LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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 prime lies in the support exactly when some element has annihilator inside it

Statement

For a left $R$ -module $M$ and a prime ideal $p$ of $R$ ,

p \in {Supp}_{R} (M) ⟺ {Ann}_{R} (m) \subseteq p for some m \in M .

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , and a prime ideal $p$ .

[L1]

The support condition $p \in {Supp}_{R} (M)$ means $M_{p} \neq 0$ , and localisation at $p$ uses denominators outside $p$ (Support of a module, Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

[L2]

The annihilator of $m \in M$ is ${Ann}_{R} (m) = {r \in R : r m = 0}$ (Annihilators, torsion elements and the torsion subset of a module).

[L3]

A localised fraction is zero exactly when one denominator kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

Proof

technique · direct

1.1

Suppose $p \in {Supp}_{R} (M)$ and choose $m / s \neq 0$ in $M_{p}$ . If $t \notin p$ satisfied $t m = 0$ , then [L3] would give $m / s = 0$ . Hence every element of ${Ann}_{R} (m)$ lies in $p$ , so ${Ann}_{R} (m) \subseteq p$ .

L1L2L3

1.2

Conversely, if ${Ann}_{R} (m) \subseteq p$ and $m / 1 = 0$ in $M_{p}$ , then [L3] gives $t \notin p$ with $t m = 0$ , so $t \in {Ann}_{R} (m) \subseteq p$ , a contradiction. Thus $m / 1 \neq 0$ , so $M_{p} \neq 0$ .

L1L2L3

2.1

Steps 1.1 and 1.2 prove the equivalence.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

10 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., Section 13 (standard reference, not scraped)