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

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.

• Semilocal here meaning that it has a finite number of maximal ideals? – rschwieb Jun 8 '12 at 1:04
• Yes, it means exactly so. – Makoto Kato Jun 8 '12 at 1:34

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.