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

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

An explicit Artin-Rees number can be computed for a submodule inside a finite module

Example

Take $R = k [x]$ , $I = (x)$ , $M = R$ , and $N = (x^{m})$ for a fixed integer $m \geq 0$ . Then for every $n \geq m$ ,

I^{n} M \cap N = (x^{n}) \cap (x^{m}) = (x^{n}) = I^{n - m} (I^{m} M \cap N) .

So in this case one may take the Artin-Rees number to be $c = m$ .

Facts & Assumptions

Given: A field $k$ , an integer $m \geq 0$ , the ring $R = k [x]$ , the ideal $I = (x)$ , the module $M = R$ , and the submodule $N = (x^{m})$ .

[L1]

Artin-Rees gives some constant $c$ 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).

Verification

technique · direct

1.1

Here $I^{n} M = (x^{n})$ and $N = (x^{m})$ . If $n \geq m$ , then $(x^{n}) \subseteq (x^{m})$ , so $I^{n} M \cap N = (x^{n}) .$

givenalgebra

1.2

Also $I^{m} M \cap N = (x^{m})$ , and therefore for $n \geq m$ , $I^{n - m} (I^{m} M \cap N) = (x^{n - m}) (x^{m}) = (x^{n}) = I^{n} M \cap N .$

algebra

2.1

So $c = m$ works, exhibiting an explicit Artin-Rees bound compatible with the abstract existence statement [L1].

L1∎

Depends on

Artin-Rees controls intersections of submodules with high ideal powers

Used by

Nothing in the library uses this result yet.

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)