Krull intersection theorem for Noetherian domains - Revision history

Vipul: 4 revisions

2008-05-12T16:26:17Z

4 revisions

← Older revision	Revision as of 16:26, 12 May 2008
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Vipul: Krull intersection theorem for domains moved to Krull intersection theorem for Noetherian domains

2008-03-15T23:53:44Z

Krull intersection theorem for domains moved to Krull intersection theorem for Noetherian domains

← Older revision	Revision as of 23:53, 15 March 2008
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Vipul at 19:18, 3 March 2008

2008-03-03T19:18:20Z

← Older revision		Revision as of 19:18, 3 March 2008
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			{{Noetherian ring result}}
	{{applicationof\|Krull intersection theorem}}		{{applicationof\|Krull intersection theorem}}

Vipul: /* Statement */

2008-03-03T19:17:49Z

Statement

← Older revision		Revision as of 19:17, 3 March 2008
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	==Statement==		==Statement==

	Let <math>R</math> be an [[integral domain]] and <math>I</math> a [[proper ideal]] in <math>R</math>. Then, we have:		Let <math>R</math> be a [[Noetherian domain]] (i.e. a [[Noetherian ring]] that is also an [[integral domain]]) and <math>I</math> a [[proper ideal]] in <math>R</math>. Then, we have:

	<math>\bigcap_{j=1}^\infty I^j = 0</math>		<math>\bigcap_{j=1}^\infty I^j = 0</math>

Vipul: New page: {{applicationof|Krull intersection theorem}} ==Statement== Let $R$ be an integral domain and $I$ a proper ideal in $R$ . Then, we have: $\bi...$

2008-03-03T19:09:43Z

New page: {{applicationof|Krull intersection theorem}} ==Statement== Let <math>R</math> be an integral domain and <math>I</math> a proper ideal in <math>R</math>. Then, we have: <math>\bi...

New page

{{applicationof|Krull intersection theorem}}

==Statement==

Let <math>R</math> be an [[integral domain]] and <math>I</math> a [[proper ideal]] in <math>R</math>. Then, we have:

<math>\bigcap_{j=1}^\infty I^j = 0</math>

==Proof==

===Applying Krull intersection theorem for modules===

The Krull intersection theorem states that if <math>M</math> is a [[finitely generated module]] over a [[Noetherian ring]] <math>R</math> and <math>I</math> is an [[ideal]] inside <math>R</math>, then there exists <math>r \in I</math> such that:

<math>(1 - r)\left( \bigcap_{j=1}^\infty I^jM \right) = 0</math>

We apply this to the case where <math>M = R</math>, to get that there exists <math>r \in I</math>, such that:

<math>(1 - r)\left( \bigcap_{j=1}^\infty I^j\right)=0</math>

===Applying the integral domain condition and properness of the ideal===

Since <math>I</math> is a ''proper'' ideal, <math>1 \notin I</math>. Hence <math>r \ne 1</math>, so the element <math>1 - r</math> cannot be zero.

Thus, by the fact that we are in an integral domain, and by the above equation, we get:

<math>\bigcap_{j=1}^\infty I^j = 0</math>

Krull intersection theorem for Noetherian domains - Revision history

Vipul: 4 revisions

Vipul: Krull intersection theorem for domains moved to Krull intersection theorem for Noetherian domains

Vipul at 19:18, 3 March 2008

Vipul: /* Statement */

Vipul: New page: {{applicationof|Krull intersection theorem}} ==Statement== Let R be an integral domain and I a proper ideal in R. Then, we have: \bi...

Vipul: New page: {{applicationof|Krull intersection theorem}} ==Statement== Let $R$ be an integral domain and $I$ a proper ideal in $R$ . Then, we have: $\bi...$