Artinian implies IZ

Statement

Verbal statement

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

Proof

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:

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

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

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

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

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

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