Krull intersection theorem for Noetherian domains

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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:

\bigcap_{j=1}^\infty I^j = 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 \in I such that:

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

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

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

Applying the integral domain condition and properness of the ideal

Since I is a proper ideal, 1 \notin I. Hence r \ne 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:

\bigcap_{j=1}^\infty I^j = 0