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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 113 / 2 / b / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 113 2 b
2026-09-28  0 By others on same topic  0 Discussions Create my own version
The scheme Pk3​ is Noetherian, integral, separated, and regular. Every open subscheme inherits these properties, so U=Pk3​∖C satisfies (⋆). Because C has dimension one, it has codimension two in Pk3​ and contains no prime Weil divisor. The localization sequence for the divisor class group therefore makes restriction an isomorphism
Cl(Pk3​)∼​Cl(U).
(1)
The hyperplane divisor generates the class group of projective space, so
Cl(U)≅Z.​
(2)

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