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ON ZERO SUBRINGS AND PERIODIC SUBRINGS
HOWARD E. BELL (Received 23 October 2000)
Abstract.We give new proofs of two theorems on rings in which every zero subring is finite; and we apply these theorems to obtain a necessary and sufficient condition for an infinite ring with periodic additive group to have an infinite periodic subring.
2000 Mathematics Subject Classification. 16N40, 16N60, 16P99.
LetRbe a ring andNits set of nilpotent elements; and callRreduced ifN= {0}. Following [4], callR anFZS-ring if every zero subring—that is, every subring with trivial multiplication—is finite. It was proved in [1] that every nilFZS-ring is finite—a result which in more transparent form is as follows.
Theorem1. Every infinite nil ring contains an infinite zero subring.
Later, in [4], it was shown that every ring withNinfinite contains an infinite zero subring. The proof relies onTheorem 1together with the following result.
Theorem 2(see [4]). If R is any semiprime FZS-ring, thenR=B⊕C, whereB is reduced andCis a direct sum of finitely many total matrix rings over finite fields.
Theorems1and2have had several applications in the study of commutativity and finiteness. Since the proofs in [1,4] are rather complicated, it is desirable to have new and simpler proofs; and in our first major section, we present such proofs. In our final section, we apply Theorems1and2in proving a new theorem on existence of infinite periodic subrings.
1. Preliminaries. LetZandZ+denote, respectively the ring of integers and the set of positive integers. For the ringR, denote by the symbolsT andP (R), respectively the ideal of torsion elements and the prime radical; and for eachn∈Z+, defineRnto be{x∈R|xn=0}. ForY an element or subset ofR, letYbe the subring generated byY; letAl(Y ),Ar(Y ), andA(Y )be the left, right, and two-sided annihilators ofY; and let CR(Y ) be the centralizer of Y. For x, y∈R, let[x, y] be the commutator xy−yx.
The subringSofRis said to be of finite index inRif(S,+)is of finite index in(R,+).
An elementx∈Ris called periodic if there exist distinct positive integersm,nsuch thatxm=xn; and the ringRis called periodic if each of its elements is periodic.
We will use without explicit mention two well-known facts:
(i) the intersection of finitely many subrings of finite index inRis a subring of finite index inR;
(ii) ifRis semiprime andIis an ideal ofR, thenR/A(I)is semiprime.
We will also need several lemmas.
Lemma 1.1is a theorem from [6];Lemma 1.2appears in [3], and with a different proof in [2];Lemma 1.3, also given without proof, is all but obvious.Lemma 1.6, which appears to be new, is the key to our proofs of Theorems1and2.
Lemma1.1. IfRis a ring andSis a subring of finite index inR, thenS contains an ideal ofRwhich is of finite index inR.
Lemma1.2. LetRbe a ring with the property that for eachx∈R, there existm∈Z+ andp(t)∈Z[t]such thatxm=xm+1p(x). ThenRis periodic.
Lemma1.3. IfRis any ring withN⊆TandHis any finite set of pairwise orthogonal elements ofN, thenHis finite.
Lemma1.4. IfRis any ring in whichR2is finite, thenRis of bounded index—that is, N=Rnfor somen∈Z+.
Proof. LetM= |R2|and letx∈Nsuch thatx2k=0 fork≥M+1; and note that xk, xk+1, . . . , x2k−1are all inR2. Sincek > M, these elements cannot be distinct; hence there existh, j∈Z+ such thath < j≤2k−1 andxh=xh+m(j−h)for allm∈Z+. It follows thatxh=0; hencey2M=0 for ally∈N.
Lemma1.5. IfRis any FZS-ring, thenN⊆T.
Proof. LetRbe a ring withN\T≠∅, and letx∈N\T. Then there exists a smallest n∈Z+such thatxn∈T, and there existsk∈Z+for whichkxn=0. Sincekxn−1∉T, kxn−1is an infinite zero subring ofR.
Lemma1.6. IfRis any FZS-ring andx is any element ofN, thenA(x)is of finite index inR. Hence, ifSis any finite subset ofN,A(S)is of finite index inR.
Proof. We use induction on the degree of nilpotence. Suppose first thaty2=0.
DefineΦ:Ry→Rbyr y[r y, y]= −yr y; and note thatΦ(Ry)is a zero subring ofR, hence finite. Thus kerΦ=Ry∩CR(y)is of finite index inRy. But it is easily seen that kerΦis a zero ring, hence is finite; consequently,Ryis finite. Now consider η:R→Ry defined byr r y, and note that kerη=Al(y)is of finite index inR.
Similarly,Ar(y)is of finite index and so isA(y)=Al(y)∩Ar(y).
Now assume thatA(x)is of finite index for allx∈N with degree of nilpotence less thank, and lety∈N be such thatyk=0. ThenA(y2)is of finite index inR.
DefineΦ:A(y2)y→R bysy[sy, y],s∈A(y2); and note that both Φ(A(y2)y) and kerΦ=A(y2)y∩CR(y)are zero rings, so thatA(y2)yis finite. Consider the map Ψ=A(y2)→A(y2)ygiven bys→sy. Now kerΨ=A(y2)∩Al(y)must be of finite index inA(y2); and sinceA(y2)is of finite index inR, kerΨis of finite index inR. It follows thatAl(y)is of finite index inR; and a similar argument shows thatAr(y)is of finite index inR. ThereforeA(y)is of finite index in R.
Lemma1.7. Letpbe a prime, and letRbe a ring such thatpR= {0}.
(i) If a∈R and apk =a, then apmk =a for allm∈Z+. Hence ifa, b∈R with apk=aandbpj=b, there existsn∈Z+such thatapn=aandbpn=b.
(ii) Ifa∈Randapk=a, then for eachs∈Z,(sa)pk=sa.
(iii) IfRis reduced andais a periodic element ofR, then there existsn∈Z+such thatapn=a.
Proof. (i) is almost obvious, and (ii) follows from the fact thatsp≡s (modp) for alls∈Z. To obtain (iii), note that if Ris reduced andais periodic, thenais finite, hence a direct sum of finite fields, necessarily of characteristicp. Since GF(pα) satisfies the identityxpα=x, the conclusion of (iii) follows by (i).
2. Proofs of Theorems1and2
Proof ofTheorem1. SupposeRis a counterexample. Note thatRis anFZS-ring, soR=TbyLemma 1.5. It is easy to see thatRcontains a maximal finite zero subring S. By Lemma 1.6, A(S) is infinite; and maximality ofS forcesA(S)2=S. Thus, by replacingRbyA(S), we may assume thatR2is finite.
ByLemma 1.6, we can construct infinite sequences of pairwise orthogonal elements;
and byLemma 1.4there is a smallestM∈Z+for whichRM contains such sequences.
Letu1, u2, . . .be an infinite sequence of pairwise orthogonal elements ofRM. Using Lemma 1.3, we can refine this sequence to obtain an infinite subsequencev1, v2, . . . such that for eachj≥2,vj∉v1, v2, . . . , vj−1. DefiningV0to be{vj2|j∈Z+}, we see thatV0⊆RM−1and henceV0is finite, so we may assume without loss of generality that there exists a singles∈Rsuch thatvj2=sfor allj∈Z+. Takem∈Z+such that ms=0; and for eachj∈Z+, definewj=mj
i=1vi. Then thewjform an infinite subset ofR2, contrary to the fact thatR2is finite. The proof is now complete.
Proof ofTheorem2. As before, sinceRis anFZS-ring, there is a maximal finite zero subringS; and byLemma 1.6A(S)is of finite index inR. ByLemma 1.1,A(S) contains an idealIofRwhich is also of finite index inR. LetC=A(I)and letB=A(C).
ThenB⊇I, soBis of finite index inR.
Next we show thatBis reduced. Letx∈B such thatx2=0. Thenx∈A(C); and sinceS⊆C, the maximality ofS forcesx∈B∩C= {0}. Therefore,Bis reduced.
The rest of the proof is as in [4]. SinceR/Bis finite and semiprime, we can write it as M1⊕···⊕Mk, where theMiare total matrix rings over finite fields. LetC=(B+C)/B and note thatCis an ideal ofR/BandCC. NowCmust be a direct sum of some of theMi, soR/B=C⊕DwhereDis the annihilator ofC. TakingDto be an ideal ofR containingBfor whichD/B=D, and noting thatCD= {0}, we haveCD⊆B.
ButCD⊆C as well, soCD⊆B∩C= {0}andD⊆A(C)=B; thereforeD= {0}and C=R/B. It follows thatR=B+Cand henceR=B⊕C; and sinceCC,Cis a direct sum of total matrix rings as required.
Remark2.1. In [5], Lanski established the conclusion ofTheorem 2under the ap- parently stronger hypothesis thatNis finite; and his proof uses induction on|N|. As we noted in the introduction, it follows from Theorems1and2thatRis anFZS-ring if and only ifNis finite.
3. A theorem on periodic subrings. We have noted that ifNis infinite,Rcontains an infinite nil subring. Since periodic elements extend the notion of nilpotent element,
it is natural to ask whether there is a periodic analogue—that is, to ask whether a ring with infinitely many periodic elements must have an infinite periodic subring. The answer in general is no, even in the case of commutative rings. The complex fieldC is a counterexample, for the set of nonzero periodic elements is the setU of roots of unity, andu∈Uimplies 2u∉U. Moreover, ifS is any finite ring,C⊕S is also a counterexample; therefore, we restrict our attention to ringsRfor whichR=T.
Theorem3.1. LetRbe a ring withR=T. Then a necessary and sufficient condition forRto have an infinite periodic subring is thatRcontains an infinite set of pairwise- commuting periodic elements.
Proof. It is known that in any infinite periodic ringR, eitherNis infinite or the centerZis infinite [4, Theorem 7]. Therefore our condition is necessary.
For sufficiency, suppose thatR has infinitely many pairwise-commuting periodic elements. Now R is the direct sum of itsp-primary components R(p); and if there exist infinitely many primesp1, p2, p3, . . .such thatR(pi)contains a nonzero periodic elementapi, then the direct sum of the ringsapiis an infinite periodic subring.
Thus, we may assume that only finitely manyR(p)contain nonzero periodic elements, so we need only consider the case thatR=R(p)for some primep. Of course we may assume thatRis anFZS-ring.
Consider the factor ring ¯R = R/P (R). Since R is an FZS-ring, it follows from Theorem 1thatP (R)is finite, in which case ¯R inherits our hypothesis on pairwise- commuting periodic elements. If ¯R has an infinite periodic subring ¯S and S is its preimage inR, then for allx∈S, there exist distinctm, n∈Z+such thatxn−xm∈ P (R)⊆N; henceSis periodic byLemma 1.2. Thus, we may assume thatR=R(p)and thatRis a semiprimeFZS-ring.
ByTheorem 2, writeR=B⊕C, whereBis reduced andC is finite; and note thatB must have an infinite subsetHof pairwise-commuting periodic elements. Note also thatpB= {0}, sinceBis reduced. Leta, b∈H, and byLemma 1.7(i) and (iii) obtain n∈Z+such thatapn=aandbpn=b. It follows at once that(a−b)pn=apn−bpn= a−band(ab)pn=apnbpn=ab; and these facts, together withLemma 1.7(ii) imply thatHis an infinite periodic subring ofR.
Acknowledgement. This research was supported by the Natural Sciences and Engineering Research Council of Canada, Grant No. 3961.
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Howard E. Bell: Department of Mathematics, Brock University, St. Catharines, Ontario, CanadaL2S 3A1
E-mail address:[email protected]
