A characterization of potent rings

Abstract An associative ring R is called potent provided that for every 
$x\in R$
 , there is an integer 
$n(x)>1$
 such that 
$x^{n(x)}=x$
 . A celebrated result of N. Jacobson is that every potent ring is commutative. In this note, we show that a ring R is potent if and only if every nonzero subring S of R contains a nonzero idempotent. We use this result to give a generalization of a recent result of Anderson and Danchev for reduced rings, which in turn generalizes Jacobson’s theorem.


Introduction
Boolean algebras play a significant role not only in mathematics but are also important tools in logic and computer science. Moreover, there is a natural correspondence between Boolean algebras and so-called Boolean rings: from a Boolean algebra, one may canonically construct a Boolean ring and vice versa (for details and an introduction to the subject, see [4]). We recall that an associative ring R is Boolean provided every member of R is idempotent, that is, x 2 = x for every x ∈ R. It is well-known (and an exercise in many undergraduate algebra texts) that every Boolean ring is commutative (see [5], for example). In fact, if one replaces the exponent 2 in the previous equation with 3, R is still commutative. Indeed, one can replace 2 with any integer greater than 1, and commutativity is guaranteed. Continuing to generalize, if one assumes only the existence of some such n > 1 for every x ∈ R (which may depend on x), R must be commutative. This beautiful result is due to Nathan Jacobson and generalizes Wedderburn's theorem that every finite division ring is a field ( [13]). Proposition 1 (Jacobson's theorem [8]). Let R be an associative ring and suppose that for every x ∈ R, there is an integer n(x) > 1 such that x n(x) = x (that is, R is a potent ring). Then, R is commutative.
We refer the reader to [2,3], and [6] for further reading, and to [1] for a recent generalization of Jacobson's theorem.
Changing gears temporarily, the theory of idempotents contributes significantly to the theory of both commutative and non-commutative rings. Indeed, the number of different kinds of idempotents defined in the literature (orthogonal, central, primitive, local, irreducible, etc.) as well as rings defined in terms of idempotents (Baer, semisimple, von Neumann regular, Zorn, Rickart, etc.) indicates the utility of this notion in the structure theory of rings (see [11] for details). The purpose of this short note is to identify potency with a natural condition on the existence of idempotents. Specifically, we prove that an associative ring R is potent if and only if every nonzero subring of R contains a nonzero idempotent. We mention the classic texts [5] and [7] as references for some standard algebraic results utilized in the paper such as the Chinese remainder theorem, the division algorithm for polynomials over a field, and Lagrange's theorem. C The Author(s), 2022. Published by Cambridge University Press on behalf of Glasgow Mathematical Journal Trust. This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0/), which permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited.
Throughout, a ring is assumed only to be associative and not to be commutative nor to contain an identity unless specified; subrings are also not assumed unital, even in rings with identity.

Main result
Before presenting the main result of this paper, we prove the following lemma. For brevity, let us agree to call a ring with the property that every nonzero subring contains a nonzero idempotent idempotent-rich. 1 Lemma 1. The following hold: (1) Every subring of an idempotent-rich ring is idempotent-rich.
If R is idempotent-rich and e ∈ R is idempotent, then the additive order of e is a square-free integer.
Proof. We establish each in succession.
(2) Suppose by way of contradiction that R is idempotent-rich but not reduced. Then, there is some nonzero α ∈ R such that α 2 = 0. But now Zα is a nonzero subring of R with no nonzero idempotent, a contradiction. (3) Let R be a nonzero reduced ring. We claim XR[X] has no nonzero idempotents. Indeed, consider an arbitrary f := a 1 X + · · · + a n X n ∈ XR[X], where a n = 0. Then because R is reduced, it is easy to see that f 2 has degree 2n and thus f 2 = f . (4) Let R be idempotent-rich and let e ∈ R be an idempotent. Then Ze is a subring of R and, moreover, Ze ∼ = Z/ n for some non-negative integer n. If n = 0, then by (1), we see that Z is idempotent-rich, but this is absurd as the subring 2Z of Z has no nonzero idempotent. Thus, n > 0 (and hence n = ord(e)). Suppose that n is not square-free, and write n = m 2 q, where m,q are integers and m > 1. Then, mq (mod n) is a nonzero nilpotent element of Z/ n , a contradiction to (1) and (2) above.
We are now ready to prove the main result of this note.

Theorem 1. Let R be a ring. Then, R is potent if and only if every nonzero subring of R contains a nonzero idempotent.
Proof. Let R be a ring. Suppose first that R is potent. We will show that R is idempotent-rich. Indeed, suppose that S is a nonzero subring of R, and let s ∈ S be nonzero. By our assumption, there is an integer n > 1 such that s n = s. Then, s n (s n−2 ) = s(s n−2 ), that is, (s n−1 ) 2 = s 2n−2 = s n−1 . As s n = s and s = 0, clearly s n−1 = 0; since n > 1, s n−1 ∈ S. Hence, S contains a nonzero idempotent.
Conversely, suppose that every nonzero subring of R contains a nonzero idempotent. We shall prove that R is potent. Suppose that α ∈ R is nonzero. We will show that the ring αZ[α] := {αf (α) : f (X) ∈ Z[X]} is a finite product of finite fields.
( Next, fix k with 1 ≤ k ≤ n, and set I := I k . Then, I is a nonzero subring of αZ[α] and thus contains a nonzero idempotent e. Let J := {ie − i : i ∈ I}. Observe that J is a subideal of I and that Ie + J = I. We claim that the sum is direct: suppose that xe = ye − y for some x, y ∈ I. Multiplying through by e and using the fact that e is idempotent, we see that xe = ye − ye = 0, proving the directness of the sum. Because e is a nonzero element of Ie, and since I is indecomposable, we have Ie = I. By (4) of Lemma 1, the additive order of e is a square-free integer. As αZ[α] is commutative, this clearly implies that the additive order of every member of Ie = I is also a square-free integer. Since k was arbitrary, it follows easily from (2.2) above that the order of every member of αZ[α] is a square-free integer; in particular, the order of α is a square-free integer m. From this fact, we deduce that 2 the additive order of every element of αZ[α] is a factor of m. (2. 3) The map a 1 X + · · · + a n X n → a 1 α + · · · + a n α n is now a well-defined ring surjection of XZ/ m [X] onto αZ [α]. Thus, (
The following equivalent formulation of Jacobson's theorem is immediate.
Corollary 1 (Jacobson's Theorem, Alternative Form). Let R be a ring. If every nonzero subring of R contains a nonzero idempotent of R, then R is commutative.
In [1], the authors show that if R is a ring with identity such that for every x ∈ R, there are positive integers m(x) and n(x) of different parity such that x m(x) = x n(x) , then R potent, and hence commutative. Note that the assumption that R is unital cannot be dropped completely, since every nil ring trivially has this property, and there are noncommutative nil rings. However, if R has no nonzero nilpotent elements, then we have the following stronger result.

Corollary 2.
Let R be a reduced ring. Suppose that for every nonzero subring S of R, there exists a nonzero x ∈ S and distinct positive integers m(x) and n(x) such that x m(x) = x n(x) . Then, R is commutative.
Proof. Suppose that R is reduced with the above property. It suffices to prove that every nonzero subring of R contains a nonzero idempotent. Thus, let S be a nonzero subring and let x ∈ S be nonzero and such that x m = x n for some distinct positive integers m and n. Without loss of generality, we may assume that m > n. Let l := m − n. Then, it follows easily by induction that for every positive integer r, we have x m+rl = x n . Thus, we may assume without loss of generality that m > 2n. Now set k := m − 2n. Then, observe that (x n+k ) 2 = x 2(n+k) = x m+k = x n+k . As R is reduced, x n+k is a nonzero idempotent of S, and the conclusion follows.
Conflicts of interest. The author declares none.