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Riemann's theorem for sufficiently positive divisors

Statement

Assume the Axiom of Choice as inherited from the vanishing theorem. Let $k$ be a field, let $C$ be a smooth proper geometrically integral curve over $k$ (Curves over a field) of genus $g = g (C)$ (Genus via the Euler characteristic), let $φ : C \to P_{k}^{1}$ be a finite $k$ -morphism and let $A$ be an effective divisor with $O_{C} (A) ≅ φ^{*} O_{P_{k}^{1}} (1)$ . Let $D_{0}$ be a divisor on $C$ (Divisors on a smooth proper curve) and let $n_{0}$ be the integer supplied for $D_{0}$ by Vanishing of H^1 in a fixed ample direction for this fixed morphism $φ$ and divisor $A$ . Then every divisor $D$ on $C$ with $D \geq D_{0} + n_{0} A$ satisfies $h^{1} (D) = 0 and l (D) = \deg_{k} (D) + 1 - g,$ where $l (D) = h^{0} (D) = \dim_{k} H^{0} (C, O_{C} (D))$ and $h^{1} (D) = \dim_{k} H^{1} (C, O_{C} (D))$ (The Riemann-Roch dimension l(D)). In particular both conclusions hold for every divisor of the form $D = D_{0} + n A + E$ with $n \geq n_{0}$ and $E$ effective.

Facts & Assumptions

Given: a field $k$ , a smooth proper geometrically integral curve $C$ over $k$ of genus $g$ , a finite $k$ -morphism $φ : C \to P_{k}^{1}$ , an effective divisor $A$ with $O_{C} (A) ≅ φ^{*} O (1)$ , a divisor $D_{0}$ on $C$ , and the integer $n_{0}$ supplied for $D_{0}$ by the vanishing theorem for this $φ$ and $A$ .

[F1]

Vanishing theorem: for the fixed curve, morphism $φ$ and effective divisor $A$ , the integer $n_{0}$ satisfies $H^{1} (C, O_{C} (D_{0} + n A + E)) = 0$ for every $n \geq n_{0}$ and every effective divisor $E$ ; equivalently $h^{1} (D) = 0$ for every divisor $D$ with $D \geq D_{0} + n_{0} A$ (Vanishing of H^1 in a fixed ample direction).

[F2]

Riemann-Roch in Euler-characteristic form: for every divisor $D$ on $C$ one has $h^{0} (D) - h^{1} (D) = χ (C, O_{C} (D)) = \deg_{k} (D) + 1 - g$ , with no Serre duality used (Riemann-Roch for curves: the Euler-characteristic form).

[F3]

Notations: $l (D) = h^{0} (D) = \dim_{k} H^{0} (C, O_{C} (D))$ and $h^{i} (D) = \dim_{k} H^{i} (C, O_{C} (D))$ for $i \geq 0$ , and $g = g (C) = h^{1} (C, O_{C}) = 1 - χ (C, O_{C})$ is the genus (The Riemann-Roch dimension l(D), Genus via the Euler characteristic, Divisors on a smooth proper curve).

[F4]

The Axiom of Choice is available and is inherited from the vanishing theorem [F1] (through ampleness and Serre vanishing), the Riemann-Roch theorem [F2], and the dimension, genus and divisor interfaces [F3]; the argument below evaluates the two statements at the given divisor and selects nothing beyond the integer $n_{0}$ supplied by [F1] (The Axiom of Choice).

Proof

technique · read the vanishing of $h^1$ off the vanishing theorem, substitute it into the Euler-characteristic form of Riemann-Roch, and record the explicit form $D_0+nA+E$

1.1F1

Vanishing above the threshold. Let $D$ be a divisor with $D \geq D_{0} + n_{0} A$ . Writing $E : = D - (D_{0} + n_{0} A) \geq 0$ exhibits $D = D_{0} + n_{0} A + E$ with $E$ effective, so [F1] gives $H^{1} (C, O_{C} (D)) = 0$ , that is $h^{1} (D) = 0$ . In particular, for every $n \geq n_{0}$ and every effective $E$ the divisor $D = D_{0} + n A + E$ satisfies $D \geq D_{0} + n_{0} A$ and therefore $h^{1} (D) = 0$ .

2.1F2F3step 1.1

The dimension formula. For every divisor $D$ with $D \geq D_{0} + n_{0} A$ , combining $h^{1} (D) = 0$ of step 1.1 with the Riemann-Roch identity [F2] gives $l (D) = h^{0} (D) = h^{1} (D) + \deg_{k} (D) + 1 - g = \deg_{k} (D) + 1 - g$ by the notation [F3]: the section space has dimension exactly $\deg_{k} (D) + 1 - g$ , with no correction term.

3.1F1F4step 1.1step 2.1∎

The explicit form and choice accounting. Every divisor $D = D_{0} + n A + E$ with $n \geq n_{0}$ and $E$ effective satisfies $D \geq D_{0} + n_{0} A$ , so step 1.1 gives $h^{1} (D) = 0$ and step 2.1 gives $l (D) = \deg_{k} (D) + 1 - g$ ; the general divisor $D \geq D_{0} + n_{0} A$ is of this form with $n = n_{0}$ , so both formulations coincide. The integer $n_{0}$ is the one supplied by the vanishing theorem for $D_{0}$ and the fixed morphism $φ$ ; it is not chosen here, and the Axiom of Choice is inherited through [F1], [F2] and [F3], as recorded in [F4].

Depends on

Used by

Why the sharp degree thresholds wait for the duality pair Remark

Dependency tree · two levels

64 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

William Fulton, Algebraic Curves (Internet Archive copy), Chs. 8 and 6 (standard reference, not scraped)
Ravi Vakil, The Rising Sea (version of October 21, 2025), Ch. 18.5 and Ch. 21 (standard reference, not scraped)
The Stacks Project, Algebraic Curves (tag 0BRV) (standard reference, not scraped)