TITLE: Chevalley-Warning-Ax for double covers
QUESTION [9 upvotes]: Let $f(x_1,\ldots,x_n)$ be a polynomial of degree $d$ with coefficients in the finite field $\mathbb F_q$ and let $V(f)\subseteq\mathbb F_q^n$ be its set of zeroes. Assume $d<n$. Then Chevalley proved that $V(f)$ cannot consist of one element alone. In the same issue, Warning showed the stronger statement that $\#V(f)$ is divisible by the characteristic $p$ of $\mathbb F_q$. Later Ax showed that $\#V(f)$ is even divisible by $q$. While Chevalley-Warning is tricky but elementary, the result of Ax seems to be much deeper.
My question is whether an analogue of Ax's theorem is known for double covers. More specifically, assume $p\ne2$ and let $g(x_0,\ldots,x_n):=x_0^2-f(x_1,\ldots,x_n)$. Assume moreover $d<2n$ (a weaker bound than above). Then Chevalley-Warning holds in this setting: the number of zeroes $V(g)\subseteq\mathbb F_q^{n+1}$ is divisible by $p$. The proof is almost the same: compute the sum $\sum_{x\in\mathbb F_q^n}f(x)^{\frac{q-1}2}$ in two different ways. The question is now whether Ax still holds:

Is it true that $q$ divides $\#V(g)$?

REPLY [3 votes]: Using the hints of Jason, I meanwhile found my own solution. It is more elementary by avoiding singular varieties but is less general.
Let $\overline f(x_0,\ldots,x_n)$ be homogeneous of even degree $2m$. Then there is a classical construction of a double cover $p:\overline X\to\mathbf P^n$ where the preimage of the affine chart $\mathbf A_i^n\subset\mathbf P^n$ is given by the equation
$$
\left(\frac y{x_i^m}\right)^2=\overline f(\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i})
$$
inside $\mathbb A_i^n\times\mathbb A^1$ (one can check that $\overline X$ sits inside the geometric realization of $\mathcal O_{\mathbf P^n}(m)$). Now let $\mathfrak Y$ be the affine space of all homogeneous polynomials $\overline f$ and $\pi:\mathfrak X\to\mathfrak Y$ be the universal family of double covers. Then I claim that for $m\le n$ all assumptions of Fakhruddin-Rajan (see Jason's answer) are satisfied: $\pi$ ic clearly projective and surjective and it is easily checked that $\mathfrak X$ and $\mathfrak Y$ are smooth. Let $\overline X$ be a generic fiber. It is smooth since the branching divisor $R:=(f)$ is smooth (Bertini). The Riemann-Hurwitz formula implies for the canonical divisor
$$
2K_{\overline X}=p^*(2K_{\mathbf P^n}+R)=p^*\mathcal O(-2n-2+2m).
$$
So $\overline X$ is Fano because $m\le n$. A theorem of Campana and Kollar-Miyaoka-Mori implies that $\overline X$ is rationally chain connected. Hence the condition on $CH_0$ holds (Bloch-Srinivas). Therefore $\#\overline X(\mathbb F_q)\equiv1\mod q\ $ for all (even singular) double covers $\overline X$.
Finally back to the original question. Let $X$ be the zero set of $y^2=f(x_1,\ldots,x_n)$ and assume $\deg f\le 2n-1$. Let
$$
\overline f(x_0,\ldots,x_n)=x_0^{2n}f(\frac{x_1}{x_0},\ldots,\frac{x_n}{x_0})
$$
be its homogenization to a polynomial of degree $2n$ (!). Then $\overline f$ is of even degree and is at least once divisible by $x_0$. This means that the corresponding double cover $\overline X\to\mathbf P^ n$ is ramified over the hyperplane at infinity. Thus we get
$$
\#X(\mathbb F_q)=\#\overline X(\mathbb F_q)-\#\mathbf P^{n-1}(\mathbb F_q)\equiv1-1\mod q\equiv0\mod q. \square
$$
By the way there is a cute Corollary to this which is due to Oliver Schnetz (Geometries in perturbative quantum field theory, arxiv: 1905.08083). For $a\in\mathbb F_q$ define the Legendre symbol as
$$
\left(\frac a{\mathbb F_q}\right):=\begin{cases}1,&{\rm if}\ a\in(\mathbb F_q^\times)^2\\
0,&{\rm if}\ a=0\\-1,&\rm otherwise.
\end{cases}
$$
Then
$$
\sum_{x\in\mathbb F_q^n}\left(\frac{f(x)}{\mathbb F_q}\right)\equiv0\mod q
$$
for every $f(x_1,\ldots,x_n)$ of degree $\le2n-1$. In fact, the sum equals $\#X(\mathbb F_q)-q^n$.