This post, the third in the mini-series so far, gives one more criterion for when a ring is Noetherian. I also discuss how prime ideals tend to crop up in commutative algebra.

**Why prime ideals are important **

As discussed in the end of my previous post and in the comments, ideals satisfying some property and maximal with respect to it are often prime. To prove these results, we often use the following convenient notation:

Definition 1If are ideals of a commutative ring , then we define

For instance, if is not prime then there are with , so

This is a frequently used fact.

We now prove the following result:

Theorem 2Let be a commutative ring. Suppose every prime ideal of is finitely generated. Then is Noetherian.

The idea here is that if is not Noetherian, then there would be ideals of which were not finitely generated, and hence ideals *maximal* with respect to being not finitely generated. The last assertion actually requires a bit of justification. The set of ideals of which are not finitely generated intois a poset, so to show (by Zorn) that there is a maximal element, we need to check that each chain has an upper bound. Now if is a chain,

is an ideal. I claim is not finitely generated. Indeed, otherwise we could pick generators , all of which would thus lie in some element of the totally ordered set . Thus so is not finitely generated, contradiction.

Now return to the theorem. Choose maximal with respect to being not finitely generated.

Claim 1is prime.

If not, we could find with . Then and both properly contain . Since was maximal, both and are finitely generated. The next lemma will show is finitely generated too, a contradiction which will complete the proof.

Lemma 3If is an ideal, , and both and are finitely generated, then is finitely generated.

To prove this, choose generators (which wlog contain ) for . By subtracting multiples of , assume . Next pick generators for .

I claim that

generate . Indeed, if , we can write

then , so . Thus we can write

putting these together gives the lemma, and the theorem.

As another illustration of this type of technique, let’s prove a result a commenter mentioned yesterday:

Theorem 4If is a ring such that each prime ideal is principal, then every ideal of is principal.

As before, we use the same Zorn-type argument to reduce this theorem to the following lemma:

Lemma 5If is maximal with respect to being not principal, then is prime.

If is not prime, then there are with . Then , and . Each of those two ideals is thus principal.

So suppose . Then I claim . Indeed, if , we have for some . Since , that is , we have for some . Thus

Also note that .

So, for the future: I want to keep these posts on commutative algebra as self-contained as possible, except in cases when the material is already covered on other blogs. I’ve already violated this principle by using integrality in the previous post. The next few will thus be on general commutative algebra, such as graded rings, filtrations, integrality, and completions.

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