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Let $K$ be a number field, $A$ the integral closure of $\mathbb Z$ in $K$. Given a valuation ring $\mathfrak o$ in $K$, $\mathfrak m$ its maximal ideal, $\mathfrak p= \mathfrak m \cap A$ the maximal ideal in $A$.

Claim $\mathfrak o=A_{\mathfrak p}$.

I can deduce $\mathfrak o\supset A_{\mathfrak p}$ but I'm not sure about the other inclusion.

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Here is one argument. $A_{\mathfrak{p}}$ is a Dedekind domain with one maximal ideal and thus is principal ideal domain. We see that it is a valuation ring. If $\pi\in\mathfrak{p}-\mathfrak{p}^2$ then every element is of the form $u\pi^k$ with $u$ a unit. It follows that for all $a\in K$ $a\in A_{\mathfrak{p}}$ or $\frac{1}{a}\in A_{\mathfrak{p}}$. The result follows.

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  • $\begingroup$ Why the result follows? $\endgroup$ – CYC Feb 4 '17 at 7:02
  • $\begingroup$ @CYC Two valuations rings of the same field can't dominate each other unless they are equal. (This is an easy exercise.) In your case $A_{\mathfrak p}\subseteq\mathfrak o$ are both valuation rings, and $\mathfrak o$ dominates $A_{\mathfrak p}$. $\endgroup$ – user26857 Feb 26 '17 at 1:14
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Let $v$ be the valuation of $\mathfrak o$ in $K$, $M=(\pi)$ be the maximal ideal of $A_\mathfrak p$, now given $x \in K$,$v(x)=v(\pi^nu)=nv(\pi)=v_{M}(x)v(\pi)$, since $A\subset \mathfrak o$ we have $v(\pi) > 0$, that is, $v(x)\geq 0$ iff $v_{M}(x)\geq 0$.

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