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In [[mathematics]], the '''Noether normalization lemma''' is a result of [[commutative algebra]], introduced by [[Emmy Noether]] in 1926.<ref>{{harvnb|Noether|1926}}</ref> A simple version states that for any [[Field (mathematics)|field]] ''k'', and any [[Finitely generated algebra|finitely generated]] commutative [[algebra over a field|''k''-algebra]] ''A'', there exists a nonnegative integer ''d'' and [[algebraically independent]] elements ''y''<sub>1</sub>, ''y''<sub>2</sub>, ..., ''y''<sub>''d''</sub> in ''A''
such that ''A'' is a finitely generated module over  the polynomial ring ''S'':=''k''[''y''<sub>1</sub>, ''y''<sub>2</sub>, ..., ''y''<sub>''d''</sub>].
 
The integer ''d'' is the [[Krull dimension]] of ''A'' (since ''A'' and ''S'' have the same dimension.) When ''A'' is an [[integral domain]], ''d'' is the [[transcendence degree]] of the [[field of fractions]] of ''A'' over ''k''.
 
The theorem has geometric interpretation. Suppose ''A'' is integral. Let ''S'' be the [[coordinate ring]] of ''d''-dimensional [[affine space]] <math>\mathbb A^d_k</math>, and ''A'' as the coordinate ring of some other ''d''-dimensional [[affine variety]] ''X''.  Then the [[inclusion map]] ''S''&nbsp;&rarr;&nbsp;''A'' induces a surjective [[finite morphism]] of [[affine varieties]] <math>X\to \mathbb A^d_k</math>. The conclusion is that any [[affine variety]] is a [[branched covering]] of affine space.
When ''k'' is infinite, such a branched covering map can be constructed by taking a general projection from an affine space containing ''X'' to a ''d''-dimensional subspace.
 
More generally, in the language of schemes, the theorem can equivalently be stated as follows: every affine ''k''-scheme (of finite type) ''X'' is [[finite morphism|finite]] over an affine ''n''-dimensional space. The theorem can be refined to include a chain of prime ideals of ''R'' (equivalently, irreducible subsets of ''X'') that are finite over the affine coordinate subspaces of the appropriate dimensions.<ref>{{harvnb|Eisenbud|1995|loc=Theorem 13.3}}</ref>
 
The form of the Noether normalization lemma stated above can be used as an important step in proving Hilbert's [[Nullstellensatz]].  This gives it further geometric importance, at least formally, as the Nullstellensatz underlies the development of much of classical [[algebraic geometry]]. The theorem is also an important tool in establishing the notions of [[Krull dimension]] for ''k''-algebras.
 
== Proof ==
The following proof is due to Nagata and is taken from Mumford's red book. A proof in the geometric flavor is also given in the page 127 of the red book and [http://mathoverflow.net/questions/42275/choosing-the-algebraic-independent-elements-in-noethers-normalization-lemma/42363#42363 this mathoverflow thread].
 
The ring ''A'' in the lemma is generated as ''k''-algebra by elements, say, <math>y_1, ..., y_m</math> such that <math>y_1, ..., y_d</math> (some ''d'') are [[Algebraic independence|algebraically independent]] over ''k'' and the rest are algebraic over <math>k[y_1, ..., y_d]</math>.<ref>cf. Exercise 16 in Ch. 5 of Atiyah-MacDonald, Introduction to commutative algebra.</ref> We shall induct on ''m''. If <math>m = d</math>, then the assertion is trivial. Assume now <math>m > d</math>. It is enough to show that there is a subring ''S'' of ''A'' that is generated by <math>m-1</math> elements and is such that ''A'' is finite over ''S'', for, by inductive hypothesis, we can find algebraically independent elements <math>x_1, ..., x_d</math> of ''S'' such that ''S'' is finite over <math>k[x_1, ..., x_d]</math>. Since <math>m > d</math>, there is a nonzero polynomial ''f'' in ''m'' variables over ''k'' such that
:<math>f(y_1, ..., y_m) = 0</math>.
Given an integer ''r'' which is determined later, set
:<math>z_i = y_i - y_1^{r^{i-1}}, \quad 2 \le i \le m.</math>
Then the preceding reads:
:<math>f(y_1, z_2 + y_1^r, z_3 + y_1^{r^2}, ..., z_m + y_1^{r^{m-1}}) = 0</math>.
Now, the highest term in <math>y_1</math> of <math>a y_1^{\alpha_1} \prod_2^m (z_i + y_1^{r^{i-1}})^{\alpha_i}, a \in k,</math> looks
:<math>a y_1^{\alpha_1 + r \alpha_2 + ... + \alpha_m r^{m-1}}.</math>
Thus, if ''r'' is larger than any exponent <math>\alpha_i</math> appearing in ''f'', then the highest term of <math>f(y_1, z_2 + y_1^r, z_3 + y_1^{r^2}, ..., z_m + y_1^{r^{m-1}})</math> in <math>y_1</math> also has the form as above. In other words, <math>y_1</math> is integral over <math>S = k[z_2, ..., z_m]</math>. Since <math>y_i = z_i + y_1^{r^{i-1}}</math> are also integral over that ring, ''A'' is integral over ''S''. It follows ''A'' is finite over ''S''.
 
If ''A'' is an integral domain, then ''d'' is the transcendence degree of its field of fractions. Indeed, ''A'' and <math>S = k[y_1, ..., y_d]</math> have the same transcendence degree (i.e., the degree of the field of fractions) since the field of fractions of ''A'' is algebraic over that of ''S'' (as ''A'' is integral over ''S'') and ''S'' obviously has transcendence degree ''d''. Thus, it remains to show the Krull dimension of the polynomial ring ''S'' is ''d''. (this is also a consequence of [[dimension theory (algebra)|dimension theory]].) We induct on ''d'', <math>d=0</math> being trivial. Since <math>0 \subsetneq (y_1) \subsetneq (y_1, y_2) \subsetneq \cdots \subsetneq (y_1, \dots, y_d)</math> is a chain of prime ideals, the dimension is at least ''d''. To get the reverse estimate, let <math>0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_m</math> be a chain of prime ideals. Let <math>0 \ne u \in \mathfrak{p}_1</math>. We apply the noether normalization and get <math>T = k[u, z_2, \dots, z_d]</math> (in the normalization process, we're free to choose the first variable) such that ''S'' is integral over ''T''. By inductive hypothesis, <math>T/(u)</math> has dimension ''d'' - 1. By [[incomparability]], <math>\mathfrak{p}_i \cap T</math> is a chain of length <math>m</math> and then, in <math>T/(\mathfrak{p}_1 \cap T)</math>, it becomes a chain of length <math>m-1</math>. Since <math>\operatorname{dim} T/(\mathfrak{p}_1 \cap T) \le \operatorname{dim} T/(u)</math>, we have <math>m - 1 \le d - 1</math>. Hence, <math>\dim S \le d</math>.
 
== Notes ==
{{reflist}}
 
==References==
*{{Citation | last1=Eisenbud | first1=David | author1-link=David Eisenbud | title=Commutative algebra. With a view toward algebraic geometry | publisher=[[Springer-Verlag]] | location=Berlin, New York | series=[[Graduate Texts in Mathematics]] | isbn=3-540-94268-8 | mr=1322960 | year=1995 | volume=150 | zbl=0819.13001 }}
*{{Springer|id=n/n066790|title=Noether theorem}}. NB the lemma is in the updating comments.
*{{citation
|last=Noether |first=Emmy |authorlink=Emmy Noether
|year=1926
|title=Der Endlichkeitsatz der Invarianten endlicher  linearer Gruppen der Charakteristik ''p''
|url=http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=63971
|journal=[[Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen]]
|volume= |pages=28–35
|doi=
}}
 
[[Category:Commutative algebra]]
[[Category:Algebraic varieties]]
[[Category:Lemmas]]
[[Category:Algebraic geometry]]

Revision as of 20:14, 29 August 2013

In mathematics, the Noether normalization lemma is a result of commutative algebra, introduced by Emmy Noether in 1926.[1] A simple version states that for any field k, and any finitely generated commutative k-algebra A, there exists a nonnegative integer d and algebraically independent elements y1, y2, ..., yd in A such that A is a finitely generated module over the polynomial ring S:=k[y1, y2, ..., yd].

The integer d is the Krull dimension of A (since A and S have the same dimension.) When A is an integral domain, d is the transcendence degree of the field of fractions of A over k.

The theorem has geometric interpretation. Suppose A is integral. Let S be the coordinate ring of d-dimensional affine space 𝔸kd, and A as the coordinate ring of some other d-dimensional affine variety X. Then the inclusion map S → A induces a surjective finite morphism of affine varieties X𝔸kd. The conclusion is that any affine variety is a branched covering of affine space. When k is infinite, such a branched covering map can be constructed by taking a general projection from an affine space containing X to a d-dimensional subspace.

More generally, in the language of schemes, the theorem can equivalently be stated as follows: every affine k-scheme (of finite type) X is finite over an affine n-dimensional space. The theorem can be refined to include a chain of prime ideals of R (equivalently, irreducible subsets of X) that are finite over the affine coordinate subspaces of the appropriate dimensions.[2]

The form of the Noether normalization lemma stated above can be used as an important step in proving Hilbert's Nullstellensatz. This gives it further geometric importance, at least formally, as the Nullstellensatz underlies the development of much of classical algebraic geometry. The theorem is also an important tool in establishing the notions of Krull dimension for k-algebras.

Proof

The following proof is due to Nagata and is taken from Mumford's red book. A proof in the geometric flavor is also given in the page 127 of the red book and this mathoverflow thread.

The ring A in the lemma is generated as k-algebra by elements, say, y1,...,ym such that y1,...,yd (some d) are algebraically independent over k and the rest are algebraic over k[y1,...,yd].[3] We shall induct on m. If m=d, then the assertion is trivial. Assume now m>d. It is enough to show that there is a subring S of A that is generated by m1 elements and is such that A is finite over S, for, by inductive hypothesis, we can find algebraically independent elements x1,...,xd of S such that S is finite over k[x1,...,xd]. Since m>d, there is a nonzero polynomial f in m variables over k such that

f(y1,...,ym)=0.

Given an integer r which is determined later, set

zi=yiy1ri1,2im.

Then the preceding reads:

f(y1,z2+y1r,z3+y1r2,...,zm+y1rm1)=0.

Now, the highest term in y1 of ay1α12m(zi+y1ri1)αi,ak, looks

ay1α1+rα2+...+αmrm1.

Thus, if r is larger than any exponent αi appearing in f, then the highest term of f(y1,z2+y1r,z3+y1r2,...,zm+y1rm1) in y1 also has the form as above. In other words, y1 is integral over S=k[z2,...,zm]. Since yi=zi+y1ri1 are also integral over that ring, A is integral over S. It follows A is finite over S.

If A is an integral domain, then d is the transcendence degree of its field of fractions. Indeed, A and S=k[y1,...,yd] have the same transcendence degree (i.e., the degree of the field of fractions) since the field of fractions of A is algebraic over that of S (as A is integral over S) and S obviously has transcendence degree d. Thus, it remains to show the Krull dimension of the polynomial ring S is d. (this is also a consequence of dimension theory.) We induct on d, d=0 being trivial. Since 0(y1)(y1,y2)(y1,,yd) is a chain of prime ideals, the dimension is at least d. To get the reverse estimate, let 0p1pm be a chain of prime ideals. Let 0up1. We apply the noether normalization and get T=k[u,z2,,zd] (in the normalization process, we're free to choose the first variable) such that S is integral over T. By inductive hypothesis, T/(u) has dimension d - 1. By incomparability, piT is a chain of length m and then, in T/(p1T), it becomes a chain of length m1. Since dimT/(p1T)dimT/(u), we have m1d1. Hence, dimSd.

Notes

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  1. Template:Harvnb
  2. Template:Harvnb
  3. cf. Exercise 16 in Ch. 5 of Atiyah-MacDonald, Introduction to commutative algebra.