Proposition 10.9.14. Let $I$ be an ideal of $A$, $S$ a multiplicative set of $A$. Then $S^{-1}I$ is an ideal of $S^{-1}A$ and $\overline{S}^{-1}(A/I)$ is isomorphic to $S^{-1}A/S^{-1}I$, where $\overline{S}$ is the image of $S$ in $A/I$.
Proof. The fact that $S^{-1}I$ is an ideal is clear since $I$ itself is an ideal. Define
\[ f : S^{-1}A\longrightarrow \overline{S}^{-1}(A/I), \quad x/s\mapsto \overline{x}/\overline{s} \]
where $\overline{x}$ and $\overline{s}$ are the images of $x$ and $s$ in $A/I$. We shall keep similar notations in this proof. This map is well-defined by the universal property of $S^{-1}A$, and $S^{-1}I$ is contained in the kernel of it, therefore it induces a map
\[ \overline{f} : S^{-1}A/S^{-1}I \longrightarrow \overline{S}^{-1}(A/I), \quad \overline{x/s}\mapsto \overline{x}/\overline{s} \]
On the other hand, the map $A \to S^{-1}A/S^{-1}I$ sending $x$ to $\overline{x/1}$ induces a map $A/I \to S^{-1}A/S^{-1}I$ sending $\overline{x}$ to $\overline{x/1}$. The image of $\overline{S}$ is invertible in $S^{-1}A/S^{-1}I$, thus induces a map
\[ g : \overline{S}^{-1}(A/I) \longrightarrow S^{-1}A/S^{-1}I, \quad \frac{\overline{x}}{\overline{s}}\mapsto \overline{x/s} \]
by the universal property. It is then clear that $\overline{f}$ and $g$ are inverse to each other, hence are both isomorphisms. $\square$
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