---
tag: 30H6B0G2FT
kind: statement
role: lemma
name: Artin–Rees lemma
msc: [13A30, "13E05"]
uses: [0670T59HGE, 096Z9MP2ME, 05QY1VEBYD, 05WDSN597C, W8ZAH8TTHS, WHME2NGANG]
about: [WHME2NGANG]
sources:
  - url: https://en.wikipedia.org/wiki/Artin–Rees_lemma
same_as:
  wikidata: Q3229329
  wikipedia: Artin–Rees lemma
---

## Statement

Let $R$ be a commutative [[Noetherian ring|0670T59HGE]] and let $I$ be an [[ideal|096Z9MP2ME]] of $R$.
Let $M$ be a [[finitely generated|05QY1VEBYD]] $R$-module, and let $N$ be a [[submodule|05WDSN597C]] of $M$.
Then there is an integer $k \ge 0$ such that
$$I^n M \cap N = I^{n-k} (I^k M \cap N) \quad \text{for all } n \ge k.$$

## Remarks

The [[Rees algebra|W8ZAH8TTHS]] $R[It]$ is Noetherian because $R$ is Noetherian and $I$ is finitely generated.
The module $\bigoplus_n I^n M$ is a finitely generated module over $R[It]$, so its graded submodule $\bigoplus_n (I^n M \cap N)$ is finitely generated too.
A finite set of homogeneous generators of degree at most $k$ gives the equality for $n \ge k$.
In words, the [[filtration|WHME2NGANG]] $(I^n M \cap N)_n$ of $N$ is a stable $I$-filtration.
Consequently the $I$-adic topology of $M$ induces the $I$-adic topology on $N$, since $I^n M \cap N \subseteq I^{n-k} N$.
Example: for $R = M = \Z$, $I = 2\Z$ and $N = 6\Z$, one has $2^n \Z \cap 6\Z = 3 \cdot 2^n \Z = 2^{n-1}(2\Z \cap 6\Z)$ for $n \ge 1$, so $k = 1$ works.

## Explanations

### L1 human

Take a big module and a smaller module inside it.
There are two ways to say that an element of the small module is tiny.
It can be a multiple of a high power of $I$ inside the big module, or inside the small module.
The lemma says that these two ways agree up to a fixed shift $k$.

### L2 human

The lemma is the main technical tool for completions of Noetherian rings.
It implies the [[Krull intersection theorem|1Q1AH2V41S]], and it shows that $I$-adic completion is exact on finitely generated modules, which gives [[flatness of completion|NPF42DJMBX]].