Krull intersection theorem for Noetherian domains

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:

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