TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Direct sums of projectives are projective, and module categories have enough projectives

Statement

Assume the Axiom of Choice. An arbitrary direct sum of projective left $R$ -modules is projective, and every left $R$ -module is the quotient in a short exact sequence $0 \to K \to P \to M \to 0$ with $P$ projective. Thus the category of left $R$ -modules has enough projectives.

For a finite direct sum, finite choice suffices; the empty direct sum is the zero module and is projective. The arbitrary free cover uses the full choice boundary recorded for free modules.

Facts & Assumptions

Given: A family $(P_{i})_{i \in I}$ of projective left $R$ -modules and a left $R$ -module $M$ .

[L1]

A family of component maps determines a unique map from the direct sum (Universal property of a direct sum of modules).

[L2]

Under AC every free module is projective; a finite basis needs only finite choice (Free modules are projective, with the exact choice boundary).

[L3]

The canonical map $R^{(M)} \to M$ is surjective (Every module is a quotient of a free module).

[F1]

AC chooses one element from each member of an arbitrary family of nonempty sets (The Axiom of Choice).

[L4]

A listed 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

Given a surjection $q : E \to N$ and a homomorphism $f : ⨁_{i} P_{i} \to N$ , each component $f_{i} = f \circ ȷ_{i}$ has a nonempty set of lifts $P_{i} \to E$ because $P_{i}$ is projective.

given

1.2

By [L3], $ε_{M} : R^{(M)} \to M$ is a canonical surjection. Under AC, [L2] makes $R^{(M)}$ projective, so with $K = \ker ε_{M}$ one obtains the asserted short exact sequence.

L2L3construct

2.1

Use [F1] to choose a lift ${\tilde{f}}_{i}$ for every $i$ ; for finite $I$ , [L4] suffices, and for $I = \emptyset$ the family is empty.

step 1.1F1L4choose

3.1

By [L1], the ${\tilde{f}}_{i}$ assemble uniquely into $\tilde{f} : ⨁_{i} P_{i} \to E$ , and equality on every summand gives $q \circ \tilde{f} = f$ . Thus the direct sum is projective.

step 2.1L1construct

4.1

Steps 1.1, 2.1, and 3.1 prove closure under arbitrary direct sums with the stated choice cost, and step 1.2 gives enough projectives.

step 1.1step 2.1step 3.1step 1.2discharge-construct∎

Depends on

Used by

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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)