$\operatorname{Hom}_{R}(M, N) \cong \operatorname{Hom}_{S}(S \otimes_{R} M, N)$

Question

Let $f: R \rightarrow S$ a ring homomorphism.

We can consider any $S$-module $N$ as $R$-module: $r \cdot n := f(r) \cdot n$

Let $S$, $R$ be rings, $M$ be a left $R$-module, $N$ a left $S$-module

How can I build an isomorphism of these two modules?

$$\operatorname{Hom}_{R}(M, N) \cong \operatorname{Hom}_{S}(S \otimes_{R} M, N)$$

Answer 1

Hint:

Observe $\operatorname{Hom}_R(M,N)$ is an $S$-module and use the universal property of tensor products: $$\operatorname{Hom}_S(S\otimes_RM,N)\simeq\operatorname{Hom}_S\bigl(S,\operatorname{Hom}_R(M,N)\bigr)\simeq\operatorname{Hom}_R(M,N).$$