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Degree 2g-1 does not force base-point-freeness

Statement refuted

Assume the Axiom of Choice; it supplies Dependent Choice through AC implies DC implies countable choice. The theorem that a line bundle of degree at least 2g on a curve of genus g is base-point-free is sharp in its degree bound: a line bundle of degree 2g−1 need not be base-point-free.

Facts & Assumptions

Given: the Axiom of Choice and its consequence Dependent Choice; a field k, a smooth proper geometrically integral curve C of genus g≥1 over k with a k-rational point p, and the invertible sheaf L=OC(KC+p) with associated divisor D=KC+p.

[F1]

deg⁡kKC=2g−2, so deg⁡kD=2g−1; and the divisor--invertible-sheaf dictionary identifies L(−p)=OC(KC) and H0(C,L(−p))⊆H0(C,L). (The canonical divisor has degree 2g - 2, Canonical bundle and canonical divisors, Degree divisor proper curve, Invertible sheaf of cartier divisor, Cartier and Weil divisors agree on a smooth curve)

[F2]

For a divisor of degree >2g−2 on a curve of genus g the nonspecial formula gives ℓ(D)=deg⁡kD+1−g and H1=0; equivalently L has h0(C,L)=deg⁡L+1−g. (Riemann-Roch in exact form for divisors of degree above 2g - 2, H^1 of a line bundle vanishes above degree 2g - 2, The full Riemann-Roch theorem for divisors on a smooth proper curve, The Riemann-Roch dimension l(D))

[F3]

ℓ(KC)=h0(C,ωC)=g, where ωC=OC(KC) is the canonical bundle; the Riemann-Roch space of KC is the space of holomorphic differentials. (The canonical bundle has exactly g independent sections, The space L(D))

[F4]

A closed point q is a base point of the complete linear system ∣D∣ of a divisor D exactly when every global section of O(D) vanishes at q, equivalently H0(C,O(D−q))=H0(C,O(D)); the sheaf is base-point-free when no closed point is a base point. (Base points and base-point-free linear systems)

[F5]

A line bundle of degree at least 2g on a curve of genus g is base-point-free; this is the statement whose degree bound is tested here. (Line bundles of degree at least 2g are base-point-free)

[F6]

On a genus-one curve, the canonical bundle is trivial and every canonical divisor is principal; hence KC∼0 and OC(KC+p)≅OC(p). The divisor-line-bundle dictionary identifies this linear equivalence with the corresponding isomorphism of invertible sheaves. (The canonical bundle of a genus-one curve is trivial, Rational sections of line bundles are Cartier divisors, Cartier and Weil divisors agree on a smooth curve)

[F7]

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

[F8]

In ZF, the Axiom of Choice implies Dependent Choice; this supplies the Dependent Choice premise of the Cartier-to-Weil divisor dictionary used in [F1] and [F6]. (AC implies DC implies countable choice, The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain)

Counterexample

Assume the Axiom of Choice; it supplies Dependent Choice through AC implies DC implies countable choice. Let k be a field, let C be a smooth proper geometrically integral curve of genus g≥1 over k with a k-rational point p, let KC be a canonical divisor and put L=OC(KC+p),D=KC+p. Then deg⁡kD=2g−2+1=2g−1, and since 2g−1>2g−2 the nonspecial formula gives ℓ(D)=deg⁡kD+1−g=g,H1(C,L)=0. On the other hand L(−p)=OC(KC) has ℓ(KC)=h0(C,ωC)=g by the canonical-sections computation, and H0(C,L(−p))⊆H0(C,L) is an inclusion of spaces of the same dimension g; hence the two spaces are equal and every global section of L vanishes at p. Therefore p is a base point of the complete linear system ∣D∣=∣L∣, and L is not base-point-free even though its degree is the largest value below the safe bound 2g of the base-point-freeness theorem. For g=1, [F6] gives KC∼0, so L≅OC(p); its unique section has zero divisor [p] and vanishes at p.

Proof technique: compute both section spaces and observe that the evaluation at p has no room to be nonzero.

1.1F1F2

By [F1], deg⁡kD=2g−1>2g−2, so [F2] applies and gives ℓ(D)=g together with H1(C,L)=0.

2.1F1F3step 1.1

Since L(−p)=OC(KC) by [F1] and ℓ(KC)=g by [F3], the space H0(C,L(−p)) has dimension g; by the inclusion of [F1] and Step 1.1, H0(C,L(−p))⊆H0(C,L) is an inclusion of k-vector spaces of the same dimension g, hence an equality.

3.1F4step 2.1

The equality of Step 2.1 says exactly that every global section of L vanishes at the k-rational point p; by the base-point criterion [F4], p is a base point of the complete linear system ∣D∣ and L=OC(KC+p) is not base-point-free.

4.1F5step 3.1

Since deg⁡L=2g−1<2g, this counterexample shows that the degree bound in the base-point-freeness theorem [F5] cannot be lowered from 2g to 2g−1: the bound is sharp.

5.1F1F2F6F7F8step 1.1step 3.1∎

For g=1, [F6] gives KC∼0 (the chosen representative need not equal the zero divisor), so L≅OC(p). Step 1.1 gives h0(C,L)=1, and this isomorphism makes H0(C,OC(p)) one-dimensional. Its canonical section has zero divisor [p], so the unique one-dimensional section space vanishes at p, in agreement with Step 3.1. The Axiom of Choice is used through the degree, duality, and divisor suppliers, and [F8] supplies the Dependent Choice premise of the Cartier-to-Weil route.

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