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

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Products of injective modules are injective, with the exact choice boundary

Statement

Assume the Axiom of Choice. An arbitrary direct product of injective left $R$ -modules is injective. Conversely, each factor is a direct summand of the product, so if a product is injective, every factor is injective.

For a finite product, finite choice suffices; the empty product is the zero module and is injective.

Facts & Assumptions

Given: A family $(I_{j})_{j \in J}$ of left $R$ -modules.

[F1]

Injectivity is extension of every map along a monomorphism (Injective modules and the extension property).

[F2]

Products have coordinatewise operations; the empty product is the zero module (The direct sum of an indexed family of modules).

[F3]

AC chooses from arbitrary families of nonempty sets, while finite choice handles a listed finite family in ZF (The Axiom of Choice, Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · constructive

1.1

Suppose every $I_{j}$ is injective. Given a monomorphism $u : A \to B$ and $f : A \to \prod_{j} I_{j}$ , let $f_{j}$ be the $j$ -th coordinate map. For each $j$ , [F1] gives a nonempty set of extensions ${\tilde{f}}_{j} : B \to I_{j}$ .

assume-hypF1F2

1.2

Conversely, suppose the product is injective and fix $j \in J$ . Given a monomorphism $u : A \to B$ and $f : A \to I_{j}$ , compose $f$ with the coordinate inclusion $I_{j} \to \prod_{k} I_{k}$ , extend to $B$ using injectivity of the product, and postcompose with the $j$ -th projection. The result extends $f$ , so $I_{j}$ is injective by [F1].

F1F2construct

2.1

Use [F3] to choose one extension for every $j$ ; finite choice suffices for finite $J$ , and no choice is needed when $J = \emptyset$ . The coordinate formula $\tilde{f} (b) = ({\tilde{f}}_{j} (b))_{j}$ defines a homomorphism $B \to \prod_{j} I_{j}$ extending $f$ .

step 1.1F2F3chooseconstruct

3.1

Thus the product is injective by [F1].

step 2.1F1

4.1

Steps 1.1, 2.1, and 3.1 prove the product theorem with its choice boundary, and step 1.2 proves the converse.

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

Depends on

Used by

Module categories have enough injectives 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)