0
$\begingroup$

Let $A$ be a commutative Noetherian ring with identity and $M,N$ two finitely generated $A$-modules.

How to show $\operatorname{Hom}_A(M,N)$ is a finitely generated $A$-module?

$\endgroup$
2

1 Answer 1

4
$\begingroup$

Consider the surjection $A^n \to M$. This induces an injection $Hom_A(M,N) \to Hom_A(A^n, N) = N^n$. So this is a submodule of a finitely generated module and by the Noetherian condition, we are done.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .