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Arithmetic genus zero forces a smooth rational curve

Statement

Assume full AC. Let C be an integral proper curve over C. Then pa(C)=0 if and only if C is smooth and isomorphic to PC1; in particular, in the zero-genus case its normalization has genus zero.

Facts & Assumptions

Given: Full AC and an integral proper curve C over the algebraically closed field C.

[A1]

Full AC is The Axiom of Choice, inherited from the normalization and proper-cohomology suppliers.

[F1]

The normalization ν:C~→C is finite, C~ is smooth and proper, and its genus is the geometric genus g(C) (Geometric genus of a singular curve, Normalization of a reduced curve is finite, Normalization of an integral finite-type curve by gluing affine integral closures).

[F2]

The normalization defect δ(C) is a nonnegative integer and δ(C)=χ(OC~)−χ(OC)=pa(C)−g(C) (Normalization defect delta of a reduced curve, The normalization defect is an Euler characteristic and a weighted sum of local lengths, Genus and arithmetic genus of a curve, Geometric genus of a singular curve).

[F3]

The normalization map is an isomorphism exactly when its defect sheaf is zero; in particular δ(C)=0 implies C is regular, and a regular curve over the perfect field C is smooth (The normalization defect is an Euler characteristic and a weighted sum of local lengths, Normalization defect delta of a reduced curve, Geometric genus of a singular curve).

[F4]

The curve C has a closed C-rational point. Choose a nonempty affine open U=Spec⁡A⊂C. Since C is an integral finite-type curve, A is a nonzero finite-type C-domain of dimension one. The zero ideal is proper, so In a nonzero commutative ring, every proper ideal is contained in a maximal ideal gives a maximal ideal m⊂A; Over an algebraically closed field, maximal ideals of an affine algebra are kernels of points identifies its residue field with C. Thus the corresponding point is closed in U and is not generic. Its closure in C cannot be all of C, since its intersection with U is just that one point whereas U has dimension one. The only proper irreducible closed subsets of an integral curve are closed points, so this point is closed in C and has residue field C (Curves over a field, The residue field at a point of an affine scheme).

[F5]

For a closed point p on a smooth proper curve, the divisor [p] has degree [κ(p):C] (Divisors on a smooth proper curve, Degree divisor proper curve); hence the point from [F4] supplies a divisor of degree one. A smooth proper genus-zero curve with a degree-one divisor is isomorphic to P1 (A genus-zero curve with a degree-one divisor is the projective line).

[F6]

For a quasi-compact separated scheme, a finite affine open cover computes the sheaf cohomology of every quasi-coherent module by its ordered Čech complex (Cech cohomology computes quasi-coherent cohomology on a separated scheme). This applies to the two standard affine opens of PC1 and its quasi-coherent structure sheaf; their intersection is Spec⁡C[z,z−1].

Proof

technique · use the nonnegative normalization defect to force both the geometric genus and every singularity defect to vanish
1.1F2givenalgebra

From [F2], pa(C)=g(C)+δ(C), where both terms are nonnegative integers. If pa(C)=0, then g(C)=0 and δ(C)=0.

2.1A1F1F3F4F5step 1.1algebra

By [F3], δ(C)=0 makes C regular, and regularity of a finite-type curve over C makes it smooth. Its smooth proper normalization therefore has genus g(C)=0 by step 1.1. By [F4]–[F5], C has a divisor of degree one. Since C is algebraically closed, the integral curve C is geometrically integral; the hypotheses of A genus-zero curve with a degree-one divisor is the projective line are therefore satisfied, and it gives C≅PC1. Full AC is inherited from the normalization/cohomology suppliers and the maximal-ideal supplier.

3.1A1F2F6algebra∎

Conversely, suppose C≅PC1. By [F6], on the standard affine cover of P1 the structure-sheaf cohomology is computed by C[z]⊕C[z−1]→C[z,z−1], with differential (f,g)↦f−g. Its kernel is the common constants C, and its cokernel is zero because every Laurent polynomial is a sum of a polynomial in z and a polynomial in z−1. Thus h0=1, h1=0, χ(OC)=1 and pa(C)=1−χ(OC)=0 by the arithmetic-genus definition in [F2]. This proves the reverse implication.

Depends on

Dependency tree · 2 levels

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