Krull intersection theorem for Noetherian domains: Difference between revisions

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{{Noetherian ring result}}
{{applicationof|Krull intersection theorem}}
{{applicationof|Krull intersection theorem}}



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=1∞Ij=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 r∈I such that:

(1−r)(⋂j=1∞IjM)=0

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

(1−r)(⋂j=1∞Ij)=0

Applying the integral domain condition and properness of the ideal

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

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

⋂j=1∞Ij=0