# Ring with finitely many zerodivisors

Show that a ring $R$ with exactly $n$ zero divisors (different from $0$) has cardinality atmost $(n+1)^2$.

I have shown that annihilator of any element among the zero divisor is a subset of the zero divisor which proves it is finite. now i think i have to show that it's coset space is finite.please help me.

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Are you counting 0 as a zero divisor? It also matters how you restrict $n$. Since a field has either 1 or 0 zero divisors, depending on how you count, any infinite field would seem to be a counterexample to your statement (unless you rule it out, somehow) – rschwieb Jan 29 '13 at 15:33
Here are some ideas: Let $x$ be a zero-divisor such that the annihilator of $x$ consists of all the zero-divisors (and $0$). Now the annihilator of $x$ has size $n+1$, so you just need to show that the quotient has size $n+1$. Can you show that any non-zero element in the quotient has the form $1 + ax$ for some $a$? – Tobias Kildetoft Jan 29 '13 at 15:38
no i am not counting – Koushik Jan 29 '13 at 15:38
Ignore the last part of my previous comment. That was obviously not right. – Tobias Kildetoft Jan 29 '13 at 15:40
It really does not sound right. Looking at $\mathbf{Z}_{4}$, for instance, you see you should count $0$ as a legitimate zero divisor. And in any case $\mathbf{Z}_{6}$ does have (nonzero) zero divisors, but its order is not a square. – Andreas Caranti Jan 29 '13 at 15:40

I assume the ring is meant to be commutative. This result is true only when $n>1$ and you can find an easy proof in the paper "Properties of rings with a finite number of zero divisors" of N. Ganesan (1964).

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Oh, good citation, thanks @Adeel. Ganesan's theorem reads like this. Suppose the commutative ring $R$ has $n \ge 1$ nonzero zero-divisors. Then $R$ is finite, of order at most $(n+1)^2$. (Spoiler: link.springer.com/article/10.1007%2FBF01362435?LI=true) – Andreas Caranti Jan 29 '13 at 16:11
There's also a noncommutative sequel link.springer.com/article/10.1007%2FBF01359907 – Andreas Caranti Jan 29 '13 at 16:56
@AndreasCaranti Why use links that require subscription when that papers are free on the web? Btw, the problem posted here is completely solved in this paper. – user26857 Jan 29 '13 at 17:15
@YACP Oh, sorry, I hadn't realised it, I accessed the link from my University, which authorised me automatically. My fault. – Andreas Caranti Jan 30 '13 at 9:51

Let the $n$ zero-divisors of $R$ be $\{z_i\}, ~~i \in \{1,2,\dots,n\}$, and $I_i$ the annhilator of $z_i$. We assume $n \ge 1$. You've already proven that each $I_i$ is finite.

For any $z$ and $r \in R$, $zr$ is a zero divisor that is either a non-zero zero-divisor or 0. All elements $r'$ that are congruent to this $r$ in $R/I_i$ are such that $r'z_i = rz_i$. Therefore every class in $R/I_i$ is represented by a distinct zero-divisor in $R$, and the index of $I_i$ is finite.

We have therefore shown that $R$ is at least finite, and we'll say that it has $|R|$ elements. Each annhiliator $I_i$ has order $|I_i|$ and $[R:I_i]$, the index of $I_i$ in $R$ is $\frac{|R|}{|I_i|}.$

Because each element of $I_i$ is a zero-divisor, it must be that $|I_i| \leq n+1$. But since each representative in $R/I_i$ is also a zero divisor, $[R:I_i] = \frac{|R|}{|I_i|} \leq n+1$ as well. This gives $$|R| \leq |I_i|(n+1) \leq (n+1)^2$$ Which proves the theorem.

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