I recently inherited a copy of Nagata’s book Local Rings, and I’ve already learned a new theorem!
Theorem of Cohen: A commutative ring is noetherian if and only if all its prime ideals are finitely generated.
This is cool because if you, like me, have ever been sad about the fact that noetherianity is not a local property of a ring, this is a sort of substitute. (NB: the axiom of choice, in the form of Zorn’s lemma, is involved.)
There is a naive reason you might hope the theorem is true, that is not the real reason, but it’s sort of the right idea: you might hope that because any non-finitely-generated (henceforth, “non-f.g.”) ideal is contained in a maximal ideal, non-f.g.ness must show up on a maximal.
This is wrong, and actually, the statement of the theorem becomes false if you reduce the quantification from all primes to just maximals. At the bottom of this post I’ll give an example to show this, due to Cory Colbert. The problem is that a non-f.g. ideal can live inside an f.g. ideal. (Dumb example: take your favorite non-f.g. ideal inside your favorite non-noetherian ring. It’s inside the unit ideal. For a more interesting case, see Cory’s example below.) Thus, the family of non-f.g. ideals is not upward-closed, and there’s no reason for a failure of noetherianity to show up on a maximal.
However, it’s approximately the right idea. Non-f.g.ness does not propagate upward because f.g.ness doesn’t simply propagate downward. But what’s true is that f.g.ness does propagate downward to an ideal from two ideals containing it that relate to it in a specific way.
Lemma: If is an ideal in a commutative ring, and is an element, and and are both finitely generated, then is finitely generated.
Proof: Choose a finite set of generators for , say , and write each one as the sum of an element of and a multiple of :
with each and each just some ring element. Also choose a finite set of generators for . Then all of lie in , and we will show that they generate . Let be arbitrary. Then there is a representation
for , where the are some ring elements. The key observation is that
is in , since both terms on the left are in , and it follows that is in ! Thus, there is a representation
where the are some ring elements. Then we have
and this is the promised expression of in terms of the proffered generators . This completes the proof.
With this lemma in hand, the proof of Cohen’s theorem can be briefly summarized as follows. Consider the family of non-f.g. ideals in a ring. It’s not upward closed, so it won’t necessarily reach the maximal ideals themselves, but if it is nonempty, it will have maximal members (this is where Zorn comes in), and what the lemma does (a sort of weaker substitute for upward-closure) is to prove that these maximal members must be prime. Thus non-f.g.ness must show up on a prime if it shows up at all. Here’s the precise argument:
Proof of Cohen’s Theorem: Obviously, a noetherian ring can have only f.g. primes. The converse is the substantive direction. We will (equivalently) prove the inverse: a non-noetherian ring must have a non-f.g. prime. So let be a non-noetherian ring, and let be the family of non-f.g. primes, which is nonempty by assumption. Note that given an ascending chain in , the union yields an upper bound; we see this as follows. If the union of an ascending chain of ideals in were not in , i.e. if it were f.g. as an ideal, then the (finite list of) generators would be found in some finite collection of members of the chain, and then the greatest of these members would contain all the generators and thus equal the whole union. This is a contradiction because the members of the chain were presumed to be in whereas the union was presumed not to be. It follows that the union of an ascending chain of non-f.g. ideals is non-f.g. Thus satisfies the hypotheses of Zorn’s lemma, so has a maximal element. Call it .
We claim is prime. Indeed, suppose , but . By the former supposition, . By the latter supposition, contains properly. Since is maximal in , this proper containment implies that is f.g. Since is not f.g. while is, the Lemma implies that is not f.g. This means that it cannot contain properly, again by maximality of in . We conclude . But recalling that , we can now conclude that ! This proves that is prime, so there is a non-f.g. prime, completing the proof.
I promised I’d tell you about Cory’s example showing that the statement cannot be quantified only over maximal ideals. Let be any field. The action will all take place inside the field of rational functions over in two variables. Let . Let . Note that , because every is a multiple of in . It follows that , so that is a maximal ideal in . Finally, consider the localization .
This is the desired ring. It is local and the unique maximal ideal is principal (generated by ), so certainly f.g. On the other hand, it is non-noetherian: the non-zero element is contained in the ideal , for all , since , but in a noetherian local ring, the intersection of the powers of the maximal ideal is zero, by the Krull intersection theorem.