ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-06

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The tensor double complex in low degrees

Example

Let $m, n \geq 1$ be integers. Label the resolutions $Q$ and $P$ , respectively, in the order $0 \to Z \overset{m}{\to} Z \to Z / m \to 0$ and $0 \to Z \overset{n}{\to} Z \to Z / n \to 0$ . Their tensor double complex has $K_{0, 0} = K_{1, 0} = K_{0, 1} = K_{1, 1} = Z$ and all other terms zero.

Verification

Given: the two displayed two-term resolutions.

1.1

In bidegrees $(p, q)$ with $p, q \in {0, 1}$ , the tensor of the two free rank-one groups is $Z$ .

given

2.1

The horizontal differential is multiplication by $m$ . Under the supplied double-complex convention the vertical differential already includes the factor $(- 1)^{p}$ , so it is $(- 1)^{p}$ times multiplication by $n$ . The total differential is therefore $d_{h} + d_{v}$ , with no second sign inserted.

step 1.1algebra

3.1

Hence ${Tot}_{0} = Z$ , ${Tot}_{1} = Z \oplus Z$ , and ${Tot}_{2} = Z$ , where the degree-one summands are ordered $K_{1, 0}, K_{0, 1}$ . The total complex is $0 ⟶ Z \overset{c \mapsto (- n c, m c)}{\to} Z^{2} \overset{(a, b) \mapsto m a + n b}{\to} Z ⟶ 0.$ The composite is $- m n c + n m c = 0$ , checking the sign and both axis labels. This also covers $m = 1$ or $n = 1$ .

step 2.1algebra∎

Depends on

The first-quadrant tensor double complex of two projective resolutions

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

Charles A. Weibel, An Introduction to Homological Algebra (standard reference, not scraped)