Rational Points on Cubic Surfaces and AG Codes from the Norm-Trace Curve

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1 Preliminaries

h Let q = p , where p is a prime and h> 0 an integer, and denote with Fq the ﬁnite ﬁeld with q elements. Fqr Fqr F r We recall that the norm NFq and the trace TFq , where r is a positive integer, are functions from q to Fq such that r − F r q 1 r−1 r−2 q q−1 q +q +···+q+1 NFq (x)= x = x and F r−1 r−2 qr q q q TF (x)= x + x + + x + x. q ··· When q and r can be derived unequivocally from the context, we will omit the subscripts.

1.1 The Norm-Trace curve

2 The Norm–Trace curve r is the curve deﬁned over the aﬃne plane A (Fqr ) by the equation N N(x)=T(y). (1)

If r = 2 the curve r is smooth, while if r 3 it can be easily seen that r has a singular point which is N ≥ 2r−1 N the point at the infinity P∞. It is then well-known that r has q affine places and a single place at the N infinity; in fact, there is exactly one place centered at each affine point of r (these are all smooth points) and the point at the infinity is either smooth (in the case r = 2) and henceN center of exactly one place, or it

2 is singular and center of only one branch of the curve (if r 3), so that r has a unique place at the inﬁnity also in this case. ≥ N If r = 2, r coincides with the Hermitian curve, and this is the only case in which r is smooth, since, N N as noted above, for r 3 it has a singularity in P∞. ≥ Moreover it is known that its Weierstrass semigroup in the place centered at P∞ is generated by r r−1 q −1 q , q−1 , see [17]. Also, the automorphism group is determined by the following result. D E r−1 r Theorem 1.1 ( [7]). The automorphism group of r Aut( r) has order q (q 1) and is a semidirect product G ⋊ C, where N N − G = (x, y) (x, y + a) T(a)=0 { 7→ | } qr −1 q−1 ∗ C = (x, y) (bx,b y) b F r . { 7→ | ∈ q } Our main aim is the study of the planar intersection (i.e. the intersection counted without multiplicity) 3 2 between r and the cubic curves of the form y = ax + bx + cx + d, where a,b,c,d Fqr and a = 0, in the case Nr = 3. The case r = 2 and a = b = 0 was investigated in [4] and the case r∈= 2 and a =6 0 was completely investigated in [14, 28]. On the other hand, the case r = 3 and a = 0 was investigated in [8]. In [8], the authors claim the following result.

3 Theorem 1.2. The number of planar intersections between 3 and cubic curves of the form y = ax + 2 2 N bx + cx + d, where a,b,c,d Fq3 , a = 0, is bounded by q +7q +1, when the surface deﬁned over Fq3 by the equation ∈ 6

3 q 3 q2 3 2 q 2 q2 2 q q2 X0X1X2 = AX0 + A X1 + A X2 + BX0 + B X1 + B X2 + CX0 + C X1 + C X2 + E

is irreducible, for A,B,C,D,E F 3 . ∈ q Still, a proof of this result is not given in the paper. The aim of this paper is to give a detailed proof of this bound in any possible case, showing that this is not correct when such surface is a cone over an elliptic curve.

1.2 Algebraic Geometry codes In this section we recall some basic facts on AG codes. For a detailed discussion we refer to [36]. Let be a projective curve over the finite field Fq and consider the function field Fq( ) of rational X X functions defined over Fq. Denote with (Fq) the set of the Fq-rationals points of . A divisor D on can be seen as a finite sum n P ,X where the n s are integers. For a functionX f F ( ), (f) denotesX P ∈X (Fq ) P P q F ∈ X q q the divisor associated toPf. A divisor D is q-rational if it coincides with its image P ∈X (F ) nP P under F q the Frobenius map over q. P F n Given an q-rational divisor D = i=1 niPi on , its support is defined as supp(D)= Pi : ni =0 . X F { 6 } The Riemann-Roch space associatedP with D is the q-vector space

L (D)= f Fq( ) (f)+ D 0 0 . { ∈ X | ≥ }∪{ }

3 It is known that L (D) is an Fq-vector space of ﬁnite dimension ℓ(D). The exact dimension of this space can be computed using Riemann-Roch theorem. n F F Consider now the divisor D = i=1 Pi where all the Pi’s are q-rational. Let G be another q-rational divisor on such that supp(G) suppP (D)= . Consider the evaluation map X ∩ ∅ n eD : L (G) F −→ q f eD(f) = (f(P ),...,f(Pn)). 7−→ 1

The map eD is Fq-linear and it is injective if n> deg(G). The AG code CL (D, G), also called functional code, associated with the divisors D and G is deﬁned as n CL (D, G) := eD(L (G)) = (f(P ),...,f(Pn)) f L (G) F . Such a code is an [n,ℓ(G) ℓ(G D), d]q { 1 | ∈ }⊆ q − − code, where d d¯= n deg(G) and d¯ is the so-called designed minimum distance (of such code). ≥ − 2 General results on the number of rational points on cubic surfaces

In this section we will exploit well-known results on cubic surfaces over finite fields to explicitly obtain some general results on the number of rational points on a cubic surface defined over a finite field Fq. In addition to being theoretically interesting, these results will be useful laterSon in the paper; in fact, we are going to study rational points on particular cubic surfaces in order to bound the intersections between the norm-trace 3 2 curve over Fq3 and the curves of the form y = ax + bx + cx + d. Our final goal consists in obtaining a bound in the form

2 (Fq) q +7q +1 |S |≤ for such surfaces, since this has interesting impact on the weight spectrum of the induced AG codes. We will obtain this result for a speciﬁc family of cubic surfaces. However, in this section we show that the bound stated in Theorem 3.3 holds more in general, being a partial generalization of the following result due to Weil.

Theorem 2.1 ( [27], Theorem 23.1). Let be a smooth irreducible cubic surface over Fq, then the number S of points of (Fq) is exactly S 2 (Fq) = q + ηq +1 |S | where η 2, 1, 0, 1, 2, 3, 4, 5, 7 . ∈ {− − } In fact, we are going to prove the following result.

Theorem 2.2. Consider an Fq-rational cubic surface and assume it is not a cone over a smooth irreducible cubic plane curve. If satisﬁes one of the following conditionsS S (i) is absolutely irreducible with non-isolated singularities, S

4 (ii) is absolutely irreducible with only isolated singularities, one of which is an Fq-rational point, S (iii) is absolutely irreducible with only two isolated singularities, S then 2 (Fq) q +7q +1. |S |≤ For brevity, in what follows we will sometimes just write irreducible instead of absolutely irreducible, 3 since we will see all the surfaces as elements of A (Fq). Note that, if is irreducible, it cannot be a cone over a reducible plane curve, otherwise it would be reducible as well.S We will discuss separately (at the end of the section) the cases in which is a cone over a smooth absolutely irreducible cubic plane curve or is reducible, which are not comprisedS in Theorem 2.2; we assume henceforth that S is not a cone over aS smooth absolutely irreducible cubic plane curve and consider only the irreducible case. We will also make some observations on the cases in which has only three or four isolated singular points. S

2.1 irreducible with non-isolated singularities S In this subsection we analyze the case in which has non-isolated singularities. Note that, if has only a ﬁnite number of singular points, then these wouldS be necessarily isolated. Therefore, in this caseS has an inﬁnite number of singular points. S We treat separately the case in which the surface is a cone or not. The possibilities (see [24, Chapter 2, Section 4] and [33]) are the following:

(i) is a cone over a singular absolutely irreducible cubic plane curve S (ii) is not a cone. S 2.1.1 is a cone over a singular cubic curve S Proposition 2.3. Let be an irreducible cubic surface deﬁned over Fq. If is a cone over a singular absolutely irreducible cubicS plane curve, then S

2 (Fq ) (q + 2)q +1= q +2q +1. (2) |S |≤ 3 Proof. Take P the vertex of the cone and choose an arbitrary plane in P (Fq) not containing P . We can think of the cone as the union of the lines between P and a singular absolutely irreducible cubic curve lying on this plane. Hence, if the cubic curve has m Fq-rational points, the obtained cone has mq +1 Fq-rational points (see [24, Chapter 2, Section 4]). This is easy to see as Fq-rational points of the cone lie on lines through P and Fq-rational points of the curve, so we count q point for every line (all Fq-rational points except the vertex P ) and ﬁnally we add P .

5 Then, due to well-known results, see [21, Chapter 11] for details, a singular absolutely irreducible cubic 2 curve in P (Fq) can have at most q +2 Fq-rational points, and this yields the thesis.

Remark 2.4. Note that, counting points as in the previous proposition, we are actually counting projective Fq-rational points, not only the aﬃne ones, so the bound we obtain is not tight; nonetheless, it is compatible with our goal bound which is q2 +7q +1.

2.1.2 is not a cone S If is not a cone and has no isolated singularities, it turns out that Theorem 2.2 is established by results inS [24, Chapter 2, Section 4] which exploit a theorem of Weil (see also [15], [38] and [27, Chapter IV] for details). Weil’s theorem (see [24, Theorem 23]) gives a suitable bound for our purposes since it ensures that the surface has necessarily a number of Fq-rational points that is S q2 + q +1+ tq (3) for some t [ 3, 6]. ∈ − Theorem 2.5 (Weil). Let be a surface defined over a finite field Fq. If S Fq is birationally trivial, then S ⊗ 2 ∗ S(Fq) = q + qTr(ϕ )+1, | | where ϕ denotes the Frobenius endomorphism and Tr(ϕ∗) denotes the trace of ϕ in the representation of Gal(Fq/Fq) on Pic(S Fq). ⊗ As noted in [24, Chapter 2, Section 4], this theorem is useful in our case because the surface is the anti-canonical model of a degree 3 weak Del Pezzo surface. S As observed, to prove the desired result in this case we need to know that the surface has no isolated singularities. This is established from the classification of cubic surfaces (see [33] and [20, proof of Proposition 6.4]), from which it is actually known even more on the singular locus of such a surface; we remark this in the following proposition. Proposition 2.6. Let P3(K) be an irreducible cubic surface with non-isolated singularities over a field K of positive characteristic.S ⊆ Then all the singular points of lie on a double line ℓ. S Proof. We start by observing that, by a result due to Bertini and proved in positive characteristic by Nakai in [32], a general hyperplane L in P3(K) cuts on an irreducible cubic curve whose singular points are exactly the singular points of lying on the hyperplaneS L. C Since the curve is an irreducibleS cubic curve, it cannot have more than one singular double point, which means that on aC general hyperplane must lie at most one singular point of the surface (see also [9, Section 2]). This leads to conclude that the singular set of is a line ℓ, which is double since Sis a cubic irreducible S S

6 surface. Indeed, if ℓ were a line of multiplicity higher than two, then would have to be the union of three (distinct or coincidental) planes through the line ℓ (see [1]). S If there were a singular point Q of not lying on ℓ, then the plane through Q and ℓ would be a component of , which is impossible since the surfaceS is irreducible. Hence, there are no singular points of not lying onSℓ and we have the thesis. S

2.2 irreducible with isolated singularities S We investigate now the case in which has only isolated singularities. We first recall that, considering the classification of cubic surfaces in positiveS characteristic, if is an irreducible cubic surface with isolated singularities, which is not a cone over a smooth plane cubic curve,S then its singular points are double points, as it is noted in [33]. Therefore, we can exploit the following useful theorem: Theorem 2.7 ( [11]). Let P3(K) be a singular irreducible cubic surface defined on the field K. Let ¯ = (K) be the surface definedS ⊂ by over K, the algebraic closure of K. Let δ be the number of isolated doubleS S points of ¯. Then δ 4 and S is birationally equivalent (over K) to S ≤ S (i) P2(K) if δ =1, 4; (ii) a smooth Del Pezzo surface of degree 4 if δ =2; (iii) a smooth Del Pezzo surface of degree 6 if δ =3. We recall that a smooth Del Pezzo surface is a smooth projective surface whose anticanonical class is ample. X In this section, we just focus on the cases in which has one singular Fq-rational point or it has two singular points, as these are the cases comprised in TheoremS 2.2. We recall the following important fact ( [8]).

Remark 2.8. As is deﬁned over Fq, if P (Fq) is a singular point then its conjugates with respect to the Frobenius automorphismS are also singular.∈ S

2.2.1 One singular Fq-rational point

In this subsection we investigate the case in which has one singular Fq-rational point. As previously noted S (see Remark 2.8), if there is only one singular point P on , then P is necessarily Fq-rational, otherwise its conjugates would be singular points of the surface. S In case has at least one Fq-rational point (which is clearly veriﬁed when has only one singularity), S S we can use the following idea to ﬁnd the desired bound for the number of Fq-rational points of . Note that this reasoning makes use of similar arguments with respect to the secant and tangent processS to generate new rational points from old ones, see [10, 27, 34] for a deeper investigation. Consider the sheaf of Fq-rational lines through the Fq-rational point P .

7 Since P is a double point of the surface, a line ℓ through P can intersect in at most one more point, 6⊂ S S which has to be Fq-rational by construction, since we are intersecting two Fq-rational varieties. Indeed, the coordinates of the points of intersection correspond to the roots of a cubic equation with coefficients in Fq, and this equation has an Fq-rational double root, corresponding to P . For this reason, also its third root has to be Fq-rational. 2 On the one hand, a direct bound for the number of Fq-rational lines through P in the affine space is q . 3 On the other hand, an Fq-rational line is a subspace of dimension 1 of (Fq) , so it has exactly q points in affine space. Therefore, an Fq-rational line ℓ through P (i.e. ℓ (Fq )) has at most q affine points on ⊂ S ⊂ S . This means that it has, apart from P , at most other q 1 Fq-rational points on the surface. S These reasonings lead to the following bound −

2 (Fq) q + h(q 1) (4) |S |≤ − where h is the number of Fq-rational lines through P contained in . S Remark 2.9. The argument we have exploited makes possible to ﬁnd all Fq-rational points on the surface S because every Fq-rational point on the surface lies on an Fq-rational line through P . In fact, the line through P and the Fq-rational point considered is an Fq-rational line (P is indeed Fq-rational). Our aim now is to determine a bound for the value of h. For this purpose, we consider a result proved in [13]. 3 Theorem 2.10 ( [13]). Let Y P be a surface of degree d deﬁned over Fq and P Y (Fq). Then one of the following holds: ⊂ ∈

(a) Y contains a plane deﬁned over Fq,

(b) Y contains a cone over a plane curve defined over Fq with center at P , (c) l P3 l is a line such that P l Y d(d 1). |{ ⊂ | ∈ ⊂ }| ≤ − Note that this result is very useful for our purposes, since is by hypothesis absolutely irreducible and not a cone over a smooth absolutely irreducible cubic plane curve.S Moreover, by the classification of cubic surfaces in positive characteristic (see [33]), we also know that cannot contain a cone over a smooth absolutely irreducible cubic plane curve. Since we are supposingS that has only isolated singularities, it cannot even contain a cone over a non-singular curve, otherwise it wouldS have a double line. Specializing this theorem to and P , and recalling that is defined by a homogeneous equation of degree 3 with coefficients S S in Fq, we get that

h l P3 l is a line such that P l 6 ≤ |{ ⊂ | ∈ ⊂ S}|≤ that, combined with (4), gives the following result.

Proposition 2.11. If has (at least) one singular Fq-rational point then S 2 2 (Fq) q + 6(q 1) = q +6q 6. (5) |S1 |≤ − −

8 2.2.2 Two singular points We consider now the case in which has exactly two singular points. Note that, by Remark 2.8, if P S and P are the singular points of , then either P and P are both 1 2 S 1 2 Fq-rational, and in this case we already have the bound determined in the previous subsection 2.2.1, or P1 and P2 are Fq2 -rational and conjugates. We have to investigate this second case, for which turns out that we can adopt the same argument used in [8], that comes from the investigation of the possible curves obtained from the intersection between and the sheaf of planes through the line between the singularities. S

Proposition 2.12 ( [8]). If has two singular F 2 -rational conjugate points then S q 2 2 q 14q + 39 (Fq) q q. (6) − ≤ |S1 |≤ − Remark 2.13. The method used for treating this case actually does not rely on the surface having exactly S two singular points, but just on having at least two singular F 2 -rational points. Hence, this case solves also S q the cases in which the surface has three or four singular points, two of which are Fq2 -rational and conjugates. Remark 2.14. Note that, according to Theorem 2.7 and Remark 2.8, for with isolated singularities there are still two other possibilities which are not covered by the results presentedS so far (and which we have not taken into account here since they are not comprised in Theorem 2.2). These are

• having three F 3 -rational and conjugates singular points; S q • having four F 4 -rational and conjugates singular points. S q We will discuss these cases in detail in section 5, for the special class of cubic surfaces deﬁned in section 3. 2 However, a trivial bound for these cases is (Fq) 3q . |S |≤ 2.3 Missing cases In this ﬁnal part of the section, we discuss the two remaining cases not comprised in Theorem 2.2, i.e., being a cone over a smooth irreducible cubic plane curve or being reducible. We investigate these twoS cases for completeness, and for the future applications to AG codS es. If the Fq-rational surface is a cone over a smooth absolutely irreducible cubic plane curve , as seen S E in section 2.1, we have that the Fq-rational points on are mq + 1, where m is the number of Fq-rational points of . S SinceE the cubic curve in this case is a smooth irreducible cubic plane curve, by well-known results 2 in [21, Chapter 11] we know that it has at most q +2√q +1 Fq-rational points in P (Fq); this yields the following result.

Proposition 2.15. Let be an Fq-rational cubic surface. If is a cone over a smooth absolutely irreducible cubic plane curve , thenS S E 2 (Fq) (q +2√q)q +1= q +2q√q +1. (7) |S |≤

9 Finally, if is reducible, then three possible situations can happen: S (i) is the union of a non-singular quadric surface and a plane; S (ii) is the union of a quadric cone and a plane; S (iii) is the union of three planes. S In the worst possible case, which corresponds to the case of a complete reducibility of the surface in three Fq-rational planes, we obtain the trivial bound

2 (Fq) 3q . |S |≤ For the other cases this bound still holds, but it is possible to reﬁne it, for instance considering well-known results on points on quadric surfaces which can be found in [22, Chapter 15].

3 Cubic surfaces from intersections of algebraic curves

We start now to investigate the intersection over F 3 of the Norm–Trace curve with the curve defined by q N3 y = (x)= A x3 + A x2 + A x + A A 3 2 1 0 where A3 = 0 and Ai Fq3 for i =0, 1, 2, 3. As already6 recalled,∈ when we refer to a planar intersection (or simply intersection) of two curves lying 2 2 in the affine space A (Fq3 ), we mean the number of points in A (Fq3 ) lying in both curves, disregarding multiplicity (see [8]). For the remaining part of this section, we exploit the same approach used in [8] to set the problem. F F q3 q3 From now on, we will write N and T instead of NFq and TFq , respectively. Moreover, throughout the paper we will always consider the curves (resp. surfaces) in the algebraic closure Fq of Fq, even when not stated explicitly. When we will need to consider smaller fields, we will point it out in the tractation. Substituting y = (x) in the equation of , and exploiting the linearity of the trace function, we get A N3 3 2 N(x)=T(A3x )+T(A2x )+T(A1x)+T(A0) (8) q q2 Consider now a normal basis = α, α , α , for a suitable α Fq3 . We know that such a basis exists, see [26, Theorem 2.35]. The vectorB space{ isomorphism} ∈

3 ΦB : (Fq) F 3 −→ q q q2 ΦB((s0,s1,s2)) = s0α0 + s1α + s2α 3 allows us to read the norm and the trace as maps from (Fq) to Fq, considering N = N ΦB and T=T ΦB. i ◦ ◦ Let Ti := T(Aix ) and Ti := Ti ΦB, for 1 i 3, then it is readily seen thateN and Ti are homogeneouse polynomials of degree respectively◦ 3 and i in≤F ≤[x , x , x ], i =0, 1, 2, 3. e q 0 1 2 e e

10 Therefore, we can rewrite (8) as

N(x0, x1, x2)= T3(x0, x1, x2)+ T2(x0, x1, x2)+ T1(x0, x1, x2)+ E (9)

e e e e3 where E = T(A0). Equation (9) is the equation of a hypersurface of A (Fq), where Fq is the algebraic closure of Fq, and both RHS and LHS have degree 3. −1 All the considerations made so far lead to the fact that ΦB induces a correspondence between Fq3 [x] q q2 and Fq[x0, x1, x2], allowing us to substitute x with x0α + x1α + x2α . We exploit this relation to write the explicit equation of the surface deﬁned by equation (9). For simpler notations, from now on we consider the equation of the cubic curve y = (x) to be written as A 3 2 y = Ax + Bx + Cx + D, A = A3,B = A2, C = A1,D = A0. We have

q q2 q q q2 q2 q2 q T1 = C(x0α + x1α + x2α )+ C (x0α + x1α + x2α)+ C (x0α + x1α + x2α ) q2 q e = x0T(αC)+ x1T(αC )+ x2T(αC ),

q q2 2 q q q2 2 q2 q2 q 2 T2 =B(x0α + x1α + x2α ) + B (x0α + x1α + x2α) + B (x0α + x1α + x2α ) 2 2 2 2q 2 2q2 q+1 q2+1 e = x0T(Bα )+ x1T(Bα )+ x2T(Bα )+2x0x1T(Bα )+2x0x2T(Bα ) q2 +q +2x1x2T(Bα ),

q q2 3 q q q2 3 q2 q2 q 3 T3 =A(x0α + x1α + x2α ) + A (x0α + x1α + x2α) + A (x0α + x1α + x2α ) 3 3 3 3q 3 3q2 2 q+2 2 q2 +2 e = x0T(Aα )+ x1T(Aα )+ x2T(Aα )+3x0x1T(Aα )+3x0x2T(Aα ) 2 q2+2q 2 1+2q 2 1+2q2 2 q+2q2 +3x1x2T(Aα )+3x0x1T(Aα )+3x0x2T(Aα )+3x1x2T(Aα ),

q2 q q q2 q q2 N =(x0α + x1α + x2α )(x0α + x1α + x2α)(x0α + x1α + x2α ) 3 3 3 2 2 2 q+2 e = (x0 + x1 + x2)N(α) + (x0x1 + x1x2 + x2x0)T(α ) 2 2 2 2q+1 3 + (x0x2 + x1x0 + x2x1)T(α )+ x0x1x2(3N(α)+T(α )).

Hence, we are now able to rewrite (9) as

11 3 3 3 2 2 2 q+2 2 2 2 2q+1 0= (x0 + x1 + x2)N(α) (x0x1 + x1x2 + x2x0)T(α ) (x0x2 + x1x0 + x2x1)T(α ) − − − 2 x x x (3N(α)+T(α3)) + x3T(Aα3)+ x3T(Aα3q )+ x3T(Aα3q ) − 0 1 2 0 1 2 2 q+2 2 q2 +2 2 q2 +2q +3x0x1T(Aα )+3x0x2T(Aα )+3x1x2T(Aα ) (10) 2 1+2q 2 1+2q2 2 q+2q2 +3x0x1T(Aα )+3x0x2T(Aα )+3x1x2T(Aα ) 2 2 2 2q 2 2q2 q+1 q2 +1 + x0T(Bα )+ x1T(Bα )+ x2T(Bα )+2x0x1T(Bα )+2x0x2T(Bα ) q2 +q q2 q +2x1x2T(Bα )+ x0T(αC)+ x1T(αC )+ x2T(αC )+ E.

Let be the surface defined by equation (10), and observe that it is defined over the field Fq. We will S1 denote with (Fq) the set of its affine Fq-rational points. S1 2 Remark 3.1. By construction, Fq-rational points of 1 correspond to the intersections in A (Fq3 ) between 3 2 S 3 and the curve y = Ax + Bx + Cx + D. In other words, our algebraic manipulations proved that there N 3 2 3 exists x Fq3 such that N(x)=T(Ax + Bx + Cx + D) if and only if exists (x0, x1, x2) (Fq) satisfying ∈ q q2 ∈ (10) and its representation on the chosen normal basis is x = x0α + x1α + x2α . Note that equation (10) can be also rewritten as

2 2 2 0= (x α + x αq + x αq )(x αq + x αq + x α)(x αq + x α + x αq) − 0 1 2 0 1 2 0 1 2 q q2 3 q q q2 3 q2 q2 q 3 + A(x0α + x1α + x2α ) + A (x0α + x1α + x2α) + A (x0α + x1α + x2α ) q q2 2 q q q2 2 q2 q2 q 2 + B(x0α + x1α + x2α ) + B (x0α + x1α + x2α) + B (x0α + x1α + x2α ) q q2 q q q2 q2 q2 q + C(x0α + x1α + x2α )+ C (x0α + x1α + x2α)+ C (x0α + x1α + x2α )+ E.

3 Consider now the affine change of variables in A (Fq) defined by t t ψ(x0, x1, x2)= M(x0, x1, x2) = (X0,X1,X2) where M is the non-singular matrix 2 α αq αq 2 M =  αq αq α  2 αq α αq    and let be the surface obtained from through ψ (i.e. = ψ( )). S2 S1 S2 S1 Proposition 3.2. is a surface defined over F 3 , has equation S2 q 3 q 3 q2 3 2 q 2 q2 2 q q2 X0X1X2 = AX0 + A X1 + A X2 + BX0 + B X1 + B X2 + CX0 + C X1 + C X2 + E

and preserves the multiplicities of the points and of the components of 1. Moreover, the points on 2 of the q q2 S S form (β,β ,β ), β F 3 , are in bijection with the Fq-rational points on . ∈ q S1

12 Proof. These properties come from straightforward computations combined with the fact that M is a non- singular affine transformation. In order to estimate the intersection between and y = Ax3+Bx2+Cx+D, our goal is the determination N3 of an upper bound for the number of (affine) Fq-rational points of 1. The final goal of this paper is to prove the following theorem: S

Theorem 3.3. Consider the Fq-rational cubic surface 1 associated to the intersections between 3 and 3 2 S N y = Ax + Bx + Cx + D, A = 0. If 1 is absolutely irreducible and it is not a cone over an irreducible smooth plane cubic curve, then6 S 2 (Fq) q +7q +1. |S1 |≤ Remark 3.4. Note that the results in [8] prove Theorem 3.3 under the assumption A =0. Considering the discussion in section 2, to prove Theorem 3.3 we are left with some of the cases in which 1 has only three or four singular points to treat, so from now on we focus on these cases. However, we firstS treat separately the particular case B = C = 0. Studying this case is interesting since it gives explicit information on the reducibility of , depending on its coefficients and the base field. Moreover, in the case S1 char(Fq) = 3, we find explicitly the form of the singular points that 1 can have, which is not possible for the general case. S

4 irreducible with isolated singularities: case B = C =0 S1 In this section we will take in account the special case in which 1 is irreducible, has only isolated singularities, and B = C = 0. S We investigate the singular points of 2, which is equivalent to studying singular points of 1, thanks to the map ψ deﬁned in precedence. For thisS case of study, the equation of is S S2 3 q 3 q2 3 X0X1X2 = AX0 + A X1 + A X2 + E and its aﬃne singular points are the solutions to the following system:

3 q 3 q2 3 X0X1X2 = AX0 + A X1 + A X2 + E 2 X1X2 =3AX0  q 2 (11) X0X2 =3A X1 2 X X =3Aq X2  0 1 2  From this system of equations, it is immediately clear that E = 0 if and only if (0, 0, 0) is a solution. We distinguish now three diﬀerent cases, depending on the characteristic of the ﬁeld Fq.

Proposition 4.1. Let B = C = 0, and char(Fq)=3. Then the only possible singularities of 2 are the following S

13 • (0, 0, 0) if and only if E =0;

ε ε ε 3 E • q 0, 0, q2 , 0, , 0 and 1 , 0, 0 if E =0 and ε solution of X = q2 . 3 2 A − A 3 − A − A 3 6 − Proof. In this case, system (11) becomes

3 q 3 q2 3 X0X1X2 = AX0 + A X1 + A X2 + E X1X2 =0  X0X2 =0 X X =0  0 1  From the last three equations it follows that at least two among X0, X1 and X2 have to be 0. Suppose q2 3 X0 = X1 = 0, obtaining A X2 + E =0. If E = 0, the unique solution is X0 = X1 = X2 = 0, otherwise we have E X3 = 2 −Aq2 E q2 and this equation is soluble if and only if 2 is a cube in F 3 . Since char(F ) = 3, A is always a cube Aq q q 3 − in Fq3 and E is always a cube since x is a permutation of the ﬁeld. This means that the above equation is always soluble and there is one possible value for X2 ε X2 = q2 −A 3

3 ε where E = ε . Hence, we have the solution 0, 0, q2 . − A 3 Iterating the above reasoning for the other two cases, we have the thesis. It is also immediate to see that there are no singular points at the inﬁnity.

Proposition 4.2. Let B = C = 0, and char(Fq)=2. Then the only possible singularity of 2 is the point (0, 0, 0). S Proof. In this case, system (11) becomes

3 q 3 q2 3 X0X1X2 = AX0 + A X1 + A X2 + E 2 X1X2 = AX0  q 2 (12) X0X2 = A X1 q2 2 X0X1 = A X2  Substituting, it is possible to see that E must be equal to zero and that (12) is equivalent to

3 q−1 3 X0 = A X1  3 q2 −q 3 X1 = A X2 (13)  2  3 q −1 3 X0 = A X2  14 Since we are looking for solutions with coordinates in Fq, note that we have all the solutions of the form

q2 −1 q2−q (ζ3,iA 3 X2, ζ3,j A 3 X2,X2)

for X2 Fq and ζ3,i, ζ3,j cubic roots of unity. ∈ q2−1 q2−q Considering now the ﬁrst equation in (12), we have that (ζ3,iA 3 X2, ζ3,j A 3 X2,X2) is a solution diﬀerent from (0, 0, 0) if and only if

q2−1 q2−q 2 2 2 3 3 3 q 3 q 3 q 3 A A X2 = A X2 + A X2 + A X2 q2+q+1 A 3 =1.

So, we have solutions if and only if N(A) = 1. We can sum up the situation as follows:

• if N(A) = 1, then there are more than 4 solutions to the system, in fact all triples of the form

q2−1 q2−q (ζ3,iA 3 X2, ζ3,j A 3 X2,X2)

satisfy the system, for every value of X2 in Fq. However, this is not possible if the surface is irreducible (as it is in our case), due to Theorem 2.7. • if N(A) = 1, there are no solutions to the system diﬀerent from (0, 0, 0) (which is a solution if and only if E = 0).6

Considering what we have just found regarding the solutions to system (13), we conclude also that there are no singular points at the inﬁnity. In fact, the only solution to the system in our case would be the point with coordinates [0 : 0 : 0 : 0]. The same result can be obtained for the general case, we do not show the proof of this proposition since it is analogue to the one just written. This can be easily adapted doing some diﬀerent reductions in system (11) and then noticing that N(A) = 27.

Proposition 4.3. Let B = C = 0, and char(Fq) = 2, 3. The only possible singularity of 2 is the point (0, 0, 0). 6 S

5 irreducible with isolated singularities S1 We investigate now the general case irreducible and with only isolated singularities. As already noted in S1 section 2.2, if 1 is an irreducible cubic surface with isolated singularities, then its singular points are double points. S

15 As pointed out above (Theorem 3.3), we wish to ﬁnd a bound of type

q2 + ηq + µ q2 +7q +1 ≤ for the four possible cases of singularities presented in Theorem 2.7 (δ =1, 2, 3, 4). We investigate the possible cases switching from the surface 2 to the surface 1 and vice-versa, thanks to the correspondence established between the two surfaces in sSection 3. Note thatS the cases in which S1 has one singular Fq-rational point or two singular Fq2 -rational points have already been treated, more in general, in section 2. Therefore, we are left to treat only some of the cases in which the surface has three or four singular points (see Remark 2.13 and 2.14). Firstly, note that the affine singular points of correspond to the solutions of S2 3 q 3 q2 3 2 q 2 q2 2 q q2 X0X1X2 = AX0 + A X1 + A X2 + BX0 + B X1 + B X2 + CX0 + C X1 + C X2 + E 2 X1X2 =3AX0 +2BX0 + C  q 2 q q X0X2 =3A X1 +2B X1 + C q2 2 q2 q2 X0X1 =3A X2 +2B X2 + C   Remark 5.1. Regarding the singular points of 2 at the infinity, we note that we can exploit the computations S 3 3 we did while investigating the case B = C = 0. In fact, considering P (Fq) and points in P (Fq) as [X0 : X1 : X2 : X3], with X3 =0 equation of the plane at the infinity, it can be seen that singular points at the infinity of are those satisfying the system: S2 3 q 3 q2 3 X0X1X2 = AX0 + A X1 + A X2 2 X1X2 =3AX0  q 2 X0X2 =3A X1 2 X X =3Aq X2  0 1 2  which is exactly system (11) (i.e. the system considered in the case B = C =0) with E =0. Note that the fact that our system is equivalent to system (11) with E =0 does not mean that we are considering the case in which the equation of 2 has E =0, it is just a matter of algebraic computations. Therefore, considering whatS we have found in section 4 regarding the solutions to this system, we conclude that there are no points at the infinity. In fact, the only solution to the system in our case ( 2 irreducible) would be the point with coordinates [0 : 0 : 0 : 0], which is not a point of the projective space. S Recall now Remark 2.8 and another important fact ( [8]).

Remark 5.2. If a singular point of is F 6 -rational, then the corresponding singularity of will be S2 q S1 Fq2 -rational, considering the way in which we have deﬁned the correspondence between 1 and 2 in section 3. S S

16 5.1 Three singular points We investigate now the case in which has exactly three singular points P , P and P . Recalling Remark S1 1 2 3 2.8, we know that there are two possible situations that may happen: either one among P1, P2 and P3 is Fq-rational, or all of them are Fq3 -rational and conjugates. If the ﬁrst situation happens, the bound given by (5) applies. Hence, we need to investigate the second possible situation. We can make use of the results in [8] to solve this case. First of all, the following proposition, which directly follows from B´ezout’s theorem, tells us that the three points cannot be collinear. Proposition 5.3. Let be a cubic curve such that it has three double points. Then is completely reducible and splits in the productC of three lines, each passing through a pair of its singular points.C

The next step is to reduce the problem of counting Fq-rational points on 1 to the problem of counting points on a certain quadric surface. S More specifically, considering the correspondence given by ψ between Fq-rational points of 1 and Fq3 - S q q2 rational points of 2 (see Remark 3.2), we will know this number by counting points of the form (α, α , α ) on a certain quadricS surface. We first use the change of coordinates given by the following proposition in order to get a new form for the equation of the cubic surface . S1 3 Proposition 5.4 ( [8]). Let be a cubic surface over P (Fq), considered with projective coordinates [r : S 0 r1 : r2 : T ], and such that it has exactly three conjugates Fq3 -rational double points, namely P1, P2 and P3. Then is projectively equivalent to the surface having affine equation, for certain β,γ F 3 S ∈ q q q2 q q2 r0r1r2 + βr0r1 + β r1r2 + β r0r2 + γr0 + γ r1 + γ r2 =0.

Thanks to Proposition 5.4, we get a new model for the surface 1, and after using a Cremona transform we get the equation of the quadric S

2 2 : βz + βqz + βq z + γz z + γqz z + γq z z 1=0. Q 3 1 2 2 3 1 3 1 2 − which is irreducible, see [8] for details. 2 By construction, the points on are in bijection with the points on in the form (δ, δq,δq ), where S1 Q δ F 3 . ∈ q Nevertheless, it is possible to show that such points (and hence the aﬃne Fq-rational points of ) are in S1 one-to-one correspondence with the Fq-rational points of an irreducible Fq-rational quadric surface (see [8] for details). Then, by well-known results on quadric surfaces (see [22, Section 15.3]), the following bound is established: 2 (Fq) = q + ηq +1, η 0, 1, 2 . (14) |S1 | ∈{ }

17 5.2 Four singular points The last case we have to deal with is the one in which has exactly four singular points P , P , P and P . S1 1 2 3 4 Note that if one among P1, P2, P3 and P4 is Fq-rational, then we already have the desired bound given by (5). If instead the four points are two couples of conjugates Fq2 -rational points, then the bound is already established by (6). Therefore, we are left with the case in which the four points P1, P2, P3 and P4 are Fq4 -rational and conjugates. To deal with this case, we introduce a change of coordinates and exploit this to transform the equation of the cubic surface in a way that will be useful for our later computations. S1 3 Proposition 5.5. Let be a cubic surface over P (Fq), considered with projective coordinates [x : x : x : S 0 1 2 x3], and such that it has exactly four conjugates Fq4 -rational double points, namely P1, P2, P3 and P4. Then is projectively equivalent to the surface having aﬃne equation S

x0x1x2 + β0x0x1 + β1x1x2 + β2x0x2 =0.

for certain β ,β ,β F 4 . 0 1 2 ∈ q Proof. We start observing that no three points among P1, P2, P3 and P4 are collinear. In fact, suppose without loss of generality that P1, P2, P3 are collinear: then a general plane through P1, P2, P3 would meet the surface in a cubic curve with three collinear double points, which means that the only possibility is that the curve is union of a double line ℓ (the line joining the three double points) and another line r (see [11]). In this case, the whole line ℓ would be double on the surface, which is impossible since by hypothesis has only isolated singularities. S Hence, we have that no three points among P1, P2, P3 and P4 are collinear and from this we can deduce that they are not coplanar. In fact, if they were on a same plane, then the intersection of this plane with the cubic surface would be a reducible cubic curve which should contain each line connecting two distinct points among P1, P2, P3 and P4, and this is clearly impossible. Therefore, we can set a new projective frame as follows: consider the four planes deﬁned by the four possible combinations of the points, that is

• the plane π through P2, P3 and P4

• the plane ρ through P1, P2 and P3

• the plane τ through P1, P2 and P4

• the plane κ through P1, P3 and P4.

Note that, since the points P1, P2, P3 and P4 are conjugates, it follows that the four planes deﬁned above are conjugates as well, in particular

πq = κ, κq = τ, τ q = ρ, ρq = π.

18 Let U be a point that does not lie in any of the planes π,ρ,τ,κ, then we have that (P1, P2, P3, P4,U) is a projective frame. With respect to this projective frame, homogeneous coordinates for the points P1, P2, P3, P4 and U are exactly:

P1 = [1 : 0 : 0 : 0], P2 = [0 : 1 : 0 : 0], P3 = [0 : 0 : 1 : 0], P4 = [0 : 0 : 0 : 1] U = [1 : 1 : 1 : 1] and projective coordinates in this new frame are [x0 : x1 : x2 : x3], where x0 = 0 is the equation of π, x3 =0 is the equation of ρ, x2 = 0 is the equation of τ and x1 = 0 is the equation of κ. Considering this projective frame, we can write the equation of the surface as

2 2 2 2 x0x1x2 + x3(α0x0 + α1x1 + α2x2 + β0x0x1 + β1x1x2 + β2x0x2)+ x3(γ0x0 + γ1x1 + γ2x2)=0 where αi,βi,γi F 4 for i 0, 1, 2 , since, as we observed, the planes π,ρ,τ and κ are conjugates. ∈ q ∈{ } Note also that in the equation there is not the constant term, since P4 , and there are no terms in 3 ∈ S xi , i =0, 1, 2, since P1 . Moreover, since P and∈ S P are double points of the surface , the line r through them is contained in . 1 4 S S So, if we write the expression of a general point on this line, we have Pλ,µ = (λ : 0 : 0 : µ). Substituting this expression in the equation of we have S

2 2 µα0λ + µ γ0λ =0 (15) µλ(α0λ + γ0µ)=0 and this equation holds for every value of µ and λ. This yields that both α0 and γ0 must be zero. Iterating this reasoning also for the lines through P2 and P4 and through P3 and P4, we get the following equation for the surface : S

x0x1x2 + β0x0x1 + β1x1x2 + β2x0x2 = 0 (16)

We apply now Proposition 5.5 to obtain an aﬃne equation in the form (16) for , namely S1

x0x1x2 + β0x0x1 + β1x1x2 + β2x0x2 = 0 (17) where β0,β1,β2 Fq4 . Applying the∈ Cremona transformation deﬁned by 1 1 1 (x0, x1, x2) , , 7−→ x0 x1 x2

1 1 1 and calling z0 = , z1 = , z2 = , we see that 1 is mapped into a plane. To see this it is suﬃcient to x0 x1 x2 S take equation (17) and divide it by x0x1x2, obtaining

19 1 1 1 1+ β0 + β1 + β2 =0; (18) x2 x0 x1 then, applying the transformation gives as result

Π:1+ β0z2 + β1z0 + β2z1 =0. (19)

We remember that our ﬁnal goal is to bound the number of Fq-rational points of 1, that is equivalent q q2 S to counting the number of points on having form (γ,γ ,γ ), γ F 3 . Hence, we count points in this S2 ∈ q form on the plane Π since these points are in direct correspondence with the ones on 2 of the same form. q q2 S Writing down γ on the normal basis we have γ = w1α + w2α + w3α . From (19), taking w1, w2 and w3 as a set of variables over Fq, we get the following equation

q2 q q q2 q q2 β0(w1α + w2α + w3α )+ β1(w1α + w2α + w3α )+ β2(w1α + w2α + w3α)+1=0. (20) ′ Direct computations show that (20) is the equation of an Fq12 -rational plane Π :

′ q2 q q q2 q q2 Π : w1(β0α + β1α + β2α )+ w2(β0α + β1α + β2α )+ w3(β0α + β1α + β2α)+1=0. (21) ′ By construction, Fq-rational points of Π are in one-to-one correspondence with the points we are looking for on our cubic surface. Hence, we want to estimate the number of Fq-rational triples (w1, w2, w3) that are solutions to equation (21). 2 Since we are looking for Fq-rational solutions, we recall that a plane can have at most q Fq-rational points. This implies that 2 S (Fq) q (22) | 1 |≤ which is consistent with the statement of Theorem 3.3. Therefore, considering all the cases discussed in this and in the previous sections, we conclude that we have finally proved Theorem 3.3. In this final part, we will give some final remarks on the remaining cases, i.e., the case of the surface being a cone over an irreducible smooth cubic plane curve or being reducible. Note that we have already discussed these two cases in general in section 2.

Remark 5.6. If 1 is a cone over a smooth absolutely irreducible cubic plane curve, we believe that it could be possible to obtainS a better bound, in the same form as Theorem 3.3, writing explicitly the equation of such cone. Moreover, due to the variety of possible diﬀerent equations of 1, it is not even clear to us if there exist choices of parameters that realize this situation. S

Remark 5.7. The cases in which 1 is reducible appear to be rare, and it is also unclear if they happen only for particular values of q. Nevertheless,S up to now, we are not able to characterize them uniquely. However, in section 4 we studied a particular case for which we were able to obtain explicit information on the reducibility of , depending on its coefficients and the base field. S1 Problem 1. Characterize the reducibility of 1, and when 1 is a cone, using only relations on its coefficients and the base field. S S

20 6 AG codes arising from Norm–Trace curves

5 2 We already know that has N = q F 3 -rational points in A (F 3 ), and that N3 q q 2 2 3 2 LF (3q P∞)= 1, x, x , x ,y,y , xy . q3 h{ }i Considering now the evaluation map

2 q5 ev : LF (3q P∞) (F 3 ) q3 −→ q 2 3 2 f = ay + bxy + cy + dx + ex + fx + g (f(P ),...,f(PN )) 7−→ 1 2 2 L ∞ LF ∞ the associated one-point code will be C (D, 3q P ) = ev( q3 (3q P )), where the divisor D is the formal 5 sum of all the q rational aﬃne points of 3(Fq3 ). The weight of a codeword associated to the evaluation of 2 N a function f LF (3q P∞) corresponds to ∈ q3 2 3 2 w(ev(f)) = (F 3 ) (F 3 ) ay + bxy + cy + dx + ex + fx + g =0 . |N3 q | − |{N3 q ∩{ }}| Using the results obtained in the previous sections, we can give some bounds in a variety of cases.

• If a = b = d = 0 then we are in a case already studied in [8]. More speciﬁcally:

2 1. if c = 0 then we have to consider the zeros of ex + fx + g that are points of (F 3 ). N3 q (a) If e = f = g = 0 then w(ev(f)) = 0; (b) if e = f = 0 and g = 0 then w(ev(f)) = q5; 6 (c) if e = 0 and f = 0 then w(ev(f)) = q5 q2; 6 − (d) if f = 0 and f 2 4eg = 0 then w(ev(f)) = q5 q2; 6 − − (e) otherwise w(ev(f)) = q5 2q2. − 2. On the other hand, if c = 0 then we have to consider the points of 3(Fq3 ) that are zeros of cy + ex2 + fx + g. 6 N (a) If e = f = g = 0 then w(ev(f)) = q5 1; − (b) if e = f = 0 and g = 0 then w(ev(f)) = q5 q2; 6 − (c) if e = 0 and f = 0 then, applying B´ezout theorem, we have that 6 w(ev(f)) q5 (q2 + q + 1); ≥ − (d) otherwise, from what we said previously, w(ev(f)) q5 (q2 +7q + 1). ≥ − • If a = b = 0 and d = 0 then we can obtain some information from our results on intersections. 6 3 2 1. If c = 0 then we have to consider the points of (F 3 ) that are zeros of dx + ex + fx + g. N3 q

21 (a) If e = f = g = 0 then w(ev(f)) = q5 q2; − (b) if e = f = 0 and g = 0 then w(ev(f)) = q5 q2; 6 − (c) otherwise w(ev(f)) q5 3q2. ≥ − 3 2 2. If c = 0 then we have to consider the points of (F 3 ) that are zeros of cy + dx + ex + fx + g. 6 N3 q We can do this exploiting our results on Fq-rational points of the cubic surface obtained from the intersection of 3 and the cubic curve we are considering. As noted, we have different bounds according to theN different shapes the surface assumes. S1 (a) If 1 is absolutely irreducible and it is not a cone over an irreducible smooth plane cubic curve,S then w(ev(f)) q5 (q2 +7q + 1); ≥ − (b) if there exist coefficients c,d,e,f,g for which is a cone over a smooth absolutely irreducible S1 cubic plane curve, then w(ev(f)) q5 (q2 +2q√q + 1); ≥ − (c) if there exist coefficients c,d,e,f,g for which is reducible, then w(ev(f)) q5 3q2. S1 ≥ − • If a =0or b = 0 unfortunately, from our results, we are not able to deduce information on the weights. 6 6 Remark 6.1. These considerations about the weight spectrum of the code tell us that, despite the fact that the dimension of the code increases with respect to the one studied in [8], the lower weights of the weight spectrum, which are those that contribute more for the computation of the PUE (Probability of the Undetected Error), do not seem to have many variations. 2 Remark 6.2. The results on the weights that we have obtained for the code CL (D, 3q P∞) hold for a more general class of AG codes. We gave lower bounds on the weights of such a code considering some monomials 2 LF ∞ that are comprised in the considered basis of the Riemann-Roch space q3 (3q P ). The same bounds hold 2 for other codes on 3 associated to divisors D and G = kP∞ with k 3q . This follows because there exists N 2 ≥ 2 a basis of LF (kP∞) (note that LF (3q P∞) LF (kP∞) for k 3q ) that contains those monomials, q3 q3 ⊆ q3 ≥ when k 3q2. Hence, our results have impact on a vast range of codes arising from . ≥ N3 As noted above, our discussion does not cover all the possible cases since the basis of the Riemann-Roch 2 2 LF ∞ space q3 (3q P ) has also the monomials xy and y . Using our approach, it seems difficult to study the intersections of 3 with curves with terms in xy or y2, as the equation of the surface corresponding to the intersections wouldN be much more complicated (for instance it can be a quartic surface). Therefore, it still remains an open problem to determine the weight 2 LF ∞ spectrum of the code q3 (3q P ). 2 Problem 2. LF ∞ Determine the weight spectrum of the code q3 (3q P ).

Acknowledgments

The results showed in this paper are included in L. Vicino’s MSc thesis (supervised by the ﬁrst and the second author).

22 References

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