When is a module a filtered colimit of finitely presented submodules?

For a (commutative, say) ring $$R$$, and an $$R$$-module $$M$$ it is known that $$M$$ is both:

• a filtered colimit of finitely generated $$R$$-submodules (by considering all finite subsets of $$M$$ and considering the submodules generated by those)
• a filtered colimit of finitely presented $$R$$-modules (by taking finitely generated submodules, and taking more and more (but always finitely many) relations). These finitely presented modules will not in general be submodules of $$M$$.

Question: For what rings $$R$$ can we present any $$R$$-module $$M$$ as a filtered colimit of finitely presented submodules (with transition maps also being injective)?

For example, this is true for Noetherian rings, since then finitely generated and finitely presented is the same thing.

Equivalently, is there a condition on $$R$$ guaranteeing such that for any $$N \subset M$$, $$N$$ being finitely generated, there is an intermediate module $$N'$$, $$N \subset N' \subset M$$, that is finitely presented? If there is such a statement only for selected modules $$M$$, I'm also interested in such partial results.

• Good question. Have you looked at $M=R/I$ for a ideal $I \subseteq R$ that is not f.g.? Commented Oct 24, 2023 at 8:38
• No, I didn't have an idea in that case. Commented Oct 24, 2023 at 8:47

Let's work with left $$R$$-modules. The condition holds iff $$R$$ is left Noetherian. You already remarked the "if" part. Conversely, assume that every left $$R$$-module is the union of its finitely presented submodules (the colimit is just a union here). This applies in particular to every cyclic module. Choose a generator. It must be contained in some finitely presented submodule, hence that submodule is the whole module. It follows that for every left ideal $$I$$ the module $$R/I$$ is finitely presented. This just means (Stacks/0519) that $$I$$ is finitely generated.