coefficient rings of isomorphic group rings

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BOL. SOC. BRAS. MAT. 7(1976), 59-63 59 Coefficient rings of isomorphic group rings M. M. Parmenter* Let R, S be commutative rings with 1 and let (x) be an infinite cyclic group. The problem of determining when the group rings R(x) and S(x) are isomorphic was previously studied in [3] and in [4]. In particular, it was shown in the latter paper that, irt a very special case (namely when R and S have no idempotents except 0 and 1), R(x) " S(x) implies R and S are subisomorphic. In this paper, we drop all conditions on R and S from the above theorem. Specifically, we' prove that R(x) " S(x) always implies R and S are sub- isomorphic. 1. Preliminary Results. In this section, we note some basic facts which will be required in the proof of the main theorem. The proofs of the first two lemmas previously appeared in [4] and will not be repeated here. Lemma 1.1.4. Let R be a commutative ring with 1. If Eaix i is a unit in R(x) and (Y~a~xl) -1 = Ebix ~, then Y aib-i = 1 and aibj is nilpotent whenever i+j~:O. Lemma 1.2.4. Let R be a commutative ring with 1. Then x ---, ~aix i in- duces an R-automorphism of R~x) if and only if the following two conditions hold: (i) Eaix i is a unit in R(x). (ii) a~ is nilpotent whenever iS 1, - 1. However, the major result required in the proof of the main theorem is the following: Lemma 1.3. Let R be a commutative ring with 1 and let Ea~x ~ be a unit in R(x) such that ao is nilpotent. Then: (i) R(Y.aix i) is a group ring. (ii) If r = g(x) ((Zaix i) - 1) where r ~ R and g(x)~ R(x), then r = O. *Recebido em Julho de 1976. *This work supported in part by National Research Council of Canada grant A-8775.

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Page 1: Coefficient rings of isomorphic group rings

BOL. SOC. BRAS. MAT. 7(1976), 59-63 59

Coefficient rings of isomorphic group rings

M. M. Parmenter*

Let R, S be commutative rings with 1 and let ( x ) be an infinite cyclic group. The problem of determining when the group rings R ( x ) and S ( x ) are isomorphic was previously studied in [3] and in [4]. In particular, it was shown in the latter paper that, irt a very special case (namely when R and S have no idempotents except 0 and 1), R ( x ) " S ( x ) implies R and S are subisomorphic.

In this paper, we drop all conditions on R and S from the above theorem. Specifically, we' prove that R ( x ) " S ( x ) always implies R and S are sub- isomorphic.

1. Preliminary Results.

In this section, we note some basic facts which will be required in the proof of the main theorem. The proofs of the first two lemmas previously appeared in [4] and will not be repeated here.

Lemma 1.1.4. Let R be a commutative ring with 1. I f Eaix i is a unit in

R ( x ) and (Y~a~xl) -1 = Ebix ~, then Y aib-i = 1 and aibj is nilpotent whenever i + j ~ : O .

Lemma 1.2.4. Let R be a commutative ring with 1. Then x - - - , ~ a i x i in- duces an R-automorphism of R ~ x ) if and only if the following two conditions hold:

(i) Eaix i is a unit in R ( x ) . (ii) a~ is nilpotent whenever i S 1, - 1.

However, the major result required in the proof of the main theorem is the following:

Lemma 1.3. Let R be a commutative ring with 1 and let Ea~x ~ be a unit

in R ( x ) such that ao is nilpotent. Then: (i) R(Y.aix i) is a group ring.

(ii) I f r = g(x) ((Zaix i) - 1) where r ~ R and g(x)~ R ( x ) , then r = O.

*Recebido em Julho de 1976. *This work supported in part by National Research Council of Canada grant A-8775.

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60 M . M . Parmentor

Remark . Result (i) in the above L e m m a simply means that powers of Za~x ~ are independent over R.

Proof. Before beginning, we note that since Eaix ~ is a unit in R ( x ) , L e m m a "1.1 says that Za~b_l = 1 and a~bj is ni lpotent whenever i + j # 0 (where Zb~x i = (Eaix~) - 1). Note that if P is a pr ime ideal of R and a~ r P, then b~ ~ P whenever i + j :/: 0 since aib~ is nilpotent. Also b_~(~P since a~b-i = 1 in R/P. This then implies that a j ~ P for all j # i since a~b_~ is nilpotent. We conclude that each of the pairs (a~, b_;} have associated disjoint sets of pr ime ideals P for which a~r and b- iCP. In part icular, we note tha t a~aj and b~b~ are ni lpotent if i # j.

N o w let us p rove (i), Assume to the cont ra ry that Ecj (Za~x~) J = 0 for cj ~ R. We will first show that each cj is ni lpotent - if not, let n be the largest integer such tha t c, is not ni lpotent and let P be a pr ime ideal of R such that c, ~ P. We k n o w also that there is a unique i such that a ~ P and b_~r P. In R/P (x ) , we have

E +. eo+ = o. j>O j<O

Note that since ao = 0, all powers of x involved are different. If n > 0, we get ~,h~' = 0 which means ~, = 0 or ~ = 0 and contradic ts the choice of P. I f n --- 0, we get Co = 0 which is impossible. I f n < 0, we get ~,b'-7 = 0 which is again a contradict ion. Hence c# is ni lpotent for all j.

Let I be the (niipotent) ideal of R generated by the c j, the ni lpotent ai and b~, all te rms a~b# where i + j # 0, and all terms a~a# and b~b# where i :/: j. Choose (if possible) k so that all c# lie in I ~ but some c# does not lie in I k+~. In R/I ~+ t ( x ) , Ecj(Za~x~) ~ = 0 becomes

y. + + Y. = o j>O j<O

since c~aiat = 0 and ?,~ibz = 0 if i # I. Let m be an integer such that am is not ni lpotent- note that m exists since Z,a~b_i = 1 and m d: 0 by assumpt ion . Mult iplying by fi= yields

e ag + ' + eoa . + x"J = o j>O j<O

since cja,,,ai -- 0 if m # i and ~ , ~ = 0 if m + i ~ O. Since m =# O, all po- wers of x involved aro different. Thus we have ~fi~+t = 0 if j > 0 and ~ , , b : ~ = 0 i f j < O. In part icular , we conclude that there exists 1 > 0 such that - - , -~ c a=b-m = 0 for all j and for all m such that am is not ni lpotent .

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Coefficient rings of group rings 61

However , Zaib- i = 1 in R/ I k+ 1. Thus (Z~ib_,) l = I and ~j(E~jb_i) * = cs-

But this becomes ~(Ealbt_~) = cs since csala ~ = 0 if i :~ s. This means, by the previous paragraph, that ~s = 0. Thus we have shown that c s ~ P + 1 for all j. Since I is nilpotent, c s = 0 for all j and R ( E a , x ~) is a g roup ring.

Next we proceed to prove (ii). Thus we have r = g(xX(Eaix i) - 1) where r e R and g(x) ~ R ( x ) . If P is a prime ideal of R, this becomes ? = O(xX~ix i) - 1)

in R / P ( x ) where i is the unique integer such that a i r P and O(x) is simply g(x) reduced m od P. Since R/P ( x ) is an integral domain and i :/: 0 by assump-

tion, this implies that g(x)~ P ( x ) . Since this is true for all P, we conclude that the coefficients of g(x) are nilpotent.

Let J be the (nilpotent) ideal of R generated by all ni lpotent a~ and bi,

the coefficients of g(x), and all a~a s and bib s where i :/: j. If g(x) = Eglx i, assu- me (if possible) that all g~eJ ~ but that some g i e J ~+1.

Choose n so that a, is not nilpotent and pass to R/J ~+~ ( x ) . Multiplying = 0(x.)((Z~ix ~) - i) by 6, we obtain ~ , = O(x)(~2x" - ?~,) since O(x)~,~i = 0

if nv~ i. Choose t maximal so that g+ r J ' + 1. Assume that n ~> 0 for the moment .

If t :/: n, we have - -2 - g+a, 0 since n > 0. If t = n, we have - -2 = -- g+ _,,a, -

- - g + a . = 0. Continuing, we obtain:

g+a~ = 0 + _ . a . ~ = 0 + - ~ . ~ . . . . = 0 + - ~ . ~ §

Since g+-:k. = 0 for some k (note that n ~ 0) we have g+a~ = 0. N o w assume n < 0. If t =~ 0, we have g+a. = 0. If t = 0, we have

- - - - - 2 O++.a. = g + a . . Also 0++z .a . = 0 + + ~ - Since g++~. = 0 for some k, we have 0 + ~ v = 0 for some p.

Thus we have 0+a~ = 0 for some p in all cases, and p can be chosen so this true for any non-ni lpotent a. . As before, (Ea,b-~F = i becomes

,~+ = 0+(EafbV_~)=0. Thus, we have g + e J "+t which is a contradict ion.

2. Main Theorem

We now prove the main result of this paper. Recall that two rings R and S are subisomorphic if R can be embedded in S and S can be embedded

in R.

Theorem 2.1. Let R, S be commutative rings with 1. Then R ( x ) ~- S ( x ) implies R and S are subisomorphic.

Proof We may as well assume R ( X ) = T(Ea~x ~) where Eaix i is a unit in R ( x ) . Hence, Lemma 1.1 tells us that (Eaixi) - t = Eb~x ~ where Yaib_~ = 1 and a~bj is nilpotent if i + j :/: 0. AS in the p roof of L e m m a 1.3, we know as well that aiaj and b~bj are nilpotent whenever i r j.

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62 M . M . Parmeatcr

First, let us consider the case where ao is niipotent (possibly even ao = 0).

In this case, part (i) of L e m m a 1.3 says that R(Eaix ~) is a g roup ring. Hence R (Ealx i) c R ( x ) = T(Eaixi). Define ct: R(Eaix i) ~ T(Eaix ' ) to be the

inclusion map and let P :.T(Eaix ~) ~ T be the augmenta t ion h o m o m o r p h i s m obtained by factoring out the augmenta t ion ideal AT of T(Eaix~). Clearly, the augmenta t ion ideal A R of R(Zaix ~) is contained in ker P o cc Conversely, if r ~ R c~ ker p oct, then r = g(x) ((Eaix i) - 1) for some g(x) ~ R (x ) . Par t (ii) of L e m m a 1.3 then tells us that r = 0.

Thus ker p o ct is exactly the augmenta t ion ideal A R of R(Za~x i) and R "" R(Eaixl) /A~ is embedded in T as required.

Next, we must consider the case where ao is taot nilpotent. Since ~a~b _~ = 1 and a~bj is nilpotent whenever i + j :/: 0, we know that aoboao - ao is nil- potent and (aobo) 2 = aobo + n where n is nilpotent. Hence there exists an idempotent e in R such that e - aobo is nilpotent ([--23 p. 72). Since central idempotents in a group ring have finite support group [-l'l, e also lies in T. Since e~R n T, e (R(x ) )= (eR)(x) =(eR)(ex) and e (T(Ea /x i ) )= (eT) (Ea ix i) = (eT) (e(Zaix~)). Similarly for the idempotent 1 - e .

But (1 - eXZaix i) - (1 - aoboXZaix i) is nilpotent. Hence (1 - eXEalx i) has

constant coefficient ao - aoboao + m where m is nilpotent. Since ao - aoboao is nilpotent, the equality ( 1 - e ) R ( x ) = ( 1 - e )T(Ea~x ~) belongs to the initial case of our proof. Hence we know that (1 - e )R can be embedded in (1 - e) T.

Also eai is nilpotent if i :~ 0, so in (eR)(x) = (eT)(Eaixl), we m a y assu- me that ai is nilpotent whenever i ~ 0. If we could conclude that e R c eT, then we would have R embedded in T since R " eR + (1 - e)R. T h u s we have reduced the problem to the following case: R ( x ) = T(Eagx i) where

ao is a unit and a~ is nilpotent if i r 0.

F r o m now on, we will deal solely wi th the above case. Since x is a unit in T(Zaix i ) ,Lemma 1.1 says that x = ~cr j for some c j e T such that

there exist dj ~ T with Y, c f l_ j = I and such that Cflk is ni lpotent whenever

j + k ~ 0. As before, we conclude that C iCk and dflk are ni lpotent whenever jv~ k.

Choose Ck V ~ O. The T-homomorphism of T(Eaix i) defined by ~ . a i . x i - - -~

Ck(Y_,aix i) + ( ~ clXZaixi)-1 is an au tomorph i sm of T(Zaix i) by L e m m a 1.2 l ~ k

since

[,ck(Za,x') + (~. c,)(Y-,a,x')-'-] ['( ~, d,)(Za,x') + d_k(Ea,x')-'] = 1 + no l ~ k IS - k

where no is nilpotent, so ck(Eaix i) + ( ~ clXEaixi) - I is a unit in T(~.aixi~. l S k

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Coefficient rings of group rings 63

Hence R(x ) = T(ck(Eaix i) + ( ~ cz)(Eaixi) - 1). We will now show that

ck(Z, alx i) + ( ~ cl)(Eaixi)-1 satisfies the condition of Lemma 1.3, namely, that l~:k

it has a nilpotent constant term (in R(x)). Once this is done, the Lemma can be used exactly as before to conclude that R can be embedded in T.

Recall that x = Ecl(Eaixi) J and that a~ is nilpotent if i ~ 0. Let P be a prime ideal of R. In R/P(x ) , we obtain x = E~'j~ since all other ai are nilpotent. In fact, since P(x) is a prime ideal of R(x) , we have x = ~fi~ for some particular s. Since ao is a unit, it follows that c~ = m~x + ~ mi,~x i

where m~,s is nilpotent for all i. This is true for any non-nilpotent cs since, in that case, a prime ideal P of R can be found such that ~', ~ 0 in R/P (x) . Therefore, the constant term of ck(Eaix i) + ( ~ clXEaix i)- ~ is nilpotent, since

l~k the only non-riilpotent entries in the c[s are coefficients of x and the only non-nilpotent a~ is ao. Thus, Lemma 1.3 is applicable and we have R em- bedded in T.

T h e identical argument in the other direction shows that T can be em- bedded in R. Hence R and T (or R and S) are subisomorphic.

References

[1] R. G. Bums, Central lderapotents in Group Rings, Canad. Math. BulL 13(4), (1970), 527-528. [2] J. Lambek, Lectures on Rings and Modules, Blaisdell, 1966. [3] M. M. Parmenter and S. K. Sehgal, Uniqueness of Coe~cient Ring in Some Group Rings,

Canad. Math. Bull. 16 (4), (1973), 551-555. [4] M. M. Parmenter, Isomorphic Group Rings, Canad. Math. BulL 18 (4), (1975), 567-576.

Memorial University of Newfoundland St. John's Newfoundland Canada