nexoncn.com

文档资料共享网 文档搜索专家

文档资料共享网 文档搜索专家

当前位置：首页 >> >> Elliptic Curves from Sextics

ELLIPTIC CURVES FROM SEXTICS

MUTSUO OKA

arXiv:math/9912041v3 [math.AG] 16 Dec 1999

Abstract. Let N be the moduli space of sextics with 3 (3,4)-cusps. The quotient moduli space N /G is one-dimensional and consists of two components, Ntorus /G and Ngen /G. By quadratic transformations, they are transformed into one-parameter families Cs and Ds of cubic curves respectively. First we study the geometry of Nε /G, ε = torus, gen and their structure of elliptic ?bration. Then√we study the Mordell-Weil torsion groups of cubic curves Cs over Q and Ds over Q( ?3) respectively. We show that Cs has the torsion group Z/3Z for a generic s ∈ Q and it also contains subfamilies which coincide with the universal families given by Kubert [Ku] with the torsion groups Z/6Z, Z/6Z + Z/2Z, Z/9Z or Z/12Z. The cubic curves Ds has torsion Z/3Z√ + Z/3Z generically but also Z/3Z + Z/6Z for a subfamily which is parametrized by Q( ?3).

1. Introduction Let N3 be the moduli space of sextics with 3 (3,4)-cusps as in [O2]. For brevity, we denote N3 by N . A sextic C is called of a torus type if its de?ning polynomial f has the expression f (x, y ) = f2 (x, y )3 + f3 (x, y )2 for some polynomials f2 , f3 of degree 2, 3 respectively. We denote by Ntorus the component of N which consists of curves of a torus type and by Ngen the curves of a general type (=not of a torus type). We denote the dual curve of C by C ? . Let G = PGL(3, C). The quotient moduli space is by de?nition the quotient space of the moduli space by the action of G. In §2, we study the quotient moduli space N /G. We will show that N /G is one dimensional and it has two components Ntorus /G and Ngen /G which consist of sextics of a torus type and sextics of a general type respectively. After giving normal forms of these components Cs , s ∈ P1 (C) and Ds , s ∈ P1 (C), we show that the family Cs contains a unique sextic C54 which is self dual (Theorem 2.8) and C54 has an involution which is associated with the Gauss map (Proposition 2.12). In section 3, we study the structure of the elliptic ?brations on the components Nε /G, ε = torus, gen which are represented by the normal families Cs , s ∈ P1 (C) and Ds , s ∈ P1 (C). Using a quadratic transformation we write these families by smooth cubic curves Cs and Ds which are de?ned by the following cubic polynomials. √ √ Ds : ?8x3 + 1 + (s + 35)y 2 ? 6x2 + 3x ? 6 ?3y ? 3 ?3x √ √ ?6 ?3x2 ? 12 ?3xy + (s ? 35)xy = 0

Date : November, 1999, ?rst version.

1

Cs : x3 ? 1 s(x ? 1)2 + sy 2 = 0 4

We show that Cs , s ∈ P1 (C) (respectively Ds , s ∈ P1 (C)) has the structure of rational elliptic surfaces X431 (resp. X3333 ) in the notation of [Mi-P]. In section 4, we study their torsion subgroups of the Mordell-Weil group of the cubic families Cs and √ Ds . The family Cs is de?ned over Q and Ds is de?ned over quadratic number ?eld Q( ?3). Both families enjoy beautiful arithmetic properties. We will show that the torsion group (Cs )tor (Q) is isomorphic to Z/3Z for a generic s ∈ Q but it has subfamilies C?6 (u) , C?6,2 (r) , C?9 (t) and C?12 (ν ) , u, r, t, ν ∈ Q for which the Mordell-Weil torsion group are Z/6Z, Z/6Z + Z/2Z, Z/9Z and Z/12Z respectively. Each of these groups is parametrized by a rational function with Q coe?cients which is de?ned over Q and this parametrization coincides, up to a linear fractional change of parameter, to the universal family given by Kubert in [Ku]. √ √ As for (Ds )tor (Q( ?3)), we show that (Ds )tor (Q( ?3) is generically isomorphic to Z/3Z + Z/3Z but it also takes Z/3Z + Z/6Z for a subfamily Dξ6 (t) parametrized by a √ rational function with coe?cients in Q and de?ned on Q( ?3). 2. Normal forms of the moduli N We consider the submoduli N (1) of the sextics whose cusps are at O := (0, 0), A := (1, 1) and B := (1, ?1). As every sextic in N can be represented by a curve in N (1) by the action of G, we have N /G ? = N (1) /G(1) where G(1) is the stabilizer of N (1) : G(1) := {g ∈ (1) (1) G; g (N ) = N }. By an easy computation, we see that G(1) is the semi-direct product (1) of the group G0 and a ?nite group K, isomorphic to the permutation group S3 where (1) G0 is de?ned by ? ? a1 a2 0 (1) 2 a1 0 ? ∈ G; a3 (a2 G0 := M = ? a2 1 ? a2 ) = 0 a1 ? a3 a2 a3 Note that G0 is normal in G(1) and g ∈ G0 ?xes singular points pointwise. The isomorphism K ? = S3 is given by identifying g ∈ K as the permutation of three singular locus O, A, B . We will study the normal forms of the quotient moduli N /G ? = N (1) /G(1) .

(1) (1) (1)

Lemma 2.1. For a given line L := {y = bx} with b2 ? 1 = 0, there exists M ∈ G0 such that LM is given by x = 0. Proof. By an easy computation, the image of L by the action of M ?1 , where M is as above, is de?ned by (a1 ? ba2 )y + (a2 ? ba1 )x = 0. Thus we take a1 = ba2 . Then 2 2 2 a2 1 ? a2 = a2 (b ? 1) = 0 by the assumption. Lemma 2.2. The tangent cone at O is not y ± x = 0 for C ∈ N (1) . Proof. Assume for example that y ? x = 0 is the tangent cone of C at O . The intersection multiplicity of the line L1 := {y ? x = 0} and C at O is 4 and thus L1 · C ≥ 7, an obvious contradiction to Bezout theorem. Let N (2) be the subspace of N (1) consisting of curves C ∈ N (1) whose tangent cone at O is given by x = 0. Let G(2) be the stabilizer of N (2) . By Lemma 2.1 and Lemma 2.2, we have the isomorphism :

2

Corollary 2.3. N (1) /G(1) ? = N (2) /G(2) . It is easy to see that G(2) is generated by the group G0 := G(2) ∩ G0 and an element τ of order two which is de?ned by τ (x, y ) = (x, ?y ). Note that ? ? a1 0 0 (2) (1) a1 0 ? ∈ G0 ; a1 a3 = 0} G0 = { M = ? 0 a1 ? a3 0 a3

(2) (1)

For C ∈ N (2) , we associate complex numbers b(C ), c(C ) ∈ C which are the directions of the tangent cones of C at A, B respectively. This implies that the lines y ? 1 = b(C )(x ? 1) and y + 1 = c(C )(x ? 1) are the tangent cones of C at A and B respectively. We have (2) shown that C ∈ Ntorus if and only if b(C ) + c(C ) = 0 and otherwise C is of a general type and they satisfy c(C )2 + 3c(C ) ? b(C )c(C ) + 3 ? 3b(C ) + b(C )2 = 0 (§4, [O2]). We consider the subspaces: √ (3) (2) (3) (2) := {C ∈ Ngen ; b(C ) = c(C ) = ?3} Ntorus := {C ∈ Ntorus ; b(C ) = 0}, Ngen Remark . The common solution of both equations: b + c = c2 + 3c ? bc + 3 ? 3b + b2 = 0 is (b, c) = (1, ?1) and in this case, C degenerates into two non-reduced lines (y 2 ? x2 )2 = 0 and a conic. Lemma 2.4. Assume that C ∈ N (2) . Then there exists a unique C ′ ∈ N (3) and an element g ∈ G(2) such that C g = C ′ . This implies that (2) (3) Ntorus /G ? /G(2) ? , Ngen /G ? =N =N = N (2) /G(2) ? = N (3)

torus torus gen gen

and we put N (3) := Ntorus ∪ Ngen .

(3)

(3)

The image Lg A is given by y ? x + xa3 ? a3 ? bxa3 + ba3 = 0. Thus we can solve the (1) equation a3 (1 ? b) ? 1 = 0 in a3 uniquely as a3 = 1/(1 ? b) as b = 1. Thus g ∈ G0 is unique if it ?xes the singular points pointwise and thus C ′ is also unique. It is easy to (3) see that the stabilizer of Ntorus is the cyclic group of order two generated by τ , as C ′ is (3) even in y (see the normal form below) and C ′ τ = C ′ for any C ′ ∈ Ntorus . Thus we have (3) (2) Ntorus /G(2) ? = Ntorus . (2) Consider the case C ∈ Ngen . Then the images of the tangent cones at A, B by the action of g are given by y ? x + xa3 ? a3 ? bxa3 + ba3 = 0 and y + x ? xa3 + a3 ? cxa3 + ca3 = 0 respectively. Assume that b(C g ) = c(C g ). Then we need to have a3 (1 ? b) ? 1 = a3 (?1 ? c) + 1, which has a unique solution in a3 , if (?) b ? c ? 2 = 0. Assume that c2 +3c ? bc +3 ? 3b + b2 = 0 and b ? c ? 2 = 0. Then we get (b, c) = (1, ?1) which is excluded as it corresponds to a non-reduced sextic. Thus the condition (?) is always satis?ed. Put (b′ , c′ ) := (b(C g ), c(C g )). They satisfy the equality c′ 2 + 3c′ ? b′ c′ + 3 ? 3b′ + b′ 2 = 0 and

3

Proof. Assume that C ∈ Ntorus , b + c = 0. Consider an element g ∈ G0 , ? ? 1 0 0 1 0? g ?1 = ? 0 1 ? a3 0 a3

(1)

(1)

√ √ b′ = c′ . Thus we have either b′ = c′ = ?3 or b′ = √ c′ = ? ?3. However in the second (3) case, (C g )τ belongs to the ?rst case. Thus b′ = c′ = ?3 and C g ∈ Ngen as desired. 2.1. Normal forms of curves of a torus type. In [O2], we have shown that a curve in (1) Ntorus is de?ned by a polynomial f (x, y ) which is expressed by a sum f2 (x, y )3 + sf3 (x, y )2 where f2 (x, y ) is a smooth conic passing through O, A, B and f3 (x, y ) = (y 2 ? x2 )(x ? 1). Proposition 2.5. The direction of the tangent cones at O , A and B are the same with the tangent line of the conic f2 (x, y ) = 0 at these points. This is immediate as the multiplicity of f3 (x, y )2 at O, A, B are 4. Assume that C ∈ (3) Ntorus , that is, the tangent cones of C at O , A and B are given by x = 0, y ? 1 = 0 and y + 1 = 0 respectively. Thus the conic f2 (x, y ) = 0 is also uniquely determined as (3) f2 (x, y ) = y 2 + x2 ? 2x. Therefore Ntorus is one-dimensional and it has the representation (2.6) Cs : ftorus (x, y, s) := f2 (x, y )3 + sf3 (x, y )2 = 0 For s = 0, 27, ∞, Cs is a sextic with three (3,4)-cusps, while C27 obtains a node. If τ g ∈ G(2) ?xes the tangent lines y ± 1 = 0, then g = e or τ . As Cs = Cs , this implies that g Cs = Cs . Thus Cs = Ct if s = t. 2.2. Normal form of sextics of a general type. For the moduli Ngen of sextics of a general type, we start from the expression given in §4.1, [O2]. We may assume b = c = √ ?3. Then the parametrization is given by fgen (x, y, s) := f0 (x, y ) + sf3 (x, y )2 , where s is equal to a06 in [O2] and f0 is the sextic given by √ √ f0 (x, y ) := y 6 + y 5(6 ?3 ? 6 ?3x) + y 4 (35 ? 76x + 38x2 ) (2.7) √ √ √ +y 3(?24 ?3x + 36 ?3x2 ? 12 ?3x3 ) + y 2 (?94x2 + 200x3 ? 103x4 ) √ √ √ +y (24 ?3x3 ? 42 ?3x4 + 18 ?3x5 ) + 64x3 ? 133x4 + 68x5 f3 (x, y ) = (y 2 ? x2 )(x ? 1)

Let Ds := {fgen (x, y, s) = 0} for each s ∈ C. Observe that D0 = {f0 (x, y ) = 0} is a sextic with three (3,4)-cusps and of a general type. For the √ computation of dual curves using Maple V, it is better to take the substitution y → y ?3 to make the equation to be de?ned over Q. Summarizing the discussion, we have Theorem 2.8. The quotient moduli space N /G is one dimensional and it has two components. (1) The component Ntorus /G has the normal forms Cs = {f (x, y, s) = 0} where f (x, y, s) = f2 (x, y )3 + sf3 (x, y )2 , f2 (x, y ) = y 2 + x2 ? 2x and f3 (x, y ) = (y 2 ? x2 )(x ? 1). The curve C54 is a unique curve in N /G which is self-dual. (2) The component Ngen /G has the normal form: fgen (x, y, s) = f0 (x, y )+ sf3 (x, y )2 where f3 is as above and the sextic f0 (x, y ) = 0 is contained in Ngen . This component has no self-dual curve. Proof of Theorem 2.8. We need only prove the assertion for the dual curves. The proof will be done by a direct computation of dual curves using the method of §2, [O2] and the above parametrizations. We use Maple V for the practical computation. Here

4

is the recipe of the proof. Let X ? , Y ? , Z ? be the dual coordinates of X, Y, Z and let (x? , y ?) := (X ? /Z ? , Y ? /Z ?) be the dual a?ne coordinates. ? ? (1) Compute the de?ning polynomials of the dual curves Cs and Ds respectively, us? ? ? ? ing the method of Lemma 2.4, [O2]. Put gtorus (x , y , s) and ggen (x , y , s) the de?ning polynomials. (2) Let Gε (X ? , Y ? , Z ? , s) be the homogenization of gε (x? , y ? , s), ε = torus or gen. Compute the discriminant polynomials ?Y ? Gε which is a homogeneous polynomial in X ? , Z ? of degree 30 (cf. Lemma 2.8, [O1]). Recall that the multiplicity in ?Y ? Gε of the pencil X ? ? ηZ ? = 0 passing through a singular point is generically given by ? + m ? 1 where ? is the Milnor number and m is the multiplicity of the singularity ([O2]). Thus ? the contribution from a (3,4)-cusp is 8. Thus if Cs has three (3,4)-cusps, it is necessary that ?Y ? (G) = 0 has three linear factors with multiplicity ≥ 8. (3-1) For the curves of a general type, an easy computation shows that it is not possible to get a degeneration into a sextic with 3 (3,4)-cusps by the above reason. (3-2) For the curves of a torus type, we can see that s = 54 is the only parameter such ? ? ? that Cs ∈ N . Thus it is enough to show that C54 = C54 . ? (4) The dual curve C54 of C54 is de?ned by the homogeneous polynomial G(X ? , Y ? , Z ? ) := 128X ? 5 Z ? + 1376X ?4 Z ? 2 ? 192X ? 3 Y ?2 Z ? + 4664X ?3 Z ? 3 ? 2X ? 2 Y ? 4 ?1584X ? 2 Y ? 2 Z ? 2 + 7090X ? 2 Z ? 4 + 58X ? Y ? 4 Z ? ? 3060X ?Y ? 2 Z ? 3 +5050X ?Z ? 5 + Y ? 6 + 349Y ? 4 Z ? 2 ? 1725Y ? 2 Z ? 4 + 1375Z ?6 We can see that C54 ? is isomorphic to C54 as ? ?4/3 ? 0 A= ?5/3

? A (C54 ) = C54 where ? 0 ?5/3 1 0 ? 0 ?13/3

2.3. Involution τ on C54 . For a later purpose, we change the coordinates of P2 so that the three cusps of Cs are at OZ := (0, 0, 1), OY := (0, 1, 0), OX := (1, 0, 0). A new normal form in the a?ne space is given by Cs : f2 (x, y )3 + sf3 (x, y )2 = 0 where f2 (x, y ) := xy ? x + y and f3 (x, y ) := ?xy . In particular, C54 is de?ned by (2.9)

? In this coordinate, C54 is de?ned by

f (x, y ) = (xy ? x + y )3 + 54x2 y 2 = 0

For a given A ∈ GL(3, C), we denote the automorphism de?ned by the right multiplication of A by ?A . Let F (X, Y, Z ) be the homogenization of f (x, y ). Then the Gauss map

5

? A1 It is easy to see that (C54 ) = C54 where ? ? ?1/3 7/3 ?1/3 A1 := ? 7/3 ?1/3 1/3 ? ?1/3 1/3 ?7/3

?28y 3 ? 17x4 y 2 ? 17x2 y 4 ? 28x3 y 3 ? 2y 5 + 1788x3 y + 1788x2 y ? 17y 4 ? 17x4 +262xy + 1788x2 y 3 ? 1788xy 2 ? 262xy 4 + 1788xy 3 ? 1788x3 y 2 ? 8166x2 y 2 + 28x3 +262x4 y ? 2x5 y ? 2xy 5 + 1 ? 17y 2 ? 17x2 + 2x5 + 2x ? 2y + x6 + y 6 = 0

? dualC54 : C54 → C54 is de?ned by

dualC54 (X, Y, Z ) = (FX (X, Y, Z ), FY (X, Y, Z ), FZ (X, Y, Z ))

where FX , FY , FZ are partial derivatives. We de?ne an isomorphism τ : C54 → C54 by the composition ?A1 ? dualC54 . Then τ is the restriction of the rational mapping: Ψ : C2 → C2 , (x, y ) → (xd , yd ) and (2.10)

y +4x ?x y +4x y ?8x y ?4x y ?8xy ?4xy ?2xy +109x y +4y +4x xd := ? ?4y 3 +x2 ?4x2 y 3 +4x3 y 2 ?8x3 y ?109x2 y 2 ?2xy ?4xy 2 ?8xy 3 +4x2 y +y 2 +4x3 4y 3 +4x2 ?4x2 y 3 +x3 y 2 ?2x3 y ?4x2 y 2 ?8xy ?109xy 2 ?8xy 3 +4x2 y +4y 2 +x3 yd := ? ? ?4y 3 +x2 ?4x2 y 3 +4x3 y 2 ?8x3 y ?109x2 y 2 ?2xy ?4xy 2 ?8xy 3 +4x2 y +y 2 +4x3

3 2 2 3 3 2 3 2 2 2 3 2 2 3

Observe that τ is de?ned over Q. C54 has three ?exes of order 2 at F1 := (1, ?1/4, 1), F2 := (1/4, ?1, 1), F3 := (4, ?4, 1) and τ exchanges ?exes and cusps: (2.11) τ (OX ) = F1 , τ (OY ) = F2 , τ (OZ ) = F3 , τ (F1 ) = OX , τ (F2 ) = OY , τ (F3 ) = OZ

Furthermore we assert that Proposition 2.12. The morphism τ is an involution on C54 . Proof. By the de?nition of τ and Lemma 2.13 below, we have (C := C54 ):

2 1 ? dualC ) = (dual A1 ? ?A ) ? (?t ?1 ? dualC ) = id τ ? τ = (?t A? C 1 A1 1

as A1 is a symmetric matrix. Let C be a given irreducible curve in P2 de?ned by a homogeneous polynomial F (X, Y, Z ) and let B ∈ GL(3, C). Then C B is de?ned by G(X, Y, Z ) := F ((X, Y, Z )B ?1 ). Lemma 2.13. Two curves (C B )? and (C ? ) B mutes. dualC ?→ C ? ?? B CB

dualC B

t ?1

coincide and the following diagram com? C ? ? ?t

B ?1

?→

(C B )?

Proof. The ?rst assertion is the same as Lemma 2, [O2]. The second assertion follows from the following equalities. Let (a, b, c) ∈ C . dualC B (?B (a, b, c)) = (GX (?B (a, b, c)), GY (?B (a, b, c)), GZ (?B (a, b, c))) = (FX (a, b, c), FY (a, b, c), Fy (a, b, c))t B ?1 = ?t B?1 (dualC (a, b, c)

In section 5, we will show that τ is expressed in a simple form as a cubic curve. 3. Structure of elliptic fibrations We consider the elliptic ?brations corresponding to the above normal forms. For this purpose, we ?rst take a linear change of coordinates so that three lines de?ned by f3 (x, y ) = 0 changes into lines X = 0, Y = 0 and Z = 0. The corresponding three cusps are now at OZ = (0, 0, 1), OY = (0, 1, 0), OX = (1, 0, 0) in P2 . Then we take the quadratic transformation which is a birational mapping of P2 de?ned by

6

(X, Y, Z ) → (Y Z, ZX, XY ). Geometrically this is the composition of blowing-ups at OX , OY , OZ and then the blowing down of three lines which are strict transform of X, Y, Z = 0. It is easy to see that our sextics are transformed into smooth cubics for which X = 0, Y = 0 and Z = 0 are tangent lines of the ?ex points. Those ?exes are the image of the (3,4)-cusps. We take a linear change of coordinates so that the ?ex on Z = 0 is moved at O := (0, 1, 0) with the tangent Z = 0. Then the corresponding families are described by the families given by {htorus (x, y, s) = 0; s ∈ P1 } and {hgen (x, y, s) = 0, s ∈ P1 } where ? 1 2 2 3 ? ? Cs : htorus (x, y, s) := x ? 4 s(x ? 1) + sy , Ds : hgen (x, y, s) := ?8x3 + 1 + (s + 35)y 2 ? 6x2 + 3x √ √ √ √ ? ? ?6 ?3y ? 3 ?3x ? 6 ?3x2 ? 12 ?3xy + (s ? 35)xy Let Hε (X, Y, Z, S, T ) = hε (X/Z, Y /Z, S/T )Z 3T for ε = torus, gen. We consider the elliptic surface associated to the canonical projection π : Sε → P1 where Sε is the hypersurface in P1 × P2 which is de?ned by Hε (X, Y, Z, S, T ) = 0.

Case I. Structure of Storus → P1 . For simplicity, we use the a?ne coordinate s = S/T of {T = 0} ? P1 and denote π ?1 (s) by Cs . We see that the singular ?bers are s = 0, 27, ∞. C∞ consists of three lines, isomorphic to I3 in Kodaira’s notation, [Ko]. C27 obtains a node and this ?ber is denoted by I1 in [Ko]. The ?ber C0 is a line with multiplicity 3. The surface Storus has three singular points on the ?ber C0 at (X, Y, Z ) = (0, 1/2, 1), (0, ?1/2, 1), (0, 1, 0). Each singularity is an A2 -singularity. We take minimal resolutions at these points. At each point, we need two P1 as exceptional divisors and let p : Storus → Storus be the resolution map. The composition π := π ? p : Storus → P1 is the corresponding elliptic surface. Now it is easy to see that C0 := π ?1 (0) is a singular ?ber with 7 irreducible components, which is denoted by IV ? in [Ko]. Here we used the following lemma. The elliptic surface Storus is rational and denoted by X431 in [Mi-P]. Assume that the surface V := {(s, x, y ) ∈ C3 ; f (s, x, y ) = 0} has an A2 singularity at the origin where f (s, x, y ) := sx + y 3 + sx · h(s, x, y ) where h(O ) = 0. Consider the minimal resolution π : V → V and let π ?1 (O ) = E1 ∪ E2 . It is well-known that E1 · E2 = 1 and Ei2 = ?2 for i = 1, 2. Lemma 3.1. Consider a linear form ?(s, x, y ) = as + bx + cy and let L′ be the strict transform of ? = 0 to V . (1) Assume that b = c = 0 and a = 0. Then (π ? ?) = 3L′ + 2E1 + E2 and L′ · E1 = 1 and L′ does not intersect with E2 , under a suitable ordering of E1 and E2 . (2) Assume that abc = 0. Then we have (π ? ?) = L′ + E1 + E2 and L′ · Ei = 1 for i = 1, 2. The proof is immediate from a direct computation. Case II. Structure of Sgen → P1 . Now consider √ the elliptic surface Sgen . Put Ds = √ ?1 π (s). The singular ?bers are at s = ?35, ?53 + 6 ?3, ?53 ? 6 ?3 and s = ∞. The ?ber s = ∞ is already I3 and Sgen is smooth on this ?ber. On the other hand, Sgen has a √ √ A2 -singularity on each ?ber Ds , s = ?35, ?53 + 6 ?3, ?53 ? 6 ?3. Let p : Sgen → Sgen be the the minimal resolution map and we consider the composition π := π ? p : Sgen → P1

7

as above. Then using (2) of Lemma 3.1, we see that π : Sgen → P1 has four singular ?bers and each of them is I3 in the notation [Ko]. This elliptic surface is also rational and denoted as X3333 in [Mi-P]. 4. Torsion group of Cs and Ds Consider an elliptic curve C de?ned over a number ?eld K by a Weierstrass short normal form: y 2 = h(x), h(x) = x3 + Ax + B . The j-invariant is de?ned by j (C ) = ?1728(4A)3 /? with ? = ?16(4A3 + 27B 2 ). We study the torsion group of the MordellWeil group of C which we denote by Ctor (K ) hereafter. Recall that a point of order 3 is geometrically a ?ex point of the complex curve C 2 2 ([Si]) and its locus is de?ned by F (f ) := fx,x fy ? 2fx,y fx fy + fy,y fx = 0 where f (x, y ) is 2 the de?ning polynomial of C ([O1]). In our case, F (f ) = 24xy ? 18x4 ? 12x2 A ? 2A. The unit of the group is given by the point at in?nity O := (0, 1, 0) and the inverse of P = (α, β ) ∈ C is given by (α, ?β ) and we denote it by ?P . For a later purpose, we prepare two easy propositions. Consider a line L(P, m) passing through ?P de?ned by y = m(x ? α) ? β . The x-coordinates of two other intersections with C are the solution of q (x) := f (x, m(x ? α) ? β )/(x ? α) which is a polynomial of degree 2 in x. Let ?x q be the discriminant of q in x. Note that ?x q is a polynomial in m. (A) When does a point Q ∈ C exist such that 2Q = P . Assume that a K point Q = (x1 , y1) satis?es 2Q = P . Geometrically this implies that the tangent line TQ C passes through ?P .

Proposition 4.1. There exists a K -point Q with 2Q = P if and only if m is a K solution of ?x q (m) = 0 and x1 is the multiple solution of q (x) = 0. If P is a ?ex point, ?x q (m) = 0 contains a canonical solution which corresponds to the tangent line at P and m = ?fx (α, β )/fy (α, β ). For any K -solution m with m = ?fx (α, β )/fy (α, β ), the order of Q is equal to 2 · order P . (B) When does a point Q ∈ C exist such that 3Q = P . Assume that a K -point Q = (x1 , y1 ) satis?es 3Q = P . Put Q′ := 2Q and put Q′ = (x2 , y2). Let TQ C be the tangent line at Q. Then TQ C intersects C at ?Q′ . Then ?3Q is the third intersection of C and the line L which passes through Q, Q′ . Thus three points ?P, Q, Q′ are colinear. Write L as y = m(x ? α) ? β . Then x1 , x2 are the solutions of q (x) = 0. Thus we have (4.2) where coe?(q, xi ) is the coe?cient of xi in q (x). Let LQ (x, y ) be the linear form de?ning TQ C and let R(x) be the resultant of f (x, y ) and LQ (x, y ) in y . Put R1 (x) := R(?coe?(q, x)/coe?(q, x2 ) ? x). Then by the above consideration, x = x1 is a common solution of q (x) = R1 (x) = 0. Let R2 (m) be the resultant of q (x) and R1 (x). Note that if ?x q (m) = 0, L is tangent to C at Q and R2 (m) = 0. In this case, 2Q = P . Proposition 4.3. Assume that there exists a K -point Q with 3Q = P and order Q = 3 · order P and let m be as above. Then R2 (m) = 0 and ?x q (m) = 0. Moreover x1 is given as a common solution of q (x) = R1 (x) = 0.

8

x2 = ?coe?(q, x)/coe?(q, x2 ) ? x1 ,

y1 = m(x1 ? α) ? β

Actually one can show that R2 (m) is divisible by (?x q )2 . 4.1. Cubic family associated with sextics of a torus type. We have observed that the family Cs for s ∈ Q is de?ned over Q. First, recall that Cs is de?ned by 1 Cs : x3 ? s(x ? 1)2 + sy 2 = 0 (4.4) 4 and the Weierstrass normal form is given by Cs : y 2 = x3 + a(s)x + b(s) where 1 1 1 6 1 4 1 3 s + s , b(s) = ? s5 + s4 + s 48 2 24 4 864 Put Σ := {0, 27, ∞}. This corresponds to singular ?bers. We have two sections of order 1 2 1 2 1 2 3: s → ( 12 s , ±1 s2 ). Put P1 := ( 12 s , 2 s ). Thus the torsion group is at least Z/3Z. By 2 [Ma], the possible torsion group which has an element of order 3 is one of Z/3Z, Z/6Z, Z/2Z + Z/6Z, Z/9Z or Z/12Z. The j-invariant of Cs is given by (4.5) a(s) = ? (4.6) j (Cs ) := jtorus (s), jtorus (s) := s(s ? 24)3 /(s ? 27)

(1) Assume that (Cs )tor (Q) has an element of order 6, say P2 := (α2 , β2 ) ∈ Cs ∩ Q2 . We may assume that P2 + P2 = P1 . By Proposition 4.1, this implies that x = α2 must be the multiple solution of As ?fx (?P1 )/fy (?P1 ) = ?s/2, we must have m = ?s/2 and thus (4.7) q (x) := s4 ? 36s3 ? 72ms2 ? 6xs2 ? 6s2 m2 + 72m2 x ? 72x2 = 0

?′x q := ?x q/(2m + s) = s3 ? 32s2 ? 2ms2 ? 4m2 s + 8m3 = 0

The curve ?′x (q ) = 0 is a rational curve and we can parametrize ?′x q = 0 as s = ?6 (u), m = ?6 (u)u where (4.8) ?6 (u) := 32/(1 + 2u)(2u ? 1)2 P2 = ( 128 ?1 + 12u2 512(6u + 1) , ) 3 (2u + 1)2 (?1 + 2u)4 (?1 + 2u)5(2u + 1)2 The point P2 is parametrized as (4.9)

where u ∈ Q. We put A6 := {s = ?6 (u); u ∈ Q} and Σ6 := ??1 (Σ). Note that Σ6 = {?1/2, 1/2, 5/6, ?1/6}. (1-2) Assume that we are given s = ?(u) and we consider the case when (4.7) has three rational solutions in m for a ?xed s. This is the case if ?6 (u) = ?6 (v ) has two rational solutions di?erent from u. This is also equivalent to (Cs )tor (Q) has Z/2Z + Z/2Z as a subgroup. The equation is given by the conic (4.10) Q: 4u2 ? 2u + 4uv ? 1 ? 2v + 4v 2 = 0 1 (r 2 + 24r ? 36) v (r ) := ? 6 (12 + r 2 )

9

By an easy computation, Q is rational and it has a parametrization as follows. (4.11) ?36 + 5r 2 u = ?2 (r ) := , 6(12 + r 2 )

The generators are P2 of order 6 and R = (γ, 0) of order 2 where γ := ? 81 (r 4 ? 48r 3 + 72r 2 ? 432)(12 + r 2 )4 4 (r 2 ? 36)4 r 4

Put ?6,2 (r ) := ?6 (?2 (r )), which is given explicitly as

1 We de?ne a subset A6,2 of A6 by the image ?6,2 (Q). Put Σ6,2 := ?? 6,2 (Σ). It is given by Σ6,2 = {0, ±2, ±6}.

?6,2 (r ) = 27(12 + r 2 )/r 2 (r ? 6)2 (r + 6)2

(2) Assume that there exists a rational point P3 = (α3 , β3 ) of order 9 such that 3P3 = P . By Proposition 4.3, this is the case if and only if R3 (m, s) := 512m9 + 768m8 s ? 512m6 s3 ? 1536m6s2 ? 192s4m5 ?6144m5 s3 ? 6528m4 s4 + 96s5 m4 ? 12288m3s4 ? 2048m3 s5 + 64s6 m3 + 480s6m2 ?15360s5 m2 ? 6144s6m + 384s7m ? 6s8 m + 56s8 ? 512s6 ? 768s7 ? s9 = 0

has a rational solution. Here R3 is R2 /(?x q )2 (s + 2m)s4 up to a constant multiplication. Again we ?nd that the curve {(m, s) ∈ C2 ; R3 (m, s) = 0} is rational and we can parametrize this curve as s = ?9 (t), m = ψ9 (t) where (4.12) ψ9 (t) :=

1 (?1+9t ?3t+3t ) ?9 (t) := ? 8 t3 (t?1)3 (t+1)3

2 3 3

1 (?1+9t2 ?3t+3t3 )2 (?t+t3 +1+7t2 ) 16 t3 (t?1)3 (t+1)3

The generator P3 = (α3 , β3 ) is given by α3 = β3 =

1 (1?18t+15t2 ?12t3 +15t4 +30t5 +33t6 )(9t2 ?1+3t3 ?3t)4 768 (t?1)6 (t+1)6 t6 1 (1+3t2 )(9t2 ?1+3t3 ?3t)6 ? 512 (t?1)5 (t+1)7 t8

1 We put A9 := {?9 (t); t ∈ Q} and Σ9 := ?? 9 (Σ) = {0, 1, ?1}.

(3) Assume that s ∈ A6 and (Cs )tor (Q) has an element P4 = (α4 , β4 ) ∈ Cs ∩ Q2 of order 12. Then we may assume that P4 + P4 = P2 . This implies that the tangent line at P4 passes through ?P2 . Write this line as y = n(x ? α2 ) ? β2 . By the same discussion as above, the equality Γ(n1 , u) = 0 holds where Γ is the polynomial de?ned by (4.13) Γ(u, n1 ) := ?786432u4 ? 98304n1 u3 ? 524288u3 + 393216u2 ? 16384n1u2 2 2 4 ?3072n2 1 u + 131072u + 24576n1 u + 4096n1 + 16384 + 256n1 + n1

and n = n1 /(2u + 1)(2u ? 1)2 . Again we ?nd that Γ = 0 is a rational curve and we have a parametrization: u = u(ν ) and n1 = n1 (ν ) where (4.14) (4.15)

1 (ν +2ν +5) , u (ν ) = ? 2 (ν 4 ?6ν 2 ?3)

4 2

n1 (ν ) = ?16 (2ν

2 ?4ν 3 ?4ν +ν 4 ?3)

s = ?12 (ν ) := ?6 (u(ν )),

ν ?3?6ν ) ?12 (ν ) := ? (ν ?( 1)4 (1+ν )4 (1+ν 2 )

(ν 4 ?6ν 2 ?3)

4

2 3

The generator of the torsion group Z/12Z is P4 = (α4 , β4 ) where α4 := β4 :=

1 (ν 8 ?12ν 7 +24ν 6 ?36ν 5 +42ν 4 +12ν 3 +36ν ?3)(ν 4 ?6ν 2 ?3)4 12 (ν ?1)8 (ν +1)8 (ν 2 +1)2 1 (ν 4 ?6ν 2 ?3)6 ν (ν 2 +3) ? 2 (ν ?1)7 (ν +1)11 (ν 2 +1)2

10

We put A12 := {?12 (ν ); ν ∈ Q}. By de?nition, A12 ? A6 . The singular ?bers Σ12 := ??1 (Σ) is given by {0, ±1}. Summarizing the above discussion, we get Theorem 4.16. The j-invariant is given by jtorus (s) = s(s ? 24)3 /(s ? 27) and the Mordell-Weil torsion group of Cs is given as follows. ? Z/3Z, s ∈ Q ? A6 ∪ A9 ∪ Σ ? ? ? ? ? ? Z/6Z, s = ?6 (u) ∈ A6 ? A6,2 ∪ A12 , u ∈ Q ? Σ6 (Cs )tor (Q) = Z/6Z + Z/2Z, s = ?6,2 (r ) ∈ A6,2 , r ∈ Q ? Σ6,2 ? ? ? Z/9Z, s = ?9 (t) ∈ A9 , t ∈ Q ? Σ9 ? ? ? Z/12Z, s = ?12 (ν ) ∈ A12 , ν ∈ Q ? Σ12

(4.17)

4.2. Comparison with Kubert family. In [Ku], Kubert gave parametrizations of the moduli of elliptic curves de?ned over Q with given torsion groups which have an element of order ≥ 4. His family starts with the normal form: E (b, c) : y 2 + (1 ? c)xy ? by = x3 ? bx2

We ?rst eliminate the linear term of y and then the coe?cient of x2 . Let Kw (b, c) be the Weierstrass short normal form, which is obtained in this way. The j-invariant is given by j (E (b, c)) = (1 ? 8bc2 ? 8cb ? 4c + 16b + 6c2 + 16b2 ? 4c3 + c4 )3 b3 (3c2 ? c ? 3c3 ? 8bc2 + b ? 20cb + c4 + 16b2 )

For a given elliptic curve E de?ned over K with Weierstrass normal form E : y 2 = x3 + ax + b and a given k ∈ K , the change of coordinates x → x/k 2 , y → y/k 3 changes the normal form into y 2 = x3 + ak 4 x + bk 6 . We denote this operation by Ψk (E ). 1. Elliptic curves with the torsion group Z/6Z. This family is given by a parameter c with b = c + c2 . 2. Elliptic curves with the torsion group Z/6Z + Z/2Z. This family is given by a parameter c1 with b = c + c2 and c = (10 ? 2c1 )/(c2 1 ? 9). 3. Elliptic curves with the torsion group Z/9Z. The corresponding parameter is f and b = cd, c = f d ? f, d = f (f ? 1) + 1. 4. Elliptic curves with the torsion group Z/12Z. The corresponding parameter is τ and b = cd, c = f d ? f, d = m + τ, f = m/(1 ? τ ) and m = (3τ ? 3τ 2 ? 1)/(τ ? 1). Proposition 4.18. Our family C?6 (u) , C?6,2 (r) , C?9 (t) , C?12 (ν ) are equivalent to the respective Kubert families. More explicitly, we take the following change of parameters to make their j-invariants coincide with those of Kubert and then we take the change of coordinates of type Ψk to make the Weierstrass short normal forms to be identical with Kw (x, y ). 1. For C?6 (u) , take u = ?(c ? 1)/2(3c + 1) and k = c2 (c + 1)/(3c + 1)2 . 2 2. For C?6,2 (r) , take r = ?12/(c1 ? 3) and k = 4(?5 + c1 )2 (c1 ? 1)2 /(c2 1 ? 6c1 + 21) /(c1 ? 3)(c1 + 3). 3. For C?9 (t) , take t = ?f /(f ? 2) and k = f 3 (f ? 1)3 /(f 3 ? 3f 2 + 1)2 . 4. For C?12 (ν ) , take ν = ?1/(2τ ? 1) and k = (τ ? 1)τ 4 (?2τ + 2τ 2 + 1)(?1 + 2τ )2 / (6τ 4 ? 12τ 3 + 12τ 2 ? 6τ + 1)2 . We omit the proof as the assertion is immediate from a direct computation.

11

4.3. Involution on C54 . We consider again the self dual curve C := C54 (see §3). The Weierstrass normal form is y 2 = x3 ?98415x+11691702. Note that 54 ∈ A6 ?A12 ∪A6,2 ∪Σ. In fact, 54 = ?6 (1/6) and 54 ∈ / A12 ∪ A6,2 . The j-invariant is 54000 and the torsion group Ctor (Q) is Z/6Z and the generator is given by P = (?81, 4374). Other rational points are 2P = (243, ?1458), 3P = (162, 0), 4P = (243, 1458), 5P = (?81, ?4374), and O = (0, 1, 0) (= the point at in?nity). Recall that C has an involution τ which is de?ned by (2.10) in §3. To distinguish our original sextic and cubic, we put The identi?cation Φ : C (3) → C (6) is given by the rational mapping: C (6) : (xy ? x + y )3 + 54x2 y 2 = 0, C (3) : y 2 = x3 ? 98415x + 11691702

and the involution τ (3) on C (3) is given by the composition Φ?1 ? τ ? Φ. After a boring computation, τ (3) is reduced to an extremely simple form in the Weierstrass normal form and it is given by τ (3) (x, y ) = (p(x, y ), q (x, y )) where (4.19) p(x, y ) := 81 2x ? 567 x ? 162 q (x, y ) := ?19683 y (x ? 162)2

Φ(x, y ) = (?2916/(27x ? 5103 ? y ), 2916/(y + 27x ? 5103))

Note that C has another canonical involution ι which is an automorphism de?ned by ι : (x, y ) → (x, ?y ). We can easily check that τ (3) ? ι = ι ? τ (3) . Note that τ (3) (P ) = 2P, τ (3) (2P ) = P, τ (3) (3P ) = O, τ (3) (O ) = 3P, τ (3) (4P ) = 5P, τ (3) (5P ) = 4P . Let η : C → C be the translation by the 2-torsion element 3P i.e., η (x, y ) = (x, y ) + (162, 0). It is easy to see that τ (3) is the composition ι ? η . That is τ (3) (x, y ) = (x, ?y ) + (162, 0) where the addition is the addition by the group structure of C54 . Thus Theorem 4.20. The involution τ on sextics C (6) is equal to the involution τ (3) on C (3) which is de?ned by (4.19) and it is also equal to (x, y ) → (x, ?y ) + (162, 0). 4.4. Cubic family associated with sextics of a general type. We consider the family of elliptic Ds curves associated to the moduli of sextics of a general type with three (3,4)-cusps. Recall that Ds is de?ned by the equation: √ √ Ds : ?8x3 + 1 + sy 2 + 35y 2 ? 6x2 + 3x ? 6 ?3y ? 3 ?3x √ √ ?6 ?3x2 ? 12 ?3xy + (s ? 35)xy = 0 √ This family is de?ned over Q( ?3). We change this polynomial into a Weierstrass normal form by the usual process killing the coe?cient of y and then by killing the coe?cient of x2 . A Weierstrass normal forms is given by y 2 = x3 + a(s)x + b(s) where ? 1 2 ? ? a(s) := ? 768 (s + 47)(s + 71)(s + 70s + 1657) 1 (4.21) (s2 + 70s + 793)(s4 + 212s3 + 17502s2 b(s) := 55296 ? ? +648644s + 9038089) √ √ The singular ?bers √ are s = ?35, ?53 + 6 ?3, ?53 ? 6 ?3 and s = ∞. Put Σ = {?35, ?53 + ±6 ?3, ∞}. In this section, we consider the Modell-Weil torsion over the

12

√ quadratic number ?eld Q( ?3). First we observe that this family has 8 sections of order three ±P3,i , i = 1, . . . , 4 where P3,i are given by (4.22) (4.23) P3,1 := (x3,1 , y3,1), P3,2 := (x3,2 , y3,2) x3,1 := 5041/48 + 71s/24 + s2 /48 y3,1 := 2917/4 + 53s/2 + s2 /4

(4.24) P3,3 := (x3,3 , y3,3), (4.25) P3,4 := (x3,4 , y3,4 ),

x3,2 := ?2209/16 ? 47s/8 ? s2 /16 √ y3,2 := ?3(s2 + 106s + 2917)(s + 35)/144 √ x3,3 := s2 /48 + 793/48 + 35s/24 + (s + 35) ?3/2 √ √ y3,3 := (?1 + ?3)(s + 35)(s + 6 ?3 + 53)/8 √ x3,4 := s2 /48 + 793/48 + 35s/24 ? (s + 35) ?3/2 √ √ y3,4 := ?(1 + ?3)(s + 53 ? 6 ?3(s + 35)/8

Thus they generate a subgroup isomorphic to Z/3Z + Z/3Z. We can take the generators √ P3,1 , P3,2 for example. Thus by [Ke-Mo], (Ds )tor (Q( ?3)) is isomorphic to one of the following. (a) Z/3Z + Z/3Z, (b) Z/3Z + Z/6Z and (c) Z/6Z + Z/6Z. The case (b) is forgotten in the list of [Ke-Mo] by an obvious type mistake. By the same discussion as in 5.1, there exists P ∈ Ds with order 6 and 2P = P3,1 if and only if ?(s, m) := s3 + 85s2 ? 4ms2 ? 568ms + 1555s ? 16m2 s ? 1136m2 ?15465 ? 20164m + 64m3 = 0

Fortunately the variety ? = 0 is again rational and we can parametrize it as (4.26) (4.27) s = ξ6 (t), ξ6 (t) := ?(27t3 ? 1304t2 + 17920t ? 71680)/(t ? 8)(t ? 16)2 m = ψ (t), ψ (t) := ?(?128t2 + 3t3 + 1536t ? 6144)/(t ? 8)(t ? 16)2

It turns out that the condition for the existence of Q ∈ Ds with 2Q = P3,2 is the same with the existence of P, 2P = P3,1 . Assume that s = ξ6 (t). Then by an easy computation, we get P = (x6,1 , y6,1) and Q = (x6,2 , y6,2 ) where

(?3072t x6,1 := ? 1 3

5 +11796480t2 +86016t4 ?1327104t3 ?56623104t+113246208+47t6 )

y6,1 := x6,2 :=

1 3 √

?4t3 (t2 ?24t+192)(7t2 ?144t+768) (t?16)5 (t?8)2

(t?8)2 (t?16)4

(37t6 ?2016t5 +40704t4 ?294912t3 ?1179648t2 +28311552t?113246208)

y6,2 := ? 8 7

(t?8)2 (t?16)4 √ √ √ √ ?3(t?12)(t?12?4 ?3)(7t?72+8 ?3)(7t?72?8 ?3)t(t?12+4 ?3) 3 3 (t?16) (t?8)

It is easy to see by a direct computation that 3P = 3Q = (α, 0) where α := ? 2 (t2 ? 48t + 384)(13t4 ? 528t3 + 8064t2 ? 55296t + 147456) 3 (t ? 8)2 (t ? 16)4

and Q ? P = P3,3 . Now we claim that √ Claim 1. (Ds )tor (Q( ?3)) = Z/3Z + Z/6Z with generators P3,3 and P .

13

In fact, if the torsion is Z/6Z + Z/6Z, there exist three elements of order two. However f0 (x) := f (x, 0) factorize as (x ? α)f0,0 (x) and their discriminants are given by ?x f0 := ?x f0,0 := 165888(t ? 12) (t ? 24t + 192) (t ? 8) (t ? 16)8

2048t6 (t?12)3 (t2 ?24t+192)3 (7t2 ?144t+768)6 (t?8)9 (t?16)18 3 2 3 7

Consider quartic Q4 : g (t, v ) := 165888(t ? 12)(t2 ? 24t + 192)(t ? 8) ? v 2 = √ 0. Thus Ds ?3)-point has three two torsion elements if and only if the quartic g ( t, v ) = 0 has Q ( √ (t0 , v0 ) with t0 = 8, 16, 12, 12 ± 4 ?3. The proof of Claim is reduces to: Assertion 1. There are no such point on Q4 . Proof. By an easy birational change of coordinates, g (t, v ) = 0 is equivalent to the elliptic curve C := {x3 + 1/16777216 ? y 2 = 0}. We see that C has two √element of order three, (0,√ ±1/4096) and three two-torsion (?1 /256, 0), (1/512 ? 1/512 ?3, 0) and √ (1/512 + 1/512 ?3, 0). Again by [Ke-Mo], Ctor (Q( ?3)) = Z/2Z + Z/6Z. As the rank of C is 0 ([S-Z]), there are exactly 12 points on C . They correspond to either zeros or poles of ?x (f0 ). This implies that the quartic Q4 has no non-trivial points and thus C does not have three 2-torsion points. This completes the proof of the Assertion and thus also proves the Claim. √ Now we formulate our result as follows. Let √ A6 = {s =√ ξ6 (t); t ∈ Q( ?3)} and ?1 Σ6 := ξ6 (Σ) is given by Σ6 = {8, 16, 0, 12, 12 ± 4 ?3, (72 ± 8 ?3)/7}. Theorem 4.28. The Mordell-Weil torsion of Ds is given by √ √ Z/3Z + Z/3Z s ∈ Q( ?3) ? A6 ∪ Σ √ (Ds )tor (Q( ?3)) = Z/6Z + Z/3Z s = ξ6 (t) ∈ A6 , t ∈ Q( ?3) ? Σ6 The j-invariant is given by j ( Ds ) = 1 (s + 47)3 (s + 71)3 (s2 + 70s + 1657)3 64 (s + 35)3 (s2 + 106s + 2917)3

4.5. Examples. (A) First we consider the case of elliptic curves Cs . In the following examples, we give only the values of parameter s as the coe?cients are fairly big. The corresponding Weierstrass normal forms are obtained by (4.5). 1. s = 54. The curve C54 with torsion group Z/6Z has been studied in §4.3. 2. Take r = 3, s = ?6,2 (3) = 343/9. Then the torsion group is isomorphic to Z/6Z + Z/2Z with generators P2 = (?55223/972, ?588245/486) and R = (88837/972, 0). The j-invariant is given by 73 · 1273 /22 · 36 · 52 . 3. Take t = ?3, s = ?9 (?3) = 1/216. Then the torsion group is isomorphic to Z/9Z and the generator P3 = (289/559872, ?7/419904). The j-invariant is 713 · 733/29 · 39 · 73 · 17. 4. Take ν = 3, s = ?12 (3) = ?27/80. Then the torsion is isomorphic to Z/12Z with generator P4 = (?2997/25600, ?6561/102400). The j-invariant is ?113 · 593 /212 · 3 · 53 . √ (B) We consider elliptic curves Ds de?ned over Q( ?3). The normal form is given by (4.21). √ 5. Take s = 1. Then (D1 )tor (Q( ?3)) = Z√ /3Z + Z/3Z and the generators are (x3,1 , y3,1) = (108, 756) and (x3,2 , y3,2) = (?144, 756 ?3). The j-invariant is 215 33 /73 .

14

6. Take t = 4 and s = ?299/9. Then the torsion is isomorphic to Z/6Z + Z/3Z. The , y6,1 ) = (?2351/243, ?532/243) and (x3,3 , y3,3 ) = √ √ generators can be taken as (x6,1 (8 ?3/9 ? 2171/243, ?680/81 + 248 ?3/81). The j-invariant is given by 53 · 173 · 313 · 22033 /26 · 36 · 73 · 196 . 4.6. Appendix. Parametrization of rational curves. Parametrizations of a rational curves are always possible and there exists even some programs to ?nd a parametrization on Maple V. For the detail, see [Ab-Ba] and [vH] for example. In our case, it is easy to get a parametrization by a direct computation. For a rational curves with degree less than or equal four is easy. For other case, we ?rst decrease the degree, using suitable bitational maps. We give a brief indication. We remark here that the parametrization is unique up to a linear fractional change of the parameter. (1) For the parametrization of s3 ? 32s2 ? 2ms2 ? 4m2 s + 8m3 = 0, put m = us. (2) For the parametrization of R3 (m, s) := 512m9 + 768m8 s ? 512m6 s3 ? 1536m6s2 ? 192s4m5 ?6144m5 s3 ? 6528m4 s4 + 96s5 m4 ? 12288m3s4 ? 2048m3 s5 + 64s6 m3 + 480s6m2 ?15360s5 m2 ? 6144s6m + 384s7m ? 6s8 m + 56s8 ? 512s6 ? 768s7 ? s9 = 0

put successively s = s1 /m1 and m = 1/m1 , then put n1 = n2 /s2 1 , then s1 = s2 ? 2 and n2 = n4 s2 . This changes degree of our curve to be 6. Then s2 + s3 ? 4 and n4 = n5 + 2 and n5 = n6 s3 . This changes our curve into a quartic. Other computation is easy. 4.7. Further remark. Professor A. Silverberg kindly communicated us about the paper √ [R-S]. He gave a universal family for Z/3Z + Z/3Z over Q( ?3), which is given by A(u) : y 2 = x3 + a0 (u)x + b0 (u) where and the subfamily, given by u = (4 + τ 3 )/(3τ 2 ), describes elliptic curves with torsion Z/6Z + Z/3Z. Again by an easy computation, we can show that by the change of parameter s = ?47 + 12u we can identify Ds and A(u). Our subfamily for Z/6Z + Z/3Z is also the same with that of [R-S] by the fractional change of parameter: t = 8(τ ? 2)/(τ ? 1). We would like to thank H. Tokunaga for the valuable discussions and informations about elliptic ?brations and also to K. Nakamula and T. Kishi for the information about elliptic curves over a number ?eld. I am also gratefull to SIMATH for many computations. References

[Ab-Ba] S. S. Abhyankar and C.L. Bajaj, Automatic parametrization of rational curves and surfaces III: Algebraic plane curves. Computer Aided Geometric Design 5 (1988), 309-321. [B-K] E. Brieskorn and H. Kn¨ orrer, Ebene Algebraische Kurven, Birkh¨ auser (1981), Basel-Boston Stuttgart. [D] A. Degtyarev, Alexander polynomial of a curve of degree six, J. Knot Theory and its Rami?cation, Vol. 3, No. 4, 439-454, 1994 [vH] M. van Hoeij, Rational parametrizations of algebraic curves using a canonical divisor, J. Symbolic Computation (1996) 11, 1-19. [Ke-Mo] M. A. Kenku and F. Momose, Torsion points on elliptic curves de?ned over quadratic ?elds, Nagoya Math. J. Vol. 109 (1988), 125-149

15

a0 (u) = ?27u(8 + u3 ),

b0 (u) = ?54(8 + 20u3 ? u6 )

K. Kodaira, On compact analytic surfaces II, Ann. of Math. 77 (1963) 563-626 and III, Ann. of Math. 78 (1963) 1-40. [Ku] D.S. Kubert, Universal bounds on the torsion of elliptic curves, Proc. London Math. Soc. (3) 33 (1976),193-237 [Ma] B. Mazur, Rational isogenies of prime degree, Invent. Math. 44 (1978) 129-162. [Mi-P] R. Miranda and U. Persson, On Extremal Rational Elliptic Surfaces, Math. Z. 193, 537-558 (1986) [N] M. Namba, Geometry of projective algebraic curves, Decker, New York, 1984 [O1] M. Oka, Flex Curves and their Applications, Geometriae Dedicata, Vol. 75 (1999), 67-100 [O2] M. Oka, Geometry of cuspidal sextics and their dual curves, to appear in Advanced Studies in Pure Math. 2?, 1999?, Singularities and arrangements, Sapporo-Tokyo 1998. [R-S] K. Rubin and A. Silverberg, Mod 6 representations of elliptic curves, 213–220 in Automorphic Forms, Automorphic Representations and Arithmetic, Proceedings of Symposia in Pure Mathematics, vol. 66, Part 1, AMS, 1999. [S-Z] U. Schneiders and H. G. Zimmer, The rank of elliptic curves upon quadratic extension, Computational number theory (1989), 239-260 [Si] J. H. Silverman, The Arithmetic of Elliptic Curves, GTM 106, Springer, New-York, 1986. [W] R. Walker, Algebraic curves, Dover Publ. Inc., New York, 1949. [Z] H. G. Zimmer, Torsion of elliptic curves over cubic and certain biquadratic number ?elds, Arithmetic geometry, 203-220, Comtemp. Math. 174, Amer. Math. Soc. Department of Mathematics, Tokyo Metropolitan University Minami-Ohsawa, Hachioji-shi Tokyo 192-03, Japan E-mail address : oka@comp.metro-u.ac.jp

[Ko]

16

更多相关文档:

更多相关标签:

相关文档

- Descent on elliptic curves
- On the rank of elliptic curves
- ID-based Signatures from Pairings on Elliptic Curves
- Complex multiplication tests for elliptic curves
- Dynamical systems arising from elliptic curves
- Quadratic twists of pairs of elliptic curves
- FAST ALGORITHMS FOR COMPUTING ISOGENIES BETWEEN ELLIPTIC CURVES
- Hyperelliptic Curve Cryptosystems Closing the Performance Gap to Elliptic Curves
- Efficient Tate pairing computation for supersingular elliptic curves over binary fields. Cr
- An Introduction to Elliptic Curves

文档资料共享网 nexoncn.com
copyright ©right 2010-2020。

文档资料共享网内容来自网络，如有侵犯请联系客服。email:zhit325@126.com