# If $R$ is an infinite ring, then $R$ has either infinitely many zero divisors, or no zero divisors

Please help me to prove that if $R$ is an infinite ring, then $R$ has either an infinite number of zero divisors, or it has no zero divisors.

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Is commutativity assumed? It would be good to add that tag, if so. I'm not sure how Zhen Lin's solution works in the noncommutative case. –  rschwieb Feb 27 '13 at 14:33
I don't know ring is commutative or not. There's no information about it in this problem. –  Jane Feb 27 '13 at 17:02

Hint. Let $a$ be a zero-divisor. Consider $\{ a b : b \in R \} \setminus \{ 0 \}$. This is a non-empty set of zero-divisors in $R$. If it is infinite, then you are done. Otherwise it is finite; what are the implications of this? (Use the pigeonhole principle.)

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so, If I understood it correctly: by pigeonhole principle, there exists an infinite subset X of R and an element x from X such that ax=ay for all y from X. so –  Jane Feb 27 '13 at 10:59
Yes, that's correct. –  Zhen Lin Feb 27 '13 at 11:06
Therefore we have infinite set y -x and I do not know, what to do with it? –  Jane Feb 27 '13 at 11:15
oh, thank you, I get it –  Jane Feb 27 '13 at 11:20

Hi here is a proof by contradiction:

Assume that there are only finitely many zero divisors in $R$ and let them be $a_1, \ldots a_n$ and denote the set of these $A$. Fix some element $b \in R, b \ne a_i$ for all $i$. Convince yourself that there are infinitely many elements $c \in R, c \ne a_i$ for all $i$ such that $(b-c) \ne a_i$ for all $i$, let $C$ be the set of these. Given a $x \in R$ that is not a zero divisor we must have that $a_i \mapsto x a_i$ is a bijection of $A$ otherwise $x$ would be a zero divisor. But then since there is only finitely many bijections of $A$ since $A$ is finite, and there are infinitely many elements in $C$, we see that we can find an elements $y$ in $C$ that give the same bijection of $A$ as $b$ , but then $(b-y)a_i = b a_i - y a_i = 0$ hence $(b-y)$ is a zero divisor a contradiction.

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