When is $\operatorname{Hom}_A(M,N)$ finitely generated? 
Let $A$ be a ring that is not a field and $M$, $N$ modules over $A$. What conditions on $M,N$ (and maybe also on $A$) guarantee that $\operatorname{Hom}_A(M,N)$ is finitely generated? Is it enough to assume $M,N$ finitely generated?

 A: This has been asked before, but I cannot find it right now, so let me just answer it.
If $M$ is finitely generated, we have an epimorphism $A^n \to M$ for some $n$. This induces a monomorphism $\hom_A(M,N) \hookrightarrow \hom_A(A^n,N) = N^n$. Submodules of finitely generated modules don't have to be finitely generated (think of non-noetherian rings), but this holds by definition for noetherian modules. It is known that noetherian modules are closed under direct sums. This proves:

If $M$ is finitely generated and $N$ is noetherian, then $\hom_A(M,N)$ is finitely generated.

In particular:

If $A$ is noetherian and $M,N$ are finitely generated, then $\hom_A(M,N)$ is finitely generated.

Without these assumptions $\hom_A(M,N)$ doesn't have to be finitely generated:


*

*If $M=A/I$ and $N=A$, then $\hom_A(M,N) \cong \mathrm{Ann}(I)$ is an ideal of $A$ which has no reason to be finitely generated when $A$ is not noetherian (Exercise: find an explicit example).

*If $M=\oplus_{n \geq 0} A$ and $N=A$, then $\hom_A(M,N) \cong \prod_{n \geq 0} A$ is not finitely generated (unless $A=0$).
