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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asked Nov 20, 2011 at 13:44
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$\begingroup$ 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. $\endgroup$– Jason StarrCommented Nov 20, 2011 at 15:54
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$\begingroup$ @JasonStarr, where can I find the proof that the normalization of an affine algebraic set is again an affine algebraic set (using Noether's normalizartion)? $\endgroup$– rmdmc89Commented May 28, 2019 at 22:39
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1$\begingroup$ @rmdmc89 "where can I find the proof that the normalization of an affine algebraic set is again an affine algebraic set (using Noether's normalizartion)" One reference is David Eisenbud's Commutative algebra. $\endgroup$– Jason StarrCommented May 29, 2019 at 9:12
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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.
answered Nov 20, 2011 at 14:05
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$\begingroup$ 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. $\endgroup$ Commented Nov 20, 2011 at 14:47
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$\begingroup$ But many hypersurfaces are not normal varieties. Most stupidly what if $X$ is defined by $z^d = 0$. $\endgroup$– mehCommented Nov 23, 2011 at 1:24
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$\begingroup$ @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 $V(z^d)=V(\surd z^d)=V(z)$. $\endgroup$ Commented Nov 23, 2011 at 7:55