#### Abstract

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

#### 1. Introduction

Let us recall a commutativity result of Herstein; here and later, denotes the (additive) *commutator* of the pair .

Theorem A (see Herstein [1]). *A ring is commutative if and only if for each there exists an integer such that .*

We have the following corresponding result for *anticommutators *. A ring is said to be *anticommutative* if for all .

Theorem B (see MacHale [2]). *A ring is anticommutative if and only if for each there exists an even integer such that . *

The restriction to even integers above is necessary: fails to be anticommutative even though it satisfies the identity for every odd .

The proofs of the above pair of results depend on Jacobson’s structure theory of rings. In the case of the much stronger identity for some , 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 , fixed, elementary commutativity proofs were given by Morita [4] for all odd , and all even , and by MacHale [5] for all even numbers that can be written as sums or differences of two powers of , but are not themselves powers of . 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 call a CP* ring*, where if, for each , there exists such that ; CP stands for commutator power. Similarly we call an ACP* ring*, where if, for each , there exists such that . We write and instead of and , respectively. A CP *ring* and an ACP *ring* mean a ring and an ring, respectively.

By the above theorems, all CP rings are commutative and all ACP rings are anticommutative. An elementary proof of the commutativity of rings was given for in [7], and for 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 is necessarily commutative if its centre is an ideal. In Section 2, we prove that rings are commutative and also give a new proof that rings are commutative; both proofs are elementary.

As for rings, an elementary proof of their anticommutativity was given in [2] for , 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 , the commutativity of rings is equivalent to the anticommutativity of rings. Thus the aforementioned proof of commutativity of rings yields an elementary proof of anticommutativity of rings, and the proof that a ring is necessarily commutative if its centre is an ideal yields an elementary proof of anticommutativity of an ring whose centre is an ideal.

#### 2. Commutators

Suppose that is a ring. We gave an elementary proof in [8, Theorem 17] that rings are commutative. Here we give a somewhat similar (but surprisingly much easier!) proof that rings are commutative. We then give a new proof that 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 that is a CP ring. Then
*

*Proof. *Let be such that for some . Then
where (if ) or (if ).

Lemma 2.2. *Suppose a ring satisfies (2.1). Then idempotents are central. Consequently, whenever satisfies for some . *

*Proof. *Let be an idempotent, let , and let . Clearly , and so . By symmetry, .

The last statement follows from the first since if , then

We now recall the following elementary result [8, Theorem 19].

Lemma C. *If and commute for all in some ring , then for all . In such a ring , a commutator satisfies an equation of the form for some if and only if . *

The commutativity of rings is now easily established.

Theorem 2.3. * rings are commutative. *

*Proof. *Let be . In view of Lemma C, it suffices to prove that whenever and , for some . Since , we have , so . It follows that is an idempotent, and so by Lemma 2.2. Since is a commutator, it follows that . In particular , and so . Since , it follows that as required.

We now give a new proof of the commutativity of 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. * rings are commutative. *

*Proof. *Consider for fixed , so that is a central idempotent. Let , , and let be the subring of generated by and . It is also clear that and, since for , is a unity for . Note that there is a symmetry between the three elements , , and : is also generated by and , and . Since is a unity, in , and any equation of the form in implies that .

We will use the fact that the following are commutators lying in : , , , , and . Thus we can apply the identity to these, and their squares are central idempotents. In particular, note that
Thus and , so commutes with (and trivially with ).

Now note that if any commutes with and , then is a commutator so
Thus , and is a central idempotent (in ). Now also commutes with and and , so . Thus , and . Also, since is a central idempotent,
Applying this to , we get that , so , and so . Thus .

Now using (2.6) and the property for the commutators , , and , we see that
But , so . Thus commutes with and , so, by our earlier analysis for , we see that .

Note that is also generated by and , since and . Thus we may replace by in the last paragraph and deduce that . Expanding this and using the fact that , we deduce that . By symmetry, both and also lie in , so
Thus , so (since is generated by and ). But , so and , as required.

#### 3. Anticommutators

In this section, we give an elementary proof of the following result.

Theorem 3.1. *Suppose that . Then all rings are commutative if and only if all rings are anticommutative. *

In view of Theorem 2.3, we thus have an elementary proof of the following result.

Corollary 3.2. * rings are anticommutative. *

Note that Theorem 3.1 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 Theorem 3.1, 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 ring satisfies (2.1); that is, whenever , . *

*Observation 1. *If and , for some , then and , where . Moreover if are both even, then so is .

We now state a simple but useful lemma.

Lemma 3.4. *Suppose that is an ACP ring, Then *(a)* for all . *(b)* for all . If instead is a ring then *(c)* for all .*

* Proof. *Both and are anticommutators. Using Observation 1 and the ACP condition, we see that that there is an even integer such that , which gives (a). We omit the proof of (c), which is very similar to that of (a).

As for (b), first let in (a) to get for all . Next use the ACP condition to deduce that there exists some even integer such that . But because of the identity . Thus , as required.

We now prove an anticommutator analogue of Lemma C.

Lemma 3.5. *Suppose that and commute for all , , in some ring . Then for all . If is additionally assumed to be an ACP ring, then is anticommutative. *

*Proof. *Suppose for some , and we also write . Note that and , so by assumption and commute. Similarly commutes with . Thus 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 . Since and are anticommutators, it follows that
Subtracting the extreme right hand side from the extreme left, we get that , as required.

By distributivity, we have . Assume now that is an ACP ring. Using Lemma 3.4(b), we see that and , and so . Thus . Now because .

*Proof of Theorem 3.1. * We suppose first that is an ring for some fixed , and we suppose that every ring is commutative. By Lemma 3.5, it suffices to prove that , where , , and are fixed but arbitrary.

Now there are even integers such that and . Since idempotents are central, it follows that , , and are all central idempotents. Thus if we write and , then , , and
Similarly . Thus if and only if and, since , it suffices to show that is anticommutative. By Lemma 3.4, so has characteristic and is an ring. But for rings of characteristic , the condition is equivalent to the condition, and so by assumption is commutative, and commutativity is also equivalent to anticommutativity in rings of characteristic , so we are done.

The converse direction is very similar. Assume that rings are anticommutative, where is fixed. Suppose is a ring. We first reduce to proving that , where , , and are fixed but arbitrary. Writing , where and , we see as before that and , so the required result follows from the commutativity of . But has characteristic , so we can finish the result as before.