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Assume that $A$ is an integrally closed integral domain, and $K$ is its fraction field. Well...this may be a stupid question, but is every overring of $A$ between $A$ and $K$ also integrally closed ? (This is known to be true if $A$ is a Dedekind domain, see e.g. Jarden, Field Arithmetic, chap. 2).

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    $\begingroup$ For Dedekind domains the overrings are even Dedekind domains. But the integrally closed property of overrings characterizes the Prufer domains. $\endgroup$ – user26857 Feb 24 '15 at 14:41
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Let $A=K+YK[X,Y]$. We have $Q(A)=K(X,Y)$. To show that $A$ is integrally closed notice that $A\subset K[X,Y]$, and the last ring is integrally closed. If an element $z∈Q(A)$ is integral over $A$, then it is integral over $K[X,Y]$, so it belongs to $K[X,Y]$. Now one can get rid of the part of $z$ which belongs to $A$ and get that $z=f(X)$ is integral over $A$ hence over $K$, so $\deg f=0$.

Now consider $R=K[X^2]+YK[X,Y]$. Clearly $R$ is not integrally closed since $X$ is integral over $R$ and $X\notin R$.

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  • $\begingroup$ Actually I'd be glad if someone can give such an example for $A=K[X,Y]$. For some reasons I didn't find it, thought it certainly do exist since $K[X,Y]$ is not Prufer. $\endgroup$ – user26857 Feb 24 '15 at 20:41

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