TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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

Field Kunneth isomorphism for homology of products Corollary
Under the Axiom of Choice, every module admits a projective resolution Corollary
Under the stated choice boundary, free modules are projective and hence flat Corollary
A projective resolution of a cyclic abelian group Example
An Ext dimension shift Example
Homology of a product of spheres by Kunneth Example
The idempotent completion of the matrix category gives the finitely generated projective modules Example
Tor term in the homology of a product of real projective spaces Example
Weak and global dimension for a field and the integers Example
ℤ/2ℤ is projective but not free over ℤ/6ℤ Example
Every projective module is free False statement
FALSE: objectwise projective-resolution choices uniquely determine a resolution functor False statement
FALSE: the Yoneda product is graded commutative for every abelian category False statement
Vanishing Tor one does not require a projective factor False statement
A free PID complex decomposes into two-term cycle-boundary pieces Lemma
Ext one of Z modulo n by Z is Z modulo n Lemma
Relative cohomological Kunneth under finite free homology hypotheses Lemma
Singular UCT extension from cycle projections Lemma
The cohomological universal-coefficient extension map Lemma
The cycle-boundary tensor sequence has the Kunneth kernel and cokernel Lemma
The Kunneth Tor map Lemma
The iterated free-module resolution is canonical in ZF Proposition
Cohomological Kunneth isomorphism under finite free hypotheses Theorem
Direct sums of projectives are projective, and module categories have enough projectives Theorem
Equivalent characterizations of projective modules Theorem
Higher Tor over the integers vanishes Theorem
The cohomology universal-coefficient sequence splits nonnaturally Theorem
The integers have global dimension one Theorem
The Kunneth sequence splits nonnaturally Theorem
The universal coefficient theorem for cohomology over a PID Theorem

Dependency tree · two levels

12 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. 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)