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This feels like something I should know but I can't find an answer in Liu or in Atiyah-MacDonald, or a counter-example.

To state the question again: let $A$ be an integral Noetherien ring of Krull dimension one and $K$ its field of fractions. Let $B$ be the set of elements of $K$ that are integral over $A$ i.e. $B$ is the normalisation of $A$ in $K$.

Is the morphism $A \to B$ finite?

Note that this is true if $A$ is excellent (or even just Nagata), and its rather difficult to construct examples of non-excellent rings.

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Isn't this a consequence of the Krull-Akizuki theorem? – Parsa Dec 16 '11 at 16:24
A note: when $A$ is local, if $A\rightarrow B$ is a finite morphism, where $B$ is the integral closure of $A$ in $K$, then $B$ is a semi-local Dedekind domain, and hence it's a PID. – Mahdi Majidi-Zolbanin Dec 16 '11 at 16:30
@Parsa, the Krull-Akizuki theorem tells you that $B$ is Noetherian, which is weaker than being finite over $A$. – Dustin Cartwright Dec 16 '11 at 18:54
up vote 4 down vote accepted

For a brief history of this question you can look at Matsumura's Commutative Ring Theory, page 264. In Ein Satz über primäre Integritätsbereiche, Math. Ann. vol. 103 (1930), p.p. 450-465 Krull proved that the integral closure of a one-dimensional Noetherian local domain $A$ is finite over $A$ if and only if the completion of $A$ is reduced. Akizuki constructed the first example of a one-dimensional Noetherian local integral domain with non-reduced completion in 1935. For another counterexample you can see Akizuki's counterexample

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Let $R$ be any non-excellent DVR with field of fractions $K$, let $L/K$ be a finite extension such that the normalization $B$ of $R$ in $L$ is not finite over $R$. We have $L=K[a_1,...,a_n]$ for some $a_i\in B$. Consider $A=R[a_1,...,a_n]\subseteq B$. Then $B$ in the integral closure of $A$, but is not finite over $A$ (because $A$ is finite over $R$).

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What's an example of a non-excellent DVR? – Keerthi Madapusi Pera Dec 16 '11 at 18:10
Let $p$ be a prime number. Let $s \in {\mathbb F}_p[[t]]$ be transcendent over ${\Bbb F}_p(t)$ and let $K=\mathbb F_p(t, s^p)\subset {\mathbb F}_p((t))$. The discrete valuation of ${\mathbb F}_p((t))$ induces a discrete valuation on $K$. The corresponding discrete valuation ring $R$ is non-excellent because its integral closure in $K[s]$ is not finite over $R$. – Qing Liu Dec 16 '11 at 18:23
More details in my book, Example 8.2.31. – Qing Liu Dec 16 '11 at 18:25
Thanks! I'll look it up in your book. – Keerthi Madapusi Pera Dec 16 '11 at 20:51

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