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 primary component is recovered by contracting its localization away from the radical

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, let $Q ⊊ M$ be a $p$ -primary submodule, and let the radical $p$ be prime. Let $S \subseteq R$ be multiplicative with $S \cap p = \emptyset$ . Then

Q = {m \in M : m / 1 \in S^{- 1} Q} .

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , a $p$ -primary submodule $Q ⊊ M$ for a prime ideal $p$ , and a multiplicative subset $S \subseteq R$ disjoint from $p$ .

[L1]

For the quotient $N = M / Q$ , every $a \notin p$ acts injectively on $N$ (Primary submodules of finite modules are characterized by a singleton associated-prime set).

[L2]

Localisation commutes with quotient modules, so $S^{- 1} (M / Q) ≅ (S^{- 1} M) / (S^{- 1} Q)$ (Localisation commutes with quotient modules and arbitrary direct sums).

Proof

technique · direct

1.1

The inclusion $Q \subseteq {m \in M : m / 1 \in S^{- 1} Q}$ is immediate, because every element of $Q$ localizes into $S^{- 1} Q$ .

given

1.2

Conversely, let $m \in M$ with $m / 1 \in S^{- 1} Q$ . Under the identification of [L2], the class of $m + Q$ in $S^{- 1} (M / Q)$ is zero. Hence some $s \in S$ satisfies $s (m + Q) = 0$ in $M / Q$ , so $s m \in Q$ . Since $s \notin p$ , [L1] makes multiplication by $s$ injective on $M / Q$ , and the equality $s (m + Q) = 0$ forces $m + Q = 0$ . Therefore $m \in Q$ .

L1L2choosealgebra

2.1

Steps 1.1 and 1.2 prove that $Q$ is exactly the contraction of $S^{- 1} Q$ .

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

16 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 (18.23) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Proposition 19.4 (standard reference, not scraped)