LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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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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A finite primary decomposition can be stripped of redundant components

Statement

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

1.1L1givenchoosealgebra

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.

2.1step 1.1algebra

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.

3.1step 2.1∎

This is the required irredundant subfamily.

Depends on

Primary decompositions, minimality, and isolated components

A finite primary decomposition can be stripped of redundant components

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources