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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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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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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