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Let $L/K$ be a finite field extension. Let $\mathcal{O}$ be a valuation ring of $K$. Let $R$ be the integral closure of $\mathcal{O}$ in $L$. Why is $R$ the intersection of all valuation rings of $L$ that lie above $\mathcal{O}$? (A valuation ring $B$ of $L$ lies above $\mathcal{O}$ if $B\cap K=\mathcal{O}$)

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up vote 5 down vote accepted

Valuation rings are integrally closed, so the integral closure of $\mathcal{O}$ is contained in each valuation ring, hence in the intersection.

Conversely, let $x\in L$ be an element that is not integral over $\mathcal{O}$; then $x\notin \mathcal{O}[x^{-1}]$. Thus, $x^{-1}$ is not a unit in $\mathcal{O}[x^{-1}]$, so there is a maximal ideal $\mathfrak{M}$ of $\mathcal{O}[x^{-1}]$ that contains $x^{-1}$.

Consider the map $\mathcal{O}[x^{-1}]\to \mathcal{O}[x^{-1}]/\mathfrak{M}$, and then map the image of $\mathcal{O}$ (via the natural inclusion) into an algebraic closure; you can extend this map to some valuation ring $B$ of $L$ that lies over $\mathcal{O}$ in such a way that the kernel is precisely the maximal ideal of $B$; and since $x^{-1}$ maps to $0$, then $x\notin B$ (otherwise, you would have that the maximal ideal of $B$ contains a unit, a contradiction). Thus, $x$ is not in the intersection of all valuation rings of $L$ that lie above $\mathcal{O}$.

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Thanks, but I don't quite get the last bit: since $x^{−1}$ maps to 0, then $x\notin B$. – Alan Lee Mar 2 '12 at 19:39
@AlanLee: Sorry about previous (deleted) comment, it's wrong. Since $x^{-1}$ lies in the kernel of the map from $B$ to the algebraic closure, it cannot be a unit of $B$, so we cannot have $x\in B$. – Arturo Magidin Mar 2 '12 at 19:57

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