CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The filtration induced on a submodule is equivalent to its intrinsic ideal-adic filtration

Statement

Let $R$ be Noetherian, 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 $c \geq 0$ such that for every $n \geq c$ , $I^{n} N \subseteq I^{n} M \cap N \subseteq I^{n - c} N .$ Equivalently, the filtration induced from the $I$ -adic filtration of $M$ and the intrinsic $I$ -adic filtration of $N$ agree up to a bounded shift.

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]

Artin-Rees gives $c \geq 0$ with $I^{n} M \cap N = I^{n - c} (I^{c} M \cap N)$ for all $n \geq c$ (Artin-Rees controls intersections of submodules with high ideal powers).

Proof

technique · direct

1.1given

For every $n$ , one always has $I^{n} N \subseteq I^{n} M \cap N$ , since $N \subseteq M$ .

1.2L1algebra

Choose $c$ as in [L1]. Then for $n \geq c$ , $I^{n} M \cap N = I^{n - c} (I^{c} M \cap N) \subseteq I^{n - c} N$ because $I^{c} M \cap N \subseteq N$ .

2.1algebra∎

Combining the preceding steps gives the two-sided eventual inclusion, hence the two filtrations are equivalent.

Depends on

Artin-Rees controls intersections of submodules with high ideal powers

Used by

Adic completion is exact on finite modules over a Noetherian ring Theorem

Dependency tree · two levels

3 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 (20.18) (standard reference, not scraped)
Stacks Project, Section 10.96 (standard reference, not scraped)