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

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.

### Like this:

Like Loading...

*Related*

or leave a trackback:

Trackback URL.

## Comments

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.

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

It should be fixed now.