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I saw this lemma in some lecture notes, there was no proof given nor a reference, only a statement that it can be found in any text-book on commutative algebra. I checked several but couldn't find it.

Let $F$ be a field, $B$ an $F$-algebra which is an integral domain, $A$ a Noetherian subalgebra of $B$, and such that $B$ is integral over $A$, and the field of fractions of $B$ is a finite extension of the field of fractions of $A$. Then $B$ is a finitely generated $F$-algebra if and only if $A$ is a finitely generated $F$-algebra. (I'm mainly interested in the 'if' part).

Has anyone seen this before, knows a proof or where to find one? Thanks.

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1 Answer 1

up vote 3 down vote accepted

The if-part:

If $A$ is a finitely generated $F$-algebra, then it is noetherian and you don't need to require this. By the Noether Normalization theorem there then exists a polynomial ring $F[x_1,\ldots ,x_r]\subseteq A$ such that the extension $A/F[x_1,\ldots ,x_r]$ is integral (and by assumption finitely generated). Consequently the field of fractions of $A$ is a finite extension of the field of fractions of $F[x_1,\ldots ,x_r]$, hence the field of fractions of $B$ is a finite extension of the field of fractions of $F[x_1,\ldots ,x_r]$. The assertion in this situation is proved for example in Matsumuras Commutative Ring Theory on page 262.

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Thanks very much Hagen. I was able to prove this lemma using a theorem in Bourbaki's Commutative Algebra, which I think is the same as what you are quoting from Matsumura. –  Nadim Rustom Apr 20 '12 at 14:42

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