A prime ideal criterion for being Noetherian

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 1

If ${I,J}$ are ideals of a commutative ring ${A}$ , then we define

$\displaystyle I\colon J = \{ x\in A : xJ \subset I \}.$

For instance, if ${I}$ is not prime then there are ${a,b \notin I}$ with ${ab \in I}$ , so

$\displaystyle I \colon a \supsetneq I, \quad I \colon b \supsetneq I.$

This is a frequently used fact.

We now prove the following result:

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

The idea here is that if ${A}$ is not Noetherian, then there would be ideals of ${A}$ 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 ${A}$ 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 ${\mathfrak{C}}$ is a chain,

$\displaystyle J:= \bigcup_{I \in \mathfrak{C}} I$

is an ideal. I claim ${J}$ is not finitely generated. Indeed, otherwise we could pick generators ${a_1, \dots, a_n \in \bigcup_{\mathfrak{C}} I}$ , all of which would thus lie in some element of the totally ordered set ${\mathfrak{C}}$ . Thus ${J \in \mathfrak{C}}$ so ${J}$ is not finitely generated, contradiction.

Now return to the theorem. Choose ${J}$ maximal with respect to being not finitely generated.

Claim 1 ${J}$ is prime.

If not, we could find ${a,b \notin J}$ with ${ab \in J}$ . Then ${(a,J)}$ and ${J\colon a := \{ s \in A\colon sa \in J \}}$ both properly contain ${J}$ . Since ${J}$ was maximal, both ${(a,J)}$ and ${J\colon a}$ are finitely generated. The next lemma will show ${J}$ is finitely generated too, a contradiction which will complete the proof.

Lemma 3 If ${I \subset A}$ is an ideal, ${a \in A}$ , and both ${(I,a)}$ and ${I\colon a}$ are finitely generated, then ${I}$ is finitely generated.

To prove this, choose generators (which wlog contain ${a}$ ) ${r_1, \dots, r_m, a}$ for ${(I,a)}$ . By subtracting multiples of ${a}$ , assume ${r_1, \dots, r_m \in I}$ . Next pick generators ${s_1, \dots, s_n}$ for ${I\colon a}$ .

I claim that

$\displaystyle r_1, \dots, r_m, s_1a , s_2 a, \dots, s_n a$

generate ${I}$ . Indeed, if ${x \in I \subset (I,a)}$ , we can write

$\displaystyle x = \sum_{i=1}^m b_i r_i + ba, \quad b_i, b \in A;$

then ${ba = x - \sum b_i r_i \in I}$ , so ${b \in I\colon a}$ . Thus we can write

$\displaystyle b = \sum_{j=1}^n c_j s_j ;$

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 4 If ${A}$ is a ring such that each prime ideal is principal, then every ideal of ${A}$ is principal.

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

Lemma 5 If ${I \subset A}$ is maximal with respect to being not principal, then ${I}$ is prime.

If ${I}$ is not prime, then there are ${a,b \notin I}$ with ${ab \in I}$ . Then ${(I,a) \supsetneq I}$ , and ${I\colon (I,a) \supset (I,b) \supsetneq I}$ . Each of those two ideals is thus principal.

So suppose ${(I,a)=(c), I\colon c = I\colon (I,a) = (d)}$ . Then I claim ${I = (cd)}$ . Indeed, if ${x \in I}$ , we have ${x = rc}$ for some ${r}$ . Since ${rc \in I}$ , that is ${r \in I\colon c}$ , we have ${r = sd}$ for some ${d}$ . Thus

$\displaystyle x = s cd.$

Also note that ${cd \in (I,a) (I\colon (I,a)) \subset I}$ .

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.

Climbing Mount Bourbaki