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.

Equal-radical primary components can be combined

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let

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

be a finite primary decomposition of a submodule $N \subseteq M$ , with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ . If several of the $p_{i}$ equal one prime $p$ , then replacing that whole group by the intersection of its components preserves the total intersection and produces one $p$ -primary component.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a finite primary decomposition $N = Q_{1} \cap \dots \cap Q_{r}$ with each $Q_{i}$ $p_{i}$ -primary for a prime ideal $p_{i}$ .

[L1]

For a prime ideal $p$ , a nonempty finite intersection of $p$ -primary submodules is again $p$ -primary (A finite intersection of primary submodules with one radical is primary).

[L2]

Minimality requires both irredundancy and pairwise distinct component radicals (Primary decompositions, minimality, and isolated components).

Proof

technique · direct

1.1

Fix a prime $p$ and let $I (p)$ be the set of indices $i$ for which $Q_{i}$ is $p$ -primary. If $I (p)$ is empty or has one element, there is nothing to combine. Otherwise set $Q (p) = ⋂_{i \in I (p)} Q_{i} .$ Fact [L1] shows that $Q (p)$ is again $p$ -primary.

L1givenconstruct

2.1

Replacing, for each prime $p$ occurring among the radicals, the whole block ${Q_{i} : i \in I (p)}$ by the single component $Q (p)$ does not change the total intersection, because intersections may be regrouped without changing their value. The resulting components have pairwise distinct radicals, which is the second minimality requirement recorded in [L2].

L2step 1.1algebra

3.1

Thus equal-radical components may be combined into one primary component with the same radical.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

6 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., Lemma (18.12) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Proposition 19.6 (standard reference, not scraped)