Isomorphism of the tensor product of homomorphisms If $M$ and $N$ are finitely generated free modules of the ring $A$ then $ (Hom_A(M,A) \otimes_A N) \cong Hom_A(M,N)$ . Ive trying to prove this and i tried creating two maps and prove that they are inverse to on another , i created one that sends $\phi \otimes_A n$ to $\phi(x) n$ but i cant seem to make the other one. I also tried using the fact that $M$ and $N$ $\cong A^k$ for some $k$ but i didnt make any progress.
Thanks in Advance.
 A: $\DeclareMathOperator{\Hom}{Hom}$
This actually works for arbitrary $M$, though $N$ still has to be finitely generated and free. $N$ is a finitely generated free module hence it has a basis $n_1,\ldots,n_k$. An arbitrary element of $\Hom_A(M,A)\otimes_A N$ is of the form
$$\sum_{i=1}^k{\phi_i\otimes n_i}$$
If you just send $\phi$ to $\phi\otimes n$, this is not the whole story. Recall that the tensor product does not usually consist only of simple tensors. So you really do end up with $k$ different homomorphisms, one for each basis element of $N$.
Now, an element of $\Hom_A(M,N)$ is a homomorphism $\phi:M\to N$. We can project it onto the cyclic submodule generated by the basis element $n_i$ to obtain a homomorphism $\phi_i:M\to A$, so that if $\phi(m)=\sum_i{a_in_i}$, then
$$\phi_i(m) = a_i$$
Hence
$$\phi(m)=\sum_i{\phi_i(m)n_i}$$
Now there is a map $\Hom_A(M,A)\otimes_A N\to \Hom_A(M,N)$ that is the homomorphism determined by
$$F\sum_i\phi_i\otimes n_i = \left(m\mapsto \sum_i{\phi_i(m)n_i}\right)$$
This is well defined by the universal property of the tensor product as we can derive it from a bilinear map $\Hom_A(M,A)\times N$, and by the characterization of $\Hom_A(M,N)$ it is surjective. What is not immediately clear is that the kernel is trivial. But this follows from the fact that $\{n_1,\ldots,n_k\}$ is a basis of $N$: if a linear combination of these elements is $0$, then all coefficients are $0$. Hence each $\phi_i$ is $0$ for all $m$, so the element in the domain is $0$ as well.
