ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

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).

[F2]

The degree- $- i$ cohomology of the module derived tensor is Tor $_{i}$ (Homology of the derived tensor product is tor).

Verification

1.1

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.

F1algebra

2.1

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 $n / g ∣ k$ . The kernel in degree $- 1$ is ${Tor}_{1}$ , and the cokernel in degree zero is ${Tor}_{0}$ . If either modulus is one both groups are zero; all other degrees have zero terms.

F2step 1.1algebra∎

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

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)