Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-113/5/ii/solution
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 113 5 ii Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-06
Its dual of a sheaf is zero. Away from its source is zero. At , a homomorphism must send one to an element annihilated by the maximal ideal. Since , that ideal contains a nonzero coordinate parameter; the local ring is an integral domain, so the image must vanish. This proves that every local homomorphism vanishes. ConsequentlyThere is no contradiction with Serre duality: the ordinary sheaf dual suffices for locally free sheaves, but general coherent sheaves require an Ext functor. This example illustrates failure of ordinary sheaf-dual Serre duality for a skyscraper sheaf.
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