ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13

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$0 \to A \to A \oplus C \to C \to 0$ is canonically split

Example

For left $R$ -modules $A, C$ , the sequence $0 \to A \overset{i}{\to} A \oplus C \overset{p}{\to} C \to 0,$ with $i (a) = (a, 0)$ and $p (a, c) = c$ , is canonically split by $s (c) = (0, c)$ and $r (a, c) = a$ .

Facts & Assumptions

Given: Left $R$ -modules $A, C$ and the displayed maps.

[F1]

The finite direct sum has coordinatewise operations (The direct sum of an indexed family of modules).

[F2]

A section satisfies $p \circ s = {id}_{C}$ , and a retraction satisfies $r \circ i = {id}_{A}$ (Split short exact sequences, sections, and retractions).

[L1]

A section or retraction splits a short exact sequence and identifies the middle module with $A \oplus C$ (The splitting lemma for short exact sequences of modules).

Verification

technique · direct

1.1

The map $i$ is injective, $p$ is surjective, and $\ker p = {(a, 0) : a \in A} = im i$ , so the sequence is short exact.

givenF1algebra

1.2

The formulas give $p (s (c)) = c$ and $r (i (a)) = a$ , so $s$ is a section and $r$ is a retraction by [F2].

F1F2algebra

2.1

By [L1] the sequence splits, and the resulting map $A \oplus C \to A \oplus C$ , $(a, c) \mapsto i (a) + s (c)$ , is the identity.

step 1.1step 1.2L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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)

0→A→A⊕C→C→00→A→A⊕C→C→0 is canonically split