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Let $G$ be a finite group acting on an integral domain $R$, and let $S$ denote the fixed point subring: $$ S=\{x\in R: gx=x \text{ for all }g\in G\}$$ I am asked to show that $R$ is integral over $S$, and that if $R$ is integrally closed then so is $S$.

The first one is easy: given $r\in R$, the polynomial $\prod _{g\in G}(t-gr)$ has $r$ as a root and is fixed by the action of $G$, so it is a polynomial in $S[t]$. But I have no idea for the second one. Any hints?

Bernard
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user302934
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    Well, it's late, so the following might contain a stupid mistake, but maybe you could try the following. Let $L$ be the quotient field of $R$, $K$ the quotient field of $S$. There is an obvious embedding $K \to L$ induced by the inclusion $S \to R$. If $s/t \in K$ is integral over $S$, then it is integral over $R$, and hence belongs to $R$, since $R$ is integrally closed in $L$. Then $s/t = r$ for some $r \in R$, i.e. $s = tr$. Moreover, since $s$ and $t$ are $G$-invariant, this relation shows that $r$ is likewise $G$-invariant, so $r$ belongs to $S$. – Alex Wertheim May 01 '20 at 10:11
  • @AlexWertheim I think it works. Thanks. – user302934 May 01 '20 at 23:20

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