Galois subcovers of the Hermitian curve in characteristic p with respect to subgroups of order dp with prime
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- 14.03.2025
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Abstract
1 Introduction
Curves with many points over a finite field have intensively been investigated, particularly for their useful connections to Coding theory, Cryptography, Finite geometry, and shift register sequences. In this context, the most important family consists of the maximal curves, that is, curves defined over the finite field \(\mathbb {F}_{q^2}\), where \(q=p^h\) and p is its charactreristic, which attain the famous Hasse-Weil upper bound. The Hermitian curve is the most known maximal curve and it is also very used for applications, especially in the study of algebraic geometry codes, shortly AG-codes. Actually, many other maximal curves derive from the Hermitian curve since any \(\mathbb {F}_{q^2}\)-subcover of a maximal curve is still maximal over the same field. If such a \(\mathbb {F}_{q^2}\)-subcover is a Galois subcover with Galois group G then the arising curve is named the quotient curve of the Hermitian curve with respect to G. Up to group isomorphism, the \(\mathbb {F}_{q^2}\)-automorphism group of the Hermitian curve is the 3-dimensional projective unitary group \(\hbox {PGU}(3,q)\) which has plenty of subgroups; see [8]. This motivated the systematic study of the quotients curves of the Hermitian curve which was eventually initiated in the seminal paper of Garcìa et al. [5]. Ever since important progress has been made in the study of the spectrum of the possible genera of the quotients of the Hermitian curve over a given finite field; see [7, Chap. 10]. Nevertheless, the problem of determining explicit equations for such curves, which is a relevant issue for applications, remains largely open. In fact, this problem has so far been solved by ad hoc methods, apart from the cases where the Galois group has either prime order; see [2], or its order equals the square of the characteristic; see [6].
In this paper we determine explicit equations for each quotient curve of the Hermitian curve whose Galois group has order dp where p is the characteristic of \(\mathbb {F}_{q^2}\) and d is a prime other than p. We also compute the Weierstrass semigroup at some \(\mathbb {F}_{q^2}\)-rational point of those curves, and discuss possible positive impacts on the minimum distance problems of AG-codes.
Anzeige
Theorem 1
In the \(\mathbb {F}_{q^2}\)-automorphism group \(G\cong \hbox {PGU}(3,q)\) of the Hermitian curve \(\mathcal {H}_q\) defined over \(\mathbb {F}_{q^2}\) with \(q=p^h\) and \(p\ge 5\), let H be a subgroup of order dp where \(d\ge 5\) is a prime number other than p. Let \(\bar{\mathcal {H}}_q=\mathcal {H}_q/H\) be the quotient curve of \(\mathcal {H}_q\) with respect to the subgroup H. Then, up to an \(\mathbb {F}_{q^2}\)-isomorphism, one of the following cases occurs.
(I)
If \(H=C_p\times C_d\) then \(\bar{\mathcal {H}}_q\) has genus and equation
$$\begin{aligned} \mathfrak {g}=\frac{1}{2d}(q-d+1)\left( \frac{q}{p}-1\right) \end{aligned}$$
(1)
(II)
If \(H=C_p\rtimes C_d\) and \(C_p\) is in the center of a Sylow p-subgroup of G, then \(\bar{\mathcal {H}}_q\) has genus and equation
where
$$\begin{aligned} \mathfrak {g}=\frac{1}{2} \frac{q}{d}\left( \frac{q}{p}-1\right) \end{aligned}$$
(2)
(III)
If \(H=C_p\rtimes C_d\) but \(C_p\) is not in the center of a Sylow p-subgroup of G, then \(\bar{\mathcal {H}}_q\) has genus and equation
where
$$\begin{aligned} \mathfrak {g}=\frac{q}{2dp}(q-1) \end{aligned}$$
(3)
2 Background
Let \(\mathcal {X}\) be a projective, non-singular, geometrically irreducible, algebraic curve of genus \(\mathfrak {g}\ge 2\) embedded in an r-dimensional projective space \(\mathrm{{PG}}(r,\mathbb {F}_\ell )\) over a finite field of order \(\ell \) of characteristic p. Let \(\mathbb {F}_\ell (\mathcal {X})\) be the function field of \(\mathcal {X}\) which is an algebraic function field of transcendence degree one with constant field \(\mathbb {F}_\ell \). As it is customary, \(\mathcal {X}\) is viewed as a curve defined over the algebraic closure \(\mathbb {F}\) of \(\mathbb {F}_\ell \). Then the function field \(\mathbb {F}(\mathcal {X})\) is the constant field extension of \(\mathbb {F}_\ell (\mathcal {X})\) with respect to field extension \(\mathbb {F}|\mathbb {F}_\ell \). The automorphism group \(\text{ Aut }(\mathcal {X})\) of \(\mathcal {X}\) is defined to be the automorphism group of \(\mathbb {F}(\mathcal {X})\) fixing every element of \(\mathbb {F}\). It has a faithful permutation representation on the set of all points \(\mathcal {X}\) (equivalently on the set of all places of \(\mathbb {F}(\mathcal {X}))\). The automorphism group \(\text{ Aut}_{\mathbb {F}_\ell }(\mathcal {X})\) of \(\mathbb {F}_\ell (\mathcal {X})\) is a subgroup of \(\text{ Aut }(\mathcal {X})\). In particular, the action of \(\text{ Aut}_{\mathbb {F}_\ell }(\mathcal {X})\) on the \(\mathbb {F}_\ell \)-rational points of \(\mathcal {X}\) is the same as on the set of degree 1 places of \(\mathbb {F}_\ell (\mathcal {X})\). Let G be a finite subgroup of \(\text{ Aut}_{\mathbb {F}_{\ell }}(\mathcal {X})\). The Galois subcover of \(\mathbb {F}_\ell (\mathcal {X})\) with respect to G is the fixed field of G, that is, the subfield \({\mathbb {F}_{\ell }}(\mathcal {X})^G\) consisting of all elements of \(\mathbb {F}_{\ell }(\mathcal {X})\) fixed by every element in G. Let \(\mathcal {Y}\) be a non-singular model of \({\mathbb {F}_{\ell }}(\mathcal {X})^G\), that is, a projective, non-singular, geometrically irreducible, algebraic curve with function field \({\mathbb {F}_{\ell }}(\mathcal {X})^G\). Then \(\mathcal {Y}\) is the quotient curve of \(\mathcal {X}\) by G and is denoted by \(\mathcal {X}/G\). The covering \(\mathcal {X}\mapsto \mathcal {Y}\) has degree equal to \(\mid G\mid \) and the field extension \(\mathbb {F}_{\ell }(\mathcal {X})|\mathbb {F}_{\ell }(\mathcal {X})^G\) is Galois. If P is a point of \(\mathcal {X}\), the stabilizer \(G_P\) of P in G is the subgroup of G consisting of all elements fixing P.
Result 2
[7, Theorem 11.49(b)] All p-elements of \(G_P\) together with the identity form a normal subgroup \(S_P\) of \(G_P\) so that \(G_P=S_P\rtimes C\), the semidirect product of \(S_P\) with a cyclic complement C.
Anzeige
Result 3
[7, Theorem 11.129] If \(\mathcal {X}\) has zero Hasse-Witt invariant then every non-trivial element of order p has a unique fixed point, and hence no non-trivial element in \(S_P\) fixes a point other than P.
A useful corollary of Result 3 is the following.
Result 4
Let \(\mathcal {X}\) be a curve defined over \(\mathbb {F}_\ell \) whose number of \(\mathbb {F}_\ell \)-rational points is \(N\ge 2\). If \(\mathcal {X}\) has zero Hasse-Witt invariant and S is a p-subgroup of \(\text{ Aut}_{\mathbb {F}_\ell }(\mathcal {X})\) then S fixes a unique point and |S| divides \(N-1\).
The following result is due to Stichtenoth [13].
Result 5
[7, Theorem 11.78(i)] Let H be a p-subgroup of \(\mathbb {F}(\mathcal {X})\) fixing a point. If |S| is larger than the genus of \(\mathbb {F}(\mathcal {X})\) then the Galois subcover of \(\mathbb {F}(\mathcal {X})\) with respect to H is rational.
From now on let \(\ell =q^2\) with \(q=p^h\) and assume that \(\mathcal {X}\) is a \(\mathbb {F}_{q^2}\)-maximal curve.
The following result follows from [12, Lemma 1].
Result 6
All \(\mathbb {F}_{q^2}\)-maximal curves have zero Hasse-Witt invariant.
The following result is commonly attributed to Serre.
Result 7
[7, Theorem 10.2] For every subgroup G of \(\text{ Aut}_{\mathbb {F}_{q^2}}(\mathcal {X})\), the quotient curve \(\mathcal {X}/G\) is also \(\mathbb {F}_{q^2}\)-maximal.
We also use the classification of all groups whose order is the product of two distinct primes.
Result 8
Suppose u and v are distinct prime numbers with \(u<v\). Then, there are two possibilities for groups G of order uv:
(I)
If \(u\not \mid (v-1)\) then G is a cyclic group.
(II)
If \(u\mid (v-1)\), then either G is a cyclic group, or G is a semidirect product \(C_v \rtimes C_u\).
2.1 The function field of the Hermitian curve
In this subsection we collect some useful results about the function field of the Hermitian curve and its Galois subcovers. The affine equation \(Y^q+Y=X^{q+1}\) of a curve defined over \(\mathbb {F}_{q^2}\) is the usual canonical form of the Hermitian curve \(\mathcal {H}_q\) whose function field is \(\mathbb {F}_{q^2}(x,y)\) where \(y^q+y-x^{q+1}=0\). The equation \(Y^q-Y+\omega X^{q+1}=0\) with \(\omega \in \mathbb {F}_{q^2}\) such that \(\omega ^{q-1}=-1\) is another useful equation of \(\mathcal {H}_q\). We exploit numerous known results on the \(\mathbb {F}\)-automorphism group \(\text{ Aut }(\mathbb {F}(\mathcal {H}_q))\) of \(\mathcal {H}_q\). For more details, the Reader is referred to [8, 9].
Result 9
[7, Theorem 12.24(iv), Proposition 11.30] \(\text{ Aut }(\mathbb {F}(\mathcal {H}_q))=\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\cong \hbox {PGU}(3,q)\). Moreover, \(\text{ Aut }(\mathbb {F}(\mathcal {H}_q))\) acts on the set of all \(\mathbb {F}_{q^2}\)-rational points of \(\mathcal {H}_q\) as \(\hbox {PGU}(3,q)\) in its natural doubly transitive permutation representation of degree \(q^3+1\) on the isotropic points of the unitary polarity of the projective plane \(PG(2,\mathbb {F}_{q^2})\).
The maximal subgroups of \(\hbox {PGU}(3,q)\) were determined by Mitchell in 1911, see Hoffer [8]. Let \(S_p\) be a Sylow p-subgroup of \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))=\hbox {PGU}(3,q)\). From Result 9, it may be assumed up to conjugacy that the unique fixed point of \(S_p\) is the point at infinity \(Y_\infty \) of \(\mathcal {H}_q\). The following result describes the structure of the stabilizer of \(Y_\infty \) in \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\).
Result 10
Let the Hermitian function field be given by its canonical form \(\mathbb {F}_{q^2}(x,y)\) with \(y^q+y-x^{q+1}=0\). Then the stabilizer G of \(Y_\infty \) in \(\text{ Aut }(\mathbb {F}(\mathcal {H}_q))\) consists of all mapswhereIn particular, \(G=S_p\rtimes C\) where \(S_p=\{\psi _{a,b,1}| b^q+b=a^{q+1},a,b\in \mathbb {F}_{q^2}\}\) and \(C=\{\psi _{0,0,\lambda }|\lambda \in \mathbb {F}_{q^2}^*\}\).
$$\begin{aligned} \psi _{a,b,\lambda }:\,(x,y)\mapsto (\lambda x+a, a^q \lambda x +\lambda ^{q+1}y +b) \end{aligned}$$
(4)
$$\begin{aligned} a \in \mathbb {F}_{q^2},\,\, \lambda \in \mathbb {F}_{q^2}^*,\,\,b^q+b=a^{q+1}. \end{aligned}$$
(5)
A direct computation by induction on i shows that for \(1\le i \le p\)For more about Result 10 see [5], Sect. 4.
$$\begin{aligned} \psi _{a,b,1}^i=\psi _{ia,a^{q+1}(i^2-i)/2+ib,1}. \end{aligned}$$
(6)
Remark 11
Changes of the generators x, y of the Hermitian function field \(\mathbb {F}_{q^2}(x,y), y^q+y-x^{q+1}=0\) provide another canonical form. For our purpose, a useful change is \(\tau :(x,y)\rightarrow (\omega x, -\omega y)\) where \(\omega ^{q-1}=-1\), and the arising canonical form is \(y^q-y+\omega x^{q+1}=0\). Then the elements in the stabilizer G of \(Y_\infty \) in \(\text{ Aut }(\mathbb {F}(\mathcal {H}_q))\) are of the form:where (5) is replaced byA direct computation by induction on i shows that for \(1\le i \le p\)For more about Result 11 see [5], Sect. 4.
$$\begin{aligned} \varphi _{a,b,\lambda }:\,(x,y)\mapsto (\lambda x+a, a^q \lambda \omega x +\lambda ^{q+1}y +b) \end{aligned}$$
(7)
$$\begin{aligned} a\in \mathbb {F}_{q^2},\,\,\lambda \in \mathbb {F}_{q^2}^*,\,\,b^q-b=-\omega a^{q+1}. \end{aligned}$$
$$\begin{aligned} \varphi _{a,b,1}^i=\varphi _{ia,a^{q+1}\omega (i^2-i)/2+ib,1}. \end{aligned}$$
(8)
From Result 10, \(S_p\) is the (unique) Sylow p-subgroup of the stabilizer of \(Y_\infty \) in \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\).
Result 12
\(S_p\) has the following properties.
(I)
The center \(Z(S_p)\) of \(S_p\) has order q and it consists of all maps \(\psi _{0,b,1}\) with \(b^q+b=0, b\in \mathbb {F}_{q^2}\). Also, \(Z(S_p)\) is an elementary abelian group of order p.
(II)
The non-trivial elements of \(S_p\) form two conjugacy classes in the stabilizer of \(Y_\infty \) in \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\), one comprises all non-trivial elements of \(Z(S_p)\), the other does the remaining \(q^3-q\) elements.
(III)
The elements of G other than those in \(Z(S_p)\) have order p, or \(p^2=4\) according as \(p>2\) or \(p=2\).
For completeness, we provide a proof for the classification of subgroups of \(\hbox {PGU}(3,q)\) of order dp. We use the canonical form \(y^q-y+\omega x^{q+1}=0\) with \(\omega ^{q-1}=-1\). The Galois subcovers of \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) with respect to a subgroup H of prime order or when its order equals the square of the characteristic were thoroughly classified in [2, 6] respectively. For the case \(|H|=dp\), the classification is reported in the following result.
Theorem 13
Let p and d be two distinct prime numbers both larger than 3. Then, up to conjugacy in \(\hbox {PGU}(3,q)\), there exist at most three subgroups of order dp in \(\hbox {PGU}(3,q)\), one is cyclic and the other two are semidirect products of \(C_p\rtimes C_d\) with \(p<d\). They are subgroups of the stabilizer of \(Y_\infty \) in \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\) with
(I)
\(G=\Sigma _p\times \Sigma _d\) with \(\Sigma _p=\langle \varphi _{0,1,1}\rangle \) and \(\Sigma _d=\langle \varphi _{0,0,\lambda }\rangle \) with \(\lambda ^d=1\), \(d|(q+1)\);
(II)
\(G=\Sigma _p\rtimes \Sigma _d\) with \(\Sigma _p=\langle \varphi _{0,1,1}\rangle \) and \(\Sigma _d=\langle \varphi _{0,0,\lambda }\rangle \) with \(\lambda ^d=1\), \(d|(p-1)\);
(III)
\(G=\Sigma _p\rtimes \Sigma _d\) with \(\Sigma _p=\langle \varphi _{1,\omega /2,1}\rangle \) and \(\Sigma _d=\langle \varphi _{0,0,\lambda }\rangle \) with \(\lambda ^d=1\), \(d|(p-1)\).
Proof
Let G be a subgroup of order pd in \(\hbox {PGU}(3,q)\). Two cases are treated separately according as \(p>d\) or \(p<d\).
Case \(p>d\). In this case, Result 8 shows that G has a unique Sylow p-subgroup \(\Sigma _p\). Moreover, \(\Sigma _p\) is a normal subgroup of G, and hence \(G=\Sigma _p\rtimes \Sigma _d\) where \(\Sigma _d\) is a Sylow d-subgroup of G. Since any non-trivial element of \(\hbox {PGU}(3,q)\) of order p has exactly one fixed point on \(\mathcal {H}_q(\mathbb {F}_{q^2})\) whereas \(\hbox {PGU}(3,q)\) acts transitively on \(\mathcal {H}_q(\mathbb {F}_{q^2})\), we may assume, up to conjugacy in \(\hbox {PGU}(3,q)\), that \(Y_\infty \) is the unique fixed point of \(S_p\). As \(S_p\) is a normal subgroup of G, the point \(Y_\infty \) is also fixed by \(\Sigma _d\). From \(|\mathcal {H}_q(\mathbb {F}_{q^2})|-1=q^3\), \(\Sigma _d\) must have a fixed point \(O\in \mathcal {H}_q(\mathbb {F}_{q^2})\) other than \(Y_\infty \). Since \(\hbox {PGU}(3,q)\) is doubly transitive on \(\mathcal {H}_q(\mathbb {F}_{q^2})\) we may assume, up to conjugacy, that \(O=(0:0:1)\). Then \(\Sigma _d\) is generated by \(t=\varphi _{0,0,\lambda }\) with \(\lambda ^d=1\) where \(d|(q^2-1)\). Furthermore, as \(\Sigma _p\) is a subgroup of the Sylow subgroup \(S_p\) of \(\hbox {PGU}(3,q)\) fixing \(Y_\infty \), two cases arise according as \(\Sigma _p\) is in the center \(Z(S_p)\) of \(S_p\) or not. Let s be a generator of \(\Sigma _p\). If \(s\in Z(S_p)\) then \(s=\varphi _{0,b,1}\) with \(b^q-b=0\). Take \(\mu \in \mathbb {F}_{q^2}^*\) such that \(\mu ^{q+1}=b^{-1}\). Then the conjugate of s by \(\varphi _{0,0,\mu }\) is \(\varphi _{0,0,1}\) while t and \(\varphi _{0,0,\mu }\) commute. Therefore, up to conjugacy, \(G=\Sigma _p\rtimes \Sigma _d\) with \(\Sigma _p=\langle s \rangle \) and \(s=\varphi _{0,0,1}\) whereas \(\Sigma _d\) and t are as before. Since \(\Sigma _p\) is a normal subgroup of G, there exists i with \(1\le i \le p-1\) such that \(st=ts^i\). A straightforward computation shows that this occurs if and only if \(i=1/\lambda ^{q+1}\). For \(d|(q+1)\), this implies \(i=1\), thus G is cyclic and Case (I) occurs. For \(d|(q-1)\), we have \(i\ne 1\) and hence G is not abelian. From Result 8, \(d|(p-1)\). Thus Case (II) occurs. If \(s\not \in Z(S_p)\) then \(s=\varphi _{a,b,1}\). For \(\mu =a^{-1}\), the conjugate of s by \(\varphi _{0,0,\mu }\) is \(\varphi _{1,b/a^{q+1},1}\) while t and \(\varphi _{0,0,\mu }\) commute. Therefore, up to conjugacy, we may assume \(G=\Sigma _p\rtimes \Sigma _d\) where \(\Sigma _p=\langle s \rangle \) and \(s=\varphi _{1,b/a^{q+1},1}\) while \(\Sigma _d\) and t are not changed. Then \(st=\phi _{1,b/a^{q+1},\lambda }\) and, from (6),Therefore, \(st=ts^i\) if and only if \(\lambda i=1\) and \(\lambda ^{q+1}({\textstyle \frac{1}{2}}\omega (i^2-i)+ib/a^{q+1})=b/a^{q+1}\). The latter condition can also be written asthat is, \(b={\textstyle \frac{1}{2}}\omega a^{q+1}\) as \(\lambda \ne 1\). Therefore, \(st=ts^i\) if and only if \(s=\varphi _{1,\omega /2,1}\) and \(i\lambda =1\). In particular, G is not abelian, and \(d|(p-1)\). This gives Case (III).
$$\begin{aligned} ts^i=\phi _{\lambda i,\lambda ^{q+1}(\omega (i^2-i)/2+ib/a^{q+1}),\lambda }. \end{aligned}$$
$$\begin{aligned} {\textstyle \frac{1}{2}}\omega (i^2-i)=(i^2-i)b/a^{q+1}, \end{aligned}$$
Case \(p<d\). In this case, a Sylow d-subgroup \(\Sigma _d\) of G is a normal subgroup of G, and hence \(\Sigma _d\) is the unique d-subgroup of G. As d divides the order of \(\hbox {PGU}(3,q)\), either \(d|(q-1)\), or \(d|(q+1)\), or \(d|(q^2-q+1)\). Assume that \(\Sigma _d\) fixes a point on \(\mathcal {H}_q(\mathbb {F}_{q^2})\). Then \(\Sigma _d\) has at least two fixed points, as \(|\mathcal {H}_q(\mathbb {F}_{q^2})|-1\) equals \(q^3\). Up to conjugacy, we may assume that \(\Sigma _d\) fixes \(Y_\infty \) and O. Then \(\Sigma _d=\langle \varphi _{0,0,\lambda }\rangle \) with \(\lambda ^d=1\). If \(\lambda ^{q+1}\ne 1\) then \(\Sigma _d\) has no any further fixed point, and hence G preserves the pair \(\{Y_\infty ,O\}\). Since \(p>2\) this yields that elements of G of order p fix two points on \(\mathcal {H}_q(\mathbb {F}_{q^2})\) which is not possible. Therefore, \(\lambda ^{q+1}=1\) and \(d|(q+1)\). This implies that \(\Sigma _d\) fixes all points \(P=(0,\eta )\) with \(\eta ^q-\eta =0\), i.e. with \(\eta \in \mathbb {F}_q\). Since \(\Sigma _d\) is a normal subgroup of G, this yields that a generator s of \(\Sigma _p\) takes O to a point \(P=(0,\eta )\) with \(\eta \in \mathbb {F}_q\). But then \(s=\varphi _{0,b,1}\) with \(b\in \mathbb {F}_q^*\). For \(\mu =b^{-1}\), the conjugate of s by \(\varphi _{0,0,\nu }\) with \(\nu ^{q+1}=\mu \) is \(\varphi _{0,1,1}\) while a generator t of \(\Sigma _d\) and \(\varphi _{0,0,\mu }\) commute. Therefore, up to conjugacy, we may assume \(s=\varphi _{0,1,1}\). Also, \(st=ts\) and Case (I) occurs. We are left with the case where \(\Sigma _d\) fixes no point on \(\mathcal {H}_q(\mathbb {F}_{q^2})\). Then either \(d|(q+1)\), or \(d|(q^2-q+1)\). We look at the action of \(\hbox {PGU}(3,q)\) as a projective group of the plane \(\mathrm{{PG}}(2,\mathbb {K})\) where \(\mathbb {K}\) is an algebraic closure of \(\mathbb {F}_{q^2}\). Then the Hermitian curve \(\mathcal {H}_q\) is left invariant by \(\hbox {PGU}(3,q)\). In particular, \(\hbox {PGU}(3,q)\) preserves both \(\mathcal {H}_q(\mathbb {F}_{q^2})\) and its complementary set in \(\mathrm{{PG}}(2,\mathbb {F}_{q^2})\) whose size equalsFurthermore, \(\mathcal {H}_q(\mathbb {F}_{q^2})\) can also be viewed as the set of all isotropic points of a unitary polarity \(\pi \) of \(\mathrm{{PG}}(2,\mathbb {F}_{q^2})\). If \(d|(q+1)\) then \(\Sigma _d\) fixes a point \(R\in \mathrm{{PG}}(2,\mathbb {F}_{q^2})\) outside \(\mathcal {H}_q(\mathbb {F}_{q^2})\).
$$\begin{aligned} q^4+q^2+1-(q^3+1)=q^2(q^2-q+1). \end{aligned}$$
Let r be the polar line of R w.r.t. \(\pi \). Then r is a chord of \(\mathcal {H}_q(\mathbb {F}_{q^2})\). Since r has as many as \(q(q-1)\) points other than those on \(\mathcal {H}_q(\mathbb {F}_{q^2})\), there are at least two fixed points on r outside \(\mathcal {H}_q(\mathbb {F}_{q^2})\) under the action of \(\Sigma _d\). Since \(\Sigma _d\) does not fix r pointwise, these two points, say \(R_1, R_2\) are the only fixed points of \(\Sigma _d\) on r. In particular, \(\Sigma _d\) fixes the vertices of the triangle \(RR_1R_2\).
We show that no more point in \(\mathrm{{PG}}(2,\mathbb {F}_{q^2})\) is fixed by \(\Sigma _d\). In fact, such a further fixed point T of \(\Sigma _d\) should lie on a side of the triangle, and that side would be fixed pointwise by \(\Sigma _d\). But this is impossible in our case, since the sides of \(RR_1R_2\) are chords of \(\mathcal {H}_q(\mathbb {F}_{q^2})\) whereas \(\Sigma _d\) is supposed not to fix points on \(\mathcal {H}_q(\mathbb {F}_{q^2})\). Since \(\Sigma _d\) is a normal subgroup of G, the triangle \(RR_1R_2\) is left invariant by G. But then G is contained in a maximal subgroup of \(\hbox {PGU}(3,q)\) whose order equals \(6(q+1)^2\). By the Lagrange theorem \(pd=|G|\) divides \(6(q+1)^2\). Since \(p>3\) and \(p\not \mid (q+1)\), this is impossible.
A similar geometric approach is used to rule out the other possibility, i.e. \(d|(q^2-q+1)\). Look at the action of \(\hbox {PGU}(3,q)\) on \(\mathrm{{PG}}(2,\mathbb {F}_{q^6})\). Fromthe Hermitian curve \(\mathcal {H}_q\) has as many as \(q^3(q+1)^2(q-1)\) points in \(\mathrm{{PG}}(2,\mathbb {F}_{q^6})\) but not in \(\mathrm{{PG}}(2,\mathbb {F}_{q^2})\). From \(d|(q^2-q+1)\) and \(d>3\), \(\Sigma _d\) fixes a point \(R\in \mathcal {H}_q(\mathbb {F}_{q^6})\) not lying in \(PG(2,\mathbb {F}_{q^2})\). The Frobenius collineation \(\mathfrak {f}\) which sends the point \(P=(a_1:a_2:a_3)\) to the pointleaves \(\mathcal {H}_q(\mathbb {F}_{q^6})\) invariant. Since \(\mathfrak {f}\) and \(\Sigma _d\) commute, \(\Sigma _d\) also fixes the points \(R_{q^2}\) and \(R_{q^3}\). Actually, \(\Sigma _d\) does not fix another point, otherwise one of the sides, say \(\ell \), of the triangle \(R R_{q^2}R_{q^4}\) would be fixed by \(\Sigma _d\) pointwise. Since \(\mathfrak {f}\) takes \(\ell \) to another side r of \(R R_{q^2}R_{q^4}\) and \(\mathfrak {f}\) and \(\Sigma _d\) commute, this would yield that r is also fixed pointwise by \(\Sigma _d\), which is impossible. As before, this implies that G leaves the triangle \(RR_{q^2}R_{q^4}\) invariant. Therefore G is a contained in a maximal subgroup of \(\hbox {PGU}(3,q)\) whose order equals \(3(q^2-q+1)\). Arguing as before, the Langrage theorem yields that this is impossible. \(\square \)
$$\begin{aligned} |\mathcal {H}_q(\mathbb {F}_{q^6})|=q^6+1+q^4(q-1)\,\, \text { and }\,\,|\mathcal {H}_q(\mathbb {F}_{q^3})|=q^3+1, \end{aligned}$$
$$\begin{aligned} P_{q^2}=\left( a_1^{q^2}:a_2^{q^2}:a_3^{q^2}\right) \end{aligned}$$
Result 14
[7, Theorem 5.74] Let H be a subgroup of \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\) of order p. The Galois subcover \(\mathbb {F}_{q^2}(\mathcal {F}')\) of \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) with respect to H is \(\mathbb {F}_{q^2}\)-isomorphic to the function field \(\mathbb {F}_{q^2}(\xi ,\eta )\) where either (I) or (II) hold:
(I)
with \(\omega ^{q-1}=-1\), \(\mathfrak {g}(\mathbb {F}_{q^2}(\mathcal {F}'))={\textstyle \frac{1}{2}}q\left( \frac{q}{p}-1\right) \), and H is in the center of a Sylow p-subgroup of \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\);
with \(\omega ^{q-1}=-1\), \(\mathfrak {g}(\mathbb {F}_{q^2}(\mathcal {F}'))={\textstyle \frac{1}{2}}q\left( \frac{q}{p}-1\right) \), and H is in the center of a Sylow p-subgroup of \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\);(II)
for \(p>2\), \(\mathfrak {g}(\mathbb {F}_{q^2}(\mathcal {F}'))={\textstyle \frac{1}{2}}\frac{q}{p}(q-1)\), and H is not in the center of a Sylow p-subgroup of \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\).
for \(p>2\), \(\mathfrak {g}(\mathbb {F}_{q^2}(\mathcal {F}'))={\textstyle \frac{1}{2}}\frac{q}{p}(q-1)\), and H is not in the center of a Sylow p-subgroup of \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\).The following result is a corollary of [5, Sect. 4].
Result 15
Let \(\mathfrak {g}\) be the genus of the Galois cover of \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) with respect to a subgroup G of \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}_q))\) of order dp. Let \(S_p\) be a Sylow p-subgroup of \(\text{ Aut }(\mathbb {F}_{q^2}(\mathcal {H}))\) containing a subgroup H of G of order p. Then eitheror
$$\begin{aligned} \mathfrak {g}=\frac{1}{2} \frac{q}{d}\left( \frac{q}{p}-1\right) \qquad \text {for}\qquad (d,q+1)=1, \end{aligned}$$
$$\begin{aligned} \mathfrak {g}=\frac{1}{2d}(q-d+1)\left( \frac{q}{p}-1\right) \qquad \text {for}\qquad (d,q+1)=d. \end{aligned}$$
3 Galois subcovers of \(\mathbb {F}_{q^2}(\mathcal {H})\) of type (I) of Theorem 13
As in Remark 11, take \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) in its canonical form \(\mathbb {F}_{q^2}(x,y)\) with \(y^q-y+\omega x^{q+1}=0\) and \(\omega ^{q-1}=-1\). The group \(\varPhi =\langle \varphi _{0,1,1}\rangle \) has order p, and it is contained in \(Z(S_p)\). Let \(\eta =y^p-y\) and \(\xi =x\). Then \(\varphi _{0,1,1}(\eta )=\varphi _{0,1,1}(y^p-y)=\varphi _{0,1,1}(y)^p-\varphi _{0,1,1}(y)=(y+1)^p-(y+1)=y^p-y=\eta \). Moreover, \(y^q-y=Tr(\eta )\). Since \(\varphi _{0,1,1}\) fixes \(\xi \), this shows that the Galois subcover \(\mathbb {F}_{q^2}(\mathcal {F}')\) of \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) with respect to \(\varPhi \) is as in (i) of Result 14. That equation can also be written asTake an element \(r\in \mathbb {F}_{q^2}\) with \(r^{d}=1\). Then \(\varphi _{0,0,r}\) commutes with \(\varphi _{0,1,1}\). Therefore, if \(d|(q+1)\) then \(\varphi _{0,0,r}\) induces an automorphism \(\varphi \) of \(\mathbb {F}_{q^2}(\mathcal {F}')\). More precisely, a straightforward computation shows that \(\varphi \) is the map \(\varphi :(\xi ,\eta )\mapsto (r\xi ,\eta )\). Let \(\varPhi _{r}\) be the \({\mathbb {F}}_q\)-automorphism group of \(\mathbb {F}_{q^2}(\mathcal {F}')\) generated by \(\varphi \). Then the Galois subcover of \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) with respect to G of Result 13 of type (I) is the same as the Galois subcover \(G_r\) of \({\mathbb {F}}_{q^2}(\mathcal {F}')\) with respect to \(\varPhi _r\).
$$\begin{aligned} \sum _{i=0}^{h-1}\ \eta ^{p^i}+\omega \xi ^{q+1}=0. \end{aligned}$$
(9)
Theorem 16
The Galois subcover \(G_r={\mathbb {F}}_{q^2}(\zeta ,\tau )\) of \({\mathbb {F}}_{q^2}(\mathcal {F}')\) with respect to \(\varPhi _r\) has genusand is given by
$$\begin{aligned} \mathfrak {g}=\frac{1}{2d}(q-d+1)(\frac{q}{p}-1) \end{aligned}$$
(10)
Proof
We show first that the fixed field F of \(\varPhi _r\) is generated by \(\tau =\eta \) together withSince \(\varphi (\tau )=\tau \) andwe have \({\mathbb {F}}_{q^2}(\zeta ,\tau )\subseteq F\). Furthermore, \([{\mathbb {F}}_{q^2}(\mathcal {F}'):{\mathbb {F}}_{q^2}(\zeta ,\tau )]=d\). Since d is prime, this yields either \({\mathbb {F}}_{q^2}(\zeta ,\tau )=F\) or \(F={\mathbb {F}}_{q^2}(\mathcal {F}')\). The latter case cannot actually occur, and hence \(F={\mathbb {F}}_{q^2}(\zeta ,\tau )\). Therefore \(F=G_r\). Now, eliminate \(\xi \) from Eqs. (9) and (11). Since d divides \(q+1\), replacing \(\xi ^{q+1}\) with
and \(\tau =\eta \) in (9) gives equation in (10). The formula for the genus follows from [7, Lemma 12.1(iii)(b)]. \(\square \)
$$\begin{aligned} \zeta =\xi ^d. \end{aligned}$$
(11)
$$\begin{aligned} \varphi (\zeta )=\varphi (\xi ^d)=\varphi (\xi )^d=r^{d}\xi ^{d}=\xi ^{d}=\zeta , \end{aligned}$$
and \(\tau =\eta \) in (9) gives equation in (10). The formula for the genus follows from [7, Lemma 12.1(iii)(b)]. \(\square \)4 Galois subcovers of \(\mathbb {F}_{q^2}(\mathcal {H})\) of type (II) of Theorem 13
We keep our notation up from Sect. 3. Assume that d divides \(p-1\), and take \(r\in \mathbb {F}_p^*\) with \(r^{d}=1\). Then \(\varphi _{0,0,r}^{-1}\circ \varphi _{0,1,1}\circ \varphi _{0,0,r}=\varphi _{0,r^2,1}\in \langle \varphi _{0,1,1}\rangle \), and hence \(\varphi _{0,0,r}\) induces an automorphism \(\varphi \) of \(\mathbb {F}_{q^2}(\mathcal {F}')\). Here, \(\varphi \) is the map \(\varphi :(\xi ,\eta )\mapsto (r\xi ,r^2\eta )\). Let \(\varPhi _{r}\) be the \({\mathbb {F}}_q\)-automorphism group of \(\mathbb {F}_{q^2}(\mathcal {F}')\) generated by \(\varphi \). Then the Galois subcover of \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) with respect to G of Result 13 of type (II) is the Galois subcover \(G_r\) of \({\mathbb {F}}_{q^2}(\mathcal {F}')\) with respect to \(\varPhi _r\).
Theorem 17
The Galois subcover \(G_r={\mathbb {F}}_{q^2}(\epsilon ,\rho )\) of \({\mathbb {F}}_{q^2}(\mathcal {F}')\) with respect to \(\varPhi _r\) has equation
where
(12)
Proof
We show first that the fixed field F of \(\varPhi _r\) is generated bytogether withSinceandwe have \({\mathbb {F}}_{q^2}(\epsilon ,\rho )\subseteq F\). Furthermore, \([{\mathbb {F}}_{q^2}(\mathcal {F}'):{\mathbb {F}}_{q^2}(\epsilon ,\rho )]=d\). Since d is prime, this yields either \({\mathbb {F}}_{q^2}(\epsilon ,\rho )=F\) or \(F={\mathbb {F}}_{q^2}(\mathcal {F}')\). The latter case cannot actually occur, and hence \(F={\mathbb {F}}_{q^2}(\epsilon ,\rho )\). Therefore \(F=G_r\). We have to eliminate \(\xi \) and \(\eta \) from Eqs. (9), (13) and (14). From (14) we have \(\eta =\rho \xi ^2\) then \(Tr(\eta )=Tr(\rho \xi ^2)\). This yields that
whence
Since d divides \(p-1\), \(Tr(\eta )\) in (15) can also be written as
Therefore
This, together with (9), giveSince \(d\mid (p-1)\) the number \(\frac{q-1}{d}\) is an integer. Thus Eq. (12) follows from (19). \(\square \)
$$\begin{aligned} \epsilon =\xi ^d \end{aligned}$$
(13)
$$\begin{aligned} \rho =\frac{\eta }{\xi ^2}. \end{aligned}$$
(14)
$$\begin{aligned} \varphi (\epsilon )=\varphi (\xi ^d)=\varphi (\xi )^d=r^d\xi ^d=\xi ^d=\epsilon \end{aligned}$$
$$\begin{aligned} \varphi (\rho )=\frac{\varphi (\eta )}{\varphi (\xi ^2)}=\frac{\varphi (\eta )}{\varphi (\xi )^2}=\frac{r^2\eta }{r^2\xi ^2}=\frac{\eta }{\xi ^2}=\rho , \end{aligned}$$
(15)
(16)
(17)
(18)
$$\begin{aligned} \omega \xi ^{q+1}=\xi ^2 A(\epsilon ,\eta ). \end{aligned}$$
(19)
5 Galois subcovers of \(\mathbb {F}_{q^2}(\mathcal {H})\) of type (III) of Theorem 13
This time, take \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) in its canonical form \(\mathbb {F}_{q^2}(x,y)\) with \(y^q+y-x^{q+1}=0\). The group
has order p, and it is not contained in \(Z(S_p)\). Let \(\xi =x^p-x\) and \(\eta =y-{\textstyle \frac{1}{2}}x^2\). A straightforward computation shows that
and
. Moreover,Since \(Tr(\xi )=x^q-x\), this givesTherefore, the Galois subcover \(\mathbb {F}_{q^2}(\mathcal {F}')\) of \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) with respect to \(\Psi \) is \(\mathbb {F}_{q^2}(\xi ,\eta )\) withReplacing \(\eta \) by \(-{\textstyle \frac{1}{2}}\eta \) this equation becomesIn particular, \(\mathbb {F}_{q^2}(\mathcal {F}')\) is the same as (II) of Result 14. Assume that d divides \(p-1\), and take \(r\in \mathbb {F}_p^*\) with \(r^{d}=1\). Then
, and hence \(\varphi _{0,0,r}\) induces an automorphism \(\varphi \) of \(\mathbb {F}_{q^2}(\mathcal {F}')\). Moreover, \(\psi \) is the map \(\psi :(\xi ,\eta )\mapsto (r\xi ,r^2\eta )\). Let \(\Psi _{r}\) be the \({\mathbb {F}}_q\)-automorphism group of \(\mathbb {F}_{q^2}(\mathcal {F}')\) generated by \(\psi \). Then the Galois subcover of \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) with respect to G of Result 13 of type (III) is the Galois subcover \(G_r\) of \({\mathbb {F}}_{q^2}(\mathcal {F}')\) with respect to \(\Psi _r\).
has order p, and it is not contained in \(Z(S_p)\). Let \(\xi =x^p-x\) and \(\eta =y-{\textstyle \frac{1}{2}}x^2\). A straightforward computation shows that
and
. Moreover,$$\begin{aligned} y^q+y-x^{q+1}= & \eta ^q+{\textstyle \frac{1}{2}}x^{2q}+\eta +{\textstyle \frac{1}{2}}x^2-x^{q+1}=\eta ^q+\eta +{\textstyle \frac{1}{2}}x^{2q}+{\textstyle \frac{1}{2}}x^2-x^{q+1}\\= & \eta ^q+\eta +{\textstyle \frac{1}{2}}(x^q-x)^2. \end{aligned}$$
$$\begin{aligned} \eta ^q+\eta +{\textstyle \frac{1}{2}}(x^q-x)^2=\eta ^q+\eta +{\textstyle \frac{1}{2}}Tr(\xi )^2. \end{aligned}$$
$$\begin{aligned} \eta ^q+\eta +{\textstyle \frac{1}{2}}\left( \sum _{i=1}^{h}\xi ^{p^{i-1}}\right) ^2=0. \end{aligned}$$
$$\begin{aligned} \eta ^q+\eta -\left( \sum _{i=1}^{h}\xi ^{p^{i-1}}\right) ^2=0. \end{aligned}$$
(20)
, and hence \(\varphi _{0,0,r}\) induces an automorphism \(\varphi \) of \(\mathbb {F}_{q^2}(\mathcal {F}')\). Moreover, \(\psi \) is the map \(\psi :(\xi ,\eta )\mapsto (r\xi ,r^2\eta )\). Let \(\Psi _{r}\) be the \({\mathbb {F}}_q\)-automorphism group of \(\mathbb {F}_{q^2}(\mathcal {F}')\) generated by \(\psi \). Then the Galois subcover of \(\mathbb {F}_{q^2}(\mathcal {H}_q)\) with respect to G of Result 13 of type (III) is the Galois subcover \(G_r\) of \({\mathbb {F}}_{q^2}(\mathcal {F}')\) with respect to \(\Psi _r\).Theorem 18
The Galois subcover \(G_r={\mathbb {F}}_{q^2}(\iota ,\nu )\) of \({\mathbb {F}}_{q^2}(\mathcal {F}')\) with respect to \(\Psi _r\) has genusand is given by
where
$$\begin{aligned} \mathfrak {g}=\frac{q}{2dp}(q-1) \end{aligned}$$
(21)
Proof
We show first that the fixed field F of \(\Psi _r\) is generated bytogether withandSinceandwe have \({\mathbb {F}}_{q^2}(\iota ,\nu ,\tau )\subseteq F\). Furthermore, \([\mathbb {F}_{q^2}(\iota ,\nu ,\tau )(\xi ):\mathbb {F}_{q^2}(\iota ,\nu ,\tau )]=d\) and \(\eta \in \mathbb {F}_{q^2}(\iota ,\nu ,\tau )(\xi )\). Therefore, \([{\mathbb {F}}_{q^2}(\mathcal {F}'):{\mathbb {F}}_{q^2}(\iota ,\nu ,\tau )]\le d\). Since d is prime, this yields either \({\mathbb {F}}_{q^2}(\iota ,\mu ,\nu )=F\) or \(F={\mathbb {F}}_{q^2}(\mathcal {F}')\). The latter case cannot actually occur, and hence \(F={\mathbb {F}}_{q^2}(\iota ,\mu ,\nu )\). Therefore \(F=G_r\). We go on by eliminating \(\xi \) and \(\eta \) from Eqs. (20–24). From the definition of the trace of \(\xi \),
By a straightforward computation,
This can also be written as
From (23), \(\xi ^2=\eta \iota \). Therefore, in (25) the square trace of \(\xi \) is equal to
Since \({\textstyle \frac{1}{2}}(p^i+p^j)-1={\textstyle \frac{1}{2}}(p^i-1)+{\textstyle \frac{1}{2}}(p^j-1)\), the sum in (26) turns out to be equal to
As d and 2 divide both \(p^i-1\) and \(p^j-1\), the sum in (26) equals
and by replacing \(\eta ^d\) with \(\nu \) we obtain
Let
Therefore, \(\eta ^q+\eta =\eta A(\iota ,\nu )\), and dividing both sides by \(\eta \) gives \(\eta ^{q-1}+1= A(\iota ,\nu )\). Since d divides \(q-1\), replacing \(\eta ^d\) by \(\nu \) shows
From (22–24), \(\nu =\frac{\tau ^2}{\iota ^d}\). Now the claim follows from (31). \(\square \)
$$\begin{aligned} \nu =\eta ^d, \end{aligned}$$
(22)
$$\begin{aligned} \iota =\frac{\xi ^2}{\eta }, \end{aligned}$$
(23)
$$\begin{aligned} \tau ={\xi ^d}. \end{aligned}$$
(24)
$$\begin{aligned} \varphi (\nu )=\varphi (\eta )^d=r^{2d}\eta ^d=\eta ^d=\nu ,\quad \varphi (\tau )=\varphi (\xi )^d=r^{d}\xi ^d=\xi ^d=\tau \end{aligned}$$
$$\begin{aligned} \varphi (\iota )=\varphi \left( \frac{\xi ^2}{\eta }\right) =\frac{\varphi (\xi ^2)}{\varphi (\eta )}=\frac{\varphi (\xi )^2}{\varphi (\eta )}=\frac{(r\xi )^2}{r^2\eta }=\frac{r^2\xi ^2}{r^2\eta }=\frac{\xi ^2}{\eta }, \end{aligned}$$
By a straightforward computation,
$$\begin{aligned} Tr(\xi )^2=\sum _{i=0}^{h-1}\sum _{j=0}^{h-1} \xi ^{p^i+p^j}. \end{aligned}$$
(25)
(26)
(27)
(28)
(29)
(30)
(31)
6 Weierstrass semigroups and application to AG-codes
We compute the Weierstrass semigroup at the unique place centred at the point at infinity of some of the maximal curves considered in the present paper.
Proposition 19
Let \(P_\infty \) be the unique point at infinity of the following two curves
Then the Weierstrass semigroup at \(P_\infty \) is generated by \(\textstyle \frac{q}{p}\), and \(q+1\), respectively by \(\textstyle \frac{q}{p}\), and \(\frac{q+1}{d}\).
(32)
Proof
For the first equation the claim follows from the remark after the proof of Lemma 12.2 in [7] applied for \(n=h-1\) and \(m=q+1\).
For the second equation the claim follows from the remark after the proof of Lemma 12.2 in [7] applied for \(n=h-1\) and \(m=\frac{q+1}{d}\). \(\square \)
Let S be a numerical semigroup. The gaps of S are the elements in \(\mathbb {N}\setminus S\). The number g(S) of gaps of S is the genus of S. If S is the Weierstrass semigroup of a curve at a point then g(S) coincides with the genus of the curve. Let \((a_1,\ldots ,a_k)\) be a sequence of positive integers such that their greatest common divisor is 1. Let \(d_0=0,\,d_i=g.c.d.(a_1,\ldots ,a_i)\) and \(A_i=\left\{ \frac{a_1}{d_i},\ldots ,\frac{a_i}{d_i}\right\} \) for \(i=1,\ldots ,k\). Let \(S_i\) be the semigroup generated by \(A_i\). The sequence \((a_1,\ldots ,a_k)\) is said to be telescopic whenever \(\frac{a_i}{d_i}\in S_{i-1}\) for \(i=2,\ldots ,k\). A telescopic semigroup is a numerical semigroup generated by a telescopic sequence.
Result 20
[11, Lemma 6.5] For the semigroup generated by a telescopic sequence \((a_1,\ldots ,a_k)\), let
$$\begin{aligned} l_g(S_k):=\sum _{i=1}^{k} \left( \frac{d_{i-1}}{d_i}-1 \right) a_i\quad then \quad g(S_k)=\frac{l_g(S_k)+1}{2}. \end{aligned}$$
Theorem 21
Let \(P_\infty \) be the unique point of infinity of the curves \(\bar{\mathcal {H}}_q\) in Theorem 1. Then the Weierstrass semigroup \(H(P_\infty )\) has the following properties:
(I)
\(H(P_\infty )=\langle \frac{q}{p},q+1\rangle \), for the curve of Equation (I);
(II)
\(\frac{q}{p},\frac{q-1}{d}\in H(P_\infty )\), for the curve of Equation (II);
(III)
\(\frac{2(q-1)}{d},q-1\in H(P_\infty )\), for the curve of Equation (III).
Proof
Case (i). The pole numbers of x and y at \(P_\infty \) are q and 2q/p, respectively. Since the curve is \(\mathbb {F}_{q^2}\)-maximal and \(P_\infty \) is an \(\mathbb {F}_{q^2}\)-rational point, \(q+1\in H(P_\infty )\); see [7, Theorem 10.6]. Let \(d_0=0\), \(d_1=2\frac{q}{p}\), \(d_2=\frac{2}{q}\) and \(d_3=1\), and \(A_1=\{1\}\), \(A_2=\{2,p\}\), \(A_3=\{2\frac{q}{p},q,q+1\}\). Then \(p\in S_1\) and \(q+1\in S_2\). Thus the sequence \(\{2\frac{q}{p},q,q+1\}\) is telescopic. Furthermore,
whence the claim follows by Result 20.
$$\begin{aligned} l_g(S_3)=-\frac{2q}{p}+q+(\frac{q}{p}-1)(q+1)=\frac{q^2}{p}-\frac{q}{p}-1, \end{aligned}$$
Case (ii). From Equation (II), \(\left[ \mathbb {F}_{q^2}(\bar{\mathcal {H}}_q):\mathbb {F}_{q^2}(x)\right] =\frac{q}{p}\) and \(\left[ \mathbb {F}_{q^2}(\bar{\mathcal {H}}_q):\mathbb {F}_{q^2}(y)\right] =\frac{q-1}{d}.\) Therefore, \(\frac{q}{p}\) and \(\frac{q-1}{d}\) are non-gaps at \(P_{\infty }\).
Case (iii). The above argument applied to the curve \(\bar{\mathcal {H}}_q\) of Equation (III), shows that \(\frac{q-1}{d}\) and \(\frac{q}{p}\) are non-gaps of \(G_r\) at \(P_{\infty }\). \(\square \)
Let C denotes any \(\mathbb {F}_{q^2}\)-maximal curve equipped with an \(\mathbb {F}_{q^2}\)-rational point P. Let D be a set of \(\mathbb {F}_{q^2}\)-rational points of C other than P. From previous work by Janwa [10] and Garcìa-Kim-Lax [3], if the divisor G is taken as multiple of P then knowledge of the gaps at \(P_\infty \) may allow one to show that the minimum distance of the resulting evaluation code \(C_L(G,D)\) or differential code \(C_\Omega (G,D)\) may be better than the designed minimum distance of that code. In particular, it is shown in [4] that t consecutive gaps at P (under some conditions on the order sequence at P) gives a minimum distance d of the code at least t greater than the designed minimum distance. This motivates to investigate large intervals of gaps at the point \(P_\infty \) of the \(\mathbb {F}_{q^2}\)-maximal curves considered in the present paper. Here we limit ourselves to show a couple of experimental results. We use Janwa’s result as stated in [4, Theorem 2] together with [4, Theorem 3] for the zero divisor \(B=0\).
Example 1
Take the curve of Eq. (1) for \(\mathcal {C}\), and let \(p=7,d=5,h=2\). Then \(d\mid (q+1)=7^2+1\). From Proposition 19, the non-gaps at \(P_{\infty }\) are \(q/p=7\) and \((q+1)/d=10\). The gap sequence at \(P_\infty \) is \(1,2,3,4,5,6,8,9,11,12,13,15,16,18,19,22,23,25,26,29,32,33,36,39,43,46,53.\) Each of the integers \(11=\gamma -2=\gamma -t\), \(12=\gamma -1\) and \(13=\gamma \) is a gap at \(P_{\infty }\). From [4, Theorem 3], the minimum distance of the code \(C_L(\gamma P_\infty ,D)\) is at least \(d^*=|D|-\gamma +t+1=5037\) whereas the designed minimum distance is \(d'=|D|-\gamma =5034\), so we have an \([n,k,d^*]=[5047, 3, 5037]-code\).
Example 2
Take the curve of Eq. (1) for \(\mathcal {C}\), and let \(p=5,d=3,h=3\). Then \(d\mid (q+1)=5^3+1\). From Proposition 19, the non-gaps at \(P_{\infty }\) are \(q/p=25\) and \((q+1)/d=42\). The gap sequence at \(P_\infty \) is \(1,\ldots ,24,26,\ldots ,41,43,\ldots ,66,68,\ldots \ldots ,920,922,\ldots ,962,964,\ldots ,981,983.\) Each of the integers \(1022=\alpha ,\ldots ,1030=\alpha +8=\alpha +t\) and \(1072=\beta ,\ldots ,1063=\beta -t=\beta -(t-1)\) is a gap at \(P_{\infty }\). From [4, Theorem 4], the minimum distance of the differential code \(C_\Omega (\gamma P_\infty ,D)\) with \(\gamma =\alpha +\beta -1\) is at least \(d^*=\alpha +\beta -1-(2\mathfrak {g}-2)+(t+1)=1120\) whereas the designed minimum distance is \(d'=\alpha +\beta -1-(2\mathfrak {g}-2)=1112\), so we have an \([n,k,d^*]=[138625, 1593, 1120]-code\).
Declarations
Conflict of interest
The authors declare no conflict of interest.
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