LemmaStatement: AI-generatedProof: AI-generated

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.

Determinants detect vanishing of the rank-one exterior-dual class

Statement

Assume AC. For a line bundle $L$ on a noetherian affine scheme, $ε (L) = 1 - [L^{\lor}]$ in the article's perfect $K_{0}$ , and $ε (L) = 0$ if and only if $L ≅ O_{X}$ . Thus the rank-one vanishing criterion uses the actual line bundle, not an abstract identification of its target groups.

Facts & Assumptions

Given: The hypotheses and data in the statement, with AC and the article's fixed universes wherever perfect K-theory is used.

Goal: Establish the asserted conclusion.

[F1]

For rank one the exterior-dual sum has only its degree-zero and degree-one terms. The actual exterior-dual class in affine perfect K zero

[F2]

The affine perfect/bundle comparison identifies object classes, and equality in the latter group is equivalent to a common finite free stabilization. Isomorphism classes of vector bundles, Equality in affine K zero and stable isomorphism

[F3]

Determinant is the top exterior power, with the increasing-wedge local bases. Exterior powers and determinant of an algebraic bundle

Proof

1.1F1F2F3construct

By [F1], vanishing is exactly $[L^{\lor}] = [O_{X}]$ . By [F2] this yields $L^{\lor} \oplus O_{X}^{N} ≅ O_{X}^{N + 1}$ for some $N \geq 0$ . Take the top exterior power of this actual isomorphism. Wedge bases give $\det (L^{\lor} \oplus O_{X}^{N}) ≅ L^{\lor}$ and $\det (O_{X}^{N + 1}) ≅ O_{X}$ ; the usual wedge-of-summands map is canonical and glues. Hence $L^{\lor} ≅ O_{X}$ , and dualizing gives $L ≅ O_{X}$ by local double-dual evaluation.

2.1F1step 1.1algebra∎

Conversely, if $L$ is trivial, its dual is trivial and the two terms of $1 - [L^{\lor}]$ cancel. This proves the equivalence with the precise spectral class.

Depends on

Used by · 2

Dependency tree · 3 levels

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