Artinian implies IZ

Statement

Any Artinian ring is IZ: every element is either invertible, or a zero divisor.

Given: An Artinian ring $A$ , and an element $x \in A$

To prove: $x$ is invertible or a zero divisor

Proof: Consider the descending chain of ideals:

$A \supset (x) \supset (x^{2}) \supset \dots$

By the Artinianness, this chain stabilizes at some point, so we have:

$x^{n} = a x^{n + 1}$

for some $a \in A$ . Rewriting, we see that:

$x^{n} (1 - a x) = 0$

If $1 - a x = 0$ , then $x$ is invertible. Otherwise, $x^{n}$ is a zero divisor, and hence $x$ is a zero divisor.