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Let $S\subset R$ be a finite ring extension, i.e. $R$ is finitely generated as an $S$-module. Assume that $S$ is integrally closed. Does this imply that also $R$ is integrally closed (in its quotient field)?

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Take $ S = \Bbb{Z}$ and take $R = \Bbb{Z}[\sqrt{-3}]$. Then $R$ is f.g. over $S$ (in fact free of rank $2$) but now $R$ is not integrally closed: The quotient field of $R$ is $\Bbb{Q}(\sqrt{-3})$ and $ \frac{1 + \sqrt{-3}}{2}$ is in here, and this is integral over $R$ being a solution of $x^2 - x + 1$. But this element is not in $R$.

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No: $S=\mathbb Z\subset R=\mathbb Z[2i]=\mathbb Z\oplus \mathbb Z\cdot 2i$

[One has $i\notin R$, although $i\in Frac(R)$ is a zero of the monic polynomial $X^2+1\in R[X]$]

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  • $\begingroup$ Nice solution Georges! I figured a counterexample would be easily furnished by looking at algebraic number theory. $\endgroup$ – user38268 Jul 4 '13 at 17:10
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    $\begingroup$ Thanks, @Benja: I'm sure you would have come up with a similar example. $\endgroup$ – Georges Elencwajg Jul 4 '13 at 17:11

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