On the missing functions of a pyramid of curves

On the missing functions of a pyramid of curves Ruud Pellikaan

∗

Appeared in Proc. 35th Allerton Conf. on Communication, Control, and Computing, Urbana-Champaign, September 29 - October 1, 1997, pp. 33-40.

Abstract We consider an asymptotically good tower (Ym )m≥1 of curves over a finite field Fq , where q is a square. The Weierstrass semigroup Hm of a rational (m) point P∞ of Ym is determined. For n ∈ Hm and m = 1, 2 and 3 we will give (m) rational functions on Ym that have pole order n at P∞ and that have no (m) poles outside P∞ . For larger m an explicit description of all these functions is still missing.

1

Introduction

A tower · · · → Xm → Xm−1 → · · · → X3 → X2 → X1 of algebraic curves over a finite field Fq is said to be asymptotically good if number of rational points of Xm /Fq > 0. m→∞ genus of Xm lim

Recently an explicit description was obtained of several asymptotically good towers by Garcia and Stichtenoth The motivation to consider these came from coding theory: such towers give rise to asymptotically good sequences of codes if this limit is larger than one. Although the existence of good codes on ∗

Department of Mathematics and Computing Science, Technical University of Eindhoven, P.O. Box 513, 5600 MB Eindhoven, The Netherlands.

1

or above the Tsfasman-Vl˘adut¸-Zink bound was guaranteed and even a polynomial construction was given, the methods used (namely, modular curves) and the degree of the complexity of the construction were such that hardly any of the resulting codes were known explicitly. Now that asymptotically good towers (Xm |m ≥ 1) of algebraic curves are known explicitly, the next step would be to give an explicit description of the vector spaces L(Gm ) resp. (m) (m) L(nP∞ ), where Gm is a divisor (resp. P∞ is a rational point of Xm . The (m) latter space L(nP∞ ) is the Fq -vector space of all rational functions on Xm (m) (m) that have no poles outside P∞ and pole order at most n at P∞ . The first attempts have been made in this direction.

2

Weierstrass semigoups in a tower of curves

In this paper we consider the tower (Ym )m≥1 over Fq , where q = r2 and r is a prime power. It is defined by giving the corresponding function fields (Tm |m ≥ 1) as follows: T1 = Fq (x1 ) and Ti+1 = Ti (xi+1 ) with xri+1 + xi+1 =

xri . xr−1 +1 i

By the form of the defining equations it is readily seen that N (Ym ), the number of rational points of Ym , is at least (r2 − r)rm−1 . The genus g(Ym ) is computed by the theory of Artin-Schreier extensions, and one finds that lim

m→∞

N (Ym ) √ = r − 1 = q − 1. g(Ym )

Hence the tower is asymptotically good and in fact optimal; it attains the Drinfeld-Vl˘adut¸ bound. This implies that geometric Goppa codes which are constructed by means of this tower lie on or above the TVZ bound, which is better than the Gilbert-Varshamov bound for all q = r2 ≥ 49. The element x1 ∈ T1 ⊆ Tm has in Tm a unique pole at the rational point (m) that we denote by P∞ or by P∞ for short. Hence it is natural to consider (m) the spaces L(nP∞ ) for all m and n. The dimension of these spaces is de(m) (m) termined. The Weierstrass semigroup Hm of P∞ , that is Hm = H(P∞ ), is the set of all n ∈ N0 such that there is some f ∈ Tm having a pole of order 2

(m)

(m)

n at P∞ and no pole outside P∞ . Theorem 2.1 Let cm =

 m m  r −r2 

rm − r

if m is even,

m+1 2

if m is odd.

Then H1 = N0 and Hm = rHm−1 ∪ { n ∈ N0 | n ≥ cm }, for all m ∈ N0 , m > 0. The number cm is called the conductor of Hm . Let Ti,j be the function field of the curve Yi,j defined by Ti,j = Fq (xi , xi+1 , . . . , xi+j−1 ) ⊂ Ti+j−1 for i, j ∈ N. Then Y1,m = Ym and Yi,m is isomorphic to Ym for all i. The inclusions Ti,j−1 ⊂ Ti,j and Ti+1,j−1 ⊂ Ti,j give corresponding maps of algebraic curves Yi,j → Yi,j−1 and Yi,j → Yi+1,j−1 , respectively. The proof is by induction on m and by considering the whole pyramid of curves Yi,j . See [1].

3

Missing functions in a pyramid of curves

The function

j Y (xr−1 + 1)xej+1 i i=1 (m)

in Tm has a pole of order rm − rm−j + erm−j−1 at P∞ and has no poles out(m) side P∞ , for all 0 ≤ e < r. But not all values of Hm are obtained in this way. For n ∈ Hm and m = 1, 2 and 3 we will give functions in Tm that have pole (m) (m) order n at P∞ and that have no poles outside P∞ . For larger m an explicit description of all these functions is still missing.

3

(1)

Example 3.1 For Y1 is the projective line and P∞ = P∞ is the point at infinity. So T1 = Fq (x1 ), H1 = N0 . The function xn1 has a pole of order n at P∞ and has no poles outside P∞ . Example 3.2 Let x = x1 and y = x2 . Then T2 = Fq (x, y) where yr + y =

xr . xr−1 + 1

(1)

(2)

The pole P∞ of x on Y1 is totally ramified in the cover Y2 of Y1 , and P∞ (1) (2) lies above P∞ . So x has a pole of order r at P∞ = P∞ . Define Ω = {α ∈ Fq | αr + α = 0} and Ω∗ = Ω \ {0}. Then Ω and Ω∗ consist of r and r − 1 elements, respectively, since q = r2 . For every α ∈ Ω∗ the function x − α has exactly one zero on Y2 that we denote by Pα . For any β ∈ Ω, there is a unique common zero of x and y − β that we denote by Qβ . The points P∞ and Pα are poles of y − β for all α ∈ Ω∗ and β ∈ Ω. Hence we have the following principal divisors P (x) = β∈Ω Qβ − rP∞ , (x − α) = rPα − rP∞ if α ∈ Ω∗ , P if β ∈ Ω. (y − β) = rQβ − α∈Ω∗ Pα − P∞ Therefore xi has a pole of order ri at P∞ and no poles outside P∞ for all i ∈ N0 . Furthermore P ((xr−1 + 1)y j ) = rjQ0 + (r − j) α∈Ω∗ Pα − [r(r − 1) + j]P∞ . So (xr−1 + 1)xi y j has a pole of order r(r − 1) + ri + j at P∞ and no poles outside P∞ if j ≤ r. In this way the functions xi , i ∈ N0 and (xr−1 + 1)xi y j , i, j ∈ N0 , 0 < j < r give all possible pole orders of the Weierstrass semigroup H2 = {0, r, 2r, . . . , (r − 1)r, (r − 1)r + 1, (r − 1)r + 2, . . .}. Finally we will treat T3 . Let z = x3 in T3 . Then T3 = T2 (z) where zr + z =

yr . y r−1 + 1

4

So we have the following tower and pyramid of function fields T3 ↑ T2 ↑ T1

Fq (x, y, z) %

-

-

%

Fq (x, y)

Fq (y, z)

% Fq (x)

-

Fq (y)

Fq (z).

There is a unique point P of Y3 that is totally ramified above P∞ or Pα , α ∈ Ω∗ , since these points are simple poles of y on Y2 . Denote this point again by P∞ or Pα , respectively. Therefore we have the following principal divisor on Y3 if α ∈ Ω∗ .

(x − α) = r2 Pα − r2 P∞ Hence (xr−1 + 1) = r2

P

α∈Ω∗ Pα

− r2 (r − 1)P∞

if α ∈ Ω∗ .

Let β ∈ Ω. Define Ω(β) = {γ | γ r + γ = β r }. Proposition 3.3 Let β ∈ Ω. Then there exists exactly one point Qβγ on Y3 above Qβ such that β r+1 −γ z− x has a zero in Qβγ and vQβγ (y − β) = r for all γ ∈ Ω(β). We refer for the proof to [2]. A consequence of this proposition is that {Qβγ | γ ∈ Ω(β)} is the set of points of Y3 lying above Qβ . So Qβ is unramified and P (x) = β∈Ω, γ∈Ω(β) Qβγ − r2 P∞ , (y − β) = r

P

(z − γ) = r2 Q0γ −

γ∈Ω(β) Qβγ

−r

P

P

− rP∞

α∈Ω∗ Pα

β∈Ω∗ , γ 0 ∈Ω(β) Qβγ 0

−

P

α∈Ω∗ Pα

if β ∈ Ω, − P∞

if γ ∈ Ω.

Definition 3.4 Let t be a local parameter at a point Q. Then we write f = g + O(te ) if vQ (f − g) ≥ e. 5

The function x is a local parameter at Qβγ and y = β + O(xr ) and z =

β r+1 + O(1), x

by Proposition 3.3. Furthermore β r+1 = −β 2 for all β ∈ Ω. Hence z+

y2 = O(1) x

at Qβγ and for all β ∈ Ω and γ ∈ Ω(β). It still has poles outside P∞ but these are cancelled by the zeros of xr−1 + 1. Therefore   y2 r−1 (x + 1) z + x is a function that has no poles outside P∞ . The functions k  y2 r−1 j (x + 1)y z + x 2

have no poles outside P∞ if j +2k ≤ r. Now the trick is to expand β j (z + βx )k first by Newton’s binomium, see which terms β 2i+j appear, replace this by a term with the right coefficient c(i, j) and the right exponent e(i, j) modulo r − 1, and use these coefficients and exponents with y instead of β. Definition 3.5 Let e(0, j) = j. Let e(i, j) be the integer in the interval [1, r − 1] that is congruent to 2i + j modulo r − 1, if i > 0. Definition 3.6 Define the constant c(i, j) by c(i, j) = (−1)b

2i+j−1 c r−1

.

Remark 3.7 Notice that β 2i+j = c(i, j)β e(i,j) , since β r = −β for all β ∈ Ω and by the definitions of e(i, j) and c(i, j). Furthermore c(i, j)y e(i,j) = β 2i+j + O(xr ) at Qβγ , since e(i, j) < r and y = β + O(xr ). 6

Definition 3.8 Let j, k be integers such that 0 ≤ j, k < r. Define the function fjk ∈ T3 by fjk =

k   X k i=0

i

c(i, j)

y e(i,j) k−i z xi

Lemma 3.9 The function fjk has no pole at Qβγ for all β ∈ Ω, γ ∈ Ω(β) and j, k < r. Proof. We will show that xk fjk = O(xk ) at Qβγ . Pk k
c(i, j)y e(i,j) (xz)k−i xk fjk = i
Pi=0 k k 2i+j (xz)k−i +O(xr ) = i=0 i β 
Pk k 2i k−i + O(xr ) = yj i=0 i β (xz) = y j (xz + β 2 )k + O(xr ) = xk y j (z +

β2 k ) x

+ O(xr ) = O(xk )

The second equality is Remark 3.7. Proposition 3.3 is used in the third and sixth equality for y and z, respectively. The fourth equality is Newton’s binomium.  Proposition 3.10 The rational function (xr−1 + 1)fjk on Y3 has a pole at P∞ of order r2 (r − 1) + jr + k and no poles outside P∞ for all j, k < r. Proof. To investigate the poles of these functions it is enough to consider the points P∞ , Pα , α ∈ Ω∗ and the points Qβγ . In Lemma 3.9 it is shown that fjk has no pole at the Qβγ , and the same holds for xr−1 + 1. So it remains to investigate the behaviour at Pα and P∞ . If i ≤ k < r and j < r, then   e(i,j) y k−i z = −e(i, j)r − (k − i) − i.0 ≥ −(r − 1)r − (r − 1) = −r2 + 1, vPα xi since e(i, j) ≤ r − 1. Hence the pole order of fjk is at most r2 − 1 at Pα . But (xr−1 + 1) has a zero of order r2 − 1 at Pα . So (xr−1 + 1)fjk has no pole at Pα . Now the pole order at P∞ is considered. If i ≥ 1, then  e(i,j)  y k−i vP∞ z = −e(i, j)r −(k −i)−i.(−r2 ) ≥ −(r −1)r −(r −1)+r2 ≥ 1, xi 7

since e(i, j) ≤ r − 1. If i = 0, then   e(i,j) y k−i vP∞ = vP∞ (y j z k ) = −jr − k. z xi The pole order of (xr−1 + 1) is r2 (r − 1) at P∞ . Therefore (xr−1 + 1)fjk has a pole of order r2 (r − 1) + jr + k at P∞ .  Corollary 3.11 Let R3 be the subring of T3 consisting of all rational functions that have no poles outside P∞ . Then {xi | ∈ N0 , i < r − 1} ∪ { (xr−1 + 1)xi fjk | i, j, k ∈ N0 , j, k < r } is a basis of R3 over Fq . Proof. These functions are elements of R3 and the pole orders of these functions form the Weierstrass semigroup H3 of P∞ , by Theorem 2.1 and Proposition 3.10, and are mutually distinct.  Example 3.12 Let q = 2. Then the conductor is c3 = 23 − 22 = 4 and H3 = {0} ∪ {n ∈ N0 | n ≥ 4}. The functions in R3 with pole orders 4, 5, 6 and 7 are x + 1, (x + 1)(z + y/x), (x + 1)y and (x + 1)y(z + 1/x), respectively. Example 3.13 Let q = 3. Then the conductor is c3 = 33 − 32 = 18 and H3 = {0, 9} ∪ {n ∈ N0 | n ≥ 18}. The functions in R3 with pole orders between 18 and 26 are in increasing order x2 + 1, (x2 + 1)(z + y 2 /x), (x2 + 1)(z 2 − y 2 /x − y 2 /x2 ), (x2 + 1)y, (x2 + 1)y(z − 1/x), (x2 + 1)y(z − 1/x)2 , (x2 + 1)y 2 , (x2 + 1)y 2 (z − 1/x) and (x2 + 1)y 2 (z − 1/x)2 , respectively.

References [1] R. Pellikaan, H. Stichtenoth and F. Torres, ”Weierstrass semigroups in an asymptotically good tower of function fields,” to appear in Finite Fields and their Applications. [2] A. Garcia and H. Stichtenoth, ”On the asymptotic behaviour of some towers of function fields over finite fields,” J. Number Theory, vol. 61, pp. 248-273, 1996.

8