Let $L, K$ be fields, $L$ an algebraic extension of $K$, $R$ a valuation ring of $K$, and $\bar{R}$ the integral closure of $R$ in $L$. Is it true that for any maximal ideal $\mathfrak{m}$ of $\bar{R}$, the localization $\bar{R}_{\mathfrak{m}}$ is a valuation ring of $L$? If not, what are some additional assumptions that allow for this statement to be true?

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    $\begingroup$ This is true: see Bourbaki's Commutative Algebra VI, §8, no. 6, Proposition 6. $\endgroup$ – abx Apr 13 at 13:02

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