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Third edge isomorphism

PDF p. 14: Proposition 6.2

Statement

Assume the Axiom of Choice. Let k be an algebraically closed field and X a normal affine threefold over k, where threefold means a finite-type k-scheme of Krull dimension three. Then the third edge homomorphism ϵ3:Hmot6,3(X,Z)⟶Fmot3K0(X) is an isomorphism. Moreover, the actual stage-to-K homomorphism π0Fmot3K(X)→K0(X) is injective.

Here K(X) is the nonconnective K-theory spectrum of perfect complexes, K0(X)=π0K(X), and Fmot3K(X) is the third stage of its decreasing motivic filtration. The group Fmot3K0(X) is the image of the stage-to-K0 homomorphism. The group Hmot6,3(X,Z) is degree-six cohomology of the weight-three motivic complex. The map ϵ3 is the inverse of the actual third-stage-to-graded isomorphism p3 followed by the stage-to-K0 homomorphism with codomain restricted to its image. A group is torsion-free if no nonzero element is killed by a nonzero integer.

Facts & Assumptions

Given: The Axiom of Choice, an algebraically closed field k, and a normal affine finite-type k-scheme X of dimension three.

Goal: Prove that the specified third edge map is an isomorphism and that the actual third-stage map to K0(X) is injective.

[F1]

A finite-type scheme over a field is Noetherian. Under the Axiom of Choice, Third-stage graded isomorphism supplies the actual isomorphism p3:π0Fmot3K(X)→Hmot6,3(X,Z). By Third edge homomorphism, the stated edge map is p3−1 followed by the stage map restricted to its image, and is surjective. The choice assumption used by these suppliers is as in The Axiom of Choice.

[F2]

For an affine finite-type threefold over an algebraically closed field, Torsion freeness of top threefold motivic cohomology states that Hmot6,3(X,Z) is torsion-free.

[F3]

Under the Axiom of Choice, Rational third-edge comparison states that the rationalization of the actual third edge map is an isomorphism for a Noetherian dimension-three scheme over a field.

[F4]

Detecting injectivity by rationalization states that a homomorphism with torsion-free domain is injective if its rationalization is injective.

Proof

technique · Detect the kernel of the edge map after rationalization, then use its relation to the actual third-stage map
1.1F1F2F3given

The given finite-type scheme X is Noetherian and has dimension three. Thus the hypotheses of [F1], [F2], and [F3] hold. In particular, the domain of ϵ3 is torsion-free, and ϵ3⊗ZQ is an isomorphism, hence injective.

2.1F1F4step 1.1

Apply [F4] to ϵ3. Step 1.1 shows that its domain is torsion-free and its rationalization is injective, so ϵ3 is injective. It is surjective by [F1], and therefore is an isomorphism.

3.1F1step 2.1∎

Let u:π0Fmot3K(X)→K0(X) be the actual stage-to-K0 homomorphism and let v be the same map with codomain restricted to its image. By [F1], v=ϵ3∘p3. Both factors are isomorphisms by [F1] and step 2.1, so v is injective. The inclusion of its image into K0(X) is injective; consequently u is injective. This proves both assertions.

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External dependencies

Jean Fasel and Maria Yakerson, Vector bundles on singular surfaces and threefolds · Proposition 6.2 and its proof, PDF p. 14

For a normal affine threefold over an algebraically closed field, the third edge map is an isomorphism and the third-stage map into K_0 is injective.

Hypotheses: X is a normal affine threefold over an algebraically closed field, with the article’s motivic filtration and conventions.

Reading limits: The proposition and proof paragraph were read in the supplied article text. The proof below invokes the completed local supplier contracts for their substantive inputs.

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