Constructing Polynomial Count Varieties I have some naive questions about polynomial-count affine varieties over $\mathbb{C}$:


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*Are all reductive algebraic groups strongly polynomial-count?

*Are products of strongly polynomial-count varieties also strongly polynomial-count?  What about (disjoint) unions?

*If X is strongly polynomial-count variety, and F is a finite group acting on X, is X//F also polynomial-count?  More generally, is the property of strongly polynomial-count invariant under étale equivalence.  

*If G is reductive algebraic group acting on a variety X, and the orbit-type stratification of X consists of strongly polynomial count quasi-projective subvarieties, then is X//G also strongly polynomial-count?   

*Are there general conditions on a variety X and algebraic group G for X//G to be strongly polynomial count?
Basically, I would like to know if there are operations that allows one to cook up polynomial count varieties from other polynomial count varieties.
See the Appendix here for the definition of polynomial count:  here
EDIT:  Just to be completely honest, when I posted this, I did have a sense for some of the questions, but I wanted to learn more about a concept I am just now learning to work with, and felt that they all fit a general theme so just included them all without detailing what I thought I knew and didn't.
 A: A combination of easy and hard questions here. The easy ones:
(1) No. For example, the group scheme $\{ (x,y) : x^2+y^2=1 \}$, with multiplication $(x_1, y_1) (x_2, y_2) = (x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1)$ has $q - (-1)^{(q-1)/2}$ points over a field with $q$ elements. Or, similarly, the group scheme $x^3=1$ has $1$ or $3$ points depending on whether $q$ is $1$ or $2$ mod $3$. However, connected split reductive groups are polynomial point count. Moreover, we can make an arbitrary reductive group be polynomial strongly point count by changing the base ring $R$: For example, switching from $\mathbb{Z}$ to $\mathbb{Z}[i]$ in the first example or from $\mathbb{Z}$ to $\mathbb{Z}[(-1+\sqrt{-3})/2]$ in the second.
(2) Yes and yes, this is immediate from the definition.
(3) Polynomial point count is not invariant under etale equivalence, because it is easy to build examples where $X$ is polynomial point count and $X/G$ is not. For example, take $X$ to be the elliptic curve $y^2=x^3-x$ with the points $(x,y) = (-1,0)$, $(0,0)$ and $(0,1)$ deleted. $X$ is not polynomial point count. Let $G = \mathbb{Z}/2$ act on $X$ by $(x,y) \mapsto (x,-y)$. Then the quotient is the affine line with $3$ points deleted, with point count $q-3$.
It is harder to find an example the other way, with $X$ polynomial point count and $X/G$ not polynomial point count. Here is one. Let $E \subset \mathbb{P}^2$ be a smooth cubic curve. Take two copies of $\mathbb{P}^2$, each with $E$ embedded in them, and glue them along $E$. Then take the disjoint union of that with another copy of $E$. Call this $\bar{X}$. So $\bar{X}$ has $2(q^2+q+1) - \#E(\mathbb{F}_q) + \#E(\mathbb{F}_q)= 2(q^2+q+1)$ points over $\mathbb{F}_q$.  Let $X$ be $\bar{X}$ with four points deleted from the copy of $E$ which is disjoint from the projective planes. So $X$ has point count $2(q^2+q+1)-4$, another polynomial. Let $\mathbb{Z}/2$ act trivially on the $2$-dimensional component of $X$ and act by a fixed point free involution on $E$ with the $4$ points deleted. So the quotient of the one dimension component is $\mathbb{P}^1$ minus $4$ points, and the point count of the whole quotient is $2 (q^2+q+1) - \#E(\mathbb{F}_q) + (q-4)$.
This mostly demonstrates that polynomial point count is a weird notion; a better condition is that all the eigenvalues of Frobenius on compactly supported cohomology be powers of $q$. I believe that this does pass to quotients as I think (but I am not an expert) that $H^{\ast}(X/G, A) = H^{\ast}(X,A)^G$ whenever $|G|$ is invertible in the coefficient ring $A$.
(4) seems harder to me.
