Dr. Katz DEq Homework Solutions 74

Dr. Katz DEq Homework Solutions 74 - finite rank and to...

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74 N.M. Katz: locally free module of finite rank. (Such a 0//always exists.) Then, over ~, the Hodge ~ De Rham spectral sequence ( Ef'~=Rqf, (s (log D)) ~ R"+qf, (f2~/s (log O)) is degenerate at E 1 . Let us recall why this is so. By Deligne's mixed Hodge theory ([8], 3.2.13), this spectral sequence degenerates at E~ over the generic point of q/. Because E1 is locally free over ~, the vanishing of dl at the generic point implies its vanishing on all of q/. Then E2 = E1 is locally free on q/, so d2 =0 on q/ because dE=0 at the generic point of ~/. .. this proves inductively that all dr = 0 over Y/. By (, after any change of base S'--, S which factors through ~, the spectral sequence ( will continue to have E1 locally free of
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Unformatted text preview: finite rank, and to degenerate at E~. (5.1) Theorem. Hypotheses as in (5.0.0), let n>O be an integer. Suppose that there is some non-void affine open set ~ c S satisfying (5.0.1), and an infinite set $ of prime numbers p, such that the following condition (5.1.0) holds (5.1.0) For every point of Twith values in (the spectrum of) a finite field Fq of characteristic peS,, after the base change Spec(Fq)---~ T, (5.1.o.o) D~ | , D~, 1 l 1 ~| , ~- 1 1 Spec(Fq) , T ~D [ ,X .,S the p-curvature of the free (Qall(~AFq module with integrable connection relative to Fq (5A.o.1) (R"f, ((2~-/s (log D))tq/, V)@Fq is zero, in other words, that the inverse image of R"f,(f2"x/s(logD)) on | Fq is spanned by its horizontal sections....
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