# 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 $a^n R$.

Solution:

Since $R$ has only finitely many ideals, we have $a^n R = a^m R$ for some $m>n$. Then, $a^n(a^{m-n}r-1)=0$ for some $r\in R$. This implies that $a^n$ is a zero divisor or $a^{m-n}$ is a unit. In turn, this implies that $a$ is a zero divisor or a unit.

Another hint: Given $a\in R$, $a$ not a zero divisor, consider the map $I\mapsto aI$ on the set of ideals of $R$.

Solution:

The maps is a bijection because it is injective. Hence, $R=aI$ for some ideal $I$, which implies that $a$ is a unit.

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Hi, I'm curious as to how you would go about answering this question using the hint you provided. Would you mind expanding on this answer at all? Thanks very much. –  Jango Nov 4 '14 at 9:54
If $a^mR=a^nR$ for some $m>n$, then $a^{n+1}R=a^nR$, and so on. –  user26857 Nov 4 '14 at 11:11
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.