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Principal divisors are endpoint boundaries

PDF p. 13: Divisor presentation for supported degree six

Statement

Assume AC. Let $T$ be any integral finite-type curve over an algebraically closed field. Put $Z_{0} (T) = ⨁_{x \in T closed} Z [x]$ , and let $B_{1} (T)$ be the free abelian group on integral one-dimensional closed subschemes $W \subset T \times A_{s}^{1}$ contained in neither endpoint $s = 0, 1$ . Such a $W$ is called admissible. Define $b [W] = [W_{1}] - [W_{0}]$ , extended linearly, with $[W_{ϵ}] = \sum_{w \in W \cap {s = ϵ}} ℓ_{O_{W, w}} (O_{W, w} / (s - ϵ)) [p (w)], ϵ = 0, 1,$ where $p : T \times A^{1} \to T$ is projection. The endpoint intersections are finite and have the indicated finite lengths; projection on each endpoint identifies its residue field with that of the corresponding point of $T$ . These definitions require no embedding of $T$ ; they agree with the supported-cycle definitions whenever such an embedding is supplied. For these groups and this boundary, $b B_{1} (T) = im ({div}_{T} : k (T)^{\times} \to Z_{0} (T)) .$ For $f \neq 1$ in $k (T)^{\times}$ , the closure in $T \times A_{s}^{1}$ of the rational graph $s = 1 / (1 - f)$ is an admissible generator whose boundary is ${div}_{T} (f)$ ; for $f = 1$ the zero chain suffices.

Statement references

The two supported cycle degrees on a curve

Facts & Assumptions

Given: AC, $T$ and $f \in k (T)^{\times}$ as stated.

Goal: Realize each principal divisor by an endpoint boundary and prove equality of subgroups.

[F1]

The length divisor map is a homomorphism, and a quotient by a nonzero element of a closed-point local ring of an integral finite-type curve has finite length. Length orders on an integral curve

[F2]

Every boundary is principal, and for a dominant generator it is the divisor of the norm of $(s - 1) / s$ . Endpoint boundaries are norm divisors

[F3]

For an integral curve embedded as a closed subscheme of a smooth threefold, the two supported cycle groups and their boundary are exactly the intrinsic endpoint groups and quotient-length formula displayed above. The two supported cycle degrees on a curve

Proof

1.1F1F3given

For any generator $V$ of $B_{1} (T)$ and $ϵ = 0, 1$ , the endpoint intersection is a proper closed subset of the noetherian integral curve $V$ , hence is finite and zero-dimensional. At each point in that intersection $s - ϵ$ is nonzero in the local domain, so [F1] gives finite quotient length. Projection identifies the endpoint subscheme with a closed subscheme of $T$ , preserving point residue fields. Thus the displayed formula defines $b$ for arbitrary $T$ and agrees with [F3] in its embedded setting.

1.2F1givenalgebra

If $f = 1$ , its divisor is zero by [F1]. Otherwise $h = 1 / (1 - f)$ is a rational function distinct from both zero and one. It is regular on a nonempty open subset of $T$ ; its graph there is integral and isomorphic to that open subset.

2.1step 1.1step 1.2construct

Its reduced closure $W$ in $T \times A^{1}$ is an integral closed curve dominating $T$ , with function field $k (T)$ . Neither endpoint contains its generic point, since $h \neq 0, 1$ ; hence each endpoint intersection is zero-dimensional or empty and $W$ is admissible.

3.1F2step 1.2step 2.1algebra∎

For $f \neq 1$ , in $k (W) = k (T)$ , $(s - 1) / s = (h - 1) / h = f$ and the norm for the identity field extension is the identity. By [F2], $b [W] = {div}_{T} (f)$ , including its length multiplicities. Thus every principal divisor is a boundary; the reverse inclusion is also [F2].

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