# Finite ring zero divisor/multiplicative inverse [duplicate]

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!

## marked as duplicate by user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 13 '17 at 10:26

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.