S Feigelstock Bar-II an University
Additive groups of rings
Pitman Advanced Publishing Program BOSTON-LONDON - MELBOURNE
PITMAN BOOKS LIMITED 128 Long Acre, London WC2E 9AN PITMAN PUBLISHING INC 1020 Plain Street, Marshfield, Massachusetts 02050 Associated Companies Pitman Publishing Pty Ltd, Melbourne Pitman Publishing New Zealand Ltd, Wellington Copp Clark Pitman, Toronto
© S Feigelstock 1983 First published 1983 AMS Subject Classifications: (main) 20K99 (subsidiary) 16, 17 Library of Congress Cataloging in Publication Data Feigelstock, S. Additive groups of rings. (Research nQtes in mathematics; 83) Bibliography: p. Includes index. 1. Abelian groups. I. Title. II. Series. 512' .2 82-22412 QA171.F34 1983 ISBN 0-273-08591-3 British Library Cataloguing in Publication Data Feigelstock, S. Additive groups of rings.-(Research notes in mathematics; 83) 1. Group rings I. Title II. Series 512'.4 QA171 ISBN 0-273-08591-3 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 and/or otherwise, without the prior written permission of the publishers. This book may not be lent, resold, hired out or otherwise disposed of by way of trade in any form of binding or cover other than that in which it is published, without the prior consent of the publishers.
Reproduced and printed by photolithography in Great Britain by Biddies Ltd, Guildford
Preface
In 1948 R.A. Beaumont [5] began investigating the additive groups of rings. Shortly thereafter L. Redei, T. Szele [541. and H.S. Zuckerman [11] joined in these investigations, to be followed by L. Fuchs [37], [38]. Since then much progress has been made in this branch of abelian group theory. Many of the results appear in Chapter 17 of L. Fuchs 1 well-known book [36]. An attempt has been made here to offer an easily accessible account of much of the work done on the additive groups of rings, with special emphasis on results not covered in [36]. This short monograph is far from being comprehensive, and many important papers have not been considered here at all. Although this work is meant to be self contained. some knowledge of abelian group theory is assumed. The prospective reader is especially urged to read Chapter 17 of Fuchs• book [36]. I wish to express my gratitude to Professor Elvira Rapaport Strasser for introducing me to this subject, for her advice and guidance during the past fifteen years, and for suggesting that I write this monograph. I also thank Professor L. Fuchs for ~is valuable comments and suggestions.
Contents
1.
Preliminaries: §1: §2: §3: §4:
2.
Definitions and structure theorems Mult G Type Examples
1: The ni 1s tufe of a group §2: Nilpotence without boundedness conditions, and generalized nilpoten~e
25 31
Other Ring Properties: §1: Semisimple, prime, semiprime, simple ring, division ring, field, radical ring §2: Principal ideal and Noetherian rings §3: Descending chain conditions for ideals §4: Subdirectly irreducible rings §5: Local rings §6: Rings with trivial left annihilator, subrings of algebraic number fields, and semisimple rings continued §]. E-rings and T-rings
5.
10 16
Additive groups of Nilpotent and Generalized Nilpotent Rings: §
4.
5 6
Ni 1 and Quasi -Ni'l Groups : §1: Nil groups §2: Quasi-nil groups
3.
4
Torsion §1: §2: §3: §4:
Free Rings: Notation, definitions, and preliminary results The Beaumont-Pierce decomposition theorem Torsion free rings with semisimple algebra type Applications
36
43 50 61 65 68
76
88 90
98 103
Notation
a group. the order of x e G. the torsion part of G. the p-primary component of G, p a prime. {nxlx E G}, n a fixed integer. {x E Gjnx = 0}, n a fixed integer. Ga v(G) N(G) r(G) lS) hlx) hp{x)
the the the the the the the
t(x) t('G) T(G) tplx) (tp(G))
the type of x E G. the type of G for G homogeneous. {t(x)jx E G, x ~ 0} =the type set of G. the p-component of t(x) (tlG)).
[(k 1 ,k 2 , ••• ,kn, •.• )J
the type containing the height vector (k 1 , k2 , ••• ,kn•···>· Additional notation concerning type which will be used may be found in [36, vol. 2, pp • 108- 111 ] •
Z(n)
cyclic group of order n.
Z(pCD) End( G) £nd(G)
p-Prufer group. the group of endomorphisms of G. the group of quasi-endomorphisms of G. quasi-isomorphism. quasi-equality a ring.
Q. :!:
R R+
a-th Ulm subgroup of G, a an ordinal. nilstufe of G. strong nilstufe of G. rank of G. subgroup of G generalized by S =G. height of x E G. p-component of hlx).
the additive group of R.
I
~
Z Q
ZP w
lSI
R
" I is an idea 1 in R" • the ideal in R generated by S c R the ring of nxn matrices with components in R, n a positive 1nteger. · the ring of integers. the field of rational numbers. the ring of p-adi c integers • the first infinite ordinal the cardinality of a set s.
Additional symbols which appear in only one section, will be defined in the section they appear in.
1 Preliminaries
§I.
DEFINITIONS AND STRUCTURE THEOREr·1S
All groups considered in this monograph are abelian, with addition the group operation. Definition: A group G is p-divisible, p a prime, if G = pG. If G is p-divisible for every prime p, then G is said to be divisible. It is readily seen that G is divisible if and only if G = nG for every positive integer n. Definition:
A subgroup H of a group G is p-pure in
G,
p a prime, if
H n pkG = pkH for every positive integer k. If H is p-pure in G for every prime p, then H is said to be pure in G. Clearly H is pure in G if and only i f H n nG = nH for every integer n. Definition: A subset {a; li E I} of a group G is said to be independent if for distinct i1' ... ,ik E I, k an arbitrary positive integer, and for n1, ... ,nk E Z, n1a11 + ... + nkaik = 0 implies that n1a11 = .•. = nkaik = 0. By Zorn's lemma, there exists a maximal independent set M in G containing only elements of infinite and prime power orders. The rank of G, r(G) = IMI. Let M 0 be maximal amongst the independent subsets of G containing only torsion free elements. Then 1r~0 1 is called the torsion free rank of G, denoted r 0 (G). Let t•lp be maximal amongst the independent subsets of G containing only elements of order a power of p, p a prime. Then lt~pl is called the p-rank of G, denoted rptG). r(G), r 0(G),
and
rptG)
are well defined [36, Theorem 16.3].
Definition: A subgroup B of a group G is a p-basic subgroup, p a prime if: 1) B is a direct sum of infinite cyclic groups, and cyclic p-groups. 2) B is p-pure in G, and 3) G/B is p-divisible.
If G is a p-group, then a p-basic subgroup of G is called a basic . subgroup. Every group G possesses a p-basic subgroup [36, Theorem 32.3], p-basic subgroups of G are isomorphic [36, Theorem 35.2].
and all
Definition: For every ordinal a, define the a-th Ulm subgroup Ga of G as follows: G1 = n nG. For every ordinal a, the a+l th Ulm n<w
subgroup of G, subgroup of G,
Ga+l = (Ga)l. G8 = n GY.
For 8 a limit ordinal, the
ath
Ulm
y
Definition: Let G be a group, and let x E G. The p-height of x, hp(x), is the greatest positive integer n, if it exists, such that x E pnG. If no such positive integer exists, then hp(x) = ~. Let {pi}7=l be the sequence of all primes in ascending order. The height (characteristic) of x, h(x) = (h (x), •.. ,h (x), ... ). .
PJ
..
Pn
A sequence (k 1 , ••• ,kn•···) with kn a non-negative integer or ~ for every positive integer n, is called a height sequence. The relation on the set of height sequences defined by (k 1, ••• ,kn•···> ~ (~ 1 ••••• ~n ... ) if kn = ~n for all but finitely many subscripts n, and kn = ~n whenever kn = ~ or ~n = ~. is an equivalence relation. The equivalence . class of a height sequence (k 1 , •.• ,kn•···) is called a type, and will be denoted [(kp···•kn•··:>L The type of x E G, t(x) = [h(x).J. Definition: The type set of a group G, T(G) = {t(x)l x E G, xI 0}. IT{G)I = 1, then G is said to be a homogeneous group with type t(G) =the singleton belonging to T(G).
If
Definition: A height sequence (k1' ... ,kn, ... ) is idempotent, if kn or ~ for every positive integer n. A type t is idempotent if , possesses an idempotent height sequence.
0
Definition: A·group G is indecomposable if G = H & K implies that H = 0 or K = 0. Definition: A torsion free group G is rigid, if End(G) a subgroup of Q+ .
2
is isomorphic to
Definition: Let G and H be subgroups of a divisible torsion free group D, in lieu of Proposition 1.1 .6, this is not a very severe restriction on torsion free groups G, H. If there exists a positive integer n such that nG ~ H, and nH ~ G, then G and H are said to be quasi-equal, G ~H. If two groups A and B are isomorphic to quasi-equal groups, G, H respectively, then A and B are said to be quasi-isomorphic, A~ B. Definition: A group G satisfying, G & H @ K implies is said to be strongly indecomposable.
H = 0 or K = 0,
Definition: A subgroup H of a group G is essential in for every subgroup K < G, K ; 0.
G if H n K ~ 0
Definition: Let G be a torsion free group, and let D be a divisible torsion free group such that G is an essential subgroup of D; such a D exists, Proposition 1.1.6. An endomorphism ~ E End(D} is a quasi-endomorphism of G, if there exists a positive integer n such that n~ e: End(G}. Definition: A group G is bounded if there exists a positive integer n such that nx = 0 for all x e: G. Definition: A subgroup H of a group G is fully invariant in ~(H) ~ H for every endomorphism ~ of G.
G if
Several structure theorems on abelian groups play a vital role in studying the additive groups of rings. These results will be stated here without proof. However, the reader will be referred to the proofs in Fuchs' book [36). Proposition 1.1.1: .Let Proof:
G be a torsion group.
Then
G=
@
p a prime
Gp.
[36, Theorem 8.4].
Proposition 1.1.2: Let T be a torsion group. T is a direct summand of every group G with Gt = T if and only if T is the direct sum of a divisible group, and a bounded group. Proof:
[36, Theorem 100.1]
Proposition 1. 1.3:
D~
mQ+ m a
Proof:
m
p a prime
A group (I)
D is divisible if and only if a, ap' arbitrary cardinal numbers.
Z( p"") ,
ap
[36, Theorem 23.1]. 3
Proposition 1. 1.4. G = Z(pk) $ H, Proof:
Let
G be a non-torsion free group.
for some prime
p,
<
k
<
Then
~.
[36, Corollary 27.3].
Corollary 1.1.5: Let G be a non-torsion free group. if and only if G = Z(pk), p a prime, 1 < k < ~.
G is indecomposable
Proposition 1.1.6: Every group G can be embedded, as an essential subgroup, into a divisible group D satisfying r 0 (D) = r 0 (G), and rp(D) = rp(G), for every prime p. Proof:
[36, Theorem 24.1, Lemma 24.3, and Theorem 24.4].
The group D in Proposition 1 .1.6 is unique up to isomorphism, and is called the divisible hull of G, [36, Theorem 24.4]. Proposition 1 .1.7: Let G be a rank one torsion free group. homogeneous and isomorphic to a subgroup of Q+. Two rank one groups are isomorphic if and only if they have the same type. type T, there exists a rank one torsion free group G with Proof:
[36, Theorem 85.1], Proposition 1.1.6 and Proposition 1.1.3.
Proposition 1.1.8: isomorphic. Proof:
§2.
A bounded group is a direct sum of cyclic groups.
[36, Theorem 17 .2].
Proposition 1.1.10: containing it. Proof:
Quasi-isomorphic rank one torsion free groups are
[36, remark preceding Proposition 92.1].
Proposition 1.1 .9: Proof:
Then G is torsion free For every t(G) = T.
A divisible
gro~p
is a direct summand of every group
[36, Theorem 21.2].
MULT G
Let G be a group, and let x be a binary operation on G inducing a ring structure on G. Clearly x may be viewed as a bilinear map from the cartesian product GxG to G via (g"l' gL) .... g 1xg 2 for all gl'g 2 E G. Hence there exists ~ E Hom(G~G, G) such that g1xg 2 = ~(g 1 ~ g2) for all g1 ,g 2 E G. Conversely for ~ E Hom(GSG, G) the multiplication 4
g 1 x~g 2
= ~(g 1 9 g2) for g1,g 2 E G induces a ring structure on G.
Therefore the ring multiplication on G form a group Mult G. Addition in Mult G is as follows: Let x1 ,x 2 E t4ult G, and let gl'g 2 E G. Then gl(xl + x2)g2 = (glxlg2) + (glx2g2). 1.2.1.
14ult G CO! Hom(GQG, G)
CO!
Hom(G, End(G)).
The second isomorphism follows from the well-known fact that the functor 8 is left adjoint to the functor Hom. The above facts as well as additional information concerning Mult G may be found in [36, section 118].
In this section all groups are assumed to be torsion free. Definition: Let t 1• t 2 be types. Then t 1 ~ t 2 means that there exist height vectors (kp···•kn, ... ), (tl' ... ,tn•···) belonging to t 1 and • 2 respectively, such that kn> t ; n = 1, 2, ... n Definition: Let • 1 = [(kl' ••. ,kn•·· .)], • 2 = [(tl' ••• ,tn•···)l be two types. The product of • 1 and • 2 is defined to be the type • 1 · • 2 = [(k 1 + t 1 , •.. ,kn + tn•· •. ) ]. The intersection of • 1 and • 2 , • 1 n • 2 = [(min(k 1, t 1), .•. ,min(kn' tn), •.. )]. If kn ~ tn for every positive integer n, then the quotient • 1:.2 = [(k 1 - t 1 , .•• ,kn- tn, ... )] with the convention ~- k = ~ for 0 < k ~~. Property 1.3.1: 1) Let g, hE G be dependent. Then t(g) = t(h). 2) t(g+h) ~ t(g) n t(h) for all g, hE G. 3) For G = H tB K, hE H, k E K, t(h+k) = t(h) n t(k). 4) For ~= G ~ H a group homomorphism, and g E G, t[~(g)] ~ t(g). Proof:
[36, p. 109].
Consequence 1.3.2: Let R be a ring, t(ab) ~ max[t(a), t(b)].
a,b E R.
Then
Proof: Property 1.3.1.4 and the fact that the maps endomorphisms of R+.
x
~ax,
Proposition 1.3.3: Let G,H be rank one torsion free groups. is a rank one torsion free group with t(G 9 H) = t(G) . t(H).
x
~
xb are
Then G 9 H
5
Proof:
[36, Proposition 85.3].
Proposition 1.3.4: Let G,H be rank one torsion free gro~ps. If t(G) ~ t(H) then Hom(G,H) = 0. If t(G) ~ t(H), then Hom(G,H) is a rank one torsion free group with t[Hom(G,H)] = t(H): t(G). Proof: §4:
[36, Proposition 85.4].
Examples:
1. 4. 1.
F
=@ a.
z+ '
a an arbitrary cardinal.
Let {a; I i E I} be a basis for F. By 1.2.1, every ~ E Hom(~F, F) determines a ring multiplication on F. Since F~F is a free abelian group with basis {ai 0 ajl i ,j E I}, any map of these basis elements into F may be linearly extended to a homomorphism mapping ~F into F. Therefore the rings with additive group F are determined by arbitrarily defining the products ai.aj, i ,j E I, and extending the multiplication linearly. 1. 4. 2.
Let
G =~a Q+ , a an arbitrary cardinal. {a; I i E I} be a basis for G,
viewed as a vector space over Q.
Then {a; ® aj I i ,j E I} is a basis over Q for ~G. The same argument used in 1.4. 1, with the exception that ~ is a free object in the category of vector spaces over Q, while F~F is free in the category of abelian groups, shows that the rings with additive group G are determined by arbitrarily defining a;·aj, i ,j E I and extending linearly. 1.4. 3.
G = (±) Z(p),
p a prime,
a.
a an arbitrary cardinal.
The same argument employed in 1.4.2, except that now G is viewed as a vector space over a field of order p, shows that the rings with additive group G are obtained by arbitrarily defining the products of basis elements, and extending linearly. 1.4.4. j
G= S
(±) j<w a.j
p a prime,
aj
an arbitrary cardinal,
= 1, 2, ••. , i.e., G is any direct sum of cyclic p-groups. Let
{aij I i E IJ.}
{a . . @a . . 1 1 I J1 1 2J 2
6
Z(pj),
be a basis for E9 Z(pJ),
i 1 E I., Jl
Clj
; 2 E I.,
J2
j
=
1, 2, .•.
Then
j 1 ,j 2 = 1, 2, ..• } is a basis for
~.
with
Ia. J. 1
11
IX\
mi nU 1,j 2) a. J. I = p • The homomorphisms 12
11> E
Hom( ~G, G)
are
2
determined by defining
~~>(a. J. (X) 11 1
a. . )
to be any element in G of order
1 2J 2
min(j 1 ,j 2) Therefore the ring multiplication on G less than or equal . p are determined by defining the product of basis elements with the sole restriction labl ~ min(lal, lbl) for every pair of basis elements a,b. In alI of the above examples, the ring multiplications which were defined are associative (commutative) if the product of basis elements is associative (commutative). 1.4.5.
G a torsion group.
G=
~ GP, Proposition 1.1.1. It is readily seen that p a prime Hom(G~, G)~ n. Hom(Gp ~ Gp, Gp) i.e., every ring R with R+ = G p a pr1me is a ring direct sum R = f.il RP with R; = Gp. Therefore the p a prime investigation of the ring structures which may be defined on a torsion group G, reduces to the p-primary case, p a prime.
1.4.6.
G a p-primary group,
p a prime.
Let B be a basic subgroup of G with basis {a; I i E I}. Then ~ = B~, [36, Theorem 61.1] and so Hom(~. G)~ Hom(~. G). As in example 1.4.4, the homomorphisms 11> from ~ into G are determined by defining ~~>(a; A aj) to be an element in G with the sole restriction that l~~>
7
1.4.7. 1.
G=H@Z(pco),
p aprime.
H a torsion group.
Let R be a ring with R+ = G, a E Zlp~), and x E R. There exists bE Z(p~) such that a= lxlb. Therefore ax= (lxJb)x = b(Jxlx) = 0. Similarly xa = 0. Therefore Z(p~) annihilates R. For every prime q ~ p, Hq is an ideal in R, and (f) Hq is a ring direct summand q a prime q
~
p
of R. It is clear from 1.4.6 how products of elements in Hq must be defined for every prime q ~ p. 2. 1)
H torsion free. H p-divisible.
Let Clearly G Hence xa = annihilates
R be a ring with R+ = G, x E R, a E Z(p~), with is p-divisible, and so there exists y E G such that (pky)a = y(pka) = 0. Similarly ax= 0, i.e., Z(pco) R.
If H2 c H, then R is a ring direct sum R = S @ T, with s+ = H, ~2 T = Z{p ), and T = 0. Otherwise, let h, h0 E H such that h·ho = h + a0 , h E H, a0 E Z(p~), a0 ~ 0. Then a0 E . For every positive integer n let hn' h~ E H such that pnhn = h0 , and pnh~ = h0. Put an = h·h n - h'. n Then a0 = pna n , i.e. -~ Z(pco). Therefore for H p-divisible, either n Z{p~ = 0, or ~ Z(p~). +
0
b)
0
H is not p-divisible.
Choose bE H with hp(b) = 0. For any positive integer n, choose an E Z(p~) with lanl = pn. The map ~~an can be extended to an epimorphism ~: ~H ~ Z(pn). Therefore a ring R with R+ = G can be constructed so that n Z(p~) is an arbitrary proper subgroup of Z(p~). Since Z(p~) is a direct summand of every group G containing it, example 1.4.7. is important. It will be studied further in chapters 2 and 4. 1.4.8.
G a subgroup of Q+ •
The subgroups of Q+ are, up to isomorphism, precisely the rank one torsion free groups. These groups are homogeneous. Two subgroups G1 , G2 of Q+ are isomorphic if and only if t(G 1J = t(G 2 ). For every type '• 8
there exists a subgroup If
t( G)
Hom(~G.
G of Q+ such that
= -r is not idempotent, then
G) = 0, is the zeroring.
t(G) = -r, > "[
Proposition 1 .3.4, i.e., the only ring
[36, Theorem 85.1]. and so R with
Suppose that t(G} = T is idempotent. Choose e E G such that e ; 0, and hp(e) = 0 or m for every prime p. Let {q; 1 i E I} be the set of primes q for which hq(e) = m, Let R be a ring with R+ = G. For X; E G there exist integers m;, and positive integers ni such that nixi = m;e• i = 1,2. Therefore n1n2 (x 1·x 2 ) = m1m2(e 2). Since G is torsion free, if X;; 0, i = 1,2, then x1x2 ; 0 i f and only if e 2 ; 0. In the latter case the product x1x2 = re 2 , r=(m 1m2)/(n 1n2 ). Suppose that i;o. Then e2 =re, r=m/n, m,n integers, (m,n)=l. Let p bea prime such that p ¢ {q;l i E 1}. Suppose that pin, i.e. n = pk, k an integer. Then (p,m) = 1, and there exist integers s,t such that sp + tm = 1. Therefore ptke 2 = tne 2 = tme = e- spe. Hence e = p(se + tke 2 ), and 0 < hp (e)< m, a contradiction. Therefore n is _the product of primes belonging to {q; I i E I}. Put e' = ne. Then e' I 0, and h (e') = 0 or m for every prime p. e• 2 = mne =me . It may therefore b~ assumed that e2 = me, m a positive integer. Write m = uv, u,v positive integers, u a product of primes belonging to {q; I i E 1}, and (q,v) = 1 for every prime q E {q; I i E I}. There exists e1 E G such that e = ue 1 , e1 ; 0, and hp(e 1 ) = 0 or m for every prime p. e~ = u- 2e 2 = vu- 1e = ve 1 • Therefore, it may be assumed that e 2 = me, m a positive integer with (q,m) = 1 for every q E {q; I i E I}. For every positive n, 1et nZ( qi 1 1 i E I) denote the subri ng of Q generated by {n/q; 1 i E 1}. The map re -+ mr for r E Z(qi 1 ; i E I) is an isomorphism from R onto mZ(qi 1 I i E I). Conversely for t(G) idempotent, 0 ; e E G, with tp(e) = 0 or m for every prime p, and m a positive integer not divisible by the primes for which e has infinite height, the product e 2 =me induces a ring structure on G isomorphic to mZ(qi 1 I i E I), with {qi 1 i E I} the set of primes for which hq(e) = m.
9
2 Nil and quasi-nil groups
Nil groups: Definit1on: A group G is said to be (associative) nil if the only (associative) ring R with R+ = G is the zeroring. Associative nil groups were first investigated by T. Szele [67, 68]. The equivalence of 2) and 3) in the following theorem was proved by him [67, Satz 1 and Satz 3]. Theorem 2. I. 1: Let G be a group which is not torsion free. The following are equivalent: 1) G is ni 1. 2) G is associative nil. 3) G is a divisible torsion group. Proof: Clearly 1) ,. 2). 2),. 3): Suppose that G is associative nil. G is either a torsion group or a mixed group. If G is a non-divisible torsion group then G = Z(pk) (f>H, p a prime 1 < k < ~; Proposition 1. 1.4 and Proposition 1. 1.3. The ring direct sum of ~he ring of integers modulo pk and the zeroring on H is an associative non-zeroring with additive group G, a contradiction. If G is mixed and Gt is not divisible, then again G has a cyclic direct summand, and the above argument shows that G is not associative nil. It may therefore be assumed that Gt is divisible, in which case G = Gt (+)H, H a torsion free group, Proposition 1.1.2. Since Z(p~) (+)H is a direct summand of G, Proposition 1.1.3, it suffices to show that Z(p~) ~H is not associative nil. Let (a 1 , a2 , a3 , •.• ,an•···; pa 1 = 0, pan+l =an for every positive integer n) be a presentation for Z(p~). Put p-na 1 = an+l for every positive integer n. This naturally induces an action of Q on Z( p~). Choose 0 'I b E H. Every element g E Z( p~) 0 H may be written in the form g ~ d + rb + c, with dE Z(p~), r E Q, c E H so that b and c are independent. Let g1 = d1 + r 1b + c1 , g2 = d2 + r 2b + c2 be elements in Z(p~)G)H written in the above form. Define g1 ·g 2 = r 1 r 2d1 • These products induce an associative non-zeroring structure on Z(p~)('i1H, a contradiction. §1:
10
3) ~ 1): Let G be a divisible torsion group, and let R be a ring with + R = G. Let 0 1 a, 0 1 b E R. There exists c E G sucn that b = jajc. Therefore a-b = a·(jajc) = (jaj.a). c = 0. In lieu of the above theorem, the investigation of (associative) nil groups reduces to the torsion free case. Unfortunately, not much progress has been made. The following is an up-to-date account of the results concerning torsion free nil groups. Theorem 2.1 . 2:
[17, Theorem 1 1.
torsion free group, and is a ni 1 group. Proof:
By 1.2. 1.,
Hom(G.~G., 1 i,j,k€I J
@
r~ul
Let
t(Gi) · t(gj)
G=
!
t G ""Hom(~, G)
Gk) = 0.
~
i€I t(Gk)
G. , with
Gi
a homogeneous
1
for all
Hom[ <±> (G. i,jEI 1
i ,j,k € I. (x) G.), J
(+)
k€I
Gk]
Then
G
c
-
A consequence of Theorem 2.1.2 is the following result of R. Ree and R.J. Wisner [55], (see also [41, Theorem 1.1]). Corollary 2.1.3: Let G be a completely decomposable torsion free group, G = (-H G. , with Gi a rank one torsi on free group for each i € I . Then i€I 1 G is (associative) nil if and only if t(Gi) • t(Gj) i T(Gk) for all i ,j ,k € I.
Proof: Since rank one torsion free groups are homogeneous, it suffices to show that if there exist i 0 ,j 0 ,k 0 € I such that t(G. )·t(G. ) < t(Gk ) , 1o Jo o then G is not associative nil. For each i € I choose 0 ; ei € Gi . The ek i f {i,j} = {io,jo} [ induce an associate ring products ei·ej = • 0 0 otherwise structure
R on
G with
R2 ; 0.
As a special case of Corollary 2. 1.3 we have that a rank one torsion free group G is (associative) nil if and only if t(G) is not indempotent, [36, Theorem 121.1 ]. This case however was completely investigated in Example 1.4.8. In the cases considered so far, associativity has not influenced the classification of nil groups. So far no group G has been constructed for which the only associative ring structure which may be defined on G is the 11
zeroring, but there exists a non-associative ring R with R2 ~ 0. One is therefore naturally led to the following: Conjecture 2.1.4:
R+ = G,
and
A Group G is nil if and only if G is associative nil.
Recently, A.E. Stratton [60, Theorem 3.3], and independently D.R. Jackett [46] made considerable progress towards classifying the nil rank two torsion free groups. Their work relies on the following result of H. Freedman [33, Theorem] which although stated here differently than in her paper [33], was in fact proved by her. Lemma 2.1.5: Let G be a rank two torsion free group. If T(G) a unique minimal element, and G is not nil, then IT(G)I ~ 3.
possesses
Proof: For G a rank n torsion free group, the length of every chain in T(G) is less than or equal n, [3, 9.1]. Let To be the unique minimal element in T(G). For every T € T(G), either T =To or T >To· In the latter case T is a maximal element in T(G). It therefore suffices to show that T(G) possesses at most two maximal elements. Let T.i be a maximal element in T(G) and let 0 ~xi € G with t(x;) = Ti, i = 1 ,2, and Tl ~ T2 • Let G; be the pure subgroup of G generated by xi, i = 1 ,2. Then G.1 is a rank one torsion free group with t(G 1.) = T., 1 i = 1 ,2, and G1 ® G2 is a rank one tors ion free group with t(G 1 ®G2 J = Tl · T2 > Ti' i = 1,2, Proposition 1.3.3. The sequence: 0
-+
G1
-+
G -+ G/ G1 .... 0
is exact.
Every torsion free group is flat, and so the sequence:
is exact.
This implies the exactness of the sequence:
[36, Theorem 44.4]. For every 0 ~ x € G, t( G1 @ G2 ) i t(x), and so Hom(G1 (X) G2 , G) = 0 by Property 1.3.1.4. Hence Hom[(G/G1 ) @.G2 , G] ~ Hom(~ 2 • G)~ Hom(G 2 , End{G)). Let T3 be a maximal element in T(G), T 3 ~ 'i' i = 1,2. Let 0 ~ x3 € G, with t(x 3 ) = T3 , and put x = the coset x + G1 for every x € G. Clearly t(x 3) = t(x 2 ), and t(x 3) ~ t(x 3). If t(x 3) = t(x 3 ), then 12
T3 = t(x 3) = t(x 3) = t(x 2J ~ t(x 2 ) = •2 . By the maximality of • 2 • • 3 = T2 , a contradiction. Hence Hom[ ( G/G1) (~ G2 ,G] = 0 by Property 1.3. 1.4) and so (A) Hom(G 2 , End(G)) = 0. A similar argument shows that (B) Hom(G1 , End(G)) = 0. Let x e: G, and let lP e: Hom(G, End(G)). There exists a positive integer n, and integers n1 ,n 2 such that nx = n1x1 + n2x2 . Hence nlP(x) = n1lP(x 1) + n2lP(x 2 ) = 0 by (A) and (B). However End(G) is torsion free, and so lP(x) = 0, i.e., Hom(G,End(G)) = 0. Byl.2.1, G isnil,acontradiction. The next result will show that stating the existence of a unique minimal element in the previous lemma is superfluous. Lemma 2.1.6: Let G be a rank two torsion free group. then T(G) possesses a unique minimal element.
If G is not nil
Proof: Let R = (G,·) be a ring satisfying R2 i 0. Suppose x,y e: R with t(x) i t(y), and x·y i 0. Clearly either t(xy) > t(x), or t(xy) > t(y). Suppose that t(xy) > t(x). Then every element in G is a linear combination of x and xy over the rationals and so t(x) is the unique minimal element in G by Property 1.3.1.2). It may therefore be assumed that (A) X·Y = 0 for all x,y e: R with t(x) i t(y). Suppose that IT(G)I ~ 3. Let x e: G. There exist y,z e: G such that t(x), t(y), t(z) are distinct types. x = ay + bz, with a,b rational numbers. Hence x2 = ~X·Y + bx-z = ~ by (A). Let 0 i x, 0 i y e: R, with t(x) = t(y). If x and y are independent, then every element in G is a linear combination of x, and y over the rationals, and so t(x) is the unique minimal element in T(G) by Property 1.3.1.2). Therefore it may be assumed that there exist non-zero integers n,m such that nx = my. This yields that nmx-y = n2x2 = 0. Since G is torsion free X·Y = 0, or (B) x-y = 0 for all x,y e: R with t(x) = t(y). Clearly (A) and (B) yield that R2 = 0, a contradiction. Therefore (A) implies that IT(G)I ~ 2. Let T(G) = £• 1• T2}, and choose 0 i X; e: G, t(x;) = T;• i = 1,2. Put x = x1 + x2 . If t(x} = Tl then the fact that x2 = x- x1 yields that T 2 = t(x 2 ) ~ Tl by Property 1.3.1.2). Similarly t(x) = T 2 implies that • 1 ~ T 2 • In either case T(G) possesses a unique minimal element. An immediate consequence of the last two lemmas is the following: 13
Theorem 2. 1.7: Let G be a rank two torsion free group. then G is ni 1. For every positive integer G with T(G) = n, [9], and infinite type set [36, vol. 2, not a statement concerning the
If
IT(G) I > 3
n, there exist rank two torsion free groups there exist rank two torsi on free groups with p. 112, Ex. 11 ). Therefore Theorem 2.1.7 is empty set.
Clearly the classification of nil rank two torsion free groups reduces to the case IT(G) I < 3. It is easy to show that the following are necessary conditions for a rank two torsion free group to be a non-nil group [60, p. 204 ]: 1) 2)
3)
IT( G) I IT(G) I IT(G)I
1, 2, 3,
i.e., G homogeneous, t(G) must be idempotent. T(G) must consist of.one minimal type and one maximal type. T(G) must consist of one minimal type, and two maximal types.
One of the maximal types must be idempotent. A classification of the nil rank two torsion free groups with will be given. First some preliminary results are required.
Definition: of G is n(G)
[10], and [60]. {a E
Let G be a torsion free group.
IT(G)j
3
The nucleus
Ql ax E G for all x E G}.
Lemma 2.1.8: [60, Lemma 2.2]. Let G be a torsion free group with r(G) > 1. If End(G) = n(G) then G is ni 1. Proof: Suppose that End(G) = n(G), and let R be a ring with R+ =G. Suppose there exist x,y E R with X·Y ~ 0. Left multiplication by x is an endomorphism of G, hence there exists 0 ~ a E n(G) such that X·Z = az for all z E G, and so x·z ~ 0 for all 0 ~ z E G. Choose z E G such that x and z are independent. Right multiplication by z is also an endomorphism of G, and so X·Z = bx, 0 ~bE n(G). Hence bx = az contradicting the fact that x and z are independent. The above lemma is actually a consequence of [36, Proposition 121.2] and is in fact contained in the statement preceding [36, Proposition 121 .2], namely that rigid groups of rank> 2 are nil.
14
Lemma 2.1.9: Let G and H be torsion free groups with G &H. is (associative} nil if and only if H is (associative} nil.
Then
G
Proof: It may be assumed, Proposition 1.1.6, that G and H are subgroups of a group G, and that G ~H. Suppose that H is not (associative} nil. Let R = (H,·J be an (associative} ring with R2 ~ 0. There exists a positive integer n such that nG ~ H, and nH ~G. For g1 ,g 2 € G define g1•g 2 = (ng 1} • (ng 2}. It is readily seen that S = (G,•} is an (associative) ring. Let h1 ,h 2 E H such that h1 -h 2 ~ 0. Then (nh 1 )•(nh 2} = n4h1h2 ~ 0. Hence s 2 ~ 0, and G is not (associative} nil. Similarly if G is not (associative) nil, then neither is H. Theorem 2.1.10: Let G be a rank two torsion free group, with jT(G} I = 3. Then G is not (associative} nil i f and only if G.; G1
~nd(G) =
x € G}
Q,
[9,
= n(G).
Now
Note: Theorem 2.1.7 is actually implicit in [9J. As part of a complete description of the quasi-endomorphism rings of rank two torsion free groups, it was shown, [9, Theorem 8.4), that i f G is a rank two torsion free group with IT(G)I > 2, then either ~nd(G} = Q, or G is quasi-isomorphic to the direct sum of two rank one groups of incomparable types. If End(G} = Q, then G is a rigid group, and so G is r.il [36, comment preceding Proposition 121.2). If G is quasi-isomorphic to the direct sum of rank one groups of incomparable types, then it is readily seen, [9, remark preceding Theorem 8.5), that IT(G}I = 3.
15
§2.
Quasi-nil Groups:
The group Q+ admits precisely two non-isomorphic ring structures, the zeroring, and Q. Szele [68, Satz 1] classified the groups allowing precisely two non-isomorphic ring structures. He conjectured that every non-nil torsion free group G ~ Q is the additive group of infinitely many non-isomorphic associative rings. This was shown not to be true by L. Fuchs, [38]. Borho, [12], showed that Szele's conjecture was not far from being true, see Theorem 2.2.4 and Corollary 2.2.5. The investigation of nil groups, and additive groups of precisely two non-isomorphic rings, initiated by Szele, was generalized by Fuchs as follows: Definition: A group G is (associative) quasi-nil if there are only finitely many non-isomorphic (associative) rings R, with R+ = G. Fuchs classified the torsion (associative) quasi-nil groups completely, and made considerable progress towards classifying the torsion free and mixed (associative) quasi-nil groups, [35], [38]. A complete description of the torsion free (associative) quasi-nil groups which are not nil was obtained by W. Borho, [11]. This in conjunction with the results of Fuchs yields a classification of the (associative) quasi-nil groups which are not nil. The main results of this section are due to Fuchs, and Borho, [12], [35], [38]. The following technical lemma is generally useful in studying the additive groups of rings: Lemma 2.2.1:
Let R be a ring with
R+ =G.
Then G1 annihilates
Gt.
Proof: Let 0 ~ x € Gt• y € G1 , and let lxl = n. Since y € nG, y = nz, z € G. Therefore xy = x(nz) = (nx)z = 0. Similarly yx = 0. Theorem 2.2,2: 1) 2) 3)
G is quasi-nil. G is associative quasi-nil. G = B ~ 0, B a finite group,
Proof: 16
Let G be a torsion group.
Clearly
1) • 2):
~
The following are equivalent:
a divisible torsion group.
2)•3): Supposethat G isassociativequasi-nil. G=H(i)D, H reduced, and D divisible. Hp is a direct summand of G for every prime p, Proposition 1.1.1, i.e., G = Hp (f) KP. For every prime p with Hp I 0, let Sp be an associative non-zeroring, with s+p = Hp , Theorem 2.1. 1, and let TP be the zeroring on KP. The ring direct sum Rp = SP@ TP satisfies R; = G, R~ I 0, and Rp + Rq for distinct primes p,q. Hence H = 0 for all but finitely many primes. It therefore suffices to show p . that Hp is finite for every prime p. Let Bp be a basic subgroup of n.
with basis {a.li € 1}. Put la-1 = p 1 , i € I. Every set of p 1 1 associative products a .. a. with la.·a-1 < min(la-1, la-1), i,j € I, 1J 1J1 J induces an associative ring structure S on Hp' [36, Theorem 120.1] and remarks following the theorem. Suppose that I is an infinite set. For every positive integer n, choose distinct elements ik € I, k = l, ... ,n.
H ,
..
For 1,J € I,
.
def1ne
ai·aj
ro
=·.. a;
i f i I j or i = j = ; k, k = 1 , ... , n i f i = j, and ; ¢ {il' i 2 , ... ,in}
These products induce an associative ring structure Un on Hp with annihilator of rank precisely n. Hence Un +Urn for distinct positive integers n,m. Again the ring direct sums Un@ Tp' TP the zeroring on KP, yield infinitely many non-isomorphic, associative ring structures on G. Therefore BP is finite, which implies that Hp = BP (f.) Ep, with EP a divisible group. Since HP is reduced, HP = BP, and so HP is finite. 1): Let G = B(+l D, B a finite group, D a divisible torsion group. By Lemma 2.2.1 every ring R with R+ = G satisfies RD =DR= 0, and for every prime p, B2p -c Bp {+) Dp . It may therefore be assumed that G is a p-group. Now t·1ul t G .,. Hom(~. G) ... Hom(B'!IlB, G), [36, Theorem 61.1]. Let r(B) = m. Then r(B~) = m2 • Let IBI = pn. Every ~ € Hom(B~. G) is a homomorphic map ~= 85<6 ... B(t) E[pn l, E a subgroup of D, with r(E) ~ m2 . A change of the group E is determined by an automorphism of D, and induces an isomorphic ring structure on G. Clearly Hom(~. B<±> E[pn]) is finite, and so G is quasi-nil. 3) ..
Lemma 2.2.3. Let G be an (associative) torsion free quasi-nil group, and let R be an associative ring with R+ = G. Then (R2)+ is a divisible group.
17
Proof: Put R = R1 , and for every positive integer n let Rn be the associative ring with R+ = G, and multiplication defined by axnb = n(a•b) for all a,b E G; the product on the right hand side of the equality being the multiplication in R. Since G is (associative) quasi-nil there exist finitely many positive integers m1 , m2 , ••. ,mk such that for every positive k
integer n,
there exists
1 < i < k sucn ti1at R .,. R • Put m = n m.• n mi i=l 1 is divisible, and so (R 2)+ is divisible.
2)+ (m.R 2)+ = (Rm An alternative proof of Lemma 2.2.3 may be found in [12, LenT11a 1 ].
Clearly
Theorem 2.2.4: 1) A torsion free group is associative quasi-nil if and only if either G is nil, G.,. Q, or G.,. Q q.'IH, H a nil rank one torsion free group. 2) A torsion free group G is quasi-nil if and only if G is nil, or G = Q+• Proof: Let G be a torsion free associative quasi-nil group. If G is notnil,then G=Q + (+)H, Lemma2.2.3. If H isnotreduced,then G = Q+ ~ Q+ {+) K. The ring direct sums of a quadratic number field on Q+ (:i') Q+ and the zeroring on K yield infinitely many non-isomorphic associative rings on G. Hence H is reduced. Suppose that r(H) > 1. Choose three independent elements bi E G, i = 0,1 ,2, with o0 E Q, b1 ,b 2 E H. Put G = G3Q. Now G may be viewed as a vector space over Q, with a natural embedding of G into G such that b0 , b1 , b2 are contained in a subspace G1 = b0Q 4) b1Ql+l b2Q of G. Let A = (aij) E M2{Q), i ,j = 1,2 be an arbitrary 2x2 matrix over Q. The products for
i
for
i = 0 or j = 0
,j
E {1 ,2 }
induce an associative Q-algebra structure (Gl)A on G1 . A Q-algebra structure AA on G may be obtained by defining all products of an element in G with. an element in a complement of G1 to be zero. The fact that b0Q ~ G yields that the above products induce an associative ring structure RA with R: = G. Hence every matrix A E 14 2(Q) determines an associative ring RA' with RA+ = G. Let 0 1 fl. = {a .. ), 0 1 B = (B .. ) E M2(Q} with lJ lJ RA.,. R8 , and let ~= RA ~ R8 be a ring isomorphism. ~ is also an automorphism of G. Since H is reduced, ~(b 0 ) E Q, i.e., ~(b 0 ) = r 0b0 , 18
of
r 0 E Q. Now ~ extends naturally to an algebra isomorphism ~: AA ~ A8 • Since ~ restricted to (G1 )A is a regular linear transformation 2
q): (Gl)A ~ (~ 1 J 8 , q}(bi) = .E riJ.bJ., riJ. E Q, J=O
i = 1 ,2;
j
= 1 ,2.
Put P = (r-J.) E M2(Q), i ,j = 1 ,2. The last identity can be written as 1 r 0A = PBP t . Therefore the determinant 1r0A1 = r 02 1AI = IPI 2 IBI, and so IAI = s 2 1BI, sEQ. It suffices to show that the equivalence relation in Q, x ~ y if and only if x = s 2y, s E Q, determines infinitely many equivalence classes. Suppose that the above relation partitions Q into finitely many equivalence classes. For x E Q let x be the equivalence class of x. There exist finitely many rational numbers r 1 , ••• ,rk such that for every rational number r, there exists 1 ~ i ~ k such that r = ri. For every integer n, the set of square roots of elements in n is precisely the field Q(lin). Therefore there are at most k quadratic number fields, a contradiction. Therefore r(HJ ~ 1. If H is not nil let S be the zeroring on Q+, and let T be a non-zeroring on H. Put R = S (-f) T. Then (R2 )+ = (T 2) c H is not divisible, contradicting Lemma 2.2.3. Conversely, if G is nil or if G ""Q + , then G is clearly quasi-nil. Suppose that G = Q+ (f) H. H a nil rank one torsion free group. Let R = (G,•) be a ring, and let nH be the natural projection of G onto H. Define axb = nH(a·b) for all a,b E H. This composition defines a ring structureon H. Since H isnil, nH(a·b)=O forall a,bEH, i.e., H2 ~ Q+. This coupled with the fact that, Q+, the maximal divisible subgroup of G, is an ideal in R, yields that R2 ~ Q+. Choose 0 f b1 E Q, 0 f b2 E H. The multiplication in R is determined completely by the products b .• b. = cx .. b1 , ex •• E Q, i,j = 1,2. For every matrix 1 J lJ lJ A = (cxij) E f,12 (Q) the above products induce a ring structure RA on G. As above, if two such rings RA and R8 are isomorphic, then IAI = s 2 1BI, s E Q, and so there are infinitely many isomorphic classes of ring structures on G. However if the ring RA is required to be associative, then the equalities (bibj)b 1 = bi(bjb 1 ), and (b 1bi)bj = b1 (bibj) yields that 19
(A)a;jall = ailajl and (B) alialj = a;jall for i ,j = 1,2. Consider two cases a) a11 = 0, b) a11 1 0. a) Suppose a11 = 0. Putting i = j in equalities (A), (B) yields that
a~ 1 = a~; = 0, and so ail = a1; = 0 for i = 1 ,2. Therefore A = (~ :) , a € Q. Conversely every such matrix determines an associative ring RA. Again map G into a vector space t' = b1Qffi b2Q. The products in RA induce a Q-algebra structure GA on G, with RA a subring of GA. The mops
b1
~ .- 1b1 ,
~ b2
b2
induce an algebra isomorphism
~·
GA ~ G(~
~)
Since G~ ~ RA' the restriction of ~ to RA is an isomorphism
~: RA ~ R(~ ~)
Therefore every associative ring
isomorphic to l~ ~)
R with
R+ = G is
or R(~ ~) •
b) Suppose that a11 ~ 0. Without loss of generality it may be assumed -1 b • Choosing i = 1, that a 11 = 1, because b1 may be replaced by a 11 1 j = 2 in (A) yields that a12 = a21 = a, and so choosing i = j = 2 in
:~
i.
(A) yields that a22 = Hence if RA is associative, then A=(: The maps b1 ~ b1 , b2 ~ b2 - ab 1 induce a group automorphism of G. Now b~ = b1, (b 2 - ab 1) 2 = 0, and b1(b 2 - ab 1) = (b 2 - ab 1)b 1 = 0. Hence ~= RA ~ R(~ ~) is a ring isomorphism. Corollary 2.2.5:
Let G be a torsion free group.
1) G is the additive group of precisely two non-isomorphic (associative) rings if and only if G ~ Q+. 2) G is· the additive group of precisely three non-isomorphic associative rings if and only if G ~ Q+~ H, H a nil rank one torsion free group. In this case every associative ring R with R+ = G is isomorphic to one of the rings
\g g) ,
Theorem 2.2.4, and G 20
R(~ i~
g) , R{g
~)
constructed in the proof of
the additive group of infinitely many,
•
non-isomorphic, non-associative rings
R.
3) Otherwise G is either nil, or the additive group of infinitely many non-isomorphic associative rings.
The proof of Theorem 2.2.4 remains valid considering only commutative rings R with R+ = G by choosing symmetric matrices A in the construction of the rings RA. Therefore: Corollary 2.2.6: A non-nil torsion free group G is the additive group of only finitely many non-isomorphic commutative, associative rings if and only if G ~o+, or G ~Q+(!)H, H a nil rank one torsion free group. Theorem 2.2.7: Let G be a mixed group. and only if either: G = B(t)H, B a finite group, group. 1)
G is (associative) quasi-nil if
H a torsion free (associative) quasi-nil
2) G = B(f) D@ H, B a finite group, D a divisible torsion group with Dp = 0 for all but finitely many primes p, H = Q+, or H a nil rank one torsion free group, satisfying pH = H for every prime p with Dp I 0. Proof: Suppose that G is a mixed (associative) quasi-nil group. Decompose Gt into B@ D, B reduced, and D divisible. G has a direct summand Z(pk), 0 < k < =, for every prime p for which BPI 0, Proposition 1. 1.1, Proposition 1.1.4, and Proposition 1 .1.3. The direct sum of the ring of integers modulo pk on Z(pk) and the zeroring on a complement of Z(pk) yields an associative ring Rp with R; = G. For primes pI q, Rp Rq. Hence Bp = 0 for all but finitely many primes p. The same argument used in proving the implication 2) • 3) in Theorem 2.2.2 shows that B is finite. Therefore G = B~D@H, H a torsion free group, Proposition 1.1.2. Clearly H must be (associative) quasi-nil.
+
Suppose that D I 0. Let p be a prime for which Dp I 0. If pH I H then for every positive integer n, pnH is a proper subgroup of H. Therefore there exists an epimorphism ~= H ~ Z(pn) (the composition of the canonical homomorphism H ~ H/pnH and a projection of H/pnH onto Z(pn)). Let dE D with Jdl = pn, and view ~ as an epimorphism of H onto the subgroup of D generated by d. Let X; E G, X; = b; + d; + hi, b; E B, 21
di E 0, hi E H, i = 1 ,2. Define x1 -x 2 = k1k2d, With kid = ll>(h;), i = 1 ,2. These products induce an associative ring structure Rn on G, in fact Rn3 = 0. Now (R2n)+ ~ Z(pn), and so Rn Rm for positive integers n ~ m. Hence there are infinitely many non-isomorphic associative rings with additive group G, a contradiction. Therefore pH = H for all primes p for which Dp ~ 0.
+
To show that required. Let
r(H) =
some facts about the endomorphisms of
Z(p~)
Z(p~)
positive integer n) p-adic integer.
For every positive integer n,
define
n(an)
This action of n on a set of generators of Z(p~) extends naturally to an endomorphism n E End(Z(pj). In fact all the endomorphi sms of Z(pj are of this type, [36, vol. 1, p. 181, Example 3]. The automorphisms of Z(p~) are clearly p-adic units, [36, vol. 2, p. 250, Example 3]. Suppose that r(H) > 1. Choose independent elements 0 ~ u, 0 ~ v E H. Every element x E G is of the form x = b + d + ru + sv + w, b E B, dE D, r, s E Q, and WE H. Let X; = bi + d; + r 1 u + siv + W;, i = 1,2 be elements of G written in the above form. The products x1 -x 2 = (r1r 2 + s 1s 2n)a 1 induce an associative ring structure Rn on G. Put pwG
=
~
n pnG, n=l x = b + d + ru + sv ( p-n x) 2 = p-2n (r2 +
and let x E pwG.
Write
x in the above canonical form,
+ w. For every positive integer n, s 2n)a 1 = (r 2 + s 2n)a 2n+l" Let n(p wG)
be the set of
p-adic integers {r2 + s 2nl obtained above for each x E pwG, and every positive integer n. Then n(pwG) is a countable set of endomorphisms of Z(p~) containing the identity automorphism. If R ~ R for two p-adic n p integers n,p then the elements of ~(pwG) are the elements of n(pwG) multiplied by a fixed automorphism of Z(p~), i.e., by a p-adic unit However the identity automorphism must belong to (pwG), and so a must be chosen from a countable set of p-adic units. Therefore the isomorphism class of Rn is countable. The non-denumerability of the p-adic integers yields that there are infinitely many non-isomorphic associative rings Rn with R+n G, a contradiction. Hence r(H) = 1. 22
Let p be a prime for which Dp ~ 0, and let 0 ~ d E Dp • Let 0 ~hE H. For x1 ,x 2 E G, put x. =b.+ d. + r.h, b. E B, d1. ED, 1 1 1 1 1 r; E Q, i = 1 ,2. The products x1x2 = r 1r 2d induce an associative ring structure Rp on G. For primes p ~ q, RP Rq. Therefore Dp = 0 for all but finitely many primes p.
+
Conversely, suppose that G is of form 1). t4ult G ""Hom{~. G) ... Hom{G~. B)(+)Hom{H~. H), since Hom(X,Y) = 0 for X a torsion group, and Y a torsion free group. If G is quasi-nil then so is H. Therefore Hom{H~. H) ""'r~ult H determines only finitely many non-isomorphic ring structures on G. Hom{G~. B) ""'Hom(B~B. B) t+) Hom{H~, B) (f) Hom(B~, B) (t) Hom{Hfti, B). The groups B, B·Sl!B, B~ are finite, and so the first three summands above are finite. Since r{~H} = 1, Proposition 1.3.3, and B is finite, Hom{H·~. B) is finite. For the associative case consider G = B t:t) H, B finite, H an associative quasi-nil torsion free group. Let lSI = m. The ring multiplications on G are determined by products of elements in B, for which there are clearly only finitely many possibilities, products of elements in B with elements of H, and products of elements in H. Since mH annihilates B in every ring on G, the products of elements in B with elements of H are determined by the products of elements of B with coset representatives of elements of H/mH. Since r{H) ~ 2, Theorem 2.2.4.1}, H/mH is finite, and there are only finitely many ways of defining products of elements of B with elements of H. Let R = (G,·) be an associative ring, and let al'a 2 E H, a 1·a 2 = a3 + b, a3 E H, bE B. If a1 is replaced by an arbitrary element in the coset a1 + mH, i = 1,2, then the element ab above remains the same. Since there are only finitely many such cosets, it suffices to show that G/B is associative quasi-nil. However G/B ""'H is associative quasi-nil. Suppose that G is of the form 2) with H nil. As in the case of form 1) it suffices to investigate Hom(~. G). Arguments similar to those used (±) Hom(~, D } in case l} reduce the investigation to Hom(H~, D) ""' p prime P Since Dp = 0 for all but finitely many primes p, it suffices to consider Hom(H~, DP), p a fixed prime, with OP ~ 0. Now r(HSltt) = 1, and so 23
every ~ € Hom(HAH, Dp) is a map ~= ~H ~ Z(p=). Changing the summand Z(p=) only alters the ring determined by ~ by a ring isomorphism. Therefore it suffices to examine Hom(H~. Z(p9j). However Homl~. Z(p~) is isomorphic to the additive group of p-adic numbers. If u1(x,y) and u2(x,y) are two ring multiplications on G with u1 a non-zero multiplication, then u2(x,y) = wu 1(x,y), w a p-adic number, for all x1,x2 € G. Multiplication by p is an automorphism of Z(p=). Therefore, the multiplications u1(x,y) and p- 2Pu 1(x,y), p a p-adic unit, define isomorphic ring structures on G. Every non-zero p-adic number is of the form pkw, k an integer, w a p-adic unit. Therefore the elements of Hom(~. Z(p=)) induce three non-isomorphic ring multiplications on G, the zero multiplication, u1(x,y), and Pu 1(x,y). If H is no·t nil, then H = Q, and it suffices to examine the multiplications determined by Hom(~. D(!) Q) ""Homl~. D)<±> Hom(Q~. Q). The second summand is isomorphic to Mult Q, and therefore determines two non-isomorphic multiplications on G. The same argument employed above to show that Hom(tfSJi, D), H nil, determines only finitely many non-isomorphic ring multiplications on G, shows the same for Hom(~. D). This proves Theorem 2.2. 7.
24
3 Additive groups of nilpotent and generalized nilpotent rings §1.
The nilstufe of a group. Another generalization of nil groups, also due to Szele is the following:
Definition: Let G ~ 0 be a group. The greatest positive integer n such that there exists an associative ring R with R+ = G, and Rn ~ 0, is called the nilstufe of G, denoted v(G). If no such positive integer n exists, then v(G) = ~. The strong nilstufe of G, N(G) is defined as above, with the associativity of R deleted. Clearly the (associative) nil groups are precisely the groups satisfying (v(G) = 1) N(G) = 1.
G
The following examples are due to Szele [68, Satz 2]: Example 3.1.1: For every positive integer n, group G satisfying v(G) = N(G) = n.
there exists a torsion free
For every positive integer i,
let G. be a rank one torsion free group n , oftype [(i,i, ••• ,i, .•• )]. Put G= (i;) G.• Clearly Rn+l =0 forevery i=l , ring R with R+ = G. For each 1 < i < n choose ei E Gi such that h(e;) = (1,i , .•• ,i, ••• ). The products
ei ·ej =
e.+. , J
for
i+j -< n
0
for i+j > n
n induce an associative ring structure R on G. Now e~ =en~ 0, and so Rn ~ 0, i.e., v(G) = n. Clearly v(G) ~ N(G) < n+l, and. so N(G) = n. ~
i, j
! r
~
Example 3.1.2: For every positive integer n, G satisfying v(G) = N(G) = n+l. For every positive integer i,
let Gi
there exists a mixed group
be a rank one torsion free group
25
of type
[(co,i,i, •.• ,i, ••• )].
n Put G = Z(2co) (i) G;·
Clearly Rn+l ~ Z(2co)
i =1
for every ring R with R+ = G. However Z(2co) annihilates R, and so Rn+ 2 = 0, i.e., N(G) ~ n+l. Choose 0 ~a € Z(2co), and ei € Gi with h(ei) = (co,i ,i, ••• ,i, ••• ). The products
r·j
ei·ej =
for i+j
~
n
a
for i+j = n
0
for i+j
>
n
for ~ i, j < n induce an associative ring structure R on G. Now n+l e 1 = a ~ 0. Hence n+l = v(G) ~ N(G) ~ n+l, and so v(G) = N(G) = n+l. Conjecture 2.1.4 can be restated as follows: For every group G, v(G) = 1 i f and only i f N(G) = 1. It is not true in general, that v(G) = N(G) for every group G. Examples will be given following Corollary 3.1.6. Theorem 3.1.3:
Let G be a torsion free group with R+ = G.
be a nil associative ring with
r(G) = n, and let R
Then Rn+l = 0.
Proof: Let 0 ~ x € R, and let m be the smallest positive integer such that xm = 0. Suppose that m > n+l. Then there exist integers m-1 · al' ••• ,am-l' not all zero, such that I: a.x 1 = 0, i.e., i =1
am-lx
m-1
m-2 - -
a xm-1 = m-1
i aix .
I:
i =1
If a1
= 0,
1
= 1 , ••• ,m-2,
then am- 1
~
0,
and
o. Since G is torsi on free xm-1 = 0, a contradiction.
Therefore a;
~
0 for some
. 1
i.e., a xmm-2
=
~
i ~ m-2.
m-2 i+l Now 0 = a xm = - I: a.x m-1 i=l 1
m-3 . 1 I: a.x 1 + • Repeating the above procedure m-3 more time . 1 1= 1
yields that a1xm-l = 0, with a 1 ~a. Hence xm-l = 0, a contradiction. Therefore xn+l = 0 for all x € R. By the Nagata-Higman Theorem 26
[48, p. 274], R is nilpotent. Let d be the smallest positive integer satisfying Rd = 0. For every positive integer k, let (O:Rk) = {x E Gl xy = 0 for all y E Rk}. Clearly (O:Rk) is a pure subgroup of G for every positive integer k, and the chain 0 < (O:R1) < (O:R 2 ) < .•. < (O:Rd-l) = G is properly ascending.
d
~
n+l,
Hence
[36, vol. 1, p. 120, Ex. 8 (b)]. r(G) = n <
3. 1.4: Let G be a torsi on free group with following are equivalent: Corollar~
1) Every associative ring 2)
Every associative ring
3)
\I(G)
~
=.
The
R with R+ = G is nilpotent. R with
R+ = G is ni 1.
n.
Proof: The implications 1) • 2) and 3) • 1) are obvious. The implication 2) • 3) is an immediate consequence of Theorem 3.1.3. Theorem 3.1.5:
Let G be a torsion free group with 2n-l be a nilpotent ring R+ = G. Then R + 1 = 0. Proof:
Put
R(l)
=
r(G) = n <
=.
Let R
R,_ and for every positive integer k inductively define
R(k+l) to be the subring of R generated by elements of the form {ax, ybl a, bE R, x,y E R(k)}. Put vk = R(k)+ ~ Q+. Now Vk is a vector space over Q for every positive integer k, and dim0v1 = n. The descending chain v1 ~ v2 ~ ... terminates at 0, and has length at most n. Hence V = 0 ' and so R(n+l) = 0. n+ 1 k
Claim:
For every positive integer k,
R2 +l
c
R(k+ 2 ).
For k
= 0 the
claim is clearly true. Suppose R2k+l c R(k+ 2 ) for some positive integer k. 2k+l + 1 2k+l 2k+1 Let x E R Then x = y·z with y E R or z E R . Hence y E R(k+ 2 ) or z E R(k+ 2 ), and so x = y.z E R(k+ 3). Therefore
R2n-l + 1
c
R( n+ 1) = 0.
27
Corollary 3.1.6: Let G be a torsion free group with following are equivalent: 1)
Every ring R with
2)
N(G) ~ 2n-l.
r(G) = n <
~.
The
R+ = G is nilpotent.
Webb, [72, Theorem], has shown that if G is a torsion free group with r(G} = n < ~. then v(G) ~ n or v(G) = ~. and N(G) ~ 2n-l or N(G) = ~. These results are consequences of ~orollaries 3.1 .4 and 3.1.6 respectively.
A group G satisfying v(G)
Example 3.1.7:
< N(G) < ~.
For i = 1 ,2,3 let Gi be a rank one torsion free group with t(G1) = [(1,1, •.• ,1 , ••• )], t(G2) = [(2,2, .•• ,2, ••• )], and -3
Put G = G.• Clearly R5 = 0 for every ring i=l 1 R with R+ = G. Let ei E G1 , i = 1,2,3, with h(e 1 ) = {1 ,1, ••• ,1 , .•• ), h{e 2 ) = {2,2, ••• ,2, ••• ). and h(e 3) = (4,4, ••. ,4, ••• ). The products t{G3) = [{4,4, ••• ,4, •.• ) ].
for i+j e. •e. 1
J
~
3
=
otherwise
induce an associative ring structure R on G with e~ = e 3 ~ 0. R3 ~ 0, and v(G) = 3 by Corollary 3.1.4. Let ei'
i
=
Hence
1,2,3, be as above and define for i = j < 3 otherwise
These products induce a non-associative ring structure S on b with (e 1 ·e 1 ) • (e 1·e 1) = e4 . Hence s4 ~ 0, and N(G) = 4 by Corollary 3.1.6. In this example tne bounds of Webb's Theorem are precisely attained. another such example see [72].
For
Example 3.1.8: A group G satisfying v(G) < N(G) = ~. For i = 1 ,2, let G. be a rank one torsion free group with 1 t( G1 ) = [ { 1 ,0, 1 ,0, ••. , 1 ,0, ••• ) ] , and t( G2) = [ l ~. 1 .~. 1 , ••••~. 1 , ••. ) ] • Let ei E Gi , i = 1 ,2, w1 th hl e 1) = (1 , 0,1 , 0, ••• , 1 ,0, •.• ) , and h(e 2) = l~,l,~,l, ••••~,l, ••• ). Put G = G1 ~ G2 • It is easily verified 28
that v(G) = 2.
The products for i
= j
2
otherwise induce a non-associative ring Suppose that e 2 E Rn. Then positive integer n, and so
structure R on G. Clearly e2 E R = R1 • e2 = e1 ·e 2 E Rn+ 1 • Hence Rn I' 0 for every NlG) = ~.
An important problem in abelian group theory is to determine when a mixed group G sp~its into a direct sum of a torsion group and a torsion free group, [36, vol. 2, p. 185]. It is interesting to note that this occurs when v(G) < ~. [18, Theorem 3]. Theorem 3.1.9: Let G be a mixed group. If vlG) <~, then G = Gt@H, with H torsion free, and Gt the maximal divisible subgroup of G. Proof: Let D be the maximal divisible subgroup of G. Then G = DG) H, Proposition 1.1.10. lf D is not torsion then Q is a direct summand of G, Proposition 1.1.3, and so vlG) =· ~. a contradiction. Suppose that H is not torsion free. Then Z(pk) is a direct summand of G, p a prime 1 ~ k < ~. Proposition 1.1.4, and Proposition 1.1.3, and so v(G) = ~. a contradiction. Corollary 3.1.10: Let G be a mixed group. v(G) < ~ if and only if G = D~ H, with D a divisible torsion group, and H a torsion free group with v(H) < ~. Proof: If v(G) < ~. then the fact that G = D~ H, D a divisible torsion group and H torsion free with 'loi(H) < ~ follows -almost immediately from Theorem 3.1.9. Conversely, suppose that G = D(f) H, D a divisible torsion group, and H a torsion free group with v(HJ = n < ~. Let R be an associative ring with H+ = G. Clearly u is an ideal in R, and (RIO)+~ H. Hence (R/D)n+l = 0, or Rn+l =D. Therefore R2n+ 2 c o2 = 0 by Theorem 2.1.1, i.e., v(G) ~2n+l. The problem of classifying the groups with finite nilstufe therefore reduces to the torsion free case. 29
Question 3.l.ll: In the proof of Corollary 3.1.10 it was shown that if D is a divisible torsion group, and H is a torsion free group 111ith v(H) = n l a positive integer, and for every positive integer k satisfying n < k < 2n+l, there exist a div1sible torsion group D, and a torsion free group H, with v(HJ = n, such that v(D (t) HJ = k? By Corollary 3.1.6 the direct sum of n rank one torsion free groups has strong nilstufe ~ 2n-l or co. This can be improved upon as follows: Theorem 3.1.12:
Let Gi
be a homogeneous torsion free group of nonn
i = 1, •.• ,n, and let G = (!) Gi. Then v(G) ~ 2" 1. i=l Proof: For n = the theorem is true, Theorem 2.1.2. Let n > 1, and su~pose the theorem is true for the direct sum of less than n homogeneous groups of non-idempotent type. If t(Gi) = t(GjJ for 1 ~ i ~ j ~ n, then G is the direct sum of less than n homogeneous groups of non-idempotent type, and v(G) < 2n-l - 1 by the induction hypothesis. After possibly relabelling the summands Gi of G, it may be assumed that t(Gi) i t(Gn) for all l ~ i ~ n-1. Let R be an associative ring with R+ = G, let n g E G, anolet gn E Gn. Then g.gn = I: h., with hiE Gi' i = l, ••• ,n. i=l 1 idempotent type,
Suppose h.1 ~ 0 for some 1 -< i -< n-1. Then by Property 1.3.1.3), and Consequence 1.3.2, t(G1.) = t(h.) > t(g·g ) > t(gnJ = t(G ), a contradiction. 1 n n Hence RG nc G , and similarly G R c Gn, i.e., G is an ideal in R. n n_2n-l n 2n- 1 Put R= R/Gn. By the induction hypothesis R 0, or R c G - n 2n 2 . n However Gn isnii,Theorem2.1.2,andso R =Gn=O, 1.e., v(G)~2-l. The conditions in Theorem 3.1.12 do not impose a bound on NtG) as was seen in Example 3.1.8. n Corollary 3.1.13: Let G = _@ G; , wi th Gi a rank one torsion free group, 1 =1 i = l, •.• ,n. The following are equivalent. t(Gi) is not idempotent for all I < i < n. 2) v(G) ~ n. Proof: 1) • 2): v(G) < 2n-l by Theorem 3.1.13, and so v(G) < n by Corollary 3.1.4.
1)
30
If t(Gi) is idempotent for some 1 ~ i ~ n, then additive group of a subring of Q, 1.48, and so v(G) = m.
2) •1):
§2:
Gi
is the
Nilpotence w1thout boundedness conditions, and generalized nilpotence.
In the previous section, groups G were considered for which there exists a positive integer n such that Rn = 0 for every (associative) ring R with R+ = G. In this section the bound on the degree of nilpotence will be deleted. Afterwards, we will go a step further by considering the generalized nilpotent rings which were introduced by Levitzki [50]. Theorem 3.2.1: lW.J. Wickless) [74J: Let G be a group. Every (associative) ring R with R+ = G is nilpotent if and only i f G = D~ H with D a divisible torsion group, and H a reduced torsion free group such that every (associative) ring S with s+ = H is nilpotent. Proof: Suppose that every (associative) ring R with R+ = G is nilpotent. If Gt is not divisible, then G has a non-trivial cyclic direct summand, Proposition 1.1.4 and Proposition 1.1.3, and so a non-nilpotent associative ring structure may be defined on G. Therefore G = D~ H, D divisible, and H torsion free, Proposition 1.1.2. Clearly every (associative) ring S with s+ = H must be nilpotent. If H is not reduced, then Q+ is a direct summand of G, Proposition 1.1 .3 and [36, Theorem 21.2], and so a non-nilpotent (associative) ring structure can be defined on G, a contradiction. Conversely, suppose that G = D~ H, D a divisible torsion group, H torsion free, and such that every (associative) ring S with s+ = H is nilpotent. Let R be an (associative) ring with R+ =G. Clearly D is an ideal in R, and R = R/D is nilpotent. Therefore there exists a positive integer n such that Rn = 0, or Rn c D. Hence R2n c o2 = 0, Theorem 2.1.1. Theorem 3.2.1 reduces the study of groups allowing only nilpotent (associative) ring structures to the torsion free case. An immediate consequence of Corollaries 3.1.4 and 3.1.6 is the following: Observation 3.2.2: (associative) ring
Let G be a torsion free group with r(G) < R with R+ = G is nilpotent if and only if
m.
Every
31
(v(G)
< ~)
N(G)
< ~.
The proof of the following theorem of W.J. Wickless [74, Corollary to Theorem 2.1] depends on results .of R.A. Beaumont and R.S. Pierce [7], (8], and will be given in Chapter 5 section 4. Theorem 3.2.3: Let Gi be a finite rank torsion free group such that every associative ring Ri with R:1 = G.1 is nil potent, = l ••.. ,k. Then every k associative ring R with R+ = (f) G.1 is nilpotent. i=l Corollary 3.2.4:
Let Gi
be a finite rank torsion free group with k
i = l , ••• , k • Then v((i;) G.)<~. i=l 1 Theorem 3.2.3 and Observation 3.2.2.
v( Gi ) < ~. Proof:
The finiteness of the number k of summands Gi in Corollary 3.2.4 is obviously necessary. The same is true for Theorem 3.2.3. Example 3.2.5:
(Wickless [74, pp. 253-254]):
Let Gi be a rank one torsion free group with t(Gi) = {i,i, ••. ,i, •.. ) , and let G = .ro Gi. For every positive integer i, let ei E Gi with 1<w
h(ei) = (i,i, ... ,i, ..• ). The products ei·ej = ei+j induce an associative ring structure R, with R+ = G. For every positive integer n, e~ = en 1 0, and so R is not nil potent. However every ring Ri with R~ = Gi, i < w, is nilpotent, Corollary 2.1.3. (The ring of polynomials constructP.d by Wickless, [74, pp. 253-254], is isomorphic to the above ring).
Question 3.2.6:
Does Corollary 3.2.4 remain true if v is replaced by N?
Definition: An associative ring R is said· to be left (right) T-nilpotent if for every sequence {a.} c R, there exists a positive integer n such . 1 1 <w that a1.a 2 ··~ an= 0 (0 =an-an-I .•. a2a1 ). This concept was initiated by Levitzki [50], and was named T-nilpotence by Bass [4]. Theorem 3.2. 7
(B.J. Gardner [42, Proposition 3.1 ]) :
Every associative ring if every associative ring 32
R with R with
R+ = G is left T-nilpotent if and only R+ = G is right T-nilpotent.
Proof: Consider the opposite ring R0 P of every associative ring R with R+ = G. Clearly R is left T-nilpotent i f and only if R0 P is right T-nilpotent. From now on T-nilpotence is meant to be left T-nilpotence. Theorem 3.2.8: Let G be a group. Every associative ring R with R+ = G is T-nilpotent if and only if G = D@H with D a divisible torsion group, and H a reduced torsion free group such that every associative ring S with s+ = H is T-nilpotent. Proof: Suppose that ever'y associative ring R with R+ = G is T-nilpotent. If Gt is not divisible then G has a non-trivial cyclic direct summand, and so there exists an associative ring R with R+ = G such that R is not T-nil potent, a contradiction. Hence G = D@ H with D a di visible torsion group, and H a torsion free group. If H is not reduced, then Q+ is a direct summand of G and again G allows a non-T-nilpotent associative ring structure, a contradiction. Conversely let G = D(.i) H, D a divisible torsion group, and H a torsion free group such that every associative ring S with s+ = H is T-nilpotent. Let R be an associative ring with R+ = G, and let {a.}. c: R. For 1 1<w a € R let a denote the ideal class of a in R = R/D. Clearly R is T-nilpotent, and so there exist positive integers n1 < n such that al-a2 ... an = 0, aud a +la +2 •.. a = 0. Therefore a 1 ..• a € D, nl n 2 n nl 1 and a +1 .•. a € D. However, D = 0, Theorem 2.1.1 and so nl n a 1 -a 2 ... an= 0. As was the case for groups with finite nilstufe, and groups which admit only nilpotent (associative) ring structures, the classification of groups which admit only T-nilpotent associative ring structures reduces to the torsion free case. Observation 3.2.9: Let Gi be a finite rank torsi on free group such that ,k every associative ring Ri with R:1 = G.1 is T-nilpotent, i = 1 k Then every associative ring R with R+ ~ Gi is nilpotent. i =1 Proof: Since every T-nilpotent ring is nil, every ring Ri with R~ = Gi G.1 is nilpotent, is ni 1. By Theorem 3.1 ,3, every ring R1. with R:1 '0
•
0
0
33
i = l, •.• ,k. +
R =
k
(:i-)
i=l
G.
Hence Theorem 3.2.3 yields that every ring
R with
is nilpotent.
1
Another generalization of nilpotence, also due to Levitzki [50], is the following: Definition:
Let R be an associative ring.
For a an arbitrary ordinal
define Ra+l to be the s~bring of R generated by {xalx ERa, a E R}. For s a limit ordinal, put R6 = n Ra. An associative ring R is said a
R witn Ra = 0 if a reduced s+ = H
Question 3.2.ll: Let G; be a finite rank torsion free group, such that for every associative ring R.1 with a. R~1 = G.1 there exists an ordinal a; (depending on
R;)
such that R11
associative ring with that
Ra = 0?
R+ =
~
i=l
G..
= 0, i = l, ••• ,k. Let R be an Does there exist an ordinal
a such
1
The concept of a-ni lpotence suggests the following: Definition: Let G be a group. The generalized nilstufe of G, denoted gv(G}, is the first ordinal a such that Ra = 0 for every associative ring R with R+ = G. If no such ordinal exists, put gv(G} = -. Fuchs, [36, problem 94], introduced the following: Definition: Let G be a group. The absolute annihilator of G = n {(O:R}jR is an associative ring with R+ = G}. Here (O:R} = {a E RlaR = Ra = 0}. Gardner, [41, p. 386], introduced the following ascending annihilator series for a group G: Put G(l} = the absolute annihilator of G. For 34
every ordinal a, put G/G(a) define G(s) = u G(a). a<S Theorem 3.2.12: Let then gv (G) ~ a+ 1.
(1)
G be a group.
G(a+l)/G(a).
For
If G = G(a)
for some ordinal
J3
a limit ordinal,
a,
~:
Let R be an associative ring with R+ = G. We wish to show that Ra is annihilated by G(a) = G. t4ore generally, it will be shown that G(a) annihilates R8 for every ordinal s. For s = 1, the definition of the absolute annihilator of G assures that G(l) annihilates R = R1 . Suppose that G(Y) annihilates RY for every ordinal y < S. 1)
Let B = y+l.
Then
R8 ·G(B) = Ry·R·G(Y+l) ~ Ry · G(Y) = 0.
2) Let B be a limit ordinal, and let a E G(B). Y < s. Hence R8·a c RY·a = 0.
Then a E G(y)
for some
Example 3.2.13, [23, Example 2]: For every positive integer i, let Gi be a rank one torsion free group with t(Gi) = [(i,i, ••• ,i, ••• )]. Put G = ·~ G. it is readily seen (Example 3.2.5) that gv(G) > n for every 1~
1
positive integer n. Let R be an associative ring with R+ = G. Every elementin Rw hastype [(oo,oo, ••• ,oo, ••• )] in G. Howevernonon-zero element in G is of idempotent type. Hence Rw = 0, and gv(G) = w. Question 3.2.14: with gv(G) = a?
For a an arbitrary ordinal, does there exist a group
G
35
4 Other ring properties
Definition: Let w be a ring property: A group G is said to be an (associative} w-group if there exists an (associative} n-ring R such that R+ = G. If G is not (associative} nil, and every (associative} ring R satisfying R2 ~ 0, and R+ = G is a n-ring, then G is said to be an (associative} strongly w-group. The above terminology was first employed by Beaumont and Lawver, (6], in studying the associative Jacobson semisimple, and the associative strongly Jacobson semisimple groups. Throughout the remainder of this section, the rings considered will be assumed to be associative. Theorem 4.1.1: Let G be a group which is not torsion free. are equivalent: 1} 2} 3}
4) 5}
G is a G is a G is a G is a G= @ a
The following
field group. division ring group. simple ring group. prime ring group. Z(p), p a prime, a an arbitrary cardinal.
Proof:
1} .. 2}: Obvious 2} • 3}: Obvious 3} • 4}: Obvious 4} • 5}: Let R be a prime ring with R+ =G. For distinct primes p,q the primary components Gp and Gq of Gt are ideals in R satisfying G ·G = 0. Hence Gt = G for some prime p. For every positive inte~erq k, I = pG[pk] is ~n ideal in R satisfying Ik = 0. Hence I = 0, and Gt is an elementary p-group, i.e., Gt = ~ Z(p}, [36, Theorem 8.5]. 36
Therefore G = Gt(i:)H, with H a torsion free group,
Proposition 1.1 .2. Let x E H, and let I = • Then Gt·I = 0, and so I = 0. Since H is torsion free, this implies that x = 0 which in turn yields that H = 0. 5) ~ 1): If G has form 5), then G is the additive group of a field extension of degree a of a field of order p, and so 5) • 1). Definition: A ring R is Jacobson semisimple (semiprimitive) if its Jacobson radical, J(R) = 0. In what follows semisimple means Jacobson semisimple (semiprimitive). The reader is referred to Chapters 1 and 2 of [45] for basic facts about semisimple rings. Theorem 4.1.2: Let G be a torsion group. The following are equivalent: 1)
G is a semisimple ring group.
2)
G is a semiprime ring group.
3) G is an elementary group, i.e., the order of every element in G is square order free. Proof: 1) • 2): Let R be a semisimple ring with R+ =G. Every nilpotent ideal in R is contained in J(R) = 0. Hence R does not possess non-zero nilpotent ideals, i.e., R is semiprime. Let R be a semiprime ring with R+ = G. Let p be a prime, and put I = pG[p 2J. Then I is an ideal in R satisfying 12 = 0. Therefore I = 0, and so the order of every element in G is square free.
2)
~
3):
3) ~ 1): Let G be an elementary group. For every prime p for which GP ~ 0, there exists a field FP with F; = Gp' Theorem 4.1.1. Put R =(t){FpiP a prime such that Gp with
~
0}.
Clearly, R is a semisimple ring
R+ = G.
The equivalence of 1) and 3) as well as a general treatment of semisimple and strongly semisimple ring groups may be found in [6).
37
Theorem 4.1.3: equivalent:
Let G be a torsion free group.
1)
G is a field group.
2)
G is a division ring group.
3)
G is a simple ring group
The following are
4) G is divisible. i.e •• G ~ (i) Q+. oc an arbitrary cardinal oc (Proposition 1 .1.3). Proof: The implications 1) • 2) • 3)
are obvious.
3) • 4): Let R be a simple ring with R+ =G. Then nR is an ideal in R for every nonzero integer n. Hence (nR)+ = nG = G. and so G is divisible. 4) • 1):
Let G =
ffi Q+. oc an arbitrary cardinal. Then G is the
oc additive group of a field extension of degree oc of the field of rational numbers. Notation: (see [1.4.8]): Let {qi li E IJ be a set of primes. is the subring of Q generated by {q- 1 li E 1}.
Z(qi 1 1i
E I)
The following theorem is due to Beaumont-Zuckerman [11]. and Redei-Szele [54].
Theorem 4.1.4:
Let R be a non-zeroring with R+ a rank one torsion free group. Then R ~ mZ(qi 1 li E I). with m a positive integer satisfying (m,qi)=l forall i€1. Proof:
38
[36. Theorem 121.11. and 1.4.8.
The following results are due to Beaumont-Lawver [6]. Lemna 4. 1.5: The maximal modular ideals in mZ(q: 1 li E I) ~ Q are precisely the ideals pmZ(qi 1 li E 1), p a prime such that 1 p ¢ {qi liE 1}, and (m,p) = 1. Proof: Let p be as in the statement of the lemma. m ¢ pmZ(qi 1 li E I), and so pmZ(q:1 1 li E I) is a proper ideal in mZ(q:1 1 1i E 1). Let
~r E mZ(qi 1 li E I) be written in reduced form, and suppose that 1 !!!!. s ¢ pmZ(q:1 1i E I). Then p+r. Hence there exist integers t,u such that tr +up= 1. Therefore m = trm + upm belongs to the ideal generated by
pmZ(qi 1 li E I), and ~r, -1
mZ( qi
Ii
Since
i.e., this ideal is the entire ring
E I). (m,p) = 1, there exist integers
x,y such that xm + py = 1.
1 Let mr element in pmZ(q:1 1 li E I), s E mZ(q:1 li E I). Then xm • .!!!!. s =.!!!!.+an s i.e., xm is a unity modulo pmZ(qi 1 li E 1), and so pmZ(qi 1 liE I) is a modular ideal.
Let I~ 0 be a proper ideal in mZ(qi 1 1i E I)~ Q, and let mr be the smallest positive integer in I. Clearly I= rmZ(qi 1 1i E 1), and (rm, qi) = 1 for all i E I. Since I. is a proper ideal in mZ(qi 1 1i E 1), r ~ 1. Let p be a prime such that plr. Then p ¢ {qi liE I}, and I= pmZ(qi 1 li E I). Hence every maximal ideal M in mZ(qi 1 1i E I) is of the form M= pmZ(qi 1 li E I). Suppose that M is modular. Then there 1
1
-
mr E mZ(q: 1i E I) such that .!!!!. • m = m + pmf with s and s 1 s 1 s s 1 1 products of primes in {qili E 1}. Hence mrs - ss = pr 1s. Now (ss 1 , p) = 1, and so (m,p) = 1. Therefore every maximal modular ideal in mZ(qi 1 1i E I) is of the form pmZ(qi.ll i E I) with p a prime such that P ¢ {qi liE 1}, and (m,p) = 1. exists
Theorem 4. 1.6: equivalent: 1)
2) 3)
Let G be a rank one torsion free group.
G is a strongly semisimple ring group. G is a semisimple ring group. t(G) = [(k 1 , k2 , ••• ,km•···)] is idempotent, with
The following are
kn = ~ for all
n 39
(i.e.
G ~ Q+)
or k = 0 for infinitely many positive integers n. n 1) • 2): Obvious.
Proof: 2) • 3):
Corollary 2. 1.3 and Theorem 4.1.4.
If k = ~ for all positive integers n, then the one and only n non-zeroring with additive group G is isomorphic to Q, and so G is a strongly semisimple ring group. Suppose that kn = 0 for infinitely many positive integers n. By Corollary 2.1.3, G is not nil. Let R be a ring satisfying R+ = G, and R2 + 0. Then R ~ mZ(qi 11i E 1), Theorem 4.1.4. The.primes {q.liEI} arepreciselythoseprimes p forwhich 1 tp(G) = ~. Therefore there are infinitely many primes p such that p ¢ {qi li E I}, and (m,p) = 1. Let ~r E J(R), with s a product of 3) • 1):
primesin {qiliEI}. Then ~rE pmZ(qi 1 1iEI),p¢£qiiEI}, and (m,p) = 1. Lemma 4.1.5. Therefore infinitely many primes p satisfying p f {qili E 1}, and {m,p) = 1, divide r. Hence ~r = 0. The equivalence of 1) and 2) in the above theorem is true in general for strongly indecomposable finite rank torsion free groups. The proof of this, and other results concerning semisimple, and strongly semisimple ring groups will be given in section 6 of this chapter, see Theorem 4.6.17. Much additional information about {strongly) semisimple ring groups may be found in [6]. A natural complement of the semisimple ring group is the radical ring group, i.e., the additive group of a ring R satisfying R2 i 0, and J(R) = R. These groups have been studied by F. Haimo [44]. Some of his results, and others, will be presented here. Lenvna 4.1.7: Let G be a radical ring group. ring group for every group H.
Then
Gf.i:)H
is a radical
Proof: Let S be a radical ring satisfying s2 i O, and s+ = G. Let T be the zeroring with T+ = H. The ring direct sum R = S (f) T, satisfies R2 iO, R+=G~H, and JlR)=R. Corollary 4.1.8: If G possesses a subgroup isomorphic to Q+ ~ Q+ , G is a radical ring group. Proof: 40
G..,.Q+@Q+@H,
Proposition 1.1.10.
then
Therefore by Lemma 4.1.7, it
suffices to show that Q+ (1) Q+ is a radical ring group. Let a,b be independent elements in Q+@ Q+. The products a2 = b, ab = ba = b2 = 0 induce a ri11g structure R on Q+ ~ Q+ satisfying R2 '! 0, but R3 = 0. Hence J(R) = R, and Q+c:i:'Q+ is a radical ring group. Lemma 4.1.9:
If Hom(OOG, H) '! 0,
then
G(t) H is a radical ring group.
Proof: Let 0 '! $ € Hom(Q®G, H). The products (g 1+h 1) • (g 2+h 2 ) = ~(g 1~g 2 ) for all gi € G, hi € H, i = 1,2 induce a ring structure R on G satisfying R2 '! 0, but R3 = 0. Hence R is associative, and J(R) = R, i.e., R is a radical ring. Theorem 4.1.10:
GroG is a radical ring group if and only if G is not nil.
Proof: If G is not nil, then Han(OOG, G) '! 0, a radical ring group by Lemma 4.1.9. Supposethat G isnil. ""
~
8 copies
Hom(~.G)=O,
1.2.1, and so G@ G is
Then Mult(G@G) ... Hom[(<JBG)@((}BG), andso G
<JBG)
Theorem 4.1.11: Let G be a divisible group. G is a radical ring group if and only if G ~ Q+ and G is not a torsion group. Proof: A divisible torsion group is nil, Theorem 2.1.1, and Q+ is a strongly semisimple ring group, Theorem 4.1.6. By Proposition 1.1.3 and Corollary 4.1.8 it suffices to show that G = Q+(t) 0, with D a divisible torsion group, is a radical ring group. Lemma 4.1.7 and Proposition 1.1.3 reduce the problem to showing that Q+@ Z(p""), p an arbitrary prime, is a radical· ring group. Now Z(p""J is a homomorphic image of Q+, and Q+(X)Q+.,.Q+. Hence Hom(Q+@Q+, Z(p""J) '! 0, and Q+@Z(p"") is a radical ring group by Lemma 4.1.9. Theorem 4.1.12: Let G be a rank one torsion free group. Then G is a radical ring group if and only if t(G) is idempotent, with finitely many zero components, but not all of its component~ = ""· Proof: If t(G) does not satisfy the conditions of the theorem, then either G is nil, or G is a strongly semisimple ring group, Corollary 2.1.3, and Theorem 4.1.6. Let G satisfy the conditions of the theorem.
There is a ring R such 41
R+=G, and R""Z(q71 1jie:I), with {q.jie:I} thesetofprimes p, 1 satisfying tp(G) = =. An immediate consequence of Lemma 4.1.6 is that J(R) = p1 ••• pkR' with {pl , ••. ,pk} the set of primes p for which tp(G) = 0. Now p1 ••• pkR+ ""G, so J(R) is a radical ring with + [J(R)] ""G. The above proof of Theorem 4.1.12 was essentially taken from the proof of [6, Theorem 3.2]. For an alternative proof, see [44, Theorem 4]. Lemma 4.1.13: Let G be a torsion group. G is a radical ring group if and only if GP is a radical ring group for some prime p • . Proof: If Gp is a radical ring group, then so is and Lemma 4.1.7.
G by Proposition 1.1.1
Conversely, suppose that R is a radical ring with R+ = G, and R2 ~ 0. Then R is a ring direct sum, R = ~. Rp• with R; = GP for every 2 p pr1me 2 prime p. Since R ~ 0, there exists a prime p such that Rp ~ 0. Now RP ~ R, and so J(Rp) = Rp n J(R) = Rp n R = Rp• i.e., Gp is a radical ring group. Lemma 4.1.13 reduces the study of torsion radical ring groups to the p-primary case, p a prime. Lemma 4.1.14: Let G be a p-primary group, p a prime. G is a radical ring group if and only if G t Z(p), and G is not divisible. Proof: If G is divisible, then G is nil, Theorem 2.1. 1, while the one and only non-zeroring with additive group Z(p), is a field. Suppose that G ~ Z(p), and that G is not divisible. Then G = Z(pk)@ H, k a positive integer, Propositions 1.1.4 and 1.1.3. Put Z(pk) = (a). If H = 0, then k > 1. The product a·a = pa induces a ring structure R on G satisfying R2 ~ O, but Rk+l = 0. Hence G is a radical ring group. If H ~ 0, choose h0 e: H with lhol = p. The products a2 ~ h0 , ah = ha = h1h2 = 0 for all h,h 1 ,h 2 e: H induce a ring structure R on G satisfying R2 ~ 0, R3 = 0. Hence G is a radical ring group. Several of the results in this section were generalized by J.D. Reid, [57], and will be presented in section 6 of this chapter.
42
§2:
Principal ideal and Noetherian rings.
Lemma 4.2.1: Let G = H@K, H '! 0, K -1 o, be an associative strongly principal ideal ring group. Then H and K are either both cyclic or both associative nil. Proof: Suppose that H is not associative nil. Let S be an associative ring with s+ = H, and s 2 '! 0, and let T be the zeroring on K. The ring direct sum R = S (+'> T is an associative ring satisfying R+ = G, and R2 '! 0. Since T is an ideal in R, T =<X>. Clearly K = T+ = (x). Therefore K is not associative nil. The above argument, interchanging the roles of H and K, yields that H is cyclic. Corollary4.2.2: Let G=H(+IK, HIO, K-!0 beanassociativestrongly principal ideal ring group. Then H and K are cyclic. Proof: It suffices to negate that H and K are both associative nil. R = (G,·) be a ring satisfying R2 '! 0. 1) R= and and
=
Suppose that R2 K. There exist h0 E H, k0 E K, such that • Let h E H. Since h € R, there exists an integer n, x E R2 such that h = n(h 0 + k0 ) + x. However x E K. Hence h = nh 0 , H is cyclic, contradicting the fact that H is associative nil.
2) Suppose that R2 ! K. For all g1 ,g 2 E G, define g1 * nH the natural projection of G onto H. Then S = (G,*) satisfying s 2 =H. The argument employed in 1) yields that contradicting the fact that K is associative nil. Theorem 4.2.3: 1) 2) 3)
Let
Let G '! 0 be a torsion group.
g2 = nH(g 1·g 2 J, is a ring K is cyclic,
The following are equivalent:
Either G is cyclic, or G""' Z(p)@ Z(p), with p a prime. G is a strongly principal ideal ring group. G is an associative strongly principal ideal ring group.
1) • 2): Non-trivial cyclic groups are clearly strongly principal ideal ring groups. Suppose that G = (x 1 ) (f) (x 2 ) with lx;l = p a prime, i = 1,2. Let R be a ring with R+ = G, and R2 '! 0, and let I be a proper ideal in R. Then III = 0, or p, and so I is generated by a single element. We may assume that R 'I <X;>, i = 1,2. Hence <X;>+= (xi) for i = 1 ,2. This implies that
~:
43
X· X.
J
l
k.x., 0 -< k.l < p if i = j, i l l {. = 0 if ; ~ j, ; ,j = 1,2.
= 1,2, either k1
~
or k2
0
~
0.
Put I = <x 1 + x2>. Suppose that k1 ~ 0. Let r,s be integers such that rk 1 + sp = 1. Then rx 1(x 1 + x2) = rk 1x1 = (l-sp)x 1 = x1 • Hence x1 € I, and so (x 1 + x2 ) - x1 = x2 E I. Therefore I = R. If k2 ~ 0, then the above argument, reversing the roles of the indices 1 ,2 again yields I = R. 2) • 3): It suffices to show that G is not associative nil. case by Theorem 2.1.1.
This is the
3) • 1): Suppose that G is an associative strongly principal ideal ring group. . If G is indecomposable, then G = Z(pn), p a prime, 1 -< n -< ~. Corollary 1.1.5. If n = ~. then G is divisible, Proposition 1.1.3, and so G is nil, Theorem 2.1.1, a contradiction. Hence G is cyclic. Suppose that G = H(t) K, H ~ 0, K ~ 0. By Lemma 4.2.1, either H and K are both cyclic, or both associative nil. If H and K are both associative nil, then they are both divisible, and so G is nil, Theorem 2. 1.1, a contradiction. Therefore G = (x 1) (f) (x 2 J, with lx; I = n1 , i = 1,2. If (n 1 , n2) = 1, then G is cyclic. Otherwise let p be a prime divisor of
m.
(n 1 ,n 2). Then G = (y 1) (t) (y 2 )@ H, with IY; 1 = p 1 , i = 1,2, and 1 ~ m1 ~ m2 • Since lY 1) (t) (y 2 J is neither cyclic nor associative nil, H = 0 by Lemma 4.2. 1. m -1 The products Y;·Yj = p 2 y2 for i,j = 1,2, induce an associative ring structure R on G with R2 ~ 0. Therefore R = <s 1y1 + s 2y2>, s 1 ,s 2 Every element x E R has the form m2-1 kxs 1y 1 + (kxs 2 + mxp )y2 , kx,mx integers.
integers. x
=
_
_
In particular
m2-1
y 1 - k s 1y 1 , and y2 - (k s 2 + m p )y 2 • Hence if m2 > 1, yl y2 y2 m -1 k s 1 = l(mod p), and k s 2 + m p 2 : l(mod p), which implies that yl y2 y2 m -1 p k , and p s 2 • However k s? + m p 2 = O(mod p), so either yl yl "" yl
t
plky 1 or pls 2 , 44
t
a contradiction.
Therefore m2
= 1 = m1•
Theorem 4.2.4: ring groups.
There are no mixed (associative) strongly principal ideal
Proof: Let G be a mixed associative strongly principal ideal ring group. G is decomposable, Corollary 1.1.5, so by Lemma 4.2.1, G = H(i) K, H ~ 0, K ~ 0, with H and K both cyclic, or both associative nil. 1) Suppose that H and K are both associative nil. There are no mixed associative nil groups, Theorem 2.1.1, so we may assume that H is a torsion group, and that K is torsion free. Let R be an associative ring with R+ = G, and R2 1 0. Clearly H is an ideal in R, and so H = . Let lhl = m. Then mH = 0. By Theorem 2.1.1, H is divisible, and therefore not bounded, a contradiction. 2) Suppose that H = (x), and K =(e) with lxl = n < ~. and lei = ~. The products x2 = xe = ex = 0, and e 2 = ne induce an associative ring structure R on G satisfying R2 ~ 0. Therefore there exist integers s,t such that R = <sx + te>. Every y € R is of the form y = mysx + (my + uyn)te, with my and uy integers. In particular, (me+ uen)t = I. Hence t = .:t. 1. Therefore mx + uxn = 0, so that nlmx. However x = mxsx = 0, a contradiction. Theorem 4.2.5: Let G be a torsion free associative strongly principal ideal ring group. Then G is either indecomposable, or is the direct sum of two associative nil groups. Proof: By Lemma 4.2.1 it suffices to negate that G = (x1)@ (x 2 ), i = 1,2. Suppose this is so. The products for i = j ,
i = 1 ,2
for i
i ,j
~
j,
xi
~
0,
1 ,2
induce an associative ring structure R on G satisfying R2 ~ 0. Therefore there exist nonzero integers k1 ,k 2 such that R = . Every x € R is of the form x = (rx + 3sx)k 1x1 + (rx + 3tx)k 2x2 , with rx, s , t X
X
integers.
r
XI
+ 3s
Xl
=
+1,
-
so that
rx
1
= -+l(rood
3).
However
+ 3t = 0, so that r = O(mod 3), a contradiction. xl xI xl Lemma 4.2.6: Let G and H be tors1on free groups with G ~H. Then G is an (associative) strongly principal ideal ring group if and only if H is. r
45
Proof: It may be assumed, w1thout loss of generality, that G; H. There exists a positive integer n such that nG ~ H, and nH ~G. Suppose that G is an (associative) strongly principal ideal ring group. Let R = (H,·) be an (associative) ring such that R2 I 0. The products g1*g 2 = (ng 1)·(ng 2 ) for all gl'g 2 E G, induce an (associative) ring structure S = (G,*). with s2 I 0. Let I , with g = nh, he: I. Clearly ~I. Let x e: I. Then nx E , and so nx = kg+ g*y + z*g• with k an integer, and y,z e: G. Therefore nx = n[kh + (nh)·(ny) + (nz)·(nh)]. Since H is torsion free, x = kh + (nh)·(ny) + (nz)·(nh) e: . Hence I= , and H is an (associative) strongly principal ideal ring group. The same argument, reversing the roles of G and H, shows that if H is an (associative) strongly principal ideal ring group, then so is G. Lemma 4.2.. 6 remains true if the word "strongly" is deleted. Corollary 4.2.7: Let G be a torsion free strongly principal ideal ring group. Then G is strongly indecomposable. Proof:
Theorem 4.2.5, Corollary 4.2.2, and Lemma 4.2.6.
Theorem 4.2.8:
Let G be a torsion group.
The following are equivalent:
G is bounded. G is an associative principal ideal ring group. G is a principal ideal ring group.
1) 2)
3)
Proof: 1) • 2): Suppose that nG = 0, n a positive integer. Then G = ~ rJ) Z(pk)], p a prime, with Pkln, and ak a cardinal number, Pin ak Proposition 1.1.9. There exists an associative principal ideal ring with unity
R k' with p
R:k = Hpk
for all
Pkln,
[36, Lemma 122.3].
The ring direct sum
R = k(£) R k is an associative principal ideal ring satisfying
P In P
and R2 '} 0. 2) 46
~
3):
Obvious.
R+ = G,
3) • 1): Let R be a principal ideal ring with Let n = lxl. Clearly nG = 0.
R+ =G.
Then
R =<X>.
corollary 4.2.9: Let G be a mixed group. If G is an (associative) principal ideal ring group, then Gt is bounded, and G/Gt is an (associative) principal ideal ring group. Conversely, if Gt is bounded, and if there exists a unital (associative) principal ideal ring with additive group G/Gt' then G is an (associative) principal ideal ring group. Proof: Let R be an (associative) principal ideal ring with R+ = G. Since Gt is an ideal in R, Gt = <x>, and nGt = 0, n = 1x1. Now G = Gt®H, H ~ G/Gt' Proposition 1 .1.2. Now R =, a E Gt, 0 ~ y E H. Suppose that R2 c Gt. Let h E H. There exists an integer kn such that 2 . 2 h = khy + b, with bE R . S1nce R Gt' b = 0, and h = khy. Therefore H = (y). Clearly H is an associative principal ideal ring group. If R2 $ Gt' then R = R/Gt is an (associative) principal ideal ring with ....+-2 R ~ G/Gt' and R ~ 0.
=
Conversely, suppose that Gt is bounded, and that there exists a unital (associative) principal ideal ring T with T+ = G/Gt. Then G"" Gt<±> G/Gt' Proposition 1.1.2. There exists a principal ideal ring S with unity, such that S+=Gt' [36,Lemmal22.3]. Let R=S(±)T, with e,f theunities of S and T respectively. Let I be an ideal in R. Then I = (I n S) (t) (I n T). Now Ins <1 S, and so I n S =<X>. Similarly I n T = . Clearly <X+ y> =I. However x = e(x+y) E <X+ y> and y = f(x+y) E <X + y>. Therefore I = <X + y>. Noetherian rings are rings satisfying the ascending chain condition for left or right ideals. The following is a departure from convention: Definition: A ring R will be said to be Noetherian, if every two sided ideal is finitely generated. The following trivic.l lemma will prove to be useful. Lemma 4.2.11. Let G = H(f) K, H ~ 0, K ~ 0 be a strongly Noetherian ring group. Then either G is finitely generated, or H and K are both nil. ~: Suppose that H is not nil. Let S be a non-zeroring with s+ = H, and 1et T be the zerori ng with T+ = K. The ring direct sum R = S (f) T satisfies R+ = G, and R2 ~ 0. Since T <1 R, T = ·
47
Clearly T = (t1 •••• ,tn), and so T is not nil. The same argument reversing the roles of H and K yields that H is finitely generated. Hence G is finitely generated. Theorem 4.2.12: Let G be a non-torsion free group. G is a strongly Noetherian ring group if and only if G is finitely generated. Proof: By Lemma 4.2.10, it suffices to show that if G is a strongly Noetherian ring group, then G is finitely generated. 1) Suppose that G is a torsion group. If G is indecomposable, then G"" Z(pn), p a prime, 1 ~ n ~ ao, Corollary 1.1.5. If n = ""• then G is nil, Proposition 1. 1.3, and Theorem 2.1.1, a contradiction. Hence H is cyclic. If G = H(f)K, H 'I 0, K 'I 0, and if G is not finitely generated, then by Lemma 4.2.11 both H and K are nil, and so G is nil by Theorem 2.1.1, a contradict1on.
2) Suppose that G is mixed. Then G is not indecomposable, Corollary 1.1.5, and so G = H<±> K, H 'I 0, K 'I 0. If G is not finitely generated, then by Lemma 4.2. 11, both H and K are nil. By Theorem 2.1.1 it may be assumed that H is a torsion group, and that K is torsion free. Let R be a Noetherian ring with R+ = G. Since ti <2 R, H = . Let n
=
k
rr lhil.
i =1
Then nH
= 0.
By Theorem 2.1.1, H is divisible, a
contradiction. Corollary 4.2.13: Let G be a strongly Noetherian ring group. and G/Gt are strongly Noetherian ring groups.
Then
Gt
i··,
Proof: If G is torsion free, the statement of the corollary is trivial. Otherwise G is finitely generated by Theorem 4.2.12, and so Gt and G/Gt are strongly Noetherian ring groups by Lemma 4.2.10. A torsion free strongly Noetherian ring group need not be finitely generated; e.g., Q+. However, we have the following: Theorem 4.2.14: Let G be a strongly Noetherian ring group. either indecomposable, or finitely generated.
Then
G is
Proof: By Lemma 4.2.11, it suffices to negate that G = H@ K, H 'I 0, K 'I 0, with H and K both nil. Suppose this is so. Let R be a ring 48
with R+ = G, and R2 ~ 0. Then Put x1 = h; + k; , i = 1 , ••• ,n.
R = <x 1 , ••• ,xn>' n a positive integer.
Suppose that R2 ~ K. Let h € H. There exist integers r 1 , ••• ,rn' and 2 n 2 x € R such that h = ~ r-{h. + k.) + x. However R c K. Hence i=l 1 1 1 n h = ~ r.h 1 , and H is finitely generated. This contradicts the fact that 1)
i =1
1
H is nil. 2) Suppose that R2 $ K. Let nH be the natural projection of G onto H. The products g1 * g2 = nH{g1 .g2) for all g1 ,g 2 € G induce a ring structure S on G such that 0 ~ s 2 ~H. The argument used in 1) shows that K is finitely generated, contradicting the fact that K is nil. For a discussion of the additive groups of Noetherian rings in accordance with the usual definition, see [36, section 126]. Fuchs, [36, Problem 97], asked whether or not free groups of large cardinality allow a Noetherian ring structure {Noetherian meaning left i~oetheri an, a1though the prob 1em is a1so open for Noetherian according to the definition given here). He noted [36, vol. 2, p. 311], that no Noetherian ring is known whose additive group is free and non-denumerable. Such a ring has been constructed by J.D. O'Neill, [52, Theorem 1], as follows: Let w1 be the first non-denumerable ordinal. Put A= Z[xi li € I], the commutative associative ring of polynomials in indeterminates {xi li € 1}, I= the set of ordinals less than ~· Let S =the set of elements in A which are relatively prime to every positive integer. The localization As is a Noetherian ring. The proof of this fact may be found in [52, Theorem 11.
49
§3.
Descending chain conditions for ideals. The additive groups of (left) Artinian rings have been completely
classified by Fuchs and Szele [70] as follows: Proposition 4.3.1:
G is the additive group of a (left)
A group
k.
G ""(t) Q+ (+) Z(p~) 0 Z(pJ.J) 1 a finite 13 p.,p. are primes, and are arbitrary cardinals,
(associative) Artinian ring if and only if
P/k lm, 0
with
where
a,l3
J
1
m is a fixed positive integer. Proof:
Let
descending chain of left ideals in k
such that
n!R = k!R
R.
Hence there exists a positive integer
n > k.
Put
m = k!
(mR)+ ""(t)Q(!) @ (£) Z(p00 ) . a p a prime ap
divisible, and so boundedjandso
for all
n:;, ...
R ::J 2! R ::J 3!
R be an Artinian ring.
Clearly
(mR)+
Clearly,
R+""(±)Q+(±) (£) ~ Z(p00)(f{i~)Z(p~). 13 J a p a prime ap
p. J
is
R+/(mR)+
is
primes,
and
p~lm,
m a fixed positive integer, Proposition 1.1.9. Put ~ (+) Z(p00) . Clearly, Dc(R+)l, andso D annihilates D= p a prime ap Lemma 2.1.
The complement of
is clearly annihilated by and so every subgroup of
D.
R;
Hence
co
1
i=2
is isomorphic to (+)Q+ which a D belongs to the annihilator of R,
Di ,
R.
If
i = 1 ,2, •.• ,
D possesses infinitely tnen
00
<±> D. :;, (f) D. :;,, • ,:;, (f) D. :;, • • • i=l
R+
D is an ideal in
many non-trivia 1 direct summands 00
in
1
of (left) ideals in
i=n R,
is. a properly descendir.g infinite chain
1
a contradiction.
Therefore
Z(p~),
®
D=
finite
primes. k
Conversely, let
k p.J m, .where
p1.
0
0
G ""® Q+ ('±) Z(p~) (f) Z(p .J) 1 a finite J
1
with
J
a,l3
are arbitrary cardinals; pi ,pj are primes; and m is a fixed positive integer. Let F be a field with F+ ""(+) Q+, S a zeroring with
a
S+ =
p~lm; 50
(±) Z(p~). 1 finite j
l, ... ,n;
k. Now (f) Z(pJ.J) 13 k
l, ..• ,nj.
n n. (+) ~ (£J j'=l k=l 13-
a prime, and
J
For
13j
finite, let
Tjk
be the sum of
~J·
copies of the ring of integers modulo p~. For ~- infinite, an J J associative commutative ring with unity Tjk can be constructed so that T;k =(f) Zlp~). and the only (left) ideals in Tjk are j n. n n. TJ.k' J:TJ.k•···•P JTJ.k = 0, [36, Lemma 122.3]. Put T = C±) d:> TJ.k" j=l k=l ring R = F(+) S 0 T is Artinian with R+ = G.
The
The ring R constructed above is commutative, associative, and possesses only finitely many ideals. These facts will prove to be useful later.· In this section it 111i 11 be shown that the additive groups of rings satisfying the descending chain condition for two sided ideals are precisely the additive groups of Artinian rings described above. The additive groups of rings possessing only finitely many ideals will be studied. The structure of the additive group of a ring possessing only finitely many ideals (satisfying the descending chain condition for two sided ideals) will be employed to obtain necessary and sufficient conditions for embedding the ring into a ring with unity possessing only finitely many ideals (satisfying the descending chain condition for two sided ideals). For the remainder of this section, ideal will mean two sided ideal and the "descending chain condition" will be abbreviated DCC. Theorem 4.3.2.
Let G be a torsion free group.
The following are equivalent: 1) G is the a~ditive group of an (associative) ring possessing only finitely many ideals.
2) G is the additive group of an (associative) ring satisfying the DCC for ideals. 3)
G ...
01
an arbitrary cardinal.
01
Proof: Clearly 1) ,. 2) and 3) ,. 1), the latter implication since <±> Q+ is the additive group of a field. It therefore suffices to show that: 2) .. 3): Let R be a ring satisfying the DCC for ideals, with R+ =G. For every prime p, and positive integer n, pnR ~ pn+lR. Hence there exists · n sue h th at a pos1· t1· ve 1nteger
pnR = pn+ l R• Th ere f ore f or a e: G, th ere 51
exists b € G such that pna = pn+lb, or pn(a - pb) = 0. Since G is torsion free, a= pb, and so G is p-divisible for every prime p. Hence G is divisible, and G ~<±> Q+, Proposition 1.1.3. a Observe that the above theorem adds two equivalent conditions to those given in Theorem 4.1.3. Let G be a non-nil torsion free group.
Corollary 4.3.3: equivalent:
1) Every (associative) ring finitely many ideals.
R with
R+ = G,
2) Every (associative) ring DCC for ideals.
R with
R+ = G, and
3)
and
The following are
R2 ~ 0 possesses only R2 ~ 0 satisfies the
G ~ Q+ •
Proof:
Again the implications 1) • 2} and 3) • 1) are obvious.
By Theorem 4.3.2, G ~(±) Q+, a and so G ~ Q+(t)H. Let S be the zeroring with s+ =H. The ring direct sum R = Q(t)S satisfies R+ ""G, and R2 ~ 0. If H ~ 0, choose 0 I a € H. The infinite chain of ideals in R, (a} ~ (2!a} ~ (3!a} ~ ••• is properly descending, a contradiction. Hence H = 0, and G ~ Q+ • 2} .. 3}:
Suppose condition 2} is satisfied.
Lemma 4.3.4:
Let G be a torsion group.
The following are equivalent:
1} G is the additive group of an (associative} ring possessing only finitely many ideals. 2) G is the additive group of an (associative} ring satisfying the ascending chain condition for ideals. m ni
3)
G
=
9
.
€) Z(p~}.
i=l j=l a.
1
pi
a prime, m,ni
positive integers,
aj
an
J
arbitrary cardinal, Proof:
= l, ... ,m;
j = l, ...
,n;.
Clearly 1} • 2}:
2) • 3}: Suppose there exist infinite1y many distinct primes {pi}7=l for which G ~ 0. Let R be an (associative} ring with R+ = G such that R pi satisfies the ascending chain condition for ideals. The infinite chain of 52
2
ideals in
R, Gp
1
contradict1on. For every p
€
0 Gp
k
G c... is properly ascending, a i=l i i=l Pi Hence Gp ~ 0 for only a finite set of primes {pi}~=l. c
c ... c (f)
{pi}~=l' Gp[p] s GP[p2 J S··· Gp[pk] s ... is an ascending
chain of ideals in
R,
positive integer ni
each contained in Gp. n. such that G = G [p. 1], Pi Pi 1
ni G = <+> <+l Z(p~), aj Pi j=l aj
Hence there exists a i
1 , ••• ,m.
Therefore
m ni . a cardinal nunber, and so G = ® (£) (±) Z{p~). i=l j=l aj
3) ~ 1): The ring T constructed in the proof of Proposition 4.3.1 is an associative, commutative ring possessing only finitely many ideals, and T+ = G.
Clearly, G satisfies condition 2) of L~mma 4.3.4 if and only if G is a Noetherian ring group in the sense of the definition in section 2 of this chapter. Condition 3) merely states that G is bounded. Therefore by Theorem 4.2.8 we have the following: Theorem 4.3.5:
Let G be a torsion group.
The following are equivalent:
1)
G is bounded.
2)
G is an associative principal ideal ring group.
3)
G is a principal ideal ring group.
G is the additive group of an associative ring satisfying the ascending chain condition for ideals. 4)
5) G is the additive group of a ring satisfying the ascending chain condition for ideals. 6) G is the additive group of an associative ring possessing only finitely many ideals. 7)
G is the additive group of a ring possessing only finitely many ideals.
Corollary 4.3.6:
Let G be a non-nil torsion group.
The following are equivalent: 53
1) Every (associative) ring finitely many ideals. m ni G = <±> <±> (±) Z(p~). P; 1 i=l j=l ()(j finite cardinal, i = l, ... ,m,
R with
R+ = G,
and
R2 1
o
possesses only
0
2)
a prime, n, ni j = l, •.• ,ni
positive integers, (i.e.
ocJO
a
G is finite).
Proof: 1) • 2): Suppose G satisfies condition 1). It may be assumed that G is of the form 3) of Lemma 4.3.4. Suppose that ocj is an infinite co cardinal for some index j. Then G =
2) • 1):
Obvious.
Theorem 4.3.7:
Let G be a torsion group.
The following are equivalent:
1) G is the additive group of an (associative) ring satisfying the DCC for ideals. m n1 2) G = @ @ G) Z(p~) (!) Z(pi), pi a prime, m, ni positive i=l j=l oco finite 0
J
integers, ocj
an arbitrary cardinal,
i = l, •.. ,m,
j = 1 , ..• ,ni.
Proof: 1) • 2): Let R be a ring satisfying the DCC for ideals with R+ = G. Suppose there exists an infinite set of primes {pi }i=l such that G 1 0, i = 1,2,... The infinite chain of ideals in R, pi co co co (±) G => G-) G => ••• => @ => ••• is propet·ly descending, a contradiction. i=l pi i=2 pi i=k m m Hence Gp 1 0 for only finitely many primes {pi }i=l • Let p E {P; }i=l, and let Dp be the maximal divisible subgroup of Gp. Now Dp = ® Z(p ()( Suppose that oc is an infinite cardinal. Then R possesses a properly co co "" descending chain of ideals @ Z(pco) => 0 Z(p => ••• => @ Z(p=) => ••• , a i=l i=2 i=k contradiction. Gp = Hp (f) Dp , Hp a reduced subgroup of Gp•
00
00
54
)
).
Proposition 1.1.10. For every positive integer n, pnGP ~ pn+lGP is an inclusion of ideals in R. Hence there exists a positive integer n, such that pnG = pn+lGP. Clearly pkG = pkH €) D for every positive integer p p p p k. Therefore pnHP = pn+lHP. Since Hp is reduced, this implies that pnH p = 0, i.e., for every 1 -< i -< m, there exists a positive integer n1.
n.
such that H pi
4
(tj j=l a.
Z(p~),
cxJ.
a cardinal number,
j = 1 , ••• ,ni.
J
Clearly G is of the form of condition 2). 2) ~ 1):
If G satisfies condition 2) then G is the additive group of an associative ring satisfying the DCC for left ideals, Proposition 4.3.1.
Corollary 4.3.8: Let G be a torsion group. G is the additive group of a ring with trivial annihilator satisfying the DCC for ideals if and only if G sati~fies one, and hence all, of the equivalent conditions in Theorem 4. 3.5. Proof: If G is the additive group of a ring with trivial annihilator, then G is reduced, Lemma 2.2.1. If in addition G is the additive group of a ring satisfying the DCC for ideals, then G is bounded by Theorem 4.3.7. Conversely, if G satisfies the equivalent conditions of Theorem 4.3.5, then the ring constructed in proving the implication 3) ~ 1) for Lemma 4.3.4, is a ring with trivial annihilator satisfying the DCC for ideals. Corollary 4.3.9: Let G be a non-nil torsion group. The following are equivalent: · 1) Every (associative) ring R with R+ = G, and R2 ~ 0 satisfies the DCC for ideals. m ni 2) G = €> (±) (£) Z(p~) (£) Z(p7), pi a prime, n, ni positive i=l J"=l ex j finite integers, cxj Proof:
1 , ••• ,m,
j = 1, .•• ,ni.
Similar to the proof of Corollary 4.3.6.
Theorem 4.3.10: 1)
a finite cardinal,
Let
G be a group.
The following are equivalent:
G is the additive group of an (associative) ring possessing only
55
finitely many ideals. m ni 2) G c.!{£)Q+ 0 (f) 0 Z(p~). pi a prime, m. ni non-negative integers. a i=l j=l aj a. aj arbitrary cardinals. i = l, .••• m; j = l, •••• ni. Proof: 1) • 2): Let R be a ring possessing only finitely many ideals. with R+ = G. Then R = R/Gt is a ring possessing only finitely many ideals. with 'It torsion free. Therefore it Co!~ Q+. a an arbitrary cardinal. Theorem4.3.2. Let I Q+. and let R be the ring constructed in proving the a
implication 3) .. 1) for Theorem 4.3.2. The ring direct sum S = F(±) R is associative. possesses only finitely many ideals. and s+ ~G. Corollary 4.3.11:
Let G be a non-nil group.
The following are equivalent:
1) Every (associative) ring R with R+ = G, and R2 '! 0 possesses only finitely many ideals. n. + m 1 . 2) G ..,9 Q (±} (t) l+l Z(p~). pi a prime. m, ni non-negative integers. a i=l .j=l a. J
a. aj
finite cardinals,
= 1 , ••• ,m;
j = 1 , ••• ,n 1 •
An argument almost identical to that used in proving Theorem 4.3.1 yields: Theorem 4.3.12:
Let G be a group.
The following are equivalent:
1) G is the additive group of an (associative) ring satisfying the DCC for ideals. n ni 2) G ... @ Q+ 0 (!) (3 a prime. m, n; a i=l j=l aj non-negative integers. a. aj arbitrary cardinals. i = 1, ••• ,m; 56
j = 1 , •••
,ni .
The following corollaries are easy to prove: Corollary 4.3.13: Let G be a group. G is the additive group of an (associatjve) ring with trivial annihilator, satisfying the DCC for ideals if and only if G satisfies one, and hence both of the equivalent conditions of Theorem 4.3.10. Corollary 4.3.14:
1) 2)
Let G be a non-nil group.
The following are equivalent:
Every ring R with R+ = G, and R2 1 0 satisfies the DCC for ideals. m ni . G ""c±)Q+ (±) (±) (±) Z(p~) 0 Z(p~), p. a prime, m, ni 1 1 1 a i=l j=l a. finite J
non-negative integers, a, aj
finite cardinals,
i = l, ••• ,m; j = l, •.• ,ni.
Lemma 4.3.15: Let R be an (associative) ring with trivial annihilator, such that R+ = D<+l (R+)t• D a divisible group. Then R is a ring direct sum R = S(+) T, with S+ = D, Proof:
Let x
€
D, 0
~
and T+ = (R+)t.
a E (R+ )t.
Then x = 1a1x 1 • x1 E D,
and so
ax = xa = 0. Hence O(R+)t = (R+)tD = 0, and it suffices to show that D2 ~D. Let x,y € D. Then xy = z+a, z E D, a E (R+)t. Suppose that a ~ 0. Since a E (R+)t• a annihilates D. Let 0 ~ b E (R+)t. There exist x1 , z1 € D such that x = lblx 1 , z = lblz 1 . Therefore ab = [lblx 1y- lblz 1 lb = 0. Similarly ba = 0. Hence a annihilates R, a contradiction. Corollary 4.3.16: Let R be an (associative) ring with trivial annihilator satisfying the DCC for ideals. Then R is the direct sum of a torsion free ring and a torsion ring. Proof:
Theorem 4.3.12 and Lemma 4.3.15.
An argument almost identical to that used in proving [36, Theorem 123.1] yields the following results: Theorem 4.3.17: An (associative) ring R possessing only finitely many ideals can be embedded in an (associative) ring with unity possessing only 57
finitely many ideals. Theorem 4.3.18: An (associative) ring R satisfying the DCC for ideals can be embedded in an (associative) ring with unity satisfying the DCC for ideals if and only if R+ has no summand Z(p~), i.e., (R+)t is reduced. Corollary 4.3.19: An (associative) ring with trivial annihilator satisfying the DCC for ideals, can be embedded in an (associative) ring with unity satisfying the DCC for ideals. Proof:
Corollary 4.3.13 and Theorem 4.3.18.
Definition: A ring R satisfying the DCC for (right) principal ideals is called an (MHR)I4HI ring. The 1-tiR and MHI rings were introduced by F.A. Szasz, [63], [64], [65], and [66]. They play an important role in the theory of radicals of rings. Most of the results presented here, are taken from [64], and [31]. Theorem 4.3.20: 1) 2) 3)
Let G be a group.
The following are equivalent:
G is an MHR ring group. G is an MHI ring group. G = D(t) K, with D a divisible group, and K a reduced torsion group.
Proof:
Clearly 1) .. 2).
2) ~ 3): Let R be an MHI ring with R+ =G. Let D be the maximal divisible subgroup of G. Then G = D(±) K, with K a reduced group, Proposition 1.1.10. Let a E K. Then ~ <2!a> ~ ~··· is a descending chain of principal ideals in R. Hence there exists a positive integer m such that = <m!a> for all n ~ m. Clearly <m!a>+ is divisible, and so <m!a>+ =D. This obviously implies that m!a = 0, or that K is a torsion group. 3) ~ 1): Let .G = D<±)K, with D a divisible group, and K a reduced torsion group. By Proposition 1.1.3, there exists a torsion free divisible group D0 such that G = D0 (±) Gt. If D0 = 0, then the zerori ng with additive group G is an MHR ring. If D0 ~ 0, let F be a field with F+ = D0 , Theorem 4.1.3, and let S be the zeroring with s+ = Gt. The ring direct sum R = F (t) S is an MHR ring with R.. = G. 58
Corollary 4.3.21:
Let
divisible subgroup of Proof:
R be an t4HI ring, and let R+.
Then
By the above theorem,
Hence
K.
R+ = D(t)K,
a € D,
Let
b € Dt.
ab"(lblc)b=c(lblb)=O.
annihilates
D be the maximal
annihilates with
D annihilates
K a reduced torsion group. annihilates
Dt
R. D a divisible group, and
K by Lemma 2.2.1, and so
Then
a = lblc
Similarly
Theorem 4.3.22:
Let
G be a group.
1)
G is a unital MHR ring group.
2)
G is a unital MHI ring group.
3)
G = D(±)B,
i.e.,
Dt
R.
The following are equivalent:
D a torsion free divisible group,
with
c € D.
for some
ba=O,
D, which yields that Dt annihilates
Dt
and
B a bounded
group. Proof:
Clearly
1) .,. 2).
2) .,. 3): Theorem 4.3.20 and Corollary 4.3.21 imply that G = D(f) B, with D a torsion free divisible group, and B a reduced torsion group. Let R be a unital MHI ring with
R+ = G,
and let
e
be the unity in
R.
Then
e = e 1 + e 2 , e 1 € D, e 2 € B. For x € B, x =ex = e 1x + e 2x. Let n a positive integer such that nx = 0. Since D is divisible, e 1 = nd, d € D.
Therefore e 1 x = (nd)·x = d·(nx) = 0. 1e 2 1x = ( 1e 2 1e 2 )x = 0, i.e., B is bounded.
3).,. 1): and
6
Suppose that
G = D(f)B,
a bounded group.
Then
with
Therefore
x = e 2x,
be
and so
D a torsion free divisible group,
m
G = u@ G)@ Z(n.), 1 i=l (X.
m, ni
positive
1
integers, and
a;
an arbitrary cardinal,
i = 1 , ••• ,m,
Proposition 1.1.9.
be a field with F+ = D, Theorem 4.1.3. Then the ring direct sum m R = F <:i:) Ef) rf) Z/n. Z] is a unita 1 t4HR ring with R+ "" G. 1 i=l (X.
Let
F
1
Corollary 4.3.23: Let R be a unital MHI (MHR) ring. Then R = R0
R+ = D(±} Rt'
with
bounded, Theorem 4.3.22. positive integer such that
D a torsion free divisible group, and
Clearly
Rt
nRt = 0.
is an ideal in Then
R.
Let
n
Rt be a
D·Rt = (nD)·Rt = D·(nRt) = 0.
59
Similarly, Rt·D = 0. Let a E R. Left (right) multiplication by a induces a homomorphism D ~ R+ via x ~ax (x ~ xa). Hence aD+ (Da+) is divisible, and so aD~D (Da~D), i.e., D is an ideal in R. Put R0 = (D,·). Clearly R = R0 <±)Rt is a ring direct sum. Let e be the unity in R. Then e = e 0 + e 1 , with e0 the unity in R0 , and e 1 the unity in Rt. It is readily seen that R0 and Rt are Mtli (MHR) rings. Lemma 4.3.24: Let R be an MHI ring, and let I be an ideal in I+ torsion free. Then I+ is divisible.
R witn
Proof: Let a E I. Then -::> <2! a> -=> <3! a> -=> ••• is a descending chain of principal ideals in R. Therefore there exists a positive integer m such that = <m!a> for every integer k ~ m. Clearly <m!a>+ is divisible. Therefore for every positive integer n there exists x E such that m!a = nm!x, or m!(a - nx). Since I + is torsion free, a= nx, and I+ is divisible. Theorem 4.3.25: An t1HI (MHR) ring R can be embedded into a unital Mtii (MHR) ring if and only if Rt is bounded. Proof: Clearly if R is a subring of a unital lv'IHI (11HR) ring, then Rt is bounded by Theorem 4.3.22. Conversely, let R be an MHI (MHR) ring with Rt bounded. By Theorem4.3.20, R+ = D(+)Rt' with D a torsion free divisible group. Put R0 = (D,·). The same argument used in proving Corollary 4.3.23 shows that R = R0 (f) Rt is a ring direct sum of MHI (MHR) rings. Put H = R~ (f) Q+. For ai E R0 , qi E Q, i = 1 ,2, define (a 1 + q 1) • (a 2 + q2 ) = a1 • a 2 + q1a2 + q2a 1 + q1q2 , where the first product, a1·a 2 , is defined by the multiplication in R0 , the second and third products by the natural action of Q on R0 (e.g. if q 1 = n/m, then q1a2 =bE R0 such that mb = na 2 ), and the fourth product by multiplication in Q. Then s0 = (H,·) is a unital f1HI (11i-IR) ring with R0 < s0 . Let n be a positive integer such that nRt = 0. Pu+ K = R;<±> (Z/nZt. Defining multiplication on K in a manner ~•milar to that in H yields a unital MHI (MHR) ring structure T with T+ = K, and Rt <1 T. Put S = s0 (!;)T. Then S is a unital MHI (MHR) ring, and R <1 S. The technique employed above in embedding R0 and Rt into unital Mtll (MHR) rings is an application of a general embedding procedure, (see 36, Lemma 12 3. 2 ) • 60
Observation 4.3.26: Let R be an MHI (ltiR) ring. If R can be embedded as a subring into a unital 14HI (MHR) ring, then R can be embedded as an ideal into a unital l.fil (MHR) ring. F.A. Szasz asked for a description of the MHR rings which may be embedded into a unital MHR ring. The problem was solved in [31]. Additive groups of rings have played an important role in the solution of several embedding problems for rings; see [39], [70], and [24].
§4.
Subdirectly irreducible rings.
Definition: A ring R is subdirectly irreducible if the intersection of its nonzero ideals is nonzero. Lemma 4.4.1: Let G be an (associative) subdirectly irreducible ring group. Then Gt is p-primary for some prime p. Proof: If a prime for irreducible GP and Gq
--
Gp = 0 for every prime p, then Gt = 0. Otherwise let p be which Gp ~ 0. Let R be an (associative) subdi rectly ring with R+ = G. For every prime q ~ p, Gp n Gq = 0. Since · are ideals in R, Gq = 0.
Theorem 4.4.2: Let G be a torsion group. G is an (associative) subdirectly irreducible ring group if and only if G is a bounded p-primary group, or G is a p-primary group with G' = Z(pk), . 1 ~ k ~ oo, p a prime. Proof: Suppose that G is a subdirectly irreducible ring group. G is p-primary for some prime p by Lemma 4.4.1. Let R be a subdirectly irreducible ring with R+ = G. Suppose that G1 = 0. Then n pnG = 0. n<W Since pnG is an ideal in R for every positive integer n, there exists a positive integer n such that pnG = 0, i.e., G is bounded. Suppose that G' ~ 0. For every subgroup H of G', RH = HR = o. Lemma 2.2.1, and so H is an ideal in R. If G1 = H® K, then H and K are idea 1s in R with H n K = 0, i .e. , G1 is indecomposable. By Corollary 1.1.5, G1 = Zlpk), 1 ~ k ~ oo. k.
Let G be a bounded p-primary group with basis {a; Ii E 1}, Ia; 1 = p 1 k1 < k for all i E I. Choose ; 0 E I such that k. k. For i ,j E I
,o
61
define for
= j
for These products induce an associative ring structure R on G. Let 0 ~ x € R. Then x = n1a. +••• + n a. , iJ. distinct elements of I, and 11 r 1r k.1 • nj an integer with p J nj for j = 1 , ••• , r. Now xa . = n1P 11 k-1 Therefore every nonzero ideal in R contains p a. , and so R is 1o subdirectly irreducible.
+
Let· G be a p-primary group with G' Z(prJ, 1 ~ r ~~. Then G' poss~sses a unique minimal subgroup generated by some a0 € G', 1a0 1 p. For every positive integer k, choose ak € G such that pkak = a0 • Let k.
B be a basic subgroup of G with basis {bi li € I}, i € I. For i,j € I define
bi•bj =
r:-1
for
i =j
for
i
~
lbi 1 = p 1 for all
j •
These products induce an associative ring structure R on G, [36, Theorem 120.1]. Let a € R, a~ 0, I a I = pj. Now B = (~ Bn, with n<W Bn=(f) Z(pn), an a cardinal number, and Bn a direct summand of G for an all n < w. Let r be the smallest positive integer such that under the natural projection nr: G ~ Br' nr(a) ~ 0. Let n be a positive integer, n ~ j+r. Since G =
(±) B. (!) B. + pnG, l
a = m1b. +.•• + m b. + g, with it distinct elements of I • mt an s 1s 11 integer, t = 1 •.•• ,s. k. = r, ml = puv, u,v integers, u < r, 11 62
va 0 • Therefore every nonzero ideal in
R contains
a0 •
Observation 4.4.3: Let G be a mixed {associative) subdirectly irreducible r1ng group. Then Gt is a p-primary group possessing nonzero elements of infinite height, p a prime. Proof: Gt is p-primary by Lemma 4.4. 1. Let R be an (associative) subdirectly irreducible ring, with R+ = G. For every positive integer n, pnGt is an ideal in R. If n pnGt ~ 0, then G possesses elements of n<(a) infinite height. Otherwise pnGt = 0 for some positive integer n. In the latter case Gt and pnG are nonzero ideals in R satisfying Gt n pnG = 0, a contradiction. Theorem 4.4.4: Let G be a torsion free group. G is an (associative) subdirectly irreducible ring group if and only if G is not reduced. Proof: Suppose that G is an (associative) subdirectly irreducible ring group. Let R be a subdirectly irreducible ring with R+ =G. The ath Ulm subgroup Ga of G is an ideal in R for every ordinal a. let a be the smallest ordinal for which G = Ga+l. If a is a limit ordinal, then Ga = n G~ is the intersection of nonzero ideals in R, and so ~
Ga ~ 0.
If a is an isolated ordinal, say a= ~+1,
the intersection of nonzero ideals nGa of R, Ga is a nonzero divisible subgroup of G. Suppose that G is not reduced. Then H a reduced group, D ~ 0. Let {eil i E for D, and let {ajJj E J} be a maximal i 0 E I. Define eiej = eiak = akei = akat
then Ga = n nG~ is n<(a) and so Ga ~ 0. Therefore
G = D(t)H, D a divisible group, I} be a maximal independent set independent set for H. Choose = ei 0 for all i ,j E I, k,t E J.
These products induce an associative ring structure on G. Every nonzero ideal in R contains e. , and so R is subdirectly irreducible.
lo
Lemma 4.4.5: Let G be an {associative) strongly subdirectly irreducible ring group, G = H(+)K, H ~ 0, K ~ 0. Then H and K are both (associative) nil. 63
Proof: Suppose that H is not (associative) nil. Let s be an + 2 (associative) ring with S = H, S ~ 0, and let T be the zeroring, with T+ = K. The ring direct sum R = S T is an (associative) ring satisfying R+ = G, and R2 ~ 0. Now S and T are ide a 1s in R, with S n T = 0, a contradiction.
m
Theorem 4.4.6: Let G be a torsion free group. G is an (associative) strongly subdirectly irreducible ring group if and only if G ~ Q+. Proof: G ~ Q+ <±' H by Theorem 4.4.4. converse is obvious.
Hence
G ~ Q+ by Lemma 4.4.5.
The
Theorem 4.4.7: Let G be a torsion group. G is an (associative) strongly subdirectly irreducible ring group if and only if G = Z(pk), p a prime, k a positive integer. Proof: Suppose that G is an (associative) strongly subdirectly irreducible ring group. If G = H(!l K, H ~ 0, K ~ 0, then H and K are (associative) nil, Lemma 4.4.5, and so G is ni 1 by Theorem 2.1.1, a contradiction. Therefore G = Z(pk), p a prime, 1 < k 1. Then r(~} = a 2 > 1, [36, vol. 1, p. 261, Ex. 8]. Let D be the divisible hull of fW1. Then D = Q1 pk. There exists a homomorphism ~= Q1 ~ Z(pco) such that ~(x) =a. Now ~(mx) = ~(x} ~ 0. Let n be the natural projection of D onto Q1 , and put ~ = ~·n. Then ~H' the restriction of ~ to ~. is a nonzero homomorphism. Let gi =a; + b;, a 1 E Z(pco}, b; E H, i = 1 ,2. Define gl*g~ = ~H(b 1 ® b2 J. Then R = (G,*) is a ring with R+ = G, and R2 i 0 Let h E H be such that ~ E (H®H) n K. Let L be the subgroup of H 64
generated by h. Then Z(p=) and L are nonzero ideals in Z(p=) n L = 0, a contradiction. Therefore r(H) = 1.
R, with
It now suffices to show that H is p-divisible. If not, there exists h € H such that hp(h) = 0. For a,b € G, define a·b = 0 if a € Z(p~ or b € Z(p~. Choose a0 € Z(p=) with 1a0 1 = p. For hi € H there exists a positive integer ni, and an integer mi such that p + ni, and nihi = mih' i = 1 ,2. There exists a unique element a € Z(p=) such that n1n2a = m,m2a0 • Define h1 ·h 2 =a. The aoove products induce a ring structure R on G with R2 'I 0. Now with Z{p=) n pH = 0, a contradiction.
Z(p~
and pH are ideals in R
Conversely, let G = Z{p~(t)H, H a rank one, p-divisible, torsion free, nil group. Clearly G is p-divisible. G is not nil by Theorem 2.1. 1. Let R be a ring with R+ = G, and R2 'I 0. Since H is nil, R/Z(p~ is a zeroring, i.e., R2 =: Z(p~. Let a0 € Z(p=), 1a0 1 = p. Every nonzero subgroup of Z(p=) contains a0 • Let I be an ideal in R, I 'I 0. Suppose that a0 I. Then In Z(p=) = 0. However RI S R2 S Z(p=), and so RI = 0, and similarly IR = 0. Let 0 'I x € I, x = a+b, a € Z{p=), b € H. Clearly b ~ 0. There exists a positive integer n such that pna = 0. Hence pnx € Hni. We may therefore assume that there exists 0 'I ho € Hni. Let h € H. There exist a non-negative integer k, a positive integer r, and an integer s such that pfr, and pkrh = sh 0 • Therefore pkrhR =: IR = 0. However pkrhR = rh(pkRJ = rhR. Now hR =: Z(p=), and for 0 'I a e: Z( pj , ra ~ 0. Hence hR = 0. Therefore 1) HR = 0, and similarly 2) RH = 0.
4
Let a € Z(p~, x € R. There exists a positive integer n such that pna = 0, and there exists y € R such that x = pny. Hence ax= a(pny) = (pna)y = 0. Therefore 3) Z(p=)·R = 0, and similarly 4) R·Z(p~ = 0. Equalities 1), 2), 3), and 4) imply that R2 = 0, a contradiction. §
5.
Local rings. In this section a ring is meant to be an associative ring.
Definition: A unital ring R is local, if its nonunits form an ideal. This ideal is clearly the unique maximal ideal in R. 65
Notation: Q*
Q - {0}.
qP
{a/bla,b E Z,
Fp
a field of order
Lemma 4.5.1: units
Let
pta,
pfb}.
p,
p
R be a local ring with maximal ideal
If
(RIM)+""(±) Q+,
2)
If
R+
a an arbitrary cardinal, then
a
is not a torsion group, and if
Proof:
R+
1) l+M
is torsion free in n.
Hence
a
n,
for a'rbitrary nonzero integers
(R/M)+,
and
1/n
and so
belong to
n = n·l ~ M for every . 1 U, and so ~ = n(iii) E U
n ,m. n
and
m to be
p.
again follows from the same argument, plus the fact that
i.e.,
an
FP ~ U.
2) follows from the same argument as for 1), assuming relatively prime to
a
p-
is a torsion group, then
nonzero integer
3)
and set of
Q* s; U.
(R/14)+ =@Z(p),
Q cU.
arbitrary cardinal, then If
M,
U.
1)
3)
a prime.
p(l+M) = M,
p E M.
Observe that
R/M
above is a simple ring, so either
( R/M) + ""'(+) Q+,
a
or
(R/M)+ =<±)Z(p), Theorems 4.1.1 and 4.1.3. Therefore the conditions a + imposed on (R/14) in Lemma 4.5.1 and in the following lemma are not res tri cti ve. Lemma 4.5.2: 1)
If
2)
If
Proof:
Let
R be a local ring with maximal ideal
R+ = G.
1)
Suppose that
(R!N)+ ,(+)Q+.
x E Gq,
a
lxl = qk.
It suffices to show that By Lemma 4.5.1,
Gq = 0 for
qk , q -k E R,
l R/t4t = (+) Z( p). Let q be a prime, q 1 p, and 1et a -k k • q X lxl = qk. By Lemma 4.5.1, q k , q -k E R, and so X = q
and
Suppose that
x E Gq,
66
and
( R/M) + ""(±) Q+, a an arbitrary ca rdi na 1 , then G is torsi on free. a (RIM)+ =G> Z(p), a an arbitrary cardinal, then Gt is p-primary. a
every prime q. Let so x = q-kqkx = 0. 2)
M,
= 0,
Theorem 4.5.3:
Let G be a torsion group. G is a local ring group if and n only if G = (+) G) Z(pk), p a prime, n a positive integer, a k an k=l ak arbitrary cardinal,
k = 1 , ••• ,n.
Proof: Let R be a local ring with R+ = G. Let 11 I = n. Clearly nx = 0 for all x E G. Hence G is a bounded group, and so G is a direct sum of cyclic groups, Proposition 1.1.9. G is p-primary, Lemma n n 4.5.2, and so G = (±) (±) Z(pk). Conversely, let G = 0 (±) Z(pk). Put k=l ak k=l ak II =(f) Z(pn). If an is an infinite cardinal, then there exists a local an ring T, with T+ = H, [36, Lemma 122.3]. If a n = r < =, then H = (a 1 ) ct'l•• if' (ar), Ia; I = pn, i = 1 , ••• ,r. The products a 1·aj = a{al = aj for j = 1 , ••• ,r, and a1 ·aj = pa 1 for i ,j = 2, ••• ,r, induce a ring structure T on H with unique maximal ideal (pa 1 )(t}(a 2 )(t)..@(ar). In either case, the unity e E T is a basis n-1
i.e., H = (e) (±) (a.). Let L = ® 0 Z(pk), and let 1 iEI k=l a k {bjlj E J} be a basis for L. Define ebj = bje = bj, and bj·bk = bk·bj = a;·bj = bj·a; = 0 for all i E I, j,k E J. Define the product of elements in H in accordance with the multiplication in T. These products induce a ring structure R on G. Let N be the maximal ideal in T. Then M = N€)L is the unique maximal ideal in R.
element for H,
Theorem 4.5.4: G is the additive group of a local ring R with maximal ideal t4 such that (RIM)+ ""(l;)Q+, a an arbitrary cardinal, if and only a
if G ""(t)Q+. Proof:
Let
R be a loca 1 ring with
( R/t4) +
""0 Q+. Let x E G and let n a
be a positive integer. Since n is a unit in R, Lemma 4.5.1, x = n(* x), and so G is divisiole. G is torsion free by Lemma 4.5.2. Therefore G ""(!) Q+, Proposition 1.1.3. Conversely, if G ""(+)Q+, a
Theorem 4.5.5:
Let
then
G is the additive group of a field.
R be a local ring with maximal ideal
M.
If
67
(RIM)+=(!) Z(p), a an arbitrary cardinal, and if R+
is not a torsion
a
group, then R+ = H(!)K, H a divisible group and K homogeneous of type [(co,,,,,l,co,,,.)], with 1 at the p-th component. Proof: Let q be a prime, q I p. By Lemma 4.5.1, q,q-l E R. Hence for every x E R+, x = q(q- 1x), and so R+ is q-divisible. Let H be the maximal divisible subgroup of R+. Then R+ = H(.+)K, K homogeneous of type [(co,.,,,l,co,,.,)] with 1 at the p-th component. Theorem 4.5.6: Let R be a Noetherian local ring. Then + n k R = H(!l (.+) (!) Z(p ), n a positive integer, p a prime, ak k=l ak cardinal, pH I H.
k = 1, •.• ,n.
and H torsion free.
If
R+
an arbitrary
is mixed, then
Proo~: R; is p-primary for some prime p, Lemma 4.5.2. Since + + 2 + n R [p] ~ R [p J ~··· is an ascending chain of ideals in R, Rt = G[p ] for some positive integer n,
i.e.,
+
Rt
n
= (±) f.±)
k
Z(p ).
Therefore
k=l ak + n k R = H(±) (±) (±) Z(p ), with k=l a k
H torsion free, Proposition 1.1.2.
Let M be the maximal ideal in R. If R+ is a mixed group, then and pnR = pnH are proper ideals in R. Hence R; M, and pnH M. pnH = H, then H~ R~ = R+ t4, a contradiction. Therefore pH I H.
=
§6:
=
=
Rings with trivial left annihilator, subrings of algebrdic number fields, and semisimple rings continued.
This section consists of the work of J.u. Reid, [56], and [57]. Several results in sect1on one of this chapter are s·pecial cases of results which will now be obtained. The above approach was advantageous in that only elementary techniques were employed. The approach in this section requires somewhat more sophisticated techniques, but yields more general results. Let G be a group, and let ~ E Hom(G, End(G)). The multiplication on G defined by X•Y = ~(x}(y) for all x,y E G induces a ring structure R on G, and in fact all rings R satisfying R+ = G may be obtained in 68
this manner, 1 .2.1. Notation: Let ~ E Hom(G, End(G)). Then above. t(A) = the left annihilator of A,
(G.~)
denotes the ring described A a subset of a ring R.
The following observation is easily verified. Observation 4.6.1: Let R = (G,~) be a ring, ~ E Hom(G, End(G)). Then is associative if and only if ~[~(x){y)] = ~(x)·~(y) for all x,y E G, i.e., ~ is a ring homomorphism, ~= R ~ End(G).
R
For the remainder of this section all rings will be assumed to be associative. Observation 4.6.2: Proof:
Let R =
(G.~),
~
E Hom(G, End{G)).
~.
Then t(R) = ker
Obvious.
Lemma 4.6.3: Let R be a ring with trivial left annihilator. Let A,B satisfying R = A+B = {a+bla E A, b E B}. If a1a2 = b1b2 = 0 for all aiEA,biEB, i=l,2, then R=O.
~
R
Proof: Clearly A c ~(A), B c t{B), and t(A) n t(B) = 0. Therefore R+ = t(A)+ + t(B)+ .-and so t(A) = A, and t{B) = B. Since t(A) is left ideal in R so is A. Let b E ~(A), x E R. Then for every a E A, (bx)a = b(xa) = 0, i.e., t{A) =A is a two-sided ideal in R. Similarly, B is a two-sided ideal in R. Hence R = A(f) B is a ring direct sum. Clearly R2 = 0 and so R t{R) = 0.
=
Theorem 4.6.4: Let G be an associative strongly trivial left annihilator group. Then G is indecomposable. Proof: Supposethat G=H(!)K, H;!O, K;!O. Let nH,nK bethe natural projections of G onto H and K respectively, and let R be a ring satisfying R+ = G, and R2 ;! 0. For a,b E R, define axHb = nH(a)·nH(b), the product on the right being multiplication in R. Then S = {G, xH) is a ring with s+ = G, and K ~ ~(S). Therefore s 2 = 0 which implies that in the ring R, H2 = 0. Similarly, K2 = 0. By Lemma 4.6.3, R = 0, a contradiction. Corollary 4.6.5: Let G be a non-torsion free group. equivalent: (1) G is an associative strongly semisimple group.
The following are
69
(2) ( 3)
G is an associative strongly trivial left annihilator group. G = Z( p) , p a pri me .
Proof: Clearly (1) • (2). If G satisfies (2) then by Theorem 4.6.4 and Corollary 1.1.5, G = Z(pk), p a prime, k a positive integer. Let G = (a), Ia I = pk. The product a· a = pa induces a ring structure R on G with pk-lG 5 ~(R). Hence pa = 0, and k = 1. The implication (3) • t 1) is obvious. It is easy to show that if G and H are quasi-isomorphic torsion free groups then G is an (associative) strongly trivial left annihilator group if and only if H is. This together with Theorem 4.6.4 yields: Corollary 4.6.6: Let G be a torsion free associative strongly trivial left annihilator group. Then G is strongly indecomposable. Let G be a finite rank torsion free group. properties of End(G) which we will need. Property 4.6.7: Proof:
For every idempotent e
E
End( G),
G ~ e(G) ® (1-e)(G).
[36, vol. 2, p. 149 te)].
Property 4.6.9: indecomposable. Proof:
is a left and right Artinian ring.
[36, vol. 2, p. 149(d)J.
Property 4.6.8: Proof:
End(G)
The following are some
End(G)
is a local ring if and only if G is strongly
[36, Proposition 92.3].
Corollary 4.6.10: free group. Then ni 1potent.
Let G be a finite rank, strongly indecomposable torsion End(G) is a local ring whose ideal of non-units is
Proof: End(G) is a local ring, Property 4.6.9. Let f4 be the ideal of non-units in End(G). By Property 4.6.8, M does not possess a nonzero idempotent. A right ideal in a right Artinian ring which does not contain a nonzero idempotent is nilpotent, [45, Theorem 1.3.2]. Hence M is nilpotent by Property 4.6. 7. Theorem 4.6.11: 70
Let R
= (G,~) G a finite rank, strongly indecomposable
torsion free group, and ~ E Hom(G, End(G)). ~(G) is a nilpotent subring of End(G).
Then either t(R) = 0,
or
Proof: If ~(G) is not nilpotent then by Corollary 4.6.10, ~(G) possesses a nonzero unit p E End(G). Since End(G) is a finite dimensional vector space over Q, p-l E Q[p], and for some positive integer m, mp-l E ~(G). Therefore m = (mP-l)P E ~(G). Put K = ker ~. Let g E G such that ~(g)= m. Define'!': ~(G) -+G via 11'[~(x)] = ~(x)(g). Suppose that ~(x) E ker '1'. Then by Observation 4.6.1, 0 = ~[~(x)(g)] = ~(x)~{g) = ~(x). Since ~(G)+ is torsion free, this implies that ~(x) = 0 and so 'I' is one-to-one. Let x E G such that ~(x)(g) E K. Again by Observation 4.6.1, 0 = ~[~(x)(g)] = ~(x), implying that ~(x) = 0, i.e., '!'[~(G)] n K = 0. Therefore we have the inclusion G ~'l'[~(g)](!}K. For x E G, ~[mx - 'l'(~(x))] = ~(x) - ~[~(x)(g)] = m~(x) - ~(x)~(g) = ~(x) - ~(x) = 0. Hence mxE'l'[~(G)]~K. andso G,;'I'[~(G)](!)K. Since G isstrongly indecomposable, K = 0, and R is a trivial left annihilator ring by Observation 4.6.1. Corollary 4.6.12: Let R be a ring with R+ a finite rank strongly indecomposable torsion free group, r(R+J = n. Then either t(R) = 0 or Rn+l = 0. Proof: R=(R+,~), ~EHom(G,End(G)) • .If t(R)fO then ~(R+) is nilpotent. Theorem 4.5.11. This clearly implies that R is nilpotent, and so Rn+l = 0 by Theorem 3.1 .3. Theorem 4.6.13: Let R = (G.~) be a ring with unity, Then ~(G) is a direct summand of End(G).
~
E Hom{G, End(G)).
is a left ~(G)-module, and ~{G) is a projective ~(G)-module. It therefore suffices to construct a ~(G)-epimorphism f:End(GJ-+~(G). Define f(i)i)=~[i)i(e)], e theunityof R, forevery ljJ E End(G). Let x E G, l/1 E End(G). Then f[~(x)i)i] = ~[~(x)l/J(e)] = ~(x)·~[w(e)] = ~(x)·f(i)i), and so f is a ~(G)-homomorphism. For x E G, f(~(x)) = ~[~(x)(e)] = ~(x)~(e) = ~(x), i.e., f is an epimorphism. Proof:
End(G)
Definition: Let G be a torsion free group. A subgroup H of G is a full subgroup of G if G/H is a torsion group. This is clearly equivalent 71
to H being an essential subgroup of G. Lemma 4.6.14: Let G be a torsion free group v its divisible hull. There is a one-to-one correspondence between the pure fully invariant subgroups of G, and the End(G)-submodules of V. Proof: Let M be an End{G)-submodule of V. Put H = GnM. Clearly H is a fully invariant subgroup of G. Since r.1 is a pure subgroup of V, H is a pure subgroup of G. Let H* be the Q-subspace of V generated by H. Since M is a Q-subspace of V, and H is a full subgroup of ~. H* = r1. Conversely, if H is a pure fully invariant subgroup of G, then H* is an End(G)-submodule of V satisfying GnH* =H. Hence the correspondence H ~ H* between pure fully invariant subgroups of G, and End{G)-submodules of V is one-to-one and onto. The above lemma motivates the following: Definition: A torsion free group G is irreducible if the only pure fully invariant subgroups of G are 0 and G. If G i H for every nonzero fully invariant subgroup H of G, then G is said to be strongly i rreduci b1e. Corollary 4.6.15: Let A be a simple Q-algebra, R a subring of A. R+ is a full subgroup of A+ then R+ is irreducible.
If
Proof: Let H be a pure fully invariant subgroup of R+, and let a EA. + + There exists a positive integer n such that na E R. The map R ~ R via x ~ nax is an endomorphism. Hence naH ~H. Similarly H{na) ~H. This together with the purity of H in R+ yields that H* is an ideal in A, and so H* = 0, or H* =A. By Lemma 4.6.14, H = 0 or H = R+. Lemma 4.6.16: Let G be a finite rank strongly indecomposable torsion free group. G is irreducible if and only if End(G) is a division ring with dimQEnd(G) = r(G). Suppose that G is irreducible, and let V be the divisible hull By Lemma 4.6.14, V is a simple End(G)-module. Therefore by Schur's ~emma HomEndtG)(V,V) is.a division ring. Since End(G) ~ HomEnd(G)tV,VJ, End(G) does not possess a non-zero nilpotent ideal, Proof: of G.
72
and so by Corollary 4.6. 10 •. tnd{G) is a division rinq. simple tnd(GJ-module, di~End{G) = dimQV = r(V) = r(G}.
Since
V is a
Conversely, suppose that End(G) is a division ring satisfying dim0End(G) = r(G). Then r(G) = r{V) = dim v = dim. V·dimQEnd(G) Q End(G) =dim. V·r(G). Hence dim. V = 1, and V is a simple End(G) End(G) End(G)-module. G is irreducible by Lemma 4.6.14. Let K be an algebraic number field, J the ring of algebraic integers in K. For P a prime ideal in J, JP = {x/ylx,y € J, y ( P}. For n a set of prime ideals in J, Jn = n JP. r~otation:
P€11
Theorem 4.6.17: Let G be a finite rank strongly indecomposable torsion free group. The following are equivalent. G is associative semisimple. (2) G is associative strongly semisimple. (3) G is quasi-isomorphic to the additive group of a subring Jrr of an algebraic number field K satisfying [K:Q] = r(G), rr an infinite set of primes, or G ~ Q+. (I)
Proof: (1) ~ (3): Let R =(G.~) be a semisimple ring, ~ € Hom(G, End(G)) ~Y Observation 4.6.2, ~ is one-to-one. Since the nonzero elements in ~(G) are units in End( G), Corollary 4.6.10, ~(G)* = Q ~~(G) is a dlVision algebra over Q. Now ~(G)+, identified with [Z lSI) ~(G)]+, is a full subgroup of ~(G)*+ and so tnd(G) is a division ring. End(G) is commutative; an easy consequence of a result in the next chapter, Theorem 5.3. 11. Hence tnd(G) is a field. Put K = tnd(G). By Lemma 4.6.16, [K:Q] = r(G). Let H be a subgroup of G, H ~ 0, satisfying ~(G)lH) s H. For x € G, y € H, ~(x)~(y) = ~[~(x)(y)], Observation 4.6.1. Since ~[~(x)(y}] € ~(H) this implies that ~(H) is a left ideal in ~(G). The same argument used in proving Theorem 4.6.11 shows that ~(H) possesses a positive integer m. Therefore ~{G) s ~{H), and so ~(G) ~ ~lH). Since ~ is one-to-one, G ~H. Clearly ~(G){H) s H for every fully invariant subgroup H of G, and so G ~ H for every nonzero fully invariant subgroup H of G, i.e., G is strongly irreducible. For every x € G, 73
x 1 0, H = ~(G)(x) and K = End(G)(x) are nonzero subgroups of G satisfying ~(G)(H) ~ H and ~(G)(K) ~ K. Hence ~(G)(x) ~ G ~ End(G)(x). This clearly implies that ~(G) ~ End(G). Since ~(GJ is a semisimple ring, the inclusion m End(G) =~(G), m a positive integer, implies that End(G) is semisimple. Let J be the ring of algebraic integers in K = End(G), and let rr be the set of prime ideals P in J satisfying P End(G). A standard procedure in algebraic number theory is to define a valuation vp on K for each prime ideal P in J as follows: Define vptO) = 0. Let p be the unique rational prime belonging to P. For x € J, xI 0, define vp(x) = p-k, where k is the highest power of P dividing <X>= the ideal in J generated by x. Extend vp multiplicatively to K. Every non-archimedian valuation on K is equivalent to vp for some prime ideal P in J. Hence the valuation rings in K are VP = {x € Klvp(x) ~ 1}, P a prime ideal in J. It is readily seen that VP = JP for every prime ideal P in J. It is well known [2, p. 911 that if S is a subring of K containing the unity of K, then the integral closure of S is the intersection of all the valuation rings in K containing S. Hence the integral closure of End(G) is n JP = Jrr.
=
P€IT
Therefore by [8, Theorem 2.7] Jrr ~ End(G). Again the semisimplicity of End(G) implies that Jrr is semisimple. This is possible only i f rr is empty or rr is infinite. If rr is empty, then Jrr = K and End(G) = K. Since G ~~(G)+~ K+, and G is indecomposable, K = Q, and so G ~ Q+. The relations G ~~(G)+~ K+ clearly imply that [K:Q] = r(G) and so (1) .. (3).
2
Let S =(G.~). ~ € Hom(G, [nd(G)) be a ring with S I 0. Since ~(G)~ End(G), and End(G) does not possess nonzero nilpotent elements, ~(S) = 0, Theorem 4.6.11, and so ~ is one-to-one by Observation 4.6.1. Therefore ~(G)+ is a full subgroup of End( G)+. Let R = (G.~) be a semisimple ring, ~ € Hom(G, End(G)). lhe map ~(G)+~ ~(G)+ via ~(x) ~ ~(x) is a ~(G) n ~(G) homomorphism. Since [~(G) n w(G)]+ is a full subgroup of K+ = End(G)+, this map extends to a K-endomorphism of K+, i.e., x ~ax, a € K. Therefore !Ji(G) = CX<.P(G), a € K. Let n be a positive integer such that na- 1 € ~(G} n w(G). Then w(G) :::> na- 1:ji(G) = na- 1CX<.P(G) = n~(G). Similarly, ~(G)~ nw(G). Hence ~(G)~ ~(G) ~ End(G). This implies that w(G) is semisimple, which in turn yields that S is semi simple. (1) .. (2):
74
(2) Q (1): (3) • (1): and H are and only if
Obvious. G Q. J~, and Jrr is semisimple. It is easily seen that if G quasi-isomorphic groups, then G is (associative) semisimple if H is.
For an alternate proof of Theorem 4.6.17 see [6, Theorem 5.51. In the next chapter, the following group theoretic analogue of the Wedderburn Principal Theorem, due to Beaumont and Pierce, wi 11 be proved: Theorem 5.2.1: Let R be a ring such that R+ is a finite rank torsion free group. Then R+ ~ S+(t)N+, where N is the maximal nilpotent ideal of R, S is a subring of R such that S* = ~S is a semisimple Q-algebra, and S®N is a ring direct sum. This result will be used in what follows. Definition: A group G is anti-radical i f J(R) 'f R for every ring with R+ = G, and R2 'f 0.
R
Theorem 4.6.18: Let G be a strongly indecomposable finite rank torsion free group. Then G is associative semisimple if and only if G is anti-radical and not nil. Proof: If G is associative semisimple, then Theorem 4.6.17, G is anti-radical.
G is not nil, and by
Conversely, let G he an anti-radical group. Let R = (G,\P), IP E Hom(G, End(G)) be a ring with R2 'f 0. Since IP(G) is not nilpotent, t(R) = 0, Theorem 4.6.11, and so by Observation 4.6.2, G ~ IP(G)+. Since G is strongly indecomposable, the maximal nilpotent ideal of IP(G) is zero by the Beaumont-Pierce Theorem, Theorem 5.2.1. This in turn yields that the nonzero elements in tP(G) are units in End(G), Corollary 4.6.10. The argument used in proving Theorem 4.6.11 shows that if I~ q>(G), I 'f 0, then I possesses a positive integer m. Hence mq>(G) c I and so tP(G)+; I+. Therefore if J[q>(G)] 'f 0, then IP(G)+; J[IP(G)]+ and so G ~ J[IP(G)]+. It is easily seen that if G and H are quasi-isomorphic torsion free groups, then G is anti-radical if and only i f H is. Therefore G Q. J[tP(GJl+ contradicts the fact that G is anti-radical. Hence J[tP(G)l = 0, i.e., q>(G) is semisimple, which implies that R is semisimple. 75
The above theorem was proved in a different manner in [6, Theorem 5.5]. Theorem 4.6.19: Let G be a finite rank strongly indecomposable torsion free group. If G is the additive group of a ring with trivial left annihilator then G is strongly irreducible, associative strongly trivial left annihilator, and End(G) is a subring of an algebraic number field K satisfying [K:Q] = r(G}. Proof: Let R = (G,~). ~ E Hom(G, End(G)) be a ring with ~(R) = 0. By Observation 4.6.2, G ~~(G)+, and the maximal nilpotent ideal of ~(G) is zero by the Beaumont-Pierce Theorem. The nonzero elements in ~(G) are units in End(G}. Corollarly 4.6.10, which implies that ~(G)+~ I+ for every nonzero ideal I in ~(G), see the proof of Theorem 4.6.11. As in the proof of Theorem 4.6.17, ~(H) is a nonzero ideal in ~(G) for every nonzero fully invariant subgroup H of G. This implies that ~(G) ~~(H) which in turn yields G ~ H, and so G is strongly irreducible. The same argument used to prove the implication (1} ~ (3) in the proof of Theorem 4.6.17, shows that K = End(G) is a field satisfying [K:Ql = r(G). Let S = (G,~), ~ E Hom(G, End( G)) be a ring satisfying s2 ~ o. Then ~(G) is a nonzero subring of End(G) and so is certainly not nilpotent. By Theorem 4.6.11, ~(S) = 0, and so G is an associative strongly trivial left annihilator group. Corollary 4.6.20: Let G be a finite rank strongly indecomposable torsion free group with r(G) = n. Then either G is an associative strongly trivial left annihilator group, or Rn+l = 0 for every ring R with R+ = G. Proof:
§7:
Theorem 4.6.19, and Corollary 4.6.12.
E-rings and T-rings:
In [35, Problem 45], L. Fuchs posed the problem of classifying the rings R satisfying R ~ End(R+). Much progress has been made on this question by P. Schultz, [59], and by R.A. Bowshell and P. Schultz [13]. Many of their results are presented here. Lemma4.7.l: If R<><End(R+), and R+=H(t)O with H reduced,and 0 divisible, then either 0 = 0, or 0 ~ Q+ and H is a torsion group. 76
Proof: Let r 0(o) = a. Then R+ has a direct summand isomorphic to G) Q+ + ++ a • and so R <><End(R) has a direct summand isomorphic to + + End ( 1'1'\ \TJ Q ) .,.. n(
R+ has a direct summand isomorphic to + r 0{R) = 1, and H is a torsion group.
Let R be a ring satisfying Rc.o End(R+ ).
Then for each
k
prime p,
R; = Z(p P) with 0 ~ kp < ~.
Proof: Let p be a prime for which R; ~ o. By Lemma 4.7.1, reduced. For every positive integer n, let rn = the number direct summands of R; of order pn. Let k be the smallest integer for which rk ~ 0. Then R; has a direct summand Bk
R; is of cyclic positive = (t) Z(pk). rk Since Bk is a pure bounded subgroup of R+, Bk is a direct summand of R+, [36, Theorem 27.5]. Therefore R+ has a direct summand isomorphic to Hom(Bk' Bk) = n (C~ Z(pk)), [36, Corollary 43.3]. Hence rk rk r[Hom(Bk' Bk)] = {k • rk if rk is infinite and r~ if rk is finite. Since r[Hom(Bk' Bk)] ~ rk, rk = 1. Let j be a positive integer, j > k, and suppose
Z(pj)
is a direct summand of R;.
Then R; has a direct
summand isomorphic to Hom(Z(pk), Z(pk) (-i) Zlpj)) = Z(pk)@ Z(pk) contradicting the fact that rk = 1. Therefore R; = Z(pk). Lemma 4.7.3: Let R be a ring satisfying R.,.. End(R+), and let p be a prime such that R; ~ 0. Then R; has a unique complement R~ = {x E Rl hp(x) = ~}, R is a ring direct sum R = Rp<±> R~, and 77
1 ... End(R1 ) Rp p Proof:
0
R; = Z(pk),
Theorem 27.5].
0 < k < ""•
pH ~ H,
If
Hom[R;(!)(H/pH), R;] = K. Therefore Then
then
R+
x = y+z,
y E R;,
rp(K) > 1 = rp(R+),
However
H=
lxl = pm,
R~.
z E H.
and let
H
~ R~.
Since
Rp
R~.
y E
xy = x(pmz) = (pmx)z = 0.
R~
and Then
This clearly yields that
R= R
y = pmz,
RPR~
R "" End( R+) "" End( R;) (t\ End(
0 R~
0 R~
Let
Proof:
p
Then
satisfying
II
Let
x E H,
x E Rp'
Hence
R~RP
and similarly R;).
= 0.
Now It is easily
Therefore
R~
and
R"" End(R+),
"" End( R~).
and let
S be the
for which
@ RP < T <
pES
-
R • p
II
-pES
ByLemma4.7.3, RP
= 0,
Rp "" l:::nd( R;) ,
R be a ring satisfying
natural projection of in
with
R~.
R I 0. Put A= {x E Rl h (x) ="" for all p p A is an ideal in R, and R/A is isomorphic to a ring T
set of primes p E S}.
z E R.
and so
R.
R~) '@Hom( R;, R~)(f) Hom( R~,
"" End( R;)
Lenrna 4.7.4:
y = 0,
is a ring direct sum.
seen that the 1ast two summands are zero. R = Rp
x E
hp(x) = hp(z) = ""•
are ideals in
Therefore
a contradiction.
Conversely, let
This clearly implies that
Clearly
R+ = Z(pk) (i)H, [36,
has a direct summand isomorphic to
H is p-divisible, and so
hp(y) = hp(x-z) = ""· i.e.,
Lenrna 4.7.2, and so
R=Rp@R~ R onto
with p-component
pES defines a homomorphism
Rp.
forevery For
np(x)
n: R ~
IT
x E R let
for every
Rp.
pES.
Clearly
Let
n(x)
p E S.
np
bethe
be the element
This clearly
kern = A
and so imn ... R/A.
pES Now
n(Rp) = Rp
for every
p E S,
and so
rH R < imn <
pES
p -
Observation4.7.5: Let n bedefinedasabove,andlet imn+ is a p-pure subgroup of u+ for every p E S. Proof: Now in
Suppose there exists
imn = R (f) C • + p p (±) Rq' and so
q€S
q~p isomorphic to
78
Since
p E S imn+
pCP I CP.
such that
imn+
R • p
IT
- pES
n R.
U=
pES
Then
p
is not p~pure in
u+.
is not p-pure in u+, C is not p-pure + p Therefore R has a direct summand
+ K = Hom(Rp@(Cp/pCP), Rp).
However
rp(K) > 1
+ rp(R ),
a contradiction.
0 G., i e:I 1 where r(Gi) = 1, and Gi is torsion free for every i e: I. Then I is t{G 1.) is idempotent finite, t(Gi) ! t(Gj) for i 'I j, G1.... End(G.)+, 1 for all i,j e: I, and R "" (t) End( G1.). i e:I Lemma 4.7 .6:
Proof:
Let R be a ring satisfying
R ... End(R+), with
R+ =
R+ has a direct summand isomorphic to
n End(G.). Hence 1 i e:I This is possible only if I is finite.
r( n End(G 1.)) ~ r(R+) = III. ie:I R+ has a direct summand isomorphic to
Now
<±> End(G.)+(t) <±> Hom(G 1., GJ.). 1 ie:I i'lj However, r((t) End(G.J+) = r(R+) = III. Therefore R"" <±> End(G"!"), and 1 1 ie:I ie:I Hom(G., G.)= 0 for all i,j e: I, i 1- j. By Proposition 1.3.4, 1
J
t(Gi),f.t(Gj) forall i,je:I, i/-j. Since t[End(Gi)+]~t(Gi) we have that Gi ""End(Gi)+. Since End{Gi) is not a zeroring, t{Gi) is idempotent for all i e: I, Corollary 2.1.3. For j e: I, let nj be the natural projection of R+ onto Gj. For i e: I, and a e: R, the maps Gi ~ Gj via x ~ nj(ax) and x ~ nj(xa) are homomorphisms. Since Hom{G;, Gj) = 0 for i 'I j, nj(ax) = nj{xa) = 0 for all x e: Gi, i 1- j, i.e., RGi: Gi, and G;R: Gi and so Gi is an ideal in R for all i e: I. Therefore R = (f) G. "" C±> End{ G.) is a ring direct sum. 1 ie:I 1 ie:I Theorem 4.7.7: Let G be a torsion group. There exists a ring with satisfying R ""End{R+J if and only if G is cyclic. Proof:
Let G = Z{n).
Then
R+ = G
(Z/nZJ+ = G, and Z/nZ ""End{G).
Conversely, let G be a torsion group, and let R be a ring with R+ = G satisfying R ""End(R+). Let S be the set of primes p for which k
GP 'I 0. By Lemma 4.7.2,
R+ = (±) Z{p P).
nonzero component in is cyclic.
for every p e: S, S is a finite set, and so R+
Theorem 4.7 .8: a ring
R with
RP
pe:S
Since the unity 1 e: R has
Let G be the direct sum of rank one groups. There exists +
+
m
R = G, and R"" End(R ) if and only if G = Z(n) (f)(!) G. i=l 1 79
with
r(Gi) = 1,
pGi = Gi
idempotent,
for all primes
pin;
and put
Conversely, let groups, and
p
R be a ring with
R "" End( R+).
Now
I j, and
for
R+ = Gt
and let
H ~ Gp
Therefore
via
RH
GtH = HGt =
~
o.
H,
CD H,
i E I.
with
Clearly Then
H = i<2 Gi, Gt
G.
np
Since
GP = Z(pk),
Hom(H, Gp)
and
HR
~
and H,
x ~ ~p(xa)
i.e.,
R.
is isomorphic Hom(H, Gp) = 0.
G onto
GP.
The
are homomorphisms.
H is an ideal in
R = Gt (!) H is a ring direct sum.
Hence
a rank
1 < k
is a p-group,
be the natural projection of
x ~ np(ax),
Gi
is an ideal in
End( Gp) (t) Hom(H, GPJ = Z( pk) (t) Hom(H, Gp)
to a direct summand of a E R,
R+ = G a direct sum of rank one
Gp I 0.
be a prime for which
Lemma 4. 7.2, and so
maps
t(Gj)
i,j = l, ••• ,m.
one torsion free group for each
Let
.$.
m G = Z(n) (!) ~ Gi satisfying the conditions of the theorem. i=I be a unital subring of Q with R~ ... Gi, i = 1, ••• ,m, see 1.4.8, m + + R = (Z/nZ) C!) {.+) R.• Then R ""G, and R ""End(R ). i=I 1
Ri
Let
t(GiJ
Let
Proof: Let
t(GiJ
R.
Clearly
Therefore
R ""Gt ® H ""End(Gt) (t) End(H) which implies that Gt ... End(GtJ, and H ""End(H). By Theorem 4.7.7, Gt"" Z(n), and by Theorem 4.7.8 it suffices
to show that pGi = Gi for all for some pin and some i E I. to
Hom( Gp (£) ( G;fpGi J, GPJ,
Theorem 4.7.9: ring
R with
Let
pin, Then
and so
and i E I. Suppose that pGi I Gi R+ has a direct summand isomorphic r P( R+ J > 1 ,
contradicting Lemma 4. 7. 2.
G be a group which is not reduced.
There exists a
if and only i f
G"" Q+(f) Z(n),
R+ = G,
and
R ""End(R+)
n
a positive integer. Proof:
+
If
R ""G,
and
G = Q+(!}z(nJ, + R ""End(R ).
then
R = Q(t) (Z/nZ)
is a ring satisfying
Conversely, let R be a ring, with R+ not reduced, satisfying + + + k R ""End(R ). By Lemmas 4.7.1 and 4.7.2, R ""Q (±) (f) Z(p P), p a prime
with
0 ~ kp
m
with
80
k
Z(n) = (t) Z(p/i). i=l
Corollary 4.7.10: Then R ~ End(R+) integer n. Proof:
Let R be a ring with R+ a finitely generated group. if and only if R ~ z or R ~ Z/nZ for some positive
Clearly if R ~ Z or if R ~ Z/nZ then
Conversely, suppose that R ~ End(R+) By Theorem 4.7.8,
+
and that R+ is finitely generated.
m
R = Z(n) (+) ff.\ G., with i =1
R ~ End(R+).
1
r(G 1.) = 1, and
t(G1.) 'I t(GJ.)
fori 'I j, i,j = l, ••• ,m. Since Gi is finitely generated, Gi ~z+ for i = l, ••• ,m, and since t(G.) 'I t(G.), R+ ~ Z(n)l+)z+. For every prime p, + + 1 J + + pZ 'I Z , so by Theorem 4.7.8 either R = Z(n) or R ~ z+. Since R is a ring with unity, either R ~ Z or R ~ Z/nZ. Theorem 4.7. 11: Let R be a ring with R+ a mixed group, S the set of primes p for which R; 'I 0, and such that R+ has no nonzero elements of infinite p-height for every p E S. Then R ~ End(R+) if and only if: k
(l)
Rp ~ Z/p Pz, 0 < kp
(2)
R is a unital subring of U = rr Rp, and pES R+ is a p-pure subgroup of u+ for every p E
(3)
pES,
s.
Proof: Suppose that R satisfies (1) - (3). The exact sequence 0 ~ Rt ~ R ~ R/Rt ~ 0 induces the exact sequence ++ + + + + +. (*) 0 ~ Hom(R /Rt' R) ~ End(R) ~ Hom(Rt' R ). Since R 1s p-pure in
+ U
for every p E S, R+/R; is p-pure in U+/R; for every p E S. Therefore the fact that U+/R; is p-divisible for every p E S implies that R+/R; is p-divisible for every p E S. This together with the fact that R+ has no nonzero elements of infinite p-height for p E S, yields that + R+ ) = 0. From l* ) we now have that t he map End( R+) ~ Hom( R+t' R+) Hom ( R+/Rt' via ~ ~ ~t = the restriction of ~ to R~ is one-to-one. Since ~(R;) ~ R; for every ~ E End(R+), this map is an embedding of End(R+) into End(R~). However End(R~) ~ U, lin [36, Corollary 43.3] it is shown that End(R;) ~ u+, and it is easy to verify that the group isomorphism constructed there is in fact a ring isomorphism). Therefore, for every f E End(R+), there exists a E U such that f(x) =ax for all x E R. Hence the map R ~ End(R+) via a~ at = left multiplication by a is an 81
epimorphism. R ~ End(R+).
Since a (1)
a,
this map is a monomorphism, and so
Conversely, let R be a ring with R+ a mixed group, and R ~ End(R+). Condition (1) follows from Lemma 4.7.2, condition (2) from Lemma 4.7.4, and condition (3) from Observation 4.7.5. Theorem 4.7.11 yields the following improvement of Lemma 4.7.4: Corollary 4.7.12: Let R be a ring with R+ a mixed group, S the set of primes p for which R; I 0, and let A= {x € R+l hp(x) = ~ for all p € S}. Then (l) A is an ideal in R, (2) R/A is isomorphic to a ring T satisfying (t) Rp < T < II R , (3) R/A ~ End[(R/A)+], and (4) i f pES
A~End(A+),
-
then
- p€S
p
R~A{+)(R/A).
Proof: (1) - (3) are precisely the conclusions of Lemma 4.7.4. Since satisfies the conditions of Theorem 4.7.11, R/A ~ End[(R/A)+]. If + A~ End(A ), then A has unity e. The natural embedding A~ R is reversiblebythemap R~A via x~ex, andso R~A(t)(R/A).
R/A
Lemma 4.7.13: Let R be a unital ring, R~ =the ring of left multiplications in R. Then R~ is a ring direct summand of End(R+). Proof: It suffices to reverse the natural embedding R~ ~ End(R+). reversal is End(K+) ~ R~ via f ~ [f(l)]~, 1 the unity of R. Definition: A commutative ring Lemma 4.7.14:
The
R is called an £-ring if R ~ End(R+).
Let R be a ring with unity.
R is an £-ring. (2) The map R ~ End(R+ ) via x f ~ f(l), l the unity of R.
The following are equivalent:
(1)
(3)
~ x~
is an isomorphism with inverse
End(R+) = R~.
Proof: (1) • (2): By Lemma 4.7.13, R~ is a direct summand of End(R+). + + + + + However R~ ~ R ""End( R ) , and so R ""R (+) K. Let f: R I+) K ~ R be an isomorphism, and let nR+ be the natural projection of R+ (:i=) K onto R+. Then for every x € K, fonR+(x} = 0. However End(R+ (+) K) ""R is commutative, and so nR+of(x) = 0. Since nR+of(x) f(x}, and f is . ( + one-to-one. x = 0, 1.e., K = 0, and so End R) = R~. Therefore the 82
map R ~ End(R+) via x ~ x1 this map is f ~ f(l). (2) • (3):
is an isomorphism.
Clearly the inverse of
Obvious.
(3) .. (1): End(R+) = R1 ""' R. Let x,y e: R, and let yr =right multiplication by y. Since y r e: End(R+) = R1 , there exists z e: R such that Yr = z1 • Hence xy = Yr(x) = z1 (x) = zx. However y = Yr(l) = z1 (1) = z. and so xy = yx, i.e •• R is commutative and therefore is an E-ring. Corollary 4.7.15: Let R be an E-ring, and let is a principal ideal in R.
~
e: End(R+). Then
~(R)
Proof: Since ~ e: End(R+) = R1 • there exists a e: R such that ~(x) = ax for all x e: R, i.e •• ~(R) = aR, and so ~(R) is a principal ideal in R. J;orollary 4.7.16: Let R be an E-ring. If H is a group direct ~ummand of R+, then H is a ring direct summand of R. and so R+ cannot be an infinite direct sum of non-trivial groups. Proof: H is a principal ideal in natural projection of R+ onto H.
R by Corollary 4.7.15.
Let wH be the
+ = wH(l)R. By Lemma 4.7.14(2), H = wH(R)
Put e = wH(l). Then H = eR, and e 2 = e·wH(l) = w~(l) = wH(l) =e. Therefore e is an idempotent. and so H = eR is a ring direct summand of R. Corollary 4.7.17: Let R be a ring with unity. commutative if and only if R is an E-ring. Proof:
Then End(R+)
is
If R is an E-ring. then End(R+) ""'R is commutative.
Conversely. let R be a ring with unity such that End(R+) is commutative. Then for f e: End(R+). x e: R+. f(x) =fox (1) = x of(l) . + r r [f(l)l 1 (x). and so f = [f(l)l 1 • 1.e., End(R) = R1 • By Lemma 4.7.14. R is an E-ring. Corollary 4.7.18: Let R be an E-ring, and let S be a ring with unity + + such that R ""' S • Then R ""' S. Proof: End(S+) ... End(R+) is commutative. so by Corollary 4.6.17, S is an E-ring. Therefore S ""'End(S+) ""'End( R+) ""' R. 83
Corollary 4.7.19: Let R and S be unital rings with torsion free additive groups. and with R+ ~ s+. If R is an E-ring then so is S. Proof: Suppose that R is an E-ring. Then End(R+) is commutative, and so -.--+ + + . + • + End(R) ~ Q ® End(R) is commutative. Now End(S) ~ End(S) ~ End(R ), and so End(S+) is commutative. Therefore S is an E-ring by Corollary 4.7.17. Lemma 4.7.20: Let G .be an E-group, and let ~ E End(G). Then ~G) an E-group. and every endomorphism of ~G) can be extended to an endomorphism of G.
is
Proof: Let R be an E-ring with R+ = G. By Corollary 4.6.15, Ill( G) = eR+, e E R. The multiplication (eX)*(ey) = exy for all x,y E R, the product on the right being multiplication in R, induces a ring structure S on ~(G). Let 1 be the unity in R. Then e = e.l is the unity in S. Clearly s. is commutative. Let f E End ~(G) = End(eR+). It is readily seen that foe~ E End(G). Hence for all x E R+ , f(ex) =foe (l)•x = f(e)·x. Since f(e) E eR, f(e) = ey for some y E R, ~ + and so flex)= (ey)·x = (ey)*(ex). Therefore f = (ey)~, End(S) = S~, and so S is an E-ring by Lemma 4.7.14. Clearly y~ is an endomorphism of G extending f. Theorem 4.7.21: Let G be a finite rank strongly indecomposable torsion free group, and let R be a unital ring with R+ = G. Then R is an E- ring. Proof: Clearly R has trivial left annihilator, so by Therrem 4.6.19, End(R+) is a subring of an algebraic number field. By Corollary 4.7. 17, is an E-ring.
R
The following result will be proved in section 4 of the next chapter. Theorem 4.7.22:
Let R be an E-ring with
R+ a finite rank torsion free
n
group. field
Then Fi,
R= and
1..._-f:\
i =1 +
Ri
Ri, with
Ri
a unital subring of an algebraic number +
a full subgroup of Fi,
i = 1 , ••• ,n.
A stronger version of Theorem 4.7.22 may be found in [13, Theorem 3.12]. Theorem 4.7.23: 84
Let
R be a unital ring.
The following are equivalent:
is an E-ring.
(1)
R
(2)
For f
Proof:
(1)
€
End(R+), f(l)
~
(2) by Lemma 4.7.14.
=0
implies that f
= 0.
(2) ~ (1): Let f € End(R+). Then ~ = f- f(l)~ € End(R+), and ~(1} Hence f = fll)~, i.e., End(R+) = Re and so R is an E-r1ng by Lemma 4. 7.14. Definition: A unital ring R is a T-ring if the homomorphism induced by ~(a~ b) = ab for all a,b € R is an isomorphism. From now on l!llz wi 11 be denoted by
~
~=
= 0.
R ®z R ~ R
•
Theorem 4.7.24: Every T-ring is an E-ring. Proof: Let R beaT-ring, and let d: R ~ R~ R be the map d(x) = 1 ® x for all x € R. It is readily seen that d is the inverse of ~· and hence is an isomorphism. For f € End(R+}, define 1 € Hom[(R ~ R)+, R+] to be the homomorphism induced by the maps f(a ®b) = f(a)·b for all a,b € R. If f(l} = 0, then fd = 0. Since d is an isomorphism, f = 0, which clearly implies that f = 0. Hence R is an E-ring by Theorem 4.7.23. Stronger versions of Theorems 4.7.23 and 4.7.24 may be found in [13, Propositions 1.2 and 1.7]. Theorem 4.7.25: (1}
(2) (3)
Let G be a torsion group.
The following are equivalent:
G is aT-ring group. G is an E-ring group. G is cyclic.
Proof: (1) .. (2) by Theorem 4.7.24. (2} .. (3) by Theorem 4.7.7. (3) • (1}: Let G = Z(n}. Then Z/nZ is aT-ring, with (Z/nZ)+ ~G. The following obvious results will soon be employed: Observation 4.7.26:
If R and S are T-rings, then so is
Observation 4.7.27:
Every quotient ring of a T-ring is a T-ring.
R~
S.
An immediate consequence of Theorem 4.7.25 is: Corollary 4.7.28: Let R be a ring with R+ a torsion group. Then R is a T-ring if and only if R ~ Z/nZ for some positive integer n. 85
The problem of classifying T-rings R therefore reduces to the case; is not a torsion group. The description of these rings is as follows:
R+
Theorem 4.7.29: Let R be a unital ring for which R+ is not a torsion group. Then R is a T-ring if and only if A) R/Rt is isomorphic to a subring of Q, and B) for every prime p for which R; f 0, R; is cyclic, and R+/R+p is p-divisible. Proof: Let R be a T-ring. Clearly, Q is a T-ring, and R/Rt is a T-ring by Observation 4.7.27. Hence Q ~ (R/Rt) is aT-ring, Observation 4.7.26. Now Q ® (R/Rt) is an E-ring, Theorem 4.7.24, and so [Q@ {R/Rt)]+ ~ Q+ by Theorem 4.7.9. Since R/Rt ~ Z ~ (R/Rt) ~ Q ~ (R/Rt)' it follows that r[(R/Rt)+] = 1, which implies that R/Rt is isomorphic to a subring of Q, Theorem 4.7.22. Since R is an E-ring, R; is cyclic for every prime p, Lemma 4.7.2. For p a prime satisfying R; t 0, R+/R; is p-divisible by Lemma 4.7.3, and so R+/R; is p-divisible. Conversely, let R be a unital ring satisfying A) and 8). commutative diagram: 0 -::>
R(i Rt - > R ~ R
1~1
-::>0
t~2 ---i>
----:::> R
where
11i
Since
Consider the
is induced by the maps
~i(a
®b)= ab,
0 •
i = 1, 2,3.
is p-divisible for every prime p for which
Rp t 0,
R®Rt=
(.j:', {RPIXIRP). Itisreadilyseenthatforeachprime p, Rp p prime k is a ring direct summand of R, and so Rp ~ Z/p Pz for some non-negative integer kp. Hence Rp ® Rp ~ Rp' and in fact, every element in Rp ~ Rp may be written in . the form 1 ~a, with l the unity in Rp • Clearly the restriction of ~l to Rp 5ll Rp is an isomorphism for every prime p, and so ~l is an isomorphism.
Again the fact that R+/R; is p-divisible for ev~ry prime p with Rp t 0, yields that R '5!1 {R/Rt) ~ (R/Rt) .x, {R/Rt) ~ R/Rt' and in fact every element in (R/Rt) •Xl {R/Rt) is of the form 1 X1 a, with 1 the unity in 86
R/Rt. that
This clearly implies that ~ 3 is an isomorphism, which now yields is an isomorphism, or that R is a T-ring.
~2
Question 4.7.30: An immediate consequence of Theorem 4.7.29 is the fact that if R is a T-ring for which R+ is not a torsion froup, then 1) r(R+/R;) = 1, 2) t~R+/R~) is idempotent, 3) R; is cyclic for every prime p, and 4) for every prime p for which RP 1 0, R+/~ is p-divisible. Is every group satisfying 1) - 4) the additive group of a T-ring? A positive answer, together with Theorem 4.7.25, would provide a complete description of the T-ring groups.
87
5 Torsion free rings
§1. Notation, definitions, and preliminary results: Most of the material in this chapter is the work of Beaumont and Pierce on torsion free rings, i.e., rings with torsion free additive group, (7], [8], and [53]. In general for n a group property, a ring R is said to be a n-ring if R+ is a n-group. A subring S of R is said to be a n-subring of R if s+ is an-subgroup of R+. Very roughly speaking, the idea of this chapter is to embed a torsion free ring R into the Q-algebra R* = Q® R. It will be shown that if R* satisfies certain properties, then· considerable information is obtained concerning R+. All rings in this chapter are assumed to be associative. Let R, S be subrings of a torsion free ring T. Then R and S are quasi-equal, R ~ S, if there exists a positive integer n such that nR S, and nS R. Two torsion free rings R, S are quasi-isomorphic, R & S, if they are isomorphic to quasi-equal rings.
=
=
Notation: dlG) = the maximal divisible subgroup of a group G. Z(R) =the center of a ring R = {x
€
Rl xy = yx for all y
€
R}.
R* = Q(x) R, R a torsion free ring. The symbol ® will often be omitted in order to simplify notation, and in order to view R as a full subring of
R* • The elements of R* will be written as
k
i = 1 •••• ,k.
Definition: A torsion free group G is quotient divisible if G possesses a free subgroup F such that G/F is a divisible torsion group.
88
Quotient divisible will be abbreviated, q.d. Definition: R* is called the algebraic type of a ring R. Two rings, . R* ~ s* . A torsion free group G admits R,S have the same algebra type 1f a multiplication of algebra type A, A a Q-algebra, if there exists a ring R such that R+ = G, and R* ~ A. Theorem 5.1.1: type.
Quasi-isomorphic torsion free rings have the same algebra
Proof: Without loss of generality we may consider the case R,; S, R,S torsion free rings, i.e., there exists a positive integer n such that nR ~ S, and nS ~ R. Then R* = Q@ R = Q ~ nR ~ S* = Q ~ S Q ~ nS ~ Q ® R = R* , and so R* = S* • An immediate consequence of Theorem 5. 1.1 is Corollary 5.1.2: Let A be a Q-algebra, and let G,H be torsion free groups, G Q. H. Then G admits a multiplication of algebra type A if and only if H does. Lemma 5. 1.3: Let R be a torsion free ring, and let P = P(x 1 , ••. ,xn) be a homogeneous polynomial with coefficients in Z, such that x1 , ••• ,xn are non-commuting variables. Then R satisfies the polynomial identity P if . R* does. and only 1f Proof: Suppose that R satisfies the polynomial identity P. Let * Since R is a full subring of R* there exists a positive a 1 , ••• ,an E R. integer m such that rna; E R, i = l, ••• ,n. Let d = deg P. Then mdP(a 1 , ••• ,a ) = P(ma 1 , ••• ,rna ) = 0. Since R* is torsion free, n * n P(a 1 , ••• ,an) = 0 and R satisfies P. The converse is obvious. Corollary 5. 1.4: Let G and H be quasi-isomorphic torsion free groups. (1) G is the additive group of a non-zeroring satisfying a homogeneous polynomial identity P = P(xl'··· ,xn) if and only i f H is. (2) Every ring with additive group G satisfies P if and only if every ring with additive group H satisfies P. Proof:
Corollary 5. 1.2 and Lemma 5. 1.3.
Lemma 5.1.5:
Let R be a torsion free ring.
R is T-nilpotent if and only 89
if R* is T-nilpotent. Proof: Suppose that
R is T-nilpotent.
= Let {ai}i=l
~
R* .
There exists a
positive integer mi such that miai E R, = 1,2, .••. Since T-nilpotent, there exists a positive integer k such that (m 1a1)(m2a2 )k ... (mkak) = 0. k
torsion free
IT
i=l
a.
1
Therefore
k
k
( n mi)( n a;) = 0. . i=l i=l
R is
Since
*
R
is
= 0, and R* is T-nilpotent.
Corollary 5. 1.6: Let G,H be quasi-isomorphic torsion free groups. (1) G is the additive group of aT-nilpotent ring which is not a zeroring if and only H is. (2) Every ring with additive group G is T-nilpotent if and only if every ring with additive group H is T-nilpotent. Proof:
Corollary 5.1.2 and Lemma 5.1.5.
Lemma 5.1.7: Let R be a torsion free ring. a an arbitrary ordinal. R is a-nilpotent if and only if R* is.
Then
Proof: An easy induction argument shows that (R*) 6 = (R6)* for every ordinal 6. Therefore if R is a-nilpotent, then (R*)a = (Ra)* = 0. The converse is obvious. Corollary 5.1.8: Let G and H be quasi-isomorphic torsion free groups, a an ordinal. (1) G is the additive group of an a-nilpotent ring whicn is not a zeroring if and only if H is. (2) Every ring with additive group G is a-nilpotent if and only if every ring with additive group H is a-nilpotent. Proof: §2.
Corollary 5.1.2 and Lemma 5.1.7.
The Beaumont-Pierce Decomposition Theorem:
The group theoretic analogue of the Wedderburn Principal Theorem, which was used in Chapter 4, section 5, will be proved here. Since the proof involves several steps, it will be stated before the preliminary results leading up to it are given. Theorem 5.2. 1: Let R be a finite rank torsion free ring, R* ~·~N a Q-space decomposition of R* , with N the nil radical of R* , and ~ a 90
semisimple subalgebra of R*. Put S = R n s, N = R n N. Then S is a subring of R satisfying s* = S, N is the maximal nilpotent ideal of R, and satisfies N* = -N. In addition S (+) N has finite index in R. Lemma 5.2.2: Then Proof:
~:
R+
Let s 1
=
I
{x e: S there exists y e: R" for which x+y e: R}.
R+I (S (t) N) + "" S~ /S +:
~
Let w
be the natural projection of R* onto S.
s+1/S+ 5 via
~(z)
= n
(z) + s+ for all
~
z e: R+.
Define
Clearly
~
is an
epimorphism, and (S ® N)+ ~ ker ~. Let z e: R, z = x+y, x e: S, y e: N. Suppose that z e: ker ~. Then x e: S, and so y = z- xe:R n N = N, i.e., r.'\+ r.'\+ + + ++ z E (S 1+1 N) , and so ker ~ = (S ':r' N) , or R /(S + N) ""S 1/S • Lemma 5.2.3:
S*
= -S, and N* = -~.
Proof: S* is the unique minimal Q-algebra containing S. It therefore suffices to show that S is a full subring of s. Let x e: S. Then x e: R* and so there exists a positive integer n such that nx e: R. Hence nx e: R n S = S and so s+ts+ is a torsion group. The same argument shows that N* = -N. Lemma 5.2.4: There is a Q-basis {x 1 , ••• ,xm} for s such that the free subgroup F of S generated by this basis is a subring of S. Proof:
Since s+ is a full subgroup of
z1 , ••• ,zm
in S is a Q-basis for
~.
s, a maximal independent set m
z.·zJ. =
~
a .. kzk with k=l lJ a.1J"k e: Q; i ,j,k = l, •.• ,m. Let n be a positive integer such that naijk e: Z for all i,j,k = 1 , ••• ,m. Then {X; = nz; , i = 1 , ••• ,m} satisfies the conditions of the lemma. Observation 5.2.5:
Hence
1
S~/F is a torsion group.
Proof: Clearly S+/F is a torsion group. It therefore suffices to show that S~/S+ is a torsion group, or by L;mma 5.2.2 that R+/(S <±> N)+ is a torsion group. Let x e: R. Then x e: R , and so x = y+z, y e: S, z e: N. By Lemma 5.2.3 there exists a positive integer n such that ny e: S, and nz e: N. Hence nx = ny + nz e: S(+)N, and R+/(Sq)NJ+ is a torsion group. Lemma 5.2.6:
For all but finitely many primes
p,
(S~/F)p, and
(S+/F)p 91
are divisible and equal. The proof of Lemma 5.2.6 involves many steps, and will be broken down into a series of claims. First we introduce some notation. The degree of nilpotence of W will be denoted by t. For p a prime, k > 0 an integer, Ik = {x € F I p-k x € S}, and Jk = {x € Fl p-k x € s1}. Claim 5.2.7: {1) F = I 0 2 I 1 2 I 2 2 ... ,
=Jl =J2 =···•
(2)
F = Jo
{3)
(4)
I k !: Jk' Ik and Jk are two sided ideals in F,
{5)
Ik. It ~ IkH '
{6)
Jk.Jt!: Jk+t' and Jkt ~ I k for a 11 k ,t
(7)
Proof:
(1) - (4)
~
0.
are obviously true.
Let x1 € Ik, x2 € I 1 . Then x1 = pky 1 , and x2 = p1y2 , y1 ,y 2 € Hence x1x2 = pk+ty1y2 , with y1y2 € S, i.e., x1x2 € Ik+t . {6): follows from the same argument used to prove (5). (5):
s.
(7): Let x € Jk. Then x = pky, y € s 1 , and there exists z € N such that y-z € R. Let x1, ••• ,xt € Jk, and let z1, ••• ,zt € N such that p-kxi - zi € R, i = 1, ••• ,t. Suppose, inductively, that p-kx. x. 11 12 for l~i 1
x. - z. z ..•• z. €R 1j 11 12 1j
( .. k
p x1x2 ••• xjxj+l - z1z2 ••• zjzj+l
x1 p x2 .•• xj+l - z2 ••• zj+l) +
(p -k x1 - z1) x2 ••• xj+l - pk (p -k x1 - z1) (p -k x2 ••• xj+l - z2 ••• zj+l) € R. Therefore p-kxl •.• xt - z 1•••• zt € R. However z1 .•• zt = 0, and so p-k x1 ••• xt € s1 n R!: -S n R = s. Hence x1 xt € Ik, i.e., Jkt ~ Ik. t_t t_t f Claim 5.2.8: Ik+t n p F- p Ik' and Jk+t n p F- p Jk or all k,t ~ o. Proof: Let x € Ik+t n p1 F. Therefore x 92
=
p1 (pky)
=
Then x
=
pk+ty
=
p1 z, y € S, z € F.
p1 z. Since R*+ is torsion free,
pky
=
z, and so
z € Ik which yields that x € p~Ik. Conversely, let x € p~Ik. Then x = p~y. y € F, and y = pkz, z € S. kH . ~ ~ Hence x = p z, and x € Ik+~· Clearly x € p F, so x € Ik+~ n p F. The same argument replacing ~
I by J,
~
and S by s 1 yields that
= P Jk. Notation: T = S+/F, T1 = st/F. Jk+~
n PF
. Cla1m 5.2.9:
(2)
k)~
+
+
~
(1) (p Tp [p ] ""lkH/[p F n Ik+~] ,
k
+
~
~
(p Tlp)[p ] ""Jk+~/[p F n Jk+~]
and
+
Proof: (1) Let x € Ik+~· Then there exists a unique element y € S such that x = p~y. Define ~= Ik+~ ~ T via ~(x) = y+F. It is easily verified that ~ is a homomorphism. x
= pk+~z.
z
€
Since
p~y
€
S, which implies that y
1;+l
=F,
~(xJ
€
TP[p~].
and so ~(x)
= pkz,
€
Now
(pkTP)[p~].
Let w € S+ be such that w + F € (p k TPJ[p ~ ]. Then w = pk z, z € S+ and R, kH kH . p w = p z € F. Let x = p z. Then x € Ik+R-" and ~(x) = w+F, 1.e., ~ is an epimorphism. Now x € 1;+t belongs to ker ~ if and only if p- 1 x € F whicn occurs if and only if x € pR-F, i.e.,. ker ~ = [p~F n Ik+t]+. Therefore (pkTP)[pR-] ""1:+~/[p~F n Ik+~l+. A similar argument yields (2). Claim 5.2.10: Let p be a prime. T1P are divisible and equal.
If F/pF
is semisimple then TP and
Proof: Let ~: F ~ F/pF be the canonical ring epimorphism. and the semisimplicity of F/pF yield: (A)
~(Ik)
=~(Jk)
(B)
~(I k)
= [~ ( I k) ]2 = ~(I 2k)
. By Cla1m 5.2.9,
= [~(Jk)lt = ~(J~) ~ ~(lk),
Claim 5.2.7
and
~ ~ ( I 2k) ~ ~ ( I k) , for a11 k > 0. k + + + p Tp[p] ""lk+l/[pF n lk+l] ""~[Ik+l] =
+ ""Jk+l/[pF n Jk+ll + ""(p k T1p) [p] for all k > o. Choosing k = 0 yields that Tp[p] ""Tlp[p]. Since Tp[p] ~ T1P[p], and Tlp[p] is finite. ~[Jk+ll
+
this implies that TP[p] = T1P[p]. r{ TP)
For every G,
r(Gp) = r(GP[p]).
Hence
= r( T1p) •
93
Claim 5.2.9 and {B) yield that {pkTP)[p] ~ I~+l/[pF n Ik+l]+ ~ ~(lk+l) + + 2k+l = ~{I 2 k+ 2 ) ~ I 2k+ 2/£pF n 12k+ 21 ~ {p TP)[p] for all k > o. This together with the finiteness of T [p] yields that Tp(p] = {pTP)[p] = {p 3Tp)[p] = •.• P This clearly implies that Tp is is divisible. Hence Tp is a divisible subgroup of Tlp with r{Tp) = r{T 1p) < ~. This implies that Tp = Tlp' In order to proceed it is necessary to introduce some facts concerning a finite dimensional algebra A over a field K of characteristic 0. The reader is referred to [14] and [47] for proofs. Defi ni ti on: Let n ax. = I: a .. x., a .. 1 j=l 1J J 1J
be a basis for A, €
K,
i,j = l, ••• ,n.
and Jet a € A.
The trace of a,
Then
t{a) = t {aij)
= the trace of the matrix Proposition 5.2.11:
t(a}
is independent of the choice of basis for A.
Definition: Let X= {x1 , ••• ,xn} be a basis for A. The discriminant of A with respect to X is the determinant of the nxn matrix (t{xixj}). Proposition 5.2.12: Let X,Y be two bases for A, and let dX' dy be the discriminant of A with respect to X and Y respectively. Then there exists a € F, a~ 0, such that dy = a 2dx· Proposition 5.2.12 makes possible the following: Definition: Let A be a finite dimensional Q-algebra. A has positive discriminant if the discriminant of A with respect to a Q-basis X of A is positive. Notation:
Let {x1 , ••• ,xn} be a basis for A over K,
n
and let L : K be
a field extension. AL is the set of all expressions { I: a. x. 1 a 1 € L}. i=l 1 1 n n For x = I: aixi' y = I: a1xi two arbitrary elements in AL' define i=l i=l n n x+y = I: {a. + a1.)x 1., and x·y = I: a.aJ.c .. kxk' where . . k-1 1 1J 1.=1 1 1 n
.J. -
x1xJ. = I: c .. kxk, c. "k € K, i ,j ,k = 1, ... ,n. k=l 1J 1J induce a ring structure on AL. 94
These operations clearly
Definition: A is separable over K, extension L ~ K.
if AL
is semisimp!e for every field
Definition: Let A be a Q-algebra with unity e. An order in A is a subring B of A satisfying: l) e E B, 2) B possesses a basis for A, 3) every element in B is a root of a monic polynomial with coefficients in Z. The above definition and the following Proposition may be generalized to algebras over arbitrary fields of characteristic 0. We will assume, from now on, that A is a finite dimensional Q-algebra. Proposition 5.2.13:
Every order in A may be embedded in a maximal order.
Proposition 5.2.14: discriminant.
A is separable over Q if and only if A has non-zero
Proposition 5.2.15: If A is semisimple then every nonzero ideal in A is a unique product of prime ideals. If I ~ J are nonzero ideals in A, and if mJ c I for ~orne positive integer m, then the same prime factors occur in the prime decomposition of I and J. Proposition 5.2.16: Let B be a maximal order in A, X a basis for A, X~ B, such that the discriminant of A with respect to X is an integer, and let p be a prime. Then B/pB is semisimple if and only if p does not divide the discriminant of A with respect to X. We are now in a position to prove Lemma 5.2.6 as follows: Proof of Lemma 5.2.6: By Claim 5.2. 10 it suffices to show that F/pF is semisimple for all but finitely many primes. The subring B of ~ generated by F and the unity e E ~ is an order in ~. By Proposition 5.2.13, there exists a maximal order ~ in ~ such that s~ B. Since is separable over Q, ~ has non-zero integral discriminant d with respect to a basis contained in ~. Proposition 5.2.14. By Proposition 5.2.16, ~/p~ is semisim~le for every prime ptd. It therefore suffices to show that F/pF ~ ~/pB for all but finitely many primes p. Let 8p: ~ ~ B/pB be the canonical epimorphism, and let 8~ be the restriction of 8P to r. It suffices to show that 8~ is onto for all but finitely many primes p. There exists a positive integer n such that n~ ~ F, see the proof of Lemma 5.2.4. Let p be a prime such that Ptn.
r
95
Then if + pif = nB" + pir ~ F + pB ~ if + plf, and so epF i s on to . concludes the proof of Lemma 5.2.6.
Th i s
In addition to being a step towards proving Theorem 5.2.1, Lemma 5.2.6 also implies the following: Corollary 5.2.17: Let G be a finite rank torsion free group. If G admits a multiplication of semisimple algebra type then G is quotient divisible. . . h R+ = G such that R* is semisimp 1e, an d Let R be a r1ng w1t let F be the subring of R = S construct~d in Lemma 5.2.4. By Lemma 5.2.6, (R/F); is divisible for all but finitely many primes p. For every prime p, {R/F); = DP C±) Cp, with DP divisible, and CP reduced. Since
P f _!QQ_:
r[(R/F);] is finite, Cp is finite for every prime p. This together with the fact that Cp = 0 for all but finitely many primes p yields that G/F = D(i) C, D a divisible torsion group and C a finite group. Let n be a positive integer such that nC = 0. The restriction of the canonical epimorphism a: G ~ G/F to nG is a homomorphism onto D with kernel nF. Hence nG is q.d. Since G ~nG, G is q.d. Observe that the argument at the end of the proof of Corollary 5.2.17 shows that if a torsion free group G possesses a free subgroup F such that G/F is the direct sum of a divisible torsion group, and a bounded group, then G is q.d. In order to complete the proof of Theorem 5.2.1 it is necessary to investigate the relationship between TP and T1P for those primes p for which Tp is not divisible. This is done in the following: Lemma 5.2.18:
T1p/TP is finite for every prime p.
Again several steps will be required to prove the lemma. As above, B" is a maximal order in ~ containing F and the unity e in ~. and n is a positive integer satisfying nB" c F. Put fk = If lklf, and Jk = B" Jklf for all k > 0. An immediate consequence of Claim 5.2.7 and the fact that the following: Claim5.2.19:
F is
{l)n 2Tk~Ik' {2)n 2JksJk' {3)Tk+lslk' {4)"Jk+ls."Jk,
{5J Tk :"Jk' {6) nTk Tt =.Tk+t' 96
n~ c
{7) nJk Jt :"Jk+t' {8) nt-l J~sTk'
Since B is semisimple, every nonzero ideal in B is a product of prime ideals. Let P1, ... ,Pu be the prime ideals in B which are factors of the ideals pB, nB, r 1 , or ~1 . The relations n ~ ~ r 2k' and nJ~ ~ n~2 k' which follow immediately from Claim 5.2.19 (6) and (7), together with Proposition 5.2.15 yield that P1 , ..• ,Pu are the only prime factors of Tk, al a2 au and "Jk for all Pl P2 Pu ~1
and
nlr = P1 6
J = p kl k
non-negative integers for all
1
k ~ 1'
i = 1 ' ••• ,u.
Claim 5.2.20: (1) (t-1)~; + toki ~ Yki' + Yki for all k ~ 1, i = l, ••• ,u. Proof:
and (2)
2~;
+
max{~a;, Yk+~,i} >~a;
Claim 5.2.19 (8) implies (1), and Claim 5.2.19 (9) implies (2).
Claim 5.2.21: There exists an integer K(t) such that for all min{Yk., ~a.}< 26. + ok 1• for all = 1 , ••• ,u. 1 1 1
k
~ K(~).
If a; = 0 or if Yki ~ 2~; for all k then the inequality clearly Assume, therefore, that ai ! 0, and that Yk ; > 2e; for some k0 .
Proof: holds.
0
Since the sequence {Yki};=l
is non-decreasing, Yki > 2~;
Therefore 5.2.20(2) implies that (2')
2~.
~ ~ ~.
k
~ ~.
1
+ Yk+~, 1.
>~a.
-
1
+ Yk.1
Yk+~,i ~~a;
for all
k
~
for all
k
k0 . Since a 1.
for all
k ~ k0 •
~
k0 and so
;
0,
so by (2') Yki ~ ~ as k ~ ~. and by Claim 5.2.20(1), This clearly implies Claim 5.2.21.
~a,. ~ ~
as
~ ~
as
oki
An immediate consequence of Claim 5.2.21 is Claim 5.2.22: For every non-negative integer ~. ~1'5" 2K(~) such that for k ~ K(~). Ik +Po~ n Jk.
there exists an integer
Proof of Lemma 5.2.18: Let p be a prime. Since r{TP) ~ r(T 1p) < ~ it suffices to show that d(Tp) = d(T 1P), or since TP .= Tlp' that d(Tlp) ~ d(TP). By Claim 5.2.21 there exists an integer K(~) such that 97
for k
~
K{R.),
4
n Jk
~
5.2.19(1), and n~ ~F.
4-.-
n Jk
~
2-
R.2
n Ik + p n B.
2-
However n Ik
{A) n4Jk ~ Ik + pR.F for all
Hence
~
Ik,
Claim
k ~ K(R.).
eR.: F ~ F/pR.F be the canonical epimorphism. By (A) eR.{n 4 J~) ~ oR.{l;) + + R. R. + R. + for all k > K{R.). Now oR.{Jk) = {Jk + p FJ/p F ~ Jk/lp F n Jk) ~ Let
{pk-R.Tlp)[pR.]
by Claim 5.2.9(2).
eR.{I~) ~ {pk-R.Tp)[pR.].
Let n4 = pjm, m a positive integer such that
Then n4(pk-R.T 1P)[pR.] = {pk+j-R.Tlp)[pR.-j]
{p,m) = 1.
subgroup of (pk-R.Tp)[pR.]. (pk-lTlpJ[p] r {pk-1 Tlp ) pk-R. Tp
=
Similarly, employing Claim 5.2.9{1),
~
Put
R. =
j+l
is isomorphic to a
and let k ~ max{K(R.), R.}.
Then
is isomorphic to a subgroup of {pk-R.TP)[pR.], and so r {pk-R. Tp ) • For
d{TP ) • Hence
d{Tlp )
k sufficiently large ~
pk-1 T1P = d{Tlp ) , and
d ( TP ) and the claim is proved.
Finally, we can prove Theorem 5.2.1. Proof of Theorem 5.2.1: All that remains to be shown is that S (+) N is of finite index in R, Lemma 5.2.3. By Lemma 5.2.2 it suffices to show that S~/S+ is finite. S~/S+ ~ (S~/F)/(S+/F) = T1/T eo~ {t). Tlp/Tp' p a prune Observation 5.2.5 and Proposition 1.1.1. Now T1p/Tp is finite for every prime p, Lemma 5.2.18, and 0 for all but finitely primes, Lemma 5.2.6. Hence S~/S+ is finite; thus concluding the proof of Theorem 5.2.1. Translating Theorem 5.2.1 to groups, we have: Corollary 5.2.23: Let G be a finite rank torsion free group. Then H$ K, where H admits a multiplication of semisimple type, and K is the additive group of a nilpotent ring N satisfying N2 ~ 0, unless K = 0, or G is nil. G c>.
§3.
Torsion free rings with semisimple algebra type.
The results of the previous section show that, up to quasi-isomorphism, the classification of the additive groups of finite rank torsion free rings reduces to two cases; the additive groups of rings with semisimple algebra type, and the additive groups of nilpotent rings, which are not zerorings. In this section a further reduction will be obtained by showing that a finite 98
rank torsion free group admitting a multiplication of semisimple algebra type is quasi-isomorphic to a direct sum of groups admitting multiplications of simple algebra type. It will further be shown that a finite rank torsion free group admitting a multiplication of simple algebra type is quasiisomorphic to a direct sum of groups admitting multiplications of field type. Therefore, upto quasi-isomorphism, the classification of the additive groups of finite rank torsion free rings with semisimple algebra type is settled by determining the additive groups of full subrings of algebraic number fie 1 ds. Theorem 5.3.1: Let R be a finite rank torsion free ring with semisimple algebra type. Then R contains a subring S = s 1 (+) .•• fi:)Sm such that S; has simple algebra type, i = 1, ... ,m, and R+ ;s+ is finite. To prove Theorem 5.3.1 we first need: Lemma 5.3.2: Let R be a torsion free ring with Q-algebra with unity e. Let R; =RnA;. Then ..... + is bounded. and R+/(Rl C-:+"J. .@ f\n)
R*
= A1 <±) ••• <±) ~ a * R; = A; , 1. = 1 , ••• ,m,
=
Proof: For each i = 1 , ••• ,m, A7/R7 = A~/(R n A;)+~ (A7 + R+)/R+ R*+/R+ which is a torsion group. Therefore for x E A., there exists a positive 1 integer n such that nx E R1., or x E-n R.. 1 Hence A. c QR. = R*.. * 1 1- 1 1 However R1. is the minimal Q-algebra containing R., and so R~ =A .• 1 1 1 Let e = e 1 +•.• + em, e; E A;, i = 1, ••• ,m. Clearly e; is the unity in Ai' i = 1, •.• ,m, and {e 1 , ••• ,em} is a set of mutually orthogonal idempotents in R* • Let n be a positive integer such that ne; E R for . * x = x1 + ••• + xm• X; E A;, l, •.• ,m. and let xE R. S1nce xE R, i = l, ..• ,m. For each i = l, ••• ,m, ne.x. = ne 1.x and so 1 1 m m neixi ERn A; = R;. Therefore nx = nex = I: ne; xi E ® R; , and so i=l i=l + m + R I (f) R; is bounded. i=l m * Proof of Theorem 5.3. 1: Since R* is semisimple, R = c+.) A;, with A. a ; =1
simple Q-algebra, Lemma 5.3.~, S;* Let S =
{+) ; ,; 1
S ..
1
= 1, ••• ,m. As in i = 1 , ••• ,m. Put S; = R n A;, 1 , ••• ,m. A; and so S; has simple algebra type, Then
R+/S+
is a finite rank bounded group, Lemma 5. 3. 2 ,
1
and hence finite. 99
Corollary 5.3.3: Let G be a finite rank torsion free group. Then G admits a multiplication of semisimple algebra type if and only if G is quasi-isomorphic to a direct sum of groups admitting multiplications of simple a 1gebra type. Proof: If G admits a multiplication of semisimple algebra type, then it follows immediately from Theorem 5.3. 1 that G is quasi-isomorphic to a direct sum of groups admitting multiplications of simple algebra type. m
Conversely let G,;, (+) G., and let R. be a ring satisfying R: i =l 1 1 m 1 and Ri* is a simple Q-algebra, i = 1, ... ,m. Put R = (+) R.. Then i =1
m * is semisimple. R* = (±) Ri
1
i=l Definition: Let R be a finite rank torsion free ring with simple algebra type. A subfield F of Z(R*) is a field of definition of R if there exists an F-basis x1 , ••• ,xk has finite index in R.
of R* in
R such that S
k
@ (R n F)x.
i =1
1
From now on R will be assumed to be a finite rank torsion free ring, R* will be denoted by A, and F will signify a subfield of Z(A). Most of the remaining results in this section appear in a more general setting in [53]. Lemma 5.3.4: + + HomF(A , A )
F is a field of definition of R if and only if • + End(R ).
=
k
be an F-basis of A in
Proof:
R.
Put S = 1+l (R n F)x .• i =1
1
Suppose that F is a field of definition of R. Then there exists a positive integer n such that nRc s. Let ~ E HomF(A+, A+). Since R is a full subring of A, there exists a positive integer m such that m.p(x.) E R, i = l, ••• ,k. and ~o ~ E End(R+).
Now mn~(R)
=m
+ + Conversely, suppose that HomF(A , A ) ni: A .... F via
n.( 1
k I:
j=l
f.x.) = fi; J J
fJ.
E
. ~(S)
k (+) (R n F)m ~(X;)~ R,
i=l . + End(R ). Define
= F,
i,j = l, ... ,k.
Clearly
ni E HomFlA+, A+), and so there exists a positive integer n such that
100
nn.(R) c R, 1
-
i = l, •.• ,k.
Let x E R.
Then
x=
k
k
~
i=l
n.(x)x., and 1
1
nx = ~ nn.(x)x. E S, i.e., nRc S. Hence R+/S+ is a finitely generated 1 1 i=l bounded group. Therefore R+/S+ is finite, and F is a field of definition of R. Lemma 5.3.5: Let A be a simple Q-algebra, A~, Ar the subrings of HomQ(A+, A+) consisting of left and right multiplications, respectively, by elements of A. Then for every subring E of Ho~(A+, A+) containing both A~ and Ar, there exists a subfield F Z(A) such that + E = HomF(A+ , A).
=
Proof: The fact that E is a simple Q-algebra, and that E ~ Ar' A~ implies that A+ is an irreducible E-module. Put F = {~ E HomQ(A+,A+)I~ ~ = ~~ for all ~ E E}. Since the elements of F commute with the elements of A~ and Ar' F is a subfield of Z(A). The Jacobson Density Theorem, and the finite dimensionality of A over Q yield that E = HomF(A+, A+). Theorem 5.3.6: Let A be a simple Q-algebra. Then F = {~ E HomQ(A+, A+) I~~=~ for all ~ E End(R+)} is the unique smallest field of definition of R, and HomF(A+, A+J = End(R+J. Proof:
Since End(R+)
is a subring of HomQ(A+, A+), and End(R+)
=A~,Ar•
End(R+) = HomF(A+, A+J, Lemma 5.3.5, and proof of Lemma 5.3.5. By Lemma 5.3.4, F is a field of definition of R. Let F' be an arbitrary field of definition of R. Again by Lemma 5.3.4, HomF,(A+ ,A+) End(R+) = + + + + HomF(A, A). Therefore F' = {~ E HomQ(A. A )I ~ = ~ for all + + + + ~ E HomF,(A , A)}~{~ E HomQ(A , A )I~~=~~ for all + + ~ E HomF(A , A )} = F.
=
In what follows, A will be assumed to be a simple Q-algebra, smallest field of definition of R, and dimF(A) = k. •
+
Corollary 5.3.7: Z[End(F )] = F. If e E R, e the unity in Z[End(R+)1 = R n F. + + • + Proof: F = Z[HomF(A , A )] = Z[End(R )] by Theorem 5.3.6. +
•
F the
A,
then
+
Suppose that e E R, and let a E Z[End(R )]. For ~ E End(R ) there exists a positive integer n such that n~ E End(R+). Since 101
a E Z[End(R+)], a(n ~) = (n ~)a. This clearly implies that n(~) = n(~). + + Since End(R ) is torsion free, ~ = ~ and so a E Z[End(R )] = F. However a= a(e) E R, and so a E R n F. Clearly R n F c Z[End(R+)]. 0
0
Corollary 5.3.8: Suppose that A= E, and that e E R, e the unity of A. Then R = End(R+). + = HomF c+ + + 1s . F , F)= F, and so End(R) Proof: By Theorem 5.3.6, End ( R) commutative. Therefore by Corollary 5.3.7, End(R+) = Z[End(R+)l = R n F = R. Corollary 5.3.9: Let H,K be subgroups of R+ such that H n K = 0. R+ = Hq.) K, then QH and QK are F-subspaces of A.
If
Proof: Clearly A = QH (i) QK. Let nH be the natural projection of A onto QH. There exists a positive integer n such that nR =: H(+) K. Therefore nnH(R) = nH(nR): nH(HG)K) = H: R, and so nH E End(R+) = HomF(A+, A+). Hence QH and similarly QK, are F-subspaces of A. Corollary 5.3.10: Let A be a field. strongly indecomposable. Proof:
Then A= F if and only if R+ is
Let x1 , ••• ,xk be an F-basis for A in
R.
Since F is a field of
k
R:! ® (R n F)xi. i=l indecomposable, k = 1, and A= F.
definition for R,
Therefore i f R+ is strongly
Conversely, if R+ is not strongly indecomposable, then by Corollary k = dimFA ~ 2, and so A ~ F.
5.39,
Piecing together the results of this section we obtain Theorem 5.3.11: Let G be a torsion free group of finite rank. admits a multiplication of semisimple type if and only if: (1)
Then G
G is quotient divisible, and k
G C. C±> G., with G. a strongly indecomposable group which admits a 1 i;,;l 1 . multiplication of field type End(Gi)' such that End(Gi) is. an algebraic n.umber field satisfying [End(Gi): Ql = r(Gi), i = 1, ... ,k. (2)
0
Proof: Suppose that G admits a multipl1cation of semisimple type. Then G is quotient divisible, Corollary 5.2.17. By Corollary 5.3.3 we may assume that G admits a multiplication of simple algebra type. Let R be a ring 102
with R+ = G and R* a simple Q-algebra. Let F s Z{R*) be the smallest field of definition of R, and let x1 , ••• ,xk be an F-basis for R* in R. Then
R,;
k
_,'fl
(R n F)xi.
*
1=1
Put Ri = (R n F)xi,
and
+
(R n F)xi = Gi,
i = l, ..• ,k. Then R. ~ F and so G. admits a multiplication of type F, 1 * * a. field. By Corollary1 5.3.8, R.1 = End(G.) 1 and so F = R.1 = End(G.) 1 = End(G 1.), i = 1, .•• ,k. The indecomposability of G., i = 1 , ••• ,k follows . * 1 from Corollary 5.3.10. Now [End(Gi): Q] = dimQRi = r(Gi), i = 1 , ••• ,k. The converse is obvious. 54.
Applications.
The results of the previous sections of this chapter will be applied to solve problems which arose in Chapter 3, and to prove Theorem 4.7.22. k
Proof of Theorem 3.2.3:
Let G = @ Gi
be a finite rank torsion free group,
i =1
and suppose that every ring Ri with R~ ~ Gi is nilpotent. Suppose there exists a ring R with R+ = G such that R is not nilpotent. Then by Corollary 5.2.23, G &. Hl+lK, H ~ 0, where H admits a multiplication of semi simple type. By Theorem 5.3.11 (2), G c.. L® M, with L a strongly indecomposable group admitting a multiplication of field type. Let k. 1 Gi ~ Gl GiJ"' G.J. strongly indecomposable, i = 1 •••• ,k; j = 1, .•• ,k .. j=l
1
.
1
By Jonsson's theorem, [36, Theorem 92.5], Gi j c.. L for some 1 ~ i ~ k, and for some 1 < j < k1.• Now G. .;. G.. @H ..• By Corollary 5.1.2, G.. admits 1 1J 1J 1J a multiplication of field type. Let Si be a ring with field algebra type, 5+1. = G.J., and let T1- be any ring with T~1 =H 1J ..• Put R.1 = S.(±)T 1 1 1.. * * * * Then Ri = Si (!) Ti with Si a field. By Lemma 5.1.3, Ri is not nilpotent, a contradiction, Corollary 5.1.4(2). In Theorem 3.1.3 it was shown that if G is a torsion free group, r(G) = n < ~. and if R is a nil ring with R+ = G, then Rn+l = 0. The same conclusion follows if R is assumed to be T-nilpotent or a-nilpotent, a an arbitrary ordinal. (Actually, the conclusion for T-nilpotence is obvious since if R is T-nilpotent, then R is nil). Theorem 5.4.1:
Let G be a torsion free group,
r(G)
=
n <
~.
Let R be a 103
ring with R+ = G. If R is T-ni lpotent or a-nilpotent for some ordinal a • then Rn+ 1 = 0. Proof: Suppose that R is not nilpotent. Then R ~ S .J) N. where S is a subring or R of semisimple algebra type. and N is the maximal nilpotent subring of R.
Theorem 5.2.1.
By Theorems 5.3.1 and 5.3.6, S
k
~ (+)
i ~1
s .• 1
k S.* (±1 N* , with Si* R* = (+t i=l 1 a field. = 1 , ••• ,n. Clearly Si* is neither T-nilpotent nor a-nilpotent. so by Lemmas 5.1.5 and 5.17, Si is neither T-nilpotent nor a-nilpotent, i = l •.•• ,k. This clearly implies that R is neither R-nilpotent nor a-nilpotent. Therefore if R is T-nilpotent, or a-nilpotent for some ordinal a, then R is nilpotent, so by Theorem 3.1.3. Rn+l = 0.
with Si
a ring of field algebra type.
Hence
Corollary 5.4.2: Let G be a finite rank torsion free group, The following are equivalent: ( 1) (2) (3) ( 4) (5)
r(G) = n.
Rn+ 1 = 0 for every ring R with R+ = G. Every ring R with R+ = G is nil potent. Every ring R with R+ = G is T-ni lpotent. Every ring R with R+ = G is nil. Every ring R with R+ = G is a-ni 1potent for some ordinal
a.
Proof: The implications (1) • (2) • (3) • (4) and (1) • (5) are obvious. (4) • (1) by Theorem 3.1.3. and (5) • (1) by Theorem 5.4.1. ProofofTheorem4.7.22: ByTheorem5.2.1, R=S(F)N, where S isaring with semisimple algebra type, and N is a nilpotent ring. Since R is unital. N = 0. Therefore R* = S* is a commutative semisimple Q-algebra, * n and so R = €J Fi, with Fi an algebraic number field, i = 1 , ••.• n. Put i =1 n Ri = R n Fi. Then R = (!) Ri , with Ri a unital subring of Fi ' an d R+.1 i =1 + a full subgroup of F1 , i = 1 , ••• ,n.
104
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109
Index
Absolute annihilator, 34 Algebra type of a ring, 89 a-nilpotent ring, 34, 35, 90, 103, 104 Anti-radical group, 75 Ascending chain condition for ideals, 47-49, 53 Associative nil group, 10-15 Associative w-group, w a ring prope_rty, 36 Associative principal ideal ring group, 44, 45, 51 Associative strongly principal ideal ring group, 43-46 Associative strongly subdirectly irreducible ring group, Associative subdirectly irreducible ring group, 61-63 Beaumont-Pierce decomposition theorem, 75, 90-98 Basic subgroup, 2 Center of a ring, 88, 101 Descending chain condition for ideals, 50-61 Discriminant, 94 Divisible group, 1, 3, 4 Division ring group, 36, 37 E-group, 84-85 E-rin~, 82-84 Essential subgroup 3, 4 Field group, 36, 38 Field of definition of a ring, 100-103 Full subgroup, 71 Generalized nilstufe, 34, 35 Height, 2 Homogeneous group, 2, 11, 14 Independent subset, 1 Irreducible group, 72 Local ring, 65-68 Local ring group, 67
63-65
Maximal order in an algebra, 95 fttii ring, 58 ffii ring group, 58-61 ffiR ring, 58 ffiR ring group, 58-61 Ni 1 group, 10-15 Nilstufe of a group, 25-31 Noetherian ring, 47 Noetherian ring group, 48 Nucleus. 14, 15 Order in an algebra, 95 p-basic subgroup. 1 p-divisible group, 1 p-height, 2 p-pure subgroup 1 p-rank. 1 w-group, 1r a ring property. 36 w-ring. 1r a group property, 88 Prime ring group. 36 . Principal ideal ring group, 46, 47, 53 Pure subgroup, 1 Quasi-endomorphism, 3 Quasi-equal, 3, 88 Quasi-isomorphic, 3, 88 Quasi-nil group, 16-24 Quotient divisible group, 88, 96, 102 Radical ring group, 40-42 Rank. 1 Rigid group, 2, 15 Semiprime ring group, 37 Semisimple ring group. 37-40, 73-75 Separable algebra, 95 Simple ring group, 36, 37 Smallest field of definition of a ring, 101 Strongly indecomposable group, 3, 46, 102 Strongly irreducible group, 72 112
Strongly Noetherian ring group, 47-49 Strongly w-group. w a ring property,· 36 Strongly principal ideal ring group, 43-46 Strongly semisimple ring group, 39, 40, 69, 70, 73-75 Strongly subdirectly irreducible ring group, 63-65 Strongly trivial left annihilator group, 69-71, 76 Strong nilstufe, 25, 26, 28 Subdirectly irreducible ring, 61 Subdirectly irreducible ring group, 61-65 T-nilpotent ring, 32-34, 89, 90, 103, 104 Torsion free rank, 1 Torsion free ring. 88 Trace, 94 T ring, 85-87 T ring group, 85, 87 Trivial left annihilator group, 69-71, 76 Trivial left annihilator ring. 69 Type, 2, 5, 6 Ulm subgroups, 2
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