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Positive square and positive ample intersection force an effective multiple

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let k be a field, let X be an integral smooth projective surface over k, let H be an ample invertible OX-module (Absolute ampleness by affine section opens) and let L be an invertible OX-module with L⋅L>0,L⋅H>0. Then there exists an integer n≥1 with H0(X,L⊗n)≠0.

Facts & Assumptions

Given: a field k, an integral smooth projective surface X over k, an ample invertible sheaf H, and an invertible sheaf L with L⋅L>0 and L⋅H>0.

[F1]

Bilinearity: Z-bilinearity of the intersection product gives L⊗n⋅H=n(L⋅H) and L⊗n⋅L⊗n=n2(L⋅L) for every n≥0; in particular L⊗n⋅H>KX⋅H for all sufficiently large n, with KX⋅H=ωX⋅H (Intersection numbers of Cartier divisors on a smooth projective surface, The surface intersection product is symmetric and bilinear, The canonical divisor of a smooth projective surface, Invertible sheaves, Tensor product of sheaves of modules, Dual of a line bundle is its tensor inverse).

[F2]

Threshold vanishing: if M⋅H>KX⋅H for an invertible sheaf M, then H2(X,M)=0 (Vanishing of top cohomology past the canonical threshold, The canonical divisor of a smooth projective surface).

[F3]

Riemann-Roch: for every invertible sheaf M, χ(X,M)=χ(X,OX)+12(M⋅M−M⋅KX) (Riemann-Roch for smooth projective surfaces). The Euler characteristic is the alternating sum χ=h0−h1+h2 of the dimensions hq=dim⁡kHq(X,M), so h0=χ+h1−h2 with h1,h2≥0 (Euler characteristic of a coherent sheaf).

[F4]

The Axiom of Choice is inherited from the Riemann-Roch and vanishing suppliers of [F2] and [F3]; the integer n is chosen below from the growth of a quadratic polynomial, no family is selected.

Proof

technique · direct: for large $n$ the top cohomology of $L^{\otimes n}$ vanishes and Riemann-Roch makes the Euler characteristic positive; the sign in $h^0=\chi+h^1-h^2$ then gives a section
1.1F1F2

Vanishing of the top cohomology for large n. By [F1], L⊗n⋅H=n(L⋅H)→+∞ as n→∞ because L⋅H>0, so for all n≫0 we have L⊗n⋅H>KX⋅H; by [F2], H2(X,L⊗n)=0 for all such n.

1.2F1F3

Growth of the Euler characteristic. By [F3] and [F1], χ(X,L⊗n)=χ(X,OX)+n22(L⋅L)−n2(L⋅KX)⟶+∞(n→∞), because the quadratic term has positive leading coefficient L⋅L>0.

2.1F3F4step 1.1step 1.2∎

A positive Euler characteristic gives a section. Choose n≫0 so large that both step 1.1 applies and χ(X,L⊗n)>0, which is possible by steps 1.1 and 1.2. By [F3], h0(X,L⊗n)=χ(X,L⊗n)+h1(X,L⊗n)−h2(X,L⊗n), and h2=0 by step 1.1 while h1≥0; hence h0(X,L⊗n)≥χ(X,L⊗n)>0 and H0(X,L⊗n)≠0. The Axiom of Choice is inherited from [F4].

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