# Krull intersection theorem for Noetherian domains

From Commalg

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

## Contents

## Statement

Let be a Noetherian domain (i.e. a Noetherian ring that is also an integral domain) and a proper ideal in . Then, we have:

## Proof

### Applying Krull intersection theorem for modules

The Krull intersection theorem states that if is a finitely generated module over a Noetherian ring and is an ideal inside , then there exists such that:

We apply this to the case where , to get that there exists , such that:

### Applying the integral domain condition and properness of the ideal

Since is a *proper* ideal, . Hence , so the element cannot be zero.

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