SOME CONDITIONS FOR FINITENESS OF A RING

Extending a result of Putcha and Yaqub, we prove that a non-nil ring must be finite if it has both ascending chain condition and descending chain condition on non-nil subrings. We also prove that a periodic ring with only finitely many non-central zero divisors must be either finite or commutative.

I. INTRODUCTION AND TERMINOLOGY.
Over the years several authors have given sufficient conditions for a ring R to be finite, among them the following: (I) (Szele, [9]) R has both ascending chain condition and descending chain condition on subrings; (II) (Ganesan, [4], [5]) R has non-trivial left zero divisors, of which there are only a finite number; (III) (Bell, [i]) R contains no infinite zero ring and no infinite subring with- out non-zero nilpotent elements; (IV) (Putcha and Yaqub, [8]) R is non-nil and has only finitely many non- nilpotent elements.
The present study, which presents some new conditions for finiteness, was motivated by the Putcha-Yaqub paper.Our first two theorems are ones suggested by that paper; the third is a new result on the old theme of commutativity and finiteness.
Throughout the paper the term zero divisor will refer to a one-sided (i.e.not necessarily two-sided) zero divisor.By a left (right) zero divisor we shall mean an element y for which there exists x # 0 such that yx 0 (xy 0).If Xl,X2,...,x k R, the subring generated by the x.lwill be denoted by fx l,x2,...,X and for each x R, the symbols A(x) and Ar(X) will denote respec- tively the left and right annihilators of x.The symbols C and N will be used for the center of R and the set of nilpotent elements of R. The symbol Z will denote the ring of integers, and Z + the set of positive integers.
Finally, the ring R is called periodic if for each x E R, there exist distinct Z + m n m,n for which x x H.E. BELL 2. TWO FINITENESS THEOREMS FOR NON-NIL RINGS.
Our first theorem, which employs (IV) in its proof, is an extension of (II).
THEOREM I. Let R be a ring, and let S be the set of non-nilpotent zero divisors of R. If S is finite and non-empty, then R is finite.
PROOF.Let x S. Applying the pigeonhole principle to the powers of x yields m n distinct m,n Z + for which x =x consequently, there exists a non-zero idempotent zero divisor e, which we assume to be a right zero divisor.Write R eR + A (e). r Since each summand consists of zero divisors of R, each has only finitely many non- nilpotent elements, hence by (IV) is either finite or nil.It is immediate that eR is finite, and to complete the proof we proceed on the assumption that A (e) is nil.r s xS-I 0, so e+x is a zero divisor.
Let 0 # x Ar(e), with x 0 # Then (e+x)x s-I Moreover, e+x is non-nilpotent, since for any k > s, we have (e+x) k k-1 i e + i x e; k and the assumption that (e+x) 0 gives, on left multiplication by e, the contradiction e=0.It follows that the set {e+xlx Ar(e)} is finite, hence Ar(e) is finite and so is R.
THEOREM 2. If R is any non-nil ring having both ascending chain condition and descending chain condition on non-nil subrlngs, then R is finite.
PROOF.Note that by (I) and (III), any infinite ring R satisfying our hypo- theses, and indeed every infinite subring of R, must contain an infinite zero ring.
Moreover, for any non-nilpotent element x, the chain <x>x2><x4 9 becomes stationary at some point, hence there exist n Z + and p(x) Z[X] for which n n+l x x p(x); and since this last condition is obviously satisfied by nilpotent elements as well, a result of Chacron ([3], [2, Theorem I]) shows that R is periodic, hence contains non-zero idempotents.The following lemma gives crucial information about the idempotents.
LEMMA.If R satisfies the hypotheses of Theorem 2 and e is any non-zero idempotent, then Ar(e) and At(e) are finite.
PROOF.Assume without loss that e is a left zero divisor; note that in any periodic ring, idempotents have finite additive order.Recall our initial remark, which implies that if A (e) is infinite, it must contain an infinite zero ring.r Let B be any zero ring contained in A (e), and let u be an arbitrary element of r B. Considering the chain ,e,u,2 ,e,2u9 <e,4u> yields k Z + such that 2ku e,2k+lu> that is, there exist p,q,t Z such that 2ku pe + q2k+lu + t2 k+lue.
t2 k+l Left-multiplying by e yields pe=0, hence <2 k q2k+l)u ue, and the fact that e has finite additive order shows that u does also.We now know that any subring E of R generated by e and a finite number of elements of B is finite.Choosing a maximal E, say E and noting that B E we see that B is finite.The proof of the lemma is now complete.
Returning to the proof of Theorem 2, suppose that e is an idempotent which is a zero divisor, say a left zero divisor; and write R =eR+A (e) =eRe+ (eR A(e)) +Ar(e). r The last two summands are finite by the lemma, and the first is a ring satisfying our original hypotheses and having a multiplicative identity element.Of course, if all idempotents of R are regular, then R has a multiplicative identity element; therefore, we have reduced the problem to proving the theorem under the additional hypothesis that R has i, in which case the periodicity of R implies that R has non-zero characteristic.
If there exists a non-zero idempotent f # I, the decomposition R fR + (1-f)R shows that R is finite, since both summands are finite by the lemma.Therefore, assume that is the only non-zero idempotent, and use the periodicity of R to obtain the property that every element is either nilpotent or invertible a property that R forces N to be an ideal [7].The factor ring has ascending chain condition and descending chain condition on subrings, hence is finite by (I).Now consider N, and let B be any zero ring contained in N.Among subrings of R generated by and finitely many elements of B I, choose M to be a maximal one.Note that M is finite and B M; hence B is finite, N is finite, and R is finite.

A THEOREM ON PERIODIC RINGS.
The final theorem may be thought of as an extension of Herstein's result ([6], [2, Theorem 2]) that periodic rings with N C are necessarily commutative.THEOREM 3. Let R be a periodic ring having only finitely many non-central zero divisors.Then R is either finite or commutative.
PROOF.Let n(R) denote the number of non-central zero divisors, and note that Herstein's result implies commutativity of R if n(R) 0. Assume henceforth that n(R) I; and consider first the case that every element of R is a left zero divisor or, more generally, the case that the set D of left zero divisors is a non-trivial additive subgroup of R. Then for d D and u D 0 C, d+u D\C; hence {d+ulu 6 D C} is finite.Thus, D is finite; and R is finite by (II).This argument covers the case R=N, so we assume that R # N and therefore R contains non-zero idempotents.
If every non-zero idempotent is regular, there exists a unique non-zero idem- potent, necessarily I; and every element is invertible or nilpotent.It follows, again by [7], that N is an ideal; and since N is equal to the set D of left zero divisors, R is finite.
Assume now that we have a counterexample R with n(R) as small as possible.Then there exists y D and therefore an idempotent e D. Thus R has a left identity element; and since we can repeat our previous arguments for right zero divisors, R has a right identity as well, hence R has I.Moreover, by the argument in the previous paragraph, R has an idempotent e which is a zero divisor.If e C, then at least one of eR and Re must be non-commutative.On the other hand, if e C, then R eR (l-e)R, where denotes a ring-theoretic direct sum; and since R was a counterexample, one of the summands must be non-commutative.Thus, in any event we may assume eR to be non-commutative.
Now eR must contain a non-central element d which is a left zero divisor in eR; otherwise, eR would be commutative by Herstein's result.For u (l-e)x ( C (l-e)R, we have eu=ue=0, hence u left-annihilates eR and d+u is a non-central left zero divisor in R. Thus, C (l-e)R is finite; and since (l-e)R consists of zero divisors H.E. BELL in R, it contains only finitely many elements not in C, hence must be finite.Now eR cannot be finite as well, since R eR + (l-e)R; therefore n(eR) n(R), and every non-central zero divisor in R must be a zero-divisor in eR.It follows that (l-e)R C.But then for any non-central zero divisor d and any element u (l-e)R, d+u is a non-central zero divisor, so both d and d+u are in eR and therefore u eR.
But this implies (l-e)R {0}, which is a contradiction.This completes the proof.
In Theorem 3 the hypothesis of finitely many non-central zero divisors cannot be replaced by the assumption that R has only finitely many non-central nilpotent elements.A counterexample is the direct sum F S, where F is an infinite periodic field and S is a finite non-commutatlve nil ring.
A plausible extension of the Putcha-Yaqub result namely, that a ring R having only a finite number of regular elements must either be finite or consist entirely of zero divisors is also false, even for commutative rings.To see this, consider the algebra A over GF(2) having basis (l,el,e 2 e n }, where the e i are pairwise orthogonal idempotents.Certainly A is not finite, and it is easily shown that is the unique regular element.