Restriction of scalars is special case of internal hom Since restriction of scalars is right adjoint to extension of scalars, which is just a tensor product, it must be (isomorphic to) the internal hom. I'm having trouble showing this is the case:
Let $f:R\rightarrow S$ be an arrow in $\mathsf{Rng}$. Let $M$ be a left $S$-module. How can I prove its restriction of scalars to $R$, denoted by $\bar M$, is (naturally isomorphic to) the internal hom $\mathsf{hom}(R,M)$? I can't even think of the isomorphism...
I am also confused because the general tensor hom adjunction for monoidal categories is an adjunction of endofunctors, and here, it seems $-\otimes S$ should somehow be adjoint to $\mathsf{hom}(R,-)$...
 A: There is a more general tensor-hom adjunction for bimodules. It is important, because if the rings are not commutative, the tensor product does not stay in a fixed category of module over a ring. More precisely, you need a right $B$-module $M$ and a left $B$-module $N$ to form the tensor product $M\otimes_B N$. And this is not a $B$-module anymore. However, if $M$ or $N$ (or both) carry another ring structure, then so does $M\otimes_B N$. Hence if $N$ is a $(B,C)$-bimodule, then we have for any ring $A$ a functor
$$\cdot\otimes_B N : (A,B)-\operatorname{bimod}\rightarrow(A,C)-\operatorname{bimod}.$$
Similarly, if $N$ and $P$ are right $C$-modules, then $\operatorname{Hom}_C(N,P)$ is not a $C$-module anymore. However, if $N$ carries a left $B$-module structure, then $\operatorname{Hom}_C(N,P)$ carries a right $B$-module structure, and if $P$ carries a left $D$-module structure, so does $\operatorname{Hom}_C(N,P)$. So in general, we have a functor for any ring $D$
$$\operatorname{Hom}_C(N,\cdot):(D,C)-\operatorname{bimod}\rightarrow(D,B)-\operatorname{bimod}$$
In this context, the tensor-hom adjunction is the following : let $M$ be a $(A,B)$-bimodule, $N$ a $(B,C)$-bimodule and $P$ a $(D,C)$-bimodule, then :
$$\operatorname{Hom}_C(M\otimes_B N,P)=\operatorname{Hom}_B(M,\operatorname{Hom}_C(N,P))$$
natural as $(D,A)$-bimodules in $M,N,P$.
The extension/restriction of scalar adjunction is indeed a special case of the tensor-hom. Let $f:R\rightarrow S$ a morphism of rings. Consider $S$ as a left $R$-module, and right $S$-module. Then for any right $S$-module $N$, $\operatorname{Hom}_S(S,N)$ is a right $R$-module and
$$ \operatorname{Hom}_S(M\otimes_R S,N)=\operatorname{Hom}_R(M,\operatorname{Hom}_S(S,N))$$
for any right $R$-module $M$.
Finally, it is easy to show that $\operatorname{Hom}_S(S,N)=N$ as right $R$-module.
