Proof needed for $\operatorname{Hom}_R(M,N) \otimes_RS \cong \operatorname{Hom}_S(M\otimes_R S,N\otimes_R S)$ I want a proof for
$$\operatorname{Hom}_R(M,N) \otimes_RS \cong \operatorname{Hom}_S(M\otimes_R S,N\otimes_R S)$$
where $\phi\colon R \to S$ is a homomorphism and $M$ is finitely generated free $R$-module and $N$ is an $R$-module.
or if anyone can say me about the isomorphism map between two side.
 A: *

*Since $M$ is a finitely generated free $R$-module, one has $M\cong R^n$ for some $n$ (this statement is equivalent to choosing a basis $e_1,\ldots, e_n$ for $M$).


*Moreover, since $M$ is free, any homomorphism $\phi\colon M\to N$ can be specified completely by specifying the images $\phi(e_1),\ldots, \phi(e_n)$ of the basis vectors. This says exactly that $\operatorname{Hom}(M,N) \cong N^n$. Thus the left hand side of the desired isomorphism is
$$\operatorname{Hom}(M,N)\otimes_RS\cong N^n\otimes_R S.$$


*Since $M\cong R^n$ and tensor product commutes with direct sums, we have that $M\otimes_RS\cong S^n$. In terms of our previously chosen basis $e_1,\ldots, e_n$ of $M$, this statement says simply that $M\otimes_R S$ is a free $S$-module with basis $e_1\otimes 1,\ldots, e_n\otimes 1$.


*Just as in point (2), since $M\otimes_R S$ is a free $S$-module, specifying a homomorphism $\phi\colon M\otimes_R S\to N\otimes_RS$ is equivalent to specifying the images $\phi(e_1\otimes 1),\ldots, \phi(e_n\otimes 1)$ of the basis vectors of $M\otimes S$. Thus $$\operatorname{Hom}_S(M\otimes_R S, N\otimes_R S)\cong (N\otimes_RS)^n.$$
Thus, combining points (2) and (4), in order to prove the desired isomorphism it suffices to show that $N^n\otimes_R S\cong (N\otimes_R S)^n$. But this is exactly the fact that tensor product commutes with direct sums. This completes the proof.
It is possible to give a simple description of the isomorphism $$f\colon \operatorname{Hom}(M,N)\otimes_R S\to \operatorname{Hom}_S(M\otimes_R S, N\otimes_R S)$$ we just constructed. It is given on simple tensors by $$f(\phi\otimes s)(m\otimes t) = \phi(m)\otimes st.$$ Despite the simplicity of this definition, it is not clear that $f$ is indeed an isomorphism. The above argument proves that $f$ is an isomorphism when $M$ is a finitely generated free $R$-module.
