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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).


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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 $i$. –  Arturo Magidin Jul 19 '12 at 22:17
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! –  oxeimon Jul 21 '12 at 19:15
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. –  user26857 Jul 21 '12 at 20:25
@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 $\overline{A_{\mathfrak{q}}} = (A-\mathfrak{q})^{-1}\overline{A}$. –  Arturo Magidin Jul 21 '12 at 22:08
@Arturo: My comment had to do with the answer he posted. No neded to explain me the difference between $\overline{A}_{\mathfrak{q}}$ and $\overline{A_{\mathfrak{q}}}$. –  user26857 Jul 21 '12 at 23:53
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2 Answers 2

First notice that without extra hypothesis, the integral closure needs not be finite over $A$, and there might be infinitely many prime ideals lying over a given one in $A$. Neverthless, the equality holds.

It is easy to see that the LHS is contained in the RHS because the latter is integrally closed and contained in the field of fractions of $A_q$. Let $f$ be an element of the RHS (possibly infinite intersection). Denote by $B=\overline{A}$. Consider the denominator ideal $$ I=\{ b\in B \mid bf\in B\}$$ of $f$. I claim that $I\cap A$ is not contained in $q$. Admitting this, let $s\in I\cap A\setminus q$, then $sf\in B$ and $f$ is integral over $A_q$.

Now we prove the claim. Suppose that $I\cap A\subseteq q$. Then, geometrically, $q\in V(I\cap A)\subseteq \mathrm{Spec}(A)$. As $A/(I\cap A)\to B/I$ is injective and integral, $\mathrm{Spec}(B/I)\to \mathrm{Spec}(A/(I\cap A))$ is surjective. Identify $V(I)$ with $\mathrm{Spec}(B/I)$, this implies that there exists $p\in V(I)$ such that $p\cap A=q$. So $p$ is a prime ideal of $B$ lying over $q$ and containing $I$. This contradicts the hypothesis $f\in B_p$.

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Nice answer for the general case! –  user26857 Aug 3 '12 at 11:20
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Consider $\dim A=1$. One can assume that $A$ is a local Noetherian domain with maximal ideal $m$ and have to prove that $\overline{A}=\bigcap_{i=1}^n\overline{A}_{P_i}$, where $P_1,\dots,P_n$ are all the prime ideals of $\overline{A}$ lying over $m$. (There are only finitely many primes in $\overline{A}$ lying over $m$ because $\overline{A}$ is Dedekind!) Now one can use the well known fact that an integral domain (in our case $\overline{A}$) is the intersection of all its localizations at maximal ideals. Note that a prime ideal $P$ of $\overline{A}$ is lying over $m$ or over $(0)$. The prime ideals $P$ lying over $(0)$ give "large" localizations, i.e. $\overline{A}_P=K$ when $P\cap A=(0)$, where $K$ is the field of fractions of $A$. In particular these primes can be removed from the intersection finally remaining only the primes $P$ lying over $m$.

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