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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.

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Example (O.Zariski?):

Consider the rational function field $k(x)$ over a field $k$ of characteristic $p>0$.

The field extension $k((x))/k(x)$ is transcendental; let $\alpha\in k[[x]]$ be transcendental and let $y:=\alpha^p$. Then $L:=k(x,\alpha )$ is a purely inseparable extension of $K:=k(x,y)$.

The discrete valuation ring $B:=k[[x]]\cap L$ is the integral closure of the discrete valuation ring $A:=k[[x]]\cap K$ and $x$ is a prime element of $A$.

For every $n\in\mathbb{N}$ the element $y$ can be written as

$ y=f_n^p + x^{pn}y_n , f_n\in k[x], y_n\in k[[x]]. $

(To see this consider the power series: $\alpha =c_0+c_1x+c_2x^2+c_3x^3+\ldots$ hence $y =c_0^p+c_1^px^p+c_2^px^{2p}+c_3^px^{3p}+\ldots$.)

Then $y_n\in A$ and one gets:

$ y_n^{1/p} =x^{-n}(-f_n+\alpha )\;\; (*). $

Now if $B$ were a finite $A$-module one could find $d\in A\setminus 0$ such that

$ dB\subseteq A+A\alpha +A\alpha^2 +\ldots +A\alpha^{p-1}=:R. $

In particular $d y_n^{1/p}\in R$, which using (*) yields that $d$ is divisible by $x^n$ for every $n$. (Note that $1, \alpha ,\ldots ,\alpha^{p-1}$ are linearly independent over $K$.)

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Thanks for this nice answer. – A.G Mar 2 '11 at 20:32

You're right: $\overline{R}$ need not be a finitely generated $R$-module. To the best of my knowledge, none of the counterexamples are "easy". Here is the one given by Kaplansky in his text Commutative Rings (I quote verbatim):

Theorem 100: Let $k$ be a field of characteristic $2$, and $T = k[[x]]$, the power series ring in an indeterminate. With $u \in T$, let $K = k(x,u^2)$ and $L = k(x,u)$ (field adjunction), $R = T \cap K$, $S = T \cap L$. Then $[L:K] \leq 2$, $R$ and $S$ are discrete valuation rings with quotient fields $K$ and $L$, and $S$ is the integral closure of $R$ in $L$. If $[L:K] = 2$, then $S$ is not a finitely generated $R$-module.

He goes on to remark how the hypothesis $[L:K] = 2$ can be ensured, and then he gives the proof, which takes almost a page. For what it's worth, I intend for this counterexample (or one like it) to be in my commutative algebra notes (in the section "Normalization Theorems"), but it is not there yet.

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Thanks for this nice answer – A.G Mar 2 '11 at 20:31

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