Let $(R,m)$ be a D.V.R with field of fraction $K$ and $L$ any finite algebraic field extension of $K$. Suppose $\bar{R}$ is the integral closure of $R$ in $L$. Then it is well known that $\bar{R}$ is a Dedekind domain and for any nonzero ideal $J$ of $\bar{R}$ the ring $\bar{R}/J$ is a finite $R$-module. My question is if $\bar{R}$ is itself a finite $R$ module. If $L$ is separable over $K$ or $(R,m)$ is essentially finite over a field, then the answer is affirmative. I think in general it is not true. But I am not getting any easy counter-example.
A question related to Krull-Akizuki theorem
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commutative-algebra