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

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Free modules are projective, with the exact choice boundary

Statement

Assume the Axiom of Choice. Every free module is projective. More precisely, if $F$ has basis $(e_{x})_{x \in X}$ , a lift of a map $F \to M$ through a surjection $E \to M$ is obtained by choosing one preimage of each basis value. For finite $X$ , finite choice suffices and no form of AC is needed; for $X = \emptyset$ , the lift is the unique map from $0$ .

Facts & Assumptions

Given: A free module $F = R^{(X)}$ , a surjection $q : E \to M$ , and a homomorphism $f : F \to M$ .

[F1]

Projectivity is the existence of a lift through every surjective homomorphism (Projective modules and the lifting property).

[L1]

A function from the basis set $X$ to a module extends uniquely to a homomorphism from $R^{(X)}$ (Universal property of the free module on a set).

[F2]

AC supplies a choice function for every family of nonempty sets (The Axiom of Choice).

[L2]

A natural-number-indexed finite family of nonempty sets has a choice function in ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · constructive

1.1

For each $x \in X$ , the fiber $q^{- 1} (f (e_{x}))$ is nonempty because $q$ is surjective.

given

2.1

Under AC, [F2] chooses $y_{x} \in q^{- 1} (f (e_{x}))$ for every $x \in X$ . If $X$ is finite with a given finite enumeration, [L2] makes this choice in ZF; if $X = \emptyset$ , there are no choices.

step 1.1F2L2choose

3.1

By [L1], the assignment $e_{x} \mapsto y_{x}$ extends uniquely to a homomorphism $\tilde{f} : F \to E$ .

step 2.1L1construct

4.1

Both $q \circ \tilde{f}$ and $f$ send each $e_{x}$ to $f (e_{x})$ , so uniqueness in [L1] gives $q \circ \tilde{f} = f$ .

step 3.1L1

5.1

Thus $F$ satisfies the lifting property [F1] and is projective. The construction records exactly where arbitrary or finite choice enters.

step 1.1step 2.1step 3.1step 4.1F1discharge-construct∎

Depends on

Used by

ℤ/2ℤ is projective but not free over ℤ/6ℤ Example
Every projective module is free False statement
Direct sums of projectives are projective, and module categories have enough projectives Theorem
Equivalent characterizations of projective modules Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 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)