Let $k$ be an algebraically closed field and $R$ be a discrete valuation ring which is also a $k$-algebra. Then, is $H^2(\mathbb{P}^n_k \times_k R,\mathbb{Z}) \cong \mathbb{Z}$?
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I assume you are considering this scheme with the Zariski topology. May I ask why you are interested in singular cohomology in this case? – bertram Aug 22 '17 at 15:46
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@bertram part of some computation problem – user43198 Aug 22 '17 at 15:49
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Every irreducible variety over an algebraically closed field of cardinality (at least) continuum is contractible (in Zariski topology). See Eric Wofsey's answer here. In particular, it will have trivial singular homology in all degrees. If your field countable then there are no nonconstant maps from simplices to your space, so 2nd homology will vanish for a trivial reason. This leaves out the case of fields of intermediate cardinality (whose existence depends on CH). I will leave somebody else to sort those out. (Can $[0,1]$ be expressed as an infinite union of less than continuum of pairwise disjoint closed subsets?)
The bottom line is that you should not use "classical" algebraic topology invariant to study Zariski topology (unless you are using cohomology with sheaf coefficients in some "interesting" sheaves).
Moishe Kohan
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+1 to the last line. That is what I was driving at in my comment above (signed bertram). – Nefertiti Aug 23 '17 at 08:30
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Dear @Nefertiti: Yes, I saw your comment. Incidentally (no offense), I like the name Bertram (an alive algebraic geometer) more than the one of a long-dead Egyptian queen. – Moishe Kohan Aug 23 '17 at 08:43
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Dear Moishe Cohen: thanks for your feedback. The name bertram happens to coincide with that of an active algebraic geometer, but that was not my motivation for choosing it: actually I had in mind another non-living non-algebraic geometer. – Nefertiti Aug 23 '17 at 08:46