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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 $A_\mathfrak{p}$?2012-01-04
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    You also end up with a field, seeing as you invert everything outside the image of $\mathfrak{p}$ which is $(0)$...2012-01-04
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    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.2012-01-04
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    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!2012-01-04
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    I would write an element of $A/\mathfrak p$ as $x + \mathfrak p$ in the first place (at least if we consider it a ring in and of itself), so the additive identity would be written as $0 + \mathfrak p$ in my book, or simply $\mathfrak p$ by abuse of notation. Writing it as $0$ would be an abuse of notation the other way, and is completely fine, but I wouldn't do it.2012-01-04
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    @Pierre-YvesGaillard Right, of course. I was thinking of $A$ as the ring containing $\mathfrak{p}$.2012-01-04
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    @Pierre-YvesGaillard: That's the equality I needed, thanks. Of course $\mathfrak{p} _\mathfrak{p} \subset A_\mathfrak{p}$ is the unique maximal ideal of $A_\mathfrak{p}$, hence we end up with a field like Alex said. Now I just have to see how exactness of localization implies that localization commutes with quotients, but I'm sure this is easy. Thanks again2012-01-04
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    Alright, I see how exactness of $S^{-1}$ implies that localization commutes with quotients. How is this technically open question handled, considering that there is no answer? Should I delete it?2012-01-04
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    Dear @Paul: You're welcome. I suggest that, when you understand everything, and if nobody has answered your question, you answer it yourself. - I've just seen your above comment. Please, don't delete your question, and answer it. It will be a service to everybody! (I upvoted your question, and am ready to upvote your answer.)2012-01-04

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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.