Proving that any finite integral domain is a field. [duplicate]

Let $R$ be an integral domain, and let $a \in R, a \neq 0$. Let $f_a: R \rightarrow R$ be defined by $f_a(r)=ar$. Prove that $f$ is injective.

Then prove that every finite integral domain is a field.

My ideas and concept:

Can someone help articulate my understandings into a proper solution?

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marked as duplicate by rschwieb 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 }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 25 '14 at 12:38

The main idea is that if $R$ is finite, then any map from $R$ to $R$ is injective iff it is surjective iff it is bijective.

We know $a\in R$ is a not a zero-divisor if $f_a$ is injective. Since $R$ is finite, then it is also surjective.

If $f_{a}$ is surjective then $a$ is a unit. Hence we have shown that any non-zero element in $R$ is a unit. Hence $R$ is a field.

(I think for future questions, you should specify that you are dealing with a commutative ring with unity)

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To show $f_a$ is injective, suppose $ar=ar'$. Then $a(r-r')=0$. Since $R$ is an integral domain, $a=0$ or $r-r'=0$. But you assumed $a\neq 0$. You are right on the other part: since $R$ is finite, any injection $\phi:R\to R$ is a bijection. This means $1$ must have a preimage... so?

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