## A nonzero finite commutative ring with no zero divisors is a field

Prove that any nonzero finite commutative ring with no zero divisors is a field. (Do not assume that the ring has a 1.)

Let be a finite commutative ring with no zero divisors.

Now let be a nonzero ideal, with nonzero. Define by . If , then , so that . Since has no zero divisors and , , so that . Thus is injective. Since is finite, is also surjective, so that every element has the form for some . In particular, . Thus has only the trivial ideals.

Now suppose that is nonzero, and let be minimal such that . Now . But then , and since , , a contradiction. Thus the nilradical of is trivial.

Now let be nonzero; then . In particular, note that since is finite, there exists such that for some . Let be minimal with this property. Now . Suppose . Then . Since is not nilpotent, we have , a contradiction. Thus and we have .

Finally, let . Now . Then satisfies the defining property of an identity element, and thus is a finite integral domain. We proved in a lemma to a previous exercise that every finite integral domain is a field.

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## Comments

good….

you don’t need the commutativity!

Just make one case with a right ideal and one with a left ideal and in both cases you can get that a^{m-1} is the identity.

Correction: We get “division ring” but not field – sorry.