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No sections in negative degree

Statement

Assume the Axiom of Choice as inherited from the divisor and degree suppliers. Let $k$ be a field, let $C$ be a smooth proper geometrically integral curve over $k$ (Curves over a field) and let $D$ be a divisor on $C$ (Divisors on a smooth proper curve). If $\deg_{k} (D) < 0$ then $L (D) = 0$ and $l (D) = 0$ (The space L(D), The Riemann-Roch dimension l(D)). The proof given is the effective-divisor argument: a nonzero $f \in L (D)$ would exhibit the effective divisor $div (f) + D$ linearly equivalent to $D$ , whose degree is therefore $\deg_{k} (D)$ , contradicting the nonnegativity of the degree of an effective divisor. It does not appeal to the Riemann inequality, which for negative degree would only give the vacuous bound $l (D) \geq \deg_{k} (D) + 1 - g$ with a nonpositive right-hand side; it cannot ensure a nonzero section.

Facts & Assumptions

Given: a field $k$ , a smooth proper geometrically integral curve $C$ over $k$ , a divisor $D$ on $C$ with $\deg_{k} (D) < 0$ , and an element $f \in L (D)$ .

[F1]

The Riemann-Roch space: $L (D) = {f \in k (C)^{\times} : div (f) + D \geq 0} \cup {0}$ is the $k$ -subspace of functions whose poles are no worse than $- D$ allows, membership being read coefficientwise as ${ord}_{x} (f) + n_{x} \geq 0$ at every closed point $x$ , and $div (f) = \sum_{x} {ord}_{x} (f) [x]$ (The space L(D)).

[F2]

The dimension: $l (D) = \dim_{k} L (D)$ is the dimension of the Riemann-Roch space, so $L (D) = 0$ exactly when $l (D) = 0$ (The Riemann-Roch dimension l(D)).

[F3]

Divisors and degree: a divisor on $C$ is a finite formal $Z$ -linear combination of closed points, the $k$ -degree is $\deg_{k} (D) = \sum_{x} n_{x} [κ (x) : k]$ , and $\deg_{k}$ is a group homomorphism on the divisor group (Divisors on a smooth proper curve, Degree divisor proper curve).

[F4]

Effective divisors: if $D = \sum_{x} n_{x} [x]$ is effective then $\deg_{k} (D) \geq 0$ , and $\deg_{k} (D) = 0$ only for $D = 0$ ; conversely an effective divisor of negative degree cannot exist (Effective divisors have nonnegative degree).

[F5]

Sections versus effective divisors: for every nonzero $f \in L (D)$ the divisor $div (f) + D$ is an effective divisor on $C$ linearly equivalent to $D$ ; in particular $L (D) = 0$ if and only if no effective divisor is linearly equivalent to $D$ (Effective divisors linearly equivalent to D are sections modulo scalars).

[F6]

The current Principal divisors on a normal proper curve have degree zero states that a principal divisor $div (f)$ of a nonzero rational function on a normal proper curve has degree $0$ , equivalently that linearly equivalent divisors have equal degree. The smooth curve $C$ is normal by the local-ring and normality clause of Divisors on a smooth proper curve, so this theorem applies at step 2.1.

[F7]

The Axiom of Choice is available and is inherited only through the suppliers named above, in particular the principal-divisor degree theorem of [F6]; the proof below makes no selection (The Axiom of Choice).

Proof

technique · direct contradiction; a nonzero section would produce an effective divisor of the same negative degree, and effective divisors have nonnegative degree

1.1F1F5

A nonzero section and its effective divisor. Suppose $f \in L (D)$ with $f \neq 0$ . By [F1] membership means $div (f) + D \geq 0$ coefficientwise, so $E : = div (f) + D$ is an effective divisor on $C$ ; by [F5] the divisor $E$ is linearly equivalent to $D$ , the difference being the principal divisor $div (f)$ .

2.1F3F6step 1.1

Its degree. By [F3] the degree is additive on the divisor group, so $\deg_{k} (E) = \deg_{k} (div (f)) + \deg_{k} (D)$ , and by the principal-divisor degree theorem [F6] the principal divisor $div (f)$ has degree $0$ ; hence $\deg_{k} (E) = \deg_{k} (D)$ , which is negative by hypothesis.

3.1F2F4step 1.1step 2.1

Contradiction and vanishing. By [F4] the effective divisor $E$ of step 1.1 has $\deg_{k} (E) \geq 0$ , contradicting $\deg_{k} (E) = \deg_{k} (D) < 0$ from step 2.1. Hence $L (D)$ contains no nonzero element, that is $L (D) = 0$ , and then $l (D) = \dim_{k} L (D) = 0$ by [F2].

4.1F3F6F7step 2.1step 3.1∎

Conclusion and choice accounting. Steps 1.1 through 3.1 show that $\deg_{k} (D) < 0$ forces $L (D) = 0$ and $l (D) = 0$ by the effective-divisor argument; the Riemann inequality is not used, and indeed for negative degree it would only bound $l (D)$ from below by the nonpositive number $\deg_{k} (D) + 1 - g$ . The principal-divisor degree theorem [F6], used at step 2.1, expresses that degree is well defined on linear-equivalence classes; the Axiom of Choice is inherited from that theorem and the divisor suppliers, and no further selection is made, since the contradiction argument chooses nothing beyond the given $f$ .

Depends on

Used by

A negative right-hand side does not contradict Riemann-Roch Counterexample

Dependency tree · two levels

57 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)
Michael Artin, MIT 18.721 Introduction to Algebraic Geometry (July 20, 2020 notes), Ch. 8 (standard reference, not scraped)
The Stacks Project, Algebraic Curves (tag 0BRV) (standard reference, not scraped)