# Every element in a ring with finitely many ideals is either a unit or a zero divisor.

I came across the above proposition on mathstackexchange If every nonzero element of $R$ is either a unit or a zero divisor then $R$ contains only finitely many ideals. the link asks a different question but assumes one i ask here. i tried proving the same but i can't. Some hint would be useful. I know basic ring theory. In particular i do not know artinian local rings as someone suggested that it follows easily from the theory of artinian rings.

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Hint: Given $a\in R$, consider the chain of ideals $a^n R$.
The approach is to examine the descending chain of ideals of the form $x^iR$ for $i\in \Bbb Z^+$, as in this solution. The Artinian condition (or the fact that there are only finitely many ideals, if you want to stick with it) means that this sequence eventually stops, and then you can show that either $x$ is a zero divisor, or $x$ is a unit.