Is $\operatorname{Hom} _A(M,N)$ a finitely generated $A$-module?

Question

If $M,N$ are finitely generated $A$-modules, is $\operatorname{Hom} _A(M,N)$ a finitely generated $A$-module?

Answer 1

If $N$ is Noetherian, then there is an injective embedding $\operatorname{Hom}_A(M,N) \rightarrow N^G$ for some finite $G \subset M$, and $N^G$ is Noetherian, so $\operatorname{Hom}_A(M,N)$ is Noetherian.

On the other hand, weird things can happen in the non-Noetherian case.

Consider $N=A$, and define, for all ideals $I \subset A$, $I^{\perp}=\{x \in A,\, \forall y \in I,\, xy=0\}$.

Now, there are isomorphisms $\operatorname{Hom}_A(A/I,A)=\{f(x)=ax,\,a \in A, aI=0\}=I^{\perp}$.

If for some $I \subset A$, $I^{\perp}$ is not finitely generated, $M=A/I$, $N=A$ is a counterexample.

Now consider the ring $A=\mathbb{Z}[(X_i)_i]/\langle (X_i^2),(X_iX_jX_k)_{i,j,k}\rangle$, and $I=\langle x_1 \rangle$, $I^{\perp}=Ax_1 \oplus \bigoplus_{j > i > 1}(Ax_ix_j)$ is not finitely generated.