# Noether's normalization lemma over a ring A

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.

• 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

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.

• 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.

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)?

• 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
• Dear Pete, the link seems to be broken. Do you have an alternative? – Arrow Jan 3 '19 at 12:24

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.

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

you can consult the paper Corrigendum to “Noether Normalization theorem and dynamical Gröbner bases over Bezout domains of Krull dimension 1” [J. Algebra 492 (15) (2017) 52-56] by Maroua Gamanda and Ihsen Yengui.