CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02

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H^1 of a line bundle vanishes above degree 2g - 2

Statement

Assume the Axiom of Choice as inherited from the duality suppliers. Let $C$ be a smooth proper geometrically integral curve over a field $k$ of genus $g$ and let $L$ be an invertible $O_{C}$ -module with $\deg (L) > 2 g - 2$ . Then $H^{1} (C, L) = 0 and h^{0} (C, L) = \deg (L) + 1 - g .$

Facts & Assumptions

Given: A field $k$ ; a smooth proper geometrically integral curve $C$ over $k$ of genus $g$ ; an invertible $O_{C}$ -module $L$ with $\deg (L) > 2 g - 2$ ; a canonical divisor $K_{C}$ .

[F1]

For a smooth proper geometrically integral curve of genus $g$ the canonical divisor has degree $2 g - 2$ . (The canonical divisor has degree 2g - 2)

[F2]

Duality identifies the index of speciality with the dual sections: $h^{1} (C, O_{C} (D)) = l (K_{C} - D)$ for every divisor $D$ . (h^1 of a line bundle equals the dimension of the space of dual sections)

[F3]

Let $L$ be an invertible sheaf on $C$ whose degree $\deg (L)$ is represented by $\deg_{k} (D)$ for any divisor $D$ with $L ≅ O_{C} (D)$ . If $\deg (L) < 0$ then $H^{0} (C, L) = 0$ . (Negative-degree line bundles have no nonzero sections)

[F4]

Riemann-Roch as $l$ minus $i$ : for every divisor $D$ one has $l (D) - i (D) = h^{0} (C, O_{C} (D)) - h^{1} (C, O_{C} (D)) = \deg_{k} (D) + 1 - g$ , and $i (D) = h^{1} (C, O_{C} (D)) \geq 0$ . (Riemann-Roch as l minus i, The index of speciality i(D))

[F5]

On a smooth proper geometrically integral curve the Cartier-to-Weil cycle map is an isomorphism and every invertible sheaf is isomorphic to $O_{C} (D)$ for a divisor $D$ well defined modulo linear equivalence, so $\deg (L) = \deg_{k} (D)$ for such a divisor; the degree is additive in divisors. (Cartier and Weil divisors agree on a smooth curve, Degree divisor proper curve, Invertible sheaf of cartier divisor)

[F6]

The Axiom of Choice: every family of nonempty sets has a choice function. (The Axiom of Choice)

Proof

Proof technique: direct; move to the dual twist, where the hypothesis forces negative degree and hence vanishing of sections, then apply Riemann-Roch.

1.1F1F5given

(Set-up.) By [F5] there is a divisor $D$ on $C$ with $L ≅ O_{C} (D)$ and $\deg (L) = \deg_{k} (D)$ , and $\deg_{k}$ is additive in divisors; by the hypothesis $\deg_{k} (D) = \deg (L) > 2 g - 2 = \deg_{k} (K_{C})$ , where the last equality is [F1].

2.1F5step 1.1

The dual twist has negative degree: $\deg_{k} (K_{C} - D) = \deg_{k} (K_{C}) - \deg_{k} (D) < 2 g - 2 - (2 g - 2) = 0$ , using additivity of $\deg_{k}$ from [F5] and step 1.1.

3.1F3step 2.1

By [F3] applied to the invertible sheaf $O_{C} (K_{C} - D)$ of negative degree one has $H^{0} (C, O_{C} (K_{C} - D)) = 0$ , that is $l (K_{C} - D) = 0$ .

4.1F2step 1.1step 3.1

(Vanishing of $H^{1}$ .) By [F2] applied to the divisor $D$ one has $h^{1} (C, L) = h^{1} (C, O_{C} (D)) = l (K_{C} - D)$ , which is zero by step 3.1; this is the first assertion.

5.1F4step 4.1

(Exact section count.) By [F4] applied to $D$ and step 4.1, $l (D) = \deg_{k} (D) + 1 - g + i (D) = \deg (L) + 1 - g + 0$ , that is $h^{0} (C, L) = \deg (L) + 1 - g$ , the second assertion.

6.1F6step 4.1step 5.1∎

Steps 4.1 and 5.1 prove both displayed statements; the Axiom of Choice [F6] is used exactly through the duality suppliers cited above, each of which assumes it.

Depends on

Used by

A genus-one curve with a rational point embeds as a plane cubic Corollary
Riemann-Roch in exact form for divisors of degree above 2g - 2 Corollary
Degree 2g does not force very ampleness Counterexample
Degree 2g-1 does not force base-point-freeness Counterexample
A degree-n line bundle on a genus-one curve has an n-dimensional space of sections for n > 0 Example
Line bundles of degree at least 2g are base-point-free Theorem
Line bundles of degree at least 2g+1 are very ample Theorem

Dependency tree · two levels

58 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), Ch. 8 (standard reference, not scraped)
Ravi Vakil, The Rising Sea (version of October 21, 2025) (standard reference, not scraped)
MIT 18.725 Algebraic Geometry (Fall 2015) course notes, Lectures 24-25 (standard reference, not scraped)