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Let $h\colon R\rightarrow S$ be a morphism of commutative ring. Let $M$ and $N$ be $R$-modules. We consider the canonical morphism of $R$-modules $$p\colon{\rm Hom}_R(S\otimes_RM,N)\rightarrow{\rm Hom}_R({\rm Hom}_R(S,M),N),\;u\mapsto(v\mapsto u(1_S\otimes v(1_S))).$$ Since source and target of this morphism can be canonically furnished with structures of $S$-modules, we may wonder:

Under which conditions is $p$ a morphism of $S$-modules?

I know for example that this is the case if $M$ is free of finite rank, or if $h$ is surjective and $M=S$, and I have the feeling (but no proof!) that it is not so in general.

(Motivation: This morphism and the question of its $S$-linearity showed up while I was trying to understand the behaviour of scalar coextension with respect to Hom functors.)

(Commutativity is probably not strictly necessary here, hence the tagging.)

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  • $\begingroup$ Dear @Uriya, I do not understand your comment. My question is not about $p$ being an isomorphism (in which category?), but about $p$ being $S$-linear. Your reduction shows that if $q$ is $S$-linear, then so is $p$. Can you clarify? $\endgroup$
    – Fred Rohrer
    Commented Aug 28, 2018 at 12:11
  • $\begingroup$ Dear @Fred. I apologize. I misunderstood and though you wanted $p$ to be an isomorphism in addition to being $S$-module homomorphism. I will delete or replace my comment. $\endgroup$
    – Uriya First
    Commented Aug 28, 2018 at 13:27
  • $\begingroup$ [Revised comment. Original deleted.] The map $p$ that you are considering is $\mathrm{Hom}_R(q,N)$, where $q:\mathrm{Hom}_R(S,M)→S\otimes_R M$ is defined by $q(v)=1_S\otimes v(1_S)$. Thus, a sufficient condition for $p$ to be $S$-linear is that $q:\mathrm{Hom}_R(S,M)→S\otimes_RM$ is $S$-linear. $\endgroup$
    – Uriya First
    Commented Aug 28, 2018 at 13:32

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