Re: Chap. 17 Polishchuk constructions and questions

  1. Taking E_0 = F(nD) for D relatively effective works because on a curve effective means ample, meaning a sufficiently large power will kill the higher cohomology.
  2. \chi(C_s, F_s) = 0 is required so that R\pi_*F has rank zero (think).
  3. In general no, the fibers C_s are not zero dimensional, S could be a point, then the fibers are just the curve.
  4. Flantess is necessary (for example) to apply cohomology with base change.
  5. As E_i are R\pi_* acyclic, it follows R\pi_*E = \pi_*E_i.  And a spectral sequence should give you the gap from going to R^1\pi_*E_0 = 0 to concluding \pi_*E_i are locally free.  See John McClear, user’s Guide to spectral sequences , specifically change of ring spectral sequence.
  6. Regarding the section of L = \det^{-1} R\pi_*F business.  Set E = \pi_*E_0 and F = \pi_*E_1 both of rank say r.  Then L = \det E^\vee \otimes \det F.  A section of \det E^\vee is a linear map \phi\colon \wedge^r E \to k where k is the base field or ring, and a section of the other is just a wedge \wedge f_1 ... \wedge f_r.  Together it gives \phi' \colon \wedge^rE \to \wedge^rF via \phi'(\wedge e_i) = \phi(\wedge e_i)\wedge f_i.  The determinant is in particular an example of such a \phi'.
  7. Finally if you have a map of smooth projective varieties a \colon J \to K of the same dimension, and you have a section b \colon K \to J.  Then certainly a is injective and J,K are compact so a should also be surjective (on geometric, or closed points), so a will be a finite morphism.  Now if either of J,K is normal then some uniqueness of integral closure argument should say a is an isomorphism as map of schemes, and that b is the inverse.
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