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Jul 20, 2016 at 16:41 comment added SashaP @ChrisDodd Probably, there is. But I doubt that there will be a complete non-conjectural solution as long as there are no resolution of singularities in positive characteristic.
Jul 19, 2016 at 19:57 comment added Chris Dodd Hi Sahsa; thanks for the comment. This was one motivation for my question. As far as I can see, no one addresses this problem one way or the other in the literature; it seems like a strange gap. My feeling was that there must be some "folklore" thinking on the problem; we'll see if the MO community offers some.
Jul 19, 2016 at 19:05 comment added SashaP Also, as you suggest rigid cohomology can be computed via the log-de Rham-Witt complex given a smooth compactification. Every such compactification gives an integral structure but it seems unclear to me how to identify these integral structures for different compactifications.
Jul 19, 2016 at 18:42 comment added SashaP There is a well-defined theory called "rigid cohomology" due to Berthelot which gives finitely generated answer for arbitrary varieties. A drawback of it is that it is defined only rationally while crystalline cohomology has a structure of $W(k)$-module. As far as I know it is still an open problem to give rigid cohomology an integral structure.
Jul 19, 2016 at 1:18 comment added Chris Dodd Well, if $X$ is a proper scheme then the De Rham cohomology groups are finite dimensional; since they are computed as the hypercohomology groups of a finite complex of vector bundles. So in that sense it is "due to the non-properness."
Jul 19, 2016 at 0:53 comment added Mariano Suárez-Álvarez Infinite dimensionality is not due, in positive characteristic, to non-properness but to the actual characteristic, no?
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Jul 18, 2016 at 15:12 history asked Chris Dodd CC BY-SA 3.0