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Is the integral closure of a local domain in a finite extension of its field of fractions semi-local?

If the answer is negative, I wonder under what conditions it would be semi-local.

EDIT Here's an example of a local domain which is not necessarily a Japanese ring. Let $A$ be a valuation ring, $K$ its field of fractions. Let $P$ be the maximal ideal of $A$. Let $L$ be a finite extension of $K$. Let $B$ be the integral closure of $A$ in $L$. It is well-known that there exist only finitely many valuation rings of $L$ dominating $A$. Let $M$ be a maximal ideal of $B$. It is well-known that $M$ lies over $P$. Hence $B_M$ dominates $A$. There exists a valuation ring $R$ of $L$ dominating $B_M$. Since $M$ is determined by $R$ and $R$ dominates $A$, there exist only finitely many maximal ideals of $B$. Hence $B$ is a semilocal ring.

Note that $B$ is not necessarily a finite $A$-module(even if $A$ is a discrete valuation ring). Hence $A$ is not necessarily a Japanese ring.

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  • $\begingroup$ Semilocal here meaning that it has a finite number of maximal ideals? $\endgroup$
    – rschwieb
    Jun 8, 2012 at 1:04
  • $\begingroup$ Yes, it means exactly so. $\endgroup$ Jun 8, 2012 at 1:34

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Here are some sufficient conditions:

Let $A$ be a local domain, $K$ its fraction field, $L$ a finite extension of $K$, and $B$ the integral closure of $A$ in $L$. If $A$ is Noetherian and integrally closed, and $L$ is separable over $K$, then $B$ is necessarily finite over $A$, and so is semi-local (by going-up/down-type theorems).

If $A$ is not integrally closed, but its integral closure in $K$ is finite over itself, then again $B$ will be finite over $A$.

The condition that the integral closure of $A$ in $K$ be finite over $A$ is (at least by some people) called N-1. More generally, the condition that $B$ be finite over $A$ is called N-2, or Japanese. (The letter N here stands for Nagata, and I believe Grothendieck coined the adjecive Japanese for these rings because these properties were studied by Nagata and the commutative algebra school around him in Japan.)

So if $A$ is a Japanese ring, then $B$ will be finite over $A$, and hence semilocal. Of course, this is rather tautological: its utility follows from the fact that many rings (indeed, in some sense, most rings --- i.e. most of the rings that come up in algebraic number theory and algebraic geometry) are Japanese. E.g. all finitely generated algebras over a field, or over $\mathbb Z$, or over a complete local ring, are Japanese.

Here are some useful wikipedia entries related to this topic: Nagata rings and Excellent rings.

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Here is a non-Noetherian result [Bourbaki, Commutative algebra, VI, $\S$8, no.3, Remarque]: For a valuation ring $R$ with fraction field $K$ and a finite field extension $L/K$, the integral closure of $R$ in $L$ is semilocal.

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