This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
(G), I 'ft 0, 0, positive integer then II possesses possesses a apositive integer m. Hence nip(G) mq>(G) cc II and and so so + +.= II+. +.= J[IP(G)]++ and tP(G)+; Therefore J[q>(G)] 'f 0, then q(G) IP(G)+; i.p(G) . Thereforeifif J[p(G)] and so It that if if CG and H H G It isiseasily easily seen seen that are quasi—isomorphic quasi-isomorphic C ~ J[IP(G)]+. onlyiiff H free groups, groups, then then CG is anti—radical anti-radical ifif and and only is. torsion free H the fact that contradicts the that G anti-radical. C is anti—radical. Therefore CG Q. J[tP(GJl+ contradicts c.p(G) is semisimple, implies that R = 0, i.e., semisimple, which which implies J[p(G)] = i.e., q>(G) Hence J[tP(G)l R is semisimple. 75
The above abovetheorem theoremwas was proved different manner in [6, 5.5]. The proved in ina adifferent manner in [6, Theorem Theorem 5.5].
be aa finite Let G be finiterank rankstrongly stronglyindecomposable indecomposable torsion G additive group groupofofa aring ringwith withtrivial trivial left left If G is the the additive group. If free group. G stronglytrivial trivial annihilator then annihilator then G is strongly strongly irreducible, associative associative strongly G field K K left annihilator, annihilator, and End(G) is aa subring subring of ofan an algebraic algebraic number number field [K:Q1 = r(G). satisfying [K:Q] r(G}. Theorem 4.6.19: Theorem 4.6.19:
be aa ring with with ~(R) = 0. By Proof: Let R == (G,~). ~ E Hom(G, End(G)) be + nilpotent ideal ideal of ~(G) Observation the maximal maximal nilpotent G ~~(G)+, p(G) , and the Observation 4.6.2, 4.6.2, G p(G) is nonzero elements elements in ~(G) are zero by by the theBeaumont—Pierce Beaumont-Pierce Theorem. Theorem. The The nonzero for units units in End(G}. End(G),Corollarly Corollarly 4.6.10, which implies that ~(G)+~ p(G) = I+ I .p(G), see every nonzero As in every nonzero ideal II in ~(G), see the proof proof ofofTheorem Theorem 4.6.11. 4.6.11. As the proof p(H) ofTheorem Theorem 4.6.17, ~(H) iss a nonzero nonzero ideal for every every the proof of ideal in ~(G) p(G) p(H) nonzerofully fully invariant H This implies implies that ~(G) ~~(H) invariantsubgroup subgroup H of G. G. This nonzero which in turn yields The same which in yields G and is strongly strongly irreducible. irreducible. The same G ~ H, and so G G argument (1} ~ (3) in inthe theproof proofofofTheorem Theorem argumentused usedtotoprove provethe theimplication implication (1) 4.6.17, shows shows that K == End(G) End(G) is aa field field satisfying satisfying [K:Ql [K:Q] = r(G). End(G)) (G,~), ~ E Hom(G, End( G)) be o. Then Let S S = (G,'4), be aa ring ring satisfying sS22 ~ 0. subring of End(G) and soisis certainly certainly not ~(G) is aa nonzero nonzero subring and so not nilpotent. By Theorem 4.6.11, 9.(S) and so G G is an stronglytrivial trivial By Theorem 4.6.11, ~(S) = 0, and an associative associative strongly left annihilator left annihilatorgroup. group.
Corollary 4.6.20: G be aa finite finite rank Corollary 4.6.20: Let G be rankstrongly stronglyindecomposable indecomposable torsion either G G is free group group with r(G) r(G) = n. n. Then Then either is an an associative associative strongly strongly trivial annihilator trivialleft left annihilatorgroup, group, or or Rn+l = 0 for every ring R with R R+ = G. G. Proof:
§7:
Theorem 4.6.19, and andCorollary Corollary 4.6.12. Theorem 4.6.19,
E—rings and and 1—rings: E-rings T-rings:
In [35, posedthe theproblem problemofofclassifying classifying the the rings In [35, Problem Problem 45], L. L. Fuchs Fuchs posed rings R by R satisfying RR ~ End(R+). Much Muchprogress progress has has been beenmade madeononthis this question by P. Schultz, [59], P. Schultz Schultz [13]. Many P. [59],and and by by R.A. R.A. Bowshell Bowshell and and P. Manyofof their their results are presented here. are presented here.
If If
H R<><End(R+), and R+=H(t)O reduced,and R and = H® D with H reduced, and 0D H divisible, then is aa torsion divisible, then either 0D == 0, or 0D ~ Q+ and H torsion group. group.
Lemma4.7.l: Lemma 4.7.1:
76
= a. Then R+ has Let rr0(D) directsummand summand isomorphic has aa direct isomorphic to G) Q+ 0(o) ++ a + • R+ and so R <><End(R) and has directsummand summand isomorphic has a direct isomorphic to + End( End ( 1'1'\ \TJ Q ) .,.. n( ®
+
Suppose that D.,.. D Suppose that Q+.. Then Q + + + + + + Hom(R Hom(R,, Q nII Q. Hence Q )coo Q . ) rr00 (( R+)
Lemma 4.7.2:
+
R+ has a direct directsummand summand isomorphic R has isomorphic to + + r 0{R)) = 1, and H is aa torsion torsion group. group. H r0(R
be ring satisfying Rc.o End(R+ ). Let R be aa ring R
Then for each Then for each
k
prime p,
= z(p) R R; = Z(p P) with with 0 <~ kp < ~.
Proof: Let pp be be aa prime R prime for for which which R; ~ o. By Lemma 4.7.1, R; is For every every positive positive integer n, let rn rn == the of cyclic cyclic reduced. For the number number of direct of R; of order the smallest directsummands summands of order pn. Let kk be be the smallest positive z(pk). integer for has a direct summand forwhich which rk ~ 0. Then R; has a direct summand Bk 6k == (t) ® Z(pk). rk p rk + R Since Bk purebounded bounded subgroup subgroup of R+, Bk is a direct summand of R+, Bk is aapure sumand isomorphic isomorphic to [36, Theorem 27.5]. Therefore R+ has aa direct summand [36, Corollary Hom(Bk' Bk) = n (C~ z(pk)), Z(pk)), [36, Corollary 43.3]. Hence Hom(Bk, Bk) rk rk
is finite. is infinite r[Hom(Bk' Bk)] = {k •• rk if rk infiniteand and r~ if rk finite. rk is rk is r[Hom(Bk, jj >> k, rk = 1. Let be a positive integer, Since r[Hom(Bk, r[Hom(Bk' Bk)] <~ rk, = 1. k, jj be a positive =
and suppose and suppose
2rk
Z(pj) Z(p3)
if
if
is aa direct directsummand summand of R;.
Then Then
has aa direct direct R; has
Z(p3)) == Z(pk) ® Z(pk) Hom(Z(pk), Z(pk) summand Z(pk) (-i) Zlpj)) Z(pk)@ Z(pk) sumand isomorphic isomorphic to Hom(Z(pk), contradicting that rk = 1. Therefore R; = Z(pk). contradicting the the fact fact that be aa Lemma 4.7.3: Let R be a ring satisfying RR.,.. End(R+), and let p be R prime such that R; ~ 0. Then R; has aa unique complement complement R == Rp<±> R~, and R R~ = {x E Rl hp(x) = ~}, R is a ring direct sum R 77
1 ... End(R1 ) Rp p
0
R; == Z(pk), Z(pk),
Proof:
If If
Theorem27.5]. 27.5]. Theorem
pH pH ~ H, H,
R] == K. Hom[R;(!)(H/pH), R;] K. x == y+z,
R+
has directsummand summand isomorphic has a direct isomorphic to
r(K) >> 11 == rp(R+), rp(K)
However
a contradiction. a
Conversely, let let x E R~. H ~ R~. H Conversely, Since hp(x) h(x) == hp(z) == ""• This clearly clearly implies 0, and and so This implies that that y == 0, so xx E H, H,
y E R;,
= ""· hp(y) == hp(x-z) =
i.e., i.e.,
then
Z(pk) (i)H, [36, R+ == Z(pk)
H is p-divisible, H is p-divisible,and and so so
Therefore Then
Lenrna 4.7.2, and so so 4.7.2, and Lemma
""• 0 << kk <
H ==
R~.
Clearly
and let let
lxl = pm,
zz
E H. H.
Rp
R~.
y E
X(PmZ) = xy == x(pmz) xy = (PmX)Z (pmx)z == 0. 0.
R~
and Then
are ideals in
y = pmz,
z E R.
R.
Let
Hence
= 0, 0, and similarly similarly R~RP R'R R R1 = RPR~ and pp pp is aa ring 0 R~ is ring direct directsum. sum. Now Now
Therefore
This clearly yields This clearly yields that that
R R == R
R "" End( R+) "" End( R;) (t\ End( R
Let
R~) '@Hom( R;, R~)(f) Hom( R~,
E S}.
R be aa ring satisfying R
It isis easily It easily
R;).
R~
End(R R"" End(R+),
"" End( R~). S be be the the
and let
R
= for all R I$ 0. Put A A= all = {x {x E Rl Rt h (x) ="" p p A is an an ideal ideal in R, and R/A is isomorphic isomorphic to A to aa ring TT
set of of primes primes p p
= 0. 0. =
1
seen that the are zero. Therefore seen that the last 1asttwo twosummands summands are End(R) with Rp "" l:::nd( R;) , and R == Rp 0 R~ "" End( R;)
x E Rp'
p p
Then
for which for which
satisfying ® R • satisfying @ RRP << T < II R pES -pES p pES pES 11
Proof: in
fl II p€S pES
for every R= R=Rp@R~ forevery
S. Let np bethe be the p pES. 11(x) onto R let be the element R Rp. For x E R let n(x) be element with p-component (x) for every pp E S. This This clearly clearly with p-component np(x)
By Lemma 4.7.3, ByLemma4.7.3,
natural projection natural projection of R RP
R
11
p
defines aahomomorphism defines homomorphism
-, n: R ~
TI IT
R . Rp.
Clearly ker11 kern == AA and so so1mm imn ... R/A.
pES Now Now
11(R = Rp R n(Rp)) = p P
for every for every
S. p EE S,
and so and
R rH R << imii imn < <
pES pES
p -
R R
H IT
- pES pES
Observation 4.7.5: Let n bedefinedasabove,andlet be defined as above, and let Observation4.7.5: + + imn+ is aa p-pure p-pure subgroup subgroup of u+ for every every pp E S. imir U S.
p
•
n R. R .
U= U=
ii
ES pES
Then
p
such Suppose there exists exists pp E S Suppose there such that imn+ is not not p-pure p~pure in in u+. S imic = R (f) C • Since im11 is not Now imn imn+ is not p-pure p-pure in u+, C is not not p-pure p-pure + p p + p : in (±) . has a direct summand G RRq', and in and so so pC pCPp I$ CCP. Therefore RR has direct summand p q€S Proof:
q$p q~p isomorphic to isomorphic
78
+
Rn). K == Hom(Rp@(Cp/pCP), Rp).
However
rp(K) > 11
=
+ rp(R ),
a a contradiction.
0 G., G., ® iiEI e:I 1 r(G1) = 1, and Gi where r(Gi) G1 where I. Then is torsion torsion free free for forevery every i e: I. Then I is finite, t(G1) i idempotent for i 'I j, G .... End(G.)+, t{G .) is idempotent finite, t(Gi) ! t(Gj) j, 1 1 1 for all i,j e: I, I, and R R "" (t) End( G1.). ® End(G.). iEI i e:I Lemma4.7 4.7.6: Lemma .6:
ring satisfying Let RR be be aa ring
R with R ... End(R+),
R+ = =
1
1
Proof: Proof:
R+ R+ has directsummand summand isomorphic has a direct isomorphic to
nfl End(G.). Hence 1 i1€! e:I This This is is possible possible only only if if II is finite.
< r(R+) = r( nn End(G 1.)) ~ Now Now = III. III. I El ie:I has aa direct direct summand isomorphic ® End(G.)+(t) <±> Hom(G Hom(G1, G.). R+ has summand isomorphic to to <±> 1., GJ.). 3 1 ie:I i'lj iEI However, r((t) = r(R+) = and = = III. III. Therefore RR"" <±> r(® End(G.J+) ® End(G"!"), 1 1 IEI ie:I ie:I i€I Hom(G., G.) = 00 for for all i,j e: I, I, i 1- j. By Proposition Proposition 1.3.4, 1.3.4, Hom(G., G.)= j. By 1
1
1
1
i,j i,j I, i,je:I,
J
t(G.) all i < t(G1) t(Gi),f.t(Gj) forall i/-j. we j. Since t[End(Gi)+]~t(Gi) $ t(G.) for have that Gi ""End(Gi)+. Since End(G1) have End{Gi) is not not aa zeroring, zeroring, t{Gi) is idempotentfor for all all i e: I, I, Corollary Corollary 2.1.3. For jj e: I, I, let nj be idempotent be the For I, natural projection projection of R+ onto G3. Gj. iI e: I, and aa e: R, the maps maps natural G1 ~ -+ Gj Gi via xx ~ nj(ax) and xx ~ nj(xa) are homomorphisms. homomorphisms. Since = nj{xa) = 0 Hom(G1, for all xx e: Gi, G1, ii 1- j, Hom{G;, Gj) = 00 for i 'I j, j, nj(ax) = for j, G1R c Gi and i.e., RGi: RG1 and so for all all i.e., Gi, and G;R: so G. Gi is an an ideal ideal in RR for Ii e: I. I. Therefore R == (f) is a ring direct sum. G. "" C±> End{ G.) a ring direct sum. ® End(G.) ® G. 1 i€I 1 ie:I i€I ie:I 1
1
be aa torsion group. Thereexists exists aa ring with 4.7.7: be group. There 4.7.7; Let G G G is cyclic. satisfying RR ""End{R+J if and only if G is cyclic.
Theorem Theorem
Proof:
Z(n). Let G == Z{n).
Then
R+ = G
End(G). (Z/nZJ+ = G, G, and Z/nZ — ""End{G).
be torsion group, let R be Conversely, let let G R Conversely, be aa torsion group, and and let be aa ring with with G for which R+ = G satisfying RR ""End(R+). Let SS be ofprimes primes pp for which be the the set of kk
4.7.2, GP 'It 0. 0. By By Lemma Lemma 4.7.2,
R+ = Z{p P). = (±) Z(p
nonzero component nonzero component in is cyclic. cyclic.
for every S every pp e: S, S is aa finite finiteset, set,and and so so R+ R+
G
Theorem4.7 4.7.8: Theorem .8:
RP
p€S pe:S
Since Since the the unity 11 e: RR has
be be the the direct directsum sum of ofrank rank one one groups. groups. There Let G G There exists exists +
+
m m
if and only if if G and only (f)(!) G. End(R )) if a R G == Z(n) Z(n)® ® G1 a ring R with R = G, and RR"" End(R i=1 1 i=l 79
r(G1) r(Gi) = = 1, 1, t(GiJ
with
Let
for i I t(G.) for .$. t(Gj)
j,j,
and
m m satisfying the G the conditions conditions of ofthe thetheorem. theorem. G == Z(n) Z(n) (!) ® ~ Gi satisfying i=I = 1, 1,... 1.4.8, be aa unital unital subring G1, i = be subring of Q with R~ ... Gi, •••,m, ,m, see see 1.4.8, Q m + + R ""End(R ). R (Z/nZ)C!)®{.+) ® R.. R == (Z/nZ) R.• Then R ""G, and R i=I 1
Let
R1 Ri
and put and
R be a ring with ring with
Conversely, let groups, and and
R— "" End( R+).
p p
i E
Gp It 0. 0.
be be a a prime prime for for which which
R+ = G R+ G aa direct directsum sum of of rank rank one one
R+ = CD H, with with = Gt Gt®H,
Now
torsion free one torsion free group group for foreach each Let
t(G.) t(GiJ
l, ••• ,m. i,j == l,...,m. pin; i,j
for all for allprimes primes
= Gi pGi =
Proof:
idempotent,
H == i<2 rank H ® Gi, Gi aa rank iEI I. Clearly Gt is an I. an ideal ideal in in R. Gt is Then
= Z(pk), GP = Z(pk),
11 < kk
= Z( Z(pk)eHom(H, is isomorphic End( Gp) (t) Hom(H, GPJ = pk) (t) Hom(H, Gp) is isomorphic is aa p-group, to aa direct directsummand summand of G. Since Hom(H, Gp) is p-group, Hom(H, Gp) = 0.
Lenina4.4.7.2, Lemma 7.2, and and so so Let
a a
maps
and let let and
E R,
-' Gp H~
via
RH RH ~cH, H,
Therefore
be the natural natural projection G be the projection of of G onto GP.
np
xx -' ~ np(ax), and
and
x -, ~
~p(xa)
The
are homomorphisms. homomorphisms.
is an HR ~EH, H, i.e., i.e., HH is an ideal ideal in in R. Clearly HR is a ring direct sum. R (!) H is a ring direct sum. Therefore R == Gt Gt®H
GtH == HGt HGt = 0. o. GtH R ""Gt ® implies that R ®HH ""End(Gt) End(H) which which implies that Gt End(GtJ, and ® End(H) — End(Gt) (t) Gt ... End(Gt), H ""End(H). End(H). By By Theorem 4.7.7, Gt and by by Theorem Theorem 4.7.8itit suffices Theorem 4.7.7, Gt"" Z(n), and 4.7.8 H to show that for all I. Suppose Suppose that that pGi pG1 I Gi show that pGi == Gi for all pin, pin, and i E I. for has aa direct forsome some pin and some some i E I. I. Then R+ R+ has directsuniuand summand isomorphic isomorphic to Hom( Gp (£) ( G;fpGi J, GPJ, and so r P( R+ J > G), and contradicting Lemma 4.7.2. >1, to 1 , contradicting Lemma 4. 7. 2. Hence
Theorem4.7.9: 4.7.9: Theorem
Let
with ring R R
G be aa group which is is not G be group which not reduced. reduced.
There exists exists aa There
if and only i f
G"" Q+(f) Z(n), G Q4®Z(n),
R+ = G,
and
R R ""End(R+)
nn
positive integer. aa positive integer.
If If
Proof:
+
R ""G,
and
G == Q+(!}z(nJ, + R ""End(R ). R
then
R R == Q(t) Q® (Z/nZ)
is ring satisfying satisfying is aa ring
Conversely, let R be aa ring, with R+ not reduced, satisfying + + + k Z(p R ""End(R ). By 4.7.1 and and 4.7.2, 4.7.2, R ""Q R By Leninas Lemmas 4.7.1 (±) (f) Z(p P), P), — ® pp a prime prime
with
for every 0 <~ kp
m
m
with 80
k
Z(n) == (t) 0 Z(p/i). Z(n) i=l i=l
Corollary 4.7.10: Corollary Then R R ~ End(R+) integer n. Proof:
R be aa ring with R be with R+ a finitely generated group. if and only if RR ~ z or RR ~ Z/nZ for some some positive positive
Let
Clearly if if R R ~ Z or ifif RR ~ Z/nZ Clearly Z or Z/nZ then
R ~ End(R+).
R
Conversely, R ~ End(R+) and that R+ is finitely finitelygenerated. generated. Conversely, suppose suppose that R and + m Theorem4.7.8, 4.7.8, R = Z(n) By Theorem Z(n) (+) (+) ff.\ G., with r(G = 1, and t(G.) t(G1.) 'I t(GJ.) r(G.) t(G.) 1.) = ii=l =1
1
J
1
for i 'I j, ,j ==1,... fori j, ii,j l, •••,m. ,m. Since Gi is finitely finitelygenerated, generated, Gi ~z+ for i = l, ••• ,m, and and since t(G.) 'I t(G.), R+ ~ Z(n)l+)z+. For every prime p, + + 1 J + + so by by Theorem 4.7.8either either R pZ 'I ZZ , so Theorem 4.7.8 Z(n) or R ~ z+. Since RR is R = Z(n) R Z . a ring with unity, either a ring with either R R ~ Z Z or RR ~ Z/nZ. Theorem4.7. 4.7.11: R be aa ring with Theorem 11: Let R be with R+ aa mixed mixed group, S the set S the set of primes p p for which R and such such that R+ has for which R; 'I 0, and nonzero elements has no no nonzero elements of infinite p—height infinite p-height for for every every pp E S. Then Then RR ~ End(R+) if and only if and only if: if: k
(1) (l)
Rp ~ Z/p Pz, 0 < kp
(2)
R
(3)
pES,
U == rrii RRp,, and R is aa unital subring subring of U pES pES R+ is aap-pure p-pure subgroup subgroup of u+ for for every every pp E
s. S.
Suppose that that RR satisfies (1) Proof: Suppose (1) — - (3). The The exact sequence exact sequence R R/Rt -' 0 induces the exact sequence Rt ~ R ~ R/Rt ~ 0 induces the exact sequence Rt + + + + +. + (*) 00 -,~ Hom(R++ is p-pure /Rt' R) ~ End(R) -*~ Hom(Rt' R ). Since R 1s p-pure in U R:). for every for every every pp E S, R+/R; p-pure in U U+/R; every pp E S. S. Therefore R /Rt is p-pure /Rt for the fact fact that U+/R; is p-divisible p-divisible for forevery every pp E S implies the S implies that R+/R; togetherwith with the the fact fact that R+ has is p—divisible p-divisible for for every every pp E S. This This together no elementsofofinfinite infinite p—height for pp E S, yields yields that no nonzero nonzero elements p-height for + R+ ) = 0. From l* (*)) we have that that the Hom ( R+/Rt' 0. From we now now have t hemap map End( R+) -. ~ Hom( R+ t' R+) = the the restriction restriction of ~p to R~ is one-to-one. via ~ ~ ~t = one-to-one. Since ~(R;) ~ R; for for every every q~ E End(R+), this map is is an an embedding embedding of this map of End(R+) (in [36, shown [36, Corollary Corollary 43.3] 43.3] ititisis shown into End(R~). However End(R~) ~ U, lin 0 0~
that End(R;) ~ and it easy totoverify verifythat that group isomorphism it isiseasy thethe group isomorphism _ u+, and constructed there in fact factaaring ringisomorphism). isomorphism). Therefore, for every every constructed there is is in Therefore, for such U a f(x) = ax for all E End(R+), there exists a E U such that =ax for x E R. ff there a - at = left multiplication by a is Hence the map map R ~ is an an -. End(R+) via a~ Hence the 81
Since a a (1) (1) = a,
epimorphism.
this map isisa amonomorphism, monomorphism, and and so so this map
R R ~ End(R+). Conversely, be aa ring with with R+ a mixed group, and R Conversely, let let RR be R ~ End(R+). Condition (1) follows 4.7.2, condition Condition followsfrom fromLemma Lemma 4.7.2, condition(2) (2)from fromLemma Lemma 4.7.4, 4.7.4, and and condition (3) (3) from from Observation Observation 4.7.5. 4.7.5. Theorem4.7.11 4.7.11yields yields the following 4.7.4: Theorem followingimprovement improvement of ofLemma Lemma 4.7.4: S the set set of 4.7.12: Let R be aa ring with with R+ a mixed mixed group, S the Corollary 4.7.12: R be for which for all primes pp for which R; I 0, and and let let A= A = {x € R+l hp(x) = ~ for to aa ring T pp € S}. Then an ideal ideal in R, (2) R/A R/A is isomorphic isomorphic to Then (l) (1) AA is an satisfying (t) Rp < (3) R/A R/A ~ End[(R/A)+], and < TI << IIn R R ,, and(4) (4)iiff R
pES p€S
A~End(A+), A—
-
then
- p€S p€S
p
R~A{+)(R/A). (R/A). R
(1) — (3) are Proof: (1) - (3) are precisely preciselythe theconclusions conclusionsofofLemma Lemma 4.7.4. Since satisfies theconditions conditionsofofTheorem Theorem 4.7.11, R/A R/A ~ End[(R/A)+]. If satisfies the If + A The natural embedding A R is has unity e. The natural embedding A~ R A~ End(A ), then A A has x - ex, andso and so R~A(t)(R/A). (R/A). reversiblebythemap R~A reversible by the map R -+ A via x~ex, R
R/A
the ring ring of of left R be aa unital unital ring, R~ = Lemma4.7.13: 4.7.13: Let R Lemma be =the multiplications in in R. Then R~ is aa ring ring direct directsumand summand of End(R+). R. Proof: It Itsuffices sufficestotoreverse reversethe thenatural naturalembedding embedding R~ ~ End(R+). the unity reversal is End(K+) ~-* R~ via ff ~ [f(l)]~, 1 the reversal unity of R. R.
The
1
ring Definition: A commutative commutative ring Lemma4.7.14: 4.7.14: Lemma
if R R is called called an an E—ring £-ring if R ~ End(R+).
R
be ring with with unity. Let R be aa ring R
R is an an E—ring. £-ring. R + (2) The map R ~ End(R ) map R via x — ~ ff ~ f(l), f(l), l the the unity unity of R. R.
Thefollowing following are The are equivalent:
(1)
x~
with inverse is an an isomorphism isomorphism with inverse
1
(3)
End(R+) = = R~.
of End(R+). By Lemma 4.7.13, R~ is aadirect (1) • (2): By Lemma 4.7.13, directsummand summand of + + + + and so f: R I+) K -.~ R+ be be an However R~ ~ R ""End( R ) , and so R ""R (+) K. Let f: an R+(I)K R+. be the the natural natural projection projection of R+ (:i=) K onto R+. isomorphism, isomorphism, and andlet let nR+ be Thenfor for every foirR÷(x) = 0. However End(R+ (+) K) ""RR is Then every xx € K, fonR+(x} Proof:
commutative, commutative, and and so so one-to-one1 x = 0, one-to-one. 82
WR+of(x) = 0. Since nR+of(x) 1TR+of(x) =f(x}, f(x), and ff is nR+of(x) .i.e., ( + End(R+) = R~. 1.e., K = 0, and and so so End R) = Therefore the Therefore
map RR—'~ End(R+) End(R+) via xx ~ x1 this this map map is ff ~ f(l). f(1). (2) • (3):
is an an isomorphism. isomorphism.
Clearly the inverse inverse of Clearly the
Obvious.
= R1 ""' R. right (3) .. (1): (1): End(R+) = Let x,y e: R, and =right R. and let let yyr = multiplication multiplication by by y. Since y r e: End(R+) = R1 , there exists z e: RR such that Yr = = z1 • Hence xy xy = Yr(x) yr(x) == z1 (x) = zx. However and so xy = yx, i.e., is commutative y = Yr(l) = z (1) = z. and xy i.e •• R is commutative and and R = zL(l) = z, 1 = therefore is an therefore an E-ring. E-ring.
Corollary 4.7.15: be an an E-ring, E—ring, and and let let R 4.7.15: Let R be is aa principal principal ideal ideal in in R. R.
e: End(R+). Then
c.p€ ~
~(R) p(R)
such Proof: Since ~,p e: End(R+) = RL, R1 • there there exists aa e: R such that ~(x) ax R p(x) == ax for all x e: R, i.e i.e., p(R) = aR, and so ~(R) •• ~(R) aR, and a principal principal ideal ideal in R. R. p(R) is a H R be an an E-ring. If is aagroup summand Corollary 4.7.16: J;orollary 4.7.16: Let R be If H is groupdirect direct ~ummand R+, R+ of R+, then H is aa ring ringdirect directsuninand summand of R. so R+ cannot be an an H and so R, and cannot be infinite direct infinite directsum sum of of non-trivial non-trivialgroups. groups. H is aa principal Proof: H principal ideal ideal in
natural projection natural projection of R+ onto H.
by Corollary 4.7.15. R R by 4.7.15.
Let wH be be the the
+ = wH(l)R. By Lemma 4.7.14(2), H By Lemma 4.7.14(2), H == wH(R) =
H = eR, e. Put e = wH(l). Then H eR, and ee22 = e.1TH(l) e·wH(l) == w~(l) == wH(l) = =e. H= summand Therefore ee is an an idempotent, idempotent. and and so so H = eR eR is is aaring ringdirect direct summand of of R. R.
Corollary 4.7.17: be aa ring ring with with unity. Corollary 4.7.17: Let RR be commutative andonly onlyif if R R commutative ifif and is an an E-ring. E-ring. Proof:
If If
Then
End(R+)
is
is an R R is an E—ring, E-ring. then then End(R+) ""'RR is commutative.
be aa ring ring with with unity Conversely. let R be unity such such that End(R+) is R f for Then for e: End(R+). x e: R+. f(x) f(x) = =fox (1) = = xrof(l) x of(l) = commutative. Then foxr(l) . + r r and so i.e., = R1 • [f(l)l 1 (x). and so f == [f(l)l 1 • 1.e., End(R) = By Lemma 4.7.14. R R is an E-ring.
E-ring, and let SS be ring with Corollary 4.7.18: Let RR be an an E-ring, and let be aa ring with unity unity + + + such that RR ""' SS •. Then such Then R R— ""' S. S. so by by Corollary Corollary 4.6.17, S commutative. so 4.6.17, S is an an Proof: End(S+) ... End(R+) is commutative, R. S — E-ring. Therefore S ""'End(S+) ""'End( R+) ""' R. 83
be unital unital rings Corollary 4.7.19: Let RR and SS be rings with torsion torsion free free additive additive Corollary 4.7.19: is an then so so is S. groups. and with R+ ~ s+. If R an E—ring E-ring then groups, and R is commutative, Suppose that is an E—ring. E-ring. Then End(R+) is commutative, and and so so R is an that R Proof: Suppose -.--+ + + . + • + c End(S) ~ End(R ), commutative. Now Now End(S) ~ End(R) ~ Q Q® ® End(R) is commutative. S and so so End(S+) is commutative. commutative. Therefore S is an an E-ring E-ring by by Corollary Corollary and 4.7.17.
be an an E-group, E-group, and and let let ~ E End(G). Then ~G) Lemma Lemma4.7.20: 4.7.20: Let GG .be an E-group, E-group. and and every everyendomorphism endomorphism of can be be extended extended to an an an of ~G) can endomorphism endomorphism of of G. G.
is
p(G) Proof: Let RR be be an G. By G) == eR+, By Corollary Corollary 4.6.15, Ill( an E-ring with R+ = G. for all x,y E R, the (ex)*(ey) == exy ee E R. R. The The multiplication exy for the product product multiplication (eX)*(ey) inducesaa ring ring structure S S on on being multiplication on on the the right right being multiplication in R, induces be the the unity unity in R. Then e ~(G). e= = e.l is the the unity unity in in S. Let 1 be It isis readily Clearly S.s. is commutative. commutative. Let ff E End '.p(G) ~(G) == End(eR+). It readily + for all all xx E R , seen that foe~ E End(G). Hence seen Hence for for some f(e) == ey f(ex) ey for some y E R, f(ex) ==foe (l)•x == f(e)·x. f(e).x. Since f(e) E eR, f(e) ~ + End(S ) == S~, and = (ey).x and so so f(ex) flex)= (ey)·x == (ey)*(ex). (ey)*(ex). Therefore f == (ey)~, End(S) and S is an is an and so so S is an E-ring E-ringbybyLemma Lemma 4.7.14. Clearly y~ is anendomorphism endomorphism G f. of G extending f. 1
be aa finite finiterank rankstrongly stronglyindecomposable indecomposable torsion Let GG be free group, let RR be free group, and and let be aa unital ring with with R+ = G. Then R is an unital ring R E- ring. Theorem 4.7.21: Theorem 4.7.21:
has trivial trivial left annihilator, so so by by Thecrem Proof: Clearly R has left annihilator, Therrem 4.6.19, 4.6.19, R By Corollary Corollary 4.7.17, End(R+) is is aa subring subring of ofan an algebraic algebraic number number field. 4.7. 17, field. By is is an an E—ring. E-ring.
R R
The be proved proved in section section 44 of ofthe thenext nextchapter. chapter. Thefollowing following result result will will be Let RR be be an an E—ring E-ring with
Theorem 4.7.22: Theorem 4.7.22:
n
group.
Then
R=
R
=
1..._-f:\
R,, Ri,
with
R1 Ri
R+ a a finite finite rank rank torsion torsion free free
aa unital unital subring subringofofananalgebraic algebraicnumber number
i =1 +
+ field Fi, and Ri aa full F1, full subgroup subgroup of Fi,
i == 1 , ••• ,n.
1
be found found in [13, A stronger version version of ofTheorem Theorem 4.7.22 may may be [13,Theorem Theorem 3.121. 3.12]. Theorem4.7.23: 4.7.23: Theorem 84
Let
R be aa unital unital ring. R be
Thefollowing following are The are equivalent:
(1)
is an R is an E—ring. E-ring. R
(2)
For ff
Proof:
(1)
€
End(R+), f(l) implies that f ==0. 0. f(l) ==00 implies
~
(2) by by Lema Lemma 4.7.14. 4.7.14.
(2) ~ (1):
Let ff € End(R+). Then ~ = = ff— f(l)~ € End(R+), and ~(1} an E—ring Hence ff = fll)~, i.e., End(R+) = Re and and so RR is an E-r1ng by by Lemma Lemma 4.4.7.14. 7.14.
unital ring R R is homomorphism is aaT—ring T-ring ififthethe homomorphism Definition: A unital induced induced by by
~(a~ 1i(a
From From now now on on
for all for
b) == ab ab
a,b
will11be l!llz wi bedenoted denoted by by
€
is an R is an isomorphism. isomorphism. R
~
•
~=
= 0.
- R R R ®z R ~
R
Every T-ring 1—ringisis an an E-ring. E-ring. Theorem4.7.24: 4.7.24: Every Theorem be the Proof: Let R beaT-ring, and let be the map map d(x) == 1 ® xx R be a 1-ring, and let d: RR ~ R~ R R for all xx € R. is the inverse It is readily seen that d R. It is readily seen that d the inverse of ~· and hence an isomorphism. isomorphism. For ff € End(R+}, define 1 € Hom[(R ~ R)+, hence is is an RYE, R+] b) = for all to be be the thehomomorphism homomorphism induced induced by by the themaps maps f(a ®b) = f(a).b f(a)·b for If f(l} is an a,b € R. If 0, then fd == 0. 0. Since dd is an isornorphism, isomorphism, f == 0, 0, R. f(l) == 0, which clearly clearly implies R which implies that f == 0. 0. Hence R is an an E-ring E-ringbybyTheorem Theorem 4.7.23.
Stronger versions 4.7.23 and and 4.7.24 4.7.24 may maybebefound foundinin [13, Stronger versions of of Theorems Theorems 4.7.23 Propositions 1.2 1.2 and 1.7]. Propositions and 1.7]. Theorem4.7.25: 4.7.25: Theorem (1} (1)
(2) (3)
Let
G be aa torsion group. G be group.
Thefollowing following are The are equivalent:
1-ring group. G is aaT-ring group. G G is an an E—ring E-ring group. group. G G is cyclic. cyclic. G
(2) .. (3) by 4.7.7. by Theorem Theorem 4.7.24. (2} by Theorem Theorem 4.7.7. Proof: (1) .. (2) by Then Z/nZ Z(n). 1—ring, with (Z/nZ)+ ~G. (3) • (1}: is aaT-ring, (1): Let G == Z(n}. G.
The following following obvious The obvious results will willsoon soonbebeemployed: employed: Observation Observation 4.7.26:
ifIf
Observation Observation 4.7.27:
Everyquotient quotient ring ring of aa T-ring Every T-ring isisa a1—ring. T-ring.
S are 1—rings, then so so is R R and S are T-rings, then
R~ S.
An immediate consequence consequenceofof Theorem Theorem4.7.25 4.7.25 is: is: An immediate
Corollary 4.7.28: 4.7.28: Let RR be be aa ring with with R+ a torsion group. Then RR is positive integer n. aa T-ring if and only if R for some some positive R ~ Z/nZ 85
The T-rings R therefore reduces reduces to the the case; case; The problem problemofofclassifying classifying 1-rings R Thedescription descriptionof of these theserings rings is is as is not not aa torsion torsion group. group. The as follows:
R+
R+
R be be aa unital ring for forwhich which R+ is not a torsion unital ring T—ringifif and andonly onlyif if A) group. Then RR is aa T-ring A) R/Rt R/Rt is isomorphic isomorphic to aa is cyclic, subring of Q, Q, and and B) B) for for every every prime prime pp for which which R; f 0, R; is cyclic, subring and R+/R+p is p-divisible. Theorem 4.7.29: Theorem 4.7.29:
Let
R
Proof: Let R be a T-ring. Clearly, Q is aa T—ring, T-ring, and and R/Rt R/Rt is aa R T-ring, Observation by Observation Observation 4.7.27. Hence QQ ~ (R/Rt) is aaT-ring, Observation T-ring by Now 4.7.24, and 4.7.26. Now Q ® (R/Rt) is an an E—ring, E-ring, Theorem Theorem 4.7.24, and so so Q 4.7.9. Since R/Rt [Q@ {R/Rt)]+ ~ Q+ by by Theorem Theorem 4.7.9. (R/Rt)' EQ (R/Rt) c~QQ ~ (R/Rt), R1Rt ~ ZZ ~ (R/Rt) it follows it follows that that r[(R/Rt)+] = 1, which which implies isomorphic to implies that R/Rt R/Rt is isomorphic Theorem 4.7.22. 4.7.22. Since RR is an is cyclic Q, Theorem an E—ring, E-ring, R; cyclic aa subring subring of Q, R p+ for every satisfying R; Lemma4.7.2. 4.7.2. For p for every prime prime p, Lemma aa prime prime satisfying tip 0, p R 4.7.3, and R+/R; is p—divisible p-divisibleby byLemma Lemma 4.7.3, and so so R+/R; is p-divisible.
unital ring Conversely, let RR be be aa unital ring satisfying satisfying A) A) and and 3). 8). commutati ye diagram: commutative -::>0 (R/Rt) —> 0
O —> RR(i Rt 0 -::> > RR ~ R —> R Rt —>
1~1
t~2
U1
----:::> >RR
O
where
11i
Since
Consider the Consider the
is induced induced by by the themaps maps
---i>>0, 0 •
>R/Rt ~i(a
-
b) = ab, ®b)= ab,
iI = 1, 2,3. 2,3.
for which is p-divisible p-divisible for forevery everyprime prime pp for which
Rp ipt 0,
It is readily seen that for each prime p, Rn). Itisreadilyseenthatforeachprime (.j:', {RPIXIRP). p. Rp R p pprime p prime k k and so is aa ring summand Zip Pz for ringdirect direct summand of R, and so Rp ~ Z/p some non-negative non-negative for some integer kp. Hence Rp ® Rp ~ Rp' and and in in fact, fact, every every element element in Rp ~ Rp may written in may bebewritten the form form 1 ~a, the a, with l the tfle unity unity in Rp • Clearly the . the restriction of to is an isornorphism for every prime restriction ~l Rp 5ll Rp an isomorphism every prime p, and U1 so U1 ~l is an an isomorphism. isomorphism. Rt = R®Rt=
1
1
.-.
Again p-divisible for for every ev~ry prime prime pp with Againthe the fact fact that R+/R; R iRt is p—divisible in fact Rp t 0, yields that RR '5!1 {R/Rt) and in fact every every (R/Rt) ~ R/Rt' (RiRt) .x, {R/Rt) (RiRt) ~ (R/Rt) RiRt, and a, element in is of the form with element (R/Rt) •Xl {R/Rt) of the form 1 X1 1 the the unity unity in (RiRt) G) (RiRt) +
+
.
1
86
1
R/Rt. that
This clearly clearly implies This implies that ~p33 is is an an isomorphism, isomorphism, which which now now yields yields or that R ~ 2 is an an isomorphism, isomorphism, or is a T-ring. R a 1-ring.
Question 4.7.30: An An immediate immediateconsequence consequence Theorem 4.7.29 thefact fact that Question of ofTheorem 4.7.29 is isthe if RR is aa 1—ring if T-ring for for which which R+ is not not aa torsion torsion froup, froup, then then 1) r(R+/R;) = 1, 2) t~R+/R~) is idempotent, idempotent, 3) RR; is cyclic cyclic for forevery every prime p, and and 4) 4) for R for which for every every prime prime pp for which RP 1 0, R+/~ is p—divisible. p-divisible. Is every every group 4) the the additive additive group group of aa groupsatisfying satisfying 1) 1) - 4) 1-ring? A positive answer, T-ring? answer, together together with withTheorem Theorem 4.7.25, would would provide aa complete description of complete description ofthe the1—ring T-ring groups. groups.
87
55 Torsion Torsion free rings
Notation, definitions, definitions, and §1. Notation, and preliminary preliminary results: Most the material in in this thischapter chapteris is work Beaumontand and Pierce Pierce Mostof of the thethe work of of Beaumont on torsion torsion free free rings, rings, i.e., i.e., rings [7], on rings with with torsion torsion free free additive additive group, group, (7], R is said it a group property, a ring [8], and and [53]. In general general for n a group property, a R said to to be aa it—ring n-ring if n-group. A A subring SS of RR is said said to to be be aa be if R+ is is aait—group. n-subring if s+ is aan-subgroup it-subgroup of R+. it-subring of R R
if
Very roughly roughly speaking, speaking,the theidea ideaof of this chapter Very chapter isistotoembed embed a torsion torsion * * into the free ring RR into that if if R R* free theQ—algebra Q-algebra RR* = Q® Q® R. R. It It will willbebeshown shown that satisfies certain satisfies certainproperties, properties,then then·considerable considerable information information is is obtained obtained concerning R+. R+.
All rings chapter are ringsininthis this chapter areassumed assumed to be be associative. associative. Let R, T. Then R and SS are be subrings subringsof of aa torsion torsion free free ring 1. R R, SS be such that that S, if ifthere there exists exists aa positive positive integer integer nn such quasi-equal, RR ~ S. torsion free free rings R, SS are quasi-isomorphic, nR C S, and nS nR R. Two Two torsion rings R, quasi-isomorphic, ifthey theyare areisomorphic isomorphictotoquasi—equal quasi-equal rings. R R & S, S. if
=
=
Notation:
d(G) divisible subgroup dlG) = the maximal maximal divisible subgroup of of aa group group G.
Z(R) Z(R) =the = the center center of of aa ring R == {x
€
RI Rl xy == yx yx
for all y for
€ R}.
R* * = Q(x) Thesymbol symbol®® will will often free ring. The R R, R aa torsion torsion free oftenbe be omitted omitted R Q®R, R as aa full full subring in order order to simplify simplify notation, notation, and and in in order order to to view view R as subring of k * * R* • R, will be be written written as as . The elements of R R* R The elements Q,
iI == 1l,...,k. •••• ,k.
divisible if if G G group GG is quotient quotient divisible possesses Definition: A torsion free free group such divisibletorsion torsiongroup. group. a free free subgroup subgroup F F such that G/F is aa divisible
88
Quotient divisible will Quotient divisible willbe beabbreviated, abbreviated, q.d. q.d. * R* Tworings, rings, Definition: R called the the algebraic algebraic type type of R. Two is called of a ring R. R* . * * the same algebratype type1f if R ~ s . A torsion free G admits R,S have have the same algebra free group group G a a multiplication multiplication ofofalgebra algebra type type A, A aa Q—algebra, Q-algebra, ififthere thereexists exists aa ring A ** + + such that R R R R such = G, R ~ A. and = G, R A. Theorem 5.1.1: Theorem 5.1.1: type.
Quasi—isomorphic torsionfree free rings have Quasi-isomorphic torsion have the the same same algebra
ofgenerality generalitywe wemay may consider consider the thecase case R,; S, Proof: Without Without loss of R 5, R,S torsion free i.e.,there there exists exists aa positive positive integer integer nn such torsion free rings, rings, i.e., such that nR c~ S, and nS R. Then R*= R* = QQ@ R R == QQ ~ nR nR ~ nS ~ER. E S* = Q ~ S = R* , Q ~ nS ~ Q ® R = R and so RR* =S = S* • An consequenceofof Theorem Theorem5.5.1.1 is An immediate immediate consequence 1.1 is
Corollary 5.1.2: Let A A be Q-algebra, and and let let G,H be torsion free Corollary be aa Q-algebra, be torsion free groups, G Q.&H. H. Then G admits of algebra algebra type type A if and G admitsaamultiplication multiplication of A if and only if if H only does. H R be torsion free free ring, and P(x1,... Lemma5.5.1.3: Lemma 1.3: Let R be aa torsion and let P P = P(x 1 , ••. ,xn) be a homogeneous polynomial withcoefficients coefficients in such that x1 , ••• ,xn are homogeneous polynomial with in Z, such non—commuting variables. Then RR satisfies the identity PP if non-commuting variables. the polynomial polynomial identity .if R * does. R* and only 1f and only
if
Proof: Suppose that R Suppose that satisfies the the polynomial polynomial identity P. Let R R*. * Since R is aa full there exists exists a positive a 1 , ••• ,an E R. full subring subring of of R* R* there R integer mm such such that rna; E R, i = = l,l,...,n. ••• ,n. Let d == deg deg P. P. Then R* mdP(a , ••• ,a ) = P(ma , ••• ,rna ) = 0. Since R* is torsion free, mdp(al,...,an) torsion free, P(mai,...,man) = 0. 1 1 n * n R The converse converse is is obvious. P(a 1 , ••• ,an) = 00 and R satisfies P. The obvious. G be quasi-isomorphic quasi-isomorphic torsion torsion free Corollary 5.1.4: 5. 1.4: Let G and HH be free groups. groups. Corollary G is a non-zeroring a homogeneous (1) G is the the additive additivegroup groupofof a non-zeroringsatisfying satisfying a homogeneous polynomialidentity identity P == P(xl'··· ,xn) if if and onlyiiff H H is. (2) Every polynomial and only Every ring with G if and with additive additive group group G satisfies PP if and only only if ifevery every ring ring with with H additive group additive group H satisfies P.
Proof:
Corollary 5.1.3. Corollary 5.1.2 5. 1.2and andLemma Lemma 5. 1.3.
Lemma 5.1.5: Lemma
be aa torsion torsion free free ring. Let R be R
R R is T-nilpotent T-nilpotent ififand and only only
89
if
R* * is T-nilpotent. if R T-nilpotent.
that Proof: Suppose Suppose that
R is 1-nilpotent. T-nilpotent.
R
Let
= {ai}i=l
R*. C ~ R* .
There exists exists aa There
such that that miai E R, i = 1,2 positive integer R is integer mi such 1,2, .••. Since R T—nilpotent,there thereexists exists aa positive positive integer T-nilpotent, integer kk such such that (m 1a1)(m2a2 )k ... (mkak) = 0. 0. (mlal)(m2a2)k. ..(mkak)
torsion free torsion free
k nIT
i=l
a. a.1 == 0,
Therefore and
kk
kk
(( iin m1)( mi)( iin a;) = 0. . 1=1 i=l i=1 i=l
Since
**
is
R
R
*
R* R is T-nilpotent. T-nilpotent.
Corollary be quasi-isomorphic quasi—isomorphictorsion torsion free groups. Corollary 5.1.6: 5. 1.6: Let G,H G,H be groups. (1) GG is the T—nilpotent ring ring which is not aa zeroring the additive additive group group of of aaT-nilpotent which is zeroring ififand and only H is. (2) Every ring with with additive additivegroup group GG is T—nilpotent T-nilpotent if and H (2) Every if and H only if every ring with additive group H is T-nilpotent. only if every ring with additive group T—nilpotent. Proof:
Corollary 5.1.2 5.1.2 and and Lema Lemma 5.1.5. 5.1.5.
be aa torsion torsion free free ring, an arbitrary arbitrary ordinal. Lemma ring. a an Lemma5.1.7: 5.1.7: Let RR be R if and only if if R* R is a-nilpotent a-nilpotent if and only R* is.
Then
An easy easy induction induction argument Proof: An argument shows shows that (R*)8 (R*) 6 == (R8)* (R6)* for every every (R*)a = (R0)* ordinal 6. a-nilpotent, then then (R*)a = (Ra)* = 0. The Thereforeifif RR is a-nilpotent, The 8. Therefore converse is is obvious. converse obvious. H be torsion free groups, Corollary 5.1.8: Let GG and H Corollary be quasi—isomorphic quasi-isomorphic torsion a an an ordinal. (1) G G is is the theadditive additivegroup groupofofanana—nhlpotent a-nilpotent ring whicn whicn is not a zeroring if and only if not a zeroring if and only if H is. (2) Every ring with with additive additive group group GG H (2) Every H is a-nilpotent a-nilpotent ififand and only only ififevery everyring ringwith withadditive additivegroup group H is a—ni lpotent. a-nilpotent.
Proof: §2. §2.
Corollary 5.1.2 5.1.7. Corollary 5.1.2and andLemma Lemma 5.1.7.
The Beaumont—Pierce Decomposition Theorem: The Beaumont-Pierce Decomposition Theorem:
The group analogue of ofthe theWedderburn Wedderburn Principal PrincipalTheorem, Theorem, which which The group theoretic theoretic analogue was usedinin Chapter Chapter4,4, section section 5, 5, will was used willbebeproved proved here. here. Since Since the the proof involves several willbe be stated stated before before the the preliminary preliminary results involves severalsteps, steps,itit will leading up uptoto it it are leading are given. given. Theorem5.2. 5.2.1: finite rank Theorem 1: Let RR be be aa finite rank torsion torsion free free ring, R* R* = ~·~N aa R*, * , and the nil nil radical Q—spacedecomposition decomposition of of R*, Q-space R* , with if N the radical of R and S~ aa 90
semisiniple subalgebra of R*. semisimple subalgebra R*. Put S ==RR nti s, N == RR n if. N. Then SS is aa satisfying s* == S, N is the subring the maximal maximal nilpotent subring of R R nilpotent ideal ideal of R, N* and satisfies N* == -if. hasfinite finite index and satisfies N. In addition addition S (+) N has index in R. Lemma5.2.2: 5.2.2: Lemma
Then
~:
R+I (S (t) N) + "" — S~ /S +:
it w be R* onto be the the natural natural projection projection of R* 5R+ -, s+1/S+ 5 via = n (z) ++ s+ for for all zz e: R+.
Proof:
'.P:
I
R). for which Let sS1 there exists y e: R" which x+y e: R}. if for Si there 1 == {x e: S
~
Let
~(z)
S. Define Clearly
~
~
is an an
c ker epimorphism, and and (S ® N)+ ~ ker p. ~. Let zz e: R, R, zz = x+y, x+y, xx e: S, 5, yy e: N. if. Suppose that that zz e: ker p. Suppose ~. Then x e: S, and =zz-- x€R xe:R fln iT= N =N, N, i.e., and so y = r.'\+ r.'\+ + + ++ zz E (S 1+1 N) , and and so so ker .p ~= = (S ':r' N) , or R /(S ++ N) ""S 1/S • Lemma 5.2.3: Lemma 5.2.3:
*
*
* == N. -~. S5 * = -S, 5, and N N —
—
*
It therefore Proof: S is the theunique uniqueminimal minimalQ—algebra Q-algebra containing containing S. It therefore S* R* * 5. Let x suffices to to show show that S is aa full full subring subring of of s. Then xx e: R S x e: S. 5-. such that nx e: R. andso sothere thereexists exists aa positive positive integer nn such and R. Hence The same nx e: R fln S = SS and and so so s+ts+ is aa torsion torsion group. group. The same argument shows shows -N. that NN* = = N. for s such such that that the There is There is aaQ—basis Q-basis {x 1 , ••• ,xm} for the free free F of 5Sgenerated generated by by this basis basis is is aa subring subring of S. F
Lemma 5.2.4: Lemma 5.2.4:
subgroup Proof:
5+
Since s+ is aa full fullsubgroup subgroup of S 5. in S is aa Q-basis Q-basis for ~.
s, maximal independent independent set 5-, a maximal m
a .. kzk with k=l lJ be aa positive integer l, •.•,m. ,m. Let nn be integer such such that ,j,k == 1,... a.1J"k e: Q; Q; i ,j,k {x1 == nz1 for all i,j,k i ,j ,k == 1,... Z nz; , iI == 1,... 1 , •••,m} ,m} naijk e: Z for 1 , •••,m. ,m. Then {X; satisfies satisfies the theconditions conditionsofofthe thelemma. lemma. z1 , ••• ,zm
Observation 5.2.5: Observation 5.2.5:
St/F S~/F
Hence
z.·zJ. = 1
~
= k=1
is aa torsion torsion group. group.
S+/F is aa torsion It therefore torsiongroup. group. It therefore suffices suffices totoshow show Proof: Clearly S+/F 5.2.2 that R+/(S <±> N)+ is aa torsiongroup, group,ororbybyLemma L;mma 5.2.2 that S~/S+ is aa torsion R,, and and so 5, zz e: N. torsion group. group. Let xx e: R. Then xx e: R so x == y+z, y+z, yy e: S, N. such that ny e: S, By 5.2.3 5, and By Lemma Lemma 5.2.3there thereexists existsaapositive positive integer nn such R+/(SmN)+ and nz e: N. Hence nx = ny + nz e: S(+)N, R+/(Sq)NJ+ is a torsion group. nz nx = ny + a torsion group. nz N.
Lemma 5.2.6: Lemma 5.2.6:
For all but many For but finitely finitely manyprimes primes
p,
(S~/F)p, and
(S+/F)p 91
are divisible divisible and are and equal. equal. The proof proof of Lema The Lemma 5.2.6 5.2.6 involves involvesmany many steps, steps, and and will willbebebroken brokendown down notation. The The degree degree of of into aa series series of of claims. claims. First Firstwe we introduce introduce some some notation. nilpotence nilpotence of W willbebedenoted denoted by by t. For pp a prime, prime, kk >>00 an an integer, integer, if will Ik == {x € F s1}. x € S}, SJ, and Jk = = {x € Fl p-k xx € s1.. F II p-k x Claim 5.2.7: 5.2.7: {1) Claim F == 10 I 0 22 IIi1 22 12 I 2 2 ... , (1) F
=Jl =J2 =···•
(2)
F= F = Jo
{3) (3)
IkE !: Jk' 'k
are two two sided sided ideals ideals in F, Ik and Jk are 'k {5) (5) 'k'it Ik. It E ~ IkH ' (4)
{6) (6)
(7)
and Jk.Jt!: E Jk+t' and Jkt ~ I k for for all a 11 k,z k ,t > ~ 0. 0.
Proof:
(1) - (4) (1) (4)
are obviously obviously true. are
Let x1 x1 € Ik, x2 == p1y2 , y1,y2 y1 ,y 2 € 'k' xX22 € I 1 . Then xx11 == pky1 , and x2 x1x2 = k+t Hence x 5, i.e., i.e., x1x2 y1y2 € S, x1x2 € Ik+t . 1x2 = p y1y2 , with y1y2 (6): follows from usedtoto prove prove (5). (5). from the the same same argument argument used {6):
(5):
s. S.
and there there exists zz € N (7): Let xx € Jk. Then xx == pky, y € s 1 , and if such that y-z € R. R. Let x1,... and let let z1, ••• ,zt € N such that p-kxi - zi z1 € R, x1, ••• ,xt € Jk, and
Suppose, inductively, that p-kx. x. 11 12 '1 12 for l~i Then < i11 < i2 <...< < t, j < t.
ii = 1, ••• ,t.
x. ... x.
1j
z. z -— z. z. ER €R z...•• ...Z. 11 12 1j '1 '2
'j
1
-k ( .. k ... zjzj+l = x1(p_kx2 ... zj+l) ++ p x1x2 ••• xjxj+l -- z1z2 z1z2 ••• x1 p x2 .•• z2 ••• ... xj+l -- z2 -k ) (p_kx1 ... xj+l - pk (p -k x1 - z1) (p -k x2 ... ... zj+l) € R. (p x1 -—zz1)x2 ••• xj+l -—zz2 2 ••• 1 x2 ••• —
—
Therefore p-kxl •.• xt - z 1•••• zt
€ R.
However
zz11 .•• and so so ... zt == 0, and i.e., Jkt ~ Ik. xt € Ik, i.e.,
p-k x1 ••• xt € s1 n RR!: -S n RR = S. s. Hence xx11 ... 1k' t_t t_t f all Claim 5.2.8: 5.2.8: Ik+t nfl ptF Claim p F-= p Ik' and and Jk+t nii p F-= p Jk for or all Proof: Let xx € Ik+t n p1 F. 1k+2, 11
Then xx
=
pk+ty
= =
k,t ~> 0. o.
s, z € F. p1 z, y € S,
Therefore x = = p1 (p (pky) z. Since RR*+ is torsion torsion free, free, y) == pp1z. 92
E 1k•
k pky = z, and p y= and so
yields that x € p~Ik. z € Ik which which yields s. Conversely, Then x == p~y. y € F, and y == pkz, z € S. Conversely, let let x € p~Ik. Then Hence x = = ppkH z, and xx € .Ik+~· Clearly xx € p~F, so xx € Ik+~ n p~F. F. The sameargument argumentreplacing replacing The same ~
p FF == Jk+~ fln P
Notation:
and S yields that that by sS11 yields S
P Jk.
St/F. = St/F, S+/F, T1 T1 = st/F. T=
. Claim Cla1m 5.2.9:
(2)
I by J,
~
k)~
+
+
~
(pkT)(pR.] (1) (p Tp [p ] ""lkH/[p F n Ik+~] ,
k
+
~
~
(p Tlp)[p ] ""Jk+~/[p F n Jk+~]
and
+
S Then there there exists aa unique Proof: (1) (1) Let Let x € Ik+~· Then unique element element y € S such that xx = p~y. Define ~= Ik+~ -'~ TT via ~(x) p(x) ==y+F. It isis easily y+F. It easily verified verified
that ~ is a homomorphism. x == pk+~z.
zz
€
S, 5,
Since
py p~y
€
=
p(x) 1;+l C F, F, ~(xJ
TP[p~].
€
which and so so ~(x) which implies implies that y == pkz, and p(x)
+ be such that w + FF € (p k TPJ[p ~ ]. be such Let w w € S
€
Now Now
(pkTP)[p~].
z € S+ and R, kH kH . p(x) ==w+F, i.e., p9'w= w+F, 1.e., p w = p z € F. Let xx == p z. Then xx € Ik+R-" and ~(x) if and only if belongs ~ is an an epimorphism. epimorphism. Now Now xx € 1;+t 1k+L belongs to ker ~ if and only if 1 p- x € FF whicn whicn occurs i.e.,.ker kerp ~= =[p9'F [p~F fln Ik+t]+. occursifif and andonly onlyif if x € pR-F, i.e.,. Therefore (pkTP)[pR-] ""1:+~/[p~F n fl Ik+~l+. A similar similar argument yields (2). (2). argument yields Claim 5.2.10: Let pp be be a prime. prime. Claim are divisible T1P are divisibleand and equal. equal.
If F/pF If
Then
k
w w == p z,
is semisimple semisimple then then TP and
-' F/pF be the Proof: Let 'P: ~: F ~ F/pF be the canonical canonical ring ringepimorphism. epimorphism. and and the the semisimplicity semisimplicity of F/pF yield:
(A)
~(Ik)
=~(Jk)
= [~(Jk)lt = = = ~(J~) ~ ~(lk),
Claim Claim 5.2.7
and
2 = = = ~(I 2k) ~c ~ ( I 2k) ~E ~ ( I k) , for for all a11 k >0. > 0. = [~ ( I k) ] . k + + + = By By Claim Cla1m 5.2.9, ppkT[p] Tp[p] ""lk+l/[pF nfl lk+l] ""~[Ik+l] = (B)
~(I k)
+ k (pkT1p)Ep] ""Jk+l/[pF n Jk+ll + ""(p T1p) [p] for for all all k > > 0. o. Choosing k ==00 Since and is finite, yields that yields that Tp[p] ""Tlp[p]. Tp[p] E ~ T1P[p], Tlp[p] is finite. T[p] == T1P[p]. For every G, r(Gp) = this implies implies that that TP[p] Hence = r(GP[p]). ~[Jk+ll
r{ TP)
+
= r( T1p) •
=
93
Claim5.2.9 5.2.9 and and{B) (B)yield yield that {pkTP)[p] ~ I~+l/[pF n Ik+l]+ ~ ~(lk+l) Claim 1k÷l' — + + (p2k+l1)[p] 2k+l ) fl for all k > o. 0. IiF n 12k+ 21 ~ {p = ~ 12k+2 I 2k+ 2/£pF TP)[p] for This = ~{I 2 k+ 2 together with the the finiteness finiteness of T [p] yields together with yields that that Tp(p] == {pTP)[p] = This clearly implies implies that Tp is = {p 3Tp)[p] = •.• ... P This
is divisible. is divisible. Hence Tp is aa divisible divisiblesubgroup subgroup of Tlp with implies that Tp = = Tlp' r{Tp) == r{T 1p) << ~. This This implies In order to In to proceed proceed ititisisnecessary necessary totointroduce introducesome some facts facts concerning concerning a over aa field field KK of characteristic a finite finitedimensional dimensional algebra algebra A over characteristic 0. The A reader referred to [14] [14] and and [47) [47] for proofs. proofs. readerisis referred Let x1,... n I: Z a .. x., a .. € K, j=l 1J J 1J
Definition: Defi ni ti on: ax. 1
= =
be a a basis be basis for A,
i,j == l,...,n. i,j l, ••• ,n.
and let aa € A. and Jet
Then
t(a) t{a) ==tt (a13) {aij)
The trace trace of a, The
the trace = the trace of the the matrix matrix Proposition 5.2.11: Proposition 5.2.11:
t(a) t(a}
of the the choice choiceof of basis basis for for A. is independent independent of
Definition: Let X= X = {x 1 , ••• ,xn} be basis for A. be aa basis A. The The discriminant discriminant of with respect X A with respect to X is the the detenninant determinant of the the nxn matrix (t{xixj}).
A
Proposition 5.2.12: Let X,V be two two bases bases for for A, and be the and let let dX' Proposition X,Y be dx, dy be discriminant of A A with respect X Then there discriminant respect to to X and VY respectively. Then exists a € F, aa~ 0, such such that that dy == a 2dx· Proposition possible the the following: Proposition 5.2.12 5.2.12 makes makes possible A be A has positive Definition: Let A be aa finite finitedimensional dimensional Q-algebra. Q-algebra. A has discriminant if if the A with respect discriminant the discriminant discriminant of A respecttotoa Q—basis a Q-basis XX of A A is positive. positive.
be aa basis basis for A A over K, Let {x •••. ,xn} {x11 ,,.. ,xn} be
Notation:
a a field field extension. extension. AL is the the set set of ofall allexpressions expressions AL For For
x == n
x+y ==
nn
Z I:
i=l
n
n
aixi' y ==
I:
i=l
a1xi
and let let L :2 KK be and
a. x. 1 aa11 € LI. L}. 11 two arbitrary arbitrary elements two elements in AL' define { I:
i=l i=1
1
n x.y = + a B1)x1, {a. + .)x 1., and x·y = I: a.aJ.c .. kxk' where 1 . . k-1 1 1J 1.=1 1 1 I:
n n
.J. —
l,...,n. E c c....kxk, xxiX. I: kxk, c.13"k € K, i,j,k i ,j ,k = = 1, ... ,n. = 1xJ. = k=l 1J 1J•k k=1 induce induce aa ring structure structure on on AL. AL. 94
nn
Theseoperations operationsclearly clearly These
A Definition: A is separable separable over over K, extension L 2~ K. K.
if
if AL AL
for every field is semisimp!e for every field
A A be aa Q-algebra with unity e. An order in A Let A be Q-algebra with An order is aa B possessesaabasis basis for for A, of AA satisfying: 1) l) ee E B, 2) BB possesses subring B B 3) every with coefficients 3) every element element in B is aa root root ofofaamonic monic polynomial polynomial with in Z.
Definition:
The abovedefinition definition and the following The above and the following Proposition Proposition may may be be generalized generalized to Wewill will assume, algebrasover overarbitrary arbitrary fields fields of characteristic characteristic 0. We assume, from from algebras nowon, on, that A A dimensional now is aa finite finite dimensionalQ—algebra. Q-algebra. Proposition 5.2.13: .2.13:
Every order order in A may be be embedded embeddedinin aa maximal maximal order. Every may A
Proposition Proposition 5.2.14: discriminant.
if and onlyif if A has non-zero A A is separable separable over over Q if and only has A Q
every nonzero ideal in A A is Proposition If AA is semisimple semisimple then then every nonzero ideal Proposition 5.2.15: If product of prime Is. If ideals in A, and aa unique unique product prime idea ideals. If I ~ J are are nonzero nonzero ideals then the same prime factors occur if mJ mJ cE II for come ~orne positive occur positive integer m, then the same prime factors of I and J. in the the prime prime decomposition decomposition of
if
be aa maximal order in A, X X basis for A, Proposition 5.2.16: Let BB be Proposition maximal order aa basis suchthat that the the discriminant discriminant of A with respect X A X~ B, such with respect to XX is an an integer. integer, onlyif if pp does and let pp be be a prime. and let prime. Then B/pB is semisimple semisimple ifif and and only not with respect A not divide the the discriminant discriminant of of A with respect to X.
Weare arenow nowininaa position position to 5.2.6 as as follows: We toprove prove Lemma Lemma 5.2.6
By Claim Claim5.2. 5.2.10 Proof Lemma 5.2.6: By 10 itit suffices to toshow show that F/pF is Proof of Lema The subring B S semisimple for all all but many B semisimple for but finitely finitely manyprimes. primes. The of ~ generated by F and S generated by F is an an order order in S. ~. By By Proposition and the the unity ee E ~ 5.2.13, order ~W in ~ such such that s~ B. Since r 5.2.13, there there exists aa maximal maximal order has non-zero non-zerointegral integral discriminant dd with respect S is separable separable over over Q, has respect Q, ~ Proposition 5.2.14. By By Proposition Proposition 5.2.16, to aa basis basis contained contained in ~. Proposition pfd. It It therefore W/pW is semisimple ~/p~ semisim~le for every every prime prime ptd. therefore suffices suffices totoshow show that that W/pW be be the many F/pF ~ ~/pB for all all but butfinitely finitely manyprimes primes p. Let e8p:: W ~ ~ B/pB be the restriction restriction of 8P 0 o It canonical and let let 8~ canonical epimorphism, epimorphism, and be the to r. It many toshow show that 8~ is onto onto for for all allbut butfinitely finitely manyprimes primes p. suffices to such that n~ ~ F, see the proof proof of There exists integer nn such see the There exists aa positive integer Lemma 5.2.4. be a a prime prime such such that Ptn. Lemma 5.2.4. Let pp be pin. 95
pW, and i s onto. on to . This Th i s nB+ Then pif = nB" + pWcF pir ~ F ++ pB ~ if + plf, and so so epF is Then if + pW= concludes the the proof concludes proof ofofLemma Lemma 5.2.6.
In addition additiontotobeing beinga astep steptowards towardsproving provingTheorem Theorem 5.2.1, 5.2.1,Lemma Lemma 5.2.6 also implies implies the also the following:
If G admits be finite rank G Corollary 5.2.17: 5.2.17: Let G be aa finite rank torsion torsion free freegroup. group. If multiplication ofofsemisimple aa multiplication semisimple algebra algebra type type then then G is quotient quotient divisible. G R* * is semisimple, . . h R+ = GG such bbe R ring with such that R Let R e aa r1ng w1t semisimp 1e, and an d the subring By Lemma 5.2.6, let FF be be the subring of R = SS constructed construct~d in inLemma Lemma 5.2.4. Lemma 5.2.6, 5.2.4. By many For every prime (R/F); is divisible divisible for forallallbut butfinitely finitely manyprimes primes p. For prime divisible, and (R/F) == DP C±) Cp, with DP divisible, p, {R/F); and CP reduced. Since
P f Proof: _!QQ_:
is finite, This together r[(R/F)] is r[(R/F);] finite, Cp is finite finitefor forevery everyprime prime p. This together with many yields that the fact fact that Cp == 0 for all the allbut butfinitely finitely manyprimes primes pp yields that a divisible torsion group and C aa finite finitegroup. group. Let nn C G/F == D(i) C, D D be aa positive integer nC = 0. The Therestriction restriction of be integer such such that nC ofthe thecanonical canonical is aahomomorphism 0: G G ~ G/F to nG is homomorphism onto DD with kernel kernel nF. epimorphism a: nG nG G Hence nG is q.d. q.d. Since G G is q.d. q.d. G ~nG, Observethat that the the argument argumentatatthe theend endofofthe the proof proof of Corollary Observe Corollary 5.2.17 5.2.17 showsthat thatif if aa torsion G possesses aa free subgroup shows torsion free free group group G possesses subgroup FF such that G/F is the torsion group, and the direct directsum sum of ofaadivisible divisible torsion group, anda bounded a bounded group, then G group, then G is q.d. q.d.
In order the proof proof of Theorem 5.2.1itit is to In order to to complete complete the Theorem 5.2.1 is necessary necessary to for those investigate the the relationship relationshipbetween between TP and T1P for those primes primes pp for This is done in the which Tp is not not divisible. This done in the following: Lemma5.2.18: 5.2.18: Lemma
T1p/TP is finite finitefor forevery everyprime prime p.
W Again several several steps steps will will be As above, B" Again be required requiredtotoprove provethe thelemma. lemma. As order in ~ S the unity e in S, is aa maximal maximal order containing FF and and the ~. and nn is positive integer satisfying nB" nW c F. aa positive Put fk B" Jklf Tk == If lklf, and Jk = = for all kk>O. for > 0.
An immediate immediateconsequence consequence Claim 5.2.7and andthe thefact fact that An of ofClaim 5.2.7 the following: the Claim 5.2.19: Claim5.2.19:
F is
F
2Tk~Ik' {2)n (4) (1) n2Tk {l)n {4)"Jk+ls."Jk, (3) Tk+1 E 'k' (2) 2JksJk' {3)Tk+lslk'
(5) Tk (6) nTk {5J Tk :"Jk' {6) nTk Tt =.Tk+t'
96
n~ c
(7) nJk Jt :"Jk+t' (8) nt-l J~sTk' {7) {8) ETk,
p1k' for all k,2. > 0.
(9) n2(Tk÷2. fl
Since Since W B is semisimple, semisimple, every every nonzero nonzero ideal B is is aaproduct product of ofprime prime ideal in W W which are factors factors of the ideals. Let P1, ... ,Pu be the prime ideals in be the prime ideals in B which are the ideals pif, The relations n pB, nB, nB, T1, r 1 , or ~T1. . The relations n ~ ~ r k' and nJ~ ~ n~ k' 2 1 2 which from Claim Claim5.2.19 5.2.19(6) (6) and and (7), (7), together which follow immediately immediately from together with with Proposition 5.2.15 5.2.15yield yield that P'l'"'1'u factors of Tk, are the the only prime prime factors Proposition 1 , ..• ,Pu are al a2 au = P1 and "Jk for all k > 1. Therefore Pl P2 P2 ••• Pu ~1
flu — nlr = P r11 r2 .5 6
k>l,
flu
— T
6
.5
J = p kl k
. ..
—0 — r1
0 r2
ap
1
an and
,
...
non-negativeintegers integersfor for all all non-negative
6k1
1,...,u. i1 = = 1 ' ••• ,u.
k ~ 1'
Claim 5.2.20: (1) (t-1)~; + toki Claim tôki ~ Yki' > 1, ii = = l,l,...,u. + Yki for all kk ~ ••• ,u.
and (2) and
2~; + max{2.a1, max{~a;, Yk+~,i} >~a;
Claim 5.2.19 and Claim Claim 5.2.19 Claim 5.2.19(8) (8) implies implies (1), and 5.2.19 (9) (9) implies implies (2).
Proof:
Claim 5.2.21: There Thereexists exists an suchthat that for for all Claim an integer K(R.) K(t) such for all I = < 26. + ok 1• = 1l,...,u. min{Yk., ~a.}< for , ••• ,u. 1 1 1 + 6ki
k >~
If If
for all then the the inequality inequality clearly if Yki ~< 2~; for a; = 0 or or if all kk then Y a. Assume, therefore, that and that for some 0, > Assume, therefore, ai ! and Yk ; > 2e; for some kk0 .
Proof: holds.
.
0
is non-decreasing, Yki > 2~;
Since the the sequence sequence {Yki};=l
Therefore 5.2.20(2) implies that (2')
2~.
~ ~ ~.
2.
kk ~ ~.
K(~).
1
+ Yk+~, 1.
>~a.
-
1
Yk+~,i ~~a;
for all for
+ Yk.1
> kk ~
for all
k >~ k0 •
for for all all kk ~ kk00 and so
k0. k0 .
Since aa11.
;
0,
so by (2') Yki ~ ~ as k ~ ~. and by by Claim Claim 5.2.20(1), This clearly clearly implies implies Claim Claim 5.2.21. 5.2.21. This
~a,. ~ ~
as
~ ~
as
oki 6ki
An An immediate immediate consequence consequence ofofClaim Claim5.2.21 5.2.21 is is integer 2., Claim 5.2.22: For every every non—negative non-negative integer ~. Claim ~1'5" 2K(L) such that that for k >K(2.), K(~) such ~ K(~). Tk Ik ++Po~ n Jk. p2.W2n2Jk.
there exists an there an integer integer
it
be aa prime. < ~ it p be prime. Since r{TP) ~< r(T 1p) < suffices to toshow show that d(Tp) == d(T 1P), or since since TP .= Tlp' that By Claim Claim5.2.21 5.2.21 there there exists exists an d(Tlp) ~ d(TP). By an integer K(Q) K(~) such such that
ofLemma Lemma 5.2.18: Proof of
Let
p
97
4 4-.2- + pn R.2 2for kK ~> K(Q), K{R.), nJk n Jk E ~ nJk n Jk E ~ n Ik + p n B. However nn2 Ik 'k ~E Ik, 'k' Claim for all k ~> K(R.). 5.2.19(1), and F. Hence {A) K(i). 5.2.19(1), and n~ ~F. n4Jk ~E Ik (A) n4Jk 'k ++ pR.F for
c oR.{l;) be the canonical By (A) eR.{n 4 J~) ~ eR.: FF ~ F/pR.F be canonical epimorphism. epimorphism. By + + + R. R. + R. + — Jk/lp F n Jk) for all k >> K{R.). Now Now oR.{Jk) == {Jk + p FJ/p F ~ ~ Let
by Claim 5.2.9(2). Similarly, by Claim 5.2.9(2). Similarly,employing employing Claim Claim 5.2.9{1), 5.2.9(1),
{pk-R.Tlp)[pR.]
eR.{I~) ~ {pk-R.Tp)[pR.]. {p,m) 1. (p,m) = 1.
Let nn44 == p3m, pjm, m aa positive integer integer such such that m
= (pk+i_9.T1)[p2._i] Then {pk+j-R.Tlp)[pR.-j] Then n4(pk-R.T 1P)[pR.] =
subgroup subgroup of of (pk-R.Tp)[pR.].
Put
R.Z = = j+l
is isomorphic isomorphic to aa
and let ~ max{K(R.), R.}. and letk >k max{K(.Q),
Then
is isomorphic isomorphic to to aa subgroup subgroup of {pk-R.TP)[pR.], so and so (p ], and k-1 T P == d{Tlp ) , and < r {pk-R. Tp ) • For {pk-1 Tlp ) ~ kk sufficiently sufficiently large large ppk_lT1 rr(pk_lT1p) 1 the claim claim is pk-R. Tp == d{TP ) • Hence d{Tlp ) ~ d ( TP ) and and the is proved. proved. (pk-lTlpJ[p] (p
Finally, Finally,we wecan can prove proveTheorem Theorem 5.2.1.
is of of Theorem Theorem 5.2.1: thatremains remains to to be be shown shown isis that S (+) N is Proof of 5.2.1: All that SC+)N finite Lemma5.2.3. 5.2.3. By By Lemma Lemma 5.2.2 suffices to finite index index in in R, R, Lemma 5.2.2 ititsuffices to show show that S~/S+ is is finite. finite. S~/S+ ~ (S~/F)/(S+/F) = T1/T T1/T eo~ {t). Tlp/Tp' ® p paprirne p a prune Observation 5.2.5 and and Proposition Proposition 1.1.1. Now Now Observation 5.2.5 T1p/Tp is finite finitefor forevery every primes, Lemma 5.2.18, and 00 for all prime p, Lemma Lemma 5.2.18, allbut butfinitely finitely primes, Lemma 5.2.6. 5.2.6. + + S1/S is finite; thus concluding Theorem Hence S~/S+ finite; thus concludingthe theproof proofofof Theorem 5.2.1. .
5.2.1 to groups, Translating Theorem Theorem 5.2.1 groups, we we have: have:
be aa finite finite rank Corollary 5.2.23: G 5.2.23: Let G be rank torsion torsion free free group. group. Then G multiplication ofofsemisimple G c>. H$ K, where HH admits admits aa multiplication semisimple type, type, and and KK is 2 ~ 0, unless K = 0, the of aa nilpotent nilpotent ring N2 the additive additive group group of ring NN satisfying N is G or G is nil. nil. §3.
Torsion free free, rings with Torsion with semisimple semisimple algebra algebra type.
The results of upupto to quasi—isomorphism, The results ofthe theprevious previoussection sectionshow showthat, that, quasi-isomorphism, the classification of of finite finite rank of the the additive additive groups groups of rank torsion torsion free free rings rings the reducesto to two two cases; cases; the the additive groups reduces groups of of rings rings with withsemisimple semisimple algebra algebra type, and the additive additive groups of nilpotent nilpotent rings, and the groups of rings, which which are are not not zerorings. zerorings. type, In this thataafinite finite In this section section aa further further reduction reduction will willbe beobtained obtained by by showing showing that 98
rank free group group admitting admitting aa multiplication multiplicationofofsemisimple semisimple algebra algebra rank torsion torsion free type is quasi-isomorphic quasi-isomorphic to direct sum sum of groups groups admitting type to a direct admitting multiplications multiplications of simple simple algebra algebra type. type. It willfurther furtherbebeshown shown that rank torsion torsion It will that aa finite finite rank free group admitting a free group admitting a multiplication multiplicationofofsimple simplealgebra algebratype typeisisquasi— quasiisomorphictoto aa direct direct sum groupsadmitting admittingmultiplications multiplicationsof of field field type. isomorphic sum ofof groups type. Therefore, the additive additivegroups groups Therefore, upto upto quasi-isomorphism, quasi-isomorphism,the theclassification classification of the of finite type is is settled finiterank ranktorsion torsionfree freerings ringswith withsemisimple semisimple algebra algebra type by the additive additive groups of full by determining determining the groups of fullsubrings subringsofofalgebraic algebraicnumber number fields. fie 1 ds.
be aa finite finite rank Theorem rank torsion torsionfree freering ringwith withsemisimple semisimple Theorem5.3.1: 5.3.1: Let RR be such contains a subring algebra subring sS = s 1 (+) ... .•• fi:)Sm such that S; algebra type. Then RR contains = has simple algebra type, i = = 1, ... ,m, and R+ ;s+ is finite. finite. has simple algebra l,...,m, To prove prove Theorem Theorem5.3.1 5.3.1we wefirst first need: To need: ring with with Let R R be aa torsion torsion free free ring be Q—algebra with unity unity e. Let R; =RnA;. Then Q-algebra with Lemma 5.3.2: Lemma 5.3.2:
and
R+/(Rl
.....
C-:+"J.
.@
f\n) +
R** R
= = A1 <±) ••• ~ aa ®Am ... <±) * . R R; = = A1, A; , 1i = 1l,...,m, , ••• ,m,
is bounded.
=
Proof: For each i = = A~/(R fl . ,m, ••• ,m, A7/R7 = n A;)+~ = 1 ,,.. (4 ++ R+)/R+ R*+/R+ — (A7 which is aa torsion group. Therefore for for xx E A., there which group. Therefore there exists positive exists aa positive 1 1 such that nx E R Hence integer nn such ., or x E-n R .. A. c QR. = R*.. R., R.. A. c QR. = 1 n 1 1i 11 1 ** However R1 R1. is the theminimal minimal Q—algebra Q-algebra containing R., and R~ = =A .• R1 1 and so 1 A1. 1 Let ee = e1 l,...,m. is the +•.• +em, em, e1 e; E A;, Ii = = 1, ••• ,m. Clearly e; is the unity unity in e 1 +...+ 1 ,... Ai' i == 1, •.• ,m, ,m, and {e •••,em} ,em} is aa set set of mutually mutually orthogonal orthogonal {e11 ,,... R*. for idempotents R idempotents in R* • Let nn be be aa positive integer integer such such that ne; E R . * R*, l,...,m, let xE Since xE i = l, x x +...+ •.• ,m. and and let R. S1nce R, x = x1 x1 + ••• + xm• x1 X; E A1, A;, and so l,...,m. = l, l,...,m, = = l, ..• ,m. For each each i = ••• ,m, ne.x. = ne 1.x and i = 1 1 m m m m ne.x. ERn R n A. = R.. Therefore nx = nex = and so Z ne.x. R.,, and nx = nex = I: ne; xi E ® R; neixi A; R;. 1 11 11 1=11 i=l i=l + m + R I (f) R; is is bounded. bounded. i=l m * Proof of 5.3.1: ofTheorem Theorem 5.3. 1: Since R* R* is semisimple, semisimple, R* R = = ® c+.) A;, with A. aa 1
I
•I
; =1 1 As in , ••• ,m. Put S; = Rn simple 1, •••,m. simple Q—algebra, Q-algebra, i = = 1l,...,m. ,m. As R n A;, I = 1,... simple algebra has simple algebra type, i = 11,... and so so S; has , •••,m. ,m. Lemma 5.3.2, S; s7* = A1 Lemma 5.3.~, A; and
Let
S ==
{+) S .. S1. ; ,; 1 1
Then
R+/S+
rank bounded 5.3.2, finite rank boundedgroup, group,Lemma is aa finite Lemma 5. 3. 2 ,
and hencefinite. finite. and hence 99
Corollary G Corollary 5.3.3: Let G be rank torsion torsion free free group. group. Then Then GG be aa finite finite rank admits aa multiplication of semisimple semisimple algebra is admits multiplication of algebratype typeifif and andonly onlyifif G G quasi-isomorphic to toa adirect of simple simple quasi—isomorphic direct sum sumofofgroups groupsadmitting admittingmultiplications multiplications of 1gebra type. aalgebra G If G admits aa multiplication ofsemisimple semisimple algebra Proof: If admits multiplication of algebra type, type, then then it it 5.3.1 follows immediately immediately from from Theorem Theorem 5.3. 1 that G is quasi-isomorphic quasi-isomorphic to aa G of groups admitting multiplications multiplications of direct sum sum of groups admitting of simple simple algebra algebra type. type.
m
Conversely let let G,;, Conversely (+) G1, G., and be aa ring G and let let R. be ring satisfying R: i =l 1 1 m 1 (') R.. Then R == (+) and Ri* is aasimple simpleQ—aigebra, Q-algebra, i = 1,... 1, ...,m. ,m. Put R
i-l
i =1
=
1
m * m ** is R semisimple. R* = ® (±)R1 Ri is sernisimple. i=l be aa finite finite rank be rank torsion torsion free free ring ringwith withsimple simplealgebra algebra Definition: Let R R Z(R*) type. A subfield FF of Z(R*) is is a field of of definition definition of RR if ifthere there
k
* exists an an F-basis F-basis x1 , ••• ,xk of RR* in has finite index has finite index in in R. R.
E1) (R (R nfl F)x. such that S = @ R such F)x.
R
i1=1 =1
1
R will be assumed assumed totobebea afinite will be finite rank rank torsion torsion free free ring, F will signify willbebedenoted denoted by by A, and F signify aa subfield subfield of Z(A). Most will of the results in setting in the remaining remaining results in this thissection sectionappear appear in in aamore more general general setting [53]. From now now on on From
R
R* * R
Lemma 5.3.4: Lemma 5.3.4:
if and onlyifif is aa field of F of definition definition of R if and only F R • + End(R ).
+ Horn (At,, A ) HomF(A F
=
Proof:
be an of AA in R. Put S == an F—basis F-basis of x1,...,x n be
+
Let
kk
1+l (R n F)x .• n F)x..
i =1 1=1
1
Thenthere there exists exists aa is aa field of R. Suppose of definition definition of of R. Then Supposethat that F F R integer nn such such that nRc s. Let ~p€E HomF(A+, A+). that nRES. At). Since R positive integer such that exists a positive m full subring subring of of A, there there exists positive integer integer m such is aa full
=
l,...,k. m.p(x.) = l, ••• ,k. nw(x.) E R, Ii = and ~o ~ E End(R+). and so
.
kk
Now mn~(R) amp(S) m ~(S) = (+) (R nn F)m ~(X;)~ R, i-l i=l + + . + Conversely, suppose that HomF(A , A ) End(R ). Define Conversely, suppose that
A-..... FF via ni: A
n.( 1
k I:
j=l
f.x.) = f1; fi; J J
=
f3 fJ.
E F,
i,j == l,...,k. i,j l, ... ,k.
Clearly
HomF(A+, A+), A+), and and so sothere thereexists exists aa positive positive integer ni E HomFlA+, integer nn such such that
100
1,... nn.(R) ccR, R, i = l, •.• ,k. ,k. 1
Let xx E R. R.
-
Then
x ==
k
kk
~
i=l i=l
n.(x)x.,, and 1
1
nx == ~Z nn.(x)x. E S, i.e., S. Hence Hence R+/S+ is aa finitely finitelygenerated generated i.e., nRc nR cS. 1 1 i=l bounded group. Therefore R+/S+ is finite, is aa field of bounded finite,and and FF is definition definition of R. R. A be Lemma5.3.5: 5.3.5: Let A Lemma be a simple Q-algebra, Q-algebra, A~, Ar Ar the the subrings subrings of consisting of of left left and HomQ(A+, A+) consisting and right right multiplications, multiplications, respectively, respectively, by by HomQ(A+, A+) elements every subring E elements of A. Then Thenfor for every of Ho~(A+, A+) containing E such that there exists aa subfield subfield F E Z(A) Z(A) such both A~ and Ar, there + At). + E == HomF(A , A).
=
fact that EE is aa simple Proof: The The fact simple Q-algebra, Q-algebra, and and that EE ~ Ar' A~ implies {i,p E HomQ(A+,A+)I~ that A+ is an an irreducible irreducibleE-module. E-module. Put F == {~ k ~ == ~~ for all ~ E E}. Since F commutewith with the the elements elements of A~ Since the elements elements of F commute and Ar' F F is aa subfield Jacobson Density Density Theorem, subfield of Z(A). The The Jacobson Theorem, and and the finitedimensionality dimensionality of A over Q that E = HomF(A+, A+). finite A yield that At). E= Q P
A be Theorem be a simple simple Q-algebra. Q-algebra. Then Theorem5.3.6: 5.3.6: Let A F = {p {~ E HomQ(A+, A+) I~~=~ the unique unique smallest smallest 'kp for for all ~ E End(R+)} is the field of of definition definition of R, and HomF(A+, A+J == End(R+J.
Proof:
Since End(R+)
At), and End(R+) is aa subring subring of HomQ(A+, A+),
=A~,Ar•
End(R+) = A+J, Lemma Lemma 5.3.5, ofLemma Lemma 5.3.5. Lemma 5.3.5, and and proof of 5.3.5. By Lemma = HomF(A+, Ak), be an anarbitrary arbitrary field field F is a 5.3.4, F is a field of of definition definition of R. Let F' be of definition Again by Lemma 5.3.4, HomF,(A+ ,A+) of definition of R. Again Lemma 5.3.4, End(R+) == + At). + + + Horn (At, A )I wP Horn (At, A). = {~ {',p€ HomF(A, Therefore F' = E HomQ(A. ~ == ~ for all + + + + = {,pEEHomQ(A for all A )} ~ E HomF,(A HomF,(A ,, A)}~{~ HomQ(A , AA )I~~=~~ for + + ~ E HomF(A , A )} = F.
=
A will willbebeassumed assumed to be be aa simple simple Q-algebra, Q-algebra,
In what follows,
FF the
A
smallest field of of definition of smallest field of R, R, and dirnF(A) dimF(A) = k. •
+
If ee E R, ee the Corollary 5.3.7: Z[End(F )] = F. If the unity unity in Corollary Z[End(R+)1 = R R n F. F. + + • + 5.3.6. Ak)] )] == Z[End(R )] by by Theorem Theorem 5.3.6. Proof: F F == Z[HomF(A , A +
•
A,
then
+
and let let aQ E Z[End(R )]. For ~ E End(R ) there Suppose Suppose that that ee E R, and such that np exists positive integer integer nn such n~ E End(R+). Since exists a positive 101
a E Z[End(R+)], a(n ~) = (n (n ~)a. This This clearly implies implies that that n(~) n(cxp) == n(~). + + Since End(R ) is torsion torsion free, free, ~ ap ==.pa ~ and and so so a E Z[End(R )] = F. F. and so a However a= a(e) E R, R, and F. Clearly R n FF c Z[End(R+)]. a = a(e) a E R n F. 0
0
A = E, and that A= and that that ee E R, ee the Corollary 5.3.8: Suppose Suppose that Corollary the unity unity of A. Then R R == End(R+). + = HomF c+ + = F, and + 1s . By Theorem 5.3.6, End ( R) Theorem 5.3.6, F , F)= Proof: By and so End(R) is =
commutative.
Therefore R n FF == R. R. Therefore by by Corollary Corollary 5.3.7, End(R+) == Z[End(R+)l = R
Corollary Corollary 5.3.9: Let H,K be be subgroups subgroups of R+ such K = 0. such that H nn K R+ = Hq.) K, then QH QH and and QK QK are F-subspaces F-subspaces of A.
If If
A Proof: Clearly A A == QHG)QK. be the the natural natural projection onto QH (i) QK. Let nH be projection of A QH. QH. There H(+) K. Therefore Thereexists exists aa positive positive integer nn such such that nR nR =: H(+)K. and so nH E End(R+) = HomF(A+, A+). nnH(R) == nH(nR): nH(HG)K) = HH: R, and At). Hence QH of A. and similarly similarly QK, are F-subspaces QH and F-subspaces of
Corollary 5.3.10: Corollary 5.3.10: Let A be A be aa field. field. strongly stronglyindecomposable. indecomposable. Proof:
A = FF if Then A= and only if and only if if R+ is
,x be an an F-basis F-basis for for AA in R. R. Let xx1,... be 1 , ••• ,xk k
definition for R, definition
k
R:! R ® (R (R n F)xi. i=l A = F. and A= F. indecomposable, k = 1, and
F is aa field of Since F of
Thereforeiiff R+ is strongly Therefore strongly
Conversely,if if R+ Conversely, R+ is not not strongly stronglyindecomposable, indecomposable, then then by by Corollary 2, and so 5.39, k == dimFA dimFA > ~ 2, and so AA ~ F.
Piecing together the the results results of this Piecing together thissection sectionwe we obtain obtain G be torsion free free group groupofof finite finite rank. G Theorem5.3.11: 5.3.11: Let G Theorem be aa torsion rank. Then G admits multiplication of semisimple typeifif and andonly onlyif: if: semisimple type admits aa multiplication
(1)
G is quotient quotient divisible, divisible, and and
G
k
G C. C±> stronglyindecomposable indecomposable group group which which admits admits aa ® G., with G.1 a strongly i;,;l 1 . such that End(Gi) is. an multiplication of field multiplication field type type End(Gi)' such an algebraic i number field satisfying = n.umber field satisfying [End(Gi): Ql Q] == r(Gi), i = 1, ... ,k. (2)
G
0
Suppose that that G Suppose admits semisimple type. type. Then Then G admits aa multipl1cation multiplication ofofsemisimple G wewemay G is quotient quotient divisible, divisible, Corollary Corollary 5.2.17. 5.2.17. By By Corollary Corollary5.3.3 5.3.3 mayassume assume that GG admits of simple simple algebra algebra type. type. Let RR be be aa ring admits aa multiplication multiplication of
Proof:
102 102
R* a simple Q-algebra. be the the smallest with R+ = GG and R* Z{R*) be simple Q—algebra. Let FF s Z(R*) R* in R. field of be an an F-basis F-basis for R let xxl,...,Xk field of definition definition of of R, and and let R. 1 , ••• ,xk be
Then
kk
_,'Cl) fl (Rn (R n
R R,;=
11=1
F)xi.
*
Put
R. Ri
= =
+
(R (Rnn F)xi (R n F)x., F)xi, and (R
= =
Gi,
and so G. i = l, ..• ,k. Then R. R. ~ FF and admits of type type F, l,...,k. G. admits aamultiplication multiplication of ** 1 1 ** a By Corollary Corollary 5.3.8, R. a. field. By R. = and so F == R. = = 1 = End(G.) 1 and 1 = End(G.) 1 = l,...,k. End(G 1.), i = 1, .•• ,k. The , ••• ,k follows The indecomposability indecomposability of of G., ii == 1l,...,k . * 1 from Corollary Corollary 5.3.10. Now = 1l,...,k. from Now [End(Gi): Q] Q] == dimQRi = r(G1), r(Gi), i = , ••• ,k. 1
1
The converse converse is is obvious. The obvious. 54.
Applications.
Theresults results of of the sectionsofof this this chapter will be The the previous previous sections chapter will be applied applied to to solve problems which arose arose in in Chapter solve problems which Chapter 3, and and to toprove prove Theorem Theorem 4.7.22. k
Proof of of Theorem Theorem 3.2.3: 3.2.3:
be finite rank be aa finite rank torsion torsion free free group, group,
G == @ Gi Let G
i =1 1=1
+
and suppose supposethat that every every ring Ri with R1 R1 Suppose there there and R~ =~ Gi is nilpotent. nilpotent. Suppose exists exists aa ring ring R is not not nilpotent. nilpotent. Then Then by R with R+ = G such that R R Corollary 5.2.23, Corollary 5.2.23, GG &. Hl+lK, HH ~ 0, where HH admits admitsaamultiplication multiplication of semisimple By Theorem 5.3.11(2), & L(T)M, semi simple type. By Theorem 5.3.11 (2), G c.. L® M, with L a strongly L indecomposable group admitting a multiplicationofoffield field type. type. Let indecomposable group admitting a multiplication k. 1 strongly indecomposable, Gi & ~ Gl GiJ"' G.J. strongly indecomposable, i == 1l,...,k; •••• ,k; jj == 1, .•• ,k .. j=l j=l
1
.
1
< k, and By Jonsson's Jonsson's theorem, theorem, [36, Theorem 92.5], Gi j c.. LL for By Theorem 92.5], forsome some 1 ~< i ~ for forsome some 1 << jj << k1.• Now Now G. .;. G.. @H ..• By .. admits By Corollary Corollary 5.1.2, G1J 1 1J 1J multiplication of of field ring with aa multiplication field type. type. Let Si be be aa ring with field fieldalgebra algebra type, type, = and let let T1- be any ring ring with T~1 =H = 5+1. = G.J., and be any ..• Put R1 R.1 == S.(±)T 1 1J 1 1.. * * * * Then Ri R field. By By Lemma 5.1.3, Ri is not = Si (!) Ti Then = with Si aa field. Lemma 5.1.3, not nilpotent, nilpotent, 1
1
a a contradiction, Corollary Corollary 5.1.4(2). 5.1.4(2). In Theorem 3.1.3itit was shownthat thatifif GG is aa torsion In Theorem 3.1.3 was shown torsion free free group, group, Rn+l andifif R nil ring ring with with R+ = G, then Rn+l == 0. The r(G) r(G) = nn << ~. and R is aa nil = G, The same conclusionfollows followsif if R R to be be T-nilpotent T—nilpotent or or x—nilpotent, same conclusion is assumed assumed to a-nilpotent, lpotence is an arbitrary arbitrary ordinal. (Actually, the a an the conclusion conclusion for for 1-ni T-nilpotence is nil). obvioussince sinceif if RR is T—nilpotent, obvious T-nilpotent, then then RR is nil). Theorem 5.4.1: Theorem 5.4.1:
Let
G be be aa torsion free free group, group,
G
r(G) = n <<
~.
Let
R R be be aa
103
ring with R+ = G. If R is 1—nilpotent T-ni lpotent or a—nilpotent a-nilpotent for for some some ordinal R Rn+l1 = 0. a • then Rn+ Suppose that that RR is not not nilpotent. nilpotent. Then R is aa Proof: Suppose R ~ S .J) N. where S S subring or RR of sernisimple N nilpotent semisimple algebra algebra type, type. and and N is the the maximal maximal nilpotent subring
subring of R. subring R, with Si
Theorem 5.2.1. Theorem 5.2.1.
By By Theorems Theorems5.3.1 5.3.1 and and 5.3.6, 5.3.6, S S
aa ring of of field fieldalgebra algebra type. type.
Hence
kk
~ (+)
s .•
~1 1 * ** * ii=l k S. R == (+t (±1 N* , with Si* R S1GN
**
k
1 'i—i i=l lpotent nor a field, field. i = = 1l,...,n. , ••• ,n. Clearly Si* is neither neither1—ni T-nilpotent nor a—nilpotent, a-nilpotent. so by byLemmas Lemmas 5.1.5 and and 5.17, S1 Si is neither neitherT—nilpotent T-nilpotent nor nor a—nilpotent, a-nilpotent, This clearly clearly implies implies that RR is neither neither R-nilpotent R-nilpotent nor nor ii=1,...,k. = l •.•• ,k. This T-nilpotent, or a-ni ipotent for some Therefore if R is T-nilpotent, or a-nilpotent for some a—ni lpotent. Therefore if R a-nilpotent. n+l = nilpotent,sosobybyTheorem Theorem 3.1.3. RRn+l ordinal a, then RR is nilpotent, 0.
*
a
be aa finite finite rank Corollary rank torsion torsion free free group, group, Corollary 5.4.2: Let GG be are equivalent: The following following are The ((1) 1) (2) (3) (4) ( 4) (5)
r(G) r(G) == n. n.
n+l1 = 0 + Rn+ = G. G. R with R+ R every ring R R = 0 for every nil potent. R with R+ = Every ring R = GG is nilpotent. R+ = R with lpotent. Every ring R = GG is 1—ni T-ni lpotent. Every ring R R with R+ = = GG is nil. nil. forsome some ordinal ordinal a-nilpotent Every ring R R with R+ = 1potent for = GG is a-ni .
.
a.
(4) and Proof: The and (1) • (5) are are obvious. obvious. The implications implications (1) • (2) • (3) • (4) by Theorem Theorem 3.1.3. and (5) • (1) by by Theorem Theorem 5.4.1. (4) • (1) by 3.1.3, and
is a ring By Theorem 5.2.1, R=S(F)N, ProofofTheorem4.7.22: where SS isaring R = S®N, where Proof of Theorem 4.7.22: ByTheorem5.2.1, Since N R is aa nilpotent ring. R is with semisimple semisimple algebra algebra type, type, and and N is aaconiuutative commutative semisimple semisimple Q-algebra, Q-algebra, unital, N = 0. unital. 0. Therefore R* R* == S* is * n R* = 1l,...,n. algebraic number field, iI = an algebraic number field, , ••.• n. Put and and so so R = €J ® Fi, with Fi an = i =1 n subring of Fi ' and R1 == R R nfl Fi. F.. Then R R= = (!) Ri , with Ri aa unital subring Ri an d R+.1 i =1 , ••• ,n. ,n. a a full fullsubgroup subgroup of F+1 , i == 11,...
104
Bibliography
1
Arnold, D., Finite FiniteRank RanKTorsion TorsionFree FreeAbelian AbelianGroups Groups and and Subrings Subrings of Arnold, D., Finite Dimensional Q-Algebras, Q-Algebras, totoappear, appear,Springer—Verlag. Springer-Verlag. Finite Dimensional
2 2
Bachman, top-adic p-adicNumbers Numbers and and Valuation Valuation Theory, Theory, Bachman,G., G.,Introduction Introduction to Academic New York-London, York-London, 1964. 1964. Academic Press, Press, New
33
order,Duke Duke Math. Math. J. Baer, R., Abelian Abelian groups groups without Baer, without elements elementsofof finite finite order,
4 4
Bass, H., Finistic generalization of semiBass, H., Finisticdimension dimension and and aa homological homological generalization semiprimary rings, Trans. Trans.Amer. Amer. Math. Math. Soc. Soc. 95 95 (1960), (1960)-, 466-488. 466-488. primary rings,
5 5
Beaumont,R.A., R.A.,Rings Ringswith withadditive additive group groupwhich whichisis the the direct direct sum sum of of Beaumont, cyclic groups, cyclic groups, Duke DuKe Math. t4ath. J. J. 15 15 (1948), (1948),367-369. 367-369.
6 6
Beaumont,R.A., R.A., Lawver, Lawyer,D.A., D.A., Strongly Strongly semisimple abelian groups, groups, Pac. Beaumont, semisimple abelian Pac. J. J. Math. Math. 53 53 (1974), (1974), 327-336. 327-336.
77
free Math. 55(1961) (1961) Beaumont, R.A., Beaumont, R.A.,Pierce, Pierce,R.S., R.S.,Torsion Torsion freerings, rings,Ill., Ill.,J.J.Math. 6 1-98. 61-98.
8 8
Beaumont, R.A.,Pierce, Pierce, R.S., R.S., Subrings Subringsof of algebraic number fields, Acta number fields, Acta Beaumont, R.A.,
99
Beaumont, R.A., Pierce, Pierce, R.S., Mem. Beaumont, R.A., R.S., Torsion Torsion free freegroups groupsofofrank ranktwo, two, Mem. Amer. Amer. Math. Soc. Soc. Nr. Nr. 38 Math. 38 (1961). (1961).
10 10
Beaumont, R.A.,W1sner, Wisner,R.J., R.J., Rings Ringswith with additive additive group which is is aa Beaumont, R.A., group which free group group of ofrank rank two, two,Acta ActaSd. Sci.Math. Math.Szeged Szeged 20 20 (1959), (1959), torsion free
3 (1937), (1937), 68-122. 68-122.
Sci. Math. Math. Szeged Szeged 22 22 (1961), (1961),202—216. 202-216.
105—116. 105-116.
11 11
Beaumont, R.A., the subgroups subgroups of Beaumont, R.A., Zuckerman, Zuckerman,H.S., H.S.,AAcharacterization characterization of the of (1951), 169—177. the additive rationals, the rationals, Pac. Pac. J.J.Math. Math. 1 (1951), 169-177. 1
W., Uber die abelschen Gruppenauf aufdenen denensich sichnur nurendlich endlich viele viele ll Borho, W., Uber die abelschen Gruppen 12 Borho, wesentlich verschiedene definieren lassen, wesentl1ch verschiedene Ringe Ringe definieren lassen, Abh. Abh. Math. Math. Sem. Sem. Univ. Univ. Hamburg 37 37 (1972), Hamburg (1972),98—107. 98-107.
rings whose endomorphisms 13 Bowshell, Bowshell, R.A., Schultz, Schultz, P., P.,Unital Unital rings whoseadditive additive endomorphisms commute, Math. Ann. commute, Math. Ann. 228 228 (1977), (1977),197—214. 197-214.
Springer-Verlag, Springer-Verlag, Berlin, Berlin, 1968. 1968.
14 14
Deuring, M., Algebren Deuring, N.,
15 15
Feigeistock, S., S., On the nilstufe nilstufe of Feigelstock, On the ofthe thedirect directsum sumofoftwo twogroups, groups, Acta Acta Math. Math. Sci. Budapest Budapest 22 22 (1971), (1971), 449-452. 449-452. 105
16 16
Feigeistock, S., S., The nilstufe of Feigelstock, The nilstufe of the the direct directsum sum of of rank rank one one torsion torsion free free groups, Acta groups, Acta Math. Math. Sci. Sci. Budapest Budapest 24 24 (1973), (1973), 269-272. 269-272.
Feigeistock, S., homogeneous 17 S.,The The nilstufe nilstufeof of homogeneous groups, groups, Acta Acta Sci. Math. 1·4ath. 17 Feigelstock, Szeged36 36 (1974), (1974), 27-28. Szeged 27-28.
Feigeistock, S., a group 18 S., The The absolute absolute annihilator annihilatorofof a groupmodulo moduloaasubgroup, subgroup, 18 Feigelstock, Math. Debr. Debr. 23 23(1976), (1976),221—224. 221-224. Publ. Math.
Feigelstock, S., Extensions of partial 19 Extensions of partial multiplications multiplicationsand andpolynomial polynomial 19 Feigelstock, identities identities ononabelian abeliangroups, groups,Acta ActaSci. Sci.Math. t4ath.Szeged Szeged 38 38 (1976), (1976),17-20. 17-20. 20 20
Comment. Feigelstock, S., S.,On On groups groups satisfying satisfyingring ringproperties, properties, Comment. Math. t•lath. Univ. Univ. Sancti Sancti Pauli Pauli2525(1976), (1976),81—87. 81-87.
21 21
Feigelstock, S., S., The nilstufe of Feigelstock, The nilstufe of rank rank two two torsion torsion free free groups, groups, Acta Acta Sci. Sci. Math. Szeged 36 {1974), (1974), 29-32. Math. Szeged 36 29-32.
22 22
groups Comment. Feigelstock, S., s.,On On the thetype typeset setof of groupsand andnilpotence, nilpotence, Comment. Math. Math. Univ. Sanct Sanct Pauli Pauli2525(1976), (1976),159—165. 159-165.
23
Feigelstock, On the of aa group, group, to to appear appear in Feigelstock, s., S., On the generalized generalizednilstufe nilstufe of Pubi. Math. Publ. Math. Debr. Debr.
24
Feigelstock, S.,An An embedding embedding theorem Feigeistock, S., theoremfor forweakly weaklyregular regularand andfully fully idempotent rings, rings, Comment. Math. Uni Univ. idempotent Conment. r~ath. v. Sancti Sancti Pauli Pauli2727(1978), (1978),101—103. 101-103.
25
Feigelstock, s., S., The additive groups groupsofof subdirectly subdirectly irreducible irreducible rings, Feigelstock, The additive rings, Bull. Aust. 164—170. Bull. Aust.Math. r~ath.Soc. Soc.2020(1979), {1979), 164-170.
26
Feigelstock, S., S., The additivegroups groupsofofsubdirectly subdirectlyirreducible irreduciblerings rings II, II, Feigelstock, The additive Bull. Aust. Bull. Aust.Math. Math.Soc. Soc.2222(1980), (1980),407—409. 407-409.
Feigeistock, S., The additive groups of rings rings possessing onlyfinitely finitely 27 Feigelstock, The additive groups of possessing only many ideals, Comment. Math. Univ. Univ. Sancti Pauli many ideals, Comment. Math. Pauli2828(1980), (1980),209—213. 209-213.
Feigelstock, S., S., The additive groups groupsofof local local rings, Archiv 28 Feigelstock, The additive Archiv for for Mat. 14at. 18 18 (1980), 49—51. (1980)' 49-51. Feigeistock, S., Comment. 29 Feigelstock, S., AAsimple simple proof proofofofa atheorem theoremofofStratton, Stratton, Comment. Math. Math. Univ. Sancti Univ. Sancti Pauli Pauli 29 29(1980), (1980),21—23. 21-23.
Feigelstock, S., S., The Theadditive additivegroups groupsofofrings ringswith withtotally totally ordered lattice 30 Feigelstock, ordered lattice of ideals, of ideals,Quaestiones QuaestionesMathematicae Mathematicae 4 (1981), (1981), 331-335. 331-335. 31 31
Feigeistock, S., Feigelstock, s., On On aa problem problem of F.A. F.A. Szasz, Szasz, submitted. submitted.
32 32
Feigelstock, s., S., Schiussel, Feigelstock, Schlussel, Z., Z., Principal Principalideal idealand andNoetherian Noetherian groups, groups, Pac. J. Math. Math. 7575(1978), {1978),87—92. 87-92. Pac. J.
Freedman, additivegroup group torsionfree free ring ring of 33 Freedman, H.,H.,OnOn thethe additive ofofa atorsion of rank rank two, two, Pubi. Publ. Math. Math. Debr. Debr. 20 20(1973), (1973),85—87. 85-87.
106
of an an abelian abelian group groupthat that are are ideals ideals in 34 Fried, E., E., On On the the subgroups subgroups of every ring, ring, Proc. every Proc. Colloq. Colloq.Ab. Ab.Groups, Groups, Budapest Budapest (1964), (1964), 51-55. 51-55. 35
Fuchs, L., L., Abelian AbelianGroups, Groups, Akade'miai Akademiai Kiadd, Kiado, Budapest Budapest (1966). {1966). Fuchs,
36
Fuchs, L., Infinite Fuchs, L., InfiniteAbelian AbelianGroups, Groups, 1 (1971), 22 (1973), (1973),Academic Academic Press, Press, New York-London. York-London. New 1
additive Gruppe, Pubi. Math. 37 Fuchs, Fuchs, L., Ringe Ringe und—ihre und-ihre additive Gruppe, Publ. r~ath. Debr. Debr. 4 (1956), {1956), 488-508. L., On Math. Szeged 38 Fuchs, Fuchs, L., On quasi-nil quasi-nilgroups, groups,Acta ActaSci. Sci. Math. Szeged1818(1957), (1957),33—43. 33-43. 39 39
Fuchs, L., Halperin, I., I.,On On the the embedding embedding ofofaaregular regular Fuchs, L., Halperin, regular ring ring in aa regular
40
Fuchs, Rangaswamy, K.t~ •• OnOn generalized Math. Z. Z. 107 107 Fuchs, L., L., Rangaswamy, K.M., generalizedregular regularrings, rings, Math. (1968), 71—81. {1968), 71-81.
41 41
Gardner, B.J., B.J., Rings torsion free groups, Gardner, Rings on on completely completely decomposable decomposable torsion groups, Comment.t~ath. Math.Univ. Univ. Carol. 15 Comment. 15(1974), (1974),381—392. 381-392.
42 42
Gardner, Gardner, B.J. 117—129. 117-129.
43
Gardner,B.J., B.J., Jackett, Jackett, D.R., oncertain certain classes classesof of torsion torsion free Gardner, D.R., Rings Rings on free abeliari groups, 493—506. abelian groups, Coment. Comment. Math. Math. Univ. Univ. Carol. Carol.1717(1976), (1976), 493-506.
withidentity, identity, Fund. Math. (1964), 285-290. ring with Fund. Math. 5454 (1964), 285—290.
Some aspectsofof T-nilpotence, T—nilpotence,Pac. Pac.J.J. Math. Math. 53 53 (1974), Some aspects
F., Radical and antiradical antiradical groups, Mt. J. Math. 44 Haimo, Haimo, F., Radical and groups, Rocky Rocky t~t. Math. 33 (1973), (1973), 91-106. 45 Herstein, I.N., I.N.,Noncommutative Noncommutative Rings, Rings, Carus Carus Monogr. Monogr. no. Math. Ass. Ass. no. 15, Math. of Amer., New New York, York, 1968. 1968. of Amer., 46
D.R., The The type torsion free freeabelian abeliangroup group of of rank rank two, two, Jackett, D.R., type set set of aa torsion J. Aust. Aust. Soc. Soc. Ser. Ser. AA27 27 (1979), (1979),507-510. 507-510.
47 Jacobson, Jacobson, N., The Theory Rings, _Math. Surveys no. 2, Amer. Amer. Math. Math. Soc., Soc., N., The Theory of of Rings, Math. Surveys no. 2, NewYork, York, 1943. New 1943. 48 48
Jacobson, N.,Structure Structure of of Rings, Jacobson, N., Rings, Colloq. Co 11 oq. Pubi. Pub l. vol. vo l.37, 37,Amer. Amer. Math. t~ath. Soc. Soc. Providence, R.I.,, 1968. Providence, R.I. 1968.
49 Kertesz, A., Volesungen Volesungen uber Ringe, uber Artinsche Ringe, Budapest ((1968). Budapest 1968).
Akademiai Akade'miai Kiado, Kiadd,
50 50
Levitzki, J., J.,Contributions Contributions to tothe thetheory theory of ofnilrings nil rings(Hebrew, (Hebrew, with with Lematimatka77(1953), (1953), 50-70. Riveon Lematimatka English summary), summary), Riveon
51 51
O'Neill, O'Neill,J.D., J.D.,Rings R1ngswhose whose additive additivesubgroups subgroups are are subrings, subrings, Pac. Pac. J. J. Math. Math. 66 (1976), 66 (1976),509—522. 509-522.
52 52
O'Neill, ring with with free O'Neill, J.D., J.D.,AnAnuncountable uncountable Noetherian Noetherian ring free additive additive group, group, Proc. Amer. Proc. Amer. Math. lvlath. Soc. Soc. 66 66 (1977), (1977), 205-207. 205-207.
107
53 Pierce, R.S., R.S., Subrings Subrings of simple simple algebras, algebras, Mich. Mich. Math. 14ath. J. 7 (1960), 241-243.
54
Redei, L., Szele, Redei, Szele, 1., T.,Die DieRinge Ringe"ersten "erstenRanges", Ranges", Acta Acta Sci. Sci.Math. Math.Szeged Szeged 12 (1950), 12 (1950).18—29. 18-29.
55
Ree, R., Wisner, Wisner, R.J., R.J., A on torsion free Ree, R., A note note on free nil nilgroups, groups,Proc. Proc.Amer. Amer. Math. Soc. Soc. 77 (1956), (1956), 6-8. Math.
56
Reid, torsion free Reid, J.D., J.D., On On the the ring ringofofquasi-endomorphisms quasi-endomorphisms ofof aa torsion free group, group, Topics in Topics inAbelian AbelianGroups, Groups,Scott—Foresman, Scott-Foresman, Chicago Chicago (1963), (1963),51—68. 51-68.
57 57
Reid, J.D., 229—237. Reid, J.D., On On rings ringson ongroups, groups,Pac. Pac.J.J.Math. f4ath.5353(1974), (1974), 229-237.
58
Schiussel, Z., Schlussel, Z., On On additive additive groups groups of ofrings rings(Hebrew, (Hebrew, with withEnglish Englishsummary), summary), Master's thesis, thesis, Bar-han Master's Bar-Ilan University University (1976). (1976).
59 59
Schuitz, P., P., The The endomorphism endomorphism ring ring, Schultz, ringofofthe theadditive additive group groupof of aa ring, J. Aust. Aust. Math. Math. Soc. Soc. 1515(1973), {1973),60—69. 60-69.
60 Stratton, Stratton, A.E., A.E., The The type rank, typeset setofoftorsion torsionfree freerings ringsofof finite finite rank, Comment.Math. Math.Univ. Univ. Sancti Sancti Pauli 27 Comment. 27 (1978), (1978),199—211. 199-211. 61 61
M.C., The Thenilpotency nilpotencyofoftorsion torsion free free rings rings with Stratton, A.E., A.E., Webb, Webb, M.C., given type 437—444. set,Comment. Comment. Math. Math. Univ. Univ. Carol. Carol.1818(1977), (1977), 437-444. type set,
62 Stratton, A.E., A.E., Type Type sets and nilpotent Acta Sci. Sci. Math. Math. sets and nilpotent multiplications, multiplications, Acta Szeged 41 (1979), {1979),209—21 209-213. Szeged 41 3. 63 63
Szasz, Uber Ringe t1inimalbedingunq fur Sza'sz,F., F., Uber Ringe mit mit Hinimalbedingunq fur Hauptrechtsideale, Hauptrechtsideale, I, I, Publ. Math. 14ath. Debr. Debr. 77(1960), (1960),54—64. 54-64.
64 64
Sza'sz, Uber Ringe Szasz, F., F., Uber Ringemit mit 1·1inimalbedingung Ilinimalbedingungfur furHauptrechtsideale, Hauptrechtsideale,II, II, Acta Acta Math. Math. Sci. Sci.Budapest Budapest12 12(1961), (1961),417—439. 417-439.
65 65
Szasz, F., Uber Uber Ringe Ringe mit mit f1inimalbedingung linimalbedingungfur fur Hauptrechtsideale, Hauptrechtsideale,III, III, Acta tiath. Acta Hath. Sci. Sci.Budapest Budapest1414(1963), (1963),447—461. 447-461.
66 66
Szasz, F., Radikale RadikaleDer DerRinge, Ringe,Akade'miai Akade'miai Kiado', Kiado', Budapest, Budapest, 1975. 1975. Szász, F.,
67
Szele, 1., Szele, T.,Zur ZurTheorie Theorieder derZeroringe, Zeroringe,Math. Math.Ann. Ann.121 121(1949), (1949),242—246. 242-246.
68
Szele, 1., T.,Gruppentheoretische GruppentheoretischeBeziehungen Beziehungen bei bei gewissen gewissen 14ath. Z.Z.5454(1951), (1951),168—180. 168-180. Ringkonstruktionen, Math.
Pubi. Math. Debr. 71—78. 69 Szele, Szele, T., T., Nilpotent NilpotentArtinian Artinianrings, rings, Publ. 14ath. Debr.4 (1955), 4 (1955), 71-78. 70 70
Szele, T., T., Fuchs, Fuchs, L., L.,On On Artinian Artinianrings, rings,Acta ActaSci. Sci.Math., Math., Szeged 17 17 Szele, Szeged (1956), (1956),30—40. 30-40.
71 71
108
Vinsonhaler, C., C.,Wickless, Wickless,W.J., W.J.,Completely Completelydecomposable decomposable groups groups which which admit only only nilpotent nilpotent multiplications, Math. 5353 (1974), 273—280. admit multiplications,Pac. Pac.J. J. Math. (1974), 273-280.
72 72
Webb,M.C., M.C.,AAbound bound thenilstufe nilstufe of of a group, Sd. Math. Webb, forforthe group, Acta Acta Sci. Math. 39 39 (1977), 185—188. (1977). 185-188.
73 Webb, and triangles of types, types, Acta Acta Webb,M.C., M.C.,Nilpotent Nilpotenttorsion torsionfree free rings rings and triangles of Sci. Math. Sci. Math.Szeged Szeged 41 41 (1979), (1979),253—257. 253-257. 74 74
Wickless, W.J., W.J., Abelian Abelian groups groupswhich whichadmit admitonly onlynilpotent nilpotentmultiplications, multiplications, Wickless, Pac. Pac. J. J. Math. Math. 40 40(1972), (1972),251—259. 251-259.
Zariski, 0., 75 Zariski, 0.,Samuel, Samuel,P., P.,Commutative Commutative Algebra, Algebra, Vol. Vol. 1,1,Van Van Nostrand, Nostrand, Princeton, Princeton, 1967. 1967.
109
Index
Absolute annihilator, 34 34 Algebra 89 Algebra type typeof of aa ring, 89 a—nilpotent ring, 34, 34, 35, 90, a-nilpotent ring, 90, 103, 103, 104 104 Anti-radical group, group, 75 75 Ascending 53 Ascendingchain chaincondition conditionfor for ideals, 47-49, 53 Associative group, 10-15 10-15 Associative nil nil group, n a Associative w-group, w prope_rty, 36 36 Associative n—group, a ring property, Associative principal principal ideal ring Associative ring group, group, 44, 45, 45, 51 51 Associativestrongly strongly principal principal ideal Associative ideal ring ring group, group, 43-46 Associative strongly subdirectly subdirectly irreducible ring Associative strongly ring group, group, 61—63 Associative subdirectly irreducible Associative subdirectly irreducible ring ringgroup, group, 61-63 Beaumont-Pierce decomposition 75, 90—98 90-98 Beaumont—Pierce decomposition theorem, 75, Basic subgroup, subgroup, 22 Centerof of aa ring, 88, 101 Center 101 Descendingchain chaincondition conditionfor for ideals, ideals, 50-61 50—61 Descending Discriminant, 94 Divisible Divisible group, group, 1, 3, 3, 44 Division ring ring group, group, 36, 37 37 E—group, 84-85 E-group, 84—85 E—ring, 82—84 E-rin~, 82-84 Essential subgroup subgroup 3, 44 38 Field group, group, 36, 38 100—103 Field of of definition definitionofofa ring, a ring, 100-103 Full subgroup, subgroup, 71 71 Generalizednilstufe, nilstufe, 34, 35 Generalized 35 Height, 22 Homogeneous 11, 14 14 Homogeneousgroup, group, 2, 11, Independent subset, 11 Independent Irreducible group, Irreducible group, 72 65—68 Local ring, 65-68 Local Local ring group, Local group, 67
63-65
Maximal order in in an Maximal order an algebra, algebra, 95 95 MHI ring, fttii ring, 58 58 ffii ring ringgroup, group, 58-61 till MIR ring, ffiR ring, 58 I4IR ring group, ffiR ring group, 58—61 58-61 Nil Ni 1group, group, 10-15 10—15 Nilstufe of ofaagroup, group, 25—31 25-31 Noetherian ring, 47 Noetherian ring, Noetherian Noetherian ring group, group, 48 Nucleus, Nucleus. 14, 15 15 Order Order in an an algebra, algebra, 95 p—basic subgroup. subgroup, 1 p-basic p-divisible group, 11 p—divisible group, p-height, 22 p—pure subgroup 1 p-pure p—rank, 1 p-rank. a ring property, w-group, yr-group, 1r a property. 36 36 w-ring. group property, 88 n-ring, 1rir a group Prime ring ring group, Prime group. 36 36 . 46, 47, Principal ideal Principal ideal ring ring group, group, 46, 47, 53 53 Pure subgroup, 1 Pure Quasi -endomorphism, 3 Quasi-endomorphism, 3 Quasi—equal, 3, 88 88 Quasi-equal, Quasi—isomorphic, Quasi-isomorphic, 3, 88 88 Quasi-nil group, 16—24 group, 16-24 Quotient divisible divisible group, Quotient group, 88, 96, 96, 102 102 Radical ring group, Radical group, 40-42 40—42 Rank, 1 Rank. Rigid group, group, 2, 15 15 Semiprimering ring group, Semiprime group, 37 Semisimple ringgroup, group.37—40, 37-40,73—75 73-75 Semisimple ring Separable algebra, 95 Separable 95 Simple ring group, Simple group, 36, 37 37 Smallestfield field of of definition definition of aa ring, Smallest ring, 101 101 Strongly indecomposable indecomposable group, 46, 102 102 group, 3, 46, Strongly Strongly irreducible irreducible group, group, 72 72 1
1
1
1
1
112
47—49 Strongly Noetherian Noetherian ring ring group, group, 47-49 w-group. itw a a ring property, property,· 36 Strongly u-group, 36 Strongly ideal ring ringgroup, group, 43—46 43-46 Strongly principal principal ideal Strongly semisimple semisimple ring group, group, 39, 69, 70, 70, 73-75 73-75 39, 40, 69, Strongly subdirectly group, Strongly subdirectly irreducible irreduciblering ring group,63—65 63-65 Stronglytrivial trivial left Strongly leftannihilator annihilatorgroup, group, 69—71, 69-71, 76 76 Strong nilstufe, nilstufe, 25, Strong 28 25, 26, 28 Subdirectly irreducible ring, Subdirectly ring, 61 61 Subdlrectly irreducible Subdirectly irreduciblering ringgroup, group,61—65 61-65 T—nilpotent ring, 32-34, T-nilpotent ring, 90, 103, 103, 104 104 32-34, 89, 90, Torsion Torsion free rank, rank, 1 Torsion free ring. ring, 88 Torsion free 88 Trace, 94 ITring, ring, 85-87 85—87 ITring ringgroup, group, 85, 87 87 Trivial left leftannihilator annihilatorgroup, group, 69-71, 76 69—71, 76 Trivial left annihilator ring, left annihilator ring. 69 Type, 2, 2, 5, 66 Ulm Ulm subgroups, 22 1
113