Two-point codes on Norm-Trace curves

Transcription

1 Two-point codes on Norm-Trace curves C. Munuera 1, G. C. Tizziotti 2 and F. Torres 2 1 Dept. of Applied Mathematics, University of Valladolid Avda Salamanca SN, Valladolid, Castilla, Spain 2 IMECC-UNICAMP, Cx.P. 6065, , Campinas-SP, Brasil Abstract. We determine the Weierstrass semigroup of a pair of rational points on Norm-Trace curves. We use this semigroup to improve the lower bound on the minimum distance of two-point algebraic geometry codes arising from these curves. 1 Introduction The theory of Weierstrass semigroups is an important part in the study of Algebraic Geometry (AG) codes. Its use comes from the theory of one-point codes; given a curve X and a (one point, AG) code C(D, mq) arising from X, there exist close conections between the parameters of C(D, mq) and its dual C(D, mq) with the Weierstrass semigroup H(Q) of X at Q, see for example [5]. Later these results were extended to codes and semigroups over two points. Matthews [10] proved that the Weierstrass gap set of a pair of points may be exploited to define a code with minimum distance greater than the Goppa bound. By using results obtained by Homma and Kim [6, 8], she determined the Weierstrass semigroup of a pair of any two points on a Hermitian curve and, as a consequence, she improved the lower bound on the minimum distance of codes defined by a linear combination of two points. Despite the great interest of these codes, its utility is limited by the difficulty of computing the Weierstrass semigroup of two points. In this paper, we focus our attention on Norm-Trace curves, which are a natural generalization of Hermitian curves. We determine the Weierstrass semigroup of a pair of points and use it to improve the Goppa bound on the minimum distance of the corresponding codes. The article is organized as follows. In section 2 we introduce some basic facts and definitions. In section 3 we determine the Weierstrass semigroup of a pair of points on Norm-Trace curves. By using this semigroup, in section 4 we are able to improve the lower bound on the minimum distance of two-point codes and show that these two-point codes can have better parameters than one-point codes over these curves. 2 Preliminaries 2.1 Curves and codes The construction of algebraic geometry codes is well known. Let X be a nonsingular, projective, geometrically irreducible, algebraic curve of genus g over a fi-

2 nite field F q. For a rational divisor E on X, we consider the vector spaces L(E) := {rational functions f : (f) + G 0} {0} and Ω(E) := {rational differential forms ω : (ω) + G 0} {0}. Let G and D = P P n be two divisors on X such that P i P j for i j and supp(g) supp(d) =. The algebraic geometry codes C L (D, G) (or simply C(D, G)) and C Ω (D, G) are defined by (see [11]) C(D, G) := {(f(p 1 ),..., f(p n )) ; f L(G)} C Ω (D, G) := {(res P1 (ω),..., res Pn (ω)) ; ω Ω(G D)}. C(D, G) and C Ω (D, G) are dual to each other. Their minimum distances verify d L n deg(g) and d Ω deg(g) 2g + 2 (the Goppa bound). If G = aq 0 for some rational point Q on X and D it is the sum of all the other rational points, then they are called one-point. Analogously, if G = aq 1 + bq 2 for two distinct rational points, then C(D, G) and C Ω (D, G) are called two-point codes. 2.2 Weierstrass semigroups Let Q be a rational point on X. It is well known that the set H(Q) := {n N 0 : there exists a rational function f with (f) = nq} is a semigroup, called the Weiertrass semigroup of X at Q. Its complement, G(Q) := N 0 \ H(Q) is the Weierstrass gap set of Q. It has precisely g elements. These definitions can be translated to the two-point case. For given two distinct rational points Q 1 and Q 2 on X the semigroup of N 2 0 H(Q 1, Q 2 ) = {(n, m) N 2 0 : there exists a rational function f with (f) = np 1 + mp 2 } is the Weierstrass semigroup of X at Q 1 and Q 2. The set G(Q 1, Q 2 ) := N 2 0 \ H(Q 1, Q 2 ) is called the Weierstrass gap set of the pair (Q 1, Q 2 ). It is always finite, and its cardinality depends on Q 1 and Q The Norm-Trace curve Let q be a prime power and let r 2 an integer. The curve X q,r defined over F q r by the affine equation x qr 1 q 1 = y q r 1 + y qr y is called the Norm-Trace curve. If r = 2 then it is a Hermitian curve. Note that the zeros of x qr 1 q 1 (y q r 1 +y qr y) = 0 in F 2 q are the pairs (α, β) r F2 q r such that N Fq r /F q (α) = T Fq r /F q (β), where N stands for the norm and T for the trace. The Norm-Trace curve has been studied in detail by Geil, [3]. X q,r has a single rational point at infinity, P = (0 : 1 : 0), plus q 2r 1 affine rational points. Its genus is g = (q r 1 1)( qr 1 q 1 1)/2.

3 3 The Weierstrass semigroup H(P 0,0, P ) for Norm-Trace curves The Weierstrass semigroup of X q,r at P is well known. Let P a,b denote the common zero of x a and y b, where a, b F q r. The divisors of x and y are given by (x) = P 0,0 + α P 0,α q r 1 P, (y) = (qr 1) q 1 P 0,0 (qr 1) q 1 P (1) where α runs over the roots of t qr t + 1 = 0. Using these functions and the fact that G(P ) = g, we can prove that H(P ) = q r 1, (q r 1)/(q 1), [3]. Let us study now the semigroup over two points. 3.1 The Weierstrass semigroup H(P 0,0, P ) for q = 2 In this section we restrict to the binary case, q = 2. Furthermore we shall assume r 3 (if r = 2 we get the Hermitian curve). Then the curve has genus g = (2 r 1 1) 2 and H(P ) = 2 r 1, 2 r 1. Proposition 1. Let γ = 2 r 2. The Weierstrass semigroup of X 2,r at P 0,0 is given by H(P 0,0 ) = γ, γ + 1, 2γ 1, 3γ 2,..., γ 2 γ (γ 2 1). Proof. From 1 we have = (2 r 2)P 0,0 and y ( ) 1 = (2 r 2)P 0,0. y Furthermore, given an integer m, 0 m < 2 r r , 2m+2+1 = ((2(m + 1) m)(2 r 1) (m + 1))P 0,0. y m+2 Thus γ, γ + 1, 2γ 1, 3γ 2,..., γ 2 γ ( γ 2 1) H(P 0,0). To see the equality it is enough to prove that the semigroup γ, γ + 1, 2γ 1, 3γ 2,..., γ 2 γ ( γ 2 1) has g = (2 r 1 1) 2 gaps. A simple computation shows that G(P 0,0 ) = (γ 1) + (γ 3) + (γ 5) = (2 r 1 1) 2 = g. Once the semigroups H(P 0,0 ), H(P ) are known, let us study the semigroups over two points. Given Q 1, Q 2 X 2,r (F 2 r), for α G(Q 1 ) we define β α := min{β N 0 ; (α, β) H(Q 1, Q 2 )}. It is known, [8], that {β α ; α G(Q 1 )} = G(Q 2 ). Let α 1 < α 2 <... < α g be the gap sequence at Q 1 and β 1 < β 2 <... < β g be the gap sequence at Q 2. The above equality implies that there exist a oneto-one correspondence between G(Q 1 ) and G(Q 2 ) so that β αi = β σ(i), where σ is a permutation of the set {1, 2,..., g}. We will often denote this permutation by σ(q 1, Q 2 ) and the graph of the bijective map between G(Q 1 ) and G(Q 2 ) by Γ (Q 1, Q 2 ), that is Γ (Q 1, Q 2 ) := {(α i, β σ(i) ) : i = 1, 2,..., g} = {(α i, β αi ) : i = 1, 2,..., g}.

4 Lemma 2. Let Γ be a subset of (G(Q 1 ) G(Q 2 )) H(Q 1, Q 2 ). If there exists a permutation τ of {1,..., g} such that Γ = {(α i, β τ(i) ) : i = 1,..., g}, then Γ = Γ (Q 1, Q 2 ). Proof. From the definition of σ = σ(q 1, Q 2 ) we have β τ(i) β σ(i) for all i = 1,..., g. Thus τ = σ. Theorem 3. Let us consider the two-point semigroup H(P, P 0,0 ). It holds that where a = 2 r 1. β 2(i j)a+j = (2a 1)j 2i, 1 j i a 1, and β (2(i j)+1)a+j = (2a 1) (2i + 1), 1 j i a 2, Proof. Let us prove the first statement. By the structure of G(P ) and G(P 0,0 ), for every pair (i, j) such that 1 j i a 1, it holds that 2(i j)a+j G(P ) and j(2a 1) 2i = 2(j 1)(a 1) + (2(a 1) 2i + j) G(P 0,0 ). Let us first show that 2(i j)a + j 2(i j )a + j if i i or j j (*). If this assertion is false, there exist two pairs (i, j) (i, j ) such that 2(i j)a+j = 2(i j )a+j. If i j = i j, then 2(i j)a + j = 2(i j )a + j hence (i, j) = (i, j ). Therefore i j i j. Assume i j > i j. Then i j = i j + k, for some 0 < k N and hence 2(i j + k)a + j = 2(i j )a + j so j = 2ak + j,which is a contradiction because 1 j, j a 1. Then (*) is proved. Let us prove now that j(2a 1) 2i j (2a 1) 2i if (i, j) (i, j ) (**). Suppose again that this assertion is false. Thus there exist (i, j) (i, j ) such that j(2a 1) 2i = j (2a 1) 2i. Since j(2a 1) 2i = j (2a 1) 2i, we have i i and j j. Write i = i + k, with k N, 1 k < a 1. Thus j(2a 1) = j (2a 1) + 2k hence (j j )(2a 1) = 2k < 2a 1, which contradicts j j. This proves (**). Now, for 1 j i a 1, we have 2i = (2(i j)a + j)p + (j(2a 1) 2i)P 0,0 y j and hence (2(i j)a+j, j(2a 1) 2i) H(P, P 0,0 ). Thus, if α l = 2(i j)a+j and β l = j(2a 1) 2i, for l, l = 1,..., g, we have (α l, β l ) (G(P ) G(P 0,0 )) H(P, P 0,0 ). Let τ be a permutation of {1,..., g} such that τ(l) = l. By Lemma 2, Γ = {(α l, β τ(l) : l = 1,..., g} = Γ (P, P 0,0 ) so β 2(i j)a+j = (2a 1)j 2i for 1 j i a 1. This proves the first statement. The second one is proved in the same way, by using that β (2(i j)+1)a+j = (2a 1) (2i+1) for 1 j i a 2. This Theorem allows us to compute Γ (P 0,0, P ), and hence H(P 0,0, P ) as follows. Given x = (α 1, β 1 ), y = (α 2, β 2 ) N 2 0, the least upper bound (or lub) of x and y is defined as lub(x, y) := (max{α 1, α 2 }, max{β 1, β 2 }). It is well known that for x, y H(Q 1, Q 2 ), we have lub(x, y) H(Q 1, Q 2 ) (see [8]).

5 Lemma 4. Let Q 1 and Q 2 be two distinct rational points. Then H(Q 1, Q 2 ) = {lub(x, y) : x, y Γ (Q 1, Q 2 ) (H(Q 1 ) {0}) ({0} H(Q 2 ))}. Proof. See [8] Lemma The Weierstrass semigroup H(P 0,0, P ) for any q Let us study now the general case. Set a = ((q r 1)/(q 1)) 1. We have the pole divisors ( ) ( ) x 1 = ap 0,0, = (a + 1)P 0,0, y y and for 0 m q r 2 + q r q 1, mq+q+1 = (((m + 1)q m)a (m + 1))P 0,0. y m(q 1)+q Proposition 5. The Weierstrass semigroup H(P 0,0 ) is given by H(P 0,0 ) = a, a+1, qa 1, (2q 1)a 2, (3q 2)a 3,..., ((λ+1)q λ)a (λ+1) where λ = q r 2 + q r q 1. Proof. Since all integers a, a+1, qa 1, (2q 1)a 2, (3q 2)a 3,..., ((λ+1)q λ)a (λ + 1) are elements of H(P 0,0 ), it is enough to show that this semigroup has g gaps. The proof of this fact is analogous to the case q = 2. Theorem 6. Let q be a prime power, r 3 and a = q r 1 + q r q 2 + q. Let us consider the semigroup H(P 0,0, P ). Then for every s such that 1 s q r q + 1 and every pair (i, j) with 1 j i a s and (s 1)q (s 1) i j sq (s + 1), we have β (i j)(a+1)+j = (q r 1 (i j + 1))(a + 1) jq r 1. Proof. Let us first prove that when (i, j) (i, j ) then (i j)(a + 1) + j (i j )(a+1)+j. Suppose that this assertion is false. Thus there exist (i, j) (i, j ) such that (i j)(a + 1) + j = (i j )(a + 1) + j. Since i j i j, we can write i j = i j + m, with 1 m q 1. Thus m(a + 1) + j = j, which is not possible, because 1 j a 1. Now, let us show that the numbers (q r 1 (i j + 1))(a + 1) jq r 1 are distinct for distinct pairs (i, j). Otherwise there exist (i, j) (i, j ) such that (q r 1 (i j + 1))(a + 1) jq r 1 = (q r 1 (i j +1))(a+1) j q r 1. As above we have i j i j. Write i j = i j +m, 1 m q 1. Thus (j j )q r 1 = m(a + 1) = m(q r q + 1), hence q r 1 divides m(q r q + 1), a contradiction. Finally, let us see that the numbers (q r 1 (i j + 1))(a + 1) jq r 1 are gaps at P. Otherwise, if (q r 1 (i 0 j 0 +1))(a +1) j 0 q r 1 q r 1, q r q +1 for a pair (i 0, j 0 ), then there exist positive integers α, β, such that αq r 1 + β(q r q + 1) = (q r 1 (i 0 j 0 + 1))(a + 1) j 0 q r 1

6 = (q r 1 (i 0 j 0 + 1))(q r q + 1) j 0 q r 1 and hence (α + j 0 )q r 1 = (q r 1 (i 0 j β))(q r q + 1). We have α + j 0 > 0 and i 0 j β > 0, so q r 1 divides (q r 1 (i 0 j β))(q r q + 1), a contradiction because q r 1 (i 0 j β) < q r 1 and q is a prime power. By Proposition 5, it holds that (i j)(a + 1) + j G(P 0,0 ). Then, by considering the pole divisor a+1 j y i j+1 = ((i j)(a + 1) + j)p 0,0 + ((q r 1 (i j + 1))(a + 1) jq r 1 )P we conclude that ((i j)(a + 1) + j, (q r 1 (i j + 1))(a + 1) jq r 1 ) (G(P 0,0 ) G(P )) H(P 0,0, P ). Let α 1 <... < α g and β 1 <... < β g be the gap sequences of P 0,0 and P. Let τ be the permutation of {1,..., g} such that β τ(l) = (q r 1 (i j+1))(a+1) jq r 1 if α l = (i j)(a+1)+j, l = 1,..., g. Then Γ = {(α l, β τ(l) ) : l = 1,..., g} (G(P 0,0 ) G(P )) H(P 0,0, P ), hence Γ = Γ (P 0,0, P ) by Lemma 2. Thus β (i j)(a+1)+j = (q r 1 (i j +1))(a+1) jq r 1 and the proof is done. 4 Codes on Norm-Trace curves In this section we will show how the knowledge of the Weierstrass semigroup H(P 0,0, P ) on Norm-Trace curves X q,r, can be used to improve the parameters of the corresponding codes. Our starting point is the following result, due to Matthews [10]. Theorem 7. Assume that (α 1, α 2 ) G(Q 1, Q 2 ) with α 1 1 and l(α 1 Q 1 + α 2 Q 2 ) = l((α 1 1)Q 1 + α 2 Q 2 ). Suppose (γ 1, γ 2 t 1) G(Q 1, Q 2 ), for all t, 0 t min{γ 2 1, 2g 1 (α 1 +α 2 )}. Let G = (α 1 +γ 1 1)Q 1 +(α 2 +γ 2 1)Q 2 and D = n j=1 P j, where Q 1, Q 2, P 1,..., P n are distinct F q -rational points. If the dimension of C Ω (D, G) is positive, then its minimum distance is at least deg(g) 2g + 3. This Theorem can be improved for Norm-Trace curves as follows. Theorem 8. Let us consider the code C Ω (D, G) arising from the curve X q,r, with G = (α 1 + γ 1 1)P 0,0 + (α + γ 1)P and D = n j=1 P j, where the points P 0,0, P, P 1,..., P n are rational and distinct. Suppose that a) α 1, (α 1, α) G(P 0,0, P ) and l(α 1 P 0,0 + αp ) = l(α 1 P 0,0 + (α 1)P ). b) (γ 1 t 1, γ), (γ 1 t 1, γ + 1), (γ 1 t 1, γ + qr 1 q 1 ), (γ 1, γ) G(P 0,0, P ), for all t, 0 t min{γ 1 1, 2g 1 (α 1 + α)}. Under these conditions, if the dimension of C Ω (D, G) is positive, then its minimum distance at least deg(g) 2g + 4.

7 Proof. By Theorem 7, the minimum distance of the code C = C Ω (D, G) is at least deg(g) 2g+3. Let d = deg(g) 2g+3. If there exists a codeword c C of weight d, then there exists a differential ω Ω(G D) with exactly d simple poles in the set {P 1,..., P n }. Let Q 1,..., Q d be such poles. Now deg(ω) = 2g 2 = deg(g) d+1, that is, (ω) G (Q Q d ). So (ω) = G (Q Q d )+P, where P is a F q r-rational point, with P Q i, for 1 i d. On the other hand, since l(α 1 P 0,0 + αp ) = l(α 1 P 0,0 + (α 1)P ), by the Riemann-Roch Theorem we have that l(w (α 1 P 0,0 + αp )) = l(w (α 1 P 0,0 + (α 1)P )) 1, that is, L(W (α 0,0 P 1 +(α 1)P )) L(W (α 1 P 0,0 +αp )), where W is a canonical divisor. Thus, there exists a rational function h such that (h) = (α 1)P + (α 1 + t)p 0,0 W + E, where E is a effective divisor whose support does not contain neither P 0,0 nor P, and 0 t 2g 1 (α 1 + α) because deg(h) = 0. Thus (ω) = G (Q Q d ) + P = (ω) W (α 1)P + (α 1 + t)p 0,0 + E and since G = (α 1 + γ 1 1)P 0,0 + (α + γ 1)P, there exists a rational function f such that (f) = γp (γ 1 t 1)P 0,0 P + (Q Q d ) + E. Let us show that this is impossible. To that end we shall consider two cases. Case 1. t γ 1 1. If P supp(e), then (f) = γp + (γ 1 t 1)P 0,0, contradicting the fact that (γ 1 t 1, γ) G(P 0,0, P ). If P = P, then (f) = (γ +1)P +(γ 1 t 1)P 0,0, contradicting again the fact that (γ 1 t 1, γ +1) G(P 0,0, P ). The same occurs if P = P 0,0, because then (γ 1 t, γ) H(P 0,0, P ). Finally, if P = P j for some P j {Q 1,..., Q d }, since (y) = ((q r 1)/(q 1))P 0,0 ((q r 1)/(q 1))P, we have (f.y) = (γ + qr 1 q 1 )P + (γ 1 t 1)P 0,0, contradicting (γ 1 t 1, γ + (q r 1)(q 1)) G(P 0,0, P ). Case 2. γ 1 1 < t 2g 1 (α 1 + α). A similar reasoning shows that when P supp(e) or P = P 0,0 then γ is a gap. If P = P then γ + 1 is a gap. Finally, if P = P j for some P j {Q 1,..., Q d }, then γ + (q r 1)(q 1) is a gap. Therefore d deg(g) 2g + 4. Let us remember that given a [n, k, d] code C, we define its information rate by R = k/n and its relative minimum distance by δ = d/n. These parameters allows us to compare codes of different length. Our final result states that twopoint codes from Norm-Trace curves can have better relative parameters than the corresponding one-point codes. More precisely, we have the following. Theorem 9. There are two-point codes C Ω (D, G) on X q,r of length X q,r 2, having relative parameters better than every one-point code C L (D, mq r 1 P ) on X q,r.

8 Proof. Let us consider the one-point code C L (D, mq r 1 P ), where D is the sum of all q 2r 1 affine rational points. As we know, these points are either of the form (0, α s (0) ), being the α s (0) s the roots in F q r of y qr 1 + y qr y = 0, or of the form (α j, α s (j) ), 0 j q r 2, where α is a generator of F qr and the α(j) s s are the roots in F q r of y qr 1 +y qr y = α j. Let d be the minimum distance of C L (D, mq r 1 P ). The function f = m i=1 (x αi ) L(mq r 1 P ), has mq r 1 zeros and hence it gives a codeword of weight q 2r 1 mq r 1. On the other hand, by the Goppa bound we have that d n deg(g) = q 2r 1 mq r 1. Thus we have equality, d = q 2r 1 mq r 1. On the other hand, it is easy to compute that the dimension of this code is k = mq r 1 g + 1. Now let us consider the two-point code C Ω (D, G), where D = q 2r 1 1 j=1 P j, P j P 0,0, P and G = P 0,0 + (q 2r 1 + 2g mq r 1 4)P. Its length is n = q 2r 1 1. Its dimension is easy to compute: according to the Goppa s estimates, and since deg(g) > 2g 2, deg(w G + D) 2g 2, where W is a canonical divisor, we have dim(c Ω (D, G)) = i(g D) = l(w G + D) = mq r 1 g + 1. Finally its minimum distance can be estimated by using Theorem 7. To see this note that (α 1, α) = (1, 2g 2) G(P 0,0, P ) by Theorem 6; furthermore for all m such that q r (q r 1 + q r q 2 + q) + q m < q r, we have (γ 1, γ) = (1, q 2r 1 mq r 1 1) G(P 0,0, P ), and these pairs satisfy the condition (γ 1, γ t 1) G(P 0,0, P ) for all t in the range 0 t min{γ 1, 2g 1 (α 1 + α)}. Therefore the minimum distance of C Ω (D, G) at least deg(g) 2g + 3 = q 2r 1 mq r 1. Then C Ω (D, G) has relative parameters R, δ, better than the corresponding of C L (D, mq r 1 P ). Example 1. Take q = r = 3 and let us consider the Norm-Trace curve X 3,3 of genus g = 48 over F 27. With the same notation as in Theorem 6, we have a = 12 and 1 s 4. The Weierstrass semigroups are as follows, H(P ) = 9, 13, hence G(P ) = {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 14, 15, 16, 17, 19, 20, 21, 23, 24, 25, 28, 29, 30, 32, 33, 34, 37, 38, 41, 42, 43, 46, 47, 50, 51, 55, 56, 59, 60, 64, 68, 69, 73, 77, 82, 86, 95}; H(P 0,0 ) = 12, 13, 35, 58, 81, hence G(P 0,0 ) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 27, 28, 29, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44, 45, 46, 53, 54, 55, 56, 57, 66, 67, 68, 69, 79, 80, 92}. Thus, by using Theorem 6, we get Γ (P 0,0, P ) = {(1, 95), (2, 86), (3, 77), (4, 68), (5, 59), (6, 50), (7, 41), (8, 32), (9, 23), (10, 14), (11, 5), (14, 82), (15, 73), (16, 64), (17, 55), (18, 46), (19, 37),(20, 28), (21, 19), (22, 10), (23, 1), (27, 69), (28, 60), (29, 51), (30, 42), (31, 33), (32, 24), (33, 15), (34, 6), (40, 56), (41, 47), (42, 38), (43, 29), (44, 20), (45, 11), (46, 2), (53, 43), (54, 34), (55, 25), (56, 16), (57, 7), (66, 30), (67, 21), (68, 12), (69, 3), (79, 17), (80, 8), (92, 4)}, and by Lemma 4, H(P 0,0, P ) = {lub(x, y : x, y Γ (P 0,0, P ) (H(P 0,0 ) {0}) ({0} H(P ))}. Now let us consider the codes C L (D, 180P ) and C Ω (D, P 0,0 +155P ), where D is the sum of all 243 affine rational points and D = P j, P j P 0,0, P.

9 The relative parameters of code C L (D, 180P ) are δ 1 = 63/243 and R 1 = 133/243. Using Theorems 7 and 9, the relative parameters of the two-point code C Ω (D, P 0, P ) are δ 2 63/242 and R 2 = 133/242. References 1. C. Carvalho and F. Torres: On Goppa codes and Weierstrass gaps at several points. Designs, Codes and Cryptography 35 (2005) A. Garcia, S. J. Kim and R. F. Lax: Consecutive Weierstrass gaps and minimum distance of Goppa codes. J. Pure Applied Algebra 84 (1993) O. Geil On codes from norm-trace curves. Finite Fields and Their Applications 9 (2003) J.P. Hansen and H. Stichtenoth: Group codes on certain curves with many rational points. Applicable Algebra Eng. Comm. Computing 1 (1990) T. Høholdt, J. van Lint and R. Pellikaan: Algebraic geometry codes. In V.S. Pless and W.C. Huffman (Eds.), Handbook of Coding Theory, vol. 1, Elsevier, Amsterdam, M. Homma and S.J. Kim: Goppa codes with Weierstrass pairs. J. Pure Applied Algebra 162 (2001) M. Homma: The Weierstrass semigroup of a pair of points on a curve. Archiv Math. 67 (1996) S.J. Kim: On index of the Weierstrass semigroup of a pair of points on a curve. Archiv Math. 62 (1994) G.L. Matthews: Codes from the Suzuki function field. IEEE Transactions on Information Theory 50 (2004) G.L. Matthews: Weierstrass pairs and minimum distance of Goppa codes. Designs, Codes and Cryptography 22 (2001) H. Stightenoth: Algebraic function fields and codes. Springer, Berlin This article was processed using the L A TEX macro package with LLNCS style