# Powers of Commutators and Anticommutators

Authors: Stephen M. Buckley; Desmond MacHale
Journal: ISRN Algebra
 (2012)

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## Abstract

Forn=3,  4, we give elementary proofs of commutativity of rings in which the identity cn=c holds for all commutators c. For even n, we show that the commutativity of rings satisfying such an identity is equivalent to the anticommutativity of rings satisfying the corresponding anticommutator equation.


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## Body

## 1. Introduction

Let us recall a commutativity result of Herstein; here and later,[x,y]:=xy-yx denotes the (additive) commutator of the pair x,y.Theorem A (see Herstein [1]).
A ringR is commutative if and only if for each x,y∈R there exists an integer n(x,y)>1 such that [x,y]n(x,y)=[x,y].We have the following corresponding result foranticommutators〈x,y〉:=xy+yx. A ring is said to be anticommutative if 〈x,y〉=0 for all x,y.Theorem B (see MacHale [2]).
A ringR is anticommutative if and only if for each x,y∈R there exists an even integer n(x,y)>1 such that 〈x,y〉n(x,y)=〈x,y〉.The restriction to even integers above is necessary:ℤ3 fails to be anticommutative even though it satisfies the identity 〈x,y〉n=〈x,y〉 for every odd n.The proofs of the above pair of results depend on Jacobson’s structure theory of rings. In the case of the much stronger identityxn(x)=x for some n(x)>1, which was proved to imply commutativity by Jacobson [3], there are elementary proofs (meaning proofs that do not require structure theory) of many special cases of the result. For instance, in the case of the identity xn=x, n>1 fixed, elementary commutativity proofs were given by Morita [4] for all odd n≤25, and all even n≤50, and by MacHale [5] for all even numbers n that can be written as sums or differences of two powers of 2, but are not themselves powers of 2. Also notable is the proof by Wamsley [6] of Jacobson’s result which uses only a weak form of structure theory, specifically the fact that a finite commutative ring can be written as a direct sum of fields.By comparison, there are far fewer elementary proofs in the literature of special cases of Theorems A and B, and they appear to be more difficult to construct. This is perhaps not surprising, since the sets of commutators or anticommutators do not in general form even additive subgroups of a ring.To aid our subsequent discussion, let us callR a CP(S)ring, where S⊂ℕ∖{1} if, for each x,y∈R, there exists n(x,y)∈S such that [x,y]n(x,y)=[x,y]; CP stands for commutator power. Similarly we call R an ACP(S)ring, where S⊂2ℕ if, for each x,y∈R, there exists n(x,y)∈S such that 〈x,y〉n(x,y)=〈x,y〉. We write CP(n) and ACP(n) instead of CP({n}) and ACP({n}), respectively. A CP ring and an ACP ring mean a CP(ℕ∖{1}) ring and an ACP(2ℕ) ring, respectively.By the above theorems, all CP rings are commutative and all ACP rings are anticommutative. An elementary proof of the commutativity ofCP(n) rings was given for n=2 in [7], and for n=3 in [8]. The only other elementary proof of a special case of Theorem A of which we are aware is the proof in [9] that a CP ring R is necessarily commutative if its centre is an ideal. In Section 2, we prove that CP(4) rings are commutative and also give a new proof that CP(3) rings are commutative; both proofs are elementary.As forACP(n) rings, an elementary proof of their anticommutativity was given in [2] for n=2, but we are not aware of any other such proofs in the literature. In Section 3, we give an elementary proof that for any given S⊂2ℕ, the commutativity of CP(S) rings is equivalent to the anticommutativity of ACP(S) rings. Thus the aforementioned proof of commutativity of CP(4) rings yields an elementary proof of anticommutativity of ACP(4) rings, and the proof that a CP(2ℕ) ring is necessarily commutative if its centre is an ideal yields an elementary proof of anticommutativity of an ACP(2ℕ) ring whose centre is an ideal.


## 2. Commutators

Suppose thatR is a ring. We gave an elementary proof in [8, Theorem 17] that CP(3) rings are commutative. Here we give a somewhat similar (but surprisingly much easier!) proof that CP(4) rings are commutative. We then give a new proof that CP(3) rings are commutative. We begin with two well-known lemmas that give useful information about all CP rings and include the short proofs for convenience.Lemma 2.1.
Suppose thatR is a CP ring. Then
(2.1)xy=0⇒yx=0,x,y∈R.Proof.
Letn>1 be such that (yx-xy)n=yx-xy for some n>1. Then
(2.2)yx=(yx)n=y(xy)u=0,
where u=x (if n=2) or u=x(yx)n-2 (if n>2).Lemma 2.2.
Suppose a ringR satisfies (2.1). Then idempotents are central. Consequently, xm-1∈Z(R) whenever x∈R satisfies xm=x for some m>1.Proof.
Lete be an idempotent, let y∈R, and let d=ey-eye. Clearly de=0, and so ed=ey-eye=0. By symmetry, ye=eye=ey.
The last statement follows from the first since ifm>2, then
(2.3)(xm-1)2=xmxm-2=xxm-2=xm-1.We now recall the following elementary result [8, Theorem 19].Lemma C.
If[x,z] and [y,z] commute for all x,y,z in some ring R, then [x,y]4=0 for all x,y∈R. In such a ring R, a commutator c satisfies an equation of the form cn=c for some n>1 if and only if c=0.The commutativity ofCP(4) rings is now easily established.Theorem 2.3.
CP
(
4
) rings are commutative.Proof.
LetR be CP(4). In view of Lemma C, it suffices to prove that cd=dc whenever c=[x,z] and d=[y,z], for some x,y,z∈R. Since -c=[-x,y], we have c=c4=(-c)4=-c, so 2c=0. It follows that f(c):=c+c2 is an idempotent, and so f(c)∈Z(R) by Lemma 2.2. Since c+d=[x+y,z] is a commutator, it follows that f(c+d)-f(c)-f(d)=cd+dc∈Z(R). In particular c(cd+dc)=(cd+dc)c, and so c2d=dc2. Since c+c2∈Z(R), it follows that cd=dc as required.We now give a new proof of the commutativity ofCP(3) rings. Like the original proof in [8], this proof uses our two lemmas above. However it avoids the reduction of Lemma C.Theorem 2.4.
CP
(
3
) rings are commutative.Proof.
Considerc=[x,y] for fixed x,y∈R, so that e:=c2 is a central idempotent. Let a=ex, b=ey, and let S be the subring of R generated by a and b. It is also clear that c=[a,b] and, since ex=xe=x for x∈{a,b}, e is a unity for S. Note that there is a symmetry between the three elements a, b, and a+b: S is also generated by a+b and b, and c=[a+b,b]. Since e is a unity, c=c-1 in S, and any equation of the form uc=vc in S implies that u=v.
We will use the fact that the following are commutators lying inS: ca=[a,ba], ac=[a,ab], w:=[a,c], cw=[ca,c]=cac-a, and wc=[ac,c]=a-cac=-cw. Thus we can apply the CP(3) identity to these, and their squares are central idempotents. In particular, note that
(2.4)wc=wc(wc)2=w(wc)2c=w2cw=-w3c=-wc.
Thus 2wc=0 and 2w=0, so 2a commutes with c (and trivially with a).
Now note that if anyd∈S commutes with a and c, then dc=[a,db] is a commutator so
(2.5)dc=(dc)3=d3c3=d3c.
Thus d=d3, and d2 is a central idempotent (in R). Now also s=d+d2 commutes with a and c and ds=d2+d3=s, so s2=2s. Thus s=s3=2s2∈Z(R), and d=s-d2∈Z(R). Also, since (ac)2 is a central idempotent,
(2.6)(ac)2=c(ac)2c-1=(ca)2.
Applying this to d=2a, we get that 2a∈Z(R), so 2c=[2a,b]=0, and so 2e=(2c)(c)=0. Thus 2S=0.
Now using (2.6) and the CP(3) property for the commutators ac, ca, and c, we see that
(2.7)w=w3=(ac)3-(ac)2(ca)-(ac)(ca)(ac)-(ca)(ac)2+(ac)(ca)2+(ca)(ac)(ca)+(ca)2(ac)-(ca)3=ac-ca-a3c-ca+ac+ca3+ac-ca=3w+ca3-ac3.
But 2w=0, so ca3=a3c. Thus a3∈S commutes with a and c, so, by our earlier analysis for d=a3, we see that a3∈Z(R).
Note thatS is also generated by e+a and b, since [e+a,b]=c and e+a-c2=a. Thus we may replace a by e+a in the last paragraph and deduce that (e+a)3∈Z(R). Expanding this and using the fact that 2S=0, we deduce that a+a2∈Z(R). By symmetry, both b+b2 and (a+b)+(a+b)2 also lie in Z(R), so
(2.8)ab+ba=(a+b)+(a+b)2-(a+a2)-(b+b2)∈Z(R).
Thus a(ab+ba)=(ab+ba)a, so a2∈Z(S) (since S is generated by a and b). But a+a2∈Z(S), so a∈Z(S) and c=[a,b]=0, as required.


## 3. Anticommutators

In this section, we give an elementary proof of the following result.Theorem 3.1.
Suppose thatS⊂2ℕ. Then all CP(S) rings are commutative if and only if all ACP(S) rings are anticommutative.In view of Theorem2.3, we thus have an elementary proof of the following result.Corollary 3.2.
ACP
(
4
) rings are anticommutative.Note that Theorem3.1 allows us to deduce Theorem B from Theorem A. Of course this is not an elementary proof, since Herstein’s result relies on structure theory, but it is an alternative to the proof in [2] which appeals both to Herstein’s result and to structure theory.To prove Theorem3.1, we first need some preparatory lemmas. We omit the simple proof of our first lemma, which follows just like that of Lemma 2.1, and which implies that idempotents are central in ACP rings.Lemma 3.3.
Every ACP ringR satisfies (2.1); that is, yx=0 whenever xy=0, x,y∈R.Observation 1.
Ifam=a and bn=b, for some m,n∈ℕ∖1, then at=a and bt=b, where t=(m-1)(n-1)+1. Moreover if m,n are both even, then so is t.We now state a simple but useful lemma.Lemma 3.4.
Suppose thatR is an ACP ring, Then (a)
2〈x,y〉=0 for all x,y∈R.(b)
2x2=0 for all x∈R. If instead R is a CP(2ℕ) ring then(c)
2[x,y]=0 for all x,y∈R.Proof.
Botha:=〈x,y〉 and -a=〈-x,y〉 are anticommutators. Using Observation 1 and the ACP condition, we see that that there is an even integer t>0 such that a=at=(-a)t=-a, which gives (a). We omit the proof of (c), which is very similar to that of (a).
As for (b), first letx=y in (a) to get 4y2=0 for all y∈R. Next use the ACP condition to deduce that there exists some even integer n>0 such that (2x2)n=〈x,x〉n=2x2. But (2x2)n=2n(xn)2=0 because of the identity 4y2=0. Thus 2x2=0, as required.We now prove an anticommutator analogue of Lemma C.Lemma 3.5.
Suppose that〈x,z〉 and 〈y,z〉 commute for all x, y, z in some ring R. Then 〈x,y〉2[x,y]=0 for all x,y∈R. If R is additionally assumed to be an ACP ring, then R is anticommutative.Proof.
Supposea=〈x,y〉 for some x,y∈R, and we also write c=[x,y]. Note that ax=〈x,yx〉=〈yx,x〉 and a=〈y,x〉, so by assumption ax and a commute. Similarly a commutes with xa=〈xy,x〉. Thus a(ax)=a(xa)=xa2 where in each case the parentheses enclose one of the factors that is commuted in the next equation. Thus squares of anticommutators are central and moreover we have a2x=axa=xa2. Since ax and ya are anticommutators, it follows that
(3.1)(a2)(xy)=(ax)(ya)=(yaa)x=a2yx.
Subtracting the extreme right hand side from the extreme left, we get that a2c=0, as required.
By distributivity, we haveac=xyxy-yxyx+yxxy-xyyx. Assume now that R is an ACP ring. Using Lemma 3.4(b), we see that -yxyx=yxyx and -xyyx=xyyx, and so a2=ac. Thus a3=a2c=0. Now a=an(x,y)=a2n(x,y)-1=0 because 2n(x,y)-1≥3.Proof of Theorem3.1.
We suppose first thatR is an ACP(S) ring for some fixed S⊂2ℕ, and we suppose that every CP(S) ring is commutative. By Lemma 3.5, it suffices to prove that ab=ba, where a=〈x,z〉, b=〈y,z〉, and x,y,z∈R are fixed but arbitrary.
Now there are even integersm,n∈ℕ such that am=a and bn=b. Since idempotents are central, it follows that am-1, bn-1, and e:=am-1bn-1 are all central idempotents. Thus if we write a′=〈ex,ez〉 and b′:=〈ey,ez〉, then a′=e2a=ea, b′=e2b=eb, and
(3.2)a′b′=(e2)(e2)ab=eab=ambn=ab.
Similarly b′a′=ba. Thus ab+ba=0 if and only if a′b′+b′a′=0 and, since a′b′∈eR=Re, it suffices to show that eR is anticommutative. By Lemma 3.4, 2e=0 so eR has characteristic 2 and is an ACP(S) ring. But for rings of characteristic 2, the ACP(S) condition is equivalent to the CP(S) condition, and so by assumption eR is commutative, and commutativity is also equivalent to anticommutativity in rings of characteristic 2, so we are done.
The converse direction is very similar. Assume thatACP(S) rings are anticommutative, where S⊂2ℕ is fixed. Suppose R is a CP(S) ring. We first reduce to proving that cd=dc, where c=[x,z], d=[y,z], and x,y,z∈R are fixed but arbitrary. Writing e=cm-1dn-1, where cm=c and dn=d, we see as before that [ex,ez][ey,ez]=cd and [ey,ez][ex,ez]=dc, so the required result follows from the commutativity of eR. But R has characteristic 2, so we can finish the result as before.



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