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 $\mathsf{hom}(R,-)$...


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.

  • $\begingroup$ So what's the (co)domain of this adjunction? Does it go between some categories of bimodules? Why does the isomorphism $\operatorname{Hom}_C(M\otimes_B N,P)=\operatorname{Hom}_B(M,\operatorname{Hom}_C(N,P))$ of hom-sets actually lift to an isomorphism of $(D,A)$-bimodules as you say? $\endgroup$ – user153312 Nov 8 '15 at 16:49
  • $\begingroup$ @Exterior Yes, but not endofunctors. I should have written this, $\otimes_B N$ is a functor from $(A,B)$-bimodules to $(A,C)$-bimodules. And $\operatorname{Hom}_C(N,\cdot)$ from $(D,C)$-bimodules to $(D,B)$-bimodules. Forget the extra-structure (the $A$ and $D$ modules structures), so you get a real adjunction. This extra-structure give then a natural isomorphism in the category of $(D,A)$-bimodules instead of just sets. $\endgroup$ – Roland Nov 8 '15 at 16:56
  • $\begingroup$ Wonderful, thanks for the answer! $\endgroup$ – user153312 Nov 8 '15 at 16:57
  • $\begingroup$ Yes they are natural in $M$, $N$ and $P$. The point is, it is natural in the category of $(D,A)$-modules, so it is even better than just sets. And $\otimes_B N$, $\operatorname{Hom}_C(N,\cdot)$ are really an adjunction between right $B$-modules and right $C$-modules. $\endgroup$ – Roland Nov 8 '15 at 17:12
  • $\begingroup$ Thanks again. I deleted my comment because I understood that right after I asked it, sorry. $\endgroup$ – user153312 Nov 8 '15 at 17:13

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