Suppose $R$ is a commutative ring and $P$ and $Q$ are projective and finitly generated $R$-modules. Prove that $\operatorname{Hom}_{R}(P,Q)$ is projective and finitely generated.
suppose I proved that $Hom_{R}(P,Q)$ is finitely generated thus it's generators will be $\left \langle \varphi_{1} ,\varphi_{2} ,...\varphi_{s} \right \rangle$,consider this diagram:
for each $\varphi _{i}$ , $1\leq i\leq s$ we have $f(\varphi _{i})=b_{i}, b_{i}\in B$.
because $g$ is surjective for these $f(\varphi _{i})$ there are $a_{i}\in A$ that $g(a_{i})=f(\varphi _{i})$ .
I define $h(\varphi _{i})=a_{i}$ so $g(h(\varphi _{i}))=g(a_{i})=f(\varphi _{i})$ thus $Hom_{R}(P,Q)$ is projective.
I think this solution is right. now I don't know how I must show that $Hom_{R}(P,Q)$ is finitely generated.
a hint or a suggestion will be great.
If $P$ is projective and finitely generated then $P$ is a direct summand in a free module of finite rank, say $n$. Then $\operatorname{Hom}_R(P,Q)$ is a direct summand of $Q^n$, so it is projective and finitely generated. (If the conclusion is too abrupt you can do the same for $Q$: it is a direct summand in a free module of finite rank, say $m$. In the end, $\operatorname{Hom}_R(P,Q)$ is a direct summand of $R^{mn}$.)