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Given a field $k$ and a finitely generated $k$-algebra $R$ without zero divisors, one knows that there exist $x_1, \ldots, x_n$ algebraically independent such that $R$ is integral over $k[x_1, \ldots, x_n]$.

Does one have a similar statement, under good assumptions, if $k$ is not a field but a ring ? In this discussion, I am also interested by geometric explanations.

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I am not sure this is really what you are looking for, but on the geometric side, let me mention the following result of Barry Green (MR1458302): let A be a Dedeking ring whose fraction field is a local or global field, then every normal projective curve over Spec(A) has a finite morphism to $\mathbf{P}^1_A$. See also arxiv.org/abs/0902.2039. –  Jérôme Poineau Oct 19 '10 at 11:02

5 Answers 5

up vote 14 down vote accepted

http://www.math.lsa.umich.edu/~hochster/615W10/supNoeth.pdf

Supplementary notes from Mel Hochster's commutative algebra class. They discuss, in particular, a generalization of Noether normalization to integral domains.

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I have not read carefully, but it seems to be the same proof? –  Martin Brandenburg Oct 17 '10 at 15:42
    
@Martin: Yes, essentially. –  Harry Gindi Oct 17 '10 at 21:09

The geometric interpretation of Noether's normalization lemma is that any affine algebraic variety has a finite surjective morphism to the affine space $\mathbb A^d_k$ of dimension $d=\dim X$. When $k$ is an integral domain, and $X$ dominates $\mathrm{Spec}(k)$, the finite surjective morphism of the generic fiber of $X$ to $\mathbb A^d_{K}$, where $K=\mathrm{Frac}(k)$ extends to a finite surjective morphism $X_V\to V$ for some dense open subset $V$ of $\mathrm{Spec}(k)$. This is the geometric interpretation of the statement in M. Hochster's note in Harry Gindi's post.

In general, such a morphism can not exist because it would imply that the fibers of $X\to \mathrm{Spec}(k)$ all have the same dimension.

Suppose that this condition is satisfied: the fibers of $X\to\mathrm{Spec}(k)$ all have the same dimension $d$. Then I think that a reasonable statement (Noether's normalization lemma over a ring $k$) would be: there exists a quasi-finite and surjective morphism $X\to \mathbb{A}^d_k$. If $k$ is noetherian, then by Zariski's Main Theorem, this implies that $X$ is an open subscheme of scheme which is finite surjective over $\mathbb{A}^d_k$. In general one can not expect better result than quasi-finite (consider the case $d=0$).

The above "reasonable statement" should be easy to prove when $k$ is a local ring.

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I recently came to want this generalization of Noether normalization for my own commutative algebra course and notes. So I just wanted to report that I found what seems to me to be the optimally efficient and clear treatment of this result, at the beginning of Chapter 8 of these commutative algebra notes of K.M. Sampath. All in all I highly recommend Sampath's notes: they are excellent.

I am starting to find it surprising that this simple and useful generalization of Noether Normalization is not the standard version: it has some important applications, e.g. finiteness of integral closure of domains which are finitely generated over $\mathbb{Z}$. Does anyone know who first came up with this version (Hochster, perhaps)?

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The Noether normalization lemma in these notes seem to be a copy of the treatment in Eisenbud's book on commutative algebra. He mentions that the case with one ideal is due to Nagata. –  Jakob Dec 7 '12 at 10:01

Abhyankar (Shreeram S.), and Kravitz (Ben), constructed counterexamples to the generalized Normalization Lemma as stated in "Commutative Algebra"(II) by Zariski/Samuel. See Proc.AMS 135 (2007)no.11, 3521-3523.

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A weak version (allowing localization upstairs) of the statement discussed in Qing Liu's answer is Lemma Tag 00QE of the Stacks project.

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