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This is a problem from Dummit & Foote.

Prove that a non-zero finite commutative ring that has no divisor is a field. (Do not assume the ring has a 1)

Evidently, one has to use the theorem proved earlier, that a finite integral domain is a field. So, I have to prove that any non-zero finite commutative ring has a multiplicative identity. I am stumped.

I had proved that the only ideals this ring has are the trivial ideals, $0$ and $R$. This is seen easily by defining the homomorphism $\phi(r): R \to R = ar $ where $a$ is a non-zero element of a non-zero ideal $I$. The surjectivity of this map (pigeon hole principle), shows that the ideals are trivial.

Is this going to help me prove it has a $1$. How do I approach the problem?

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@YACP: No. integral domains necessarily have unity, which is not allowed to be assumed here. –  Cameron Buie Apr 11 '13 at 11:20
If you read through it again, you'll see that the OP is trying to prove that it has unity. –  Cameron Buie Apr 11 '13 at 11:30
@YACP: There we go. –  Cameron Buie Apr 11 '13 at 12:24
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marked as duplicate by YACP, Davide Giraudo, Amzoti, Cameron Buie, azimut Apr 11 '13 at 12:28

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HINT: Fix a non-zero $a\in R$. The map $\varphi_a:R\to F:r\mapsto ar$ is surjective, so there is some $e\in R$ such that $ae=a$. Now use the surjectivity of $\varphi_a$ again to show that $e=1_R$.

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