Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 101 / 6

Restriction of scalars makes every $S^{-1}A$-module an $A$-module, and multiplication by $s\in S$ is invertible with inverse multiplication by $1/s$.

Conversely, suppose each multiplication map $s_M:m\mapsto sm$ is an automorphism of the $A$-module $M$. DefineThe universal property of localization shows that this is well-defined and gives the unique $S^{-1}A$-module structure extending the $A$-action. These constructions are inverse.

Under this structure the natural map $M\to S^{-1}M$ has inverseso it is an isomorphism.

Localization of the inclusion $N\hookrightarrow M$ gives an injection $S^{-1}N\hookrightarrow S^{-1}M$, whose image lies in $N^{\prime }$. Conversely, take $m/s\in N^{\prime }$. Since $s/1$ is a unit and $N^{\prime }$ is an $S^{-1}A$-submodule,By the definition of the inverse image, $m\in N$, and hence $m/s$ lies in the image of $S^{-1}N$. Therefore