Let $R$ be an infinite commmutative ring which contains a zero divisor. Show that there exists infinite many zero divisors. [duplicate]

I'm making exercises to prepare for my ring theory exam.

Let $R$ be an infinite commmutative ring which contains a zero divisor. Show that there exists infinite many zero divisors.

Let $a\in R$ a zero divisor. Then $a⋅b=0$ for some element $b≠0$ in $R$. If $a$ or $b$ has infinite order, then I can find infinite zero divisors. But otherwise, I don't see why this should be true. I think I need to do something with the fact that $R$ is commutative. A hint or a detailed solution are both appreciated.

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marked as duplicate by Martin Brandenburg, TZakrevskiy, Cameron Buie, Peter Taylor, Lord_FarinOct 17 '13 at 16:16

Let $a$ be a zero divisor. Consider two cases:

• $Ra$ is an infinite set.
• $Ra$ is a finite set.

(If $R\to Ra$ has finite range, $xa=ra$ has infinitely many solutions $x$ for some $r\in R$, so...)

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Suppose $Ra$ is an infinite set. Then all elements in $Ra-\{0\}$ are zero divisors. So there are infinite zero divisors. Thinking about the other case. – Kasper Oct 17 '13 at 15:32
$xa=ra$ has to have infinite many solutions for some $r\in R$, otherwise there would be finite many elements in $R$. We know that $xab=rab=0$. therefore there are infinite zero divisors of the form $xa$ ? No this isn't true because all $xa$ are equal. I don't see what to do. – Kasper Oct 17 '13 at 15:42
Or can I choose $r$ as the element $r≠0$ with $ra=0$ ? – Kasper Oct 17 '13 at 15:44
@Kasper Consider rewriting $xr=ar$ as $(x-a)r=0$. – anon Oct 17 '13 at 16:00
Aah, I see it now. $xa=ra$ has infinitely many solutions $x$ for some $r\in R$. Therefore I can choose $r≠0$. And $(x-a)r=0$ for infinitely many $x$. So $(x-a)$ is infinitely many times not zero. So we have infinite zero divisors. – Kasper Oct 17 '13 at 16:05

If $ab=0$ for $b\neq 0$ then it implies that $a^nb=0$ for all $n\in \mathbb N$. Thus if $a$ is not nilpotent all $a^n$ are zero divisors.

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If $a$ is nilpotent it only has finitely many distinct powers. – anon Oct 17 '13 at 15:30
@ anon: yes, you are right. My answer is true for only non-nilpotent $a$. – Anupam Oct 17 '13 at 15:38