Let $M,N$ be finite-rank free $R$-modules and Hom$(M,N)$ the set of homomorphisms from $M$ to $N$, and $R$ a commutative ring with unity. I know that Hom$(M,N)$ is an $R$-module where sums and scalar products of functions are defined in terms of sums and scalar products of their function values. I need to further prove that Hom$(M,N)$ is a free module and I'm trying to come up with a basis.
Let's call a basis of $M$ the ordered collection of elements $\mathcal{B}=\{v_1,v_2,...,v_n\}$ and $\mathcal{C}=\{w_1,w_2,...,w_m\}$ a basis for $N$. My first hypothesis is that every homomorphism can be generated as a linear combination of the homomorphisms $\phi$ such that the value of $\phi(v_i)$ is chosen from $\mathcal{C}$. To prove this consider an arbitrary homomorphism $\psi$ and consider its mapping on the bases, $\psi(v_i)=x_i=a_{i,1}w_1+...+a_{i,m}w_m$. The set of all $n$ such mappings determines an $n\times m$ matrix of coefficients, $A$. Somehow I need to use this matrix to determine a linear combination of homomorphisms as described above, but I just can't see how to do that. Or perhaps I'm just fundamentally working down a dead-end path, I can't really tell.
You lose no generality by taking $M=R^m$ and $N=R^n$. Then $\textrm{Hom}(R^m,R^n)$ can be expressed as the collection of $m\times n$ matrices over $R$, which is indeed a free rank $mn$ $R$-module.