Let $R$ be a ring and $S\subset R$ a multiplicative set. Then one can define the localization $S^{-1}R$ which is also a ring. Now let $M$ be an $R$-module and $N$ be an $S^{-1}R$ module. Then our prof said that we can view $N$ also as an $R$-module.
I don't see why this is true because $S^{-1}R$ is not necessarily a subring of $R$. Does it mean that we can identify $R$ with $\{1\}^{-1}R$?
Thanks for your help.
Given a homomorphism of rings $\rho: R_1\rightarrow R_2$, then any $R_2$-module $M$ can be viewed as a $R_1$-module through the action $ a\cdot x :=\rho(a) \cdot x$ for $a\in R_1, x\in M$. In this particular case, we do have a canonical morphism $R\rightarrow S^{-1}R$, hence any $S^{-1}R$-module is naturally a $R$-module.
When $\rho$ is injective, we can view $R_1$ as a subring of $R_2$. Surely a subring $R_1$ naturally acts on $R_2$-modules, but this is just a special case of the above fact.