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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

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A refinement of the above result will provide us with a
beautiful crystalline interpretation of the Hodge filtration...
The essential point of the calculation is the cocycle condi
tion for Eg «
First we must give a more explicit formula for the HPD-
stratification of the L-construction...
Because the isomorphism of G.3) is natural, we
can use it to compute the crystalline cohomology of maps, as
well as spaces...
(The fact that the arrow in the second line is a quasi-
isomorphism depends on the injectivity of the modules j .)
This gives us the desired arrow...
Suppose, in the diagram below, that f is
quasi-separated and smooth, that Y is quasi-compact, and that
X1 = XxyY'...
To define the cohomology we want, we construct a site
Cris(X/'S) which computes the limit automatically...
We had best begin by remarking that we have a
language in the second statement, since we have written IR
in a derived categroy...
Since
e-(k'-l) > c(k'), this can only be because e(k'+l) < c(k
Now we can repeat the argument, and we obtain the double
absurdity: e is strictly decreasing beyond к and
e * e1 beyond к ...
Neverthel
has a left derived functor, in a slightly extended sense,
ause of the following simple result:
9 Proposition...
Since TD/S is a
morphism of complexes, (8.21.1) follows, and (8.21.2) is ал
immediate consequences...
For simplicity, we restrict attention to the case in which
S is the formal spectrum of the p-Witt ring W of a perfect
field k, and X is smooth and proper over S...
Then if
t = bQ+--.+b , one sees from the above that A 2L В if
30+"-+8т^ V'+tm for a11 " *
8.37.2 If M'—>M is a span with Hodge numbers e ,ех.....
(We leave this verification for the reader.)
f.'ow suppose that the span MlC—*M of T has Kodge nujnbers
e.",»1,.....
Roby [1,2] in which the many interesting proper-
properties of the functor Гд(М) are developed...
Now write P^_^ as a quotien
of a free An-module P^ and take K^ = P^ © P^ with the
obvious boundaries...



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