Recently I mentioned that Noetherian rings can be characterized either by the fact that their finitely generated modules don’t have non-finitely generated submodules, or by the a priori weaker condition that their ideals are all finitely generated. In other words, if you want a counterexample to the statement that a submodule of a finitely generated module is again finitely generated, it’s sufficient to consider the case .
I also claimed that the function sending an -module to its minimum number of generators is not generally increasing: for -modules we don’t necessarily have , even when is Noetherian. For example, over , the ideal has , but .
These two facts led me to ask the following question:
If we want a commutative ring to have the property that is increasing, is it again sufficient to look for counterexamples among the ideals of ? In other words, are the rings whose ideals can all be generated by single elements (the principal ideal rings) exactly the rings with this property?
Answer: Yes! For example, is a principal ideal ring (any ideal is generated by the gcd of its elements), so this tells us that any subgroup of an abelian group with generators also has at most generators. I imagine this can be proven directly from the structure theorem for finitely generated abelian groups, but dealing with arbitrary submodules of a direct sum sounds a little fiddly.
The key to the proof I’ll show below is to use exact sequences to build arbitrary finitely generated modules out of very simple ones, using this lemma reminiscent of the rank-nullity theorem for vector spaces:
Lemma: Let be any exact sequence of -modules. If and can be generated by and elements, respectively, then can be generated by elements. In other words, .
Proof of lemma: Let and be generating sets for and . Let be arbitrary elements of mapping to the (which is possible since is surjective). We show that is a generating set for .
Let be an arbitrary element of , and let be its image in . Use our generating set for to write
for some . We don’t know immediately that equals , but we know that their difference is sent to zero in . Therefore this difference actually belongs to , so call it and use our generating set to write
for some . Therefore
represents as a linear combination of as desired.
Now on to the proof that if is a principal ideal ring, then is increasing. There are two steps to the proof: reducing to the case of free modules, and inducting on the number of generators. Since these two steps can be done in either order, we present both possibilities.
Proof 1: Suppose is a principal ideal ring. We prove by induction on that if is an -module with generators, so is any submodule of .
Base cases: and . For , the only module generated by zero elements is the zero module, which has only itself as a submodule! For , the single generator gives us a surjection of -modules . Let be the inverse image of under this surjection; since is a principal ideal ring, can be generated by one element, and the image of that element maps to the desired single generator of .
Induction step: Suppose and suppose can be generated by . Let be the submodule generated by alone, and let be the quotient , which can be generated by the classes of . Therefore we have a short exact sequence
with generated by a single element and generated by elements. Let be the inverse image of in (i.e. the intersection of submodules ) and let be the image of in . We get another short exact sequence
because surjects onto , and the only elements sent to zero are those that also belong to . Now since and , by the base case and induction hypothesis, we know that and can be generated by and elements, respectively. But then by the lemma about adding generating sets in short exact sequences, we find that can be generated by elements too! This completes the induction step, and the proof.
Proof 2: This time, we first prove by induction on that arbitrary submodules of can be generated by elements, and then we generalize to arbitrary finitely generated -modules.
Base cases: and . Again, is the zero module, which is generated by zero elements and only has itself for a submodule. The case, that submodules of need at most one generator, is exactly our assumption that is a principal ideal ring.
Induction step: Let and set up the short exact sequence
where the map sends and the map sends . Let be any submodule of , and as in Proof 1 let be its inverse image in and let be its image in , so we get the short exact sequence
Then by the base case and induction hypothesis, and can be generated by and elements, so can be generated by elements as desired.
Now let be an arbitrary -module with a generating set of size , and let be a submodule of . The generators give us a surjection , and let be the inverse image of in . We have already shown that this can be generated by generators; their images then form a size- generating set for , as desired.
These proofs are inspired by the proof that over rings whose ideals are all finitely generated, submodules of arbitrary finitely generated modules are again finitely generated. That proof again shows by induction on that submodules of are finitely generated—the case is the hypothesis on ideals—and then generalizing straightforwardly to arbitrary finitely generated modules as quotients of finitely generated free modules.
So why use Noetherian rings, which only distinguish between finitely- and infinitely-generated modules, if we have a notion of principal ideal ring with much better behavior of generating set sizes? Simply put, it’s because Noetherian rings are common and principal ideal rings are not. The Hilbert Basis Theorem, that is Noetherian if is, tells us in particular that all finitely generated algebras over fields are Noetherian, which allows us to do a lot of high-dimensional algebraic geometry without leaving the Noetherian setting. On the other hand, principal ideal rings only capture some one-dimensional behavior, and that’s just not general enough for most purposes.