$\text{Hom}_S(S\otimes_R M,N)\cong\text{Hom}_R(M,N)$

Question

Given homomorphism $\phi:R\to S$ and $M$ is an $R$-module and $N$ is an $S$-module. We know that an $S$-module can be turn into a an $R$-module naturally by $\phi$. So the role of modules may alter in the equation. It that true? I can easily find a map from the latter to the former but it is not very clear how to find an inverse.

Answer 1

Given $\DeclareMathOperator{\Hom}{Hom} \psi \in \Hom_S(S \otimes_R M, N)$, consider the map $\iota: M \to S \otimes_R M$ given by $\iota(m) = 1 \otimes m$. Then we get a map \begin{align*} F: \Hom_S(S \otimes_R M, N) &\to \Hom_R(M,N)\\ \psi &\mapsto \psi \circ \iota \, . \end{align*}

Conversely, given $\rho \in \Hom_R(M,N)$, define $\widetilde{\rho}: S \otimes_R M \to N$ by $\widetilde{\rho}(s \otimes m) = s \rho(m)$ and extending linearly. Denote this map $G: \Hom_R(M,N) \to \Hom_S(S \otimes_R M, N)$. I'll leave it to you to show that these associations are mutually inverse.