A new family of maximal curves over a finite field
aa r X i v : . [ m a t h . AG ] N ov A new family of maximal curves overa finite field
M. Giulietti ∗ and G. Korchm´aros ∗ November 1, 2018
Abstract
A new family of F q -maximal curves is presented and some of theirproperties are investigated. Let q be a power of a prime number p . A maximal curve defined over afinite field F q with q elements, briefly an F q -maximal curve, is a projective,geometrically irreducible, non-singular algebraic curve defined over F q whosenumber of F q -rational points attains the famous Hasse-Weil upper bound q + 1 + 2 gq where g is the genus of the curve. Maximal curves have also beeninvestigated for their applications in Coding theory. Surveys on maximalcurves are found in [11, 14, 12, 13, 36, 37], see also [10, 9, 31, 35].By a result of Serre, see Lachaud [27, Proposition 6], any non-singularcurve which is F q -covered by an F q -maximal curve is also F q -maximal.Apparently, the known maximal curves are all Galois F q -covered by one ofthe curves below, see [1, 2, 3, 4, 5, 6, 7, 8, 15, 16, 17, 18, 19, 20, 21, 22, 28, 29].(A) for every q , the Hermitian curve over F q ;(B) for every q = 2 q with q = 2 h , h ≥
1, the DLS curve (the Deligne-Lusztig curve associated with the Suzuki group) over F q ; ∗ Research supported by the Italian Ministry MURST, Strutture geometriche, combi-natoria e loro applicazioni, PRIN 2006-2007 q = 3 q with q = 3 h , h ≥
1, the DLR curve (the Deligne-Lusztig curve associated with the Ree group) over F q ;(D) for every q = p h , the GS-curve (the Garcia-Stichtenoth curve) over F q .It seems plausibile that each of the known F q -maximal curve is Galois F q -covered by exactly one of the above curves, apart from a very few possibleexceptions for small q ’s. This has been investigated so far in three specialcases: The smallest GS-curve, q = 8, is Galois F q -covered by the Hermitiancurve over F , but this does not hold for q = 27, see [16], while an unpub-lished result by Rains and Zieve states that the smallest DLR-curve, q=3, isnot Galois F -covered by the Hermitian curve over F .In this preliminary report, a new F q -maximal curve X is constructed forevery q = n . For q >
8, the relevant property of X is not being F q -coveredby any of the four curves (A),(B),(C),(D); we stress that this even holds fornon Galois F q -coverings. The case q = 8 remains open.The automorphism group Aut( X ) of X is also determined; its size turnsout to be large compared to the genus X . For curves with large automorphismgroups, see [23, 30, 33]. Throughout this paper, p is a prime, n = p h and q = n with h ≥ F n [ X ] concerning the polynomial h ( X ) = n X i =0 ( − i +1 X i ( n − . (1) Lemma 2.1. X n − X = ( X n + X ) h ( X ) , (2) and X n + X − ( X n + X ) n − n +1 = ( X n + X ) h ( X ) n +1 , (3) Proof.
A straightforward computation shows (2). Also,( X n − X ) n ( X n − X + ( X n − X ) n − n +1 ) = ( X n − X ) n +1 . (4)2ow, choose ρ ∈ F q with ρ n = − ρ and replace X by ρX . From (4),[( ρX ) n − ρX ] n [( ρX ) n − ρX + (( ρX ) n − ρX ) n − n +1 ] = [( ρX ) n − ( ρX )] n +1 . Since ρ n = ρ and ρ n = − ρ , the assertion (3) follows.In the three–dimensional projective space PG(3 , q ) over F q , consider thealgebraic curve X defined to be the complete intersection of the surface Σwith affine equation Z n − n +1 = Y h ( X ) , (5)and the Hermitian cone C with affine equation X n + X = Y n +1 . (6)Note that X is defined over F q but it is viewed as a curve over the algebraicclosure K of F q . Moreover, X has degree n + 1 and possesses a uniqueinfinite point, namely the infinite point X ∞ of the X -axis.A treatise on Hermitian surfaces over a finite field is found in [24, 32].Our aim is to prove the following theorem. Theorem 2.2. X is an F q -maximal curve. To do this, it is enough to show the following two lemmas, see [26].
Lemma 2.3.
The curve X lies on the Hermitian surface H with affine equa-tion X n + X = Y n +1 + Z n +1 . (7) Proof.
Clearly, X ∞ ∈ H . Let P = ( x, y, z ) be any affine point of X . From(5), z n +1 = y n +1 h ( x ) n +1 . On the other hand, (3) together with (6) implythat y n +1 h ( x ) n +1 = x n + x − y n +1 . This proves the assertion. Lemma 2.4.
The curve X is irreducible over K .Proof. Let Y be an irreducible component of X defined over K . Let K ( Y )be the function field of Y . Let x, y, z, t ∈ K ( Y ) be the coordinate functionsof the embedding of Y in PG(3 , K ). Since Y lies on H , x n + x − y n +1 − z n +1 = 0 . (8)3ake a non-singular affine point P = ( x P , y P , z P ) on Y , and let ξ = x − x P , η = y − y P , ζ = z − z P . From (7), ξ − ηy n P − ζ z n P = − ξ n + η n y P + η n +1 + ζ n z P + ζ n +1 , whence v P ( ξ − ηy n P − ζ z n P ) ≥ n , where, as usual, v P ( u ) with u ∈ K ( X ) \ u at P .Since the tangent plane π P to H at P has equation X − x P − y n P ( Y − y p ) − z n P ( Z − z P ) = 0 , the intersection number I ( P, Y ∩ π P ) is at least n . Therefore, if X 6 = Y ,then either deg Y = n or Y lies on π . Since the equation of π P may also bewritten as X − y n P Y − z n P Z + x n P = 0 , (9)and x n P + x P − y n +1 P − z n +1 P = 0implies that x n P + x n P − y n + n P − z n + n P = 0 , we see that the point, the so-called Frobenius image of P , ϕ ( P ) = ( x q P , y q P , z q P )also lies on π P .Now, in the former case, X splits into Y and a line. In particular, Y isdefined over F q . Now, if the above point is not defined over F q , that is P ∈ Y but P ∈ PG(3 , K ) \ PG(3 , F q ), then the point ϕ ( P ) of Y is distinctfrom P . Also, π P contains ϕ ( P ). From this, the intersection divisor of Y cutout by π has degree bigger than n ; a contradiction with deg Y = n .It remains to consider the case where Y lies on π for every non-singularaffine point P . Since the tangent planes to H at distinct points of X aredistinct, Y must be a line lying on H . But this contradicts the fact that thelines of C contain the vertex of C which is a point outside H .From [26] and Theorem 2.2, X is a non-singular curve, and the linearseries | qP + ϕ ( P ) | with P ∈ X is cut out by the planes of PG(3 , K ).4 heorem 2.5. X has genus g = ( n + 1)( n −
2) + 1 .Proof.
Every linear collineation (
X, Y, Z ) → ( X, Y, λ Z ) with λ n − n +1 = 1preserves both Σ and C . For λ = 1, the fixed points of such a collineation g λ are exactly the points of the plane π with equation Z = 0. Since π containsno tangent to X , the number of fixed points of g λ with λ = 1 is independentfrom λ and equal to n + 1.The above collineation g λ defines an automorphism of X . Let Λ be thegroup consisting of all these automorphisms. Since p ∤ | Λ | , the Hurwitz genusformula gives2 g − n − n + 1)(2¯ g −
2) + ( n + 1)( n − n ) , where ¯ g is the genus of the quotient curve Y = X / Λ. From the definitionof X and Λ, this quotient curve Y is the complete intersection of C and therational surface of equation Z = Y g ( X ). This shows that Y is birationallyequivalent to the Hermitian curve of equation X n + X = Y n +1 . Since thelatter curve has genus ( n − n ), we find that ¯ g = ( n − n ). Now, from theabove equation, 2 g − n + 1)( n −
2) whence the assertion follows. F q -coverings of the Hermitian curves We show that if q > X is not F q -covered by any of the curves(A),(B),(C),(D). Actually, this holds trivially for (B),(C),(D), as the genusof each of the latter three curves is smaller than the genus of X . Therefore,we only need to prove the following result. Proposition 3.1. If q > , then X is not F q -covered by the Hermitian curvedefined over F q .Proof. Assume on the contrary that X is F q -covered by the Hermitian curve H q over F q . Let m denote the degree of such a covering ϕ . Since H q hasgenus q ( q −
1) = n ( n − ϕ gives: n − n − ≥ m ( n + 1)( n − . This yields that m ≤ n for n > q + 1 = n + 1 F q -rational point of H q lies over an F q -rational point of X and the number of F q -rational points of5 q lying over a given F q -rational points of X is at most m . Since X hasexactly n − n + n + 1 F q -rational points, this gives: n + 1 ≤ m ( n − n + n + 1) . For this m > n , a contradiction. F q Let Aut( X ) be the F q -automorphism group of X . In terms of the associatedfunction field, Aut( X ) is the group of all automorphisms of K ( X ) which fixesevery element in the subfield F q of K .First we point out that Aut( X ) contains a subgroup isomorphic to thespecial unitary group SU(3 , n ). This requires to lift SU(3 , n ) to a collineationgroup of PG(3 , q ).If the non-degenerate Hermitian form in the three dimensional vectorspace V (3 , n ) over F n is given by X n T + XT n − Y n +1 then SU(3 , n ) isrepresented by the matrix group of order ( n + 1) n ( n −
1) generated by thefollowing matrices:For a, b ∈ F n such that a n + a − b n +1 = 0, and for k ∈ F n , k = 0, Q ( a,b ) = b n a b , H k = k − n k n −
00 0 k , W = − . The subgroup of SU(3 , n ) consisting of its scalar matrices λI , with λ ∈ F n is either trivial or has order 3 according as gcd(3 , n + 1) is either 1 or 3.From each of the above matrices a 4 × , , , Q ( a,b ) , ˜ H k , ˜ W are the 4 × , n ).By the same lifting procedure, each 3 × λI defines a4 × D λ with diagonal [ λ, λ, , λ ]. If λ ranges over the set ofall ( n − n +1)–st roots of unity, the matrices ˜ D λ form a cyclic group C n − n +1 .Obviously, ˜ D λ commutes with every matrix in T , and hence the group M generated by T and C n − n +1 is T C n − n +1 . Here, T ∩ C n − n +1 is either trivialor a subgroup of order 3, according as gcd(3 , n + 1) = 1 or gcd(3 , n + 1) = 3.In the latter case, let C ( n − n +1) / be the subgroup of C n − n +1 of index 3. Note6hat if gcd(3 , n + 1) = 3 then 9 ∤ ( n − n + 1). Therefore, M can be writtenas a direct product, namely M = ( T × C n − n +1 when gcd(3 , n + 1) = 1; T × C ( n − n +1) / when gcd(3 , n + 1) = 3 . In PG(3 , q ) equipped with homogeneous coordinates ( X, Y, Z, T ), everyregular 4 × , , ,
0, the group M can be viewed as a collineation group of PG(3 , q ). Our aim is to provethat M preserves X . This will be done in two steps. Lemma 4.1.
The group T preserves X .Proof. Let P = ( x, y, z, ∈ X . The image of P under ˜ Q ( a,b ) is ( x , y , z, x = x + b n y + a, y = y + b . From (6), x n + x = y n +11 . (10)Furthermore, if x n + x = 0, then by (2) yh ( x ) = y x n − xx n + x = y ( x n + x ) n − ( x n + x ) x n + x = y y ( n +1) n − y n +1 y n +1 = − y + y n . Since b ∈ F n , this implies that yh ( x ) = y ( y n − − y n − = ( x n + x ) n − . Therefore, if x n + x = 0, then yh ( x ) = y (( x n + x ) n − −
1) = y (cid:18) ( x n + x ) n x n + x − (cid:19) = y h ( x ) . Since x n + x = 0 only holds for finitely many of points of X , and the sameholds for x n + x = 0, this implies that ˜ Q ( a,b ) ∈ Aut( X ).Similar calculation works for ˜ H k showing that ˜ H k ∈ Aut( X ).To deal with ˜ W , homogeneous coordinates are needed. Note that (6)reads X n T + XT n = Y n +1 in homogeneous coordinates. Let P = ( x, y, z, t )be a point of X . Then the image of P is the point P ′ = ( t, − y, z, x ). Since7 n t + xt n = t n x + tx n and x n t + xt n − y n +1 = 0, we have that P ′ ∈ C . Further,if x n + xt n − = 0 and t = 0, then yh ( x ) = y x n − xt n − x n + xt n − = − y t n − tx n − t n + tx n − = − yh ( t ) . From this ˜ W ∈ Aut( X ), as x n + xt n − = 0 and t = 0 only hold for finitelymany points of X . Lemma 4.2.
The group C n − n +1 preserves X .Proof. A straightforward computation shows the assertion.Lemmas 4.1 and 4.2 have the following corollary.
Theorem 4.3.
Aut( X ) contains a subgroup M such that M ∼ = ( SU(3 , n ) × C n − n +1 when gcd(3 , n + 1) = 1;SU(3 , n ) × C ( n − n +1) / when gcd(3 , n + 1) = 3 . Actually, Aut( X ) = M when gcd(3 , n + 1) = 1, but Aut( X ) is a bit largerwhen gcd(3 , n + 1) = 3. To show this, the following bound on | Aut( X ) | willbe useful. Lemma 4.4. | Aut( X ) | ≤ ( n + 1) n ( n − n − n + 1) . Proof.
From the remark before Theorem 2.5, Aut( X ) is linear, that is, it con-sists of all linear collineations of PG(3 , K ) preserving X . Obviously, Aut( X )fixes Z ∞ , the vertex of C . Further, Aut( X ) preserves H as X lies on H , andAut( X ) is a subgroup of PGU(4 , q ), see [26, Theorem 3.7]. Also, Aut( X )must preserve the plane π of equation Z = 0, as π is the polar plane of Z ∞ under the unitary polarity arising from H . Therefore, Aut( X ) inducesa collineation group S of π preserving the Hermitian curve of π of equa-tion (6). Hence, S is isomorphic to a subgroup of PGU(3 , n ). In particular, | S | ≤ ( n + 1) n ( n − U of Aut( X ) fixing π pointwise pre-serves every line through Z ∞ . From (5), all, but finitely many, lines through Z ∞ meeting X contain each exactly n − n + 1 pairwise distinct commonpoints from X . Therefore, | U | ≤ n − n + 1. Since | Aut( X ) | = | S || U | , theassertion follows.For gcd(3 , n + 1) = 1, Theorem 4.3 together with Lemma 4.4 determineAut( X ). 8 heorem 4.5. If gcd(3 , n + 1) = 1 , then Aut( X ) ∼ = SU(3 , n ) × C n − n +1 . Inparticular, | Aut( X ) | = n ( n + 1)( n − n − n + 1) . Furthermore, Aut( X ) is defined over F q but it contains a subgroup isomorphic to SU(3 , n ) definedover F n . For gcd(3 , n + 1) = 3, we exhibit one more linear collineation preserving X . To do this choose a primitive n + 1 roots of unity in F q , say ρ , anddefine ˜ E to be the diagonal matrix[ ρ − , ρ n − n , , ρ − ] . It is straightforward to check that the associated linear collineation ofPG(3 , q ) preserves X , and that it induces on π the collineation α asso-ciated to the diagonal matrix [1 , ρ n − n +1 , π , the Hermitian curve H of equation (6) is preserved by α which also fixes every common point point of H and the line of equation Y = 0. Since α has order n + 1 but the stabiliserof three collinear points of H has order ( n + 1) / , n + 1) = 3, itturns out that α ∈ PGU(3 , n ) \ PSU(3 , n ). Therefore, the group generatedby M together with ˜ E is larger than M and, when viewed as a collineationgroup of PG(3 , q ), it preserves X . This together with Theorem 4.3 andLemma 4.4 give the following result. Theorem 4.6.
Let gcd(3 , n + 1) = 3 . Then
Aut( X ) has a normal subgroup C n − n +1 such that Aut( X ) /C n − n +1 ∼ = PGU(3 , n ) . In particular, | Aut( X ) | = n ( n +1)( n − n − n +1) . Also, Aut( X ) is defined over F q but it containsa subgroup isomorphic to SU(3 , n ) defined over F n . Furthermore, Aut( X ) has a subgroup M index such that M ∼ = SU(3 , n ) × C ( n − n +1) / . Since Aut( X ) is large, X produces plenty of quotient curves. Here we limitourselves to point out that some of these curves X have very large automor-phism groups, that is, | Aut( X ) | > g where g is the genus of X .For a divisor d of n − n + 1, the group C n − n +1 contains a subgroup C d of order d . Let X = X /C d the quotient curve of X with respect to C d .Since C d fixes exactly n + 1 points of X , and C d is tame, the Hurwitz genus9ormula gives( n + 1)( n −
2) = 2 g − d (2 g −
2) + ( d − n + 1) , whence g = 12 (cid:18) ( n + 1)( n − d − d + 2 (cid:19) . Furthermore, since C d is a normal subgroup of Aut( X ), see Theorems 4.5and 4.6, Aut( X ) /C d is a subgroup G of Aut( X ) such that | G | = n ( n + 1)( n − n − n + 1) d . Comparing | G | to g shows that if d ≥ | G | > g . F q -rational place As we observed in Section 2, X ∞ = (1 , , ,
0) is the unique infinite point of X . Our aim is to compute the Weierstrass semigroup H ( X ∞ ) of X at X ∞ .For this purpose, certain divisors on X are to consider. From Section 2, thefunction field K ( X ) of X is K ( x, y, z ) with z n − n +1 = yL ( x ) , x n + x = y n +1 .Let ( ξ ) denote the principal divisor of ξ ∈ K ( X ) , ξ = 0. Note that( x ) ∞ = ( n +1) X ∞ , ( y ) ∞ = ( n − n + n ) X ∞ , ( yh ( x )) ∞ = ( n ( n − n +1)) X ∞ , whence ( z ) ∞ = n X ∞ .A useful tool for the study of H ( X ∞ ) is the concept of a telescopic semi-group, see [25, Section 5.4]. Let ( a , . . . , a k ) be a sequence of positive integerswith greatest common divisor 1. Define d i = gcd ( a , . . . , a i ) and A i = { a /d i , . . . , a i /d i } for i = 1 , . . . , k . Let d = 0. If a i /d i belongs to the semigroup generated by A i − for i = 2 , . . . , k , then the sequence ( a , . . . , a k ) is said to be telescopic .A semigroup is called telescopic if it is generated by a telescopic sequence.Recall that the genus of a numerical semigroup Λ is defined as the size of10 \ Λ. By Proposition 5.35 in [25], the genus of a semigroup Λ generatedby a telescopic sequence ( a , . . . , a k ) is g (Λ) = 12 k X i =1 (cid:18) d i − d i − (cid:19) a i ! (11) Lemma 6.1.
The genus of the numerical semigroup generated by the threeintegers n − n + n, n , n + 1 is ( n + 1)( n − Proof.
The sequence ( n − n + n, n , n + 1) is telescopic. Then (11) applies,and the claim follows from straightforward computation. Proposition 6.2.
The Weierstrass semigroup of F at X ∞ is the subgroupgenerated by n − n + n, n , n + 1 .Proof. The numerical semigroup Λ generated by n − n + n, n , n + 1 isclearly contained in H ( X ∞ ). As g ( H ( X ∞ )) = g (Λ), the claim follows.As a corollary, we have the following result. Proposition 6.3.
The order sequence of X at X ∞ is (0 , , n − n + 1 , n + 1) . Lemma 5.34 in [25] enables us to compute a basis of the linear space L ( mX ∞ ) for every positive integer m . Lemma 6.4 (Lemma 5.34 in [25]) . If ( a , . . . , a k ) is telescopic, then for every m in the semigroup generated by a , . . . , a k there exist uniquely determinednon-negative integers j , . . . , j k such that ≤ j i < d i − d i for i = 2 , . . . , k and m = k X i =1 j i a i . Proposition 6.5.
For a positive integer m , a basis of the linear space L ( mX ∞ ) is { y j z j x j | j ( n − n + n )+ j n + j ( n +1) ≤ m, j i ≥ , j ≤ n − n, j ≤ n − } . Proof.
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