Let $M, P$ be two finitely generated $A$-modules ($A$ is a commutative ring with $1$). Let also $P$ be projective. I have to prove that Hom$_A(P,M)$ is finitely generated as $A$-module.
Before to face this exercise I had to prove that for each $n\in\mathbb{N}_+$ the $A$-module Hom$_A(A^n,M)$ is finitely generated.
I proved that observing that Hom$_A(A^n,M)\cong M^n$ wich is obviously finitely generated.
Now being $P$ projective i thought to exploit exacr sequences.
If I had an exact sequence
$$0\rightarrow \mbox{Hom}_A(P,M)\rightarrow \mbox{Hom}_A(A^k,M)\rightarrow P\rightarrow 0$$
where $k$ is the number of generator of $P$, I would conlcude that
$$\mbox{Hom}_A(A^k,M)\cong P\oplus\mbox{Hom}_A(P,M)$$
and it would follow that Hom$_A(P,M)$ is finitely generated.
Unfortunately I have not been able to define the exact sequence that I stated above.
Am I on the wrong way?
We have a splitting short exact sequence
$$0 \to Q \to A^n \to P \to 0.$$
Apply $\operatorname{Hom}(-,M)$ to get a splitting short exact sequence
$$0 \to \operatorname{Hom}(P,M) \to \operatorname{Hom}(A^n,M) \to \operatorname{Hom}(Q,M) \to 0.$$