Let $R$ be a domain and $K$ be it's field of fractions. Let $S$ be the integral closure of $R$ in $K$. Let $M$ be a maximal ideal of $R$. If $S\subset R_{M}$ is $R_{M}$ integrally closed in $K$?
My thoughts on this are it is false. But I can't find a counter example or find it in any literature. Here are my thoughts. Let $M'$ be a lift of $M$ in $S$. Then the integral closure $(R_{M})'=S_{M'}$. If $x\in S_{M'}$ we have $x=a/b$ with $b\in S-M'\subset S-M$. There is no guarantee that $b$ is also in $R$.
Any counter example or idea towards the proof would be nice. To answer the geometry question I'm working on I would really like it to be true. Thanks in advance.
I think it is.
First observe that by the transitivity of integrality $S$ is integrally closed in $K$.
Write $W = R \setminus M$. Since $R \subseteq S$ is integral, $W^{-1}R = R_M \subseteq W^{-1}S$ is integral. By the assumption $S \subseteq R_M$ we have $R_M = W^{-1}S$. Hence $R_M$ is integral.