LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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.

Degree-one evaluation for a principal degree-zero boundary ideal

Statement

Let $R$ be an associative unital ring with $1 \neq 0$ and no zero divisors, embedded as a unital subring in a division ring $K$ , put $N = K / R$ as a right $R$ -module, and let $C_{*}$ be a right $R$ -module chain complex concentrated in nonnegative degrees. Supply an identification $C_{0} = R$ . If the right ideal $im (d_{1})$ is principal, restriction to cycles gives an isomorphism $H^{1} ({Hom}_{R} (C_{*}, N)) \tilde{\to} {Hom}_{R} (H_{1} (C_{*}), N) .$ Neither a principal ideal domain hypothesis on $R$ nor a choice principle is required.

Facts & Assumptions

Given: The specified complex and division-ring embedding. Put $Z_{1} = \ker d_{1}$ , $B_{1} = im d_{2}$ , and $B_{0} = im d_{1} = a R$ for one generator $a$ .

Proof

1.1givenconstructalgebra

A degree-one cocycle is a map $f : C_{1} \to N$ vanishing on $B_{1}$ , and a coboundary is a map of the form $g d_{1}$ . Restriction of a cocycle to $Z_{1}$ descends to $Z_{1} / B_{1} = H_{1}$ ; coboundaries restrict to zero. Thus restriction defines the stated map. If $a = 0$ , then $Z_{1} = C_{1}$ and all coboundaries vanish, so the map is already an isomorphism.

2.1step 1.1givenconstructalgebra

Suppose $a \neq 0$ . Choose $z \in C_{1}$ with $d_{1} z = a$ . Because $R$ has no zero divisors, $a R \to C_{1}$ , $a r \mapsto z r$ , is well-defined and a section of $d_{1}$ . Hence $C_{1} = Z_{1} \oplus z R$ . Any map $H_{1} \to N$ pulls back to a map on $Z_{1}$ vanishing on $B_{1}$ and extends by zero on $z R$ to a degree-one cocycle. This proves surjectivity.

3.1step 1.1givenconstructalgebra∎

If a degree-one cocycle restricts to zero on $Z_{1}$ , it factors as $v d_{1}$ for a right-linear map $v : a R \to N$ . Let $v (a) = q + R$ with $q \in K$ and put $u = q a^{- 1} + R$ . Then $u a = v (a)$ in $N$ . The right-linear map $g : R \to N$ , $g (r) = u r$ , extends $v$ , since $g (a r) = v (a) r = v (a r)$ . The cocycle is consequently the coboundary $g d_{1}$ . This proves injectivity and the result. All selections comprise one ideal generator, one lift and one fraction representative.

Used by

Classical and localized PTFA Blanchfield nonsingularity criteria Lemma

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Cochran, Orr and Teichner, Knot concordance, Whitney towers and L2-signatures, degree-one evaluation in the proof of Theorem 2.13 (standard reference, not scraped)