Grrr. At some point I'm going to have transcribed down all the content of my abstract algebra textbooks and then some. I've got a little confession, I don't know what I'm doing, but if you want you can poke holes in these proofs so I can get closer to understanding. What?

I proved Hilbert's basis theorem a few posts ago while sidestepping the alternate definition that "every ideal contained in [a Noetherian ring] is finitely generated as a module". Then I forgot all about that characterization and was unable to understand when it popped up in discussions of finiteness criteria. If you don't know what the finiteness criteria I'm referring to yet, I just mean when rings are generated as modules or rings by adding finitely many elements over other rings, but it doesn't matter.

It turns out furthermore that stuff we prove is more useful if we generalize the idea of Noetherian-ness to modules.

Let be a commutative ring (I am not brave enough to handle noncommutative ones) and let be an -module (which means we have a function satisfying various clear multiplication axioms that I will irresponsibly not list here.)

1. is Noetherian_1 if any infinite chain of submodules is eventually constant.
2. is Noetherian_2 if every submodule of is finitely generated as a module.

When we speak of Noetherian rings, that means they're Noetherian when we treat them as modules of themselves. Another note to self: in a ring , an ideal and an -module are the same zarking thing and it means the same thing for them to be finitely generated, which is the thing the Noetherian-ness criterion is all about. I am unbelievably obtuse sometimes. So you can just read the proofs below while mentally replacing "module" or "submodule" by "ideal".

First, let's prove the equivalence.

(standard, ~ DF 12.1 Theorem 1) An -module is Noetherian_1 iff it is Noetherian_2.

Proof. () Let be a submodule of . Starting from the zero module, attempt to pick a sequence of elements such that is in but not in the module generated by . This cannot go on forever since the modules generated by are an infinite chain of ascending modules; the only way it stops is if at some point , at which point we have proven that generate .

() Let be an infinite chain of submodules of . Then the union of all of them is a module (verify the axioms), which is finitely generated. Each of the generators appears in one of the modules ; if the maximum of those then the chain is constant after .

— end proof —

Okay, weirder stuff:

Let be an -module, and let be a submodule. Then is Noetherian iff and are both Noetherian.

Proof. () If is Noetherian, then any submodule of is a submodule of and is thus finitely generated, so is Noetherian. As for , any submodule of it corresponds to a submodule of containing . That submodule is also finitely generated, and the image of those generators in finitely generate it, so is Noetherian.

Now assume and are both Noetherian; we want to prove is Noetherian. Let be a submodule of ; we want to prove it's finitely generated. Consider . Then is a submodule of and is finitely generated, say by . Also, observe that the obvious mapping from into is well-defined, a homomorphism, and injective, so is isomorphic to a submodule of and is also finitely generated. Take preimages of those generators in and call them ; those elements and the generators of together generate because you can pick elements from to reach the equivalence class in of any element and then reach the actual element in that class using .

— end proof —

Corollary 1. If is a Noetherian -module, then (the obvious -module of -tuples of ) is Noetherian.

Proof. Induct on . It's trivial when . If you let be the submodule of tuples of with only the first element nonzero, then is isomorphic to and is a Noetherian -module, and is isomorphic to and is a Noetherian -module by the induction hypothesis, so is Noetherian.

Corollary 2. If is a Noetherian -module, then any module of the form for is also a Noetherian -module. (The condition is just what we mean when we say is module-finite over .)

Proof. There's a surjective homomorphism from to sending the tuple with th element 1 and all others 0 to the element . So is isomorphic to where is the kernel of , and is Noetherian by Corollary 1, so is Noetherian by the theorem.

Corollary 3. If is a Noetherian ring, then an -module is Noetherian iff it is finitely generated.

Proof. If is Noetherian then any submodule of , including itself, is finitely generated. On the other hand, if it's finitely generated by , that means , so it's Noetherian by Corollary 2.

— end corollaries —

Separately:

If is a Noetherian ring and is an ideal, then is a Noetherian ring.

Proof. Any ideal in is for some ideal of . Then is finitely generated and the image of the generators in finitely generate . Alternatively, any chain of ascending ideals in likewise corresponds to a chain of ascending ideals in , which is eventually constant. — end proof —

(Note that this does not instantly follow from the above theorem, which proves that is Noetherian if you treat it as a -module. It appears you can deduce the consequence you want from this, but it's so easy by itself anyway that I don't think it's worth it.)

And, finally: If is a Noetherian ring and is ring-finite over , then is Noetherian.

Proof. Let be ring-generated over by elements . There is the obvious homomorphism from the polynomial ring in indeterminates to ; if is the kernel of this then . By the Hilbert basis theorem is Noetherian and by the above theorem is Noetherian. — end proof —