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Over a Noetherian ring, an ideal filtration is stable exactly when its Rees module is finite, and the Rees algebra is Noetherian

Statement

Let $R$ be a Noetherian commutative ring, let $I \subset R$ be an ideal, and let $M = M_{0} \supseteq M_{1} \supseteq \dots$ be a filtration of a finite $R$ -module $M$ such that $I M_{n} \subseteq M_{n + 1}$ for all $n$ .

The filtration is $I$ -stable when $M_{n + 1} = I M_{n}$ for all sufficiently large $n$ . Then:

the filtration is $I$ -stable if and only if its Rees module $R (M_{∙})$ is a finite graded module over the Rees algebra $R (I)$ ;
the Rees algebra $R (I)$ is Noetherian.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , an ideal $I \subset R$ , and a finite $R$ -module $M$ with filtration $M_{∙}$ as above.

[L1]

The Rees algebra and Rees module are the graded objects $R (I) = ⨁_{n \geq 0} I^{n} t^{n}, R (M_{∙}) = ⨁_{n \geq 0} M_{n} t^{n}$ (The Rees algebra of an ideal and the Rees module of a filtered module, Nonnegatively graded rings and modules, homogeneous elements, and twists).

[L2]

Over a Noetherian ring, every finitely generated module is Noetherian, so each of its submodules is finitely generated (Over a Noetherian ring a module is Noetherian exactly when it is finitely generated, exactly when it is finitely presented).

[L3]

In a Noetherian ring every ideal is finitely generated (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

[L4]

A polynomial ring in finitely many variables over a Noetherian commutative ring is Noetherian (If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ ).

Proof

technique · direct

1.1

If the filtration is $I$ -stable from some index $c$ onward, then $R (M_{∙}) = \sum_{n = 0}^{c} R (I) \cdot (M_{n} t^{n}),$ because for $n > c$ every element of $M_{n} t^{n}$ is a product of an element of $I^{n - c} t^{n - c}$ with one of $M_{c} t^{c}$ . Since $R$ is Noetherian and $M$ is finite, [L2] makes each submodule $M_{n} \subseteq M$ finitely generated. Therefore finitely many homogeneous elements in degrees at most $c$ generate $R (M_{∙})$ over $R (I)$ .

L1L2givenalgebra

1.2

Conversely, suppose $R (M_{∙})$ is generated over $R (I)$ by homogeneous elements lying in degrees at most $c$ . For $n \geq c$ , every element of $M_{n + 1} t^{n + 1}$ is therefore a sum of products $(I^{n + 1 - d} t^{n + 1 - d}) (m_{d} t^{d})$ with $d \leq c$ , hence lies in $I M_{n} t^{n + 1}$ . Thus $M_{n + 1} \subseteq I M_{n}$ . The reverse inclusion is part of the filtration hypothesis, so $M_{n + 1} = I M_{n}$ for all $n \geq c$ .

L1givenalgebra

1.3

By [L3], choose generators $I = (f_{1}, \dots, f_{s})$ . Sending $X_{i}$ to $f_{i} t$ defines a surjective graded map $R [X_{1}, \dots, X_{s}] ↠ R (I) .$ By [L4] its source is Noetherian, so its quotient $R (I)$ is Noetherian.

L3L4construct

2.1

Steps 1.1 and 1.2 prove the equivalence, and step 1.3 proves that $R (I)$ is Noetherian.

step 1.1step 1.2step 1.3∎

Depends on

Used by

Artin-Rees controls intersections of submodules with high ideal powers Theorem

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, Lemma (20.17) (standard reference, not scraped)
Stacks Project, Sections 10.51 and 10.70 (standard reference, not scraped)