Krull intersection theorem for Jacobson radical

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 for modules
View other applications of Krull intersection theorem for modules OR Read a survey article on applying Krull intersection theorem for modules

This fact is an application of the following pivotal fact/result/idea: Nakayama's lemma
View other applications of Nakayama's lemma OR Read a survey article on applying Nakayama's lemma

Statement

Let $R$ be a Noetherian ring and $I$ an ideal contained inside the Jacobson radical of $R$. Then, we have:

$\bigcap _{j=1}^{\mathrm{\infty }}{I}^j=0$

In particular, when $R$ is a local ring, then the above holds for any proper ideal $I$.

Proof

Applying the Krull intersection theorem for modules

We apply the Krull intersection theorem for modules, which states that if $R$ is a Noetherian ring and $M$ is a finitely generated module over $R$, and $I$ is an ideal in $R$, we have:

$I(\bigcap _{j=1}^{\mathrm{\infty }}{I}^{j}M)=\bigcap _{j=1}^{\mathrm{\infty }}{I}^{j}M$

We apply it to the case $M=R$. We thus get:

$I(\bigcap _{j=1}^{\mathrm{\infty }}{I}^j)=\bigcap _{j=1}^{\mathrm{\infty }}{I}^j$

Applying Nakayama's lemma

Consider the ideal $N=\bigcap _{j=1}^{\mathrm{\infty }}{I}^j$ as a $R$-module. Since $IN=N$, and $I$ is contained in the Jacobson radical of $R$, Nakayama's lemma tells us that $N=0$. This is precisely what we want.