False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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.

Every projective module is free

Statement

False. Every projective module is free.

Facts & Assumptions

Given: $R = Z / 6 Z$ and the ideal $P = 3 R = {0, 3}$ .

[L1]

Under AC, projective modules are exactly the direct summands of free modules (Equivalent characterizations of projective modules).

[L2]

A free module with a finite basis is projective using only finite choice, which is provable in ZF (Free modules are projective, with the exact choice boundary).

[F1]

A free module is a direct sum of copies of the regular module (The free module on a set and its standard basis).

Refutation

technique · direct

1.1

Modulo six, $3$ and $4$ are orthogonal idempotents with sum one, so $R = 3 R \oplus 4 R$ . The rank-one free module $R$ is projective without AC by [L2], and the direct lifting argument for a summand makes $P$ projective; this is the finite instance of [L1].

givenL1L2algebra

1.2

The module $P$ has two elements. By [F1], a free module on no basis vectors has one element, one on a nonempty finite set of size $k$ has $6^{k}$ elements, and one on an infinite set is infinite. Therefore $P$ is not free.

F1algebra

2.1

The projective nonfree module $P$ refutes the statement.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 29 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

A. Kleshchev, Lectures on Abstract Algebra for Graduate Students, sections 3.6, 3.14, and 3.15 (standard reference, not scraped)
The Stacks Project, Algebra (standard reference, not scraped)
P. Hekmati, Homological Algebra, section 3.1 (standard reference, not scraped)