Proposition 10.9.10. Let $\overline{S}$ be the image of $S$ in $S'^{-1}A$, then $(SS')^{-1}A$ is isomorphic to $\overline{S}^{-1}(S'^{-1}A)$.

**Proof.**
The map sending $x\in A$ to $x/1\in (SS')^{-1}A$ induces a map sending $x/s\in S'^{-1}A$ to $x/s \in (SS')^{-1}A$, by universal property. The image of the elements in $\overline{S}$ are invertible in $(SS')^{-1}A$. By the universal property we get a map $f : \overline{S}^{-1}(S'^{-1}A) \to (SS')^{-1}A$ which maps $(x/t')/(s/s')$ to $(x/t')\cdot (s/s')^{-1}$.

On the other hand, the map from $A$ to $\overline{S}^{-1}(S'^{-1}A)$ sending $x\in A$ to $(x/1)/(1/1)$ also induces a map $g : (SS')^{-1}A \to \overline{S}^{-1}(S'^{-1}A)$ which sends $x/ss'$ to $(x/s')/(s/1)$, by the universal property again. It is immediately checked that $f$ and $g$ are inverse to each other, hence they are both isomorphisms. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: