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 $M$. Define
$$
\frac as\cdot m=a\,s_M^{-1}(m).
$$
The 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 inverse
$$
\boxed{\frac ms\longmapsto s_M^{-1}(m),}
$$
so it is an isomorphism.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-101/6/a/solution