I am looking for an example of a smooth irreducible quasiprojective variety $X$ over ${\mathbb C}$, such that when reduced over finite fields ${\mathbb F\_q}$, the number of its points is a polynomial $P(q)$ of $q$ with nonnegative (integer) coefficients, but $X$ has some odd cohomology. 


Background: as discussed in an answer to [this question](https://mathoverflow.net/questions/92255/poincare-polynomial-and-counting-rational-points), if a variety $X$ is paved by affine spaces, then it only has $(p,p)$ cohomology, and the number of its ${\mathbb F\_q}$-points equals $P(q)$, where $P(t)$ is the Poincare polynomial of (compactly supported cohomology of) $X$. Note that the coefficients of $P$ are necessarily non-negative, given by the number of affine cells in the paving of a fixed dimension. In the appendix to  [this paper](http://xxx.lanl.gov/abs/math/0612668), N.Katz proves a kind of converse to this statement: if the number of points of $X$ over a finite field is given by a polynomial $P(q)$ of $q$, then this polynomial determines the so-called $E$-polynomial $E(x,y)$ of $X$ by the formula $E(x,y)=P(xy)$. The $E$-polynomial is a partial Euler characteristic, where we remember the weights of (compactly supported) cohomology but not the degrees. 


Of course varieties of the latter type can have odd cohomology; the typical example is $X={\mathbb C}^\*$, with point count polynomial $P(q)=q-1$. $X$ of course has odd cohomology. A slighly more complicated example, due to N.Katz, shows that $X$ can also have non-$(p,p)$ cohomology. But in these examples, the polynomial $P$ has some negative coefficients. 


Hence the question: can $X$ have positive polynomial count, but still some odd cohomology (which cancels in the $E$-polynomial)? Note that $X$ can't be smooth projective, since then its cohomology would be pure so any odd cohomology would have to show up in the $E$-polynomial. 


There may of course be a trivial example which I am missing. 



Smooth variety with positive point-count polynomial and odd cohomology
The answer to your question is yes:


Let $X$ be the blowup of $\mathbb{A}^1 \times (\mathbb{A}^1 - \{0\})$ in a point. The number of points over a field of $q$ elements is $q(q-1) + q = q^2$ and $X$ has non-trivial $H^1$.


