In a commutative ring, all prime ideals are maximal

Let R be a finite commutative ring with 1 \neq 0. Prove that every prime ideal of R is maximal.

We begin with a lemma.

Lemma: If R is a finite integral domain, then R is a field. Proof: Let u \in R be nonzero, and define \varphi_u : R \rightarrow R by \varphi_u(r) = ur. We claim that \varphi_u is injective. To see this, suppose \varphi_u(r) = \varphi_u(s); then ur = us, so that u(r-s) = 0. Since u \neq 0, r-s=0, so that r=s. Hence \varphi_u is injective. Since R is finite, \varphi_u is also surjective. In particular, there exists v \in R such that \varphi_u(v) = uv = 1, so that u is a unit. Thus every nonzero element of R is a unit, and R is a field. \square

Now if P \subseteq R is prime, then R/P is a finite integral domain. Then R/P is a field, and thus P \subseteq R is maximal.

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  • Eri  On February 12, 2011 at 1:16 pm

    “r-s” should be “r=s” (before the word “hence”)

    • nbloomf  On February 12, 2011 at 1:30 pm

      Fixed. Thanks!

  • maestra  On July 16, 2011 at 8:04 pm

    ese R es real o es culaquier conjunto

    • nbloomf  On July 18, 2011 at 12:59 pm

      If I understand correctly, the question is whether R is the ring of real numbers or an arbitrary ring. Here R is an arbitrary (commutative) ring.

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