LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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.

A finite primary decomposition can be stripped of redundant components

Statement

N = Q_{1} \cap \dots \cap Q_{r}

is a finite primary decomposition of a submodule $N \subseteq M$ , then some subfamily of the $Q_{i}$ has the same intersection and is irredundant.

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , a submodule $N \subseteq M$ , and a finite primary decomposition $N = Q_{1} \cap \dots \cap Q_{r}$ .

[L1]

Minimality requires both that no component be redundant and that the component radicals be pairwise distinct (Primary decompositions, minimality, and isolated components).

Proof

technique · direct

1.1

If the displayed decomposition has no redundant component, then it is already irredundant, which is the first condition recorded in [L1]. Otherwise choose an index $j$ such that $Q_{j} \supseteq ⋂_{i \neq j} Q_{i}$ . Then $N = ⋂_{i = 1}^{r} Q_{i} = ⋂_{i \neq j} Q_{i},$ so removing $Q_{j}$ preserves the intersection.

L1givenchoosealgebra

2.1

Each removal in step 1.1 shortens the finite list of components by one. Repeating step 1.1 therefore terminates after finitely many deletions and produces a decomposition with the same intersection and no redundant component.

step 1.1algebra

3.1

This is the required irredundant subfamily.

step 2.1∎

Depends on

Primary decompositions, minimality, and isolated components

Used by

Dependency tree · two levels

4 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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §18 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §19 (standard reference, not scraped)