TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-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.

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Assuming the Axiom of Choice, Nakayama's lemma

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring, let $I ⊴ R$ satisfy $I \subseteq J (R)$ , and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then $M = 0$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ with $I \subseteq J (R)$ , and a finitely generated left $R$ -module $M$ with $I M = M$ .

[L1]

An element $x$ lies in $J (R)$ exactly when $1 - r x$ is a unit for every $r \in R$ (Assuming the Axiom of Choice, an element lies in the Jacobson radical exactly when one minus any multiple is a unit).

[L2]

If $I M = M$ for finite $M$ , then $(1 - a) M = 0$ for some $a \in I$ (Determinant trick for Nakayama).

Proof

technique · direct

1.1

By [L2], choose $a \in I$ with $(1 - a) M = 0$ . Since $a \in I \subseteq J (R)$ , [L1] makes $1 - a$ a unit.

L1L2givenchoose

2.1

Multiplying the equality $(1 - a) m = 0$ by $(1 - a)^{- 1}$ shows $m = 0$ for every $m \in M$ . Therefore $M = 0$ .

step 1.1algebra∎

Depends on

Used by

Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators Corollary

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., Exercise 10.12 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Lemma 3.9 (standard reference, not scraped)