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Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 25 / 5 / a / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 25 5 a
Created 2026-10-03 Updated 2026-10-14  0 By others on same topic  0 Discussions Create my own version
The divisors have no common irreducible component, since a common curve would give infinitely many intersection points. The Bézout theorem therefore gives total intersection multiplicity degCdegD=3⋅3=9. As all intersection is supported at p,
Ip​(C,D)=9.​
(1)

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