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I would like to study the ordered Cech cohomology with respect to a Zariski covering of a variety. I can pass to the limit with respect to refinements; the components of the 'limit covering' will be local schemes. Unfortunately, it seems that the intersection of two components of this type is not local. Is there a way to fix that? I would be satisfied with replacing the Cech cohomology by the cohomology with respect to 'something like a Zariski hypercovering'; here I need this to be similar to the ordered Cech cohomology.

Does the situation become nicer if we consider ordered Cech cohomology with respect to certain Nisnevich covers (there are cases when ordered Cech cohomology is reasonable; at least this is so for the so-called elementary distinguished squares)?

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    $\begingroup$ I think that by using hypercoverings instead you should get all local schemes in the limit. I don't see that using Nisnevich coverings instead would change anything, the tensor product of two Henselian local rings will still not be local I believe. $\endgroup$
    – Torsten Ekedahl
    Commented Oct 26, 2011 at 5:07
  • $\begingroup$ Yes, probably the limit of all hypercoverings will be local (though I wonder whether a reference for this exists). Could the following be true: for any Zariski hypercovering the corresponding 'ordered Cech-like complex' computes cohomology correctly? The problem is that I do not want to consider unordered Cech complexes. $\endgroup$
    – Mikhail Bondarko
    Commented Oct 26, 2011 at 9:50
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    $\begingroup$ I'm not sure what an ordered hypercovering means. However, have you looked at the answers at mathoverflow.net/questions/10056/… $\endgroup$
    – David E Speyer
    Commented Oct 26, 2011 at 12:36
  • $\begingroup$ Yes, I did; possibly I should do this once more.:) $\endgroup$
    – Mikhail Bondarko
    Commented Oct 26, 2011 at 13:09

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