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 R be a finite commutative ring with no zero divisors.

Now let I \subseteq R be a nonzero ideal, with a \in I nonzero. Define \varphi_a : R \rightarrow R by \varphi_a(r) = ar. If \varphi_a(r) = \varphi_a(s), then ar = as, so that a(r-s) = 0. Since R has no zero divisors and a \neq 0, r-s=0, so that r=s. Thus \varphi_a is injective. Since R is finite, \varphi_a is also surjective, so that every element r \in R has the form as for some s \in R. In particular, R \subseteq I. Thus R has only the trivial ideals.

Now suppose that a \in \mathfrak{N}(R) is nonzero, and let m be minimal such that a^m = 0. Now m \geq 2. But then aa^{m-1} = 0, and since a \neq 0, a^{m-1} = 0, a contradiction. Thus the nilradical of R is trivial.

Now let a \in R be nonzero; then (a) = R. In particular, note that since R is finite, there exists m \geq 2 such that a^m = a^n for some n < m. Let m be minimal with this property. Now a^m-a^n = 0. Suppose n \geq 2. Then a^{n-1}(a^{m-n+1} - a) = 0. Since a is not nilpotent, we have a^{m-n+1} = a, a contradiction. Thus n=1 and we have a^m = a.

Finally, let x = ra \in R. Now a^{m-1}x = a^{m-1}ra ra^m = ra = x. Then a^{m-1} satisfies the defining property of an identity element, and thus R 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

  • Arjun Paul  On August 9, 2011 at 7:26 am

    good….

  • ki  On April 15, 2012 at 11:43 am

    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.

    • ki  On April 15, 2012 at 10:39 pm

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

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