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 \(\max \{\deg (f_1),\deg (f_2)\}<(\frac{1}{6.3}q)^{\frac{3}{13}}-1\).
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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:
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:
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
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.
Note that Condition (1) reads
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
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
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
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:
As already mentioned, we will consider the hypersurface \(\mathcal {W}_{f_1,f_2}\subset \textrm{PG}(6,q)\) defined by
where \(F_{f_1,f_2}(1,X_1,X_2,X_3,X_4,X_5,X_6)\) equals
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:
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
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
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
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,
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
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
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
where
where \(f_2\) satisfies (6). Suppose that
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,
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
that is
So, if \(O_6(f_1,f_2)\) is an ovoid then
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
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.
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
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
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
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
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
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
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
To see this, it is enough to observe that any such linear combination cannot vanish since each of the following monomials
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
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
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
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
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
is the zero-polynomial.
If \(\delta _2\ne 0\), then, considering \(X_4=0\), we get
which does not vanish unless \(C_1=C_2=1\). In this case \(\delta _1=\delta _2=\delta _4=1\) and
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
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
To see this, it is enough to observe any such linear combination cannot vanish since each of the following monomials
is contained in a unique \(t_i\).
Thus
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
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
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
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
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
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
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
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
To see this, it is enough to observe any such linear combination cannot vanish since each of the following monomials
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
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
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
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
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 \(\Delta (X_1,X_2,0,0)\)
is not a square.
If \(C_2=C_1=1\) then the highest-degree homogeneous part in \(\Delta (X_1,X_2,X_4,X_5)\)
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
Proof
Similar as in Proposition 18, we have that
Thus,
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
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
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
In particular
which is a square only if \(C=-E\). In this case
where
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
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
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
In this case the ovoid \(\textrm{O}_6(f_1,f_2)\) reads
which can be written as
For any \(E\in {\mathbb {F}}_q^*\) the above example is equivalent to Thas-Kantor via the projectivity
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
Hence \(O_6(f_1,f_2)\) is a Thas-Kantor ovoid.
References
Aubry Y., McGuire G., Rodier F.: A few more functions that are not APN infinitely often. In: Finite Fields: Theory and Applications, vol. 518 of Contemporary Mathematics American Mathematical Society, pp. 23–31. Providence, RI (2010).
Bartoli D.: Hasse-Weil Type Theorems and Relevant Classes of Polynomial Functions. London Mathematical Society Lecture Note Series, pp. 43–102. Cambridge University Press, Cambridge (2021).
Bartoli D., Durante N.: On the classification of low-degree ovoids of \({Q}(4,q)\). Combinatorica, to appear (2021).
Bombieri E.: Counting points on curves over finite fields (d’après S. A. Stepanov). In: Séminaire Bourbaki, 25ème année (1972/1973), Exp. No. 430. 1974, vol. 383, pp. 234–241. Lecture Notes in Mathematics.
Cafure A., Matera G.: Improved explicit estimates on the number of solutions of equations over a finite field. Finite Fields Appl. 12, 155–185 (2006).
Ghorpade S.R., Lachaud G.: Étale cohomology, Lefschetz theorems and number of points of singular varieties over finite fields. Mosc. Math. J. 2, 589–631 (2002).
Ghorpade S.R., Lachaud G.: Number of solutions of equations over finite fields and a conjecture of Lang and Weil. In: Number Theory and Discrete Mathematics (Chandigarh, 2000). Trends Math, vol. 2002, pp. 269–291. Birkhäuser, Basel (2000).
Ghorpade S.R., Lachaud G.: Corrigenda and addenda: Étale cohomology, Lefschetz theorems and number of points of singular varieties over finite fields. Mosc. Math. J. 9, 431–438 (2009).
Gunawardena A., Moorhouse G.E.: The non-existence of ovoids in \(O_9(q)\). Eur. J. Comb. 18, 171–173 (1997).
Hartshorne R.: Algebraic Geometry. Graduate Texts in Mathematics, vol. no. 52. Springer, New York (1977).
Kantor W.M.: Ovoids and translation planes. Can. J. Math. 34, 1195–1207 (1982).
Lang S., Weil A.: Number of points of varieties in finite fields. Am. J. Math. 76, 819–827 (1954).
Lidl R., Niederreiter H.: Finite Fields, 2nd ed., vol. 20 of Encyclopedia of Mathematics and Its Applications. Cambridge University Press, Cambridge (1997). With a foreword by P. M. Cohn.
Penttila T., Williams B.: Ovoids of parabolic spaces. Geom. Ded. 82, 1–19 (2000).
Schmidt W.: Equations Over Finite Fields: An Elementary Approach, 2nd edn Kendrick Press, Heber City (2004).
Thas J.A.: Ovoids and spreads of finite classical polar spaces. Geom. Dedicata 10, 135–144 (1981).
van der Geer G.: Counting curves over finite fields. Finite Fields Appl. 32, 207–232 (2015).
Acknowledgements
This research was supported by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA - INdAM).
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Bartoli, D., Durante, N. & Grimaldi, G.G. Ovoids of Q(6, q) of low degree. Des. Codes Cryptogr. 92, 2287–2306 (2024). https://doi.org/10.1007/s10623-024-01388-9
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DOI: https://doi.org/10.1007/s10623-024-01388-9