# Re: Chap. 17 Polishchuk constructions and questions

- Taking for relatively effective works because on a curve effective means ample, meaning a sufficiently large power will kill the higher cohomology.
- is required so that has rank zero (think).
- In general no, the fibers are not zero dimensional, could be a point, then the fibers are just the curve.
- Flantess is necessary (for example) to apply cohomology with base change.
- As are acyclic, it follows . And a spectral sequence should give you the gap from going to to concluding are locally free. See John McClear, user’s Guide to spectral sequences , specifically change of ring spectral sequence.
- Regarding the section of business. Set and both of rank say . Then . A section of is a linear map where is the base field or ring, and a section of the other is just a wedge . Together it gives via . The determinant is in particular an example of such a .
- Finally if you have a map of smooth projective varieties of the same dimension, and you have a section . Then certainly is injective and are compact so should also be surjective (on geometric, or closed points), so will be a finite morphism. Now if either of is normal then some uniqueness of integral closure argument should say is an isomorphism as map of schemes, and that is the inverse.

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- Published:
- October 21, 2009 / 7:50 pm

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