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

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Equivalent characterizations of injective modules

Statement

For a left $R$ -module $I$ , the following are equivalent:

$I$ is injective;
every short exact sequence $0 \to I \to E \to C \to 0$ splits;
${Hom}_{R} (-, I)$ takes every short exact sequence to a short exact sequence.

These equivalences use no choice principle.

Facts & Assumptions

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

[F1]

Injectivity is the extension property along every module monomorphism (Injective modules and the extension property).

[L1]

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

[L2]

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

[F2]

Quotient modules have the usual coset operations (Quotient module $M / N$ with scalar multiplication on additive cosets).

Proof

technique · direct

1.1

If $I$ is injective and $0 \to I \overset{j}{\to} E \to C \to 0$ is short exact, extend ${id}_{I}$ along $j$ using [F1]. The extension is a retraction, so [L1] makes the sequence split.

assume-hypF1L1

1.2

Conversely, assume every short exact sequence beginning in $I$ splits. Given a monomorphism $u : A \to B$ and $f : A \to I$ , let $S = {(f (a), - u (a)) : a \in A} \leq I \oplus B$ and $P = (I \oplus B) / S$ .

assume-hypF2construct

1.3

If $I$ is injective, every map $A \to I$ extends across the monomorphism in a short exact sequence $0 \to A \to B \to C \to 0$ , so the final precomposition map in [L2] is surjective; hence ${Hom}_{R} (-, I)$ is exact.

assume-hypF1L2

1.4

Conversely, if ${Hom}_{R} (-, I)$ takes short exact sequences to short exact sequences, apply it to $0 \to A \overset{u}{\to} B \to B / u (A) \to 0$ . Surjectivity of $u^{*}$ extends every $A \to I$ across $u$ , so [F1] makes $I$ injective.

assume-hypF1F2L2

2.1

The map $j : I \to P$ , $j (t) = [(t, 0)]$ , is injective: if $(t, 0) = (f (a), - u (a))$ , injectivity of $u$ gives $a = 0$ and $t = 0$ . Hence $0 \to I \overset{j}{\to} P \to P / j (I) \to 0$ is short exact and splits by hypothesis; let $r : P \to I$ retract $j$ .

step 1.2L1F2

3.1

Define $\tilde{f} : B \to I$ by $\tilde{f} (b) = r ([(0, b)])$ . In $P$ , $[(0, u (a))] = [(f (a), 0)]$ , so $\tilde{f} (u (a)) = r (j (f (a))) = f (a)$ . Thus $I$ is injective by [F1].

step 1.2step 2.1F1construct

4.1

Steps 1.1, 1.2, 2.1, and 3.1 prove $1 \Leftrightarrow 2$ , while steps 1.3 and 1.4 prove $1 \Leftrightarrow 3$ .

step 1.1step 1.2step 2.1step 3.1step 1.3step 1.4∎

Depends on

Used by

Every short exact sequence of modules splits False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 27 results over 11 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)