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

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?

$$\operatorname{Hom}_C(R_1, L_1)$$ 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 $$N$$.
• Thank you for your answer. If $(a_i)_{1\leq i\leq m}$, $(b_i)_{1\leq i\leq n}$ are the bases of $M$ and $N$, respectively, can you please tell me what the isomorphism between $\text{Hom}_C(M,N)$ and $C^{m\cdot n}$ is? I am having difficulty constructing it. – alf262 May 23 '20 at 22:05
• As usual with vectors spaces, the correspondence is an $m{\times}n$ matrix, with column vectors the coordinates of the $f(a_i)$s in basis $(b_j)$. – Bernard May 23 '20 at 22:09