LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A unit differential entry splits a contractible two-term summand

Statement

Let $R$ be a commutative ring and $F_{∙}$ a complex of finite free $R$ -modules. If one matrix coefficient of $d_{i} : F_{i} \to F_{i - 1}$ is a unit, then $F_{∙}$ is isomorphic as a complex to the direct sum of a shorter complex and the contractible two-term complex $0 \to R \overset{1}{\to} R \to 0$ in degrees $i, i - 1$ . The shorter complex has ranks one smaller in those two degrees and the same ranks elsewhere.

Facts & Assumptions

Given: The finite free complex and one invertible matrix coefficient of its differential.

[F1]

Elementary row and column operations using a unit preserve free bases. The complex identities are $d_{i - 1} d_{i} = 0$ and $d_{i} d_{i + 1} = 0$ .

Proof

technique · clear the unit row and column, then use the complex identities to separate adjacent maps

1.1F1

Permute bases to move the unit coefficient to the first row and column of $d_{i}$ , and scale the source basis vector to make it $1$ . Subtract its multiples from the remaining target basis vectors to clear the first column, then subtract multiples of the first source basis vector to clear the first row. These are invertible basis changes, and in the resulting decompositions $F_{i} = R \oplus F_{i}^{'}$ and $F_{i - 1} = R \oplus F_{i - 1}^{'}$ the map is $d_{i} = 1_{R} \oplus d_{i}^{'}$ .

2.1F1step 1.1∎

The equation $d_{i} d_{i + 1} = 0$ forces the component of $d_{i + 1}$ landing in the displayed $R \subseteq F_{i}$ to be zero, because $d_{i}$ is identity there. Likewise $d_{i - 1} d_{i} = 0$ forces $d_{i - 1}$ to vanish on the displayed $R \subseteq F_{i - 1}$ . All other differentials already avoid these two summands. Thus the displayed identity pair is a direct summand as a complex, and the complement is the shorter complex in the Statement. No choice principle beyond finite basis operations is used.

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

The Stacks Project, Algebra, Lemma 10.102.2 (tag 00MT), unit-entry splitting (standard reference, not scraped)