
Proposition 10.9.11. View $S'^{-1}M$ as an $A$-module, then $S^{-1}(S'^{-1}M)$ is isomorphic to $(SS')^{-1}M$.

Proof. Note that given a $A$-module M, we have not proved any universal property for $S^{-1}M$. Hence we cannot reason as in the preceding proof; we have to construct the isomorphism explicitly.

We define the maps as follows

\begin{align*} & f : S^{-1}(S'^{-1}M) \longrightarrow (SS')^{-1}M, \quad \frac{x/s'}{s}\mapsto x/ss'\\ & g : (SS')^{-1}M \longrightarrow S^{-1}(S'^{-1}M), \quad x/t\mapsto \frac{x/s'}{s}\ \text{for some }s\in S, s'\in S', \text{ and } t = ss' \end{align*}

We have to check that these homomorphisms are well-defined, that is, independent the choice of the fraction. This is easily checked and it is also straightforward to show that they are inverse to each other. $\square$

There are also:

• 2 comment(s) on Section 10.9: Localization

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).