This is the problem I want to solve :
Show that for any finitely generated projective module $P$ over a ring $R $, $\mathrm{Hom}(P,M)$ is isomorphic with $\mathrm{Hom}(P,R)\otimes M $.
This is what I've done:
I define a map : $Hom (P,R) × M \rightarrow Hom (P,M) $, with the law: $(\phi,m)\rightarrow \phi_m $, $\phi_m (x):=\phi(x) m $.
This map is obviously bilinear, so we have a map $Hom (P,R) \otimes M \rightarrow Hom (P,M) $, with the law $\phi\otimes m \rightarrow \phi_m $, $\phi_m (x):=\phi(x) m $.
Now I have to construct the inverse homomorphism to end the proof.
Because $P $ is finitely generated, we can assume $P=R^n $. I define a map:
$Hom (P,M) \rightarrow Hom (P,R)\otimes M $ that sends $f $ to $f'\otimes f (1) $, when $1$ is the identity element of $ P=R^n $. I don't know how to define $f'$.
Is there any hint?
If you don't agree with these, do you have another idea?
Thanks.
Hint: This is where you need to use finite generation and projectiveness of $P$. Since $P$ is finitely generated, there is a surjective $R$-map $p : R^n \to P$ for some $n > 0$. Projectiveness of $P$ yields a section $s : P \to R^n$ of $p$ (so that $p\circ s = \operatorname{id}_P$). For $1\le i\le n$, let $s_i$ be the composition $P \xrightarrow{s} R^n \xrightarrow{\pi_i} R$, where $\pi_i$ is the projection onto the $i$th coordinate. Now we define an $R$-map $$\operatorname{Hom}(P,M) \to \operatorname{Hom}(P,R) \otimes M$$ by sending a $g\in \operatorname{Hom}(P,M)$ to $\sum s_i \otimes g(p(e_i))$, where $e_1,\ldots, e_n$ is the standard basis of $R^n$.