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-hom-of-cyclic-abelian-groups.md

Example

For $m, n > 0$ , $R {Hom}_{Z} (Z / m, Z / n)$ is represented by $(Z / n \overset{m}{\to} Z / n)$ in degrees $0, 1$ . Its $H^{0}$ and $H^{1}$ are isomorphic to $Z / \gcd (m, n)$ ; other cohomology is zero.

Facts & Assumptions

Given: For $m, n > 0$ , $R {Hom}_{Z} (Z / m, Z / n)$ is represented by $(Z / n \overset{m}{\to} Z / n)$ in degrees $0, 1$ . Its $H^{0}$ and $H^{1}$ are isomorphic to $Z / \gcd (m, n)$ ; other cohomology is zero.

[F1]

Derived Hom can be represented using a supplied projective source and the cochain Hom differential (Derived hom in the bounded setting).

[F2]

Derived Hom cohomology of degree-zero objects is classical Ext in nonnegative degrees (Cohomology of derived hom is ext).

Verification

1.1

Apply the Hom complex to the free resolution $(Z \overset{m}{\to} Z)$ in degrees $- 1, 0$ and the target $Z / n [0]$ . Its two terms are $Z / n$ in degrees zero and one with differential $- m$ . Multiplication by $- 1$ in degree one and identity in degree zero gives a complex isomorphism to the displayed $+ m$ model.

F1algebra

2.1

For $g = \gcd (m, n)$ , the kernel of $m$ is generated by $n / g$ modulo $n$ and is cyclic of order $g$ ; the cokernel is $Z / (m Z + n Z) = Z / g$ . These are respectively $Hom (Z / m, Z / n)$ and ${Ext}^{1} (Z / m, Z / n)$ by derived Hom cohomology. Modulus one makes both zero, and all other degrees vanish.

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

10.7.2–10.7.5 and Exercise 10.7.1, pp. 399–400 (standard reference, not scraped)