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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 24F / a / i

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 4 24F a
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i
Let H=V(L) for a nonzero homogeneous linear polynomial L. Because H does not contain X, the restriction L∣X​ is a nonzero section of OX​(1). Its zero divisor is the hyperplane-section divisor of a projective plane curve
[X∩H]=∑p∈X∩H​Ip​(X,H)p,
(1)
where Ip​(X,H) is the local intersection multiplicity; equivalently, it is the order of vanishing of L∣X​ at p.
The curves X and H have no common component and have degrees d and 1. The Bézout theorem therefore gives
deg[X∩H]=p∈X∩H∑​Ip​(X,H)=d.​
(2)

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