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On the Jacobian Varieties of Picard curves: explicit Addition Law and Algebraic Structure

Jorge Estrada Sarlabous Ernesto Reinaldo Barreiro Jorge Alejandro Pi~neiro Barcelo

Department of Geometry and Combinatorics.

ICIMAF. Ministry of Sciences.

Calle E No.309, esquina a 15 Vedado 4, C. Habana. Cuba.

Abstract

In this paper a system of coordinates for the e ective divisors on the Jacobian Va- riety of a Picard curve is presented. These coordinates possess a nice geometric inter- pretation and provide us with an unifying environment to obtain an explicit structure of algebraic variety on the Jacobian as well as an ecient algorithm for the addition of divisors.

1 Introduction

In the middle '80, D. Mumford laid a corner stone for the study of families of special curves and their Jacobian Varieties. In his book "Tata Lectures on Theta II", 15], he presented a coordinate system on an Zariski open subset of the Jacobian Variety of an hyperelliptic curve, which facilitated him and several authors obtainig many concrete and explicit infor- mation: a projective model of hyperelliptic jacobians, a characterization of period matrices arising from hyperelliptic curves, algebraic and dierential identities for hyperelliptic theta functions, among others. In particular, D. Cantor 1] used these coordinates to develop an algorithm to compute the addition of divisors on hyperelliptic jacobians .

More recently, another special families of curves has been focused with increasing interest:

the Picard curves (and a generalization containing both Picard and hyperelliptic curves, the

Supp orted partiallybyaDFGgrant

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n-gonal cyclic curves, also called cycloelliptic or superelliptic). These curves play a central role in some approaches to generalizations of Hilbert problems 7, 12, 21 and 22, in special dierential equations and many other researches 4], 5], 6], 9], 10], 11], 12].

The authors felt motivated to nd out, to which extent the Picard curves (or more general, the n-gonal cyclic curves) could share with the hyperelliptic curves the nice property of mak- ing things to become explicit. In this paper, we show our rst results in this direction.

We wish to thank R.-P. Holzapfel for his valuable comments, discussions and encouragement.

2 Preliminaries

Let k be an arbitrary eld. we write k for its algebraic closure. The ane space Ank con- sists of all the points f(x1 x2 ::: xn) xi 2 kg. Points in the projective space Pnk consist of equivalence classes of points in Ank+1nf(0 0 :: 0)g where (x0 x1 :: xn) and (y0 y1 :: yn) are equivalent if there is c 2k (c 6= 0), such that xi =cyi for all i = 0 1 :: n. Note that A2k is naturally embedded in Pk2 by the map (x y)!(x y 1).

If G 2 kx y] is an homogeneous polynomial then, G dene an algebraic subset of Pk2. If H is other polynomial, we write G = H if G and H are equivalent up to a non-constant factor. An irreducible plane projective curve of degree n is dened as the zero-set of a given irreducible, homogeneous polynomial of the same degree. We say that a curve is non-singular if its polynomial has a well dened tangent in every point. We let1 denote a xed point of C.We use the notation of Fulton 7]. A divisor D on C is a formal sum D = PP2CnPP of points in C where nP = ordP(D) is an integer and nP = 0 for all but a nite number of points P. The set of points supp(D) = fP 2 C=nP 6= 0g is called the support of D. The degree of D is PP2CnP and D is eective if nP 0 for all P. We say D D1 if D;D1 is eective and put D0 =D;n11. A divisor D is called ane if D = D0.

The set of all divisors on C form an additive group Div(C) of which the divisors of degree t form a subsetDivt(C) and Div0(C) is a subgroup. Div+t(C) is the subset of efective divisor of Divt(C)

The eldk(C) of rational functions on C is the eld of fractions of the graded ring kx y z]=(F), where F is the homogeneous polynomial denig C. The local ring OC P of C at P is the collection of rational functions dened at P. If P is non-singular then, OC P is a discrete valuation ring and there exist a local parameter t 2OC P, such that every z 2OC P can be written as z = utnwhereu is a unit and n2Z is the order ofz at P, written ordCP(z). With help of this representation we can express everyz 2K(C) as z =Pli;1=0aiti+ml (ai 2K(C) and ml 2 ftlK(C)g). To require z to have order at least l at P is then equivalent to the system of equations ai= 0 (for i = 0 to l;1), which are linear in the coecients ofz.

The divisor of f = F1=F2 is uniquely dened by : (f) = X

P2CordCP(f)P 2

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where :

ordCP(f) = ordCP(F1);ordCP(F2)

It follows from Bezout's theorem that every such divisor has degree 0. A divisor of the form (f) for some rational function f is called principal. We write P for the subgroup of Div0(C) of principal divisors and two divisors D and D1 are called linearly equivalent (D = D1) whenever D and D1 are in the same coset of Div0(C)=P i.e. when there is a rational function f such that (f) + D = D1.

For any divisor D one dene :

L(D) =ff 2K(C)=(f);Dgf0g

Let k(C) be the space of dierentials of k(C) over k. Let ! 2 be a dierential (! 6= 0) and P 2 C. If we choose a local parameter t 2 OC P we may write ! = fdt for some f 2k(C), and so dene ordCP(!) = ordCP(f).

Similar to (f) the divisor of ! is uniquely dened by : (!) = X

P2CordCP(!)P and to any divisor D we may associate :

(D) =f! 2=(!)Dgf0g

If we put l(D) = dimk(L(D)) and (D) = dimk((D)) then the Riemann-Roch Theorem states :

l(D);deg(D) = (D) + 1;gC

where gC =(0) is called the genus of C.

De nition 2.1

(Cyclic n-gonal curves.)

An algebraic curve C dened over a eld k is called n;gonal cyclic is there exists a non trivial 2Autk(C), such that n=idC, and C=() = Pk1.

Let k be an algebraic closed extension of k and C=k be an n;gonal cyclic curve then, (C k k)=() = Pk1 and if char(k) = 0 or n 6 j char(k) 6= 0, the eld of rational functions of C=k, R(C) is a Galois extension of the eld of rational functions of Pk1, k(x), therefore (see 4], 6]) there exists an element y 2 R(C), such that (y) = y, R(C) = k(x)(y) and yn2k(x), for primitive n-th root of unity. So, C=k has the ane model:

C : yn =pm(x) = a0(x;a1)m1:::(x;as)ms (x y) = (x y)

with m| 2 Z, a| 2 k, a{ 6= a| for { 6= |, m = Pm|. Here, we may assume that n 6 j m (otherwise, we can construct a birational equivalent ane model ofC=k with this property).

Ifn > char(k), m|= 1 (i.e.,pm(x) has not double roots) and m = n+1, the projective model 3

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of C, Ch :zm;nyn =zmpm(x=z) is non singular. Since there is not wild ramication in this case, by the Riemman-Hurwitz formula we can compute the genus ofC : g(C) = n(n;1)=2 .

Local parameters 2.1

Let C be a n;gonal cyclic curve, with non singular ane model as above (m = n + 1).

At P = (x0 y0) y0 6= 0, t = x;x0 is a local parameter.

At P = (a| 0), t = y is a local parameter

At the point 1= (0 : 1 : 0) on zyn=Qm{=1(x;za{), t = yn;1=xn is a local parameter

De nition of Picard curves 2.1

A Picard curve is a genus three trigonal curve, i.e., n = 3 m = 4.

In what follows we deal with non-singular Picard curves, so we assume char(k)6= 2 3. More details can be found in 5] and 6].

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3 Reduction algorithm

Let C be a non-singular cyclic curve of degree n (n > 2) with an ane model yn;1 =pn(x) and D2J(C). Using the fact: (x;xP) =Pni=0;2iP;(n;1)1we may obtain an equivalent divisor of the form Pai=1Pi ;a1. Such a divisor is called a semireduced representant and it is called reduced representant if agC. From Riemann-Roch theorem follows that every divisor D has a reduced representant. Given a divisor D, the problem of obtaining a reduced equivalent is known as The Reduction Problem. A general solution can be found in 13], nevertheless, in the case of a non-singular cyclic curve, we present in this paper a solution admiting a nice geometric interpretation. In the particular case of Picard curves, we obtain an algorithm for the reduction problem using O(deg(D)) operations.

Let's begin assigning to every efective divisor D with D0 2 Div+t , vD 2 kx y] of lowest degree at 1 such that (vD)0 D0 and :

ord1(vD)

( 0 ift m(m + 3)=2

m(m + 3)=2;t if t < m(m + 3)=2 where m is the degree of vD and :

deg1(X

i j aijxiyj) =maxi j(nj + (n;1)i) for every Pi jaijxiyj polynomial function overC.

The vector space of plane curves of degree m has dimensionm(m + 3)=2 over k, so vD

always exist, and by denition, if there are two curves through Pi its dierence has lower degree at 1. Hence vD is unique.

Let's see howvD above permits us to translate the reduction problem to the computation of several curve intersections.

If t (n;1)(n + 2)=2 then m < n and we can assure that gcd(C vD) = 1, since C is irreducible.

(vD) =D0+D1;(ndeg(vD);ord1(vD))1 (1) where deg(D1)mn;m(m + 3)=2 . Due to (2n;3)2 ;(2n;4)(2n + 2) < 0 if n > 2 we have mn;m(m + 3)=2 < (n;2)(n + 1)=2 and so deg(vD1)n;2 and :

(vD1) =D1+D2;(ndeg(vD1);ord1(vD1))1 (2) Substracting (1) and (2) we have :

D0;(ndeg(vD);ord1(vD))1= D2;(ndeg(vD1);ord1(vD1))1 On the other hand, from the fact that the function :

F(m) = mn;m(m + 3)=2;(n;1)(n;2)=2 5

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has its zeros in m = n ;1 , m = n ; 2 follows that if m n ; 2 is F(m) 0 so deg(D2)(n;1)(n;2)=2 = gC:

An algorithm for reduction could be the following. Take an eective divisor D, if deg(D0) gC we have nished, and if it not the case, take an efective ane divisor D0 (D D0) with gC < deg(D0)(n;1)(n + 2)=2 and put :

Da:=D;D0 +D2+ (ndeg(vD0);ord1(vD);ndeg(vD1) +ord1(vD1))1

D0 , D2 are ane with deg(D0)> deg(D2) hencedeg(Da0)< deg(D0) and return to the rst step with D := Da.

Let's illustrate the reduction algorithm for the case of the reduction of a degree 4 ef- fective divisor on a Picard curve to an linearly equivalent divisor of degree 3: Put D0 = P1+P2+P3+P4and consider the conicvD0, which interpolates the Picard curveC at D0+1. This conic interpolates C in three further points H1 H2 and H3. Set D1 = H1 +H2 +H3 and consider the conic vD1 interpolating C at D1+ 21. The conic vD1 cuts C in three ad- ditional pointsP10 P20andP30, and the divisorD2 =P10+P20+P30is the desired reduction ofD0.

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In the general case, the algorithm denes a sequence of divisorsD0 D1 D2 :: D3s+2 such that :

D3s+2 ;(ndeg(vD3s+1);ord1(vD3s+1))1= D3s;(ndeg(vD3s);ord1(vD3s))1 The sequence is nite, so this procedure give us a solution for the reduction problem, never- theless, the computation of Di involve factorizations in each step fromv3i to v3i+1 and from v3i+1 to v3i+2, and this may be computationally expensive. With the objetive to give to the sequence Di a more implicit treatment we associate to D in a natural way the polynomial :

uD =Ys

i=1(x;xi) with D0 =Psi=1Pi(xi yi) and the polynomial :

wD =Ry(vD C)=uD

where Ry( ) denotes the resultant with respect to y.

We denote (D) = (uD vD wD)2kx]kx y]kx].

Suppose now thatC is a Picard Curve, that is gC = 3 anddeg(C) = 4. We will concen- trate in what follows in efective ane divisors D3i of degree 4 and put D3i+3 :=D3i+Ei to form the sequence Di associated to D.

For D eective, the algorithm will consist in the iteration of the following steps : 1)Take D0 D (eective and ane of degree four). Put D := DnD0.

2)Compute (D0).

3)If D0 is generic then, compute (D1) and (D2) from (D0) after lemma 3.3 and lemma 3.5.

4)Take a new divisor E1 =P1+P10 D of degree 4;deg(u2). If this E1 does not exist then nish.

5)If D2+E1 is generic then, based on (D2) compute (D2+E1) which represents the coor- dinates of an eective divisor (D3) of degree 4 and return to step 1) with (D0) := (D3).

With help of Lemma 3:2 we nd that non-generic cases are easier than generic ones, since in these cases we may explicitly compute the coordinates of the points, without factorization.

lemma3.1 give a usefull relationship betweenv(D) and D that characterize non-generic cases.

Lemma 3.1

Given an eective divisorD of degre4the following propositions are equivalent : i) v(D) is linear or factorizes in linear factors.

ii) there exist among Pi 2Supp(D +1) three collinear points.

iii) v(D) = a20x2+a10x + a00+a11xy + a01y anda00a211;a10a01a11+a20a201 = 0. 7

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Proof: ii)) i) If P1 +P2 +P3 2 Supp(D +1) and (r)0 P1 +P2 +P3 for some line r then, r and v(D) have more than two points in common (counting multiplicity) because deg(v(D)) > 1 )ord1(v(D))5;deg(D0). By Bezout Theorem r divides v(D).

i)) iii) If deg(v(D)) = 1 ) a20 = a11 = 0 ) iii). If deg(v(D)) > 1 we have already seen ord1(v(D)) 1 so v(D) = r1r2 with ord1(r1) 1. This imply r1 = (x;x) with_ a20x_2+a10x + a_ 00 =a11x + a_ 01 = 0 , so :

Rx(a20x2+a10x + a00 a11x + a01) =a00a211;a10a01a11+a20a201= 0 iii))ii) If a11 =a20= 0, ii) is clear. If a11=a01= 0 then :

v(D) = a20x2+a10x + a00 =r1r2 Else, it can be shown :

v(D) = (a20x + a11y + a10;a20a01=a11)(x + a01=a11) =r1r2

In any case Supp(D +1) has ve points, so at least three of then belong to r1 or to r2. 2 From lemma 3:1 we obtain deg1(v3i)< 8 and by (1) is deg(D3i+1)< deg(D3i) = 4.

Furthermore, if deg(v3i)> 1 the degree at1 of v3i is not less than 6 so deg(D3i+1)> 1 and we can show v3k+2 =v3k+1 as follows :

If deg(u3k+1) = 2 ) deg(v3k+1) = deg(v3k+2) = 1 and so v3k+1 = v3k+2. If the points in support of D3k+1 are collinear, by lemma 3.1 there exist three points P1 P2 P3 inSupp(D3i) that belong to Supp((x;x)_ 0) , so v3i+1 = (x;x)(x_ ;x4) with P4 =D3i;(P1 +P2+P3) and so deg(u3k+1) = 2. That is, if deg(u3k+1) = 3 , the points in support of D3k+1 are not collinear and deg(v3k+2) =deg(v3k+1) = 2, hence v3k+1 =v3k+2.

Lemma 3.2

Suppose that we know E1 and E2 in the sequence Di associated to an eective divisor D. Ifdeg(v0)> 1 , deg(v3)> 1 and Supp(D5)have a pair of conjugate then, we may compute D5 andD6 without making any factorization in each of the following cases :

a)We know the vectors (Di) for i = 0 to 5. b)we have (D5) and explicitly know D3. Proof:

D5 =P + P + Q v5 =v4 = (x;xP)(x;xQ)

D4 =2P + Q + 2Q v3 =r3(x;xQ)

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We obtain the x-coordinate of 2P as the root of the linear equation u5=v5 = 0 and the y-coordinate is obtained fromr3(2P) = 0. In order to nd Q we proceed as follows :

1) Take the valuexQ as the root of the linear polynomialv25=u5.

2) If we explicity know D3 take the point in D3 that does not belong to r3 and nish.

Else, supose that Q is the set of ane points of C over xQ then, take the set : S = Q\(Supp(E1)Supp((v2)0))

Q 2S so, 0 < card(S) < 4 and if L is a point of S nfQg ,L2D3 )r3(L) = 0.

3) If card(S) = 3 and v2 does not depend on y then for k6= 3p is : D2 =T + kT + R

If T 6=Q then (r3)0 T + kT )2P =1 so,T = Q , r3(kQ) = 0 and : D2 =Q + kQ + R

D1 =2R + R + 2kQ v0 =r0(x;xR)

Then r0(2kQ) = 0 since otherwise D0 = (r0)0. We can take Q as the point L of S such that r0(L)6= 0 and r3(L)6= 0.

4) If card(S) = 3 and v2 does depend on y then : E1 =kQ + k+1Q If k = 3p + 1 we have :

D3+D4 Q + 2Q + 22Q)(r3)0 Q + 2Q)2P =1 Hence, we can take Q as the point L in Supp(E1) withr3(L)6= 0.

5) If card(S) = 1 take Q2S.

6) if card(S) = 2 then take Q1 2Supp(E1) laying over xQ. Ifr3(Q1)6= 0 then Q = Q1, else take Q as the point in SnfQ1g.

Given Q , 2P and E2 we can construct D5 and D6. 2

Lemma 3.3

Suppose that we know the vectors (Di) for i = 0 to 6 and E1, E2 and E3 in the sequence Di associated to an eective divisor D. If deg(v3i)> 1 for 0 i < 3 then, we may compute (D7), (D8) and (D9) without making any factorization.

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Proof: Denote

u7 =degX(u7)

l=0 (;1)lslxl u8 =degX(u8)

l=0 (;1)ls0lxl

Now note that ifi = 7 8 ui =wi;1 wherewi;1 means the quotient ofwi;1 by its leading coecient. On the other hand, using u8 and the points in E3 we can obtain u9, so let's concentrate in the vi.

Let's state several cases for the step fromv6to v7 (1 :: 5) and from v8tov9 (i;1 :: i;3):

Case 1: D6 =P1+P2+P3+P4 (without three collinear points or pairs of conjugate).

Case 1.1: a116= 0, so the curves :

v6 =a20x2+a10x + a00+a11xy + a01y v7 =b20x2+b10x + b00+b01y

are sharing three points besides 1, which is equivalent to the equation :

Ry(v6 v7) =b01(a20x2+a10x + a00);(a11x + a01)(b20x2+b10x + b00) =u7 (3) and solving the system we obtain :

b20=;=a11

b10=(s2a20a01+a20a11s3;a11a00(s1+a01=a11)+a01a10(s1+a01=a11))=(;a00a211+a10a01a11;a20a201) b00=(s2a11a00;a20a01s3+a11a10s3+a01a00(s1+a01=a11))=(;a00a211+a10a01a11;a20a201)

b01=(s2a11a01+a211s3+a201(s1+a01=a11))=(;a00a211+a10a01a11;a20a201) Case 1.2: a11= 0, that is :

v6=a20x2+a10x + a00+a01y v7 =b10x + b00+b01y

b01(a20x2+a10x + a00);(a01)(b10x + b00) =u7 b01==a20

b10 =(;s1=a01+a10=a20a01) 10

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b00 =(;s2=a01+a00=a20a01)

Case 2: D6 =P + P + Q1+Q2 (withQ1 6=Q2 ,2Q2 and Qi 6=2P).

By lemma 3.1, v6 =r6(x;xP) and, ifv5 factorizes, we may apply lemma 3.2 in order to nd the coordinates of 2P. If is not the case we may obtain these coordinates from Supp(E2) and evaluating xP inv5. Now ifr6(P) = 0 or r6(P) = 0 then :

v7= (x;xP)Ry(C r6)(x;xP)=u6

Otherwise, if there are not collinear points in Supp(D6) proceed as in case 1 :

Ry(r6 v7) =b01(;a20=a11x + a20a01=a211;a10=a11) + (b20x2+b10x + b00) =u7=(x;xP) b20= b00=s2+(a10;a20a01)b01 b10=s1;a20b01. Ifr6 andv7 have order of contacts in 2P, b01 is chosen such that v7 and C have at least contact s+1 at this point (It is possible due to the dimension over k of the space of plane conics). Case 3: D6 =Q1+Q2+Q3+Q4 (without pair of conjugate, but with three collinear points).

v6 =r6(x;xP) and Ry(C r6)(x;xP)=u6 is linear. Hence : v7= (x;xP)Ry(C r6)(x;xP)=u6 Case 4: D6 =P + P + 2P + Q.

v6 = (x;xQ)(x;xP) , u6=v6 =ax2+bx + c and : v7 =v6=(x + b=2a) Case 5: D6 =P + P + Q + Q.

Either we may apply lemma 3:2, or we can construct D6 as the divisor E2 + the points on v5 with the same x-coordinates as the points of E2. Consider the line r through 2P and 2Q then, v7=r:

Case i-1: D8 =Q1+Q2+Q3 (without pairs of conjugate).

If (v8)0 P3 =E3 and u8(x3)6= 0 , then v9 =v8. Else, proceed as in case 1.1 to obtain : a020= (a011(b10+s01b20) +b20a001)=b01

a010= (a011(b00;s02b20) +b10a001)=b01 a000= (a011s03b20+b00a001)=b01

Now, proceed as in case 2 takinga001 and a011 (a0116= 0), such that (v9)0 E3+D8. Case i-2: D8 =P + P + Q (with Q6=2P).

Use lemma 3.2 to nd the line r through P3 and Q, then : v9 =r(x;xP)

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Case i-3: D8 =Q1+Q2.

In this case E3 = P3 +P30. If Q1 = kQ2 and r is the line through P3 and P30 then v9 =r(x;xQ1). Otherwise ifQ1 andQ2 are not conjugate we solve some particular subcases : i)IfP3 P30 are conjugate then,v9 =v8(x;x3)

ii)Ifu8(x3) = 0 andv8(P3) = 0, due to deg(u8) = 2 , we can compute the roots of u8 = 0 and using v8, the points in supp(D8 +E3).

iii)In the remaining cases, we compute :

Ry(v8 v9) =b01(a20x2+a10x + a00);(a11x + a01)(b10x + b00) =u8 From here, we obtain :

a010= (b10a001+a011b00;s01)=b01 a010= (b00a001+s02)=b01

a010 = (a011b10+)=b01

And now, we select the free parameters a001 , a011, such that (v9)0 P3 +P30 . This lead us to one of the solvable systems :

( a011(b10x23+b00x3 +b01x3y3) +a01(b10x3+b00+b01y3) =s01x3;s02;x23 a011(b10x032+b00x03 +b01x03y03) +a01(b10x03+b00+b01y03) =s01x03;s02;x032

( a011(b10x23+b00x3 +b01x3y3) +a01(b10x3+b00+b01y3) =s01x3;s02;x23 a011(2b10x3+b00+b01y3+x3b01@y=@x)+ a01(b10+b01@y=@x) = s01;2x23

Depending onP3 =P30 or not. 2

Lemma 3.4

Suppose that we know the vectors (Di) for i = 0 to 6 and E1, E2, E3 in the sequence Di associated to an eective divisor D. If deg(v0) > 1 and deg(v3) > 1 then, we may decide in which case of lemma 3.3 D6 and D8 are.

Proof: Let's consider D6 :

v6 =a20x2+a10x + a00+a11xy + a01y

Ifv6 does not factorizes and a116= 0 then, D6 is in case 1.1. If v6 does not factorizes and a11= 0 , we are in case 1.2. If a11=a01= 0 then, we may use of lemma 3.2 to decide if we are in case 4 or 5.

If a11 =a20 = 0 then, the four points are collinear, that is, D6 = 0 and we begin with new 12

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points.

If a11 6= 0, but v6 factorizes, we proceed as follows : v6 =r(x;x0)

First if deg(E2) = 2 and the points P2, P20 in Supp(E2) are conjugate, we are in case 2 and if it is not the case we may go on.

If v5 does not depend ony, or if v5 passes through a conjugate ofP2 or of P20 andu5(x0) = 0 then, it is case 2. Else it is case 3.

Let's considerD8 :

v7 =b20x2+b10x + b00+b01y

Ifb20= 0 we are in case i-3, if b01= 0 we are in case i-2, otherwise we are in case i-1. 2

Lemma 3.5

If we knowD0 , E1 and E2 in the sequence Di associated to an eective divisor D then, we can compute(Di) for i = 0 to 6.

The precedent lemmas permit us to show the theorem :

Theorem 3.1

Given an eective ane divisor D, it is possible to compute a reducedDf;

31equivalent toD;deg(D)1performing only one factorization and O(deg(D))operations.

Proof: 1)Take an eective ane divisor D.

2)If deg(D) < 4 then D is already reduced Df := D and go to 20). Else, we can take D0 D an eective divisor of degree four.

3)Compute (D0) and put D := D;D0.

4)If deg(v0) = 1 then go to 2), else compute (D1) and (D2) fromD0 and lemma 3:5.

5)If 4;deg(u2)> deg(D) > 0 compute (D3) using D and lemma 3:5, put (Df) = (D3) and go to 19).

6)If deg(D) = 0 then (Df) = (D2) and go to 19).

7) Take an eective divisor E1 = 1P1 +1P10 (D E1) with deg(E1) = 4;deg(u2).

Using lemma 3:5 we can construct (D3). D := D;E1.

8)If deg(v3) = 1 then go to 2), else compute (D4) (D5) applying lemma 3:5.

9)If 4;deg(u5)> deg(D) > 0 compute (D6) using D and lemma 3:5, put (Df) = (D6) and go to 19).

10)If deg(D) = 0 then (Df) = (D5) and go to 19).

11) Take an eective divisorE2 = 2P2 +2P20 (D E2), with deg(E2) = 4;deg(u5).

Using lemma 3:5 we can construct (D6). D := D;E2. 13

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12)If deg(v6) = 1 then go to 2), else compute (D7) (D8) applying lemma 3:3.

13)If 4;deg(u7)> deg(D) > 0 compute (D9) using D and lemma 3:3, put (Df) = (D9) and go to 19).

14)If deg(D) = 0 then (Df) = (D8) and go to 19).

15) Take an eective divisor E3 = 3P3 +3P30 (D E3) with deg(E3) = 4;deg(u8).

Using lemma 3:3 we can construct (D9). D := D;E3. 16)Put (Di) := (Di+3) for i = 0 1 2 3 4 5 6.

17) Pi :=Pi+1 i :=i+1 i:=i+1 for i = 1 2.

18)Go to 12).

19)If vf does not depend on y apply lemma 3:2 to obtain Df, if not, factorize uf and evaluate in vf to obtain the points in Supp(Df).

20)Finish.

If we assume unit cost for all operation in the given eldk, in each step of the algorithm we perform O(1) operations. We iterate these steps O(Deg(D)) times so, this give a total

of O(Deg(D)) operations to compute Df. 2

3.1 Numerical examples

Example 1.

If k = Z23 ,C : y3 =x4;1 and D = (1 0) + (22 0) + (8 1) + (18 12)

(D) = (x4+ 20x3+ 5x2+ 3x + 17 10 + 14y + 8xy + 13x3 17(x3+ 16x2+ 22x + 5)) and the algorithm gives the result :

(Df) = (x3+ 12x2+x + 8 21 + 17x + 7y + x2 4(x3 + 16x2+ 22x + 5)) Df = (21 21) + ( 14 + 22) + (13 + 22 1 + )

where 2 + 10 + 4 = 0.

Example 2.

If k = Z13 and C : y3 =x4;1 then,

(0 4) + (4 2) + (6 2) + ( 11 + 4) + (12 + 6 9(1 + ));51= 2(1 0);21 where 52 + 9 + 9 = 0.

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4 Algebraic Structure

Let C be a Picard curve, J(C) its Jacobian Variety, both dened over the eld k. In the previous section of this paper, it is shown that for any eective divisorD = Q1+Q2:::+Qnof degree n, an eective divisor D0 =P1+P2+P3 can be constructed, such that P| are dened on an algebraic extension of the eld of denition ofD, on J(C) holds Q1+Q2:::+Qn;n1= P1+P2 +P3 ;31 and D0 may be explicitly computed in terms of the coordinates of the points in Supp(D) .

If we set Div+n(C) :=fQ1+Q2::: + Qng, then the map : { : Div+n(C) ! J(C)

Q1+Q2::: + Qn P1+P2+P3 ;31

is surjective. Moreover, on Div0+3(C) :=fD = Q1+Q2+Q3=Q| 6=1 L(D) = kg, { is also injective, since for D1 6=D2,{(D1) ={(D2) impliesD1;D2 = (h) h2L(D2) h 6= constant.

For any smooth projective curve C, it is possible to describe Div+n(C) as a projective variety using the bijection Div+n(C)= Symmn(C), the orbit space of Cn under the action of the symmetric group permuting the factors. Given an embedding C ,!PN, we have the associated Segre embedding

j : Cn,!P(N+1)n;1

In the case of Picard curves embedded in P2, j : C3 ,!P26, and the dening equations in such high dimensional projective space have not been explicitly obtained. An alternative approach should be to take as coordinates for the elements of Div+3(C) the coecients of its Chow form (see 2], 3] and 13]), which are elements ofP9. Nevertheless, the construction of an explicit set of dening equations (consisting at least of 6 polynomial equations) for the image of Div+n(C) in P6 happens to be more complicated than in the ane model that we present in this paper.

Using as coordinates the coecients (u{ v|) of the conics and cubics associated to a di- visor in Div+3(C) in the previous section of this paper, we are able to construct a model of algebraic variety for the Jacobian Variety of a Picard curve dened over an algebraic closed eld of characteristic greater than three or equal to zero. The open sets of its atlas are iso- morphic to an open ane subset of A6, which is a complete intersection of three polynomial equations in the variables (u{ v|). These dening equations are computed explicitly.

The following result is strongly motivated by Mumford's works on hyperelliptic curves 15]. Therefore, many steps of the proof happen to be analogous to the hyperelliptic case and here we will simply refer Mumford's book for a detailed argumentation. Nevertheless,

15

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those aspects which needed to be generalized will be meticulously explained, since in these places special features of the geometry of Picard curves have been exploited.

Theorem 4.1

(Explicit structure of algebraic variety for J(C).)

The Jacobian Variety J(C) of a Picard curve contains a subgroup of 6;torsion points, T and a Zariski open subset, Z, such that:

a)Z is isomorphic to an algebraic variety. Furthermore,Z is complete intersection of three polynomial equations in A6, which may explicitly given.

b) T = (Z=2Z)2(Z=3Z)3 and J(C) =t2T ( Z + t )

c) The family A = fZ+t=t2 Tg is the atlas of a structure of algebraic variety on J(C) . Moreover, A endows J(C) with a structure of abelian variety.

Proof. a) First we will prove:

Proposition 4.1

Div+0 3(C) =fD = P1+P2+P3=P| 6=1 P| not collinear g

Proof. The canonical class K of the Picard curve C is the class of divisors of zeroes arising from the intersection of straight lines with C, since B =fdx=y2 xdx=y2 dx=yg is a basis of abelian dierentials on C (see 4]). Applying the Riemann-Roch theorem to D = P1+P2+P3 (deg(D) = 3 g = 3) ,

l(D) = l(K ;D) + 1

Therefore, l(D) > 1 (i.e. L(D)6=k) if and only if K D, hence the points P| are collinear.

Our aim is to parametrize a convenient subset of the {-image of Div0+3(C), since on Div0+3(C) the injectivity of { holds. The coecients of the conics and cubics associated to a divisor in the rst section of this paper may be a suitable set of coordinate functions, once a bijection between open subsets can be ensured.

SetDiv1+3(C) :=fD 2Div+0 3(C) / there are not three collinear points in Supp(D+1)g. To any divisor D 2Div1+3(C) we have associated a conic

v(D) = v00+v10x + v20x2+v01y (v(D) C)D + 21 and a cubic

u(D) = u3x3+u2x2+u1x + u0 (u(D) C)D + D + 2D Since D2Div+13(C), we may assume v01 = 1 andu3 = 1:

This set up a map

Div1+3(C) ,! A6k

D = P1 +P2+P3 (D) = (u0 u1 u2 v0 v1 v2) 16

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