ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)

How statement and proof provenance work

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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ex-derived-tensor-of-two-cyclic-abelian-groups.md

Example

For $m, n > 0$ , $Z / m \otimes_{Z}^{L} Z / n$ is represented by $(Z / n \overset{m}{\to} Z / n)$ in degrees $- 1, 0$ . Both $H^{- 1}$ and $H^{0}$ are isomorphic to $Z / \gcd (m, n)$ , and all other cohomology vanishes.

Facts & Assumptions

Given: For $m, n > 0$ , $Z / m \otimes_{Z}^{L} Z / n$ is represented by $(Z / n \overset{m}{\to} Z / n)$ in degrees $- 1, 0$ . Both $H^{- 1}$ and $H^{0}$ are isomorphic to $Z / \gcd (m, n)$ , and all other cohomology vanishes.

[F1]

A supplied projective replacement represents the bounded derived tensor (Derived tensor product in the bounded above setting).

Verification

1.1F1algebra

Use the free resolution $(Z \overset{m}{\to} Z) \to Z / m$ and tensor with $Z / n$ . This gives exactly the displayed two-term complex; since $m > 0$ the resolution is exact at its left endpoint. Its cohomology is its kernel at $- 1$ and cokernel at zero.

2.1step 1.1algebra∎

Put $g = \gcd (m, n)$ . The cokernel is $Z / (m Z + n Z) = Z / g$ . The kernel consists of the multiples of $n / g$ modulo $n$ , and $Z / g \to \ker (m)$ , $\overset{ˉ}{k} \mapsto \overline{k n / g}$ , is an isomorphism: $n ∣ m k$ iff $n / g ∣ k$ . Thus the cohomology in degrees $- 1$ and zero is $Z / g$ in each case. If either modulus is one both groups are zero; all other degrees have zero terms.

Depends on

Derived tensor product in the bounded above setting

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

10.6.1–10.6.4 and Exercise 10.6.1, p. 395; elementary finite-diagonal replacement for spectral sequence proof (standard reference, not scraped)