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This is probably a pretty basic question but I'm an applied math guy trying to understand these basics: If a nonzero element of a finite ring does not have a multiplicative inverse, must that element be a zero divisor?

I've tried proving this to myself to no avail, and searches have only turned up some deep pure math stuff (as far as I'm concerned, anyways). Thanks in advance!

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This is indeed an extremely easy question, and this is the wrong place for it. –  Steven Landsburg Sep 20 '13 at 3:00
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Left multiplication by that element is an additive homomorphism... –  Ryan Budney Sep 20 '13 at 3:28
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migrated from mathoverflow.net Sep 20 '13 at 3:38

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2 Answers

If $R$ is finite, and $r \in R$ is not a zero divisor (and not itself zero... I don't remember if we count zero as a zero divisor at the moment), then for each $x \in R$, define $f:R \rightarrow R$ by $f(x) = rx$. This map is injective, since $f(x) = f(y)$ means $rx = ry$, so $r(x-y) = 0$ and so $x-y=0$. But the trick is this: any injective map of a finite set to itself must be a surjection! So $f$ hits one, and so there is some $s \in R$ with $f(s) = 1$. But this just means $rs = 1$, and we have found our inverse.

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Finite ring has zero divisors

This was a post a few days ago. It has a simple solution using the pigeonhole principle.

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