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Show that any commutative ring $R$ having only $n$ non-zero zero divisors ($n\geq 1$) is finite and doesn't contains more than $(n+1)^2$ elements.

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This is a theorem by Ganesan. You show that for any non-zero zero divisor $x$, its annulator $Ann(x)$ is finite and has finite index in $R$. – Cocopuffs Jun 26 '13 at 18:07
This is a curious theorem and I'm tempted to upvote it, but OP put so little effort into the question that I can't really do that... – tomasz Dec 6 '13 at 21:35
up vote 13 down vote

Let $x$ be a nontrivial zero divisor. Then the annulator $Ann(x)$ is finite: every element of it is either zero or a nontrivial zero divisor itself, so it has at most $n+1$ elements.

Also, it has index at most $n+1$ in $R$: each coset $r + Ann(x)$ corresponds uniquely to the zero divisor $rx$.

So $R$ must have at most $(n+1)^2$ elements.

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I notice for myself that second paragraph means the map $R/ann(x)\to R$ induced by the multiplication-by-$x$ map is injective and lands in the set of zero divisors. – Cantlog Sep 12 '13 at 22:50
@Cantlog Yes, if $rx = sx$ then $(r-s)x = 0$, i.e. $r-s \in Ann(x)$ – Cocopuffs Sep 13 '13 at 6:08

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