f J (R) = 0.
The set N(R) of all nilpotent elements in R constitutes an ideal called the nilradical of R. R is reduced if and only if N(R)
Note that N(R) k J(R) and that =
0.
1.1
Notation and Terminology
3
A prime ideal P of a ring R is minimal if there is no prime ideal P' of R such that P' c P.
Note that N(R) = intersection of
all prime ideals of R = intersection of all minimal prime ideals of R. For an arbitrary set I, we denote by R[(Xi)iEI] the polynomial ring in the indeterminates Xi with coefficients in a ring R.
For
I= {1,2, ... ,n}, we ~rite R[X 1 ,x 2 , ... ,Xn] instead of R[(Xi)iEI]. The elements \EIX~(i) of R[(Xi)iEI] [v(i) ~ 0 and v(i) = 0 for all but finitely many i] are called monomials in the indeterminates X.. l
Any nonzero polynomial in R [X] can be written in the form n f = r 0X
+
n1 r 1X
+
...
+
rnlx + r n
r . E R, ro f. 0 l
We call n the degree of f and write n = deg f· thus the polynomials ' of degree 0 are just the nonzero constants. The leading coefficient off is r 0 and r 0 Xn is its leading term. manic if its leading coefficient is 1. Let R be a subring of a ring S.
A polynomial f of R[X] is
An element a ES is integral
over R if a is a root of a monic pQlynomial in R[X]; the monic polynomial of least degree satisfied by a is called the minimal polynomial for a over R.
The set of all elements in S that are integral
over Risa subring of S containing R. integral closure of R in S.
This subring is called the
We say that Sis integral over R (or
that Sis an integral extension of R) if Sis the integral closure of R in S; R is integrally closed in S if R is the integral closure of R in S.
An integral domain is integrally closed if it is inte
grally closed in its field of fractions. Let R be a subring of S.
For any subset A = {a. l
I
i E I} of
S, denote by R[A] the smallest subring of S containing Rand A (for A= {a 1 ,a 2 , ... ,an}' we write R[a 1 ,a 2 , ... ,an] instead of R[A]). The ring R[A] consists of all polynomials in a. (i EI) with coeffi1
cients in R.
We say that Sis finitely generated over R (or simply
finitely generated if R is the prime subring of S) if there is a
finite subset A of S such that S Let R be a ring.
R[A].
For any r ER, the ideal Rr [also denoted by
(r)] generated by r is called a principal ideal.
We say that an
4
Chap. 1.
Prerequisites
,ideal J of R is finitely generated if J = Ra 1 + ···+Ran for suitable a 1 ,a 2 , ... ,an ER. Let S be a subset of a ring R.
We denote by ann(S) the anni
hilator of Sin R, i.e., ann(S) = {r ER I rs= 0 for every s ES}.
Clearly, ann(S) is an ideal of R. Let M be an Rmodule, and let m EM. is the submodule of M generated by m. generated if there exist m1 ,m 2 , ...
RI\:.
Then Rm= {rm Ir ER}
We say that Mis finitely
,1\:
EM such that M = Rm 1 + ··· + A module Mis completely reducible if every submodule of Mis
a direct summand of M.
A nonzero Rmodule Mis irreducible if its
only submodules are O and M. A direct sum of a family of modules (Mi)iEI is written $Mi whenever the index set I is not pertinent to the discussion. Let M be a free Rmodule with the set Sas a basis.
Then the
cardinality of Sis said to be the rank of M; the rank of M will be denoted by (M:R).
A group G is called free if G (written additively)
is a free Zmodule. A diagram of Rmodules and Rhomomorphisms (or rings and ring homomorphisms)
f L
g
s is commutative if g
O
f
h.
The same terminology applies to more
complicated diagrams. A sequence fi fi+l ...  M i +Mi+l  M i + 2   + ... of Rmodules and Rhomomorphisms is exact if for each i, Im fi
1. 2
Background Sketch
Ker fi+l·
5
The sequence
oM
f
(*)
is exact if and only if f is injective, g is surjective, and Im f Ker g.
A sequence of this type is called a short exact sequence.
An exact sequence(*) splits if Im f
=
Ker g is a direct summand of
N. 2.
BACKGROUND SKETCH
In this section we list explicitly those basic results of the theory of rings, modules, and abelian groups which will be assumed without proof.
Details may be found in standard texts covering these topics,
for example, Bourbaki [1,2], Curtis and Reiner [l], Fuchs [1,2], Kaplansky [1,2], Lambek [l], and Lang [l]. Ordered Sets
A relation Son a set Sis said to be a partial order if it is reflexive, antisymmetric, and transitive, i.e., if (a)
s x for all x ES.
x
y and y s x, then x
(b)
If x,y ES with x
(c)
If x,y,z ES with x Sy and y S z, then x S z.

Jl
iEI
The injective homomorphism
G.
1
sending any gi E Gi onto the vector whose ith component is gi and whose other components are 1 is called canonical. If we identify each Gi with its image in llGi' then the group G
=
llGi contains a family of subgroups (Gi)iEI such that each ele
ment of G can be uniquely written in the form TiiEJgi for a suitable finite subset J of I.
Conversely, any group G which possesses a
family of subgroups (Gi)iEI that satisfies the above property is isomorphic to llGi; by an abuse of notation, we write G G
=
llGi (or
= G1 x ··· x Gn if JIJ = n) whenever a family of subgroups (G.) 1
of G'satisfies the abovementioned property. In the future, unless explicitly stated otherwise, "direct product" will always mean restricted direct product. The group llGi enjoys the following universal property: 2.36
PROPOSITION
Let H be a group, and for all i EI, let Fi :
Gi> H be a homomorphism. morphism
1jJ :
Then there exists one and only one homo
llGi > H which for all i E I renders commutative the
diagram
Uc.l
Chap. 1.
24
Prerequisites
A pgroup is defined to be a group the orders of whose elements are powers of a fixed prime p.
An elementary pgroup is a dire.ct
product of cyclic groups of order p. whose order is a power of the prime p.
Let G consist of all g E G p
Then G is a subgroup of G, p
called the pcomponent of G. 2.37
PROPOSITION G
Let G be a torsion group.
Then
=Jl p Gp
For any integer n > 0, the mapping G + G sending every element g of G to gn is a homomorphism, whose image and kernel are denoted by Gn and G[n}, respectively. some n ~ 1. every n (a)
~
1.
A group G is bounded if' Gn = 1 for
A subgroup Hof G is pure in Gin case Gn
nH
~ Hn for
The following is an easy consequence of the definition:
Any direct factor of G is pure in G.
(b)
If G/H is torsion free, then His pure in G; in particular, G0 is a pure subgroup of G. (c)
If G is torsion free, then a subgroup Hof G is pure if and
only if G/H is torsion free. (d)
Purity is transitive, i.e., if K is pure in Hand His pure in
G, then K is pure in G. A group G is called divisible if Gn = G for any n ~ 1, and pdivisible if Gp= G.
pdivisible.
A pgroup is divisible if and only if it is
Let p be a prime.
The pnth complex roots of I, with
n running over all integers~ 0, form an infinite multiplicative group.
This group is called quasicyclic or a group of type p
is denoted by Z(p
00
),
and the additive group
2.38
and
).
The groups Z(p groups.
00
00
U(K) where K is an algebraically closed field,
a(
of 01 are typical examples of divisible
The group 01+ is also called the full rational group.
PROPOSITION
Any divisible group is a direct product of quasi
cyclic and full rational groups. It is easy to verify that the product Gd of all divisible subgroups of G is a divisible group; the group Gd is called the maximal
1.2
25
Background Sketch
divisible subgroup of G.
2.39
PROPOSITION
if and only if G[p]
A group G is reduced if Gd"= 1.
Let G and H be divisible pgroups. ~
Then, G  H
H[p].
A group G is noetherian if every ascending chain of subgroups of G breaks off. 2.40
PROPOSITION
The fol lowing conditions are equivalent:
G is finitely generated.
(i) (ii)
G is a direct product of finitely many cyclic groups.
(iii)
G is noetherian. ct
Let p be a fixed prime.
For an ordinal
ct,
the group Gp
is
defined as follows: G
n
if
is a limit ordinal
ct
Sl®g is an injective homomorphism. (ii)
Now apply Proposition 2.1.
Special case of (i).
(iii)
Follows from (i) by application of K ®R·
•
Another consequence of Proposition 2.1 is the following: 2.3
PROPOSITION RG
Let G be a cyclic group of order n.
= R[X]/(Xn
JG>
 1)
Let g be a generator for G.
Proof:
Then
Then the mapping
R[XJ/(:n  1)
1g I> X
+
(X
 1)
is an injective homomorphism whose image is an Rbasis for R[X]/ (Xn  1).
Hence the desired isomorphism follows by virtue of
Proposition 2.1.
•
We also have the following result related to Proposition 2.2.
2.2
Elementary Properties
2.4
PROPOSITION
31
Suppose His a group such that RG
there exists a field' K such that KG= KH.
= RH.
Then
Furthermore, if a ra
tional prime pis a nonunit of R, then as Kone can choose a field of characteristic p. Let J be a maximal ideal of Rand let K
Proof:
=
R/J.
Then K
is a field, and by Proposition 2.2(iii), we deduce that KG= KH.
If
pis a nonunit of R, then as J we can choose a maximal ideal of R containing p.
Since in this case char K
p, the result follows.
=
•
We shall next show that the class of group algebras is closed under the formation of arbitrary tensor products. 2.5
PROPOSITION R(_il G.)
iEI Proof:
1
=
Let (Gi)iEI be a family of groups.
Then
RG.
©
1
iEI
For any i E I, let Tii : RGi> R(UGi) be the homomorph
ism induced by the canonical homomorphism Tii : Gi> UGi, and let fi : RGi> A be a homomorphism of Ralgebras.
The restriction of
fi'
is a homomorphism of groups.
Therefore, thanks to the universal
property of UGi' there exists a homomorphism 1j.,
:
UGi > U(A)
which for all i EI renders commutative the diagram
Uc.
1
G. 1
Chap. 2.
32
Preliminary Results
Let~: R(UGi) +Abe the homomorphism induced by ij,. Because any homomorphism RGi + A of Ralgebras is uniquely determined by its restriction to Gi' the diagram R(UG.) 1.
RG.
1.
is commutative.
The desired conclusion is now a consequence of the
universal characterization of the tensor product of algebras.
•
Given an arbitrary set I 1 denote by 1
R[(X.), (X. ) . EI] l l l the ring of fractions of R[(Xi)iEI] associated with the multiplicative set of monomials in the indeterminates X.. 1 1
.
If I is finite, 1
1
.
say I= {1,2, ... ,n}, we write R[X 1 ,x 2 , ... ,Xn,Xl ,x 2 , ... ,Xn] instead of R[(X.) ,(x~ 1 ) "EI]. The ring R[X,X 1 ] is called Laurent polynom1
ial ring.
2.6
l
l
We are now ready to prove:
PROPOSITION
Let I be an arbitrary set and let G be a free
group with (gi)iEI as a basis. RG = R[(X.) l
Then
'ex.l1 ) l. EI]
as Ralgebras. Proof:
Put R1 = R[(Xi)iEI] and observe that R1 is a free R
module on the set of monomials. ij,:
v(i)
R1 + RG of the map TIXi
Therefore the Rlinear extension v(i) + Tigi is a homomorphism of Ralge
bras such that the image of any monomial is a unit.
Let f: R1 + B be a homomorphism of Ralgebras which maps monomials into units. Since G is a free group, the map g. + f(X.) can be uniquely extended l
l
2. 2
Elementary Properties
to a homomorphism A: G
+
33
U(B).
Leth: RG
+
B be the homomorphism
induced by A.
Because any Ralgebra homomorphism R1 + B is determined by its restriction to the set of monomials, we infer that h is a unique homomorphism which renders commutative the diagram
1
An appeal to the universal characterization of R[(X.),(X. ).EI]' l
now completes the proof.
g
l
•
Let T be an ideal of R. TG = {Ix g ERG [ x
l
g
Then the set
ET
for all g}
is the ideal of RG generated by T. The proof of the following observation is straightforward and therefore will be omitted. 2.7
LEMMA
Let~ : R
let T =Ker~
+
S be a surjective homomorphism of rings and
Then the mapping~: RG
+
SG defined by ~(Zx g) g
Z~(xi?g is a surjective homomorphism whose kernel is TG. As an application of the preceding lemma, we now prove: 2.8
PROPOSITION
By Lemma 2.7, the canonical homomorphism ~S : R
to a ring homomorphism~: RG
+
RSG.
+
+
RS extends
Furthermore, since ~(S)
~S(S) it follows that any element in ~(S) is a unit. that f: RG
Then
Again, we use universal characterization of quotient
Proof:
rings.
Let S be a multiplicative subset of R.
Suppose next
A is a homomorphism of rings such that all elements
in f(S) are units.
Denote by f the restriction off to R.
Because
all elements in f(S) are also units a universal characterization of
Chap. 2.
34
Preliminary Results
RS tells us that there is a unique homomorphism rr
RS+ A which
renders commutative the diagram
It is now readily verified that the mapping h: RS~+ A defined by h(Ix g) = Irr(x )f(g) is a unique homomorphism which renders commug g tative the diagram
RG
A
Consequently, the result follows by virtue of the universal characterization of (RG) S.
•
There is one concept relating to group algebras that is of particular importance. 2. 9
DEF IN IT ION
This is embodied in the following.
The augmentation ideal I (R, G) of the group algebra
RG is the kernel of the homomorphism from RG to R induced by collapsing G to 1.
In other words, I(R,G) consists of all x
for which aug(x) = Ix
g
=0
We shall refer to aug(x) as the augmentation of x.
=
Ix g ERG g
2.2
35
Elementary Properties
It follows from the equality l:x g = l:x (g  1) + l:x that as g g g an Rmodule, I(R,G) is a free module with the elements g  1 (1 'f g E G) as a basis.
In the future we shall often suppress ref
erence to Rand simply denote the augmentation ideal of RG by I(G). The proposition that follows demonstrates that for any subgroup N of G, the group algebra R(G/N) inherits all identities enjoyed by RG. 2.10
PROPOSITION
groups and let N
Let 1/J : G + H be a surjective homomorphism of =
Ker 1/J.
Then the mapping f: RG
+
RH which is
the Rlinear extension of 1/J is a surjective homomorphism of Ralgebras whose kernel is RG•I(N). RG/RG•I(N)
In particular,
R(G/N)
~
That f is a surjective homomorphism of Ralgebras is a
Proof:
consequence of Proposition 2.1.
It is plain that RG•I(N)
Consequently, f induces a homomorphism f: RG/RG•I(N)
+
~
Ker f.
RH.
The
restriction off, X: [G
+
RG•I(N)]/RG•I(N)
is an isomorphism.
+
H
Thanks to Proposition 2.1, \
1
can be extended
to a homomorphism RH+ RG/RG·I(N) which is inverse to£. Ker f
=
RG•I(N), as desired.
Thus
•
Let J be an ideal of RG and let G
n
o
Then G n (1
+ J) = {g E G +
I
g  1 E J }
J) is the multiplicative kernel of the natural map
G + RG/J, and hence a subgroup of G.
In view of this observation,
the next corollary arises from Proposition 2.10 by taking J = RG•I(N). 2.11
COROLLARY
Let N be a subgroup of G.
G n Il + RG•I(N)J
=
Then
N
In particular, if N1 ,N 2 are subgroups of G such that RG•I (N 1) RG•I(N 2), then N1 = N2 .
Chap. 2.
36
Preliminary Results
Let J be an ideal of R, and let N be a subgroup of G.
We
shall often in the future consider one of the homomorphisms RG
+
RG
(R/J)G
+
R(G/N)
RG
+
(R/J)(G/N)
induced in the obvious manner by reduction modulo J, N, or both J and N.
We shall refer to these homomorphisms as natural maps.
Occasionally we shall consider homomorphisms into a direct product. We shall also refer to them as natural maps provided each homomorphism into a factor is a natural map. It is a wellknown fact that any group G is isomorphic to F/N for a suitable free group F and a suitable subgroup N of F.
The
following classical result shows that the additive groups I(l,G) and I(l,G) 2 fulfill the roles of F and N, respectively. To whom this result should be ascribed is a mystery.
One can easily re
cover it from the unpublished thesis of Higman {l].
To the best
of our knowledge, the first proof which appeared in the literature is due to P. Cohn {l]. 2 .12
THEOREM
Let G be an arbitrary group.
Then
G ~ I (l , G) / I (l , G) 2 Proof:
{g  1
I
Since I(l,G) is a free abelian group on the set
1 / g E G}, the mapping g  1 + g determines a surjective
homomorphism I (l ,G) (a  1) (b  1)
L =
G.
The identity
(ab  1)  (a  1)  (b  1)
a,b E G
shows that I(l,G) 2 S Ker f, while the identity ab  1
= (a
a,b E G
 1) (b  1) + ( a  1) + (b  1)
ensures that the mapping
w:
G + I(l,G)/I(l,G) 2 defined by t(g)
(g  1) + I(l,G) 2 is a homomorphism.
Since
w is
homomorphism induced by f, the result follows.
the inverse to the •
As an immediate consequence, we derive the following. 2 .13 (i)
(ii)
COROLLARY l:x (g  1)
g
G
n [1
Let G be a group.
=
Then
X
(ITg g)  1 [mod I(l,G) 2 ] (x
+ I (l , G) 2 ] = 1.
g
El).
2.2
Elementary Properties
37
We keep the notation of the preceding theorem. The kernei of the homomorphism f is I(Z,G) 2 . Hence (i)
Proof:
(i)
follows by virtue of the equality X
£Cix Cg  1)) g
(ii)
=
X
f(Tig g  1)
ng g
This is a direct consequence of the fact that¢ is an
isomorphism.
•
While on the subject, we also observe the following interesting fact. 2.14
PROPOSITION
Let G be a group.
Then G x
~z
is the unit group
of a ring. We shall show that G x z 2 is isomorphic to the unit 2 group of ZG/I(Z,G) . It follows from Corollary 2.13(ii) that the Proof:
natural map ±G
+
ZG/I(Z,G) 2
is an injective homomorphism. image ±G
+
assume x
Therefore we need only show that the I~,G/ of this map is the unit group of ZG/I(Z,G) 2 . So LX g is a unit modulo I(Z,G)
2
g
unit in Z, hence aug(x)
=
±1.
.
Certainly, aug(x) is a
Replacing x by x, if necessary, we
may assume that aug(x) = 1, in which case x = 1 + LX (g  1) . now follows from Corollary 2.13(i) that x g E G, hence the result.
= g[mod
g
2
It
I(Z,G) ] for some
•
Let H be a subgroup of G.
Because RH is a subring of RG, we
can view RG as an RHmodule by way of ordinary multiplication. will next be shown that RG is a free RHmodule.
It
As an immediate
consequence of this, we shall infer that RH 2.15
n U(RG)
=
PROPOSITION
U(RH) Let H be a subgroup of G and let T be a trans
versal for Hin G. (i) (ii) of RG.
RG is a free RHmodule with T as a basis. An element u E RH is a unit of RH if and only if it is a unit
Chap. 2.
38
(i)
Proof:
Prel.iminary Resul. ts
Observe that for any t ET, (RH)t is the Rlinear
Accordingly, for any t 1 ,t 2 , ... ,tn ET, (RH) t 1 + •.. + (RH) tn is the R~linear span of U~= 1Hti. As is apparent from the definition of RG, if s 1 and s 2 are disjoint subsets of of G, then their Rlinear spans meet at 0. Hence RG = $tET(RH)t, span of the coset Ht.
proving (i) . Only the "if" part needs a verification.
(ii)
such that uv
=t
o(t) 11.
:
RG
+
1.
Thanks to (i), the mapping o : T
fort EH and o(t)
= O fort 1 H
RH of RHmodules.
Therefore ?.(uv) = u?.(v)
asserted.
Let v E RG be +
RH defined by
extends to a homomorphism =
1, as
•
The following result will frequently be applied in investigating properties of single elements (e.g., idempotents, nilpotent elements, units). 2.16
PROPOSITION
Let a group G be the direct product of its sub
2.
groups G1 and G2 . Then RG = (RG 1)G Proof: By choosing H = G1 and T = G2 in Proposition 2.lS(i), we conclude that each element of RG can be uniquely written in the form t g
g
Since the multiplication in RG is the RG 1 linear extension of that in G2 , the result follows. • As a further consequence of Proposition 2.lS(i), we have 2.17
PROPOSITION
and let H
=
G1
n
Let G be the product of its subgroups G1 and G2 , G2 . Then
RG :: RG . © RG 1 RH 2 as RHalgebras. Proof:
l, 2) .
Let T. be a transversal for Hin G. containing 1 (i = 1
1
Then, by Proposition 2. lS(i) , RGi is a free RHmodule with
2.2
39
Elementary Properties
T. as a basis. Consequently, RG 1 ®mi RG 2 is a free RHmodule with 1 {a 18> S J a E T1 , S E T 2 } as a basis. Because T1 T 2 is a transversal for Hin G, it is an RHbasis for RG, again by Proposition 2.lS(i). Therefore, the mapping f : RG
+
RG 1 ®RH RG 2 which is the RHlinear
extension of aS + a ® S (a E T1 , S E T) is an RHisomorphism, preserving identity elements. Since f(xy) = f(x)f(y) for any x,y E T1T 2 , f is in fact an isomorphism of RHalgebras, as required.
•
We next record some elementary properties concerning annihilators of augmentation ideals. 2.18
PROPOSITION
(i)
If His an infinite subgroup of G, then
ann(_I (}!}} = 0. (ii}
If His a finite subgroup of G, then ann(I(H)) = RG EhEHh.
(iii)
If g E G has infinite order, than ann(l
g)
(iv)
If g E G has finite order n, then ann(l
g)  RG(l + g +
+ g
n1
0.
) . (i)
Proof:
Suppose that O ,f, a = Ea g E ann(I (H)) . Then for g Consequently, h permutes the
h EH, a(h  1) = 0 and so ah= a.
elements of Supp a by way of ordinary multiplication.
It follows
that H admits an injective homomorphism into the permutation group of a finite set, hence His finite. (ii) therefore a
The equality ah= a implies a g
=ah for any g E G and g g is constant on the cosets of H. This proves that
ann(I(H)) S RG
l
h
hEH The reverse inclusion is obvious. (iii)
For any n E Z, we have the identity 1  gn n1 g + •·· + g ) and so
(1  g)(l +
ann(l  g) = ann(I) Hence (iii) follows from (i) by taking H (iv)
Apply (ii) for H = .
.
•
We now turn our attention to group algebras of ordered groups. Recall that a group G is an ordered group if the elements of Gare
40
Chap. 2.
Preliminary Results
linearly ordered with respect to relations and, if for all x,y,z E G,x s y implies xz s yz. Let us remark 0n a trivial, but useful fact.
Since this fact
will be used repeatedly, we shall isolate it in the following. 2.19
LEMMA
Let G be an ordered group and let x = Ex g and y = g
Ey g be nonzero elements of RG. g
(i)
If a and bare maximal (or minimal) elements in Supp x and
Supp y, respectively, then (xy)ab = xayb. (ii) If R is an integral domain and if xy is a monomial, then so are x and y. (i)
Proof:
The hypothesis ensures that for g E Supp x and
h E Supp y, gh
ab implies g = a and h = b.
Accordingly,
(xy} ab (ii)
Suppose the contrary, say xis not a monomial.
Then we
may choose the greatest and the smallest elements a,c E Supp x where a f c.
Denote by band d their counterparts in Supp y.
by (i), (xy)ab f O and (xy)cd f 0. !Supp xy\ > 1, a contradiction.
Then
Since ab> cd, we must have
•
The next proposition will enable us to take full advantage of the preceding lemma. 2.20
PROPOSITION Proof:
Every torsionfree group is an ordered group. 1
For any subset T of a torsionfree group G, put T
{tl It ET}.
Let L be the class of all subsets T of G that sat
isfy the conditions: x,y ET implies xy ET
1
f
T
By Zorn's lemma, L contains a maximal element T.
We claim that
G = T U {1} U Tl Deny the statement.
(*)
Then there exists a nonidentity gin G such
that neither g nor g 1 belongs to T.
With
S =TU {tgn It~ T, n ~ 1} U {gn In~ 1}
2.2
Elementary Properties
41
it follows that SJ T and that Sis closed under multiplication. The maximality of T now yields 1 ES and since G is torsion free tgn for some t ET and some n ~ 1.
we see that 1
Thus gn ET
and replacing g by gl in the above argument, we infer that gm E T for some m ~ 1.
This substantiates our claim, as 1 = (gm)n(gn)m E
T, contrary to our assumption. We now define x
0.
(ii)
dim RG is finite if and only if both dim Rand a are finite.
(iii)
Suppose that both dim Rand a are finite and R is noetherian.
Then dim RG
=
(dim R) + a.
3.5
Krull Dimension
(i)
Proof:
81
If a = 0, then G is torsion, so RG is integral
over R; hence dim RG a
>
dim R, by virtue of Corollary 5.7.
0 and choose a maximal independent system M among systems con
sisting of elements of infinite order in G, F
=
So assume
Then JMJ
is free of rank a and G/F is a torsion group.
a, the group
=
Since G/F is
torsion, every element in G is integral over RF; hence RG is integral over RF. dim RG
=
We may therefore apply Corollary 5.7 to conclude that dim RF
(*)
Consequently, there is no loss of generality in assuming that G is free of rank a> 0. Let P0 J P1 J ••• J Pn be a prime chain in R. Since RG/PiG (R/Pi)G and since (R/Pi)G is an integral domain, it follows that ...
J
p G
n
is a prime chain in RG. Furthermore, P0G is not a maximal ideal since (R/P 0 )G is not a field [every element in (R/P 0 )G with zero augmentation is a nonunit]. Thus RG possesses a prime chain of length n + 1, as asserted. (ii)
Suppose that dim RG is finite.
virtue of (i).
By(*) we may assume for the rest of the proof of
the theorem that G is free. a is infinite.
Then so is dim R, by
Suppose by way of contradiction that
Then there exists an infinite strictly ascending
chain of subgroups
such that for any i, G/Hi is torsion free.
Consequently,
is an infinite strictly ascending prime chain in RG, contrary to our assumption that dim RG is finite.
Thus a is finite.
Conversely, assume that both dim Rand a are finite. that dim RG is finite, we argue by induction on a. the result is true by virtue of Lemma S.S(ii).
If a
To prove 1, then
If G has rank n
+
1,
Chap. 3.
82
then G
RingTheoretic Properties
G1 x G2 with G1 of rank n and G2 of rank 1. Since RG (RG 1)G 2 and since dim RG 1 is finite (by induction hypothesis), it follows that dim RG is also finite, again invoking Lemma 5.S(ii). =
The property (ii) is therefore established. (iii)
Suppose first that
cyclic group.
a=
1, so that G is an infinite
Then dim RG = dim R[X], by virtue of Lemma 5.S(i).
Since R is noetherian, Theorem 5. 5 tells us that dim RG proving (iii) for
a=
1.
on a, by an argument similar to that in (ii). true.
•
=
(dim R)
+ 1.
The general case now follows by induction So the theorem is
4 Integral Domains
Throughout this chapter, R denotes an integral domain and Ga torsionfree group.
We start by determining necessary and sufficient
conditions on Rand Gin order that RG should be a GCD domain. Unique factorization domains are precisely the GCD domains in which the ascending chain condition for principal ideals is satisfied. Hence our results on GCD domains are useful to us in.considering RG as a UFD. In Sec. 2 we find necessary and sufficient conditions on G and R so that RG should be a UFD, PID, Dedekind domain, and a euclidean domain. Using Pontryagin's construction of a group G of rank 2 such that each rankone subgroup of G is cyclic, but G is not finitely generated, we enrich the supply of nonnoetherian UFOs.
Namely, we
construct examples of nonnoetherian unique factorization domains of arbitrary characteristic and arbitrary Krull dimension~ 2. 3
In Sec.
we determine necessary and sufficient conditions under which RG is
a Krull domain. 1.
GCD DOMAINS
We start by recalling the notion of divisibility in R.
As in Z we
define alb (for any a,b ER) to mean b = ar
for some r ER
This is equivalent to the requirement (b)
~
element u ER is a unit if and only if ull.
(a).
Observe that an
The units are trivial
divisors since they are divisors of every element of R. 83
84
Chap. 4.
Integral Domains
If alb and bla [or equivalently, if (a) = (b)], then we shall say that a and bare associated.
It is easy to see that a and b
are associated if and only if a= bu for some unit u of R. A nonzero element p of R is said to be a prime if (p) is a prime ideal.
Expressed otherwise, a nonzero p E R is a prime if p
is a nonunit such that plab implies pla or plb By an irreducible element we understand a nonunit which is not a product of two nonunits. A prime is always irreducible:
if a prime p satisfies p = ab,
then by definition, pla or plb, say pla.
It follows that a
abr for some r ER, and a~ 0 (as divisor of p).
pr=
Hence br = 1, by
cancellation, sob is a unit. A greatest common divisor (GCO) of elements a 1 , ... ,an ER is an element r ER such that r divides each ai and r is divisible by each common divisor of a 1 , The GCO is not unique:
a . n
if r is a GCD, then so is any r' asso
ciated to r.
Conversely, if r and r' are both GCOs of a 1 , ... , an, then r' is associated tor. We shall ignore the distinction between
associates and denote any one of the GCDs of a 1 , ... , an by (al, ... ,an). 1.1
DEFINITION (Kaplansky)
A GCD domain is an integral domain in
which each pair of nonzero elements has a greatest common divisor. GCD domains have proved to be of interest at several points in the literature, notably in the work of P. Cohn [2], Gilmer [l], Jaffard [l] and Prufer [2].
The concept of a GCO domain provides a
useful generalization of that of a UFO, since several of the standard results for UFOs can be proved in this more general setting. Probably, the simplest examples of GCD domains are Zand F[X] where Fis a field (although both of them belong to a much narrower class of rings, namely, the euclidean domains).
The main result of this
section (Theorem 1.9) provides necessary and sufficient conditions on Rand Gin order that RG should be a GCD domain.
4 .1
GCD Domains
85
We start our investigations by establishing some elementary properties of GCD domains.
The proof of the following lemma is
obvious and therefore will be omitted. 1.2
LEMMA
(i)
Let a,b,c ER.
If (ca,cb) exists, then (a,b) exists and (ca,cb)
(ii)
If (a,b)
(iii)
If (a)
(iv)
c(a,b).
(a,c) = 1, then (a,bc) = 1.
n (b) = (ab), then ajbc implies ajc.
If R is a GCD domain, then any finite number of elemen'ts in
R have a GCD.
Furthermore
(a,b,c) = ((a,b),c) = (a,(b,c)) (a 1 , ... ,a) = ((a 1 , ... ,a 1),a) n nn
a. E R l
As we have remarked earlier, a prime is always irreducible. The following lemma shows that the converse is also true, provided Risa GCD domain. 1.3
LEMMA
Let R be a GCD domain.
Then every irreducible element
of Risa prime. Suppose pis irreducible and that pjab for some a,b in
Proof:
R.
Since pis irreducible and (p,a) is a divisor of p, either
(p,a)
p or (p,a) = 1.
(p,a)
(p,b) = 1 would contradict (p,ab)
l.2(ii).
Similarly, (p,b)
p or (p,b) = 1.
Now
p, by virtue of Lemma
Thus either pja or pjb, as asserted.
•
The next two lemmas exhibit some important properties of GCD domains that we shall need later. 1.4
LEMMA
Any GCD domain is integrally closed.
Proof:
Let R be a GCD domain and let K be its quotient field.
Suppose k EK is such that k
n
Write k
+
r 1k
n1
+ ··· +
rn
a/b with a,b ER.
= 0
r. l
E R
(*)
Then, by Lemma l.2(i), there exist
s,t ER such that a= (a,b)s, b = (a,b)t, and (s,t) 1. Substin n1 n tuting k = s/t in(*), we obtains + r 1s t + ··· + rnt = 0, whence t divides sn. Since, by Lemma l.2(ii), (sn,t) = 1, it
86
Chap. 4.
follows that t is a unit of R. follows. 1.5
Integral Domains
Hence k E R and the result
•
LEMMA
Let S be a multiplicative subset of R.
If Risa GCD
domain, then so is RS. Proof:
such that a
Let a,b E RS. r 1/s and b
Then there exist r 1 ,r 2 ER ands ES r 2/s. By hypothesis, (r 1 ,r 2) (as,bs)
exists, hence (a,b) also exists, owing to Lemma l.2(i).
•
Next we exhibit a useful characterization of GCD domains. 1. 6
PROPOSITION
R is a GCD domain. if and only if the intersection
of any two principal ideals of Risa principal ideal. Proof:
To prove sufficiency, fix nonzero elements a,b of R.
By hypothesis, (a)
n
(b) = (c) for some c ER; hence ab= er for a
suitable r E R. We claim that r = (a,b). Indeed, since C = ar 1 br 2 for some rl,r2 E R, we have ab = ar 1r = br 2r and so rla, rib by cancellation. If dis another common divisor of a and b, say a = ds 1 and b = ds 2 , put m = ds 1s 2 . say, and so er= ab= dm = eds.
Then m E (a)
n (b), hence m = cs
It follows, by cancellation, that
r = ds and sufficiency is therefore established. To prove necessity, let d = (a,b) and write a
dx, b = dy.
By Lemma l.2(i), (x,y) = 1 and (xz ,yz) = z for any z ER. ter implies that if xlyz, then xlz.
If xz 1 = yz2 E (x)
The lat
n (y) ' then
xlyz2 and so xi z 2 . This implies z2 E (x) and yz2 E (xy) so that (x) n (y) = (xy). Consequently, (a)
n (b) = d((x) n (y)) = (dxy)
as asserted.
•
Let R be a GCD domain.
We know, by Lemma l.2(iv), that any
finite number of elements in R have a GCD.
We shall say that a non
zero polynomial fin R[X] is primitive if the GCD of its coefficients is 1.
The next lemma shows that primitive polynomials con
stitute a multiplicative subset of R[X]. 1.7
LEMMA
(Gauss's Lemma)
Let R be a GCD domain.
primitive polynomials of R[X], then so is fg.
If f and g are
4.1
GCD Domains
Proof:
Write f and gin the form
=
l
i=O
n
s.
m
f
87
a.X
1
g
1
l i=O
t.
b.X
1
1
where s 0 < s 1 < < sm' t 0 < t 1 < ·•· < tn and each ai,bj is nonzero. We must show that if d f O is a nonunit of R, then d fails to divide some coefficient of fg. Suppose first that (a 0 ,d) = (b 0 ,d) = 1. Then, by Lemma 1.2(ii), (a 0b 0 ,d) = 1 and sod does not divide a 0b 0 , a coefficient of fg. We may therefore assume that either (a0 ,d) f 1 or (b 0 ,d) f 1; say for definiteness (a 0 ,d) f 1. dm defined by
We now consider the sequence d 0 , d 1 , ... ,
d.
J
Then d0 f 1 and since f is primitive, there is a smallest integer i
~
1 such that di= 1.
Since dil is a nonunit divisor of d it
suffices to prove that dil fails to divide some coefficient of fg. Consequently we may assume that d divides a 0 , a 1 , ... , ail and that (a.,d) = 1. 1
Again, we choose k which is minimal with respect
to the property (b 0 ,b 1 , ... ,bk,d) = 1. Replacing d by d' = (b 0 ,b 1 , ... , bk~·l, d) (if k = 0, then d = d'), we can also assume that d divides b 0 , b 1 , ... ,bkland that (bk,d) = 1. Let  : R[X] + [R/(d)][X] be the natural homomorphism. d divides ao, al' m
f =
l
j=l
s. a.X J J
... ,
Since
ai1' bo, bl, ... ' bk1' we have g
n
l
t. fi .X J
j=k
J
Taking into account that (ai,d) = (bk,d) = 1, we must also have (aibk,d) = 1. Consequently, d J aibk' so that the coefficient aibk of fg is nonzero. It follows that fg = f•g f O and hence d fails to divide some coefficient of fg.
So the lemma is true.
We now apply the foregoing to provide a large class of GCD domains.
88
1.8
Chap. 4.
THEOREM
Let R be a GCD domain.
Integral Domains
Then any quotient ring of
R[X 1 ,x 2 , ... ,Xn] is also a GCD domain. By Lemma 1.5 and equality R[X 1 , ... ,Xn] = R[X 1 , ... , Xn_ 1 J [Xn] it suffices to show that R[X] is a GCD domain. Denote by Proof:
F and T the quotient field of Rand the set of primitive polynomials in R[X], respectively.
We argue first that for any s ES= R  {O}
and for any t ET, R[X]s
n R[X]t
RIX]st
The inclusion R[X]st S R[X]s
(*)
n R[X]t always holds, so let h = fs =
n R[X]t. We write g = s 1 t 1 where s 1 is the GCD of the coefficients of g and t 1 ET. Since tt 1 ET, by Gauss's lemma, it follows that s 1 is the GCD of the coefficients of hand that sJs 1 .
gt E R[X]s
This shows that h = s 1t 1 t E R[X]st and therefore proves (*). Regarding R[X] as a subring of F[X] we argue next that F[X]t
n
R[X] = R[X]t
(**)
for all t ET
For let g E R[X] be such that g = ft for some f E F[X].
Then there
exists s ES and f 1 E R[X] such that sg = tf 1 . It follows from(*) and Lemma l.2(iii) that sJf 1 , so g E R[X]t and hence F[X]t n R[X] ~ R[X]t.
The opposite inclusion being trivial (**) is established.
Let f and g be nonzero polynomials in R[X].
Owing to Proposi
tion 1.6, the result will follow provided we show that the ideal R[X]f
n R[X]g is principal.
To this end, we write f = s 1t 1 and g = s 2t 2 for some s 1 ,s 2 ES and t 1 ,t 2 ET. Because Risa GCD domain, Proposition 1.6 tells us that Rs 1 n Rs 2 = Rs for some s ES. Now the elements of Sare units of F[X], so F[X}f = F[X]t 1 and F[X]g F[X]t 2 . Bearing in mind that F[X] is a GCD domain, we conclude that there is at ET such that F[X]t 1
n F[X]t 2 = F[X]t
(***)
It will next be shown that R[X]f will finish the proof.
n
R[X]g = R[X]st and this
Since t E F[X]ti for i = 1,2, we deduce
from(**) that t E R[X]t.; hence st E R[X]st. i
fore R[X]st
s R[X]f n R[X]g.
1
~
R[X]s.t. and there1
1
To prove the reverse inclusion, write
4.1
GCD Domains
h E R[X]f
89
n R[X]g in the form h
xy where x ES and y ET.
=
silxy for i = 1,2, and, by(*), R[X]si n R[X]y = R[X]siy. lows, from Lemma l.2(iii), thats.1 Ix and hence six.
Then It fol
On the other
hand, thanks to (**) and· (***) we have F [X]h
n R[X]
so that tly. RIX]f
n R[X]
R[X]y
s:
F[X]t
n R[X]
FIX]f
n F[X]g n R[X] R[X]t
=
Accordingly, stlxy and so
n R[X]g
as required.
F[X]y
~
R[X]st
•
We come now to the demonstration for which this section has been developed. 1. 9
THEOREM
(Gilmer and Parker [1])
The group algebra RG is a
GCD domain if and only if Risa GCD domain. Proof:
Suppose RG is a GCD domain.
To prove that Risa GCD
domain, fix r 1 ,r 2 ER. By Proposition 1.6, it suffices to show that Rr 1 n Rr 2 is a principal ideal. Clearly, we may assume that r 1 f O and r 2 f O in which case r
=
r 1r 2 f 0.
Since
Proposition 1.6 tells us that there is a nonzero x ERG such that
It follows that r E RGx and hence xy
=
r for some nonzero y ERG.
Consequently, Lemma 2.2.19(ii) may be employed to infer that x for some A ER and for some g E G.
=
Ag
This implies at once that
(RA) G and so Rr 1 n Rr 2 = RA, as required. Conversely, assume that Risa GCD domain.
Given any two non
zero elements x,y iri RG, we wish to prove that (RG)x principal ideal.
n (RG)y is a
To this end, we denote by H the subgroup of G
Chap. 4.
90
generated by Supp x U Supp y.
Integral Domains
Then x,y E RH where His a free
group of finite rank, say of rank n.
By Proposition 2.2.6, RH is
the quotient ring of the polynomial ring R[X 1 ,x 2 , ... ,Xn]' so RH is a GCD domain, by virtue of Theorem 1.8.
It now follows, from
Proposition 1.6, that there is a z E RH such that (RH)x
n (RH)y
(RH) z. Let T be a transversal for Hin G.
Then, by Proposition
2. 2 .15 (i), RG is a free RHmodule with T as a basis.
(RG)x
n (RG)y
Accordingly,
I (RH)tJx n { I [(RH)t]y} [tET tET I (RH)tx n I (RH)ty tET
tET
I
[(RH)x
n (RH)y]t
I
(RH) zt
=
tET tET
[
I
(RH) t] z
(RG)
z
tET
This completes the proof, by appealing to Proposition 1.6. 2.
•
UNIQUE FACTORIZATION DOMAINS
Our main concern in this section is to find necessary and sufficient conditions on Rand Gunder which the group algebra RG is a unique factorization domain. some machinery.
To accomplish this we shall have to set up
We start by recalling some fundamental definitions
and elementary properties. 2.1
DEFINITION
A unique factorization domain (UFD) is an integral
domain in which every element not zero or a unit can be written as a product of irreducible elements, and given two complete factorizations of the same element, y.,z. irreducible
X
1
then r
J
t, and after suitably renumbering the z's, yi is associated
to z .. 1
Our first aim is to show that the UFDs are precisely the GCD domains in which the ascending chain condition for principal ideals (ACCP) is satisfied.
4.2
Unique Factorization Domains
2.2
PROPOSITION
(i)
91
The following conditions are equivalent:
Every nonzero nonunit of Risa product of primes.
(ii)
Risa UFD.
(iii)
Risa GCD domain in which ACCP is satisfied. Proof:
(i)
~
(ii)
Since any prime is irreducible, we need
only prove uniqueness of factorization.
Let c
b 2b 2 ···bs be two factorizations into primes.
=
a 1a 2 ··ar
=
We must prove that
r =sand that after suitably renumbering the b's, bi is associated to ai.
We use induction on r; for r = 1 there is nothing to prove
(because s must then also be 1), so let r
>
1.
A simple inductive
argument shows that, if a product of more than two factors is divisible by a prime p, then so is one of the factors.
Therefore
1 (by renumbering the b's) and so a 1 = b 1u for some unit u of R. Dividing a 1 ···ar = b 1 ···bs by b 1 , we obtain ua 2 ···ar = b 2 ···bs. It therefore follows, by induction, a 1 \bk for some k, say k
that r  1 = s  1 and that we can renumber the b's so that a. is l
associated to bi (2 S i s r).
But this also holds for i = 1, hence
the uniqueness of factorization. (ii)~ (iii)
Given two nonzero elements x,y ER we can write 0
X =
where u and v are units and where p 1 , ... , pt are pairwise nonassociated irreducible elements. It follows that any common divisor of 0 1 °2 °t x and y is of the form wp 1 p 2 ···pt where w is a unit and cSi s min{a.,S.}. l
l
Setting
y. l
= min{a.,S.} we infer that
proving that Risa GCD domain.
l
l
Bearing in mind that there are only
finitely many of pairwise nonassociated divisors of x, we also conclude that (x) is contained in only finitely many principal ideals of R.
Consequently, ACCP is satisfied in R. (iii)~ (i)
prime.
By Lemma 1.3, every irreducible element of Risa
Therefore, given a nonunit x
is a produc~ of irreducible elements.
r
0, we need only prove that x We shall first show that x
92
Chap. 4.
has an irreducible divisor. to prove.
Otherwise let
Integral Domains
If xis irreducible, there is nothing
= xlyl where x 1 ,y 1 are nonunits. Then either x 1 is irreducible or xl = x 2y 2 where ~ 2 ,y 2 are nonuni ts. Continuing this process, we obtain the ascending chain X
By hypothesis, this chain must break off; if is irreducible, and x
X
n
is the last term,
x. n and write X = P1C1. If cl is irreducible, = P1 n then the proof is complete. Otherwise we have c 1 = p 2c 2 where p 2 X
n
We now set
is irreducible.
J
X
Continuing in this way, we obtain the ascending
chain
This breaks off with an irreducible element en= pn+l"
Accordingly,
there exist irreducible elements p 1 , p 2 , ... , pn+l such that
and the result follows.
•
Recall that a principal ideal domain (PID) is an integral domain in which all ideals are principal. Every ascending chain of ideals
in a PID must terminate.
For the union is an ideal generated by b,
say, and if b E (ak)' then Un(an) = (b) = (ak)' i.e., (ak) = In view of this observation the following corollary arises from Propositions 1.6 and 2.2. 2.3
COROLLARY
Every PID is a UFD.
The next lemma is in preparation for Proposition 2.5. 2.4
LEMMA
Let S be a multiplicative subset of R.
then so is R8 .
If Risa UFD,
4.2
Unique Factorization Domains
Proof:
93
Because the canonical homomorphism ~S : R
jective, we may regard Ras a subring of RS. nonunit of RS.
+
RS is in
Let x = r/s f O be a
By Proposition 2.2, we need only show that xis a
product of primes.
To this end, observe that the elements of S
(and hence their divisors) are units in RS.
Since by hypothesis r
is a product of primes, say r = p 1p 2 ···pn' then by renumbering the p/s, if necessary, we may write x = p 1p 2 : · ·pt•u where u is a unit of RS and where none of the pi' s divide an element of S. we are left to prove that each pi stays prime in RS.
Therefore
So let pi lab
in RS; here a,b may be taken in R, by passing to associates.
Then
ab= pid where d = r 1/s 1 (r 1 ER, s 1 ES); hence abs 1 = pir 1 . By hypothesis, p. t s 1 , but p. is a prime in R; hence·p. la or p. lb. l
l
l
l
This shows that pi is a prime in RS and completes the proof. 2.5
PROPOSITION
Let R be a UFD.
•
Then any quotient ring of
R[X 1 , ... ,Xn] is also a UFD. Proof:
By the preceding lemma, it suffices to prove that
R[X 1 , ... ,Xn] is a UFD.
Since R[X 1 , ... ,Xn] = R[X 1 , ... ,Xn_ 1 l[Xn] we may harmlessly assume that n = 1. By Proposition 2.2 and Theorem 1.8, the result will follow, provided we show that any ascending chain
of principal ideals of R[X] breaks off.
Clearly, we can assume that
all fn f 0, in which case deg f 1 ~ deg f 2 ~···~deg fn ~ Therefore there is an integer m ~ 1 such that deg fn = deg fm for all n
~
m.
Consequently, we may assume that deg f 1 = deg f 2 = · · · in which case there is a nonzero rn+l in R such that for all n ~ 1,
In particular, if b is the leading coefficient of rn+l fn+l = f. n n rn+lbn+l = bn (n = 1, 2, ... ), so we obtain an ascending fn' then chain:
This breaks off, by Proposition 2.2; hence there is an integer m ~ 1 such that (bn) = (bm) for all n
~
m.
Accordingly, (r
.b .) = (b .) m+J m+J m+J
Chap. 4.
94
,for all j "". 1 , and so for all j "". 1, r (fm+l) = · · ·, as desired.
Integral Domains
. is a unit. m+J
•
At this point we shall digress from the main investigation to establish a few facts that will prove useful when we make a detailed study of divisibility properties in group algebras. We need first an elementary lemma. 2.6
LEMMA Let P, P1 , P2 , ... , Pn be prime ideals of R. If n PS Ui=lpi' then PS Pi for some i. Proof: The lemma being trivial for n = 1, we argue by induction on n "". 2.
We claim that there is an index j such that
n P.J
o,
(i) Let B be a pure subgroup of a group A. A/Bn splits over B/Bn.
Then for
Let G0 be bounded and let G0 x H be a pure subgroup of G. Then there exists a subgroup L of G containing H and such that (ii)
G
=
G0 x L. Proof:
(i)
We shall show that if C is a subgroup of B, then
B/C is pure in A/C.
Since B/Bn is bounded, the required assertion
will follow by appealing to Lemma 3.4(ii). (A/C)m n B/C.
Let x = amC = bC E
Since am EB and since Bis pure in A, am= b~ for
132
Chap. 5.
Study of Units
some b 1 EB. It follows that x = (b 1C)m E (B/C)m, proving that B/C is pure in A/C. (ii)
By hypothesis, there is an n :". 1 such that G~ = 1.
Set
ting D = G0 x H, it follows from (i) that there is a subgroup K? Dn such that
Let X = hk EL n Go (h EH, n n K = Dn = H, so x EH n G0 = 1 and hence
Put L = H•K; then obviously G = Go·L. k EK).
G
G0
Then xh x
1
ED
L. •
We are now ready to prove 3.8
THEOREM
(May [6))
A group G splits over its torsion subgroup
G0 if and only if there exists a countable ascending sequence Al S. A2 S · · · S: An c · · · of subgroups of G such that the following three conditions hold: (1)
G = U A n n
(2)
For every n
1, 2, ... the torsion subgroup T(An) of An is
bounded. (3)
For every n = 1, 2, ... , T(G/An) = (G 0 •An)/An. Proof: Suppose there is a subgroup F of G such that G = G0 x F. Then the groups An= G0 [n!] x F, n = 1, 2, ... satisfy conditions (1)(3).
Conversely, assume that (1)~(3) hold.
Then owing to (2)
and Theorem 3.6, we infer that there is a subgroup G1 of A1 such that A1 T(A 1) x G1 . It therefore suffices to exhibit a chain
of torsionfree subgroups of G such that Ai = T (Ai)
x
Gi (i = 1, 2,
... ) (for then G = G0 x B where B = UGn). Suppose that for all i s n we have already obtained decompositions A. = T(A.) x G. such 1
that G1 £ G2 £ ··· S Gn.
1
1
Set Bn+l = T(An+l) x Gn and observe that
5.3
Splitting Unit Groups
133
Consequently, we may apply (3) to deduce that the group An+l/Bn+l  (Go·An+l)/(Go·An)  [(G 0 •An+l)/An]/T(G/An) is torsion free. of An+l"
The latter implies that Bn+l is a pure subgroup
Invoking Lemma 3.7(ii), we conclude that there is a sub
group Gn+l 2 Gn such that An+l = T(An+l) x Gn+l. true. •
So the theorem is
Having examined some properties of infinite groups, we shall now confine our attention to the unit group of RG. Our first goal is to find sufficient conditions in order that U(RG) be free modulo torsion. Let A be an Ralgebra.
By the idempotent subalgebra of A we
understand the Rsubalgebra of A generated by the idempotent elements.
Every element of the idempotent subalgebra is an Rlinear
combination of idempotent elements in A. 3.9
THEOREM
(May [S])
Let R be a finitely generated integral do
main, and let G be a group which is free modulo torsion and of finite torsionfree rank.
If R has characteristic p, assume that G
has no elements of order p. Proof:
Then U(RG) is free modulo torsion.
By hypothesis, there exists a free group F of finite
rank such that G = G0 x F. We know that RG = (RF) G0 and that RF is a finitely generated integral domain such that G0 has no elements of order p if char RF= p.
Consequently, without restricting gen
erality, we may assume that G is torsion.
Let R be the integral
domain obtained by adjoining all roots of unity to R, let K be the quotient field of R, and let A be the idempotent Rsubalgebra of KG.
Suppose x ERG.
Setting H = it follows that His fi
nite, say of order m.
By looking at the image (y 1 , ... ,ym) of x under the isomorphism KH + Km described in Theorem 3.2.3, we infer that
y i E R (1 s i s m).
This implies at once that RG £ A, so it suf
fices to prove that U(A) is free modulo torsion.
Thanks to Proposi
tion 1.2.33, the latter will follow provided we show that there is an index set I such that
134
Chap. 5.
Study of Units
The proof of(*) rests on the following observation.
Let H c H' be
subgroups of G such that H has finite index n in H'.
Let U' (re
spectively, U) be the unit group of the idempotent subalgebra of KH' (respectively, KH).
U'
+
Then there exists an isomorphism
u Xu X ... Xu
( **)
n copies
whose restriction to U is an isomorphism U + U x {1} x · · · x {1}. To prove(**), we first invoke Proposition 3.2.6 to deduce that KH' and (KH) n are isomorphic KHalgebras.
From this we see that (x 1 ,x 2 , ... ,xn) E (KH)n is a unit of the idempotent subalgebra of (KH)n if
and only if xi E U (1 ::: i ::: n).
It follows that U' is isomorphic
to U x ··· x U (n copies) by an isomorphism which maps U (as a subgroup of U') to the diagonal in U x ··· x U.
Since U x ··· x U
admits an automorphism which ~aps its diagonal to U x {l} x ·•· x {l},
(**) is verified. It is now an easy matter to prove(*).
Consider the set X of
all ordered pairs (H , {U . } . EI ) that satisfy the following condia a,1 1 a tions: (a)
Ha is a subgroup of G.
(b)
If Ua is the unit group of the idempotent Rsubalgebra of KHa,
then {U
.}.EI
a,1 1
a
is a family of subgroups of U that give an inner u.
direct decomposition of it. ( c)
U
.
a,1
= U(R)
for every i EI . a
The set Xis nonempty:
({l}, {U(R) }) E X.
(H
specifying that (H ,{U .}.EI)::: a a,1 1 a H
a
 Hs
C
{U
. }. EI
a,1 1
a
C,: {Uo µ
0.
µ
Partially order X by
,{U 0 .}.EI) if and only if µ,J J S
.}.EI
'J J
s
Observe that a finite number of elements of KG involve only finitely many elements of G nontrivially and that the order relation respects the inner direct decompositions.
It therefore follows that
Xis inductively ordered and so we may choose a maximal element
5.3
Splitting Unit Groups
(H,{Ui}iEI)' say.
135
We now apply(**) to deduce that H cannot be con
tained properly in a subgroup H' of G such that (H' :H) is torsion; hence H = G and the result follows.
2.
(ii)
(iii) KG= KG 3 if and only if G/G 00 is a direct product of cyclic groups with infinitely many factors of order >2. (iv)
KG
(v)
KG

KG4 if and only if G is divisible. KG(s) if and only if G = A X B where A is divisible 5
and B a direct product of s cyclic groups. REMARK
If G is an arbitrary countably infinite torsion group, then
lRG is isomorphic to a group algebralRH where His one of the 2groups listed in Theorem 5.6 (Berman [3]). 6.
REAL GROUP ALGEBRAS
In this section we provide necessary and sufficient conditions under which the real group algebras lRG and lRH are isomorphic.
We begin by
assembling some facts that will be useful later. 6.1
LEMMA
Let G be a torsion group.
Then the torsion subgroup of
U(RG) is a direct product of a divisible group and a group of exponent 2. Let u0 be the torsion subgroup of U(RG) and let D be the maximal divisible subgroup of u0 . Then there is a subgroup E Proof:
of u0 such that u0 = D x E. It therefore suffices to show that u0 /D is of exponent 2. Fix x E u0 . By looking at the supporting subgroup of x, we may harmlessly assume that G is finite. SinceJRG is a direct product of finitely many copies of JR and[, U(IRG) is a direct product of finitely many copies of U(IR) and U([).
Taking into
account that the torsion subgroup of U(IR) [respectively, U([)] is of order 2 {respectively, divisible), we conclude that x some y ED and some element z of order 2. of exponent 2 and completes the proof.
•
yz for
This shows that u0 /D is
6.6
Real Group Algebras
191
With the aid of Lenuna 6.1, we now derive the following reduction step. 6. 2
JRGO
LEMMA
=lRHO
For arbitrary groups G and H, JRG
=lRH
if .and only if
and G/Go = H/Ho. The "only if" part is a consequence of Proposition 1. 6
Proof:
and Corollary S,. 1. 4.
To prove the "if" part, it suffices to show
=
=
that JRG JRG 0 ~ JR (G/G 0 ) or, equivalently, that JRG JR [G 0 x (G/G 0 )] . Let B be the torsion subgroup of UQRG 0 ). Since B? G0 it follows that Bis also the torsion subgroup of G•B.
We know, from Lemma 6.1,
that Bis a direct product of a divisible and a bounded group. we may apply Theorem 5.3.6 to deduce that G•B splits over B. Lemma 1.7 (with A= G0 ) we now conclude that RG •
=JR[G 0
x
Hence Invoking
(G/G 0 )], as
required.
We now direct our attention to real group algebras of torsion groups. 6.3
=
LEMMA
(i)
Let JRG 0 JRH 0 . 2 (Go:Go) = (Ho:Ho)
(ii)
!Gil= IH~I provided
Then the following two properties hold:
2
Proof:
(i)
Gi
= 1 or G~ is infinite.
Let I be the ideal ofRG 0 generated by all g  1
with g E G~, and let J be the intersection of kernels of all real characters X : JRG 0
+
JR.
We wish to show that I is determined by
JRG 0 in an invariant way, namely, that I= J. it will then follow thatJR(G 0 /Gi)
=JR(H 0/Hi)
SincelRG 0 /I
=JR(G 0 /Gi)
and hence (i), by com
paring dimensions. Let X be any real character of JRG 0 • Then X(g) = 1 for all 2 g E G0 , so IS Ker X and hence IS J. Conversely, assume x E J. Then the image of x inJR(G 0 /Gi) belongs to the counterpart of Jin 2 2 R(Go/Go). But the latter is obviously O since G0 /G 0 is an elementary 2group. Hence x EI and I J. 2
Suppose first that G0 1. Then the image of every JRhomomorphism JRG 0 + a:: is JR. It follows that the same is tTue for (ii)
JRH 0 ; hence Hi= 1.
Suppose now that G~ is infinite.
By Lemma
3.1.3, the ideal I has a generating set of cardinality s !Gil.
Chap. 6.
192
Isomorphism Questions
We shall show that every generating set for I has cardinality~ !Gil. It will then follow that IG;I is the smallest of the cardinalities of all generating sets for I. This will finish the proof, since I is determined bylR.G 0 in an invariant way.
Let X = {x} be Cl
a generating set for I, let
and let Y be the set of all distinct g. (for all a). With F = , 2 let 2 it follows that I= lR.G 0 •I(G 0 ) = lR.G 0 •I(F); hence F = G0 by Corollary 2.2.11. Since is infinite, we must have !Gil= IYI. On the
Gi
other hand, since for any a there are only finitely many gia' we have IY I s IX I.
Thus IX I
~
Gi I,
I
and the result follows.
•
The next two lemmas dispose of most of the technicalities which occur in the proof of the main result. 6.4
LEMMA
Let A be a subgroup of a torsion group G.
Suppose that
one of the following three conditions hold: (i) (ii) (iii)
G/A is cyclic of order n with a generator gA such that gn E A2 . G/A is cyclic of odd order. G/A  Z(2
and there exists a generating set {g 1 ,g 2 , ... } 2 2 2 for G modulo A such that g 1 = a 1 , g 2 = g 1 a 2 , g 3 = g 2 a 3 , ... , where a 1 , a 2 , a 3 , ... are in A2 . ThenlRG has a group basis of the form Ax C with C  G/A. Proof:
00
)
Let B denote the torsion subgroup of U(IRA).
By Lemma
1.7, it suffices to show that G•B splits over B. Suppose (i) holds. Owing to Lemma 6.1, B2 is divisible; hence gn has an nth root, say 1 n . B2 x, 1n Then (gx ) = 1, G•B/B = G/B n G = G/A, and so G•B =Bx .
Suppose that (ii) holds. Let G/A = ; it follows that , so (ii) is a consequence of (i). Suppose now that (iii) holds.
The argument employed in the 2
proof of case (i) shows that there is an x 1 EB such that (g 1x 1) 1. If y 1 g 1x 1 it follows, from the relation g; = g 1a 2 , that g; 1 2 y 1 (x 1 a 2) y 1 •u 2 where u 2 = x1 1 a 2 EB. Assume, by induction,
2
6.6
Real Group Algebras
193
that we have chosen xi E B2 (1 s i s t) such that for yi 1
2 1 2 2 Since B is divisible, ut+l = dt+l for a suitable dt+l EB. Therefore, setting xt+l = dt+l' we obtain xt+l E B2 , y~+l = yt, and 2
1 E B2. gt+2 = gt+lat+2 = Yt+lut+2 where ut+ 2 = xt+lat+ 2 2 2 have chosen yi (for all i ~ 1) such that y 1 1 and yi ~
Let D = ; it now follows that G•B Dis an ascending union of its cyclic 2groups (n (i
2).
Hence we
D•B where 1, 2, ... ) .
The latter implies that y 1 is the only element of order 2 in D. Hence y 1 is a power of every nonidentity element in D. But y 1 since otherwise g 1 EB n G and the result follows. 6.5
LEMMA
A, a contradiction.
f
B,
Thus DnB= 1
•
Let a torsion group G satisfy one of the following two
conditions: (i) (ii)
G has trivial 2component. G is a reduced 2group.
ThenRG has a group basis which can be written as a direct product of cyclic groups or as a direct product of cyclic groups and groups of type 2
00 •
Let G1 S ··· S Ga S ··· S GA= G be an ascending transfinite chain of subgroups of G such that G1 and Ga+l/Ga Proof:
(i)
(1 s a 1, and assume that we have constructed G8 , for all S < a, such that GS n G2 = G~. If a is a limit ordinal, Put Ga= US