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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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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
up vote 4 down vote accepted

Here are some sufficient conditions:

Let $A$ be the local domain, $K$ its fraction field, $L$ the 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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Needn't A be integrally closed? – Makoto Kato Jun 8 '12 at 2:35
@Makoto: Dear Makoto, Yes, you're right; I wrote this too hastily. Let me edit the answer, and add more detail. Regards, – Matt E Jun 8 '12 at 2:41
Thank you, Matt. – Makoto Kato Jun 8 '12 at 4:03

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