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Berthelot P., Ogus A. Примечания относительно прозрачной когомологии (Принстон, 1978) MAh

Berthelot P., Ogus A. Notes on crystalline cohomology (Princeton, 1978)(T)(264s)_MAh_.djvu

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Date Jun 22, 2005

Cites: There is a functor Ly from the category
Oy-modules and HPD differential operators to the category of
HPD stratified Oy-moduies and 0v-linear horizontal maps...
Suppose S is quasi-compact, f: X + (S,I,Y) is
quasi-compact and quasi-separated, and fV/o: (X/S) ...
However, in the next chapter, we shall need to
work with the crystalline cohomology of nonproper schemes, and
therefore we have to give a different construction...
The same is
П '" I" —
true of «.* and k* , and it follows that they take injective
to injectives...
In this chapter we shall study
this action, in particular, its relationship to the Hodge filtra-
filtration on crystalline cohomology (as determined from the ideal Jy/s^'
The main global applications are Mazur's theorem (8.26), which says
that (with suitable hypotheses on X) the action of Frobenius de-
determines the Hodge filtration on H* (X/k) , and Katz's conjecture
(8.39), which says how the Hodge filtration limits the possible
"slopes" of Frobenius...
The Frobenius auto
morphism of W is a PD morphism, covered by the absolute Frobenius
endomorphism F....
Indeed, since we
are working with p-adic formal schemes, it is enough to check
this mod p, and we have Wv /Q о Fv ,<, = Fv and
*0/b0 0 0 x0
Fv /o ° Wv .„ = Fvt ...
It suffices to check
torsion free complexes, with tha triangles replaced by s
exact sequences...
Because K* is bounded above, the following is true for all
s >> 0:
L(s): The map H*(Kr)) + H*(KC) is injective if
all i <...
Indeed, it is easy to see that if the
"divisibility" held for all e , then (equivalently, in fac
Fv /c;FkM:l"(Y',«*,) С pk n^y fi' ) for all к - and thi
1 n / О I/O I/O
false, in general...
nr n0
p Hn = Pr Hcr.s n ImU) + p
To prove (8.26.2), first note that
ImU) n рГ Hcris(X/S) = Hn П Hr = Hc +r , and
r c +r ? +r ,
So it suffices to prove that ir maps H onto И*(Х,Т ^С/с
(Recall that the image of M1(X,Tr) in HJjR(X/SQ) is
F1".^ H^R(X/SQ).) But (8.23.S) gives us an exact sequence:
Since Proposition (8.24) implies that the p's are injective,
the tt^s are surjective...
To prove that we get an isomorphism if M is protective
and of finite rank, we may assume that M is free, because both
sides are compatible with localization...
This filtrat
* n m n
is compatible with the J-adic filtration, and gr_H (L*) is
finitely generated grTA-module...
Thus, P is fully fait
To prove that our functor is essentially surjective, su
pose {D : n S K} is an object in Pr(A.)...

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