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Artin-Rees controls intersections of submodules with high ideal powers

Statement

Let $R$ be a Noetherian commutative ring, let $I \subset R$ be an ideal, let $M$ be a finite $R$ -module, and let $N \subseteq M$ be a submodule. Then there exists an integer $c \geq 0$ such that $I^{n} M \cap N = I^{n - c} (I^{c} M \cap N)$ for every $n \geq c$ .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , an ideal $I \subset R$ , a finite $R$ -module $M$ , and a submodule $N \subseteq M$ .

[L1]

For the $I$ -adic filtration on $M$ and any induced filtration on a finite submodule, Rees-module finiteness is equivalent to eventual stability, and the Rees algebra is Noetherian (Over a Noetherian ring, an ideal filtration is stable exactly when its Rees module is finite, and the Rees algebra is Noetherian).

[L2]

A finite module over a Noetherian ring is Noetherian, so each submodule of it is finite (Over a Noetherian ring a module is Noetherian exactly when it is finitely generated, exactly when it is finitely presented).

Proof

technique · direct

1.1L1given

The $I$ -adic filtration on $M$ is already stable, since $I (I^{n} M) = I^{n + 1} M$ for every $n \geq 0$ . Hence [L1] makes $R_{I} (M) : = ⨁_{n \geq 0} I^{n} M t^{n}$ a finite module over the Noetherian ring $R (I)$ .

2.1L1L2step 1.1algebra

The induced filtration $N_{n} : = I^{n} M \cap N$ defines a graded submodule $R_{N} : = ⨁_{n \geq 0} N_{n} t^{n} \subseteq R_{I} (M) .$ The module $R_{I} (M)$ is finite over the Noetherian ring $R (I)$ by step 1.1 and [L1], hence is Noetherian by [L2]. Therefore its submodule $R_{N}$ is finite.

3.1L1step 2.1

Applying the stability direction of [L1] to the finite Rees module established in step 2.1 yields an index $c$ with $N_{n} = I^{n - c} N_{c} (n \geq c) .$ Since $N_{n} = I^{n} M \cap N$ and $N_{c} = I^{c} M \cap N$ , this is exactly the displayed Artin-Rees equality.

4.1step 3.1∎

Therefore the required constant $c$ exists.

Artin-Rees controls intersections of submodules with high ideal powers

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources