Krull intersection theorem for Noetherian domains

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Revision as of 19:09, 3 March 2008 by Vipul (talk | contribs) (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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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 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