Let be a commutative ring with . Prove that the set of prime ideals in has an inclusion-minimal element (possibly zero).
Let denote the set of prime ideals. Note that is partially ordered by the superset relation, and is nonempty since every maximal ideal is prime. Let be a chain in ; that is, for all . We know that is an ideal in by this previous exercise; we now show that this ideal is prime. Suppose . Then for all . Suppose that (without loss of generality) there exists such that . Then , and moreover, for all . Now let ; since , either or . If , then , a contradiction. So . Hence for all , and we have . Thus is prime. So , and thus the chain has an upper bound.
Since the chain is arbitrary, by Zorn’s Lemma, contains an inclusion-minimal element.