Krull intersection theorem for modules - Revision history

Vipul: 9 revisions

2008-05-12T16:26:32Z

9 revisions

← Older revision	Revision as of 16:26, 12 May 2008
(No difference)

Vipul: Krull intersection theorem moved to Krull intersection theorem for modules

2008-03-03T19:00:39Z

Krull intersection theorem moved to Krull intersection theorem for modules

← Older revision	Revision as of 19:00, 3 March 2008
(No difference)

Vipul at 19:00, 3 March 2008

2008-03-03T19:00:04Z

← Older revision		Revision as of 19:00, 3 March 2008
Line 1:		Line 1:
	{{Noetherian ring result}}		{{Noetherian ring result}}
	{{applicationof\|Artin-Rees lemma}}		{{applicationof\|Artin-Rees lemma}}
	{{~~aplpicationof~~\|Cayley-Hamilton theorem}}		{{applicationof\|Cayley-Hamilton theorem}}
	==Statement==		==Statement==

Vipul at 18:59, 3 March 2008

2008-03-03T18:59:53Z

← Older revision		Revision as of 18:59, 3 March 2008
Line 1:		Line 1:
	{{Noetherian ring result}}		{{Noetherian ring result}}
	{{applicationof\|Artin-Rees lemma}}		{{applicationof\|Artin-Rees lemma}}
	{{~~applicationof~~\|~~Nakayama's lemma~~}}		{{aplpicationof\|Cayley-Hamilton theorem}}
	==Statement==		==Statement==

Line 12:		Line 12:

	* [[Artin-Rees lemma]]		* [[Artin-Rees lemma]]
	* [[~~Nakayama's lemma~~]]		* [[Cayley-Hamilton theorem]]

			==Applications==

			* [[Krull intersection theorem for Jacobson radical]], also covers the case of a [[local ring]]
			* [[Krull intersection theorem for domains]]

	==Proof==		==Proof==
Line 42:		Line 47:
	Since <math>IN = N</math>, we can find an element <math>r \in I</math> such that <math>(1 - r)N = 0</math>. This is an application of the [[Cayley-Hamilton theorem]]: we first find the Cayley-Hamilton polynomial, then observe that <math>1</math> is a root of the polynomial, and then take the negative of the sum of all coefficients of higher degree terms.		Since <math>IN = N</math>, we can find an element <math>r \in I</math> such that <math>(1 - r)N = 0</math>. This is an application of the [[Cayley-Hamilton theorem]]: we first find the Cayley-Hamilton polynomial, then observe that <math>1</math> is a root of the polynomial, and then take the negative of the sum of all coefficients of higher degree terms.

	~~===Proving the intersection is trivial for an integral domain===~~

	If <math>R</math> is an integral domain, and we set <math>M = R</math> in the above, we find an element <math>r \in I</math> such that <math>(1 - r)(\bigcap_j I^j) = 0</math>. Thus <math>1 - r</math> is a zero divisor on the intersection. Since <math>R</math> is an [[integral domain]], one of these must hold:

	* <math>1 - r = 0</math>. This forces <math>1 \in I</math>, contradicting the assumption that <math>I</math> is a [[proper ideal]]
	* <math>\bigcap_j I^j = 0</math>

	~~This completes the proof.~~

	~~===Proving that the intersection is trivial for a local ring===~~

	In this case, <math>I</math> is contained inside the unique maximal ideal <math>\mathfrak{m}</math>, which is the Jacobson radical. As above, let <math>N = \bigcap_j I^j</math>. Then, we have <math>IN = N</math>. Since <math>I</math> is contained in the [[Jacobson radical]], this forces <math>N = 0</math>.
	==References==		==References==

Vipul: /* Statement */

2008-03-03T18:58:35Z

Statement

← Older revision		Revision as of 18:58, 3 March 2008
Line 4:		Line 4:
	==Statement==		==Statement==

	Let <math>R</math> be a [[Noetherian ring]] and <math>I</math> be an [[ideal]] inside <math>R</math>.		Let <math>R</math> be a [[Noetherian ring]] and <math>I</math> be an [[ideal]] inside <math>R</math>. Suppose <math>M</math> is a [[finitely generated module]] over <math>R</math>. Then, we have the following:

	* If <math>M</math> ~~is a finitely generated~~ <math>R</math>~~-module, then there~~ exists <math>r \in I</math> such that:		# Let <math>N = \bigcap_{j=1}^\infty I^j M</math>. Then, <math>IN = N</math>
			# There exists <math>r \in I</math> such that <math>(1 - r)N = 0</math>
	<math>(1-r)~~(\bigcap_1^\infty I^jM) = 0</math>~~

	* If <math>R</math> is an [[integral domain]] or a [[local ring]] and <math>I</math> is a [[proper ideal]] then:

	~~<math>\bigcap_1^\infty I^j~~ = 0</math>

	==Results used==		==Results used==

Vipul at 18:21, 3 March 2008

2008-03-03T18:21:57Z

@@ Line 1: / Line 1: @@
 {{applicationof|Artin-Rees lemma}}
 {{applicationof|Nakayama's lemma}}
@@ Line 18: / Line 19: @@
 * [[Nakayama's lemma]]
 ==References==

Vipul: /* References */

2008-03-03T17:58:39Z

References

← Older revision		Revision as of 17:58, 3 March 2008
Line 20:		Line 20:
	==References==		==References==

	* ''''Dimensionstheorie in Stellenringen'' by [[Wolfgang Krull], 1938		* ''''Dimensionstheorie in Stellenringen'' by [[Wolfgang Krull]], 1938

	===Textbook references===		===Textbook references===
	* {{booklink\|Eisenbud}}, Page 152		* {{booklink\|Eisenbud}}, Page 152

Vipul at 17:58, 3 March 2008

2008-03-03T17:58:27Z

← Older revision		Revision as of 17:58, 3 March 2008
Line 17:		Line 17:
	* [[Artin-Rees lemma]]		* [[Artin-Rees lemma]]
	* [[Nakayama's lemma]]		* [[Nakayama's lemma]]

			==References==

			* ''''Dimensionstheorie in Stellenringen'' by [[Wolfgang Krull], 1938

			===Textbook references===
			* {{booklink\|Eisenbud}}, Page 152

Vipul at 17:51, 3 March 2008

2008-03-03T17:51:48Z

← Older revision		Revision as of 17:51, 3 March 2008
Line 1:		Line 1:
			{{applicationof\|Artin-Rees lemma}}
			{{applicationof\|Nakayama's lemma}}
	==Statement==		==Statement==

Vipul at 12:20, 7 August 2007

2007-08-07T12:20:46Z

New page

==Statement==

Let <math>R</math> be a [[Noetherian ring]] and <math>I</math> be an [[ideal]] inside <math>R</math>.

* If <math>M</math> is a finitely generated <math>R</math>-module, then there exists <math>r \in I</math> such that:

<math>(1-r)(\bigcap_1^\infty I^jM) = 0</math>

* If <math>R</math> is an [[integral domain]] or a [[local ring]] and <math>I</math> is a [[proper ideal]] then:

<math>\bigcap_1^\infty I^j = 0</math>

==Results used==

* [[Artin-Rees lemma]]
* [[Nakayama's lemma]]