# Chevalley-Warning-Ax for double covers

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 $$d. Then Chevalley proved that $$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 $$\#V(g)$$?

• Are you asking whether the proof by James Ax still works in this setting, or are you asking whether the divisibility result is known (by any method)? The results of Fakhruddin-Rajan and Esnault imply the divisibility, cf. Theorem 2.1 of Esnault's appendix to Fakhruddin-Rajan: arxiv.org/pdf/math/0403265.pdf Mar 29 at 17:58
• Thanks! I was indeed asking about the result. I‘ll check out Esnault. Mar 29 at 18:03
• Certainly a step in the right direction. I would have never found it on my own.. Unfortunately "smooth irreducible varieties" is a showstopper for me. I have no control over $f$. Could be $0$ for all I know. Mar 29 at 18:26
• The morphism $f$ is projection from a "universal double cover" to the parameter space $Y$ of all degree $d$ polynomials. Thus, although many fibers of $f$ will be singular, and even reducible, the domain and target will be smooth. Some care is required to "homogenize" to get projective fibers (the correct way to do this is with a weighted projective space -- I will try to write more details soon). Then the beauty of these theorems is that, provided at least one closed fiber is smooth and has "Frobenius coniveau one", then this holds for every fiber (no matter how bad) . . . Mar 29 at 19:43
• Now I understand your point. Thanks again! Mar 30 at 7:53

## 2 Answers

Congruences for rational points on fibers. Let $$\mathbb{F}_q$$ be a finite field with $$q$$ elements.

Definition. A quasi-projective $$\mathbb{F}_q$$-scheme has congruence $$1$$, respectively congruence $$0$$ if for every positive integer $$r$$, the set of $$\mathbb{F}_{q^r}$$-points has cardinality congruent to $$1$$ modulo $$q^r$$, resp. congruent to $$0$$ modulo $$q^r$$. A quasi-projective morphism of quasi-projective $$\mathbb{F}_q$$-schemes has congruence $$1$$, respectively congruence $$0$$, if for every positive integer $$r$$, the fiber of the morphism over every $$\mathbb{F}_{q^r}$$-point of the target has congruence $$1$$, resp. congruence $$0$$.

Let $$\pi:X\to Y$$ be a proper, surjective morphism to a finite type $$\mathbb{F}_q$$-scheme. Generalizing the technique of Esnault (which generalization was later generalized again by Esnault and her coauthors), Fakhruddin and Rajan proved the following result as a corollary of their theorem in the following article.

MR2195144 (2006h:14028)
Fakhruddin, N.; Rajan, C. S.
Congruences for rational points on varieties over finite fields.
Math. Ann. 333 (2005), no. 4, 797–809.
https://arxiv.org/abs/math/0402230

Fakhruddin-Rajan Result. If $$X$$ and $$Y$$ are regular then $$\pi$$ has congruence $$1$$ if every degree-$$0$$ zero-cycle in the fiber of $$\text{pr}_1:X\times_Y X\to X$$ over the geometric generic point is torsion in the Chow group (by Bloch-Srinivas, this holds if the geometric generic fiber of $$\pi$$ is smooth and rationally chain connected).

Now let $$Z$$ be a closed subvariety of $$X$$ such that $$\pi(Z)$$ equals $$Y$$. Denote by $$U$$ the open complement of $$Z$$ in $$X$$. Denote by $$\pi|_Z$$, respectively by $$\pi|_U$$, the restriction of $$\pi$$ to $$Z$$, resp. the restriction of $$\pi$$ to $$U$$.

Corollary. If $$X$$, $$Y$$ and $$Z$$ are regular, then $$\pi|_U$$ has congruence $$0$$ if the condition above holds for both the geometric generic fiber of $$\pi$$ and $$\pi|_Z$$, e.g., if the geometric generic fibers of both $$\pi$$ and $$\pi|_Z$$ are smooth and rationally chain connected.

Proof. For every $$\mathbb{F}_q$$-point of $$Y$$, by applying the Fakhruddin-Rajan Result to both the fiber of $$\pi$$ and the fiber of $$\pi|_Z$$, each fiber has congruence $$1$$. The relative complement of these fibers is the fiber of $$\pi|_U$$, and this has congruence $$1-1=0$$. QED

Remark. There are generalizations of the Fakhruddin-Rajan theorem where the triviality of the degree-$$0$$ part of the $$\mathbb{Q}$$-Chow group is replaced by a direct hypothesis on the étale or rigid cohomology. This is the main result of Esnault's appendix to the Fakhruddin-Rajan theorem. This is relevant in trying to extend the theorem to situations where not all of the schemes are regular. If there are appropriate resolutions of singularities (e.g., there is a combinatorial method for resolving singularities of toric varieties such as weighted projective spaces), then one can work directly with these regular $$\mathbb{F}_q$$-varieties.

Desingularization of weighted projective spaces. Let $$n$$ be a nonnegative integer. Let $$\underline{e} =(e_0,\dotsc,e_n,e_{n+1})$$ be an ordered $$(n+1)$$-tuple of positive integers whose greatest common divisor equals $$1$$. Denote by $$S = S(\underline{e})$$ the polynomial ring $$\mathbb{Z}[t_0,\dotsc,t_n,t_{n+1}]$$ with the $$\mathbb{Z}_{\geq 0}$$-grading such that each variable $$t_i$$ is homogeneous of degree $$e_i$$. For every integer $$e\geq 0$$, denote by $$S_e$$ the degree-$$e$$ graded piece of $$S$$.

$$\DeclareMathOperator\Spec{Spec}$$Denote by $$\mathbb{P}(\underline{e})$$ the projective scheme over $$\Spec \mathbb{Z}$$ obtained as Proj of this $$\mathbb{Z}_{\geq 0}$$-graded ring. This is a projective toric scheme. There is an associated fan. An appropriate subdivision of this fan defines a torus-equivariant desingularization of the weighted projective space, $$\rho:\widetilde{\mathbb{P}}(\underline{e})\to \mathbb{P}(\underline{e}),$$ whose exceptional set is a simple normal crossings divisor. Note, this works over $$\Spec \mathbb{Z}$$, not just in characteristic $$0$$.

Proposition. When restricted over $$\Spec \mathbb{F}_q$$, the morphism $$\rho$$ has congruence $$1$$.

Proof. The singular set of $$\mathbb{P}(\underline{e})$$ is stratified by irreducible, torus-invariant, locally closed subsets that are smooth, and over which $$\rho$$ is a torus-equivariant morphism that is Zariski locally a product. Each irreducible component of each fiber of $$\rho$$ over each stratum is a projective toric variety, hence of congruence $$1$$. The intersections are unions of projective toric varieties. Chasing through, it suffices to prove that the dual complex of the fiber over a stratum is contractible. This is a combinatorial statement, thus independent of the characteristic. In characteristic $$0$$, for the cyclic quotient singularities that arise on weighted projective space, this has been proved in Corollary 0.3 of the following article of Kerz and Saito.

MR3119100
Kerz, Moritz; Saito, Shuji
Cohomological Hasse principle and resolution of quotient singularities.
New York J. Math. 19 (2013), 597–645.
https://arxiv.org/pdf/1111.7177.pdf

I learned of this article from an article of de Fernex-Kollár-Xu that generalizes the Kerz-Saito result. QED

Now let $$Y$$ be a quasi-projective $$\mathbb{F}_q$$-scheme $$Y$$, and let $$(\tau,\pi):X\to \mathbb{P}(\underline{e})\times_{\Spec \mathbb{F}_q} Y,$$ be a morphism of quasi-projective $$\mathbb{F}_q$$-schemes. Denote the fiber product $$\widetilde{\mathbb{P}}(\underline{e})\times_{\rho,\tau} X$$ with its projection to $$Y$$ by $$\widetilde{\pi}:\widetilde{X}\to Y.$$

Corollary. The morphism $$\pi$$ has congruence $$1$$ if and only if the morphism $$\widetilde{\pi}$$ has congruence $$1$$.

Proof. Every fiber of $$\rho$$ has congruence $$1$$. Thus the number of points of $$\pi$$ and $$\widetilde{\pi}$$ are congruent modulo the size of the residue field. QED

Fano complete intersections in weighted projective space. Let $$m = (m_1,\dotsc,m_c)$$ be a $$c$$-tuple of positive integers, each of which is a common multiple of each integer $$e_i$$. Let $$\mathbb{A}_m$$ be the affine space over $$\Spec \mathbb{F}_q$$ associated to the $$\mathbb{F}_q$$-vector space $$S_{m_1} \times \dotsb \times S_{m_c}$$. This affine space parameterizes ordered $$c$$-tuples of homogeneous polynomials on $$\mathbb{P}(\underline{e})$$ of respective degrees $$(m_1,\dotsc,m_c)$$. For every $$c$$-tuple of polynomials parameterized by a point of $$\mathbb{A}_m$$, consider the zero locus in $$\mathbb{P}(\underline{e})$$. With its second projection, the product scheme $$V_m \mathrel{:=} \mathbb{A}_m\times_{\Spec \mathbb{F}_q} \mathbb{P}(\underline{e}),$$ is a smooth, affine space bundle over $$\mathbb{P}(\underline{e})$$. There is a codimension-$$c$$, smooth, affine space subbundle over $$\mathbb{P}(\underline{e})$$, $$X_m \subset \mathbb{A}_m\times_{\Spec \mathbb{F}_q}\mathbb{P}(\underline{e}),$$ such that for every point of $$\mathbb{A}_m$$, the fiber of the first projection in $$X_m$$ equals the associated zero scheme in $$\mathbb{P}(\underline{e})$$.

In particular, the pullback $$\widetilde{X}_m$$ of this affine space bundle over $$\widetilde{\mathbb{P}}(\underline{e})$$ is a smooth morphism to a smooth $$\mathbb{F}_q$$-scheme. Thus, also $$\widetilde{X}_m$$ is a smooth $$\mathbb{F}_q$$-scheme. Denote the projection to $$\mathbb{A}_m$$ from $$X_m$$, respectively from $$\widetilde{X}_m$$, as follows, $$\pi:X_m \to \mathbb{A}_m, \ \ \widetilde{\pi}_m:\widetilde{X}_m \to \mathbb{A}_m.$$ The following hypothesis is automatic in characteristic $$0$$ by Bertini's smoothness theorem. By the analysis of Lefschetz pencils in SGA 7, it also holds in positive characteristic if each $$m_i$$ is at least $$2$$ times the least common multiple of all $$e_i$$.

Hypothesis. The geometric generic fiber of $$\widetilde{\pi}_m$$ is smooth.

Under this hypothesis, rational chain connectedness of the geometric generic fiber of $$\widetilde{\pi}_m$$ is implied by the Fano condition.

Fano Condition. Under the hypothesis, the geometric generic fiber of $$\pi$$ is $$\mathbb{Q}$$-Fano if and only if the integer $$m_1+\dotsb + m_c$$ is strictly less than the integer $$e_0+\dotsb+e_{n+1}$$. In this case, the geometric generic fiber of $$\widetilde{\pi}_m$$ is rationally chain connected, and thus both $$\pi_m$$ and $$\widetilde{\pi}_m$$ have congruence $$1$$.

Proof. The first assertion is a straightforward computation by the adjunction formula. The geometric generic fiber of $$\widetilde{\pi}_m$$ over $$\Spec \mathbb{F}_q$$ is the specialization from characteristic $$0$$ of the analogous geometric generic fiber. In characteristic $$0$$, rational connectedness of desingularizations of $$\mathbb{Q}$$-Fano varieties was proved by Qi Zhang.

MR2208131 (2006m:14021)
Zhang, Qi
Rational connectedness of log Q-Fano varieties.
J. Reine Angew. Math. 590 (2006), 131–142.
https://arxiv.org/pdf/math/0408301.pdf

Since the specialization of a rationally chain connected variety is again a rationally chain connected variety, the geometric generic fiber of $$\widetilde{\pi}_m$$ is rationally chain connected. Now apply the Fakhruddin-Rajan Result and the application above of the Kerz-Saito Result. QED

Finally, let $$\overline{c}>c$$ be a positive integer, and let $$\overline{m}$$ be a $$\overline{c}$$-tuple of the form $$\overline{m}=(m_1,\dots,m_c,\overline{m}_{c+1},\dots,\overline{m}_{\overline{c}})$$. As above, let $$\mathbb{A}_{\overline{m}}$$ be the corresponding affine variety parameterizing complete intersections of type $$\overline{m}$$. There is a projection morphism that is a smooth, linear morphism of affine vector bundles, $$\text{pr}_{\overline{m},m}:\mathbb{A}_{\overline{m}} \to \mathbb{A}_m.$$ The pullback $$\text{pr}^*_{\overline{m},m}\widetilde{X}_m$$ is a closed subscheme of $$\mathbb{A}_{\overline{m}}\times_{\Spec \mathbb{F}_q}\widetilde{\mathbb{P}}(\underline{e})$$ that contains $$\widetilde{X}_{\overline{m}}$$.

Stronger Hypothesis. Both the geometric generic fiber of $$\widetilde{\pi}_m$$ and the geometric generic fiber of $$\widetilde{\pi}_{\overline{m}}$$ are smooth.

Corollary. Under the stronger hypothesis, if the integer $$m_1+\dotsb + m_c + \widetilde{m}_{c+1}+\dots + \widetilde{m}_{\widetilde{c}}$$ is strictly less than the integer $$e_0+\dots+e_{n+1}$$, then both $$\text{pr}_{\overline{m},m}^*\widetilde{X}_m$$ and the closed subscheme $$\widetilde{X}_{\overline{m}}$$ have congruence $$1$$ relative to $$\mathbb{A}_{\overline{m}}$$. Thus, the open complement of this closed subscheme has congruence $$0$$ over $$\mathbb{A}_{\overline{m}}$$.

Now let $$\widetilde{Y}$$ be the inverse image of the open subset $$Y$$ of $$\mathbb{A}_m$$ under the obvious projection from $$\mathbb{A}_{\widetilde{m}}$$ to $$\mathbb{A}_m$$. Denote by $$\widetilde{Z}$$ the restriction of $$X_{\widetilde{m}}$$ over this open subset. Let $$\widetilde{X}$$ be the fiber product $$X\times_Y \widetilde{Y}$$. Thus, each of $$\widetilde{X}$$, $$\widetilde{Z}$$ and $$\widetilde{Y}$$ are smooth, and the geometric generic fiber over $$Y$$ of $$\widetilde{X}$$, resp. $$\widetilde{Z}$$, is the complete intersection of type $$(m_1,\dots,m_c)$$, resp. of type $$(m_1,\dots,m_c,\widetilde{m}_{c+1},\dots,\widetilde{m}_{\widetilde{c}})$$. Denote by $$\widetilde{U}$$ the open complement in $$\widetilde{X}$$ of $$\widetilde{Z}$$.

Corollary. If the integer $$m_1+\dotsb + m_c + \widetilde{m}_{c+1}+\dots + \widetilde{m}_{\widetilde{c}}$$ is strictly less than the integer $$e_0+\dots+e_{n+1}$$, then every fiber of $$\widetilde{U}$$ over a $$\mathbb{F}_q$$-point of $$\widetilde{Y}$$ is congruence $$0$$.

This follows by applying the previous result to both $$\widetilde{\pi}_m$$ and $$\widetilde{\pi}_{\overline{m}}$$.

Ax type statement. The first hypothesis holds if some geometric point of $$\mathbb{A}_m$$ parameterizes a complete intersection that is disjoint from the singular set of $$\mathbb{P}(\underline{e})$$. In this case, we can eliminate every "linear polynomial" of degree $$m_i=1$$ by reducing the number of homogeneous coordinates of degree $$1$$. Then, for the remaining polynomials with $$m_i\geq 2$$, the results of SGA 7 imply that the geometric generic fiber of the linear system is smooth (away from the singularities of the weighted projective space). Iterating this, the geometric generic fiber of $$\pi_m$$ is even smooth and disjoint from the singular locus of the weighted projective space.

The same argument applied to the further complete intersection of hypersurfaces of degrees $$(m_1,\dots,m_c,\overline{m}_{c+1},\dots,\overline{m}_{\overline{c}})$$ then implies the stronger hypothesis as well.

Application. In the homogeneous variables $$(t_0,\dots,t_n)$$ of specified degrees $$(e_0,\dots,e_n)$$, let $$(f_1,\dots,f_c)$$ be (not necessarily homogeneous) polynomials of total degrees $$(m_1,\dots,m_c)$$. Let $$t_{n+1}$$ be a homogeneous variable of degree $$1$$, and denote by $$(F_1,\dots,F_c)$$ the $$t_{n+1}$$-homogenizations. Assume that for a generic such $$c$$-tuple, the common zero locus is disjoint from the singular locus of the weighted projective space. If $$m_1+\dots+m_c$$ is strictly less than $$e_0+\dots+e_n$$, then the number of solutions of $$(f_1,\dots,f_c)$$ in $$\mathbb{F}_q^{n+1}$$ is divisible by $$q$$.

Indeed, form $$\overline{c}=1+c$$ and $$\overline{m}=(m_1,\dots,m_c,1)$$. Then the degree hypothesis above is precisely the Fano inequality from $$\overline{m}$$.

In particular, if $$\underline{e}$$ is either $$(d,2,\dots,2)$$, in case $$d$$ is odd, or $$(d/2,1,\dots,1)$$, in case $$d$$ is even, then the polynomial $$x_0^2 - f(x_1,\dots,x_n)$$ above has total degree $$2d$$, resp. $$d$$, whereas $$e_0+\dots+e_n$$ equals $$d+2n$$, resp. $$d/2+n$$. In both cases, the Fano condition is that $$d$$ is strictly less than $$2n$$. Moreover, the polynomial $$x_0^2 + x_1^d + \dots + x_n^d$$ vanishes at none of the singular points of the weighted projective space. Thus, the congruence holds.

• I am not sure what you are after. You are considering intersections of $c$ hypersurfaces. The question was about branched double covers. Am I missing something? Still, I understand the gist of your argument. I'll try to adapt it to my situation. Mar 30 at 12:18
• I need the congruence not only for the projective hypersurface, but also for the codimension-2 complete intersection whose complement is the affine hypersurface. Once codimension-2 complete intersections are involved, might as well explain the argument for all complete intersections. I will try to finish the answer soon ... Mar 30 at 18:58
• I eliminated the kludge of avoiding the singular locus of the weighted projective space. Since weighted projective spaces have cyclic quotient singularities, these admit resolution of singularities that preserve the congruence. Apr 6 at 18:01
• I corresponded with Will Sawin who, in particular, pointed out Corollary 1.6 of the following article of Berthelot-Bloch-Esnault: arxiv.org/pdf/math/0510349.pdf This leads to more uniform proofs and a more general result (all toric varieties, not only weighted projective spaces). I will try to add this to the answer above soon (I will keep the original post as well). Apr 9 at 17:26

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 $$\#X(\mathbb F_q)-q^n$$.