How is the canonical localization isomorphism $M_U\cong M\otimes_R R_U$ an isomorphism of $R_U$-modules?

Question

Let $M$ be an $R$-module, and let $U\subset R$ be a multiplicatively closed subset. Let $M_U$ be the localization of $M$ at $U$, and let $R_U$ be the localization of $R$ at $U$. Then we have the "well known" isomorphism $$ M_U\cong M\otimes_R R_U. $$ My question is, how is this an isomorphism of $R_U$ modules? I see how it is an isomorphism of $R$ modules, but not as $R_U$ modules. Mainly, how is $M\otimes_R R_U$ an $R_U$-module? If $\frac{s}{v}\in R_U$ and $m\otimes \frac{r}{u}\in M\otimes_R R_U$, we should be able to scale $m\otimes\frac{r}{u}$ by $\frac{s}{v}$, if this were an $R_U$-module. But by properties of the tensor product, $$ \frac{s}{v}\cdot(m\otimes\frac{r}{u})=(\frac{s}{v}m)\otimes\frac{r}{u}. $$ This makes no sense to me, because $\frac{s}{v}m\not\in M$. What am I missing?

Answer 1

The $R_U$-module structure is given by acting on the right factor. It may look a little nicer to write it as a right action

$$\left( m \otimes \frac{r}{u} \right) \frac{s}{v} = m \otimes \frac{rs}{uv}$$

although of course since all rings are commutative here left and right actions are the same.

In general, if $f : R \to S$ is a ring homomorphism and $M$ is a (right) $R$-module, the extension of scalars $M \otimes_R S$ canonically acquires the structure of a (right) $S$-module. This construction gives a functor $\text{Mod}(R) \to \text{Mod}(S)$ left adjoint to the restriction of scalars functor given by pulling back along $f$.