2. 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.Footnote ¹

Proof. We establish each in succession.

1. (1) Trivial.

2. (2) Suppose by way of contradiction that R is idempotent-rich but not reduced. Then, there is some nonzero $\alpha\in R$ such that $\alpha^2=0$ . But now $\mathbb{Z}\alpha$ is a nonzero subring of R with no nonzero idempotent, a contradiction.

3. (3) Let R be a nonzero reduced ring. We claim XR[X] has no nonzero idempotents. Indeed, consider an arbitrary $f\;:\!=\;a_1X+\cdots+a_nX^n\in XR[X]$ , where $a_n\neq 0$ . Then because R is reduced, it is easy to see that $f^2$ has degree 2n and thus $f^2\neq f$ .

4. (4) Let R be idempotent-rich and let $e\in R$ be an idempotent. Then $\mathbb{Z}e$ is a subring of R and, moreover, $\mathbb{Z}e\cong\mathbb{Z}/\langle n\rangle$ for some non-negative integer n. If $n=0$ , then by (1), we see that $\mathbb{Z}$ is idempotent-rich, but this is absurd as the subring $2\mathbb{Z}$ of $\mathbb{Z}$ has no nonzero idempotent. Thus, $n>0$ (and hence $n=\mathrm{ord}(e)$ ). Suppose that n is not square-free, and write $n=m^2q$ , where m,q are integers and $m>1$ . Then, mq (mod n) is a nonzero nilpotent element of $\mathbb{Z}/\langle n\rangle$ , a contradiction to (1) and (2) above.

We are now ready to prove the main result of this note.

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 $s\in S$ 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 $s\neq 0$ , clearly $s^{n-1}\neq 0$ ; since $n>1$ , $s^{n-1}\in S$ . 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 $\alpha\in R$ is nonzero. We will show that

(2.1) \begin{equation}\textrm{the ring} \ \alpha\mathbb{Z}[\alpha]\;:\!=\;\{\alpha f(\alpha)\colon f(X)\in\mathbb{Z}[X]\} \ \textrm{is a finite product of finite fields}.\end{equation}

We first verify that the ring $X\mathbb{Z}[X]$ is Noetherian (that is, every ascending chain of ideals stabilizes). Indeed, notice that every ideal of $X\mathbb{Z}[X]$ is also an ideal of $\mathbb{Z}[X]$ . As $\mathbb{Z}[X]$ is Noetherian, $X\mathbb{Z}[X]$ is as well. Next, since $\alpha\mathbb{Z}[\alpha]$ is a homomorphic image of $X\mathbb{Z}[X]$ , we conclude that $\alpha\mathbb{Z}[\alpha]$ is a Noetherian ring. It now follows immediately that $\alpha\mathbb{Z}[\alpha]$ does not contain an infinite internal direct sum of nonzero ideals. Therefore (see Lemma 4 of [Reference Oman and Stroud12]),

(2.2) \begin{equation}\alpha\mathbb{Z}[\alpha]=I_1\oplus I_2\cdots\oplus I_n \ \textrm{for some nonzero indecomposable ideals} \ I_1,\ldots,I_n \ \textrm{of} \ \alpha\mathbb{Z}[\alpha].\end{equation}

Next, fix k with $1\leq k\leq n$ , and set $I\;:\!=\;I_k$ . Then, I is a nonzero subring of $\alpha\mathbb{Z}[\alpha]$ and thus contains a nonzero idempotent e. Let $J\;:\!=\;\{ie-i\colon i\in I\}$ . 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 $x,y\in I$ . 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 $\alpha\mathbb{Z}[\alpha]$ 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 $\alpha\mathbb{Z}[\alpha]$ is a square-free integer; in particular, the order of $\alpha$ is a square-free integer m. From this fact, we deduce thatFootnote ²

(2.3) \begin{equation}\textrm{the additive order of every element of} \ \alpha\mathbb{Z}[\alpha] \ \textrm{is a factor of} \ m.\end{equation}

The map $\overline{a}_1X+\cdots+\overline{a}_nX^n\to a_1\alpha+\cdots+a_n\alpha^n$ is now a well-defined ring surjection of $X\mathbb{Z}/\langle m\rangle[X]$ onto $\alpha\mathbb{Z}[\alpha]$ . Thus,

(2.4) \begin{equation}\alpha\mathbb{Z}[\alpha]\cong X\mathbb{Z}/\langle m\rangle[X]/K \ \textrm{for some ideal} \ K \ \mathrm{of} \ X\mathbb{Z}/\langle m\rangle[X].\end{equation}

Write $m=p_1p_2\cdots p_n$ , where the $p_i$ are distinct primes (recall above that $\alpha\neq 0$ , and hence $m>1$ ). As is well-known, a simple application of The Chinese Remainder Theorem yields that, as rings, $\mathbb{Z}/\langle m\rangle\cong\mathbb{Z}/\langle p_1\rangle\oplus\cdots\oplus\mathbb{Z}/\langle p_n\rangle$ . We conclude that $X\mathbb{Z}/\langle m\rangle[X]\cong X\mathbb{Z}/\langle p_1\rangle[X]\oplus\cdots\oplus X\mathbb{Z}/\langle p_n\rangle[X]$ . We may now view K as an ideal of $X\mathbb{Z}/\langle p_1\rangle[X]\oplus\cdots\oplus X\mathbb{Z}/\langle p_n\rangle[X]$ . Via this identification,

(2.5) \begin{equation}\alpha\mathbb{Z}[\alpha]\cong(X\mathbb{Z}/\langle p_1\rangle[X]\oplus\cdots\oplus X\mathbb{Z}/\langle p_n\rangle[X])/K.\end{equation}

Next, for $1\leq i\leq n$ , let $I_i\;:\!=\;K\cap X\mathbb{Z}/\langle p_i\rangle[X]$ . We claim that $I_i$ is nontrivial. If $I_i$ is trivial, then the mapping $y\mapsto K+y$ is a ring isomorphism of $X_i\mathbb{Z}/\langle p_i\rangle[X]$ into $(X\mathbb{Z}/\langle p_1\rangle[X]\oplus\cdots\oplus X\mathbb{Z}/\langle p_n\rangle[X])/K\cong\alpha\mathbb{Z}[\alpha]$ . But then by (1) of Lemma 1, $X\mathbb{Z}/\langle p_i\rangle[X]$ is idempotent-rich, contradicting (3) of Lemma 1. So we see that each $I_i$ is a nonzero ideal of $X\mathbb{Z}/\langle p_i\rangle[X]$ . But it is easy to see that $I_i$ is also a nonzero ideal of the PID $\mathbb{Z}/\langle p_i\rangle[X]$ . The Division Algorithm for polynomials over a field shows that $\mathbb{Z}/\langle p_i\rangle[X]/I_i$ is finite and thus also

(2.6) \begin{equation}\textrm{each} \ X\mathbb{Z}/\langle p_i\rangle[X]/I_i \ \textrm{is finite}.\end{equation}

Now observe that

(2.7) \begin{equation}(X\mathbb{Z}/\langle p_1\rangle[X]\oplus\cdots\oplus X\mathbb{Z}/\langle p_n\rangle[X])/(I_1\oplus\cdots\oplus I_n)\cong X\mathbb{Z}/\langle p_1\rangle[X]/I_1\oplus\cdots\oplus X\mathbb{Z}/\langle p_n\rangle[X]/I_n.\end{equation}

Invoking (2.6), it follows that $(X\mathbb{Z}/\langle p_1\rangle[X]\oplus\cdots\oplus X\mathbb{Z}/\langle p_n\rangle[X])/(I_1\oplus\cdots\oplus I_n)$ is finite. Since $I_1\oplus\cdots\oplus I_n\subseteq K$ ,Footnote ³ it is immediate that

(2.8) \begin{equation}\alpha\mathbb{Z}[\alpha]\cong(X\mathbb{Z}/\langle p_1\rangle[X]\oplus\cdots\oplus X\mathbb{Z}/\langle p_n\rangle[X])/K \ \textrm{is finite}.\end{equation}

Thus, finally, we see that $\alpha\mathbb{Z}[\alpha]$ is a finite, reduced commutative ring. As is well-known (see p. 22 of [Reference Jans9], for example), this implies that $\alpha\mathbb{Z}[\alpha]$ has an identity. By the Chinese Remainder Theorem, $\alpha\mathbb{Z}[\alpha]=F_1\oplus\cdots\oplus F_k$ for some finite fields $F_1,\ldots, F_k$ , establishing (2.1). We claim that

(2.9) \begin{equation}\alpha^{(|F_1|-1)(|F_2|-1)\cdots(|F_k|-1)+1}=\alpha,\end{equation}

showing that R is potent. To see this, fix i with $1\leq i\leq k$ , and let $\pi_i\colon\alpha\mathbb{Z}[\alpha]\to F_i$ be projection onto the ith coordinate. It clearly suffices to establish that $(\pi_i(\alpha))^{(|F_1|-1)(|F_2|-1)\cdots(|F_k|-1)+1}=\pi_i(\alpha)$ . If $\pi_i(\alpha)=0_i$ , this is patent, so assume that $\pi_i(\alpha)\neq 0$ . Lagrange’s Theorem implies that $\pi_i(\alpha)^{|F_i|-1}=1_i$ . Raising both sides to the power $\prod_{j\neq i}|F_j|-1$ and then multiplying through by $\pi_i(\alpha)$ yields the desired equation, and (2.9) is established. The proof is now complete.

The following equivalent formulation of Jacobson’s theorem is immediate.

In [Reference Anderson and Danchev1], the authors show that if R is a ring with identity such that for every $x\in R$ , 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.

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 $x\in S$ 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.