等價編碼及等價因子

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(1)Equivalent Codes and Equivalent Divisors Ming-Huei Chou.

(2) Abstract. A geometric Goppa code is associated with a divisor. Under certain conditions, we prove that two geometric Goppa codes are equivalent if and only if the associated divisors are equivalent..

(3) Contents. Chapter 1.. Introduction. 1. Chapter 2.. Basic Properties of Divisors and Weil Differentials. 3. §2.1.. Divisors. 3. §2.2.. Weil Differentials. 8. Chapter 3.. Codes. 13. §3.1.. Introduction to Codes. 13. §3.2.. Geometric Goppa Codes. 15. Chapter 4.. Main Results. 22. §4.1.. Equivalent codes. 22. §4.2.. Equality of Geometric Goppa Codes. 23. §4.3.. Equivalent Codes and Equivalent Divisors. 29. Bibliography. 32. iii.

(4) Chapter 1. Introduction. Coding theory is a related aspect of the problem how to transmit information efficiently and detect or even correct these errors. However, if we expect our codes with any meaningful structure, we must generally assume that the alphabet has some structure, which is usually taken to be that of a finite field. One of the great advantage of using a finite field as code alphabet is that we can perform vector space operations on the codewords. Such codes are called linear codes. In linear codes, geometric Goppa codes are well-known. One of the reasons for the interest in geometric Goppa codes is that for this large class of codes a good lower bound for the minimum distance is available. This paper is devoted to geometric Goppa codes which combines coding theory and algebraic geometry. Goppa used algebraic curves over finite fields to define linear codes, so called the geometric Goppa codes, which are determined by divisors. When two geometric Goppa codes are equivalent, we find their parameters are equal. And we know that two codes which have the same parameters will have the same performance. Since geometric Goppa codes are associated with divisors, naturally, we observe the relation between two codes whose divisors are equivalent. It is easy to see that eqivalent divisors imply eqivalent geometric Goppa codes. However, for any nonnegative integer n, the number of the divisor classes of degree n is finite. Hence we obtain that geometric Goppa codes can be divided into finite parts by difference of parameters. A question is raised as to whether codes in an equivalence class come from equivalent divisors? We have a clue from the following. If we have two equivalent codes then there exists a divisor which is equivalent to a divisor of a code and constructs a code equal to the other code. Hence we consider equality of geometric Goppa codes first. From Chao-Ping, Xing [4] we know under certain condition the same. 1.

(5) 1. Introduction. 2. codes must come from the same divisors. Moreover Carlos Munuera and Ruud Pellikaan [1] show that under certain condition, two equal codes have to come from two equivalent divisors. In this paper, we use a similar approach as Carlos Munuera and Ruud Pellikaan [1] to obtain under certain conditions equivalent codes must come from equivalent divisors. We provide two counterexamples, which show the restriction of the conditions is sharp. In chapter 2 we recall some important definitions and properties about divisors and Weil differentials. In chapter 3 we will introduce the linear codes and geometric Goppa codes. Finally, we will show under certain condition two codes are equivalent if and only if two divisors are equivalent in chapter 4..

(6) Chapter 2. Basic Properties of Divisors and Weil Differentials. Before introducing algebraic geometric codes, we recall some important definitions and properties in this chapter. Some theorems or lemmas would not be proven in this chapter, one can find the details in [2].. 2.1. Divisors Let Fq denote the finite field with q elements. From now on, F/Fq will always denote an algebraic function field of genus g. Definition 2.1.1. A place P of the function field F/Fq is the maximal ideal of some valuation ring O of F/Fq . The set of all places of F/Fq denote by PF . If O is a valuation ring of F/Fq and P is its maximal ideal, then O is uniquely determined by P . OP is called the valuation ring of the place P . Note that in a function field F/Fq a place P is correspondent with a discrete valuation vP : F → Z ∪ {∞}. Definition 2.1.2. Let OP be the valuation ring of a place P . 3.

(7) 2.1. Divisors. 4. 1. FP := OP /P is the residue class field of P . The natural map φ : OP → FP by setting x → 7 x + P is called the residue class map with respect to P . We always denote x(P ) := x + P . 2. deg P := [FP : Fq ] is called the degree of P . Next we introduce the Weak Approximation Theorem. Theorem 2.1.3 (Weak Approximation Theorem). Let F/Fq be a function field, P1 , . . . , Pn pairwise distinct places of F/Fq , x1 , . . . , xn ∈ F and r1 , . . . , rn ∈ Z. Then there is some x ∈ F such that vPi (x − xi ) = ri ,for i = 1, . . . , n. In the following we will focus on divisors. Definition 2.1.4. The free abelian group which is generated by the places of F/Fq is denoted by DF , the divisor group of F/Fq . The elements of DF are called divisors of F/Fq . In other words, for D ∈ DF X D= nP P with nP ∈ Z almost all nP = 0. P ∈PF. The support of D is defined by suppD := {P ∈ PF | nP 6= 0}. Two divisors D = P and D0 = P ∈PF n0P P are added coefficientwise X D + D0 = (nP + n0P )P.. P. P ∈PF. nP P. P ∈PF. P. For Q ∈ PF and D = P ∈PF nP P we define vQ (D) := nQ . If a divisor D with vP (D) = 0 for all P ∈ PF then we call such D the zero divisor. A partial ordering on DF is defined by D1 ≥ D2 ⇐⇒ vP (D1 ) ≥ vP (D2 ) for any P ∈ PF . A divisor D ≥ 0 is called effective(or positive). The degree of a divisor is defined by X deg D := vP (D) · deg P. P ∈PF. For any nonzero element x ∈ F it has only finitely many zeros and poles in PF . Thus we have the following definition. Definition 2.1.5. Let x ∈ F \ {0}. Then we define the principle divisor of x X (x) = vP (x)P. P ∈PF.

(8) 2.1. Divisors. 5. Note that by the definition of principal divisor we have vP ((x)) = vP (x). By the definition of discrete valuation, for x ∈ F \ {0} we have that x ∈ Fq ⇐⇒ (x) = 0. In the following we derive equivalent divisors and their properties. Definition 2.1.6. Two divisors D, D0 ∈ DF are said to be equivalent, written D ∼ D0 , if D = D0 + (x) for some x ∈ F \{0}. This is easily verified to be an equivalent relation. Moreover, one can find that if D ∼ D0 then deg D = deg D0 . Definition 2.1.7. For a divisor D ∈ DF , we set L(D) := {x ∈ F \ {0} | (x) + A ≥ 0} ∪ {0}. It is easy to check that L(D) is a vector space over Fq . In this paper, we shall consider various Fq -vector spaces. In particular, we denote dim D := dim L(D) for D ∈ DF . Lemma 2.1.8. Let D ∈ DF . If D0 is a divisor equivalent to D then L(D) ∼ = L(D0 ) (as vector spaces over Fq .) Proof. Since D ∼ D0 , D = D0 + (x) for some 0 6= x ∈ F . Consider the map φ : L(D) → F by φ(z) = zx. It is easy to check that φ is a Fq -linear and Imφ ⊆ L(D0 ). Similarly, take φ0 : L(D0 ) → F by φ0 (z) = zx−1 and it is also Fq -linear from L(D0 ) to L(D). For z ∈ L(D) φ0 (φ(z)) = φ0 (zx) = (zx)x−1 = z. Therefore φ0 ◦ φ = 1L(D) . Likewise, z ∈ L(D0 ), φ(φ0 (z)) = φ(zx−1 ) = zx−1 x = z implies that φ ◦ φ0 = 1L(D0 ) . Thus φ is an isomorphism between L(D0 ) and L(D). Naturally, we want to calculate dim D. First we consider the case of deg D < 0. Lemma 2.1.9. Let D ∈ DF . If deg D < 0, then dim D = 0. Proof. Suppose dim D > 0. Then there exists 0 6= x ∈ L(D). Since D0 = (x) + D ≥ 0, it implies D0 ∼ D. Hence deg D = deg D0 ≥ 0, which gives a contradiction. Next, we will consider the situation of deg D > 2g − 2. Riemann-Roch Theorem tells us how to calculate the dimension of a divisor when deg D > 2g − 2..

(9) 2.1. Divisors. 6. Theorem 2.1.10 (Riemann-Roch Theorem). There exists a divisor W such that for any D ∈ DF dim D = deg D + 1 − g + dim(W − D). When we introduce Weil differentials, we will know the divisor W is a canonical divisor. We can get some properties of canonical divisor by Riemann-Roch Theorem. Corollary 2.1.11. Let W be a canonical divisor then we have dim W = g and deg W = 2g − 2. Proof. By Riemann-Roch Theorem, we have dim D = deg D + 1 − g + dim(W − D) for any divisor D. First, setting D = 0 we obtain 1 = dim 0 = deg 0 + 1 − g + dim(W − 0). Hence dim W = g. Take D = W , we can get g = dim W = deg W + 1 − g + dim(W − W ) = deg W + 2 − g. Thus deg W = 2g − 2.. . Using Corollary 2.1.11, we can calculate dim D when deg D > 2g − 2. Corollary 2.1.12. If D is a divisor of F/Fq with deg D > 2g − 2, then dim D = deg D + 1 − g. Proof. We have dim D = deg D + 1 − g + dim(W − D) by Riemann-Roch Theorem. Since deg W = 2g − 2 and deg D > 2g − 2 which imply deg(W − D) < 0. From Lemma 2.1.9, we obtain dim(W − D) = 0. Finally, we want to calculate dim D when 0 ≤ deg D ≤ 2g − 2. Although we can not calculate dim D by deg D. But Clifford’s Theorem gives a bound by deg D. Theorem 2.1.13 (Clifford’s Theorem). For any divisor D ∈ DF with 0 ≤ deg D ≤ 2g − 2 holds deg D . dim D ≤ 1 + 2 In some case, deg G > deg H implies dim G > dim H. Lemma 2.1.14. Let G and H be divisors of F/Fq with deg G > deg H. If deg G > 2g − 1 then dim G > dim H..

(10) 2.1. Divisors. 7. Proof. Since deg G > 2g − 1, dim G = deg G + 1 − g > g. If deg H < 0, then by Lemma 2.1.9 we have dim H = 0 ≤ g < dim G. If 0 ≤ deg H ≤ 2g − 2, then dim H ≤ 1 +. deg H ≤ g < dim G 2. from Clifford’s treorem. If deg H > 2g − 2, then dim H = deg H + 1 − g < deg G + 1 − g = dim G from Riemann-Roch theorem.. . We naturally have the following definition. Definition 2.1.15. For G, H ∈ DF with G =. P. vP (G)P , H =. P. G∩H =. X. P. vP (H)P , define. P. min{vP (G), vP (H)}P.. P. Let G and H be two divisors of F/Fq . From the definition, we can immediately get deg(G ∩ H) ≤ deg G and deg(G ∩ H) ≤ deg H. Moreover, suppose deg G = deg H, then G 6= H if and only if deg(G ∩ H) < deg G. Lemma 2.1.16. Let D, G and H be divisors of F/Fq . Then we have (D − G) ∩ (D − H) = D − G − H + (G ∩ H) = (D − G) + (D − H) − D + (G ∩ H). Proof. For P ∈ PF vP ((D − G) ∩ (D − H)) = min{vP (D − G), vP (D − H)} = vP (D) − max{vP (G), vP (H)} = vP (D) − vP (G) − vP (H) + vP (G ∩ H). Hence (D − G) ∩ (D − H) = D − G − H + (G ∩ H) = (D − G) + (D − H) − D + (G ∩ H). Next we want to show that the relation among L(G ∩ H), L(G) and L(H). Of course, one may interest in dim(G ∩ H). Obviously, G ∩ H ≤ G implies dim(G ∩ H) ≤ dim G . Moreover, in some case we have dim(G ∩ H) < dim G..

(11) 2.2. Weil Differentials. 8. Lemma 2.1.17. Let G and H be two divisors of F/Fq . 1. L(G) ∩ L(H) = L(G ∩ H). 2. Suppose deg(G ∩ H) < deg G. If deg G > 2g − 1, then dim G > dim(G ∩ H). Proof. 1. The inclusion L(G ∩ H) ⊆ L(G) ∩ L(H) follows from the inequalities G ∩ H ≤ G and G ∩ H ≤ H. Conversely, if f ∈ L(G) ∩ L(H), then vP (f ) ≥ −vP (G) and vP (f ) ≥ −vP (H). Therefore vP (f ) ≥ max{−vP (G), −vP (H)} = − min{vP (G), vP (H)} = −vP (G ∩ H) for all places P ∈ PF . 2. Since deg G > 2g − 1 and deg G > deg(G ∩ H), we have dim G > dim(G ∩ H) by Lemma 2.1.14. Note that if G 6= H then deg G > deg(G ∩ H).. 2.2. Weil Differentials In this section, we introduce the concept of Weil differentials which has close relation to genus. First, we introduce the notion of an adele. Definition 2.2.1. An adele of F/Fq is a mapping α : PF P. −→ F, 7−→ αP ,. such that αP ∈ OP for almost all P ∈ PF . (αP is the P -component of an adele.) We Q regard an adele as an element of the direct product P ∈PF F and therefore use the notation α = (αP )P ∈PF . The set AF := {α | α is an adele of F/Fq } is called the adele space of F/Fq ..

(12) 2.2. Weil Differentials. 9. It is easy to check that AF is a vector space over Fq . Given an element x in F we introduce two definitions associated with x. Definition 2.2.2. For x ∈ F the principal adele of x is defined by x ˜ = (˜ xP )P where x ˜P = x for all P ∈ PF . For P ∈ PF define P -component x by ( ip (x) = (αQ )Q with αQ =. x Q = P, 0 otherwise.. This gives an embedding from F to AF . The valuations vP of F/Fq extend naturally to AF by setting vP (α) := vP (αP ). By definition, vP (α) ≥ 0 for almost all P ∈ PF . Definition 2.2.3. For D ∈ DF we define AF (D) := {α ∈ AF | vP (α) ≥ −vP (D) for all P ∈ PF }. It is easy to check that AF (D) is a Fq -subspace of AF . Hence we are interested in their dimensions. Althought the vector spaces AF , AF (D) and F over Fq are infinite-dimensional, we know the quotient space AF /(AF (D) + F ) is finite dimension over Fq . In fact, we know that dim(AF /(AF (D) + F )) = dim D − deg D + g − 1. Next we use adele to introduce the concept of Weil differentials which will lead to a special divisor. Definition 2.2.4. A Weil differential of F/Fq is a Fq -linear map w : AF → Fq vanishing on AF (D) + F for some divisor D ∈ DF (Note that F means princial adele). We call ΩF := {w | w is a Weil differential of F/Fq } the module of Weil differentials of F/Fq . For A ∈ DF let ΩF (D) := {w ∈ ΩF | vanishes on AF (D) + F }. Note for x ∈ F w(x) := w(˜ x). Hence if x ∈ F then w(x) = 0. We regard ΩF as a Fq -vector space, clearly ΩF (D) is a subspace of ΩF . Hence we also want to know the dimension of ΩF (D). Lemma 2.2.5. For D ∈ DF we have dim ΩF (D) = dim D − deg D + g − 1. We have known that the dimension of ΩF (D) over Fq . Next we want ΩF (D) to be a vector space over F ..

(13) 2.2. Weil Differentials. 10. Definition 2.2.6. For x ∈ F and w ∈ ΩF define x · w : AF → Fq by (x · w)(α) := w(xα). It is easily checked that x · w is also a Weil differential of F/Fq . In fact, if w vanishes on AF (D) + F then x · w vanishes on AF (D + (x)) + F . From definition 2.2.6 we find ΩF (D) indeed a vector space over F . In fact, one can verify that dimF ΩF = 1. Next we will use Weil differential to define a special divisor. Definition 2.2.7. Given a non-zero Weil differential w. 1. We call the divisor (w) canonical divisor of F/Fq . The divisor (w) is the uniquely determined divisor of F/Fq satisfying (a) w vanishes on AF ((w)) + F . (b) If w vanishes on AF (D) + F then D ≤ (w). 2. For P ∈ PF we define vP (w) := vP ((w)). In fact, the canonical divisor is the divisor W which is mentioned in Riemann-Roch Theorem. It follows immediately from the definitions that ΩF (D) = {w ∈ ΩF | w = 0 or (w) ≥ D}. Lemma 2.2.8. For 0 6= x ∈ F and 0 6= w ∈ ΩF we have 1. (x · w) = (x) + (w). 2. vP (x · w) = vP (x) + vP (w) for all P ∈ PF . Proof. 1. If w vanishes on AF (D) + F , then x · w vanishes on AF (D + (x)) + F . Hence w vanishes on AF ((w)) + F and x · w vanishes on AF ((w) + (x)) + F . From the definition of canonical divisor (x · w), we have (x) + (w) ≤ (x · w). Likewise (x · w) + (x−1 ) ≤ (x−1 · (x · w)) = ((x−1 x) · w) = (w). Combining these inequalities we obtain (x) + (w) ≤ (x · w) ≤ −(x−1 ) + (w) = (x) + (w). Hence (x) + (w) = (x · w). 2. By (1) we know (x) + (w) = (x · w). Hence vP (x · w) = vP (x) + vP (w) for all P ∈ PF ..

(14) 2.2. Weil Differentials. 11. Let D be an arbitrary divisor and W = (w) a canonical divisor of F/Fq . We can construct a map from L(W − D) to ΩF (D) by setting x 7→ x · w. One can check that it is an isomorphism of Fq -vector spaces. Hence dim(W − D) = dim ΩF (D). From Lemma 2.2.5 we get Riemann-Roch Theorem immediately. Next we will introduce the local components of a Weil differential. Definition 2.2.9. For a Weil differential w ∈ ΩF define its local component wP by wP (x) := w(iP (x)) where iP (x) is the P −component x. It is easy to check that wP is Fq -linear. Lemma 2.2.10. Let w ∈ ΩF and α = (αP ) ∈ AF . Then X w(α) = wP (αP ). P ∈PF. The following Lemma will show the relation between Weil differential and valuation. Lemma 2.2.11. Let w 6= 0 be a Weil differential of F/Fq and P ∈ PF . Then for an integer r∈Z vP (w) ≥ r ⇐⇒ wP (x) = 0 for all x ∈ F with vP (x) ≥ −r. Lemma 2.2.12. Let P ∈ PF be a place of degree one and w be a Weil differential with vP (w) ≥ −1. Then we have 1. If vP (x) ≥ 0 then wP (x) = x(P ) · wP (1). 2. wP (1) = 0 if and only if vP (w) ≥ 0. Proof. 1. Let x ∈ F with vP (x) ≥ 0, that is, x ∈ OP . We can write x = x(P ) + z with vP (z) > 0. By assumption vP (w) ≥ −1 and vP (z) ≥ 1. Therefore from Lemma 2.2.11 we have wP (z) = 0. Hence we obtain wP (x) = wP (x(P ) + z) = wP (x(P )) + wp (z) = wP (x(P )) = x(P ) · wP (1)..

(15) 2.2. Weil Differentials. 12. 2. Suppose vP (w) ≥ 0. we have vP (1) = 0. By Lemma 2.2.11, we get wP (1) = 0. Conversely, suppose that wP (1) = 0. We claim that for all x ∈ F with vP (x) ≥ 0 then wP (x) = 0. Let x ∈ F with vP (x) ≥ 0. It follows from (1), we obtain wP (x) = x(P ) · wP (1) = 0. Therefore for all x ∈ F with vP (x) ≥ 0 such that wP (x) = 0. Then by Lemma 2.2.11 we have wP (1) = 0. By the Weak Approximation Theorem we can create a special Weil differential. Lemma 2.2.13. Let P1 , . . . , Pn be pairwise distinct places of F/Fq of degree 1. There exists a Weil differential η such that vPi (η) = −1 and ηPi (1) = 1 for i = 1, . . . , n. Proof. Choose an arbitrary Weil differential w 6= 0. Since we want vPi (η) = −1. By the Weak Approximation theorem, there is an element z ∈ F with vPi (z) = −vPi (w) − 1 for i = 1, . . . , n. Setting w0 = z · w we obtain vPi (w0 ) = vPi (z · w) = vPi (z) + vPi (w) = −1 ∀i = 1, . . . , n, by Lemma 2.2.8. Since vPi (w0 ) = −1 and vPi (1) = 0, by Lemma 2.2.12 we have wP0 i (1) 6= 0. Take ai = wP0 i (1) 6= 0 ∈ Fq from the definition of Weil differential. Again by the Weak Approximation theorem we can find y ∈ F such that vPi (y − ai ) > 0. Hence we have vPi (y) = vPi ((y − ai ) + ai ) = 0 by ai ∈ Fq . However y(Pi ) = (y − ai )(Pi ) + ai (Pi ) = ai . We put η = y −1 · w0 . Therefore we obtain vPi (η) = vPi (y −1 · w0 ) = vPi (y −1 ) + vPi (w0 ) = vPi (w0 ) = −1, by vPi (y) = 0. Since vPi (y −1 ) = 0 and vPi (w0 ) = −1 we can write wP0 i (y −1 ) = y −1 (Pi )·wP0 i (1) by Lemma 2.2.11. Hence we get ηPi (1) = (y −1 · wP0 i )(1) = wP0 i (y −1 ) = y −1 (Pi ) · wP0 i (1) = a−1 i · ai = 1. Since we concentrate on codes, we consider the function field F/Fq which is over finite field Fq . In fact, what we mention in this section all work for function field F/K of one variable such that K is full constant field of F/K..

(16) Chapter 3. Codes. 3.1. Introduction to Codes In this paper, we focus on geometric Goppa codes. First, we start with a brief survey of the concept of coding theory, and only deal with the situation of finite field. Let Fq denote the finite field with q elements. The set Fnq = Fq × · · · × Fq (n-tuples) is equipped with the Hamming metric. Definition 3.1.1. For a = (a1 , . . . , an ), b = (b1 , . . . , bn ) ∈ Fnq , we define the Hamming distance from a to b d(a, b) := #{i | ai 6= bi }. It is easy to check that the Hamming distance is a metric on Fnq . Definition 3.1.2. The weight of an element a ∈ Fnq is defined as w(a) := d(a, 0) = #{i | ai 6= 0}. Since d(a, b) = d(a − b, 0), we have d(a, b) = w(a − b). Definition 3.1.3. A linear code C (over the alphabet Fq ) is a vector subspace of Fnq ; the elements of C are called codewords. We call n the length of C and dimFq C the dimension of C. An [n, k] code is a linear code of length n and dimension k. From the following examples, we know that two different codes have the same length n and dimension k. Let C1 = {0000, 0001, 0010, 0011}, C2 = {0000, 0100, 1000, 1100} ⊆ F42 . 13.

(17) 3.1. Introduction to Codes. 14. We can know that C1 and C2 are [4, 2] codes, but C1 6= C2 . Hence we call n and k the parameters of an [n, k] code. Except length n and dimension k, minimum distance d of linear code is also an important parameter. When data transmit, d tells us how many errors can be corrected. Definition 3.1.4. The minimum distance of a code C is d := d(C) = min{d(a, b) | a, b ∈ C and a 6= b}. An [n, k] code with minimum distance d will be referred to as an [n, k, d] code. Suppose C is a linear code. According to the definition of minimum distance, if d(C) = d then there exist x, y ∈ C such that d(x, y) = d. However x − y ∈ C follows from C is linear and we have d = d(x, y) = d(x − y, 0) = w(x − y). Hence by definition we have d = min{w(c) | c ∈ C, c 6= 0}. The dimension and minimum distance of a code are important. Suppose C is an [n, k, d] code, we may think of each codeword as having k information symbols and n − k checks. Thus, we want k large (with respect to n) to make our code efficient. On the other hand, the value of minimum distance determines how many errors our code can correct. We set n t := [ d−1 2 ]. Then C is called to be t-error correcing. The following is obvious: if u ∈ Fq and d(c, u) ≤ t for some c ∈ C, then c is the only codeword with d(c, u) ≤ t. So we also hope that d is large. However, there are some restrictions. The simplest bound is the following. Proposition 3.1.5 (Singleton Bound). For an [n, k, d] code C holds k + d ≤ n + 1. Proof. Consider the vector subspace W ⊆ Fnq given by W := {(a1 , . . . , an ) ∈ Fnq | ai = 0, ∀i ≥ d}. Any a ∈ W has weight ≤ d − 1, hence W ∩ C = 0. Since dim W = d − 1, we obtain k + (d − 1) = dim C + dim W = dim(C + W ) + dim(C ∩ W ) = dim(C + W ) ≤ n. Codes with k + d = n + 1 are called MDS codes (maximum distance separable codes). However, a much harder problem is to obtain lower bounds for the minimum distance of a.

(18) 3.2. Geometric Goppa Codes. 15. given code. One of the reasons for the interest in geometric Goppa codes is that for this large class of codes a good lower bound for the minimum distance is available. Now, we introduce the dual of a code in the end of this section. Definition 3.1.6. The canonical inner product on Fnq is defined by ha, bi :=. n X. ai bi. i=1. for a = (a1 , . . . , an ), b = (b1 , . . . , bn ) ∈ Fnq . Obviously, the canonical inner product is a non-degenerate symmetric bilinear form on Fnq . Suppose a = (a1 . . . , an ) ∈ Fnq and at 6= 0 for some t ∈ {1, . . . , n}. Then take b = (b1 , . . . , bn ) ∈ Fnq with bt = 1 and bi = 0, for all i 6= t. Then ha, bi = at 6= 0. Definition 3.1.7. If C ⊆ Fnq is a code then C ⊥ := {u ∈ Fnq | hu, ci = 0, ∀c ∈ C} is called the dual of C. C is called self-dual if C = C ⊥ . Linear algebra tells us that the dual of an [n, k] code is an [n, n−k] code, and (C ⊥ )⊥ = C.. 3.2. Geometric Goppa Codes In the beginning, we consider Reed Solomon codes over Fq . This important class of codes is well-known in coding theory for a long time. Geometric Goppa codes are a very natural generization of Reed Solomon codes. Definition 3.2.1. For any non-negative integer r, define Lr := {f ∈ Fq [x] | deg(f ) ≤ r − 1} ∪ {0} Note that Lr is a vector space over Fq with dim Lr = r and we can find an explicit basis {1, x, . . . , xr−1 }. Definition 3.2.2. Let n = q − 1 and fix a primitive element β ∈ Fq of the multiplicative group F∗q = Fq \ {0}. Pick an integer k with 1 ≤ k ≤ n. Then the Reed Solomon code RS(k, q) is defined to be RS(k, q) := {(f (β 1 ), f (β 2 ), . . . , f (β n )) | f ∈ Lk }.

(19) 3.2. Geometric Goppa Codes. 16. We consider the evaluation map ev : Lk → Fnq given by ev(f ) := (f (β 1 ), f (β 2 ), . . . , f (β n )) ∈ Fnq Obviously, the map is Fq -linear. Since f ∈ Lk \ {0} implies deg(f ) ≤ k − 1 < n. Thus f has less than n zeros. Hence RS(k, q) is an [n, k] code over Fq . The weight of a codeword 0 6= c = ev(f ) ∈ RS(k, q) is given by w(c) = n − #{ i ∈ {1, . . . , n} | f (β i ) = 0} ≥ n − deg f ≥ n − (k − 1) Hence the minimum distance of RS(k, q) satisfies the inequality d ≥ n + 1 − k. On the other hand, d ≤ n + 1 − k by Singleton Bound. Thus Reed Solomon codes are MDS codes over Fq . However, Reed Solomon codes are short in comparison with the size of alphabet Fq since n = q − 1. Now we introduce the notation of a geometric Goppa code. Let us fix some notation valid for the entire section. F/Fq is an algebraic function field of genus g. P1 , . . . , Pn are pairwise distinct places of F/Fq of degree 1. D = P1 + · · · + Pn . G is a divisor of F/Fq such that supp G ∩ suppD = ∅. Definition 3.2.3. The geometric Goppa code (or algebraic geometric code) CL (D, G) associated to D and G is defined by CL (D, G) := {(x(P1 ), . . . , x(Pn )) | x ∈ L(G)} ⊆ Fnq Similiar to Reed Solomon codes, the geometric Goppa code CL (D, G) is the image of the evaluation map evD : L(G) → Fnq which is given by evD (x) := (x(P1 ), . . . , x(Pn )) ∈ Fnq and the evaluation map is Fq -linear. Let F = Fq (z) be a rational function field and β be the primitive element of F∗q . Let Pi be the zero of z − β i (i = 1, . . . , n) and P∞ be the pole of z. Then we know Reed Solomon codes really are a special case of geometric Goppa codes by taking D = P1 + · · · + Pn and G = (k − 1)P∞ . The next theorem will show that why these codes are interesting: one obtains a lower bound for their minimum distance in a very general situation..

(20) 3.2. Geometric Goppa Codes. 17. Theorem 3.2.4. Suppose deg D = n. CL (D, G) is an [n, k, d] code with parameters k = dim G − dim(G − D) and d ≥ n − deg G. If, in addition, deg G < n, then k = dim G. Proof. We know the evaluation map is a surjective linear map from L(G) to CL (D, G). We claim ker(evD ) = L(G − D). Let x ∈ ker(evD ) then x(Pi ) = 0 for all i = 1, . . . , n. Therefore vPi (x) > 0 for all i = 1, . . . , n. Since supp G ∩ suppD = ∅, x ∈ L(G − D). We obtain ker(evD ) ⊆ L(G − D). Conversely, if x ∈ L(G − D) then vP (x) ≥ −vP (G − D) for all P ∈ PF . Hence vPi (x) ≥ −vPi (G) + vPi (D) = 1 for all i = 1, . . . , n. We get x(Pi ) = 0 for all i = 1, . . . , n immediately. Thus ker(evD ) = L(G − D). We can get that k = dim CL (D, G) = dim G − dim(G − D). Moreover, if deg G < n, then we can find k = dim G immediately by dim(G − D) = 0. Since the minimum distance does not make sense if CL (D, G) = 0. Without loss of generality, we assume CL (D, G) 6= 0. Choose a non-zero element x ∈ L(G) with weight w(evD (x)) = d. Then exactly n − d places Pi1 , . . . , Pin−d in suppD are zeros of x, so 0 6= x ∈ L(G − (Pi1 + · · · + Pin−d )). Since dim(G − (Pi1 + · · · + Pin−d )) > 0. Then by Lemma 2.1.9 0 ≤ deg(G − (Pi1 + · · · + Pin−d )) = deg G − n + d. Hence d ≥ n − deg G.. . Since we hope that the dimension of a code is large ( with respect to n ), we often assume that deg G < n = deg D. Hence under this assumption, we have dim CL (D, G) = dim G. Furthermore, by Riemann-Roch Theorem we can get immediately that if 2g−2 < deg G < n then dim CL (D, G) = deg G + 1 − g. In the proof of Theorem 3.2.4, we didn’t use the properties of minimum distance. One may ask whether n − deg G is the best lower bound or not. In fact, in some cases n − deg G is indeed the best lower bound of minimum distance. Lemma 3.2.5. Suppose deg D = n and CL (D, G) is an [n, k, d] code. Then d = n−deg G if and only if there exists a divisor D0 with 0 ≤ D0 ≤ D, deg D0 = deg G and dim(G −D0 ) > 0..

(21) 3.2. Geometric Goppa Codes. 18. Proof. Suppose d = n − deg G. Then there is an element x ∈ L(G) \ {0} such that weight w(evD (x)) = d. Say x(Pij ) = 0 for j = 1, . . . , deg G. Take D0 =. deg XG. Pij .. j=1. Then 0 ≤ D0 ≤ D and deg D0 = deg G. Since x ∈ L(G − D0 ), dim(G − D0 ) > 0. Conversely, By assumption we choose y ∈ L(G − D) \ {0}. Then we have weight w(evD (y)) ≤ n−deg G. Since by Theorem 3.2.4 we have d ≥ n−deg G. Hence d = n−deg G. The idea of a geometric Goppa code goes back Goppa, who actually introduced the dual code of CL (D, G) which can be described in terms of local components of Weil differentials. Definition 3.2.6. Let D = P1 + · · · + Pn where P1 , . . . , Pn are pairwise distinct places of F/Fq of degree 1 and G is a divisor of F/Fq such that supp G ∩ suppD = ∅. Then we define the linear code CΩ (D, G) ⊆ Fnq by CΩ (D, G) := {(wP1 (1), . . . , wPn (1)) | w ∈ ΩF (G − D)}. Let µD : ΩF (G − D) → Fnq by setting w 7→ (wP1 (1), . . . , wPn (1)). Then we can consider CΩ (D, G) as the image of µD . Next we will show that the bound of the parameters of CΩ (D, G). Theorem 3.2.7. Suppose deg D = n. CΩ (D, G) is an [n, k 0 , d0 ] code with parameters k 0 = dim(G − D) − dim G + deg D and d0 ≥ deg G − (2g − 2). If moreover, 2g − 2 < deg G < n then k 0 = n + g − 1 − deg G. Proof. We have that CΩ (D, G) is the image of µD . Now we claim that ker µD = ΩF (G). Let w ∈ ΩF (G − D) then (w) ≥ G − D. Moreover we have vPi (w) ≥ vPi (G − D) = −1 for all i = 1, . . . , n. If w ∈ ker µD then wPi (1) = 0. Hence by Lemma 2.2.12 we have vPi (w) ≥ 0. Thus w ∈ ΩF (G). Conversely, suppose w ∈ ΩF (G) then vPi (w) ≥ vPi (G) = 0 for all i = 1, . . . , n. We have vPi (w) ≥ vPi (G − D) = −1 by definition of w ∈ ΩF (G − D). Hence by Lemma 2.2.12, we have wPi (1) = 0. Thus ker µD = ΩF (G)..

(22) 3.2. Geometric Goppa Codes. 19. Hence k 0 = dim ΩF (G − D) − dim ΩF (G). Therefore by Lemma 2.2.5 k 0 = dim ΩF (G − D) − dim ΩF (G) = dim(G − D) − dim G + deg D. Let µD (w) ∈ CΩ (D, G) with weight m > 0. Then we have wPi (1) = 0 for some indices i = i1 , . . . , in−m , so by Lemma 2.2.12 w ∈ ΩF (G − (D −. n−m X. Pij )).. j=1. P Let A = G − (D − n−m j=1 Pij ). Since 0 6= w ∈ ΩF (A) implies dim ΩF (A) > 0. Hence by Lemma 2.2.5 dim ΩF (A) = dim A − deg A + g − 1 > 0. By Riemann-Roch Theorem, we know that if deg A > 2g − 2 then dim A − deg A + g − 1 = 0. Therefore deg A ≤ 2g − 2. We obtain 2g − 2 ≥ deg G − (n − (n − m)) = deg G − m. Hence the minimum distance d0 of CΩ (D, G) also satisfies the inequality d0 ≥ deg G−(2g−2). Assume now that 2g − 2 < deg G < n, then k 0 = dim(G − D) − dim G + deg D = 0 − dim G + n = −(deg G + 1 − g) + n by Riemann-Roch Theorem.. . From Theorem 3.2.4 and Theorem 3.2.7 we have dim CL (D, G) = dim G − dim(G − D) and dim CΩ (D, G) = dim(G − D) − dim G + deg D. Observe dim CL (D, G) + dim CΩ (D, G) = deg D = n. In fact, they are dual to each other. Theorem 3.2.8. Let D = P1 + · · · + Pn where P1 , . . . , Pn are pairwise distinct places of F/Fq degree 1 and G be a divisor with supp G ∩ suppD = ∅. The codes CL (D, G) and CΩ (D, G) are dual to each other, that is, CΩ (D, G) = CL (D, G)⊥ . Proof. Since we have known that dim CΩ (D, G) = dim CL (D, G)⊥ , next we claim that CΩ (D, G) ⊆ CL (D, G)⊥ . Let w ∈ ΩF (G − D) and x ∈ L(G). Since w ∈ ΩF (G − D), we.

(23) 3.2. Geometric Goppa Codes. 20. have vPi (w) ≥ −1. By the definition of x ∈ L(G), we have vPi (x) ≥ −vPi (G) = 0 for all i = 1, . . . , n. By Lemma 2.2.12 we obtain wPi (x) = x(Pi ) · wPi (1). Thus hµD (w), evD (x)i = h(wP1 (1), . . . , wPn (1)), (x(P1 ), . . . , x(Pn ))i =. n X. x(Pi ) · wPi (1) =. i=1. n X. wPi (x).. i=1. For P ∈ PF \ {P1 , . . . , Pn }, from (x) ≥ −G and (w) ≥ (G − D) we obtain vP (x) ≥ −vP (G) = −vP (G − D) ≥ −vP (w). Hence we have wP (x) = 0 by Lemma 2.2.11. Therefore n X i=1. wPi (x) =. X. wP (x) = w(˜ x). P ∈PF. by Lemma 2.2.10. Since x ˜ is principal adele, w(˜ x) = 0. Thus h(wP1 , . . . , wPn ), (x(P1 ), . . . , x(Pn ))i = w(˜ x) = 0. Hence CΩ (D, G) ⊆ CL (D, G)⊥ . Thus CL (D, G)⊥ = CΩ (D, G).. . Now we want to prove that CΩ (D, G) can be represented by CL (D, H) for some appropriate divisor H. Proposition 3.2.9. Suppose that deg D = n and deg G < n with supp G ∩ suppD = ∅. Let η be a Weil differential such that vPi (η) = −1 and ηPi (1) = 1 for i = 1, . . . , n. Then CL (D, G)⊥ = CL (D, H) where H = D − G + (η). Proof. Let H = D − G + (η). Since vPi (η) = −1 for i = 1, . . . , n, suppH ∩ suppD = ∅. CL (D, H) is defined. Cinsider a map ψ : L(D−G+(η)) → ΩF (G−D) given by ψ(x) := x·η. First we claim CL (D, H) ⊆ CΩ (D, G). Since by assumption vPi (η) = −1 and x ∈ L(H), we have vPi (x) ≥ −vPi (H) = −vPi (D − G + (η)) = −1 + 1 = 0. Hence by Lemma 2.2.12 we can write (x · η)Pi (1) = ηPi (x) = x(Pi ) · ηPi (1) = x(Pi ) Thus CL (D, H) ⊆ CΩ (D, G). Conversely, for w ∈ ΩF (G − D) there exists an element x ∈ F such that w = x · η by dimF ΩF = 1. Since (w) ≥ (G − D), (x · η) = (x) + (η) ≥ (G − D). Hence (x) ≥ G − D − (η) = −(D − G + (η)) = −H. Therefore x ∈ L(H). Hence CL (D, H) = CΩ (D, G). By Theorem 3.2.8 we obtain CL (D, H) = CL (D, G)⊥ . .

(24) 3.2. Geometric Goppa Codes. 21. In this paper we will always use W to be the canonical divisor instead of (η). From the Proposition 3.2.9 we define G⊥ := D − G + W . Lemma 3.2.10. For D, W, G and H ∈ DF , then we have 1. G⊥ ∩ H ⊥ = D − W + (G ∩ H). 2. G⊥ ∩ H ⊥ = G⊥ + H ⊥ − D − W + (G ∩ H). Proof. 1. By Proposition 3.2.9 we have G⊥ = D + W − G and H ⊥ = D + W − H. Then by Lemma 2.1.16 we obtain G⊥ ∩ H ⊥ = ((D + W ) − G) ∩ ((D + W ) − H) = (D + W ) − G − H + (G ∩ H). 2. Follow (1), we can get G⊥ ∩ H ⊥ = ((D + W ) − G) ∩ ((D + W ) − H) = ((D + W ) − G) + ((D + W ) − H) − D + (G ∩ H) = G⊥ + H ⊥ − D + (G ∩ H). The dual codes of geometric Goppa codes will play important role in our main work..

(25) Chapter 4. Main Results. In this chapter, we will show our main result. We want to know the relation between divisors and geometeric Goppa codes.. 4.1. Equivalent codes First we introduce the equivalence of codes. Definition 4.1.1. Two codes C1 , C2 ⊆ Fnq are said to be equivalent if there is a vector a = (a1 , . . . , an ) ∈ (Fq \ {0})n such that C2 = a · C1 , that is, C2 = {(a1 c1 , . . . , an cn ) | (c1 , . . . , cn ) ∈ C1 }. If a code C is linear, then we know the codes which are equivalent to C also are linear codes and have the same parameters with C. Furthermore, if a code C is equivalent to a geometric Goppa code CL (D, G) with deg G < deg D then C must also be a geometric Goppa code. Let us recall some notation which is fixed in preceding sections. F/Fq is an algebraic function field of genus g. DF is the divisor group of F/Fq . P1 , . . . , Pn are pairwise distinct places of F/Fq of degree 1. D = P1 + · · · + Pn is a divisor of F/Fq of degree n. CL (D, G) is the geometric Goppa code associated with D and G. 22.

(26) 4.2. Equality of Geometric Goppa Codes. 23. where G ∈ DF such that supp G ∩ suppD = ∅. Proposition 4.1.2. Suppose deg G < deg D = n and supp G ∩ suppD = ∅. If C ⊆ Fnq with C ∼ CL (D, G), then there exists G0 which is equivalent to G such that supp G0 ∩ suppD = ∅ and C = CL (D, G0 ). Proof. Let C = a · CL (D, G) with a = (a1 , . . . , an ) ∈ (Fq \ {0})n . By the Weak Approximation Theorem, choose x ∈ F with x(Pi ) = ai for i = 1, . . . , n and set G0 = G − (x). Since deg G < n, dim CL (D, G) = dim G. By G ∼ G0 , dim CL (D, G) = dim G = dim G0 = dim CL (D, G0 ). Following the proof of Lemma 2.1.8, for z ∈ L(G), there exists a unique element zx ∈ L(G0 ). Hence there is a one-to-one correspondence between (z(P1 ), . . . , z(Pn )) and (zx(P1 ), . . . , zx(Pn )) = a · (z(P1 ), . . . , z(Pn )). Hence CL (D, G0 ) = a · CL (D, G). Thus C = CL (D, G0 ). We know geometric Goppa codes are associated with divisors. When two divisors are equivalent, we find the codes which associated with the divisors are equivalent. Proposition 4.1.3. Let both deg G and deg H be less than deg D = n. Suppose G, H ∈ DF with G ∼ H and supp G ∩ suppD = suppH ∩ suppD = ∅, Then CL (D, G) ∼ CL (D, H). Proof. Since G ∼ H and supp G ∩ suppD = suppH ∩ suppD = ∅, suppose H = G + (x) with vPi (x) = 0, ∀i = 1, . . . , n. It is similar to the proof of Proposition 4.1.2. Hence we have CL (D, G) = a · CL (D, H). Next we ask whether the converse of Proposition 4.1.3 is true or not. From Proposition 4.1.2, we have the following : if CL (D, G) ∼ CL (D, H) then there exists a divisor H 0 such that H 0 ∼ G and CL (D, H 0 ) = CL (D, H). Hence we will talk about equal codes in next section.. 4.2. Equality of Geometric Goppa Codes First we can find when two equal codes in some case imply the same degree of divisors. Note that when we write 2g − 1 < deg G < n − 1 which automatically implies that we assume 2g < n..

(27) 4.2. Equality of Geometric Goppa Codes. 24. Lemma 4.2.1. Suppose deg D = n and 2g − 1 < deg G < n − 1. If CL (D, G) = CL (D, H), then deg G = deg H. Proof. Since deg G < n − 1 < n, we have dim CL (D, G) = dim G by Lemma 3.2.4. Since deg G > 2g − 1 > 2g − 2, dim CL (D, G) = deg G + 1 − g by Riemann-Roch Theorem. From the bound of deg G we obtain g = 2g − 1 + 1 − g < dim CL (D, G) = deg G + 1 − g < deg D − 1 + 1 − g = deg D − g. From CL (D, G) = CL (D, H), dim G = dim CL (D, G) = dim CL (D, H) = dim H − dim(H − D). Hence g < dim H − dim(H − D) = dim G < deg D − g. Case 1: deg H ≥ deg D Since deg H ≥ deg D > 2g > 2g − 2 then dim H = deg H + 1 − g, by Riemann-Roch Theorem. Suppose deg H > deg D + 2g − 2 then deg(H − D) = deg H − deg D > 2g − 2. Hence again by Riemann-Roch Theorem, dim(H − D) = deg(H − D) + 1 − g = deg H − deg D + 1 − g. Therefore, in this case we obtain dim H − dim(H − D) = deg H + 1 − g − (deg H − deg D + 1 − g) = deg D. Since dim H − dim(H − D) = dim CL (D, H) = dim CL (D, G) = dim G < deg D − g, we have deg D < deg D − g. Thus g < 0 which gives a contradiction. Suppose deg H ≤ deg D + 2g − 2 then deg(H − D) ≤ 2g − 2. In this case we assume deg H ≥ deg D which implies deg(H − D) ≥ 0. Hence we obtain 0 ≤ deg(H − D) ≤ 2g − 2 by combining these inequalities. From Clifford’s Theorem we have dim(H − D) ≤ 1 +. deg(H − D) . 2. Therefore dim H − dim(H − D) ≥ deg H + 1 − g − (1 +. deg(H − D) deg H deg D )= −g+ . 2 2 2. Since dim H − dim(H − D) = dim CL (D, H) = dim CL (D, G) < deg D − g, it implies that deg H deg D −g+ ≤ dim H − dim(H − D) < deg D − g. 2 2 Thus deg H < deg D which contradics to deg D ≤ deg H..

(28) 4.2. Equality of Geometric Goppa Codes. 25. Case 2: deg H < deg D In this case the assumption implies deg(H − D) < 0. Then we get dim(H − D) = 0 immediately, which implies dim CL (D, H) = dim H − dim(H − D) = dim H. Hence dim H = dim CL (D, H) = dim CL (D, G) = dim G. Assume deg H < 0 then dim H = 0 = dim G. Since dim G > g, we get 0 > g which gives a contradiction. Suppose 0 ≤ deg H ≤ 2g − 2 then deg H ≤ g, 2 by Clifford’s theorem. Since in this case we have dim H = dim G ≤ g, which contradicts to dim G > g. If deg H > 2g − 2 then dim H = deg H + 1 − g by Riemann-Roch theorem. Then by the assumption of deg H < deg D, we have dim H ≤ 1 +. deg H + 1 − g = dim H = dim CL (D, H) = dim CL (D, G) = dim G = deg G + 1 − g. Hence deg G = deg H. If the assumption of deg G does not satisfy then we can find two divisors with different degrees while they lead to the same code. We will give some examples later. If the degrees of two divisors are different, then these two divisor are impossible to be equivalent. Hence the assumption of deg G in Theorem 4.2.9 is necessary. We have known that in some cases equal codes imply equal degree of divisors. If two divisors are equivalent, then they have the same degree. We are interested in whether or not equal codes imply equivalent divisors or not. If deg G > 2g − 1 and G 6= H then we have dim G > dim(G ∩ H) by Lemma 2.1.17. Latter we will show an upper bound of dim G. Lemma 4.2.2. Let D = P1 + · · · + Pn and supp G ∩ suppD = suppH ∩ suppD = ∅. Suppose both deg G and deg H be less than deg D = n and CL (D, G) = CL (D, H). Then for all f ∈ L(G) there exists a unique hf ∈ L(H) such that f (Pi ) = hf (Pi ) for all i = 1, . . . , n. Moreover, f − hf ∈ L(G + H − (G ∩ H) − D). Proof. Since deg G and deg H are less than n, which implies dim G = dim CL (D, G) = dim CL (D, H) = dim H..

(29) 4.2. Equality of Geometric Goppa Codes. 26. For f ∈ L(G) then (f (P1 ), . . . , f (Pn )) ∈ CL (D, G) = CL (D, H). There exists an element hf ∈ L(H) such that f (Pi ) = hf (Pi ) for all i = 1, . . . , n. By dim G = dim H we know the element is unique. If P ∈ {P1 , . . . , Pn } vP (f − hf ) ≥ 1 = −vP (G + H − (G ∩ H) − D). If P ∈ / {P1 , . . . , Pn } vP (f − hf ) ≥ min{vP (f ), vP (hf )} ≥ min{−vP (G), −vP (H)} = −vP (G) − vP (H) + min{vP (G), vP (H)} = −vP (G) − vP (H) + vP (G ∩ H) = −vP (G + H − (G ∩ H)). Hence f − hf ∈ L(G + H − (G ∩ H) − D).. . Lemma 4.2.3. Suppose deg D = n. Let deg G < deg D and deg H < deg D. Suppose CL (D, G) = CL (D, H), then dim G ≤ dim(G ∩ H) + dim(G + H − (G ∩ H) − D). Proof. By Lemma 4.2.2 we have that for all f ∈ L(G), there exists a unique element hf ∈ L(H) such that f (pi ) = hf (pi ), for all i = 1, . . . , n. Consider the Fq -linear map φ : L(G) → L(G + H − (G ∩ H)) by setting f 7→ f − hf . If f ∈ ker φ, then we obtain f = hf ∈ L(G) ∩ L(H) = L(G ∩ H) by Lemma 2.1.17. Hence ker φ ⊆ L(G ∩ H). Conversely, if f ∈ L(G ∩ H) then f ∈ L(H). Therefore φ(f ) = f − f = 0. Hence ker φ = L(G ∩ H). By Lemma 4.2.2 Imφ ⊆ L(G + H − (G ∩ H)). Hence dim G ≤ dim(G ∩ H) + dim(G + H − (G ∩ H) − D). . Corollary 4.2.4. Suppose that deg D = n. Let 2g − 1 < deg G < n − 1 and CL (D, G) = CL (D, H). If 0 ≤ deg(G ∩ H) ≤ 2g − 2 and 0 ≤ deg(G + H − (G ∩ H) − D) ≤ 2g − 2, then n ≤ 2g + 2..

(30) 4.2. Equality of Geometric Goppa Codes. 27. Proof. Clifford’s theorem gives dim(G ∩ H) ≤ 1 +. deg(G ∩ H) 2. and dim(G + H − (G ∩ H) − D) ≤ 1 +. deg(G + H − (G ∩ H) − D) . 2. By Lemma 4.2.3 we have deg G + 1 − g = dim G ≤ dim(G ∩ H) + dim(G + H − (G ∩ H) − D). Since 2g − 1 < deg G < deg D − 1 we have deg G = deg H by Lemma 4.2.4. Combining these inequalities we obtain deg G + 1 − g ≤ 2 −. deg D + deg G. 2. Thus n ≤ 2g + 2.. . Lemma 4.2.5. Suppose deg D = n. Let 2g −1 < deg G < n−1 and CL (D, G) = CL (D, H). If deg(G ∩ H) > 2 deg G − deg D, then G = H. Proof. Suppose G 6= H, then dim(G ∩ H) < dim G by Lemma 2.1.17. By Lemma 4.2.3 we have dim(G ∩ H) < dim G ≤ dim(G ∩ H) + dim(G + H − (G ∩ H) − D). Hence dim(G + H − (G ∩ H) − D) > 0. But deg(G ∩ H) > 2 deg G − deg D implies deg(G + H − (G ∩ H) − D) < 0. This gives a contradiction. By the dual codes, we can get the following Corollary immediately. We will replace deg(G ∩ H) > 2 deg G − deg D of a condition which only associated with genus. Corollary 4.2.6. Suppose that deg D = n. Let 2g − 1 < deg G < n − 1 and CL (D, G) = CL (D, H). If deg(G ∩ H) > 2g − 2, then G = H. Proof. Since CL (D, G) = CL (D, H), CL (D, G)⊥ = CL (D, H)⊥ . Note that G⊥ = D+W −G where W is the canonical divisor and deg G⊥ = deg(D + W − G) = deg D + 2g − 2 − deg G. Since 2g − 1 < deg G < n − 1, we have deg D + 2g − 2 − (n − 1) < deg G⊥ < deg D + 2g − 2 − (2g − 1)..

(31) 4.2. Equality of Geometric Goppa Codes. 28. Hence we also get 2g − 1 < deg G⊥ < n − 1. Thus deg G⊥ = deg H ⊥ by Lemma 4.2.3. From Lemma 3.2.10 we have (G⊥ ∩ H ⊥ ) = G⊥ + H ⊥ − D − W + (G ∩ H). Consider deg(G⊥ + H ⊥ − D − W + (G ∩ H)) = 2 deg G⊥ − deg D − (2g − 2) + deg(G ∩ H). By deg(G ∩ H) > 2g − 2 we get deg(G⊥ ∩ H ⊥ ) > 2 deg⊥ −n. Hence Lemma 4.2.5 implies G⊥ = H ⊥ , that is, D + W − G = D + W − H. Therefore G = H. Proposition 4.2.7. Suppose deg D = n. Let 2g−1 < deg G < n−1, CL (D, G) = CL (D, H) and deg D > 2g + 2. If deg(G ∩ H) > deg G − deg D, then G = H. Proof. We consider deg(G∩H) > 2g−2. Then G = H follows from Corollary 4.2.6. Next if deg(G ∩ H) < 0 then dim(G ∩ H) = 0. By Lemma 4.2.3 dim G ≤ dim(G + H − (G ∩ H) − D). However, deg(G + H − (G ∩ H) − D) = 2 deg G − deg(G ∩ H) − deg D < deg G by assumption of deg(G ∩ H) > deg G − deg D. Since deg G > 2g − 1, dim G > dim(G + H − (G ∩ H) − D) by Lemma 2.1.14. Then it gives a contradiction. Now we may assume 0 ≤ deg((G ∩ H) ≤ 2g − 2. Case 1: 0 ≤ deg((G ∩ H) ≤ 2g − 2 and deg(G⊥ ∩ H ⊥ ) > 2g − 2 Since 2g − 1 < deg G < n − 1, 2g − 1 < deg G⊥ < n − 1 and CL (D, G) = CL (D, H) implies CL (D, G⊥ ) = CL (D, H ⊥ ). Then G⊥ = H ⊥ follows from Lemma 4.2.6. Hence G = H. Case 2: 0 ≤ deg((G ∩ H) ≤ 2g − 2 and deg(G⊥ ∩ H ⊥ ) ≤ 2g − 2. By Lemma 3.2.10 we have G⊥ ∩ H ⊥ = D + W − G − H + (G ∩ H). Then it implies that deg(D + W − G − H + (G ∩ H)) ≤ 2g − 2. We obtains deg(G + H − (G ∩ H) − D) ≥ 0 since deg W = 2g − 2. Suppose 0 ≤ deg(G + H − (G ∩ H) − D) ≤ 2g − 2. Since we have 0 ≤ deg(G ∩ H) ≤ 2g − 2, deg D ≤ 2g + 2 by Corollary 4.2.4, which contradicts to the assumption. Suppose deg(G + H − (G ∩ H) − D) > 2g − 2. Using Lemma 4.2.3, Cliffird’s theorem and Riemann-Roch theorem, we can find deg G + 1 − g = dim G ≤ dim(G ∩ H) + dim(G + H − (G ∩ H) − D) ≤1+. deg(G ∩ H) + deg(G + H − (G ∩ H) − D) + 1 − g. 2. Then. deg(G ∩ H) ≤ deg G − n + 1 < 0 from deg G < n − 1. 2 It contradicts to the assumption 0 ≤ deg(G ∩ H) ≤ 2g − 2.. .

(32) 4.3. Equivalent Codes and Equivalent Divisors. 29. Corollary 4.2.8. Suppose deg D = n. Let 2g − 1 < deg G < n − 1, CL (D, G) = CL (D, H) and deg D > 2g + 2. If both G and H are effective divisors, then G = H. Proof. Since both G and H are effective, deg(G ∩ H) ≥ 0 > deg G − deg D. Thus G = H by Proposition 4.2.7. Under the restrictions of deg G and deg D, if two codes are the same, then the associated divisors will be equivalent. Theorem 4.2.9. Suppose deg D = n and supp G ∩ suppD = suppH ∩ suppD = ∅. Let 2g − 1 < deg G < n − 1 and deg D > 2g + 2. If CL (D, G) = CL (D, H), then G ∼ H. Proof. We know that supp G only has finite places. Since supp G ∩ suppD = ∅ and D = P1 + · · · + Pn , by the Weak Approximation Theorem there exist an element f ∈ F such that for P ∈ supp G vP (f ) = −vP (G) and for P ∈ {P1 , . . . , Pn } vP (f ) = 0. Hence f ∈ L(G) with vPi (f ) = 0. Take G0 = G + (f ). Then G0 ≥ 0 by the definition of L(G). And set c = (f (P1 ), . . . , f (Pn )) ∈ (Fq \{0})n . Since CL (D, G) = CL (D, H), we obtain h ∈ L(H) such that h(Pi ) = f (Pi ), ∀i = 1, . . . , n. Setting H 0 = H + (h) implies H 0 ≥ 0. We have two isomorphisms φG : L(G0 ) → L(G) by φG (z) = zf and φH : L(H 0 ) → L(H) by setting φH (z) = zh. Since 2g − 1 < deg G < deg D − 1, by Lemma 4.2.1 we have deg H = deg G < deg D − 1. So dim G0 = dim G = dim CL (D, G) = dim CL (D, H) = dim H = dim H 0 . Hence CL (D, G) = c·CL (D, G0 ) and CL (D, H) = c·CL (D, H 0 ). Since CL (D, G) = CL (D, H) we have CL (D, G0 ) = CL (D, H 0 ). Moreover both G0 and H 0 are effective. By Corollary 4.2.8 we obtain G0 = H 0 . In other words, G + (f ) = H + (h). Hence G ∼ H. . 4.3. Equivalent Codes and Equivalent Divisors In the preceding chapters we have the notions of equivalent divisor and equivalent code. And previous section tells us the relation between equal codes and equivalent divisors. In this section, our goal is to find the condition to imply equivalent codes is the sufficient and necessary condition of equivalent divisors..

(33) 4.3. Equivalent Codes and Equivalent Divisors. 30. Theorem 4.3.1. Suppose deg D = n, both deg G and deg H are less than n and supp G ∩ suppD = suppH ∩ suppD = ∅. Let 2g − 1 < deg G < n − 1 and deg D > 2g + 2. Then CL (D, G) ∼ CL (D, H) if and only if G ∼ H. Proof. By Proposition 4.1.3, we have known that two equivalent divisors lead to the same code. Next we will prove another side. Since CL (D, G) ∼ CL (D, H), there exists a divisor H 0 ∼ G such that CL (D, H 0 ) = CL (D, H) by Proposition 4.1.2. From H 0 ∼ G we get deg H 0 = deg G immediately. Since 2g − 1 < deg G = deg H 0 < n − 1, deg D > 2g + 2 and CL (D, H 0 ) = CL (D, H), by Theorem 4.2.9 we have H 0 ∼ H. Hence G ∼ H 0 ∼ H. Thus G ∼ H. Finally, we will give two examples to show when deg G = 2g − 1 this theorem will be fail and when deg D = 2g + 2 this theorem is wrong. Example 4.3.2. Consider the projective plane curve over F4 with equation Y 2Z + Y Z 2 = X 3. F/F4 is the function field associated with the curve. Take F4 = F2 [t]/(t2 + t + 1) with α2 + α + 1 = 0 Hence the genus of function field F/Fq is 1. Then C(F4 ) = {P∞ = [0, 1, 0], P1 = [0, 0, 1], P2 = [0, 1, 1], P3 = [1, α, 1], P4 = [1, 1 + α, 1], P5 = [α, α, 1], P6 = [α, 1 + α, 1], P7 = [1 + α, α, 1], P8 = [1 + α, 1 + α, 1]}. Setting D = P1 + P2 + P3 + P4 + P5 , then deg D = 5 > 2g + 2 = 4. Take G = P∞ and H = 0, then we obtain supp G ∩ suppD = suppH ∩ suppD = ∅. And we know deg G = 1 = 2g − 1 and deg H = 0 which imply dim G = dim H = 1. In particular, we observe 1 ∈ L(G) and 1 ∈ L(H). Hence L(G) and L(H) have the same basis {1}. Moreover, CL (D, G) = CL (D, H). But G H since 1 = deg G 6= deg H = 0. The previous example shows that if deg G = 2g−1 then there exists a divisor H such that deg G 6= deg H but CL (D, G) = CL (D, H). If deg G0 = n−1, we can use the dual code of the previous code to get it. We set G0 = G⊥ = D + W − G and H 0 = H ⊥ = D + W − H. Since CL (D, G) = CL (D, H), it implies CL (D, G⊥ ) = CL (D, G)⊥ = CL (D, H)⊥ = CL (D, H ⊥ ). Hence we obtain deg G0 = n − 1 and deg H 0 = n construct the same code. If a code does not satisfy the condition of deg D > 2g + 2 then the divisors of equall codes don’t always be equivalent..

(34) 4.3. Equivalent Codes and Equivalent Divisors. 31. Example 4.3.3. Consider the projective plane curve over F4 with equation Y 2Z + Y Z 2 = X 3. F/F4 is the function field associated with the curve. Take F4 = F2 [t]/(t2 + t + 1) with α2 + α + 1 = 0 Then C(F4 ) = {P∞ = [0, 1, 0], P1 = [0, 0, 1], P2 = [0, 1, 1], P3 = [1, α, 1], P4 = [1, 1 + α, 1], P5 = [α, α, 1], P6 = [α, 1 + α, 1], P7 = [1 + α, α, 1], P8 = [1 + α, 1 + α, 1]}. Setting D = P1 + P3 + P5 + P7 and let G = 2P∞ and H = 2P2 , then we obtain supp G ∩ suppD = suppH ∩ suppD = ∅. A generator matrix of a code C is a k × n matrix whose rows are a basis of C. Since both deg G = 2 and deg H = 2 are less than deg D = 4, by Lemma 3.2.4 we have dim CL (D, G) = dim G = dim H = dim CL (D, H). Hence the codes CL (D, G) and CL (D, H) are determinied by L(G) and L(H) respectively. X The basis of L(G) is {1, X Z } and the basis of L(H) is {1, Y +Z }. Hence the generator matrix of CL (D, G) is ! 1 1 1 1 . MG = 0 1 α 1+α Similarly, the generator matrix of CL (D, H) is MH =. 1 0. 1. 1. 1. 1 1+α. α 1+α. 1+α 1+α. ! =. 1 0. 0 1 1+α. ! MG .. Hence MH and MG span the same vector space over F4 . Thus we have CL (D, G) = CL (D, H) from their generator matrices. Suppose G ∼ H, then there exists an element f ∈ F \{0} such that (f ) + 2P∞ = 2P2 . So f ∈ L(2P∞ ) and the basis of L(2P∞ ) is {1, X Z }. Hence X f = a1 · 1 + a2 · Z for some a1 , a2 ∈ F4 . However we have f (P2 ) = f ([0, 1, 1]) = a1 and (f ) + 2P∞ = 2P2 . Therefore, P2 is the zero of f implies a1 = 0. Hence f = a2 · xz . Thus (f ) = P∞ + P1 + P2 − 3P∞ = P1 + P2 − 2P∞ . But (f ) + 2P∞ = P1 + P2 6= 2P2 , which gives a contradiction..

(35) Bibliography. [1] Carlos Munuera and Ruud Pellikaan, “Equality of geometric Goppa codes and equivalence of divisors,” Journ. Pure Appl. Algebra, vol. 90, pp. 229–252, 1993. [2] Henning Stichtenoth, “Algebraic Function Field and Codes,” Springer-Verlag, New York, 1993. [3] Judy L. Walker, “Codes and curves,”Springer, 1999. [4] Chao-Ping, Xing, “When are two geometric Goppa codes equal ?” IEEE Transctions of information theory, vol. 38, No. 3, May 1992.. 32.

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