Let $R$ be a ring (commutative and with unity), $S\subset R$ be a subring. Consider three $R$-modules $M$, $N$ and $Z$.

Let $\operatorname{Hom}_R(M\otimes_RN;Z)$ be the R-module of $R$-bilinear maps $\phi:M\otimes_R N\rightarrow Z$.

Let $\operatorname{Hom}_S(M_S\otimes_SN_S;Z_S)$ be the $S$-module of $S$-bilinear maps $\varphi:M_S \otimes_S N_S \rightarrow Z_S$, where $M_S$ denotes scalar restriction.

-There always exist a canonical way to take "scalar restriction" of a $R$-bilinear maps? More precisely, there exists some canonical $S$-linear map $$\operatorname{Hom}_R(M\otimes_RN;Z)_S \rightarrow \operatorname{Hom}_S(M_S\otimes_SN_S;Z_S) ?$$

  • if not, what are the most general situations in which such canonical map exists?

P.S: we clearly have a function $$ \operatorname{Hom}_R(M\otimes_RN;Z) \rightarrow \operatorname{Hom}_S((M\otimes_R N)_S;Z_S) $$ arising from the fact that scalar restriction is functorial. Thus, my main question should be:

  • under which conditions there is a canonical map $$ M_S\otimes_S N_S \rightarrow (M\otimes_R N)_S? $$
  • In other words, under which conditons the scalar restriction functor $\mathbf{Mod}_R\rightarrow \mathbf{Mod}_S$ is a lax monoidal functor?

P.S: Since scalar restriction is a right-adjoint to extension of scalars, by very fact that in an adjunction if the left-adjoint is strong monoidal then the right-ajoint is lax monoidal (see here), it is suffices to prove that scalar extension $\mathbf{Mod}_S\rightarrow \mathbf{Mod}_R$ is strong monoidal. Since we are in commutative rings, the tensor product is commutative up to isomorphisms and it seems to me that this is enough to ensure that scalar extension is strong monoidal. Thus, a priori I have a proof of the desired fact in my mind. It happens that looking at the first paragraph of page 382 of this book, the author adds a strong requirement on the rings $R$ and $S$ in order to take the restriction of a bilinear map. This make me a bit confusing and insecure with my arguments.

If this is obvious an I'm forgetting something, I'm sorry and just let me know.


Edit: By "canonical" I meant some map arising from a natural transforation.


1 Answer 1


This is much easier if you are dealing with the (equivalent) actual bilinear maps $M\oplus N\to Z$, for if such a map is $R$-bilinear, via the identification $(M\oplus N)_S=M_S\oplus N_S$, it is easily seen to be $S$-bilinear, too. So there is your natural map $\mathrm{Bil}_R(M\oplus N,Z)\to \mathrm{Bil}_S(M_S\oplus N_S,Z_S)$.

If you don't want to do the elementary verification, here's another proof: We have to show that the obvious map $M_S\otimes_S N_S\to (M\otimes_RN)_S$ is well-defined. Considering the co-unit of the scalar extension- and restriction-adjunction, $R\otimes_SM_S\to M$ and analogously for $N$, we get an $R$-linear map $$R\otimes_SM_S\otimes_SN_S=(R\otimes_SM_S)\otimes_R (R\otimes_S N_S)\to M\otimes_RN,$$ mapping $1_R\otimes m\otimes n$ to $m\otimes n$. Thus, via the adjunction $\hom_R(R\otimes_SM_S\otimes_SN_S,M\otimes_RN)=\hom_S(M_S\otimes_SN_S,(M\otimes_RN)_S)$, we find the obviously defined $S$-linear map $M_S\otimes_SN_S\to(M\otimes_R N)_S$, as required.

  • $\begingroup$ Than you very much @Ben. $\endgroup$ Jun 18, 2020 at 14:57

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