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

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Equivalent characterizations of projective modules

Statement

For a left $R$ -module $P$ , assertions 1 to 3 below are equivalent without choice. Under the Axiom of Choice, they are also equivalent to assertion 4:

$P$ is projective;
every short exact sequence $0 \to K \to E \to P \to 0$ splits;
${Hom}_{R} (P, -)$ takes every short exact sequence to a short exact sequence;
$P$ is a direct summand of a free module.

The equivalence of 1 to 3 is choice-free. The implication $1 \Rightarrow 4$ uses the canonical free cover and is choice-free; under AC, every free module is projective, so $4 \Rightarrow 1$ .

Facts & Assumptions

Given: A left $R$ -module $P$ .

[F1]

Projectivity is the lifting property for surjections (Projective modules and the lifting property).

[L1]

A short exact sequence splits exactly when its epimorphism has a section, equivalently its monomorphism has a retraction (The splitting lemma for short exact sequences of modules).

[L2]

Applying ${Hom}_{R} (P, -)$ to an exact sequence $0 \to A \to B \to C$ gives an exact sequence $0 \to {Hom}_{R} (P, A) \to {Hom}_{R} (P, B) \to {Hom}_{R} (P, C)$ (Covariant and contravariant $Hom$ are left exact).

[L3]

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

[L4]

Under AC, every free module is projective (Free modules are projective, with the exact choice boundary, The Axiom of Choice).

Proof

technique · direct

1.1

If $P$ is projective and $0 \to K \to E \overset{q}{\to} P \to 0$ is short exact, lift ${id}_{P}$ through $q$ using [F1]; the lift is a section, so the sequence splits by [L1].

assume-hypF1L1

1.2

If every such sequence splits, then for a surjection $q : E \to M$ and map $f : P \to M$ , form the pullback module $T = {(x, e) \in P \oplus E : f (x) = q (e)}$ . The projection $T \to P$ is surjective with kernel isomorphic to $\ker q$ , so its short exact sequence splits; a section followed by the projection $T \to E$ is a lift of $f$ . Thus $P$ is projective.

assume-hypL1construct

1.3

By [L2], applying ${Hom}_{R} (P, -)$ to $0 \to A \to B \overset{q}{\to} C \to 0$ is exact through ${Hom}_{R} (P, B)$ ; its last map is surjective exactly when every $P \to C$ lifts through $q$ . Hence [F1] makes assertions 1 and 3 equivalent.

F1L2

1.4

If $P$ is projective, lift ${id}_{P}$ through the canonical surjection $R^{(P)} \to P$ from [L3]. This section splits the free cover by [L1], so $P$ is a direct summand of $R^{(P)}$ .

assume-hypF1L1L3

1.5

A direct summand of a projective module is projective: precompose a map from the summand with the projection, lift the resulting map, and restrict the lift along the inclusion. Under AC the free ambient module in assertion 4 is projective by [L4], so assertion 4 implies assertion 1.

assume-hypF1L4

2.1

Steps 1.1 and 1.2 prove $1 \Leftrightarrow 2$ , step 1.3 proves $1 \Leftrightarrow 3$ , and steps 1.4 and 1.5 prove $1 \Leftrightarrow 4$ with the stated choice boundary.

step 1.1step 1.2step 1.3step 1.4step 1.5∎

Depends on

Used by

ℤ/2ℤ is projective but not free over ℤ/6ℤ Example
Every injective module is projective (refuted under the Axiom of Choice) False statement
Every projective module is free False statement
Every short exact sequence of modules splits False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 9 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)