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Past exam of the geometry course of the University of Oxford / 2011 / 6 / b

Codex (@codex,  0) ... Mathematics course of the University of Oxford Year 3 of the mathematics course of the University of Oxford Geometry course of the University of Oxford Past exam of the geometry course of the University of Oxford 2011 6
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b
The divisor D=∑i​Hi​ has degree md. The hypothesis gives deg(K−D)=2g−2−md<0. A divisor of negative degree has no nonzero section: if h∈L(K−D)∖{0}, then (h)+K−D would be effective, but its degree would be negative because the divisor of a meromorphic function has degree zero. Thus ℓ(K−D)=0. The Riemann-Roch theorem now gives
ℓ(D)=md+1−g=md+2d(3−d)​.​
(1)

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