Krull intersection theorem for Noetherian domains: Difference between revisions

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(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...)
 
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{{Noetherian ring result}}
{{applicationof|Krull intersection theorem}}
{{applicationof|Krull intersection theorem}}


==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>

Latest revision as of 16:26, 12 May 2008

This article defines a result where the base ring (or one or more of the rings involved) is Noetherian
View more results involving Noetherianness or Read a survey article on applying Noetherianness

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

Statement

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

j=1Ij=0

Proof

Applying Krull intersection theorem for modules

The Krull intersection theorem states that if M is a finitely generated module over a Noetherian ring R and I is an ideal inside R, then there exists rI such that:

(1r)(j=1IjM)=0

We apply this to the case where M=R, to get that there exists rI, such that:

(1r)(j=1Ij)=0

Applying the integral domain condition and properness of the ideal

Since I is a proper ideal, 1I. Hence r1, so the element 1r cannot be zero.

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

j=1Ij=0