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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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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.

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The zero-dimensional critical-rank boundary

PDF p. 1: Introduction: critical-rank splitting criterion (boundary supplied locally); PDF p. 3: Notation and conventions: algebraically closed base field and finite-type affines

Statement

Assume AC. Let $k$ be an algebraically closed field, let $X$ be a nonempty smooth affine finite-type $k$ -scheme of dimension zero, and let $E$ be a vector bundle of constant rank zero on $X$ . Then $E = 0$ and $c_{0} (E) = 1 \neq 0$ in ${C H}^{0} (X)$ . There is no vector bundle $E'$ with $E ≅ E' \oplus O_{X}$ . Thus the two conditions in the critical-rank splitting equivalence are both false in dimension zero.

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]

The smooth affine scheme is a finite disjoint union of integral components. Smooth affine components and gluing a trivial line

[F2]

By convention the zeroth Chern class is 1 for every bundle, including the zero bundle. Chern classes of a vector bundle on a smooth scheme

[F3]

The codimension-zero cycle group is freely generated by the irreducible components, and there are no rational-equivalence relations coming from larger-dimensional integral subschemes in dimension zero. Rational equivalence and the Chow group of cycles

[F4]

A rank-zero locally free module is locally zero, hence zero. Local bundle ranks are nonnegative integers and add under direct sums. Isomorphism classes of vector bundles

Proof

1.1F1F2F3F4

By [F4], $E$ is the zero bundle. By [F1] and [F3], ${C H}^{0} (X)$ is the free abelian group on its finitely many nonempty components. The unit class 1 has coefficient 1 on each component; since $X$ is nonempty, it is not zero. By [F2] this unit is exactly $c_{0} (E)$ .

2.1F4step 1.1∎

If $E ≅ E' \oplus O_{X}$ , taking a fiber at any point would give $0 = rank (E') + 1$ , impossible for a bundle rank. Thus splitting and top-Chern vanishing both fail, as stated.

Depends on

Used by · 1

Smooth critical-rank splitting criterion

Dependency tree · 3 levels

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