Weak nullstellensatz for arbitrary fields: Difference between revisions

Latest revision as of 18:05, 12 May 2008

This fact is an application of the following pivotal fact/result/idea: Artin-Tate lemma
View other applications of Artin-Tate lemma OR Read a survey article on applying Artin-Tate lemma

This fact is an application of the following pivotal fact/result/idea: Noether normalization theorem
View other applications of Noether normalization theorem OR Read a survey article on applying Noether normalization theorem

Statement

Here are two equivalent formulations:

Suppose $k$ is a field and $K$ is a field extension of $k$ , such that $K$ is finitely generated as a $k$ -algebra. Then, $K$ is algebraic over $k$ , and in fact, is a finite field extension of $k$ .
Suppose $M$ is a maximal ideal in a polynomial ring in finitely many variables over $k$ . Then the quotient field for $M$ is a finite field extension of $k$

Applications

Weak nullstellensatz for algebraically closed fields

Proof using Artin-Tate lemma

Facts used

Steinitz theorem: This states that any field extension can be expressed as an algebraic extension of a purely transcendental extension
Artin-Tate lemma
The fact that a purely transcendental field extension cannot be finitely generated as an algebra over the field

Proof outline

We use Steinitz theorem to show that we can find a subfield $k(T)$ of $K$ , which is the field of fractions of a subset $T$ of $k$ , such that $K$ is algebraic over $k(T)$ . In our case, since $K$ is finitely generated over $k$ , it is also finitely generated over $k(T)$ , so in fact $K$ is a finite field extension of $k(T)$ . Further information: Finitely generated and integral implies finite
We use Artin-Tate lemma and the fact that fields are Noetherian, to deduce that $k(T)$ is finitely generated as a $k$ -algebra (here $A=k,B=k(T),C=K$ )
We now use the fact that if $T$ is nonempty, $k(T)$ can never be finitely generated over $k$ . Thus, $T$ is empty, forcing $K$ to be a finite field extension of $k$ (This uses the fact that the polynomial ring is a unique factorization domain with infinitely many irreducibles).

Proof using Noether normalization theorem

Facts used

Noether normalization theorem

Proof outline

Suppose $K$ is a finitely generated algebra over $k$ , that happens to be a field. Then, by the Noether normalization theorem, there exists a polynomial algebra $k[x_{1},x_{2},\ldots ,x_{n}]$ inside $K$ such that $K$ is finite over this polynomial algebra.

But $K$ being a field, must at any rate contain the field of fractions of $k[x_{1},x_{2},\ldots ,x_{n}]$ , and this yields a contradiction if $n\geq 1$ .

Alternatively, we can observe that the injective map from $k[x_{1},x_{2},\ldots ,x_{n}]$ to $K$ being a finite morphism, must give a surjective map on spectra, but the spectrum of $K$ has one element, and so can surject to the spectrum of a polynomial ring only if $n=0$ .

This yields that $K$ is finite-dimensional over $k$ .

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+{{applicationof|Artin-Tate lemma}}
+{{applicationof|Noether normalization theorem}}
 ==Statement==
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 ==Proof using Artin-Tate lemma==
-==Facts used==
+===Facts used===
 * [[Steinitz theorem]]: This states that any field extension can be expressed as an algebraic extension of a purely transcendental extension
@@ Line 22: / Line 26: @@
 * We use [[Steinitz theorem]] to show that we can find a subfield <math>k(T)</math> of <math>K</math>, which is the field of fractions of a subset <math>T</math> of <math>k</math>, such that <math>K</math> is algebraic over <math>k(T)</math>. In our case, since <math>K</math> is finitely generated over <math>k</math>, it is also finitely generated over <math>k(T)</math>, so in fact <math>K</math> is a finite field extension of <math>k(T)</math>. {{further|[[Finitely generated and integral implies finite]]}}
 * We use [[Artin-Tate lemma]] and the fact that fields are Noetherian, to deduce that <math>k(T)</math> is finitely generated as a <math>k</math>-algebra (here <math>A = k, B = k(T), C = K</math>)
-* We now use the fact that if <math>T</math> is nonempty, <math>k(T)</math> can never be finitely generated over <math>k</math>. Thus, <math>T</math> is empty, forcing <math>K</math> to be a finite field extension of <math>k</math>.
+* We now use the fact that if <math>T</math> is nonempty, <math>k(T)</math> can never be finitely generated over <math>k</math>. Thus, <math>T</math> is empty, forcing <math>K</math> to be a finite field extension of <math>k</math> (This uses the fact that the polynomial ring is a [[unique factorization domain]] with infinitely many irreducibles).
+==Proof using Noether normalization theorem==
+===Facts used===
+* [[Noether normalization theorem]]
+===Proof outline===
+Suppose <math>K</math> is a finitely generated algebra over <math>k</math>, that happens to be a field. Then, by the Noether normalization theorem, there exists a polynomial algebra <math>k[x_1,x_2,\ldots,x_n]</math> inside <math>K</math> such that <math>K</math> is finite over this polynomial algebra.
+But <math>K</math> being a field, must at any rate contain the field of fractions of <math>k[x_1,x_2,\ldots,x_n]</math>, and this yields a contradiction if <math>n \ge 1</math>.
+Alternatively, we can observe that the injective map from <math>k[x_1,x_2,\ldots,x_n]</math> to <math>K</math> being a finite morphism, must give a surjective map on spectra, but the spectrum of <math>K</math> has one element, and so can surject to the spectrum of a polynomial ring only if <math>n = 0</math>.
+This yields that <math>K</math> is finite-dimensional over <math>k</math>.