Confusado — April 30, 2012



People had issues with my proof. I’m a bit confused – perhaps I will update it tomorrow if I figure out what the issue is.

[edit] So I think this proof actually does work. The key part is that U is arbitrary and there is some P(U) which is nonzero so that map P(U) -> k(x)(U) is nonzero, yet we can factor it through 0 = Ov(U).

Hartshorne III.7.3 — April 29, 2012
Projective Space Cohomology — April 9, 2012