Question about integral closures and localizations 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.
 A: 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 $f\in B_p$.
A: 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 $m$.
