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As far as I can tell, Noether normalization uses the term "normalization" in the English sense, that something has been given a standard form. And as such it's not intrinsically related to normalization in the sense of "take the integral closure in the ring of fractions". I would leave the matter at that, except that in book after book, the two are introduced and studied in very close proximity. Am I missing something?

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Another aspect is that Noether's normalization theorem is the main step in proving that the normalization of an affine algebraic set is again an affine algebraic set, i.e., the normalization of a finitely generated k-algebra is again finitely generated as a k-algebra. –  Jason Starr Nov 20 '11 at 15:54
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1 Answer 1

What is the statement of Noether normalization theorem you are referring to?

The one I know is the following, that you can find for instance in Shafarevich's book "Basic Algebraic Geometry I", page 65.

Theorem (Noether normalization). For an irreducible, affine variety $X$ there exists a finite map $$\varphi \colon X \longrightarrow \mathbb{A}^n$$ to an affine space.

That means that any integral domain $A$ which is finitely generated over the field $k$ is integral over a subring isomorphic to a polynomial ring.

In particular, let $X \subset \mathbb{A}^{n+1}$ be any degree $d$ hypersurface. Then its "normal form" is given by a monic equation $$z_{n+1}^d + f_{d-1}z_{n+1}^{d-1}+ \ldots + f_1 z_{n+1}+f_0=0,$$ with $f_i \in k[z_1, \ldots, z_n]$, and the map $\varphi \colon X \longrightarrow \mathbb{A}^n$ in Noether theorem is simply the projection onto the first $n$ coordinates $z_1, \ldots, z_n$.

This shows how the two concepts of "normalization" are closely related.

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Okay, so in this interpretation, Noether normalization says "$A$ is contained in the normalization of a polynomial ring inside some field (not its field of fractions)". I guess that is pretty close. –  Allen Knutson Nov 20 '11 at 14:47
    
But many hypersurfaces are not normal varieties. Most stupidly what if $X$ is defined by $z^d = 0$. –  aginensky Nov 23 '11 at 1:24
    
@agineski: "But many hypersurfaces are not normal varieties". This is true, but in fact Noether normalization does not tell you that $A$ is integrally closed in its ring of fractions (see Allen Knutson's comment). The ideal $(z^d)$ is not ratical; then $V(z^d)=V(\surd z^d)=V(z)$. –  Francesco Polizzi Nov 23 '11 at 7:55
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