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I'm trying to prove or disprove the following fact:

Let $R$ be a finite ring and $x \in R$ be a zero divisor. If $x$ is not nil-potent, then for some $n > 1$ we have $x^n = x$, i.e., $x$ is $n$-potent.

I've already proved it for the modular rings $\mathbb{Z}/m\mathbb{Z}$, and my colleague found a counterexample for infinite rings, the matrix $\pmatrix{2 &2\\0 &0}$ over the integers. The proof for $\mathbb{Z}/m\mathbb{Z}$ used the explicit characterization of zero divisors in terms of the prime factorization of $m$, and I can't see how I could adapt it for the general case.

Edit: found a flaw in my proof. A counter example is $4\pmod{48}$, which satisfies $4^n \equiv 16 \pmod{48}$ for $n \geq 2$. I'm changing the question.

New claim: let $R$ and $x$ be as above. Seems to me, so far, that $x$ can be nil-potent and stationary ($x^n = y$ for $n > n_0$). Can something weirder go on? Like, can we have $x^n = x^m \neq x$ for $1 < m < n$?

Henrique Augusto Souza
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  • Perhaps start here https://math.stackexchange.com/questions/60969/every-nonzero-element-in-a-finite-ring-is-either-a-unit-or-a-zero-divisor – lhf Aug 04 '17 at 20:46

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Can something weirder go on? Like, can we have $x^n = x^m \neq x$ for $1 < m < n$?

Sure, why not? Consider $F_2[X]/(X^4-X^2)$ in which the image of $X$ (call it $x$) satisfies $x^4=x^2\neq x$.

And $x$ isn't stationary: the powers go $x, x^2, x^3, x^2, x^3\ldots$

Of course, every element in a finite ring must satisfy $a^n=a^m$ for some $m$ and $n$. That is what it means to say that $\{a^n\mid n\in \mathbb N\}$ is finite.

rschwieb
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