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.