TITLE: restriction of scalars, reference or suggestion for proof
QUESTION [6 upvotes]: Let $f:R \rightarrow R'$ be a ring homomorphism that is epic in the category of rings. Let $M,N$ be $R'$-modules. Why is it that a homorphism $h:M \rightarrow N$ is $R'$-linear if and only if it is $R$-linear? I have used this result in specific cases in the past (and know why it's true),e.g., if $f$ is the usual map of $R$ to $S^{-1}R$,$S$ a multiplicatively closed subset of $R$, but never in the generality I stated. But now I need the general result and would like to know why it works. A reference or a sketch for a proof would be appreciated.

REPLY [2 votes]: Here's an elementary approach, following this answer.
Claim. For any ring morphism $f:R\to S$, restriction of scalars is faithful.
Proof. We have a commutative triangle formed by restriction of scalars $f^\ast$ and each forgetful functor from modules to abelian groups. Each of the latter is faithful, and if $G\circ F$ is faithful so is $F$.
Proposition. If $f:R\twoheadrightarrow S$ is a commutative ring epimorphism then restriction of scalars $f^\ast$ is full.
Proof. Let $X,Y$ be two $S$-modules and write $f^\ast X,f^\ast Y$ for their pulled back $R$-module structures. We want to prove any $R$-linear map $\varphi:f^\ast X\to f^\ast Y$ satisfies $s\varphi(x)=\varphi (sx)$. We shall do this universally by considering all such morphisms at once.
Recall ${}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y)$ is an abelian group. As any abelian group, we may consider its endomorphism ring.
This endomorphism ring admits two ring morphisms from $S$, $$S\rightrightarrows \mathsf{Ab}({}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y),{}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y))$$
with one given by $s\mapsto (\varphi(x)\mapsto s \varphi(x))$ and the other by $s\mapsto (\varphi(x) \mapsto \varphi (sx))$. These are ring morphisms because $ X,Y$ are $S$-modules. We want to show these two ring morphisms coincide.
I claim these ring morphisms coincide upon precomposing with $f:R\to S$.
$$R\to S\rightrightarrows \mathsf{Ab}({}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y),{}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y))$$
Indeed this amounts to the equation $f(r)\varphi(x)=\varphi(f(r)x)$ which holds because $X,Y$ are $S$-modules. We now apply the fact $f$ is epic to conclude the two of $S$-actions were already equal, as desired.
Corollary. If $f:R\twoheadrightarrow S$ is a commutative ring epimorphism then restriction of scalars $f^\ast$ is fully faithful (consequently also conservative).