Is the normalisation of an integral noetherien dimension one ring a finite morphism? - MathOverflow

Is the normalisation of an integral noetherien dimension one ring a finite morphism?

2011-12-16T15:03:21Z

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.

Answer by Qing Liu for Is the normalisation of an integral noetherien dimension one ring a finite morphism?

2011-12-16T18:08:23Z

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 $R$).

Answer by Mahdi Majidi-Zolbanin for Is the normalisation of an integral noetherien dimension one ring a finite morphism?

2011-12-16T18:27:17Z

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