# Localizing a quotient ring $A/ \mathfrak{p}$

Let's assume $A$ is a commutative ring with $1$ and $\mathfrak{p} \subset A$ is a prime ideal. We shall consider $A/ \mathfrak{p}$ as an $A$-module, so there is a localization $(A/ \mathfrak{p})_\mathfrak{p}$.

What can we say about this? I know it is isomorphic to $A_\mathfrak{p} \otimes _A A/\mathfrak{p}$, but I was hoping to somehow directly relate it to a quotient of the $A$-module $A_\mathfrak{p}$ (if this is even possible).

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Localization is exact: $$(A/\mathfrak{p})_\mathfrak{p}=A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p}.$$ What do we know about $\mathfrak{p}_\mathfrak{p}$ as an ideal of $A_\mathfrak{p}$? – Pierre-Yves Gaillard Jan 4 '12 at 15:10
You also end up with a field, seeing as you invert everything outside the image of $\mathfrak{p}$ which is $(0)$... – Alex Becker Jan 4 '12 at 15:13
I agree with Alex. Moding out by $\mathfrak p$ kills the prime ideals below $\mathfrak p$. Localizing with respect to $\mathfrak p$ kills the prime ideals above $\mathfrak p$. These two operations commute. – Pierre-Yves Gaillard Jan 4 '12 at 15:18
Dear @Alex: $A/\mathfrak p$ is an $A$-module. As such, it can be localized at $\mathfrak p$. We have $$(A/ \mathfrak{p})_\mathfrak{p}=(A/ \mathfrak{p})_{(0)}.$$ (Canonical isomorphism.) - But your whole point is a great one! – Pierre-Yves Gaillard Jan 4 '12 at 15:37

This is a consequence of the fact that localization is an exact functor (see the comment section for this and other helpful remarks).

Exactness implies that localization (at arbitrary multiplicative subsets) commutes with quotients, which in this case gives us $(A / \mathfrak{p})_\mathfrak{p} \cong A_\mathfrak{p} / \mathfrak{p}_\mathfrak{p}$. This (considered as a ring, not an $A$-module) is also a field.

For the sake of completeness, let $\mathfrak{a} \subset A$ be an ideal and $S \subset A$ some multiplicative subset. The exact sequence

$$0 \rightarrow \mathfrak{a} \rightarrow A \rightarrow A/\mathfrak{a} \rightarrow 0$$

localized at $S$ yields the exact sequence

$$0 \rightarrow S^{-1}\mathfrak{a} \rightarrow S^{-1}A \rightarrow S^{-1}(A/\mathfrak{a}) \rightarrow 0$$

which immediately gives $$S^{-1}(A/\mathfrak{a}) \cong S^{-1}A/S^{-1}\mathfrak{a},$$ as needed.

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