TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passverified 2026-09-23 (gpt-6-sol)

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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: The Axiom of Choice (The Axiom of Choice), 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]

Under AC, 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). This is the only use of AC in the proof.

[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

1.1L1L2givenchoose

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

2.1step 1.1algebra∎

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

Assuming the Axiom of Choice, Nakayama's lemma

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources