---
tag: 1Q1AH2V41S
kind: statement
role: theorem
name: Krull intersection theorem
msc: ["13E05", 13A15]
uses: [0670T59HGE, 096Z9MP2ME, 05QY1VEBYD, 30H6B0G2FT, XA0DCNH7WY, MZ02BKPHJA, 0QVD7WBFGN]
about: [096Z9MP2ME]
sources:
  - url: https://en.wikipedia.org/wiki/Artin–Rees_lemma
---

## Statement

Let $R$ be a commutative [[Noetherian ring|0670T59HGE]], let $I$ be an [[ideal|096Z9MP2ME]] of $R$, and let $M$ be a [[finitely generated|05QY1VEBYD]] $R$-module.
Then
$$\bigcap_{n \ge 0} I^n M = \set{x \in M : (1 + a) x = 0 \text{ for some } a \in I}.$$
In particular:

1. if $I$ is contained in the [[Jacobson radical|XA0DCNH7WY]] of $R$, then $\bigcap_{n} I^n M = 0$; this applies when $R$ is a [[local ring|MZ02BKPHJA]] and $I \neq R$;
2. if $R$ is an [[integral domain|0QVD7WBFGN]] and $I \neq R$, then $\bigcap_{n} I^n = 0$.

## Remarks

Let $N = \bigcap_n I^n M$.
By the [[Artin–Rees lemma|30H6B0G2FT]], $N = I^n M \cap N = I^{n-k}(I^k M \cap N) \subseteq I N$ for $n > k$, so $N = IN$.
As $N$ is finitely generated, the [[determinant trick|A8HRT9H3XA]] gives $a \in I$ with $(1 + a) N = 0$.
Conversely, if $(1 + a) x = 0$ then $x = -a x = (-a)^n x \in I^n M$ for every $n$.
Example: in $R = \Z/6\Z$ with $I = (2)$, one has $I^2 = I$, so the intersection is $I = \set{0, 2, 4}$, and indeed $3 = 1 + 2$ kills $2$ and $4$.
The Noetherian hypothesis is needed.
In the local ring of germs at $0$ of smooth functions $\R \to \R$, consider the germ of $e^{-1/x^2}$, extended by $0$ at $0$.
It is nonzero, but it lies in every power of the maximal ideal.

## Explanations

### L1 human

An element divisible by $p$, by $p^2$, by $p^3$ and so on forever must be $0$ in the integers.
The theorem says that the same holds in any Noetherian local ring and in any Noetherian domain: an element that lies in every power of a proper ideal is $0$.
In other words, the $I$-adic topology separates points.

### L2 human

Clause 1 shows that a Noetherian local ring embeds into its $\mathfrak m$-adic [[completion|KBN13NPYJQ]].
It also shows that two elements of a Noetherian local ring are equal if they agree modulo every power of $\mathfrak m$.
This is the basis of many arguments by successive approximation.