In a commutative unital ring, if a nontrivial prime ideal contains no zero divisors, then the ring is an integral domain

Let R be a commutative ring with 1 \neq 0. Prove that if P \subseteq R is a prime ideal containing no zero divisors (in R), then R is an integral domain.


Let a,b \in R such that ab = 0. Since P is a prime ideal and ab \in P, without loss of generality a \in P. If a \neq 0, then since P contains no zero divisors in R, b = 0.

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Comments

  • ai  On July 3, 2011 at 11:23 am

    How do we know that R has a prime ideal P to begin with? I have a longer proof that doesn’t assume R has a prime ideal, but this proof is nicer, assuming of course that we can always assume an arbitrary, commutative ring with a 1 has prime ideals.
    Thanks.

    • ai  On July 3, 2011 at 11:26 am

      How do we know that R has a prime ideal P to begin with? I have a longer proof that doesn’t assume R has a prime ideal, but this proof is nicer, assuming of course that we can always assume an arbitrary, commutative ring with a 1 has prime ideals.
      Thanks.

      Oh, never mind, I got it:

      If R has a 1, then every proper ideal is contained in a maximal ideal, and if R also commutative, then every maximal ideal is a prime ideal.

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