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Apr 22, 2016 at 7:38 comment added მამუკა ჯიბლაძე It would be useful to see this comment promoted to an answer I think
Dec 22, 2015 at 15:15 comment added Lennart Meier Ah, I see. I check the vanishing in every fiber and then know that the rank of the first cohomology of $\omega^{\otimes k}$ is constant over all fibers - then I apply cohomology and base change to obtain that $R^1f_*\omega^{\otimes k}$ is locally free and hence also zero (for $f\colon \overline{\mathcal{M}}_n \to \mathrm{Spec}\mathbb{Z}[\frac1n]$). Thanks!
Dec 22, 2015 at 14:25 comment added Jason Starr You should probably read about the semicontinuity theorem and compatibility of cohomology and base change. This is Section 12 of Chapter III of Hartshorne's book.
Dec 22, 2015 at 14:16 history asked Lennart Meier CC BY-SA 3.0