Singular cohomology of projective space

Question

Let $k$ be an algebraically closed field and $R$ be a discrete valuation ring which is also a $k$-algebra. Then, is $H^2(\mathbb{P}^n_k \times_k R,\mathbb{Z}) \cong \mathbb{Z}$?

Answer 1

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).