Weak nullstellensatz for arbitrary fields - Revision history

Vipul: 6 revisions

2008-05-12T18:05:36Z

6 revisions

← Older revision	Revision as of 18:05, 12 May 2008
(No difference)

Vipul: /* Proof outline */

2008-02-08T22:21:48Z

Proof outline

← Older revision		Revision as of 22:21, 8 February 2008
Line 26:		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 [[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 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> (This uses the fact that the polynomial ring is a [[unique factorization]] ~~domain~~ with infinitely many irreducibles).		* 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==		==Proof using Noether normalization theorem==

Vipul: /* Proof outline */

2008-02-08T22:21:33Z

Proof outline

← Older revision		Revision as of 22:21, 8 February 2008
Line 26:		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 [[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 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> (This uses the fact that the polynomial ring is a unique factorization domain with infinitely many irreducibles).		* 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==		==Proof using Noether normalization theorem==

Vipul at 20:42, 2 February 2008

2008-02-02T20:42:53Z

← Older revision		Revision as of 20:42, 2 February 2008
Line 1:		Line 1:
			{{applicationof\|Artin-Tate lemma}}

			{{applicationof\|Noether normalization theorem}}

	==Statement==		==Statement==

Vipul at 20:42, 2 February 2008

2008-02-02T20:42:16Z

@@ Line 12: / Line 12: @@
 ==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 22: @@
 * 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).

Vipul at 20:51, 19 January 2008

2008-01-19T20:51:00Z

← Older revision		Revision as of 20:51, 19 January 2008
Line 1:		Line 1:
	==Statement==		==Statement==

	Suppose <math>k</math> is a [[field]] and <math>K</math> is a field extension of <math>k</math>, such that <math>K</math> is finitely generated as a <math>k</math>-algebra. Then, <math>K</math> is algebraic over <math>k</math>, and in fact, is a finite field extension of <math>k</math>.		Here are two equivalent formulations:

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

	==Applications==		==Applications==

Vipul at 20:49, 19 January 2008

2008-01-19T20:49:54Z

New page

==Statement==

Suppose <math>k</math> is a [[field]] and <math>K</math> is a field extension of <math>k</math>, such that <math>K</math> is finitely generated as a <math>k</math>-algebra. Then, <math>K</math> is algebraic over <math>k</math>, and in fact, is a finite field extension of <math>k</math>.

==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 <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>.