The module $\text{Hom}_C(E,F)$ of two finitely generated projective $C$-modules

Question

Let $C$ be an abelian ring and $E,F$ two finitely generated projective modules. Then $\text{Hom}_C(E,F)$ is a finitely generated projective $C$-module.

First of all, since $C$ is abelian, the abelian group $\text{Hom}_C(E,F)$ is a $C$-module. As $E,F$ are finitely generated projective $C$-modules, there exist free $C$-modules $M,N$, with finite bases, such that $E,F$ are isomorphic, respectively, to direct factors of $M,N$.

So, let $R_1,R_2$ be supplementary submodules of $M$ such that $E\simeq R_1$. On the other hand, let $L_1,L_2$ be supplementary submodules of $N$ such that $F\simeq L_1$. Furthermore, $\text{Hom}_C\left(R_1\oplus R_2,L_1\oplus L_2\right)$ is isomorphic to $$\text{Hom}_C(R_1,L_1)\oplus\text{Hom}_C(R_1,L_2)\oplus\text{Hom}_C(R_2,L_1)\oplus\text{Hom}_C(R_2,L_2).$$

Also, $\text{Hom}_C(E,F)\simeq\text{Hom}_C(R_1,L_1)$.

I am not sure how I can deduce that (i) $\text{Hom}_C(E,F)$ is finitely generated and (ii) $\text{Hom}_C(E,F)$ is projective from the above.

Any hints?

Answer 1

Hint:

$\operatorname{Hom}_C(R_1, L_1)$ is a direct summand of $\operatorname{Hom}_C(M, N)$ , which is isomorphic to $C^{m\cdot n}$, if $m$ is the rank of the free module $M$ and $n$ the rank of the free module $N$.