The set of prime ideals of a commutative ring contains inclusion-minimal elements

Let R be a commutative ring with 1 \neq 0. Prove that the set of prime ideals in R has an inclusion-minimal element (possibly zero).

Let \mathcal{P} denote the set of prime ideals. Note that \mathcal{P} is partially ordered by the superset relation, and is nonempty since every maximal ideal is prime. Let \{P_k\}_K be a chain in \mathcal{P}; that is, P_\ell \supseteq P_k for all \ell \leq k. We know that \bigcap P_k is an ideal in R by this previous exercise; we now show that this ideal is prime. Suppose ab \in \bigcap P_k. Then ab \in P_k for all k. Suppose that (without loss of generality) there exists m \in K such that a \notin P_m. Then b \in P_m, and moreover, b \in P_k for all \ell \leq m. Now let \ell > m; since ab \in P_\ell, either a \in P_\ell or b \in P_\ell. If a \in P_\ell, then a \in P_m, a contradiction. So b \in P_\ell. Hence b \in P_k for all k, and we have b \in \bigcap P_k. Thus \bigcap P_k is prime. So \bigcap P_k \in \mathcal{P}, and thus the chain \{P_k\}_K has an upper bound.

Since the chain \{P_k\}_K is arbitrary, by Zorn’s Lemma, \mathcal{P} contains an inclusion-minimal element.

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  • ai  On January 13, 2011 at 6:33 pm

    I just wanted to point out that an arbitrary chain need not be indexed by the natural numbers, which was assumed in this proof. And in order to appeal to Zorn’s Lemma, one needs to establish that every chain has a lower bound in your set P of prime ideals.

    • nbloomf  On January 13, 2011 at 8:48 pm

      You are correct on both counts; not sure what I was thinking. I’ll mark this “incomplete” for now and fix it later. Thanks!

      • nbloomf  On February 28, 2011 at 10:25 am

        It should be fixed now.

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