Let $f:X\rightarrow Y$ be a projective map of schemes, $Y$ is finite type over $\mathbb Z[1/N]$ for some big $N$ and moreover $R^1 f_* \mathcal O_X=0$. As far as I understand all this implies that $\mathrm{Pic}_{X/Y}$ is actually a scheme and has dimension 0 over $Y$. Is it true then that $p$torsion $\mathrm{Pic}_{X/Y}[p]$ is zero for prime $p$ big enough? Over $\mathbb C$ I can say that the Picard group will be a locally constant sheaf of finetely generated groups, so everything follows, but I'm not sure about what is true in purely algebraic setting, especially when the base is not a field.
1 Answer
You need more conditions in order for the functor $\underline{\mathrm{Pic}}_{X/Y}$ to be representable: for instance $f$ flat, and the fibers geometrically integral. You should look at the 2 lectures by Grothendieck on the Picard scheme in the Bourbaki seminar (they can be found in FGA, i.e. "Fondements de la Géométrie algébrique").

$\begingroup$ I remembered from Milne's Etale Cohomology that projectivity is enough, but I obviously can be mistaken or there also could be some additional assumptions in the statement. Thank you for the reference, I will try to find answer there! $\endgroup$ Nov 10, 2013 at 15:40

$\begingroup$ But anyway Picard being only an algebraic space does not change the situation much, it still has dimension 0 and I suppose there should be some bound on its torsion $\endgroup$ Nov 10, 2013 at 18:35