Ovoids of Q(6, q) of low degree

Abstract

Ovoids of the parabolic quadric Q(6, q) of $\textrm{PG}(6,q)$ have been largely studied in the last 40 years. They can only occur if q is an odd prime power and there are two known families of ovoids of Q(6, q), the Thas-Kantor ovoids and the Ree-Tits ovoids, both for q a power of 3. It is well known that to any ovoid of Q(6, q) two polynomials $f_1(X,Y,Z)$, $f_2(X,Y,Z)$ can be associated. In this paper we classify ovoids of Q(6, q) with $\max \{\deg (f_1),\deg (f_2)\}<(\frac{1}{6.3}q)^{\frac{3}{13}}-1$.

1 Introduction

Let $\mathbb {F}_q$, $q=p^h$, be the finite field with q elements. Denote by ${\mathbb {P}}$ a finite classical polar space, i.e. the set of absolute subspaces of either a polarity or a non-singular quadratic form of a projective space $\textrm{PG}(m,q)$. The maximal dimensional projective subspaces contained in ${\mathbb {P}}$ are the generators of ${\mathbb {P}}$.

Definition 1

Let ${\mathbb {P}}$ be a finite classical polar space of $\textrm{PG}(m,q)$. An ovoid of ${\mathbb {P}}$ is a set of points of ${\mathbb {P}}$ that has exactly one point in common with each generator of ${\mathbb {P}}.$

Regarding ovoids of the parabolic quadric Q(m, q) of $\textrm{PG}(m,q)$, m even, it is known that there are no ovoids if $m>6$ (see [9, 16]). Hence for parabolic quadrics the only open problems concern $m=4$ and $m=6$.

In this paper we will focus on ovoids of Q(6, q). Denote by $(X_0,X_1,X_2,X_3,X_4,X_5,X_6)$ the homogenous coordinates of $\textrm{PG}(6,q)$ and by $H_\infty$ the hyperplane at infinity $X_0=0$.

Up to equivalence, in $\textrm{PG}(6,q)$ there is a unique non-degenerate quadric, the parabolic quadric Q(6, q) and in the sequel we fix the following parabolic quadric:

$$\begin{aligned} \textrm{Q}: X_0X_6+X_1X_5+X_2X_4+X_3^2=0.\end{aligned}$$

It is well known that there are no ovoids of Q(6, q) for q even [16], and thus from now on, we will consider only $p>2$. By Definition 1, an ovoid of $\textrm{Q}$ is a set of $q^3+1$ pairwise non-collinear points on the quadric. We can always assume that an ovoid $\textrm{O}_6$ of Q(6, q) contains the points (1, 0, 0, 0, 0, 0, 0) and (0, 0, 0, 0, 0, 0, 1) and hence it can be written in the following form:

$$\begin{aligned} \textrm{O}_6(f_1,f_2)= & {} \{(1,x,y,z,f_1(x,y,z),f_2(x,y,z),-z^2-yf_1(x,y,z)-xf_2(x,y,z))\}_{x,y,z\in \mathbb {F}_q}\\{} & {} \quad \cup \{(0,0,0,0,0,0,1)\}, \end{aligned}$$

for some functions $f_i:\mathbb {F}_q^3 \longrightarrow \mathbb {F}_q$ with $f_i(0,0,0) =0$ for $i\in \{1,2\}$.

The set $\textrm{O}_6(f_1,f_2)$ is an ovoid if and only if

$$\begin{aligned} \langle{} & {} (1,x_1,y_1,z_1,f_1(x_1,y_1,z_1),f_2(x_1,y_1,z_1), -z_1^2-y_1f_1(x_1,y_1,z_1)-x_1f_2(x_1,y_1,z_1)), \nonumber \\{} & {} (1,x_2,y_2,z_2,f_1(x_2,y_2,z_2),f_2(x_2,y_2,z_2), -z_2^2-y_2f_1(x_2,y_2,z_2)-x_2f_2(x_2,y_2,z_2))\rangle \ne 0 , \end{aligned}$$

(1)

for every $(x_1,y_1,z_1) \ne (x_2,y_2,z_2)$ in ${\mathbb {F}}_q^3$, where $\langle \cdot ,\cdot \rangle$ is the symmetric form associated to the quadratic form of the quadric $\textrm{Q}$.

The only known ovoids of $\textrm{Q}$ are the two classes of ovoids listed in the following table.

Table 1 Known ovoids of Q(6, q)

Note that Condition (1) reads

$$\begin{aligned}{} & {} (z_1-z_2)^2 + (x_2-x_1)(f_2(x_2,y_2,z_2)-f_2(x_1,y_1,z_1))\nonumber \\{} & {} \quad +(y_2-y_1)(f_1(x_2,y_2,z_2)-f_1(x_1,y_1,z_1)) \ne 0,\end{aligned}$$

(2)

for every $(x_1,y_1,z_1) \ne (x_2,y_2,z_2)$ in ${\mathbb {F}}_q^3$.

In order to obtain non-existence results and partial classifications for the ovoids $\textrm{O}_6(f_1,f_2)$ of $\textrm{Q}$, we will consider the hypersurface $\mathcal {W}_{f_1,f_2}\subset \textrm{PG}(6,q)$ defined by $F_{f_1,f_2}(X_0,X_1,X_2,X_3,X_4,X_5,X_6)=0$, where $F_{f_1,f_2}(1,X_1,X_2,X_3,X_4,X_5,X_6)$ equals

$$\begin{aligned} (X_3-X_6)^2 + (X_4-X_1)(f_2(X_4,X_5,X_6)-f_2(X_1,X_2,X_3))\nonumber \\ +(X_5-X_2)(f_1(X_4,X_5,X_6)-f_1(X_1,X_2,X_3)). \end{aligned}$$

(3)

Lemma 2

If $f_1,f_2\not \equiv 0$, then $\textrm{O}_6(f_1,f_2)$ is an ovoid of $\textrm{Q}$ if and only if $\mathcal {W}_{f_1,f_2}$ contains no affine ${\mathbb {F}}_q$-rational points off the solid $X_1-X_4=X_2-X_5=X_3-X_6=0$.

The main result of this paper is the following.

MainTheorem

Let $f_1(X,Y,Z)=\sum _{ijk}a_{i,j,k}X^iY^jZ^k$ and $f_2(X,Y,Z)=\sum _{ijk}b_{i,j,k}X^iY^jZ^k$. Suppose that $q>6.3 (d+1)^{13/3}$, where $\max \{\deg (f_1),\deg (f_2)\}=d$. If $\textrm{O}_6(f_1,f_2)$ is an ovoid of $\textrm{Q}$ then

$$\begin{aligned} q=3^h, \quad f_1(X,Y,Z)=X^2Y-nY^3-XZ, \quad f_2(X,Y,Z)=-1/n X^3+XY^2+YZ. \end{aligned}$$

Hence $O_6(f_1,f_2)$ is a Thas-Kantor ovoid.

It is worth pointing out that the degree of the functions $f_1(X,Y,Z)$ and $f_2(X,Y,Z)$ associated with the Ree-Tits ovoid does not satisfy the bound $q>6.3 (d+1)^{13/3}$ in our Main Theorem.

2 Classification of low degree ovoids

The aim of this section is to provide a partial classification of the ovoids of $\textrm{Q}$ via the investigation of specific algebraic varieties; see [2] for a survey on links between algebraic varieties over finite fields and relevant combinatorial objects.

An algebraic hypersurface $\mathcal {S}$ is an algebraic variety that may be defined by a single implicit equation. An algebraic hypersurface defined over a field ${\mathbb {K}}$ is absolutely irreducible if the associated polynomial is irreducible over every algebraic extension of ${\mathbb {K}}$. An absolutely irreducible ${\mathbb {K}}$-rational component of a hypersurface $\mathcal {S}$, defined by the polynomial F, is simply an absolutely irreducible hypersurface such that the associated polynomial has coefficients in ${\mathbb {K}}$ and it is a factor of F. For a deeper introduction to algebraic varieties we refer the interested reader to [10].

A fundamental tool to determine the existence of rational points in an algebraic variety over finite fields is the following result by Lang and Weil [12] in 1954, which can be seen as a generalization of the Hasse–Weil bound.

Theorem 3

(Lang–Weil Theorem) Let $\mathcal {V}\subset {\mathbb {P}}^N({\mathbb {F}}_q)$ be an absolutely irreducible variety of dimension n and degree d. Then there exists a constant C depending only on N, n, and d such that

$$\begin{aligned} \left| \#\mathcal {V}({\mathbb {F}}_q)-\sum _{i=0}^{n} q^i\right| \le (d-1)(d-2)q^{n-1/2}+Cq^{n-1}, \end{aligned}$$

(4)

where $\#\mathcal {V}({\mathbb {F}}_q)$ denotes the number of ${\mathbb {F}}_q$-rational points of $\mathcal {V}$.

Although the constant C was not computed in [12], explicit estimates have been provided for instance in [4,5,6,7,8, 13, 15] and they have the general shape $C=f(d)$ provided that $q>g(n,d)$, where f and g are polynomials of (usually) small degree. We refer to [5] for a survey on these bounds. Excellent surveys on Hasse–Weil and Lang–Weil type theorems are [7, 17]. We include here the following result by Cafure and Matera [5].

Theorem 4

[5, Theorem 7.1] Let $\mathcal {V}\subset \textrm{AG}(n,q)$ be an absolutely irreducible ${\mathbb {F}}_q$-variety of dimension $r>0$ and degree $\delta$. If $q>2(r+1)\delta ^2$, then the following estimate holds:

$$\begin{aligned} |\#(\mathcal {V}\cap \textrm{AG}(n,q))-q^r|\le (\delta -1)(\delta -2)q^{r-1/2}+5\delta ^{13/3} q^{r-1}. \end{aligned}$$

As already mentioned, we will consider the hypersurface $\mathcal {W}_{f_1,f_2}\subset \textrm{PG}(6,q)$ defined by

$$\begin{aligned} F_{f_1,f_2}(X_0,X_1,X_2,X_3,X_4,X_5,X_6)=0, \end{aligned}$$

where $F_{f_1,f_2}(1,X_1,X_2,X_3,X_4,X_5,X_6)$ equals

$$\begin{aligned}{} & {} (X_3-X_6)^2 + (X_4-X_1)(f_2(X_4,X_5,X_6)-f_2(X_1,X_2,X_3))\\{} & {} \quad +(X_5-X_2)(f_1(X_4,X_5,X_6)-f_1(X_1,X_2,X_3)). \end{aligned}$$

For our purposes, the existence of an absolutely irreducible ${\mathbb {F}}_q$-rational component in $\mathcal {S}_f$ is enough to provide asymptotic non-existence results for ovoids of $\textrm{Q}$.

Theorem 5

Let $\mathcal {W}_{f_1,f_2}: F_{f_1,f_2}(X_0,X_1,X_2,X_3,X_4,X_5,X_6)=0$, where $F_{f_1,f_2}$ is defined as in (3). Suppose that $q>6.3 (d+1)^{13/3}$, $\max \{\deg (f_1),\deg (f_2)\}=d$, and $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component $\mathcal {V}$ defined over ${\mathbb {F}}_q$. Then $\textrm{O}_6(f_1,f_2)$ is not an ovoid of $\textrm{Q}$.

Proof

Since $q>6.3 (d+1)^{13/3}$, by Theorem 4, with $\delta =d+1$ and $r=5$, it is readily seen that $\mathcal {V}$ contains more than $q^3$ affine ${\mathbb {F}}_q$-rational points. This yields the existence of at least one affine ${\mathbb {F}}_q$-rational point $(a_1,b_1,c_1,a_2,b_2,c_2)$ in $\mathcal {V}\subset \mathcal {W}_{f_1,f_2}$ (defined as in (3)) such that $a_1\ne a_2$ or $b_1\ne b_2$ or $c_1\ne c_2$ and therefore $\textrm{O}_6(f_1,f_2)$ is not an ovoid of $\textrm{Q}$. $\square$

We also include, for seek of completeness, the following result which will be useful in the sequel.

Lemma 6

[1, Lemma 2.1] Let $\mathcal {H},\mathcal {S}\subset \textrm{PG}(n,q)$ be two projective hypersurfaces. If $\mathcal {H} \cap \mathcal {S}$ has a reduced absolutely irreducible component defined over ${\mathbb {F}}_q$ then $\mathcal {S}$ has an absolutely irreducible component defined over ${\mathbb {F}}_q$.

2.1 Ovoids in Q(4, q)

A starting point in our investigation is the main result of [3] which characterizes ovoids of Q(4, q) of “small degree". Similarly to what happens in the dimension-6 case, any ovoid $\textrm{O}_4$ of Q(4, q) can be written in the following form:

$$\begin{aligned} \textrm{O}_4(f) = \{(1,x,y,f(x,y),-y^2-xf(x,y))\}_{x,y\in \mathbb {F}_q} \cup \{(0,0,0,0,1)\}, \end{aligned}$$

for some function $f:\mathbb {F}_q^2 \longrightarrow \mathbb {F}_q$ with $f(0,0) =0$ and $\textrm{O}_4(f)$ is an ovoid if and only if

$$\begin{aligned} \langle (1,x_1,y_1,f(x_1,y_1),-y_1^2-x_1f(x_1,y_1)), (1,x_2,y_2,f(x_2,y_2),-y_2^2-x_2f(x_2,y_2))\rangle \ne 0, \end{aligned}$$

for every $(x_1,y_1) \ne (x_2,y_2)$ in ${\mathbb {F}}_q^2$, where $\langle \cdot ,\cdot \rangle$ is the symmetric for q odd (or alternating for q even) form associated to the quadratic form of the quadric Q(4, q).

Theorem 7

[3, Main Theorem] Let $f(X,Y)=\sum _{ij}a_{i,j}X^iY^j$, with $a_{0,1}=0$ if $p>2$. Suppose that $q>6.3 (d+1)^{13/3}$, where $\deg (f)=d$. If $\textrm{O}_4(f)$ is an ovoid of Q(4, q) then one of the following holds:

• 1. $p=2$ and $f(X,Y)=a_{1,0}X+a_{0,1}Y$. Hence $O_4(f)$ is an elliptic quadric.

• 2. $p>2$ and $f(X,Y)=a_{p^j,0}X^{p^j}$, $j\ge 0$. Hence $O_4(f)$ is either an elliptic quadric or a Kantor ovoid.

Corollary 8

Let $f(X,Y)=\sum _{ij}a_{i,j}X^iY^j$, with $p>2$. Suppose that $q>6.3(d+1)^{13/3}$, where $d=\deg (f)$. If

$$\begin{aligned} f(X,Y)\ne a_{p^j,0}X^{p^j}+\frac{a_{0,1}^2}{4}X+a_{0,1}Y, \end{aligned}$$

for some $j\ge 0$, then $\mathcal {S}_f$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$, and so $O_4(f)$ is not an ovoid of Q(4, q).

Proof

Let $\mathcal {S}_f: F(X_1,X_2,X_3,X_4)=0$, where

$$\begin{aligned} F(X_1,X_2,X_3,X_4)= (X_2 -X_4)^2 + (X_1-X_3) \Big ( f(X_1,X_2)- f(X_3,X_4)\Big ), \end{aligned}$$

be the surface associated with an ovoid in Q(4, q); see [3]. As already observed in [3], up to a change of variables we can always suppose that $a_{0,1}=0$. Suppose that this is not the case, then consider $\mathcal {S}_{f^{\prime }}:F(X_1,X_2-a_{0,1}X_1/2 ,X_3,X_4-a_{0,1}X_3/2)=0$ and so the coefficient of $(X_1-X_3)(X_2-X_4)$ in $F(X_1,X_2-a_{0,1}X_1/2 ,X_3,X_4-a_{0,1}X_3/2)$ vanishes, i.e. the coefficient $a_{0,1}^{\prime }$ in $f^{\prime }$ is 0. Therefore, by Theorem 7,

$$\begin{aligned} F(X_1,X_2-a_{0,1}X_1/2 ,X_3,X_4-a_{0,1}X_3/2)=(X_2 -X_4)^2+ (X_1-X_3)^{p^h+1}, \end{aligned}$$

otherwise $\mathcal {S}_{f^{\prime }}$ (and so $\mathcal {S}$) contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

So, $F(X_1,X_2,X_3,X_4)$ equals

$$\begin{aligned} (X_2 -X_4)^2 + (X_1-X_3)\left( (X_1-X_3)^{p^h}+a_{0,1}(X_2-X_4)+\frac{a_{0,1}^2}{4}(X_1-X_3)\right) . \end{aligned}$$

(5)

The claim follows. $\square$

2.2 Ovoids in Q(6, q)

Corollary 8 provides constraints on the shape of the functions $f_1,f_2$ associated with putative ovoids $O_6(f_1,f_2)$ of $\textrm{Q}$.

Proposition 9

Suppose that

$$\begin{aligned}{} & {} (f_1(X,Y,Z), f_2(X,Y,Z))\ne \left( a_1(X)Y^{p^{h_1}}+\frac{c_1(X)^2Y}{4}+c_1(X)Z,a_2(Y)X^{p^{h_2}}\right. \nonumber \\{} & {} \quad \left. +\frac{c_2(Y)^2X}{4}+c_2(Y)Z\right) , \end{aligned}$$

(6)

for some $a_1,c_1 \in {\mathbb {F}}_{q}[X]$, $a_2,c_2\in {\mathbb {F}}_q[Y]$, $h_1,h_2\ge 0$. If $q>6.3 (d+1)^{13/3}$, $\max \{\deg (f_1),\deg (f_2)\}=d$, then $O_6(f_1,f_2)$ is not an ovoid.

Proof

Consider the variety $\mathcal {W}^{\prime }_{f_1,f_2}:= \mathcal {W}_{f_1,f_2}\cap (X_5=X_2=\lambda )$, $\lambda \in {\mathbb {F}}_q$, defined by

$$\begin{aligned} G^{\prime }_{f_1,f_2,\lambda }(X_0,X_1,X_3,X_4,X_6)=F_{f_1,f_2}(X_0,X_1,\lambda ,X_3,X_4,\lambda ,X_6)=0, \end{aligned}$$

where

$$\begin{aligned} G^{\prime }_{f_1,f_2,\lambda }(1,X_1,X_3,X_4,X_6):= & {} (X_3-X_6)^2 + (X_4-X_1)(f_2(X_4,\lambda ,X_6)-f_2(X_1,\lambda ,X_3)), \end{aligned}$$

where $f_2$ satisfies (6). Suppose that

$$\begin{aligned} f_2(X,\lambda ,Z)\ne \alpha _{\lambda } X^{p^j}+\frac{\beta _{\lambda }^2}{4}X+\beta _{\lambda } Z, \end{aligned}$$

(7)

for any $\alpha _{\lambda },\beta _{\lambda } \in {\mathbb {F}}_q$. Then, by Corollary 8, $\mathcal {W}^{\prime }_{f_1,f_2}$ possesses an absolutely irreducible component defined over ${\mathbb {F}}_q$. Such a component, since $q>6.3 (d+1)^{13/3}$, contains at least an ${\mathbb {F}}_q$-rational point $(1,\overline{x}_1,\overline{x}_3,\overline{x}_4,\overline{x}_6)$, with $\overline{x}_1\ne \overline{x}_4$ or $\overline{x}_3\ne \overline{x}_6$. Therefore there exists a six-tuple $(1,\overline{x}_1,\lambda ,\overline{x}_3,\overline{x}_4,\lambda ,\overline{x}_6)$, with $(\overline{x}_1,\lambda ,\overline{x}_3)\ne (\overline{x}_4,\lambda ,\overline{x}_6)$ satisfying $F_{f_1,f_2}(X_0,X_1,X_2,X_3,X_4,X_5,X_6)=0$ (defined as in (3)) and therefore $O_6(f_1,f_2)$ is not an ovoid.

Suppose now that $f_2(X,Y,Z)$ contains a term $a_{i,j,k}X^iY^jZ^k\ne 0$, with $(i,k)\notin \{ (p^{h_2},0),(1,0),(0,1)\}$. The coefficient of $X^iZ^k$ in $f_2(X,\lambda ,Z)$ is determined by $A_{i,k}(\lambda ):=\sum _{j}a_{i,j,k}\lambda ^j$ and, by (7), we have that $A_{i,k}(\lambda )=0$ for each $\lambda \in {\mathbb {F}}_q$. Since $\deg A_{i,k}(Y)\le d<q$, $A_{i,k}(Y)\equiv 0$ for each $(i,k)\notin \{ (p^{h_2},0),(1,0),(0,1)\}$. Also, $(A_{0,1}(\lambda ))^2=4A_{1,0}(\lambda )$ for each $\lambda \in {\mathbb {F}}_q$. Since $\deg (A_{0,1}(Y))^2\le 2d<q$, we have that $(A_{0,1}(Y))^2=4A_{1,0}(Y)$. Summing up,

$$\begin{aligned} f_2(X,Y,Z)=a_2(Y)X^{p^{h_2}}+\frac{c_2(Y)^2}{4}X+c_2(Y)Z, \end{aligned}$$

for some $a_2,c_2\in {\mathbb {F}}_q[Y]$.

A similar argument, for any $\lambda \in {\mathbb {F}}_q$, holds for $\mathcal {W}^{\prime \prime }_{f_1,f_2}:= \mathcal {W}_{f_1,f_2}\cap (X_4=X_1=\lambda )$ defined by

$$\begin{aligned} G^{\prime \prime }_{f_1,f_2,\lambda }(X_0,X_2,X_3,X_5,X_6)=F_{f_1,f_2}(X_0,\lambda ,X_2,X_3,\lambda ,X_5,X_6)=0, \end{aligned}$$

that is

$$\begin{aligned}G^{\prime \prime }_{f_1,f_2,\lambda }(1,X_2,X_3,X_5,X_6):= & {} (X_3-X_6)^2 + (X_5-X_2)(f_1(\lambda ,X_5,X_6)-f_1(\lambda ,X_2,X_3)). \end{aligned}$$

So, if $O_6(f_1,f_2)$ is an ovoid then

$$\begin{aligned} f_1(X,Y,Z)=a_1(X)Y^{p^{h_1}}+\frac{c_1(X)^2}{4}Y+c_1(X)Z, \end{aligned}$$

for some $a_1,c_1\in {\mathbb {F}}_q[X]$. $\square$

By Proposition 9, if $O_6(f_1,f_2)$ is an ovoid then the polynomials $F(1,X_1,X_2,X_3,X_4,X_5,X_6)$ defining $\mathcal {W}_{f_1,f_2}$ read

$$\begin{aligned}{} & {} (X_3-X_6)^2 + (X_4-X_1)\Bigg (a_2(X_5)X_4^{p^{h_2}}-a_2(X_2)X_1^{p^{h_2}}+\frac{c_2^2(X_5)}{4}X_4\nonumber \\{} & {} \quad -\frac{c_2^2(X_2)}{4}X_1 +c_2(X_5)X_6-c_2(X_2)X_3\Bigg )\nonumber \\{} & {} \quad +(X_5-X_2)\Bigg (a_1(X_4)X_5^{p^{h_1}}-a_1(X_1)X_2^{p^{h_1}}+\frac{c_1^2(X_4)}{4}X_5\nonumber \\{} & {} \quad -\frac{c_1^2(X_1)}{4}X_2 +c_1(X_4)X_6-c_1(X_1)X_3\Bigg ). \end{aligned}$$

(8)

Proposition 10

Suppose that $\deg (\mathcal {W}_{f_1,f_2})=2$. Then $O_6(f_1,f_2)$ is not an ovoid.

Proof

In this case (8) defines a quadric in $\textrm{PG}(6,q)$ and, by Theorem 4, it contains at least $q^4$ affine ${\mathbb {F}}_q$-rational points. Thus, there exists at least an affine ${\mathbb {F}}_q$-rational point off the solid $X_1-X_4=X_2-X_5=X_3-X_6=0$ and, by Theorem 5, $O_6(f_1,f_2)$ is not an ovoid. $\square$

In the sequel, we will make use of the following notations.

$$\begin{aligned} \deg (a_1)=A_1, \quad \deg (a_2)=A_2, \quad \deg (c_1)=C_1, \quad \deg (c_2)=C_2. \end{aligned}$$

(9)

When one of the above polynomials vanishes, we use $-\infty$ to indicate its degree.

Proposition 11

Let $p^{h_1}=p^{h_2}=1$. Then $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

Proof

By way of contradiction, suppose that $O_6(f_1,f_2)$ is an ovoid. By Proposition 10, we can suppose that $\deg (\mathcal {W}_{f_1,f_2})\ge 3$.

Now, $F(1,X_1,X_2,X_3,X_4,X_5,X_6)$ reads

$$\begin{aligned}{} & {} (X_3-X_6)^2 + (X_4-X_1)\Bigg (\alpha _1(X_5)X_4-\alpha _1(X_2)X_1+ \beta _1(X_5)X_6-\beta _1(X_2)X_3\Bigg )\nonumber \\{} & {} \quad +(X_5-X_2)\Bigg (\alpha _2(X_4)X_5-\alpha _2(X_1)X_2 +\beta _2(X_4)X_6-\beta _2(X_1)X_3\Bigg ), \end{aligned}$$

(10)

for some polynomials $\alpha _1,\alpha _2,\beta _1,\beta _2$ of degrees $A_1,A_2,B_1,B_2$ (possibly $-\infty$). Note that at least one among $A_1,A_2,B_1,B_2$ is positive, since $\deg (\mathcal {W}_{f_1,f_2})\ge 3$.

We distinguish a number of cases, analyzing the homogeneous component $M(X_1,X_2,X_3,X_4,X_5,X_6)$ of highest degree, say $\ell$, in $F(1,X_1,X_2,X_3,X_4,X_5,X_6)$.

• C.1. $\ell =A_1+2\ge 3$.

  Then $A_1>0$. The component of highest degree $N(X_1,X_4,X_5,X_6)$ in $M(X_1,1,1,X_4,X_5,X_6)$ is

  $$\begin{aligned} u (X_4-X_1)X_5^{A_1}X_4+v (X_4-X_1)X_5^{B_1}X_6+ wX_4^{A_2} X_5^2+z X_4^{B_2}X_5X_6, \end{aligned}$$

  for some $u,v,w,z\in {\mathbb {F}}_q$, $u\ne 0$. It is already seen that $\deg _{X_1}(N(X_1,X_4,X_5,X_6))=1$ and therefore $N(X_1,X_4,X_5,X_6)=0$ contains a nonrepeated absolutely irreducible component defined over ${\mathbb {F}}_q$. Such a component extends to an absolutely irreducible component defined over ${\mathbb {F}}_q$ in $\mathcal {W}_{f_1,f_2}$ by applying twice Lemma 6.

• C.2. $\ell =B_1+2\ge 3$ and $B_1>A_1$.

  Then $B_1>0$. The component of highest degree $N(X_1,X_4,X_5,X_6)$ in $M(X_1,1,1,X_4,X_5,X_6)$ is

  $$\begin{aligned} v (X_4-X_1)X_5^{B_1}X_6+ wX_4^{A_2} X_5^2+z X_4^{B_2}X_5X_6, \end{aligned}$$

  for some $v,w,z\in {\mathbb {F}}_q$, $v\ne 0$, and $\deg _{X_1}(N(X_1,X_4,X_5,X_6))=1$. The claim follows similar as in Case C.1.

• C.3. $\ell =B_2+2\ge 3$, and $B_2>A_1$, $B_2>B_1$ .

  Then $B_2>0$. The component of highest degree $N(X_2,X_4,X_5,X_6)$ in $M(1,X_2,1,X_4,X_5,X_6)$ is

  $$\begin{aligned} (X_5-X_2)(w X_4^{A_2}X_5+ zX_4^{B_2} X_6), \end{aligned}$$

  for some $w,z\in {\mathbb {F}}_q$, $z\ne 0$, and $\deg _{X_6}(N(X_2,X_4,X_5,X_6))=1$. The claim follows similar as in Case C.1.

• C.4. $\ell =A_2+2\ge 3$ and $A_2>A_1$, $A_2>B_1$, $A_2>B_2$.

  Then $A_2>0$. The component of highest degree $N(X_2,X_4,X_5,X_6)$ in $M(1,X_2,1,X_4,X_5,X_6)$ is $w(X_5-X_2)X_4^{A_2}X_5$, with $w\in {\mathbb {F}}_q^*$. The claim follows similar as in Case C.1.

In all the cases above we obtain that $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$. $\square$

From now on we will make use a number of times of the following lemma to show that a polynomial $F(X_1,\ldots ,X_n)$ is not square in $\overline{{\mathbb {F}}_q}[X_1,\ldots ,X_n]$.

Lemma 12

Let $F(X_1,\ldots ,X_n)$ be a polynomial in ${{\mathbb {F}}_q}[X_1,\ldots ,X_n]$. Assume that $F(X_1,\ldots ,X_n)$ is a square in $\overline{{\mathbb {F}}_q}[X_1,\ldots ,X_n]$. Then, the following hold:

• any specialization $X_i=\alpha _i$ of $F(X_1,\ldots ,X_n)$ is a square.

• the homogeneous part in $F(X_1,\ldots ,X_n)$ of highest degree is a square too.

• if the homogeneous parts of highest and second highest degree in $F(X_1,\ldots ,X_n)$ are $G(X_1,\ldots ,X_n)^{\ell }$ and $H(X_1,\ldots ,X_n)$, for some even number $\ell$, then $G(X_1,\ldots ,X_n)^{\ell /2}$ divides $H(X_1,\ldots ,X_n)$.

Proof

By hypothesis, there is a polynomial $W(X_1,\ldots ,X_n)$ such that $F(X_1,\ldots ,X_n)=(W(X_1,\ldots ,X_n))^2$. It follows immediately that any specialization $X_i=\alpha _i$ of $F(X_1,\ldots ,X_n)$ is a square. Now, write

$$\begin{aligned} W(X_1,\ldots ,X_n)=\sum _{i \le m}L_i(X_1,\ldots ,X_n), \end{aligned}$$

where m denotes the degree of $W(X_1,\ldots ,X_n)$ and $L_i(X_1,\ldots ,X_n)$ is the homogeneous part of $W(X_1,\ldots ,X_n)$ of degree i. Then, we have

$$\begin{aligned} F(X_1,\ldots ,X_n)=\sum _{i \le m}(L_i(X_1,\ldots ,X_n))^2 + 2 \sum _{i \ne j}L_i(X_1,\ldots ,X_n)L_j(X_1,\ldots ,X_n), \end{aligned}$$

and so $F(X_1,\ldots ,X_n)$ has degree 2m. The homogeneous parts of highest and second highest degree in $F(X_1,\ldots ,X_n)$ are given by

$$\begin{aligned} (L_m(X_1,\ldots ,X_n))^2 \,\,\, \textrm{and} \,\,\, 2L_m(X_1,\ldots ,X_n)L_{k}(X_1,\ldots ,X_n), \end{aligned}$$

respectively, where $k=\max \{ i : L_i(X_1,\ldots ,X_n) \not \equiv 0, i \ne m\}$. The claim follows. $\square$

Proposition 13

Let

$$\begin{aligned} \Delta (X_1,X_2,X_4,X_5):= & {} \Big ((X_4-X_1)c_2(X_5)+(X_5-X_2)c_1(X_4)\Big )^2\nonumber \\{} & {} -4(X_4-X_1)\left( a_2(X_5)X_4^{p^{h_2}}-a_2(X_2)X_1^{p^{h_2}}+\frac{c_2^2(X_5)}{4}X_4-\frac{c_2^2(X_2)}{4}X_1 \right) \nonumber \\{} & {} -4(X_5-X_2)\left( a_1(X_4)X_5^{p^{h_1}}-a_1(X_1)X_2^{p^{h_1}}+\frac{c_1^2(X_4)}{4}X_5-\frac{c_1^2(X_1)}{4}X_2 \right) \nonumber \\ \end{aligned}$$

(11)

If $\Delta$ is not a square in $\overline{{\mathbb {F}}_q}[X_1,X_2,X_4,X_5]$ then $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

Proof

Consider the hypersurface $\mathcal {W}^{\prime }_{f_1,f_2}:= \mathcal {W}_{f_1,f_2}\cap (X_3=0)$. Note that, since the polynomial $F(1,X_1,X_2,0,X_4,X_5,X_6)$ is of degree 2 in $X_6$, $\mathcal {W}^{\prime }_{f_1,f_2}$ is absolutely irreducible if and only if the discriminant $\Delta (X_1,X_2,X_4,X_5)$ with respect to $X_6$ is a non-square in $\overline{{\mathbb {F}}_q}[X_1,X_2,X_4,X_5]$. Therefore, if $\Delta (X_1,X_2,X_4,X_5)$ is not a square in $\overline{{\mathbb {F}}_q}[X_1,X_2,X_4,X_5]$ then $\mathcal {W}^{\prime }_{f_1,f_2}$ is absolutely irreducible and by Lemma 6 the hypersurface $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$. $\square$

Proposition 14

Let $p^{h_1}=1$ and $p^{h_2}\ge 3$ or $p^{h_1}\ge 3$ and $p^{h_2}=1$. Then $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

Proof

Suppose by way of contradiction that $\mathcal {W}_{f_1,f_2}$ does not contain an absolutely irreducible component defined over ${\mathbb {F}}_q$.

Without loss of generality, we can suppose that $p^{h_1}\ge 3$ and $p^{h_2}=1$. Let $\widetilde{a_2}(X)=a_2(X)+\frac{c_2^2(X)}{4}$, with $\deg (\widetilde{a_2})=\widetilde{A_2}$. Now, the polynomial $F(1,X_1,X_2,X_3,X_4,X_5,X_6)$ reads

$$\begin{aligned}{} & {} (X_3-X_6)^2+ (X_4-X_1)\Bigg (\widetilde{a_2}(X_5)X_4-\widetilde{a_2}(X_2)X_1 +c_2(X_5)X_6-c_2(X_2)X_3\Bigg )\nonumber \\{} & {} \quad +(X_5-X_2)\Bigg (a_1(X_4)X_5^{p^{h_1}}-a_1(X_1)X_2^{p^{h_1}}+\frac{c_1^2(X_4)}{4}X_5 -\frac{c_1^2(X_1)}{4}X_2 +c_1(X_4)X_6-c_1(X_1)X_3\Bigg ). \end{aligned}$$

First, we claim that the homogeneous part $M(X_1,X_2,X_3,X_4,X_5,X_6)$ of highest degree in the polynomial $F(1,X_1,X_2,X_3,X_4,X_5,X_6)$ is a linear combination (with not all the coefficients zero) of

$$\begin{aligned} t_1= & {} (X_4-X_1)(X_5^{\widetilde{A_2}}X_4-X_2^{\widetilde{A_2}}X_1),\\ t_2= & {} (X_4-X_1)(X_5^{C_2}X_6-X_2^{C_2}X_3),\\ t_3= & {} (X_5-X_2)(X_4^{A_1}X_5^{p^{h_1}}-X_1^{A_1}X_2^{p^{h_1}}),\\ t_4= & {} (X_5-X_2)(X_4^{2C_1}X_5-X_1^{2C_1}X_2). \end{aligned}$$

To see this, it is enough to observe that any such linear combination cannot vanish since each of the following monomials

$$\begin{aligned} X_1X_4X_5^{\widetilde{A_2}}, X_1X_6X_5^{C_2},X_1^{A_1}X_2^{p^{h_1}}X_5,X_1^{2C_1}X_2X_5 \end{aligned}$$

is contained in a unique $t_i$. Also, if $t_1$, $t_2$ or $t_4$ are nonzero then $\widetilde{A_2}$, $C_2$, or $C_1$ (respectively) are nonnegative, since $\deg (\mathcal {W}_{f_1,f_2})>3$.

Thus $M(X_1,X_2,X_3,X_4,X_5,X_6)=u_1t_1+u_2t_2+u_3t_3+u_4t_4$ for some $u_i\in {\mathbb {F}}_q$, not all vanishing.

If $u_2\ne 0$, then $M(X_1,X_2,X_3,X_4,X_5,X_6)$ is of degree one in $X_6$ and therefore the hypersurface $M(X_1,X_2,X_3,X_4,X_5,X_6)=0$ contains a nonrepeated absolutely irreducible component defined over ${\mathbb {F}}_q$. Such a component extends to an absolutely irreducible component defined over ${\mathbb {F}}_q$ in $\mathcal {W}_{f_1,f_2}$ by applying Lemma 6, a contradiction.

From now on we can suppose that $u_2=0$. If $u_1\ne 0$, then the part of the highest degree in $M(X_1,X_2,X_3,X_4,X_5,X_6)$ is of degree one in $X_1$ and it contains a nonrepeated absolutely irreducible component defined over ${\mathbb {F}}_q$. Such a component extends to an absolutely irreducible component defined over ${\mathbb {F}}_q$ in $\mathcal {W}_{f_1,f_2}$ by applying twice Lemma 6, a contradiction.

Thus, only $u_3$ and $u_4$ can be nonzero. If $u_4\ne 0$ or $u_4=0$ and $A_1\ne 0$ then (since $C_1>0$) $X_5-X_2$ is a nonrepeated absolutely irreducible factor defined over ${\mathbb {F}}_q$ of $M(X_1,X_2,X_3,X_4,X_5,X_6)$. Arguing as above we get a contradiction.

So $u_1=u_2=u_4=0$, $u_3\ne 0$, $A_1=0$, and $M(X_1,X_2,X_3,X_4,X_5,X_6)=u_3(X_5-X_2)^{p^{h_1}+1}$. Note that $p^{h_1}+1>\max \{2C_1+2,C_2+2\}.$

We now investigate the discriminant $\Delta (X_1,X_2,X_4,X_5)$ of $F(1,X_1,X_2,0,X_4,X_5,X_6)$ with respect to $X_6$. Recall that, since $\mathcal {W}_{f_1,f_2}$ does not contain an absolutely irreducible component defined over ${\mathbb {F}}_q$, $\Delta (X_1,X_2,X_4,X_5)$ must be a square.

By direct computations, $\Delta (X_1,X_2,X_4,X_5)$ reads

$$\begin{aligned}{} & {} ((X_4-X_1)c_2(X_5)+(X_5-X_2)c_1(X_4))^2-4(X_4-X_1)(a_2(X_5)X_4-a_2(X_2)X_1)\\{} & {} \quad -(X_4-X_1)( c_2^2(X_5)X_4-c_2^2(X_2)X_1)-4\alpha (X_5-X_2)^{p^{h_1}+1} \\{} & {} \quad -(X_5-X_2)( c_1^2(X_4)X_5-c_1^2(X_1)X_2)\\= & {} -4\alpha (X_5-X_2)^{p^{h_1}+1}+(X_4-X_1)X_1(c_2^2(X_2)-c_2^2(X_5))+(X_5-X_2)X_2(c_1^2(X_1)-c_1^2(X_4))\\{} & {} \quad +2(X_4-X_1)(X_5-X_2)c_2(X_5)c_1(X_4)-4 (X_4-X_1)(a_2(X_5)X_4-a_2(X_2)X_1), \end{aligned}$$

for some $\alpha \ne 0$.

Clearly, the homogeneous component of highest degree in $\Delta$ is $-4\alpha (X_5-X_2)^{p^{h_1}+1}$, which is a square. Now we claim that the homogeneous part $N(X_1,X_2,X_4,X_5)$ of the second highest degree in $\Delta (X_1,X_2,X_4,X_5)$ is a linear combination of

$$\begin{aligned} z_1:= & {} (X_4-X_1)X_1(X_2^{2C_2}-X_5^{2C_2}),\\ z_2:= & {} (X_5-X_2)X_2(X_1^{2C_1}-X_4^{2C_1}),\\ z_3:= & {} (X_4-X_1)(X_5^{A_2}X_4-X_2^{A_2}X_1),\\ z_4:= & {} (X_4-X_1)(X_5-X_2)X_5^{C_2}X_4^{C_1}.\\ \end{aligned}$$

We can suppose that $C_1,C_2>0$, otherwise we just exclude $z_1$ and $z_2$.

If $C_1=0$ and $C_2>0$, then $z_2\equiv 0$, $\deg (z_1)>\deg (z_4)$, and any linear combination of $z_1$ and $z_3$ does not vanish. A similar argument applies to the case $C_1>0$ and $C_2=0$. If $C_1=C_2=0$ then $z_1\equiv z_2\equiv 0$ and any linear combination of $z_3$ and $z_4$ does not vanish.

So we consider only $C_1,C_2>0$.

• 1. $(C_1,A_2)\ne (1,C_2+1)$. The four monomials $X_1^2X_5^{2C_2}, X_2^2X_4^{2C_1},X_4^2X_5^{A_2},X_1X_2X_4^{C_1}X_5^{C_2}$ belong to a unique $z_i$ and therefore a nontrivial linear combination of $z_i$ cannot vanish.

• 2. $(C_1,A_2)=(1,C_2+1)$. If $C_2>1$ the degree of $z_1$ is the highest among the $z_i$’s.

• 3. $(C_1,C_2,A_2)=(1,1,2)$. If $\lambda _1z_1+\lambda _2z_2+\lambda _3z_3+\lambda _4z_4=0$ then $\lambda _3=0$ since $(X_5-X_2)$ divides $z_1,z_2,z_4$ but not $z_3$. Now, it can be readily seen that $\lambda _1z_1+\lambda _2z_2+\lambda _4z_4$ cannot vanish for any $(\lambda _1,\lambda _2,\lambda _4)\ne (0,0,0)$.

Thus we know that

$$\begin{aligned} N(X_1,X_2,X_4,X_5)=\delta _1 w_1^2 z_1+\delta _2w_2^2 z_2+2w_1w_2 \delta _4z_4+\delta _3w_3z_3, \end{aligned}$$

for some $w_1,w_2,w_3\in {\mathbb {F}}_q^*$, and $\delta _1,\delta _2,\delta _3,\delta _4\in \{0,1\}$, where $\delta _i=1$ only if the corresponding $z_i$ has the highest degree.

Recall that since $\Delta (X_1,X_2,X_4,X_5)$ is a square, $(X_5-X_2)^{(p^{h_1}+1)/2}\mid N(X_1,X_2,X_4,X_5)$ and $(p^{h_1}+1)/2\ge 2$; see Lemma 12.

If $C_1=C_2=0$ then $N(X_1,X_2,X_4,X_5)=\delta _3w_3 (X_4-X_1)(X_5^{A_2}X_4-X_2^{A_2}X_1)+2\delta _4 w_1w_2(X_4-X_1)(X_5-X_1)$ which is never divisible by $(X_5-X_2)^2$.

If $C_1=0$ and $C_2>0$ then $N(X_1,X_2,X_4,X_5)$ is

$$\begin{aligned}{} & {} \delta _1w_1^2(X_4-X_1)X_1(X_2^{2C_2}-X_5^{2C_2})+\delta _3 w_3 \\{} & {} \quad (X_4-X_1)(X_5^{A_2}X_4-X_2^{A_2}X_1)+2\delta _4w_1w_2(X_4-X_1)(X_5-X_1)X_5^{C_2}, \end{aligned}$$

which is divisible by $(X_5-X_2)$ only if $\delta _3=0$. Also $C_2>0$ yields $\delta _4=0$ since $\deg (z_1)>\deg (z_4)$. This forces $p\mid 2C_2$.

Analogously, $C_1>0$ and $C_2=0$ yields a contradiction.

Suppose now that $C_1,C_2>0$. Since $(X_5-X_2)$ must divide $N(X_1,X_2,X_4,X_5)$, $\delta _3=0$. Now $(X_5-X_2)^2$ divides $N(X_1,X_2,X_4,X_5)$ only if

$$\begin{aligned}{} & {} \phi (X_1,X_2,X_4):=2C_2\delta _1 w_1^2(X_4-X_1)X_1X_2^{2C_2-1}+\delta _2 w_2^2X_2(X_1^{2C_1}\\{} & {} \quad -X_4^{2C_1})+2\delta _4w_1w_2(X_4-X_1)X_2^{C_2}X_4^{C_1} \end{aligned}$$

is the zero-polynomial.

If $\delta _2\ne 0$, then, considering $X_4=0$, we get

$$\begin{aligned} \phi (X_1,X_2,X_4)=-2C_2\delta _1w_1^2X_1^2X_2^{2C_2-1}+\delta _2w_2^2X_2X_1^{2C_1}, \end{aligned}$$

which does not vanish unless $C_1=C_2=1$. In this case $\delta _1=\delta _2=\delta _4=1$ and

$$\begin{aligned} \phi (X_1,X_2,X_4)=2 w_1^2(X_4-X_1)X_1X_2+ w_2^2X_2(X_1^{2}-X_4^{2})+2w_1w_2(X_4-X_1)X_2X_4 \end{aligned}$$

vanishes only if $2w_1^2=2w_1w_2$, $2w_1^2=w_2^2$, which yields $w_1=w_2=0$, a contradiction.

So we can suppose that $\delta _2=0$ and

$$\begin{aligned} \phi (X_1,X_2,X_4)=2C_2\delta _1 w_1^2(X_4-X_1)X_1X_2^{2C_2-1}+2\delta _4w_1w_2(X_4-X_1)X_2^{C_2}X_4^{C_1}. \end{aligned}$$

It is readily seen that $\phi$ vanishes only if $\delta _4=0$ and $p \mid 2C_2$.

Summing up, $N(X_1,X_2,X_4,X_5)$ is divisible by $(X_2-X_5)^{(p^{h_1}+1)/2\ge 2}$ only if $N(X_1,X_2,X_4,X_5)=(X_4-X_1)X_1(X_2^{2C_2}-X_5^{2C_2})$ with $p\mid 2C_2$. Let $\ell$ be the highest power of p dividing $2C_2$. From $p^{h_1}+1>\max \{2C_1+2,C_2+2\}$, $\ell \le h_1-1$ and $(X_5-X_2)^{p^\ell +1}$ does not divide $N(X_1,X_2,X_4,X_5)$. A final contradiction arises from $(p^{h_1}+1)/2\le p^{\ell }\le p^{h_1-1}$. The claim follows from Proposition 13. $\square$

By Proposition 14, if $\mathcal {W}_{f_1,f_2}$ does not contain an absolutely irreducible component defined over ${\mathbb {F}}_q$ then both $p^{h_1}>1$ and $p^{h_2}>1$ and $\deg (\mathcal {W}_{f_1,f_2})\ge 4$.

Proposition 15

Let $\mathcal {W}_{f_1,f_2}$ be the hypersurface defined as in (8). Suppose that $\deg (\mathcal {W}_{f_1,f_2})\ge 4$ and $p^{h_1},p^{h_2}>1$. If $A_1A_2> 0$ then $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

Proof

We proceed as in the proof of Proposition 14. Consider the variety $\mathcal {W}^{\infty }_{f_1,f_2}:= \mathcal {W}_{f_1,f_2}\cap (X_0=0)$. It is defined by the homogeneous part $L(X_1,X_2,X_3,X_4,X_5,X_6)$ of highest degree in (8) and it arises from a linear combination (with not all the coefficients zero) of

$$\begin{aligned} t_1= & {} (X_4-X_1)(X_5^{A_2}X_4^{p^{h_2}}-X_2^{A_2}X_1^{p^{h_2}}),\\ t_2= & {} (X_4-X_1)(X_5^{2C_2}X_4-X_2^{2C_2}X_1),\\ t_3= & {} (X_5-X_2)(X_4^{A_1}X_5^{p^{h_1}}-X_1^{A_1}X_2^{p^{h_1}}),\\ t_4= & {} (X_5-X_2)(X_4^{2C_1}X_5-X_1^{2C_1}X_2). \end{aligned}$$

To see this, it is enough to observe any such linear combination cannot vanish since each of the following monomials

$$\begin{aligned} X_1^{p^{h_2}}X_2^{A_2}X_4, X_1X_2^{2C_2}X_4,X_1^{A_1}X_2^{p^{h_1}}X_5,X_1^{2C_1}X_2X_5 \end{aligned}$$

is contained in a unique $t_i$.

Thus

$$\begin{aligned} L(X_1,X_2,X_3,X_4,X_5,X_6)=\delta _1u_1t_1+\delta _2u_2t_2+\delta _3 u_3 t_3+\delta _4 u_4 t_4, \end{aligned}$$

with $\delta _i \in \{0,1\}$, $u_i \in {\mathbb {F}}_q^*$, and not all the $\delta _i$’s vanish. We distinguish two cases.

• (i) $(\delta _1,\delta _2)\ne (0,0)$. The homogeneous part in $L(X_1,1,X_3,X_4,X_5,X_6)$ of highest degree is

  $$\begin{aligned} M{} & {} := \delta _1 u_1(X_4-X_1)X_5^{A_2}X_4^{p^{h_2}}+\delta _2u_2(X_4-X_1)X_5^{2C_2}X_4 +\delta _3 u_3 X_4^{A_1}X_5^{p^{h_1+1}}\\{} & {} \quad \ +\delta _4 u_4 X_5^2 X_4^{2C_1}, \end{aligned}$$

  which is of degree 1 in $X_1$. Thus there exists a non-repeated ${\mathbb {F}}_q$-rational factor of M (depending on $X_1$) which is absolutely irreducible. Therefore $L(X_1,1,X_3,X_4,X_5,X_6)$ contains a non-repeated absolutely irreducible factor defined over ${\mathbb {F}}_q$, by Lemma 6. Again by Lemma 6 the corresponding variety extends to a non-repeated ${\mathbb {F}}_q$-rational absolutely irreducible component in $\mathcal {W}^{\infty }_{f_1,f_2}$.

• (ii) $(\delta _1,\delta _2)= (0,0)$. So, $(\delta _3,\delta _4)\ne (0,0)$. The homogeneous part in $L(1,X_2,X_3,X_4,X_5,X_6)$ of highest degree is

  $$\begin{aligned} M:= \delta _3 u_3 (X_5-X_2) X_4^{A_1}X_5^{p^{h_1}} +\delta _4 u_4 (X_5-X_2) X_4^{2C_1}X_5, \end{aligned}$$

  which is of degree 1 in $X_2$. Thus we obtain the same conclusion as above.

$\square$

Proposition 16

Let $\mathcal {W}_{f_1,f_2}$ be the hypersurface defined as in (8). Suppose that $\deg (\mathcal {W}_{f_1,f_2})\ge 4$ and $p^{h_1},p^{h_2}>1$. If $(A_1,A_2)\notin \{(0,-\infty ),(-\infty ,0), (-\infty ,-\infty ),(0,0)\}$ then the hypersurface $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

Proof

Suppose that $A_1> 0$, the case $A_2> 0$ follows completely similar. In view of Proposition 15, we only have to consider $A_2\le 0$.

Following the same notation as in Proposition 15, let $L(X_1,X_2,X_3,X_4,X_5,X_6)$ be the part of the highest degree in (8). It arises from a linear combination (with not all the coefficients zero) $\delta _1 u_1 t_1+\delta _2u_2 t_2+\delta _3u_3t_3+\delta _4 u_4 t_4$, where

$$\begin{aligned} t_1= & {} (X_4-X_1)^{p^{h_2}+1},\\ t_2= & {} (X_4-X_1)(X_5^{2C_2}X_4-X_2^{2C_2}X_1),\\ t_3= & {} (X_5-X_2)(X_4^{A_1}X_5^{p^{h_1}}-X_1^{A_1}X_2^{p^{h_1}}),\\ t_4= & {} (X_5-X_2)(X_4^{2C_1}X_5-X_1^{2C_1}X_2). \end{aligned}$$

Note that if $A_2=-\infty$, that is $a_2\equiv 0$, we may assume $\delta _1=0$.

To see this, it is enough to observe any such linear combination cannot vanish since each of the following monomials

$$\begin{aligned} X_1^{p^{h_2}+1}, X_1X_2^{2C_2}X_4,X_1^{A_1}X_2^{p^{h_1}}X_5,X_1^{2C_1}X_2X_5 \end{aligned}$$

is contained in a unique $t_i$.

Suppose that $(\delta _3,\delta _4)\ne (0,0)$. The homogeneous part in $L(1,X_2,X_3,X_4,X_5,X_6)$ of highest degree is

$$\begin{aligned} M:= \delta _1 u_1X_4^{p^{h_2}+1}+\delta _2u_2X_4^2X_5^{2C_2} +\delta _3 u_3 (X_5-X_2) X_4^{A_1}X_5^{p^{h_1}} +\delta _4 u_4 (X_5-X_2) X_4^{2C_1}X_5, \end{aligned}$$

which is of degree 1 in $X_2$. Thus there exists a non-repeated ${\mathbb {F}}_q$-rational factor of M (depending on $X_1$) which is absolutely irreducible. Therefore $L(1,X_2,X_3,X_4,X_5,X_6)$ contains a non-repeated absolutely irreducible factor defined over ${\mathbb {F}}_q$, by Lemma 6. Again by Lemma 6 the corresponding variety extends to a non-repeated ${\mathbb {F}}_q$-rational absolutely irreducible component in $\mathcal {W}^{\infty }_{f_1,f_2}$. Thus $(\delta _3,\delta _4)= (0,0)$.

If $\delta _1=0$ or $\delta _1\delta _2\ne 0$ then $C_2>0$ and $(X_4-X_1)$ is a non-repeated ${\mathbb {F}}_q$-rational absolutely irreducible factor of $L(X_1,X_2,X_3,X_4,X_5,X_6)$ and, arguing as above, $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

Thus, we have only to deal with the case $\delta _1\ne 0$ (which implies $A_2= 0$) and $\delta _2=\delta _3=\delta _4=0$. Note that $p^{h_2}+1>\max \{ 2C_2+2,A_1+p^{h_1}+1,2C_1+2\}$.

Consider now $\Delta (X_1,X_2,X_4,X_5)$ defined as in (11) and we will prove that it is not a square in $\overline{{\mathbb {F}}_q}[X_1,X_2,X_4,X_5]$ using Lemma 12.

Suppose by way of contradiction that $\Delta (X_1,X_2,X_4,X_5)$ is a square in $\overline{{\mathbb {F}}_q}[X_1,X_2,X_4,X_5]$.

Hence, $\Delta (X_1,X_2,X_4,0)$ is a square in $\overline{{\mathbb {F}}_q}[X_1,X_2,X_4]$. Recalling that $p^{h_1}\ge 3$, it is readily seen that its homogeneous component of the second largest degree is

$$\begin{aligned} N(X_1,X_2,X_4)= & {} \delta _1 u_1 X_2^2 X_4^{2C_1}+\delta _2 u_2(X_4-X_1)X_1X_2^{2C_2}+\delta _3 u_3 X_2^{p^{h_1}+1}X_1^{A_1}-\delta _1 u_1 X_2^2X_1^{2C_1}, \end{aligned}$$

for some $\delta _1,\delta _2,\delta _3 \in \{0,1\}$ not all vanishing, and $u_1,u_2,u_3\in {\mathbb {F}}_q$. By Lemma 12, $(X_1-X_4)^{(p^{h_2}+1)/2}$ must divide $N(X_1,X_2,X_4)$. This immediately forces $\delta _3=0$ and thus $p^{h_1}+1+A_1<\min \{2+2C_1,2+2C_2\}$.

Also, $N^{\prime }(X_1,X_2,X_4):=N(X_1,X_2,X_4)/(X_4-X_1)$ reads

$$\begin{aligned} \delta _1 u_1 X_2^2 \left( \sum _{i=0}^{2C_1-1}X_4^{i}X_1^{2C_1-1-i}\right) +\delta _2 u_2X_1X_2^{2C_2}. \end{aligned}$$

Now, $(X_4-X_1)^2\mid N(X_1,X_2,X_4)$ if and only if $N^{\prime }(X_1,X_2,X_1)\equiv 0$, that is

$$\begin{aligned} \delta _1 u_1 (2C_1)X_2^2X_1^{2C_1-1} +\delta _2 u_2X_1X_2^{2C_2}\equiv 0. \end{aligned}$$

The case $C_2=1=C_1$ must be excluded, since $4\le p^{h_1+1}+A_1<\min \{2+2C_1,2+2C_2\}=4$ yields a contradiction. The only possibility is then $\delta _2=0$, $\delta _1=1$, $p\mid C_1$. In this case $N(X_1,X_2,X_4)=u_1X_2^2(X_4^{2C_1}-X_1^{2C_1})=u_1X_2^2(X_4^{C_1}-X_1^{C_1})(X_4^{C_1}+X_1^{C_1})$, which is not divisible by $(X_1-X_4)^{(p^{h_2}+1)/2}$, since $2C_1<p^{h_2}-1$. The claim follows from Proposition 13. $\square$

Proposition 17

Let $\mathcal {W}_{f_1,f_2}$ be the hypersurface defined as in (8). Suppose that $\deg (\mathcal {W}_{f_1,f_2})\ge 4$ and $p^{h_1},p^{h_2}>1$. If $(A_1,A_2)= (0,-\infty )$ or $(A_1,A_2)= (-\infty ,0)$ then the hypersurface $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$ or $O(f_1,f_2)$ is not an ovoid.

Proof

We consider only the case $(A_1,A_2)= (0,-\infty )$, the case $(A_1,A_2)= (-\infty ,0)$ being similar.

Note that $C_2\ne -\infty$, otherwise $f_2\equiv 0$.

If $C_1=-\infty$ and $C_2=0$ then $\mathcal {W}_{f_1,f_2}$ reads

$$\begin{aligned} (X_3-X_6)^2 + (X_4-X_1)\Bigg (\frac{c_2^2}{4}(X_4-X_1) +c_2(X_6-X_3)\Bigg ) +a_1(X_5-X_2)^{p^{h_1}+1}=0 \end{aligned}$$

and contains all the points on $X_5=X_2$, $2(X_6-X_3)+c_2(X_4-X_1)=0$ and $O(f_1,f_2)$ is not an ovoid.

Following the same notation as in Proposition 15, let $L(X_1,X_2,X_3,X_4,X_5,X_6)$ be the part of the highest degree in (8). It arises from a linear combination (with not all the coefficients zero) $\delta _2u_2 t_2+\delta _3u_3t_3+\delta _4 u_4 t_4$, where

$$\begin{aligned} t_2= & {} (X_4-X_1)(X_5^{2C_2}X_4-X_2^{2C_2}X_1),\\ t_3= & {} (X_5-X_2)^{p^{h_1}+1},\\ t_4= & {} (X_5-X_2)(X_4^{2C_1}X_5-X_1^{2C_1}X_2). \end{aligned}$$

To see this, it is enough to observe any such linear combination cannot vanish since each of the following monomials

$$\begin{aligned} X_1X_2^{2C_2}X_4,X_2^{p^{h_1}+1},X_1^{2C_1}X_2X_5 \end{aligned}$$

is contained in a unique $t_i$.

If $\delta _2\ne 0$ then $C_2>0$ since $\deg (\mathcal {W}_{f_1,f_2})>3$ and the homogeneous part in $L(X_1,1,X_3,X_4,X_5,X_6)$ of highest degree is of degree 1 in $X_1$. Arguing as in the previous propositions $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$. So $\delta _2=0$. If $\delta _4\ne 0$ then $C_1>0$ and $(X_5-X_2)$ is a nonrepeated factor of $L(X_1,X_2,X_3,X_4,X_5,X_6)$. Arguing as above, $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

Thus $\delta _3\ne 0$ and $\delta _2=\delta _4=0$, which yields $p^{h_1}+1>\max \{2C_2+2,2C_1+2\}$.

We consider now $\Delta (X_1,X_2,X_4,X_5)$ as in (11) and we show that it is not a square in $\overline{{\mathbb {F}}_q}[X_1,X_2,X_4,X_5]$ by means of Lemma 12. We have that

$$\begin{aligned} \Delta (X_1,X_2,X_4,X_5)\hspace{-0.3 cm}&=\hspace{-0.3 cm}&\Big ((X_4-X_1)c_2(X_5)+(X_5-X_2)c_1(X_4)\Big )^2\\{} & {} -4(X_4-X_1)\left( \frac{c_2^2(X_5)}{4}X_4-\frac{c_2^2(X_2)}{4}X_1 \right) \\{} & {} -4(X_5-X_2)\left( a_1X_5^{p^{h_1}}-a_1X_2^{p^{h_1}}+\frac{c_1^2(X_4)}{4}X_5-\frac{c_1^2(X_1)}{4}X_2 \right) \\= & {} X_1(X_4-X_1)(c_2^2(X_2)-c_2^2(X_5))+X_2(X_5-X_2)(c_1^2(X_1)-c_1^2(X_4))+\\{} & {} \alpha (X_5-X_2)^{p^{h_1}+1}+2(X_5-X_2)(X_4-X_1)c_1(X_4)c_2(X_5). \end{aligned}$$

If $C_1=-\infty$ and $C_2>0$, $\Delta (X_1,X_2,X_4,X_5)$ is not a square and, by Proposition 13, $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

So we can suppose that $C_1,C_2\ge 0$.

If $\Delta (X_1,X_2,X_4,X_5)$ is a square in $\overline{{\mathbb {F}}_q}[X_1,X_2,X_4,X_5]$, then the homogeneous part of the highest degree in $\Delta (X_1,X_2,X_4,X_5)$ is clearly $(X_5-X_2)^{p^{h_1}+1}$ and $\Delta (0,X_2,X_4,X_5)$ is a square in $\overline{{\mathbb {F}}_q}[X_2,X_4,X_5]$. Recalling that $p^{h_1}\ge 3$, it is readily seen that its homogeneous component of the second largest degree is

$$\begin{aligned} N(X_2,X_4,X_5)= & {} \delta _1 u_1 X_2(X_5-X_2)X_4^{2C_1}+\delta _2 u_2(X_5-X_2)X_4^{C_1+1}X_5^{C_2}, \end{aligned}$$

for some $\delta _1,\delta _2 \in \{0,1\}$ not all vanishing, and $u_1,u_2\in {\mathbb {F}}_q$. By Lemma 12, $(X_5-X_2)^{(p^{h_1}+1)/2}$ must divide $N(X_2,X_4,X_5)$. This immediately forces $p^{h_1}=3$, $C_1=C_2=1$, a contradiction to $4=p^{h_1}+1>\max \{2C_2+2,2C_1+2\}$. Thus $\Delta (X_1,X_2,X_4,X_5)$ is not a square in $\overline{{\mathbb {F}}_q}[X_1,X_2,X_4,X_5]$ and, by Proposition 13, $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$. $\square$

Proposition 18

Let $\mathcal {W}_{f_1,f_2}$ be the hypersurface defined as in (8). Suppose that $\deg (\mathcal {W}_{f_1,f_2})\ge 4$, $(A_1,A_2)= (-\infty ,-\infty )$, and $p^{h_1},p^{h_2}>1$. Then the hypersurface $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$.

Proof

Since $(A_1,A_2)=(- \infty , -\infty )$, the polynomial $F(1,X_1,X_2,X_3,X_4,X_5,X_6)$ defining $\mathcal {W}_{f_1,f_2}$ reads

$$\begin{aligned}{} & {} (X_3-X_6)^2 + (X_4-X_1)\Bigg (\frac{c_2^2(X_5)}{4}X_4-\frac{c_2^2(X_2)}{4}X_1 +c_2(X_5)X_6-c_2(X_2)X_3\Bigg )\nonumber \\{} & {} \quad +(X_5-X_2)\Bigg (\frac{c_1^2(X_4)}{4}X_5-\frac{c_1^2(X_1)}{4}X_2 +c_1(X_4)X_6-c_1(X_1)X_3\Bigg ). \end{aligned}$$

(12)

We can consider $C_1,C_2\ge 0$ otherwise $f_1$ or $f_2$ vanishes. Also, either $C_1$ or $C_2$ is positive otherwise $\mathcal {W}_{f_1,f_2}$ has degree 2.

Thus

$$\begin{aligned} \Delta (X_1,X_2,X_4,X_5)= & {} \Big ((X_4-X_1)c_2(X_5)+(X_5-X_2)c_1(X_4)\Big )^2\\{} & {} -4(X_4-X_1)\left( \frac{c_2^2(X_5)}{4}X_4-\frac{c_2^2(X_2)}{4}X_1 \right) \\{} & {} -4(X_5-X_2)\left( \frac{c_1^2(X_4)}{4}X_5-\frac{c_1^2(X_1)}{4}X_2 \right) \\= & {} X_1(X_4-X_1)(c_2^2(X_2)-c_2^2(X_5))+X_2(X_5-X_2)(c_1^2(X_1)-c_1^2(X_4))\\{} & {} +2(X_5-X_2)(X_4-X_1)c_2(X_5)c_1(X_4). \end{aligned}$$

If $C_2>C_1$ (resp. $C_1>C_2$) then the highest-degree homogeneous part in $\Delta (X_1,X_2,X_4,X_5)$ is $\alpha ^2 X_1(X_4-X_1)(X_2^{2C_2}-X_5^{2C_2})$ (resp. $\alpha ^2 X_2(X_5-X_2)(X_1^{2C_1}-X_4^{2C_1})$) which is not a square. If $C_2=C_1>1$ then the highest-degree homogeneous part in $\Delta (X_1,X_2,0,0)$

$$\begin{aligned} -\alpha ^2X_1^2X_2^{2C_2}-\beta ^2X_2^2X_1^{2C_1}=X_1^2X_2^2(-\alpha ^2X_2^{2C_1-2}-\beta ^2X_1^{2C_1-2}) \end{aligned}$$

is not a square.

If $C_2=C_1=1$ then the highest-degree homogeneous part in $\Delta (X_1,X_2,X_4,X_5)$

$$\begin{aligned}{} & {} \alpha ^2X_1(X_4-X_1)(X_2^2-X_5^2)+\beta ^2X_2(X_5-X_2)(X_1^2-X_4^2)+2\alpha \beta (X_5-X_2)(X_4-X_1)X_4X_5 \\{} & {} \quad =(X_4-X_1)(X_5-X_2)(-\alpha ^2X_1(X_2+X_5)-\beta ^2X_2(X_1+X_4)+2\alpha \beta X_4X_5) \end{aligned}$$

is not a square, since $(X_5-X_2)$ does not divide $(-\alpha ^2X_1(X_2+X_5)-\beta ^2X_2(X_1+X_4)+2\alpha \beta X_4X_5)$ unless $\alpha =\beta =0$, which is not possible. In all these cases, by Lemma 12, $\Delta (X_1,X_2,X_4,X_5)$ is not a square and, by Proposition 13, $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$. $\square$

Proposition 19

Let $\mathcal {W}_{f_1,f_2}$ be the hypersurface defined as in (8). Suppose that $\deg (\mathcal {W}_{f_1,f_2})\ge 4$, $(A_1,A_2)= (0,0)$, and $p^{h_1},p^{h_2}>1$. Then the hypersurface $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$ unless

$$\begin{aligned} p^{h_1}=p^{h_2}=3, \text { and } C_1,C_2\le 1. \end{aligned}$$

Proof

Similar as in Proposition 18, we have that

$$\begin{aligned} \Delta (X_1,X_2,X_4,X_5)= & {} \Big ((X_4-X_1)c_2(X_5)+(X_5-X_2)c_1(X_4)\Big )^2\nonumber \\{} & {} -4(X_4-X_1)\left( a_2 X_4^{p^{h_2}}-a_2 X_1^{p^{h_2}}+\frac{c_2^2(X_5)}{4}X_4-\frac{c_2^2(X_2)}{4}X_1 \right) \nonumber \\{} & {} -4(X_5-X_2)\left( a_1X_5^{p^{h_1}}-a_1X_2^{p^{h_1}}+\frac{c_1^2(X_4)}{4}X_5-\frac{c_1^2(X_1)}{4}X_2 \right) \\= & {} X_1(X_4-X_1)(c_2^2(X_2)-c_2^2(X_5))+X_2(X_5-X_2)(c_1^2(X_1)-c_1^2(X_4))\\{} & {} +2(X_5-X_2)(X_4-X_1)c_2(X_5)c_1(X_4)-4a_2(X_4-X_1)^{p^{h_2}+1}\\{} & {} -4a_1(X_5-X_2)^{p^{h_1}+1}. \end{aligned}$$

Thus,

$$\begin{aligned} \Delta (0,X_2,1,X_5)= & {} X_2(X_5-X_2)(c_1^2(0)-c_1^2(1)) +2(X_5-X_2)c_2(X_5)c_1(1)\\{} & {} -4a_2-4a_1(X_5-X_2)^{p^{h_1}+1}, \end{aligned}$$

where $a_1a_2\ne 0$.

We distinguish a few cases:

• (i) $c_1(1)=0$.

  Then $\Delta (0,X_2,1,X_5)$ reads $-4a_2-4a_1(X_5-X_2)^{p^{h_1}+1}+c_1^2(0)(X_5-X_2)X_2$ and it cannot be a square.

• (ii) $c_1(1)\ne 0$.

• If $C_2= p^{h_1}>1$ then $\Delta (0,X_2,1,X_5)$ cannot be a square since its highest-degree homogeneous part is $-4(X_5-X_2)^{p^{h_1}+1}+2c_1(1)c_2(X_5)(X_5-X_2)$, a nonsquare.

• If $C_2> p^{h_1}$, then $2c_1(1)c_2(X_5)(X_5-X_2)$, the highest-degree homogeneous part in $\Delta (0,X_2,1,X_5)$ is not a square.

• If $C_2< p^{h_1}$, the highest-degree homogeneous part in $\Delta (0,X_2,1,X_5)$ is $-4a_1(X_5-X_2)^{p^{h_1}+1}$. Since no other homogeneous part in $\Delta (0,X_2,1,X_5)$ is divisible by $(X_5-X_2)^{(p^{h_1}+1)/2}$ if $p^{h_1}+1>4$, $\Delta (0,X_2,1,X_5)$ is not a square if $p^{h_1}>3$. Also, when $p^{h_1}=3$, and $C_2<p^{h_1}$, we must have $C_2\le 1$, otherwise the second highest-degree homogeneous part in $\Delta (0,X_2,1,X_5)$ is $X_5^2(X_5-X_2)$ which is not divisible by $(X_5-X_2)^{(p^{h_1}+1)/2}= (X_5-X_2)^{2}$. By Lemma 12$\Delta (0,X_2,1,X_5)$ is not a square.

This shows that $\Delta (0,X_2,1,X_5)$ is not a square unless $p^{h_1}=3$ and $C_2\le 1$.

A similar argument applied to $\Delta (X_1,0,X_4,1)$ shows that this is not a square unless $p^{h_2}=3$ and $C_1\le 1$.

So, unless $p^{h_2}=3=p^{h_1}$ and $C_1,C_2\le 1$, by Proposition 13, $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$. $\square$

In view of the previous propositions, if we assume that $O(f_1,f_2)$ is an ovoid and thus $\mathcal {W}_{f_1,f_2}$ does not contain any ${\mathbb {F}}_q$-rational absolutely irreducible components, then

• 1. $p^{h_1}= p^{h_2}=3$;

• 2. $C_1,C_2\le 1$ (possibly $-\infty$),

• 3. $a_1(X)=a_1$, $a_2(X)=a_2$, $a_1a_2\ne 0$.

Secondly, from now on we assume that

$$\begin{aligned} c_1(X)=CX+D, \quad c_2(X)=EX+F, \end{aligned}$$

for $C,D,E,F \in {\mathbb {F}}_q$.

Proposition 20

If $p^{h_1}= p^{h_2}=3$, $C_1,C_2\le 1$ (possibly $-\infty$), and $A_1=A_2=0$, then $\mathcal {W}_{f_1,f_2}$ contains an absolutely irreducible component defined over ${\mathbb {F}}_q$, unless $(D,F)=(0,0)$, $C=-E$, and $a_2=E^4/a_1$. In this case, $O_6(f_1,f_2)$ is an ovoid if and only if $-a_1$ is not a square in ${\mathbb {F}}_q$.

Proof

We want to determine whether $\Delta (X_1,X_2,X_4,X_5)$ is a square by applying Lemma 12. We have

$$\begin{aligned} \Delta (0,X_2,X_4,0)= & {} -(4a_1X_2^4 -C^2 X_2^2X_4^2 + 4a_2X_4^4) +2CX_2X_4(DX_2 -FX_4) -2FDX_2X_4\\= & {} -(a_1X_2^4 + 2C^2 X_2^2X_4^2 + a_2X_4^4) -CX_2X_4(DX_2 -FX_4) +FDX_2X_4. \end{aligned}$$

First note that $C\ne 0$, otherwise $(4a_1X_2^4 +2C^2 X_2^2X_4^2 + 4a_2X_4^4)$ cannot be a square. Since $FDX_2X_4$ must be a square, $FD=0$. Now, in order to have that $\Delta (0,X_2,X_4,0)$ is a square, $-CX_2X_4(DX_2 -FX_4)$ must be a square and this can happen only if it vanishes, i.e. $F=D=0$. Also, if $(a_1X_2^4 + 2C^2 X_2^2X_4^2 + a_2X_4^4)$ is a square, then $C^4= a_1a_2$.

Now, $\Delta (X_1,X_2,X_4,X_5)$ reads

$$\begin{aligned}{} & {} ( E(X_4-X_1) X_5+C(X_5-X_2)X_4)^2\\{} & {} \quad - (X_4-X_1) \left( a_2 (X_4- X_1)^3+E^2 X_5^2 X_4-E^2 X_2^2 X_1\right) \\{} & {} \quad -(X_5-X_2) \left( a_1 (X_5- X_2)^3+C^2 X_4^2 X_5-C^2 X_1^2 X_2\right) . \end{aligned}$$

In particular

$$\begin{aligned} a_1\Delta (X_1,0,X_4,1)= & {} -a_2(X_1 -X_4)^4 +Ea_1(E X_1 + C X_4)(X_1 -X_4) -a_1, \end{aligned}$$

which is a square only if $C=-E$. In this case

$$\begin{aligned}{} & {} -a_1F(1,X_1,X_2,X_3,X_4,X_5,X_6)\\{} & {} =F_1(1,X_1,X_2,X_3,X_4,X_5,X_6)F_2(1,X_1,X_2,X_3,X_4,X_5,X_6), \end{aligned}$$

where

$$\begin{aligned} F_1(1,X_1,X_2,X_3,X_4,X_5,X_6):= & {} E^2 X_1^2 + E^2 X_1 X_4 + \eta ^2 \sqrt{a_1} E X_1 X_5 + a_1 X_2^2 - \eta ^2 \sqrt{a_1} E X_2 X_4\\{} & {} + a_1 X_2 X_5 - \eta ^2 \sqrt{a_1} X_3 + E^2X_4^2 + a_1 X_5^2 + \eta ^2 \sqrt{a_1} X_6;\\ F_2(1,X_1,X_2,X_3,X_4,X_5,X_6):= & {} E^2 X_1^2 + E^2 X_1 X_4 - \eta ^2 \sqrt{a_1} E X_1 X_5 + a_1 X_2^2 + \eta ^2 \sqrt{a_1} E X_2 X_4\\{} & {} + a_1 X_2 X_5 + \eta ^2 \sqrt{a_1} X_3 + E^2X_4^2 + a_1 X_5^2 - \eta ^2 \sqrt{a_1} X_6, \end{aligned}$$

with $\eta \in {\mathbb {F}}_9$ satisfying $\eta ^2 -\eta -1=0$.

If $q=3^{2h}$ and $a_1\in \square _q$ or $q=3^{2h+1}$ and $a_1\notin \square _q$ then both the factors $F_1$ and $F_2$ are defined over ${\mathbb {F}}_q$ and both $F_1=0$ and $F_2=0$ are absolutely irreducible quadrics. Therefore, $O_6(f_1,f_2)$ is not an ovoid.

If $q=3^{2h}$ and $a_1\notin \square _q$ or $q=3^{2h+1}$ and $a_1\in \square _q$ then both the factors $F_1$ and $F_2$ are not defined over ${\mathbb {F}}_q$ and both $F_1=0$ and $F_2=0$ are absolutely irreducible quadrics. Therefore the ${\mathbb {F}}_q$-rational solutions satisfy $F_1+F_2=0$ and $F_1-F_2=0$, that is

$$\begin{aligned} {\left\{ \begin{array}{ll} E X_1 X_5 + 2EX_2 X_4 + 2X_3 + X_6&{}=0,\\ (EX_1 + \eta ^2 \sqrt{a_1} X_2 + 2E X_4 -\eta ^2\sqrt{a_1} X_5)(EX_1 -\eta ^2 \sqrt{a_1} X_2 + 2E X_4 +\eta ^2\sqrt{a_1} X_5)&{}=0.\\ \end{array}\right. } \end{aligned}$$

Since both the factors $G_1$ and $G_2$ of the second equation in the system above are not defined over ${\mathbb {F}}_q$, the ${\mathbb {F}}_q$-rational solutions also satisfy $G_1+G_2=0$ and $G_1-G_2=0$, and thus

$$\begin{aligned} {\left\{ \begin{array}{ll} E X_1 X_5 + 2EX_2 X_4 + 2X_3 + X_6&{}=0,\\ X_1-X_4&{}=0,\\ X_2-X_5&{}=0.\\ \end{array}\right. } \end{aligned}$$

It is readily seen that the system above has only the solutions $X_1-X_4=X_2-X_5=X_3-X_6=0$ and therefore $O_6(f_1,f_2)$ is an ovoid. The claim follows by observing that the two conditions $q=3^{2h}$ and $a_1\notin \square _q$ or $q=3^{2h+1}$ and $a_1\in \square _q$ are equivalent to $-a_1$ not being a square in ${\mathbb {F}}_q$. $\square$

2.3 Conclusion

Summing up, the only choice of $f_1$ and $f_2$ for which $\textrm{O}_6(f_1,f_2)$ is an ovoid is

$$\begin{aligned} f_1(x,y,z)= & {} a_1 y^3+E^2x^2y-Exz;\\ f_2(x,y,z)= & {} \frac{E^4}{a_1}x^3+E^2y^2x+Eyz.\\ \end{aligned}$$

In this case the ovoid $\textrm{O}_6(f_1,f_2)$ reads

$$\begin{aligned}{} & {} \Bigg \{\Bigg (1,x,y,z,a_1 y^3+E^2x^2y-Exz,\frac{E^4}{a_1}x^3+E^2y^2x+Eyz, -z^2-a_1 \left( y^2+\frac{E^2}{a_1}x^2\right) ^2\Bigg )\Bigg \}_{x,y,z\in \mathbb {F}_q}\\{} & {} \quad \cup \{(0,0,0,0,0,0,1)\}, \end{aligned}$$

which can be written as

$$\begin{aligned}{} & {} \Bigg \{\Bigg (1,x/E,y,z,a_1 y^3+x^2y-xz,E\left( x^3/a_1+y^2x+yz\right) , -z^2-a_1 \left( y^2+x^2/a_1\right) ^2\Bigg )\Bigg \}_{x,y,z\in \mathbb {F}_q}\\{} & {} \quad \cup \{(0,0,0,0,0,0,1)\}. \end{aligned}$$

For any $E\in {\mathbb {F}}_q^*$ the above example is equivalent to Thas-Kantor via the projectivity

$$\begin{aligned} (X_0,X_1,X_2,X_3,X_4,X_5,X_6) \mapsto (X_0,E X_1,X_2,X_3,X_4,X_5/E,X_6) \end{aligned}$$

that fixes the quadric $\textrm{Q}$.

Combining the previous propositions and with Theorem 5, we have proved the following

Theorem 21

Let $f_1(X,Y,Z)=\sum _{ijk}a_{i,j,k}X^iY^jZ^k$ and $f_2(X,Y,Z)=\sum _{ijk}b_{i,j,k}X^iY^jZ^k$. Suppose that $q>6.3 (d+1)^{13/3}$, where $\max \{\deg (f_1),\deg (f_2)\}=d$. If $\textrm{O}_6(f_1,f_2)$ is an ovoid of Q(6, q) then

$$\begin{aligned} q=3^h, \quad f_1(X,Y,Z)=X^2Y-nY^3-XZ, \quad f_2(X,Y,Z)=-1/n X^3+XY^2+YZ. \end{aligned}$$

Hence $O_6(f_1,f_2)$ is a Thas-Kantor ovoid.