Let $k$ be a field and $A$ a finitely generated algebra over $k$ that doesn't have zero divisors. Why is the integral closure of $A$ a finitely generated module over $A$ ?
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1This is not true in general. But it holds when $A$ is f.g. over $k$, by a famous theorem of Emmy Noether. – Martin Brandenburg May 14 '13 at 09:16
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As Martin said in the comments, the result is not true in general but holds if $A$ is finitely generated over $k.$ This follows from Proposition 16 (page 46) of Serre's Local Algebra:
Proposition 16. Let $A$ be a domain which is a finitely generated algebra over a field $k,$ let $K$ be its field of fractions, and let $L$ be a finite extension of $K.$ Then the integral closure $B$ of $A$ in $L$ is a finitely generated $A$-module (in particular it is a finitely generated $k$-algebra).
The Google Books preview of Serre's book lets one see page 46 (and surprisingly, most of the book).
Ragib Zaman
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This is also contained in any other complete text on commutative algebra, for example Eisenbud's book. – Martin Brandenburg May 14 '13 at 09:47
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@MartinBrandenburg You are of course correct. I should have mentioned that this in many commutative algebra books and I linked to Serre's book simply because I wanted to provide an online reference and that was the first one I found. – Ragib Zaman May 14 '13 at 09:59
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Remark: the statement Proposition 16 is the same as saying "fields are universally Japanese", see 032R, 032F. For a SP based proof, note that fields are Nagata (see definition in first link), so it suffices to apply 0334. – Elías Guisado Villalgordo Aug 29 '23 at 16:16