SOME CONDITIONS FOR FINITENESS AND COMMUTATIVITY OF RINGS

We present several new sufficient conditions for a ring to be finite; we give two conditions which for periodic rings R imply that R nst be either finite or cumutative; and we study cumutativity in rings with only finitely many non-central subrings.

Over the years, several authors have given sufficient conditions for a ring to be finite, these conditions typically involving restrictions on subrings or zero divisors. More recently, Putcha and Yaqub [i provided a sufficient condition for finiteness of a non-nil ring-specifically that the set of non-nilpotent elements be finite; and Bell [2] presented conditions implying that a ring is either cumutative or finite.
In this paper, we offer scme new conditions for a ring to be finite, and continue the developrent of the cummtative-or-finite theme. Scme of our results are extensions of known results; others, particularly those in the final section, are of a quite different character.
In what follows, R is an associative ring with center C. The set of nilpotent elements is denoted by N; and for a subset S of R, the subring generated by S is denoted by <S>. The term zero divisor will mean a one-sided zero divisor (i.e. not necessarily a two-sided zero divisor), and 0 will be considered a zero divisor. For x e R, the symbols Ar(X A (x), and A(x) denote respectively the right, left, and two-sided annihilators of x. Finally, the symbols Z, Z n, and C(p) denote respectively the ring of integers, the ring of integers mod. n, and the Prefer p-group. We shall frequently have use for direct-sun deccmpositions of R, both as a ring and as an additive group. To make the distinction clear, we use the symbol to denote a ring-theoretic direct sun and the symbol to denote an additive-group direct sum.

A FINITENESS RESULT.
We begin by discussing rings in which certain subrings of zero divisors are assuned to be finite.

536
H. E. BELL AND F. GUERRIERO THC4 i. Let R be a ring with at least one non-nilpotent zero divisor, and suppose that every proper non-nil subring of zero divisors is finite. Then either (i) R is finite, or (ii) there exist primes p and q, not necessarily distinct, and a positive integer k, such that R Zqk T, where T is the zero ring on C(p ).
PROOF. Let x be a non-nilpotent zero divisor. If <x2> R, then <x2> is finite and there exist distinct positive integers n and m, with n > m, such that x n xm. A standard tation (see [3]) shows that e x m(n-m) is a non-zero idempotent zero divisor. On the other hand, if <x2> R, then x x2p(x), where p(X) 6 Z[X]-and xp(x) is a non-zero idempotent zero divisor. In any case, R contains a non-zero idempotent zero divisor e, which wit/xt loss we assune to be a left zero divisor.
Both of the subrings eR and Ar(e are proper and we have the grouptheoretic direct sn R eR + Ar(e ).

If A t (e)
{0), then eR is a proper non-nil subring of zero divisors.
If A t (e) {0), then eR ere and again er is a proper non-nil subring of zero divisors. Therefore er is always finite. Moreover, if A r (e) is non-nil, then it is finite and so is R. Thus, assne that Ar(e is nil; and consider <e, A r (e) >, which is clearly a non-nil subring of R. If y E <e, Ar(e)>, then y ne + x where n is an integer and x E Ar(e)-and choosing m > 1 such that x m 0 xm-l, we see that yx "I 0, so that <e, Ar(e)> is a non-nil subring of zero divisors. If it is a proper subring, it is finite, in which case Ar(e is finite and R is finite. Therefore we can suppose that <e, A r (e) > R, and that A r (e) is nil and infinite. We now have the group-theoretic direct sn Ar(e Ar(e)e + A(e). (2.1) We claim that Ar(e)e is finite. Omsider S <e, Ar(e)e>, which is a non-nil subring of zero divisors. Since <e> is a subring of the finite ring eR, e nst have finite additive order. Moreover, e is a right identity element for S, and hence (S,+) is a periodic abelian group of bounded order; therefore, every subgroup of (S,+) is a direct sn of cyclic groups [4, p. 44]. Let B {x a 6 A} be a basis for the group (Ar(e)e, +), and hence {e} U B a basis for S. Choose x0 6 B, and define B 1 to be B\x0.
Then S 1 <e, BI> is a proper non-nil subring of zero divisors; therefore B 1 is finite, B is finite, and Ar(e)e is finite. Since Ar(e was infinite and nil, we know by (2.1) that A(e) is nil and infinite; consequently, A(e) SOME CONDITIONS FOR FINITENESS AND COMMUTATIVITY OF RINGS 537 contains an infinite zero ring T [5, Proposition 5].
Recall that R eR Ar(e)e A(e). If Ar(e)e {0}, then <e> T is a proper infinite non-nil subring of zero divisors, contrary to our hypothesis. Thus R eR A(e). Moreover eR <e> and A(e) T, for otherwise <e> T violates our hypothesis again. It is now i,mediate that R is a ring-theoretic direct man of <e> and A(e).
Our basic finiteness hypothesis on R forces every proper subring of A(e) to be finite, and a result of Laffey [6] implies that A(e) ast be the zero ring on C(p) for scme prime p. It is clear that <e> Z n for scme positive integer n. If n jk, where j and k are relatively prime and greater than I, then Z n Zj Z k, and Z k T violates our hypothesis on R. Therefore <e> Z for scme prime q and positive integer k.
An imneiate consequence of Theorem 1 is the following: CEROLLARY i. Let R have at least one non-nilpotent zero divisor, and assne that every non-nil subring of zero divisors is finite. Then R is finite.

FINITE CR ATIVE PHRICDIC RINGS.
We nw Dam our attention to periodic rings, and prove two results on the t]lene of ccmatativity and finiteness. The first is motivated by Bell's result that a periodic ring with only a finite nurser of non-central zero divisors ast be tative or finite 2, Theorem 3].
THECR4 2. A periodic ring R with only finitely many non-central subrings of zero divisors is finite or ccmutative. Before beginning the proof, we recall a useful fact about periodic rings, namely that they are either nil or possess non-zero idempotents. Indeed scme power of each element is idempotent 3], so that a periodic ring having non-nilpotent zero divisors has a non-zero idempotent zero divisor.
There does not seem to be a short proof of Theorem 2, so we separate out scme of the details in four initial lemnas. The first is a well-known result of Herstein; the seccnd is due to Szele. 1 ([7], [8]). If R is a periodic ring all of whose nilpotent elements are central, then R is ccmutative.
IZMMA 2 [9 ]. A ring R having both ascending chain condition and descending chain condition on subrings ast be finite. In particular, any ring with only a finite number of subrings is finite. iZMMA 3. Let R be a ring with only finitely many non-central subrings of zero divisors. Then any non-central nilpotent element has finite additive order.
PROOF. Let u be a non-central nilpotent element of R with u m 0 u m-I for scme positive integer m > i. Note that there is at most one prime p for which pu is central, hence there exists a prime p such that for all primes p _> P, pu C. There must exist distinct primes p and q such that p, q P and <pu> <qu>, and thus there exist integers n l, n 2 n k such that 2 k pu nlqu + n2u + + nku Hence, tu n2 u2 + + nkuk, where t p nlq 0. Cnsequently, t2u n2u(tu + n3u2(tu + + nkuk-l(tu ).
(3.1) The right side of (3.1) is a polynunial in u, with each term of degree at least three. By continuing in this manner, we see that tm-lu 0.
4. Let the periodic ring R have only finitely many non-central subrings of zero divisors.
(i) If e is an idempotent and eR is not cummtative, then A(e) is finite.
(ii) If R is the ring-theoretic direct sum of R 1 and R 2, with R 1 non-ccmtative, then R 2 is finite.
(iii) If e is a non-central idempotent of R, then Ar(e)e is finite.
(iv) If e is a non-central idempotent of R such that eR is ccmatative and Ar(e)e q C, then eR is finite.
PROOF. (i) Since eR is not cc,mutative, there exists a nilpotent element u in eR which is not central in eR, hence not central in R. Let u k 0 uk-l, for scme positive integer k > i. For any subring S of A(e), <u,S > u k-I [0), hence <u,S> is a non-central subring of zero divisors, and therefore there are only finitely many such subrings <u,S>. Suppose that S 1 and S 2 are subrings of A(e) such that <u, SI> <u, S2>. Then for arbitrary s I in S I, we can write r s I p(u) + s 2 + qi(u)ti for some positive integer r, i=l and the t i are in S 2, and p(X) and the qi(X) are in XZ[X].
where s 2 Thus, r s I s 2 ). It follows that S 1 c_ S2, and similarly one shows that S 2 c_ S 1 Therefore, A(e) has only a finite number of subrings, and hence is finite by lanna 2.
(ii) The argunet is similar to that of (i) and is unitted.
(iii) We may assune that A r(e)e [0); moreover, A r(e)e is a zero ring. If S is any subring of Ar(e)e, then <e, S> Ar(e)e [0), so <e,S> is a non-central subring of zero divisors. By hypothesis, there are only finitely many such subrings; and furthermore, it is even easier than in part (i) to show that different S give rise to different <e,S>. Thus Ar(e)e has only finitely many subrings and must therefore be finite, again by lanna 2. SOME CONDITIONS FOR FINITENESS AND COMMUTATIVITY OF RINGS 539 (iv) Choose w e A r (e)e\C. Since eR is cumutative, eR ere, so that eRA r(e) (0); therefore, for any subring T of er, <w, T> w [0) and <w, T> is non-central. As usual, this implies there are only finitely many such T; and er is finite. F THECPaM 2. In view of lanna 1 we may assune that R contains a non-central nilpotent element x. Then <x> is a non-central sabring of zero divisors, and it is finite by Lemna 3. Since R has only finitely many non-central subrings of zero divisors, we see that if each non-central zero divisor is nilpotent, then the set of non-central zero divisors is finite and by Bell's result [2] R is finite or cumutative.
Henceforth we may assune that R contains a non-central, non-nilpotent zero divisor. Therefore, R contains a non-zero ide,10otent zero divisor e, which we assume without loss to be a left zero divisor.
Let n(R) denote the nunber of proper non-central subrings of zero divisors. Assune now that the conclusion is false and that R is a counterexemple with n(R) minimal emong ccunterexamples.
If e e C, then R is the ring-theoretic direct sun of er and A(e), with one of these non-cummtative. Since both eR and A(e) are proper subrings of zero divisors, we have that n(eR), n(A(e)) < n(R). Whichever of these is non-cumutative must be finite; and by Lemna 4 (ii), the other is finite as well. Therefore e C, and we assune first that A(e) (0). Then eR and Re are both proper non-central subrings of zero divisors; moreover, if both were crmmatative, we would have ex exe xe for all x e R, contrary to the fact that e C. Thus we may assune that eR is not cumtative and n(eR) < n(R), so the minimality of n(R) implies that eR is finite. By Lemna 4 (i) and (iii), both Ar(e)e and A(e) are finite, hence R eR + Ar(e)e + A(e) is also finite.
Finally we assune that e C and A(e) (0). Then R eR Ar(e)e, with the latter being finite by Lemna 4 (iii). If Ar(e)e C_ C, then Ar(e)e [0), contradicting cur assunption that A(e) (0). Thus, assune that Ar(e)e C, in which case n(eR) < n(R) again. Therefore, eR is again finite or cumutative; and in view of Lenna 4 (iv), eR is finite in any case, shwing that no counterexample can exist. THRCREM 3. Let R be a periodic ring. If every proper non-central sabring of zero divisors is finite, then R is finite or cumutative.
. Henceforth we may assume that R contains regular elements. If all zero divisors are nilpotent, the fact that R N guarantees that R has non-zero idempotents, all of which are regular. Therefore R contains a unique non-zero idempotent, which must be I-and it follows that every element of R is either nilpotent or invertible, a sufficient condition for N to be an ideal [ii]. If N c_ C, then R is ccmatative by Lema i. Otherwise, N is a proper non-central subring, hence is finite; therefore R has only a finite number of zero divisors and is finite by an old theorem of Ganesan ([12], [13]).
We can now assize that R contains a non-nilpotent zero divisor and hence a non-zero idempotent zero divisor e, which we assue to be a left zero divisor.
As usual we write R eR Ar(e ). If A(e) {0}, then eR consists of zero divisors" and if A(e) [0}, then eR eRe and eRA r(e) [0). In any case eR and Ar(e are both proper subrings of zero divisors. If both are central, then R is ccnutative; otherwise, at least one is finite by hypothesis, and arguments similar to those used in proving Lemna 4 establish finiteness of the other as well.

RINGS WITH FINITELY MANY NN-CNTRAL SUBRINGS.
Theorem 2 of this paper suggests that we should exanine arbitrary rings with only a finite nunber of non-central subrings. We now present scme results on such rings, and note that the first is a refinement of a theorem of Bell 14 ], which asserts that a ring with fewer than three proper subrings is ccmutative. THCR4 4. If R has fewer than four non-central subrings, then R is cumtative.
PROOF. Let R be a non-cumtative ring. Then there exist x, y E R\C such that [x,y] # 0. Note that <x,C> and <y,C> are cumutative and hence proper; moreover <x,C> <y,C>, for otherwise we would have [x,y] 0.
Since (R,+) cannot be the union of two proper subgroups, there exists z <x,C> U <y,C>. Clearly <z,C> is a proper subring different frcm both <x,C> and <y,C>; hence, we have now exhibited four non-central subrings, including R itself.
As the following example demonstrates, Theorem 4 is best possible. Let (R,+) {0,a,b,c} be the Klein four-group and define nitiplication by 0x x0 cx 0, ax bx x for all x E R. Clearly R is not cumutative and C {0}; moreover, R has precisely four non-central subrings, <a>, <b>, <c>, and of course, R itself. THCRM 5. If R has no non-zero nilpotent elements and only finitely many non-central subrings, then R is cumtative.
We shall make use of the following lesma. SOME CONDITIONS FOR FINITENESS AND COMMUTATIVITY OF RINGS 541 5. Let R be a ring wit/xt non-zero divisors of zero. If y E R and there exist relatively prime integers m, n such that E C and C, then y C. PRCDF. Let y be a non-zero element of R that satisfies ur hypothesis.
Assne without loss that n > m, and write n mq + r with 0 < r < m and q a positive integer. Observe that for all x in R, 0 r + rl.
Since q e C, it follows imnediately that q[x, yr] 0 [x, yr]. Therefore yr e C; and repeating the above argnent for m ql r + r I and 0 < r I < r, we get that yrl E C. Continuing with the Euclidean algorit/n until the remainder is i, we have that y C. PROOF OF THEC 5. We first note that R is a subdirect product of rings with no non-zero divisors of zero [15], hence we may asstm that R is also without non-zero divisors of zero. Assne that there exists y R\C. By Lesna 5, there exists an infinite sequence of integers n i n I < n 2 < n 3 < denoted by A, such that y C for all n i in A. Since n. the set [< y 1 > nie A contains only finitely many distinct subrings, there exist n and m in A, with n < m, such that <yn> <. Hence p(), where p(X) e XZ[X]. Using the fact that R has no non-zero divisors of zero, we see that y y2q(y) for some q(X) E Z[X]. Thus, for each y R\C there exists q(X) Z[X] such that y y2q(y) is central-and since central elements obviously have the same property, R is ccmmtative by a well-theorem of Hersteln 16 ]. THEC4 6. Let R be a ring with (R,+) torsion-free. If R has only finitely many non-central subrings, then R is cc, ntative.
PRO(. Assuming that there exists a non-central element y, we again claim that there exists an infinite increasing sequence A of positive integers such that C for all n A. To see this, we need only suppose that our claim is false and hence there exists a positive integer K such that for all integers k _> K, ? C. An art similar to that used in proving lanwa 3 implies the existence of a non-zero integer t, and a polyncmial g(X) 6 z[x] such that tK-ly g(y). Therefore tK-ly C, which contradicts our asstmption that y C.
Oonsideration of (<y n e A) again implies the existence of integers n and m in A, with n < m, and a polyncmial p(X) Z[X], such that (y). Therefore -l(y n+Ip(y)) 0, and a standard computation (cf. [5], proof of lanna 3) shows that y n+ip(y) is in N. But lanna 3 and the torsion-freeness of (R,+) imply that N c_ C, and once again R is ccnmtative by Herstein's result. 542 H. E. BELL AND F. GUERRIERO . Although a proof has eluded us thus far, we suspect that any ring with only a finite number of non-central subrings is finite or ccmutative.

RINGS WITH A FINITE MAXIMAL SUBRING.
T. J. Laffey has asked whether a ring with a finite maximal subring must be finite. The general question, which appears to be difficult, remains unanswered; bat in interesting special cases, we have shown that the answer is affirmative. and hence the index of I is finite and bounded by IM Note that R is finitely-generated, since for any x M, R <x,M>. A theorem of Lewin [17, Theorem 1 ], which asserts that a finitely-generated ring has only finitely many subrings of index a given integer, implies that R can have only finitely many ideals not contained in M. Since M is finite, we conclude that R contains only finitely many ideals.
Oonsider next a subring S of finite index. Each such S contains an ideal of finite index 17 ]-and since there are only finitely many such ideals, the index of S is bounded by the maximum of the set of indices of the ideals. Using Theorem i of [17] again, we have that R has only finitely many subrings of finite index. Hence R has only finitely many subrings, and is therefore finite.
We conclude with 8. If R is a ccmutative ring with a finite maximal subring S, then R is finite.
PROOF. If S is an ideal, then R/S has no proper subrings; it is therefore finite, implying that R is also. let us asstne that R is a counterexample to the theorem with SI minimal mong counterexnples. Consider the Jacobson radical J(S), which is nilpotent, and J(S)R + S, which is a subring containing S. Maximality of S implies that either J(S)R + S S or J(S)R + S R. If we assume the latter, then R (J(S)R + S) c_ (J(S)(J(S)R + S) + S) c_ J(S) + S.
Repetition of this argument shows that R C_ J(S)nR + S for all positive integers n, and the nilpotency of J(S) then implies the ridiculous assertion that R c_ S. Therefore J(S)R + S S, implying that J(S)R is an ideal of S.
Since R is ccmmtative, J(S)R is nilpotent, which then implies that SOME CONDITIONS FOR FINITENESS AND COMMUTATIVITY OF RINGS 543 J(S)R c_ J(S) and hence J(S) is an ideal of R.
Note that R/J(S) has a finite maximal subring S/J(S). If IS/J(S)I < ISl, then, (by the minimality of ISl), R/J(S) is finite and R is also. Therefore we must have that J(S) {0}, so S is non-nil and contains a non-zero idempotent e. Assume temporarily that e is a zero divisor in R, and suppose also that eR S. Then R eR + S and hence R/eR (eR + S)/ea S/(S n eR).
(5.1) is finite. Since R eR @ A(e), (5. i) implies that A(e) is finite. Oonsidering eS as a subring of eR, we observe that eS eR or eS is a maximal subring of eR; and since we have assumed that eR S, the latter nst hold. If eSl < IS I, then again by the minimality of IS l, eR is finite and so is R. Thus asstrne that eS S, in which case the fact that S eS @ (A(e) N S) implies that A(e) N S {0}. Because A(e) {0} and S is a maximal subring, we have that R S + A(e), which implies that R is finite.
Therefore we may assne that eR c_ S implying, of course, that eR is finite. Since eR is an ideal of R, cr initial statement implies that eR mast be properly contained in S. Then S/eR is a maximal subring of R/eR with S/eR < Sl. Minimality of Sl implies that R/eR is finite and so is R.
It remains only to consider the case that all idempotents in S are regular in R. Since J(S) (0}, S is a direct product of finite fields. therefore, the regularity of idempotents forces S to be a finite field. For any x e RkS, we have <x,S> R (p(x):p(X) e SIX]}; hence, if x is algebraic over S, it is clear that R is finite. Cn the other hand, if x is not algebraic over S, then R is iscmorphic to SIX], which does not contain a finite maximal subring. In any event, we have contradicted ur assumption that R was a ccunterexample.