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Suppose $A$ is an integral domain with integral closure $\overline{A}$ (inside its fraction field), $\mathfrak{q}$ is a prime ideal of $A$, and $\mathfrak{P}_1,\ldots,\mathfrak{P}_k$ are the prime ideals of $\overline{A}$ lying over $\mathfrak{q}$. Show that $\overline{A_\mathfrak{q}} = \bigcap\overline{A}_\mathfrak{P_i}$ (note that the LHS is the integral closure of a localization, whereas the RHS is the intersection of localizations of integral closures of $A$).

If it would help, I suppose we could assume that $A$ has dimension 1, so that $\overline{A}$ is Dedekind, though I don't think that assumption is required.

(Geometrically, we're comparing the integral closure of a local ring at a singular point Q of some variety with the intersection of local rings at points in the normalization mapping to Q).

Thanks.

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    Don't know if this will suffice, but remember that if $A\subseteq C$, and $B$ is the integral closure of $A$ in $C$, then for any multiplicative subset $S$ of $A$, the integral closure of $S^{-1}A$ in $S^{-1}C$ is $S^{-1}B$. Thus, $$\overline{A_{\mathfrak{q}}} = (A-\mathfrak{q})^{-1}\overline{A}.$$ This gives you the $\subseteq$ inclusion easily, since $A-\mathfrak{q}\subseteq \overline{A}-\mathfrak{p}_i$ for each $i$.2012-07-19
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    Actually Arturo, that's all I need, since if $S = A\setminus\mathfrak{q}$, then $S^{-1}\overline{A}$ is an integral domain, and hence it's the intersection of all its localizations at all maximal ideals. But in this case, it's easy to see that the maximal ideals of $S^{-1}\overline{A}$ are exactly the maximal ideals of $\overline{A}$ lying above $A$. thanks!2012-07-21
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    It seem that you confused us by writing $\overline{A_{\mathfrak{q}}}$ instead of $\overline{A}_{\mathfrak{q}}$. But if you really meant in the LHS the integral closure of $A_{\mathfrak{q}}$, then your answer is wrong.2012-07-21
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    @navigetor23: I think he was quite explicit on what he meant, even added the comment to note that it was the integral closure of the localization, and not the localization of the integral closure; it's important, because $\overline{A}_{\mathfrak{q}} = (\overline{A}-\mathfrak{q})^{-1}\overline{A}$, but $\overline{A_{\mathfrak{q}}} = (A-\mathfrak{q})^{-1}\overline{A}$.2012-07-21
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    @Arturo: My comment had to do with the answer he posted. No neded to explain me the difference between $\overline{A}_{\mathfrak{q}}$ and $\overline{A_{\mathfrak{q}}}$.2012-07-21
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    @oxeimon: Why call this question "stupid"?2012-07-22
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    @navigetor23: Well, *I'm* confused now, at any rate...2012-07-22

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