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Spectral K-zero equality gives an actual projective stabilization

Statement

Assume AC and the suitable universes of the article's perfect K-theory construction. Let A be a commutative unital noetherian ring, put X=Spec⁡A, and let E,F be vector bundles on X, meaning finite locally free OX-modules. Equality [E]=[F] in the article's perfect spectral K0(X) is equivalent to an isomorphism of finite projective A-modules Γ(X,E)⊕AN≅Γ(X,F)⊕AN for some integer N≥0. This is the bundle/perfect K-zero interface, without a passage through coherent-sheaf G-theory.

Facts & Assumptions

Given: AC, the stated universes, a commutative unital noetherian ring A, the affine scheme X=Spec⁡A, and vector bundles E,F on X.

Goal: Establish the stated conclusion.

[A1]

AC is assumed through the supplied affine perfect-K-zero comparison and affine bundle/module interfaces; their universe bounds are also assumed. No choice principle stronger than those supplier hypotheses is introduced. The Axiom of Choice

[F1]

The spectral group identifies with vector-bundle K0 and preserves bundle classes. Isomorphism classes of vector bundles

[F2]

Equality in K0 is equivalent to stabilization by the same finite free bundle. Equality in affine K zero and stable isomorphism

[F3]

Stabilization is carried to and from modules. Free stabilization passes through global sections

Proof

1.1A1F1F2F3given

Suppose [E]=[F] in spectral K0(X). The affine comparison identifies these classes with the classes of the bundles E,F in K0(X). Since X is noetherian affine and E,F are vector bundles, [F2] gives E⊕OXN≅F⊕OXN for an integer N≥0. By [F3], global sections give the asserted isomorphism of finite projective modules. The choice and size hypotheses required by these suppliers are included in the given assumptions.

2.1A1F1F2F3step 1.1∎

Conversely, suppose the displayed module isomorphism holds for some integer N≥0. By [F3] it gives a bundle isomorphism E⊕OXN≅F⊕OXN. By [F2], E and F have equal classes in K0(X), and [F1] identifies these with their spectral classes. Together with step 1.1 this proves the equivalence.

Depends on

Used by · 1

Dependency tree · 3 levels

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