# 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.