How Should I Prove that the Localization of Quotients Commute from here?

Question

Let S be a multiplicatively closed subset of R, M an R-module, and N a submodule of M. I want to prove that:

S$^{-1}$(M/N) is isomorphic to (S$^{-1}$M)/(S$^{-1}$N).

Attempt: Since $0$$\rightarrow$ N$\xrightarrow{f}$ M$\xrightarrow{g}$ M/N$\rightarrow$ $0$ is exact, then I can show that $0$$\rightarrow$ S$^{-1}$N$\xrightarrow{f^{‘}}$ S$^{-1}$M$\xrightarrow{g^{‘}}$ S$^{-1}$M/N$\rightarrow$ $0$ is exact as well. However, I don’t know how to go from here. That is,

Question: How can I prove that S$^{-1}$(M/N) is isomorphic to (S$^{-1}$M)/(S$^{-1}$N) from here?

Your help would be much appreciated.

Answer 1

Let $0\to M'\xrightarrow{i} M\xrightarrow{e} M''\to 0$ be an exact sequence of modules. Then know that:

• $i$ is injective because its kernel is the image of $0\to M'$, which is the zero submodule of $M'$.
• $e$ is surjective because its image is the kernel of $M''\to 0$, which is all $M''$.

The exactness in $M$ says that $\operatorname{ker}e=\operatorname{im}i$, hence the first isomorphism theorem yields $M''\cong M/\operatorname{im}i$, as $e$ is surjective. But $i$ is injective, so $M'\cong \operatorname{im}i$: you can then view $M'$ itself as a submodule of $M$ and write $M/M'\cong M''$.