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Why is a finite integral domain always field?

How do you prove this;

Let R be a finite commutative domain. Prove that R is a field.

I need to know this because doing a course in non commutative rings. Yet, my algebra has gone from summer holidays. Also, another list of exercise I can't do.

I know it's simple.

I have a, b, 1, 0 in R. ab=ba. Might need to use it finite. So for groups if you have finite then you apply 1 element over and over again you get back to it. Just need to find the order of the group. Yet, I don't see how I can use it here without proving a lot of other stuff. $a^{n-1}a=1$ Where n is the order of the ring???

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    We have discussed this a few times already here on math.SE. Have you tried searching a bit?2011-09-22
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    Yes, but the only proof that makes sense is one where you assume that commutative domain has unit in it. Might just try and get old algebra notes. However, I know I've seen an easy proof of it. looks like there isn't an easy proof of it.2011-09-22
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    Lol that proof is wrong. Hmm.2011-09-22
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    @algebears: *what* proof is wrong? If you browse a little bit the site, you will see that people don't LOL around here...2011-09-22

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