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I have the following question:

Let $A$ be an integrally closed Noetherian domain, $K$ its field of fractions. let $L$ be a finite extension of $K$, and $B$ the integral closure of $A$ inside $L$. Is it true then that $B$ is finite over $A$?

If $L/K$ is separable, it is true (there is a usual proof with considering the non-degenerate bilinear form $tr(xy)$). What about a non-separable extension?

Thanks, Sasha

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See mathoverflow.net/questions/59762/… –  ACL Nov 8 '11 at 10:31

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up vote 6 down vote accepted

No, it's not true. You can say some things about B even if L/K is not separable : B is a dvr if A is, B is a Dedekind ring if A is (that is called the Krull-Akizuki theorem). But it's not true that B is always finite over A, even if A is a dvr. There is a counterexample in theorem 100 of Kaplansky's "Commutative rings".

By the way, an integral domain A such that the integral closure of A in any finite extension of its fields of fractions is finite over A is called a Japanese ring. The wikipedia article on Nagata rings gives examples of Japanese rings. Searching for "non Japanese discrete valuation rings" will give you other counterexamples to your question (or at least other references).

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Perhaps it should be said that essentially all rings appearing in geometry are excellent, and thus Japanese. –  Karl Schwede Nov 9 '11 at 12:54
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Depends on your definition of "geometry". Discrete valuation rings of equal characteristic $p$ don't have to be excellent, and if you do analytic geometry in positive characteristic you can encounter them. –  Alex Nov 9 '11 at 19:25

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