LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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The tensor-total differential is balanced, well defined, and squares to zero

Statement

Let $P$ be a chain complex of right $R$ -modules and $Q$ a chain complex of left $R$ -modules. On the finite-diagonal total module $Tot (P \otimes_{R} Q)$ , the formula $d (p \otimes q) = d_{P} p \otimes q + (- 1)^{p} p \otimes d_{Q} q$ is balanced and satisfies $d^{2} = 0$ .

Proof

Given: homogeneous $p \in P_{p}$ , $q \in Q_{q}$ , and the tensor-total convention.

1.1given

For $r \in R$ , $d ((p r) \otimes q) = d_{P} p r \otimes q + (- 1)^{p} p r \otimes d_{Q} q = d (p \otimes r q)$ , since both differentials are $R$ -linear; thus the formula descends from elementary tensors.

2.1step 1.1algebra

Applying $d$ again gives $d_{P}^{2} p \otimes q + (- 1)^{p - 1} d_{P} p \otimes d_{Q} q + (- 1)^{p} d_{P} p \otimes d_{Q} q + p \otimes d_{Q}^{2} q$ .

3.1step 2.1algebra∎

The first and last terms vanish and the two middle terms cancel. Linearity then proves $d^{2} = 0$ on every finite sum in each total degree.

Depends on

Dependency tree · 2 levels

Sources

Charles A. Weibel, An Introduction to Homological Algebra