A complete diophantine characterization of the rational torsion of an elliptic curve

arXiv:math/0703578v1 [math.NT] 20 Mar 2007 A complete diophantine characterization of the rational torsion of an elliptic curve Irene Garc´ıa–Selfa Jos´e M. Tornero March, 2007 Abstract We give a complete characterization for the rational torsion of an elliptic curve in terms of the (non–)existence of integral solutions of a system of diophantine equations. MSC 2000: 11G05 (primary); 11Dxx (secondary). Keywords: Elliptic curves, Diophantine equations. 1 Introduction In this paper we consider elliptic curves defined over Q. As it is known [1, 15], each one of such curves is birationally equivalent to one, say E; given by an equation of the type E:Y2=X3+AX +B, with A, B, ∈Z; called short Weierstrass form, where it must hold ∆ = 4A4+27B26= 0. The set of rational points of its projective clausure, noted E(Q), is a finitely generated abelian group (Mordell–Weil Theorem [9, 17]) and its torsion part, noted T(E(Q)) has been exhaustively described by Mazur [8, 7] as isomorphic to one of the following groups: Z/nZfor n= 1,2, ..., 10,12; Z/2Z×Z/2nZfor n= 1,2,3,4. We will give a procedure for finding out which of these groups is the actual torsion subgroup of our curve, by characterizing the existence 1 of points of order nin terms of the existence of an integral solution for a system of diophantine equations. Note that only orders which are prime or pure prime powers have to be considered; for the rest of the cases may be solved by joining systems, according to the factorization of the order we are interested on. Although the actual system depends on n, we will see that they are fairly similar systems, in more than one sense. The main result can be stated as follows: Theorem.– Let E:Y2=X3+AX +Bbe an elliptic curve, with A, B ∈Z. For every n∈ {3,4,5,7,8,9}there are, at most, 4 quasi– homogeneous polynomials Pn, Qn, Rn, Sn∈Z[z1, ..., z4] such that E has a rational point of order nif and only if there exists an integral solution for the system Σn:         Pn(z1, ..., z4) = 62·A Qn(z1, ..., z4) = 63·B Rn(z1, ..., z4) = 0 Sn(z1, ..., z4) = 0 Remark.– The case n= 2 will be treated separatedly, as not only the existence, but the number of solutions is important for determining whether the torsion group is cyclic or not. However, we will get systems of the above type anyway, which we will call Σ(1) 2and Σ(3) 2, for detecting the existence of one or three points of order two, respectively. Remark.– Actually, for most cases, less equations and variables have to be considered, according to the following table: Case Equations and variables Max. degree n= 3 2 equations in 2 variables 6 n= 4 2 equations in 2 variables 3 n= 5 3 equations in 3 variables 4 n= 7 4 equations in 4 variables 4 n= 8 3 equations in 3 variables 6 n= 9 3 equations in 3 variables 10 The remarkable equality between the number of equations and variables may be of some help if we try to use some numerical algorithm to find integral solutions. 2 Remark.– Now and then we will use the Nagell–Lutz Theorem [11, 6, 1, 15] which, among more specific results, states that the coordinates of torsion points must be integers, under our assumptions. There exist some previous descriptions of torsion structures in terms of diophantine equations, notably by Ono [12], for the non– cyclic case and by Qiu and Zhang [13, 14] for the even cyclic case (strongly influenced by Ono’s ideas). Both papers took advantage of the possibility of choosing (0,0) to be a 2–torsion point, which is not suitable in the general case. Instead, while trying to cope with the odd order case we eventually arrived to a global strategy for finding equations which led to the previous theorem. Both Ono’s and Qiu–Zhang’s results have been used to study the behaviour of torsion structures when a bigger ground field is considered (typically a quadratic extension of Q), for instance in [5, 2, 3]. We hope that our characterization will shed some light also in this matter, where much is yet to be known. As a final remark, we should note that the resulting systems seem to be suitable for a computational diophantine attack, either classical and/or modern (see for instance [10, 16] for a good account on each approach). However, it seems difficult to us that this strategy can beat the existing algorithms [4] which can compute torsion subgroups of curves with discriminant ∆ ∼220000 in only a few minutes. 2 Points of order 3and 9 Assume our curve Ehas a point (x, y) of order three. As it is known [15], (x, y)∈E(Q) is a point of order three if and only if Ψ3(x) = 3x4+ 6Ax2+ 12Bx −A2= 0. If (x, y) = (0, z2), then we must have A= 0 and therefore, B=z2 2. So we will suppose x6= 0 from now on, and divide Ψ3(x) by x4to obtain the conic 3 + 6X+ 12Y−X2= 0, where X=A/x2and Y=B/x3.Thus we can parametrize the conic by X=3(t+ 1) t−3, Y =−t2+ 6t+ 3 (t−3)2, t ∈Q\ {3} 3 Now if we set t=p/q for p, q ∈Zwith gcd(p, q) = 1 and q6= 0, p −3q6= 0, then we have A x2=3(p+q) p−3q,B x3=−p2+ 6pq + 3q2 (p−3q)2, Writing u=p−3qwe get A= 3(u+ 4q)x2/u, B = (12q2−u2)x3/u2. Thus, y2=3(u+ 2q)2 u2x3, and, since (x, y)∈E(Q), there exist z1, z2∈Zsuch that x= 3z2 1and y=±(9z3 1+z2) with z2= 18z3 1q/u. Therefore, A= 27z4 1+ 6z1z2, B =z2 2−27z6 1, which also fits the case x= 0. Hence we have found: Σ3:   62·A=P3= 6227z4 1+ 6z1z2 63·B=Q3= 63z2 2−27z6 1 Remark.– The polynomials obtained are, in fact, quasi–homogeneous. If we get to compute the solutions for the systems, we obtain all points of order three, which are {(3z2 1,9z3 1+z2),(3z2 1,−(9z3 1+z2))}. Conversely, we can easily check that (3z2 1,±(9z3 1+z2)) has order three in the curve Y2=X3+ (27z4 1+ 6z1z2)X+ (z2 2−27z6 1). This shows that the conditions are also sufficient. Assume now that Ehas a rational point of order nine. Then there must be rational points of order three in the curve, thus we already have P9= 62(27z4 1+ 6z1z2) Q9= 63(z2 2−27z6 1) Also, if (x, y) is a point of order nine, the first coordinate of [3](x, y) must be 3z2 1, as [3](x, y) has order three. 4 Using the chord–tangent and duplication formulae, the first coordinate in [3](x, y) is 3z2 1=N(A, B, x) (3x4+ 6Ax2+ 12Bx −A2)2. where N(A, B, x) = x9−12Ax7−96Bx6+30A2x5−24ABx4+(36A3+48B2)x3 +48A2Bx2+ (96AB2+ 9A4)x+ 8B(A3+ 8B2). So (x, y)∈E(Q) has order nine if and only if R9=x9−12Ax7−96Bx6+30A2x5−24ABx4+(36A3+48B2)x3+48A2Bx2 +(96AB2+9A4)x+8B(A3+8B2)−3z2 1(3x4+6Ax2+12Bx−A2)2= 0. For writing it our way, we should make the appropriate substitutions A= 27z4 1+ 6z1z2, B =z2 2−27z3 1; to obtain a quasi–homogeneous polynomial, for instance with weights w(z1) = 1, w(z2) = 3, w(x) = 2; weighted degree 18 (with the previous choice) and 69 monomials. This polynomial is the most complicated we have to deal with, but it turns out it can be written in a more amusing way. If we perform the Tchirnhausen transformation w.r.t. x; that is, we make z3=x−3z2 1, we find out our polynomial R9can be written as R9=z9 3−72z1(9z3 1+z2)z7 3−48(27z3 1+ 2z2)(9z3 1+z2)z6 3 −864z2 1(9z3 1+z2)2z5 3−144z1(9z3 1+z2)3z4 3 +48(45z3 1+z2)(9z3 1+z2)3z3 3+ 1728z2 1(9z3 1+z2)4z2 3 +576z1(9z3 1+z2)5z3+ 64(9z3 1+z2)6 which, taking into account the given parametrization of points of order three, proves the following funny fact: Corollary.– Let (x, y) be a point of order nine, and write [3](x, y) = (x′, y′). Then (x−x′)|(2y′)6. In order to get the easiest possible system we will perform z27−→ z2−9z3 1 5 and our system is then Σ9:                    62·A=P9= 6227z4 1+ 6z1z2 63·B=Q9= 63z2 2−27z3 1 0 = R9=z9 3−72z1z2z7 3−48 9z3 1+ 2z2z2z6 3 −864z2 1z2 2z5 3−144z1z3 2z4 3+ 48 36z3 1+z2z3 2z3 3 +1728z2 1z4 2z2 3+ 576z1z5 2z3+ 64z6 2 Remark.– If one wants to use this system for deciding whether a given curve has a torsion group of order nine, it is wise to solve first 62·A=P9,63·B=Q9, which is a system of two equations in two variables. One then moves on to solving R9as a polynomial in z3. Notice then that this polynomial, which has degree nine, can only have three different integer roots corresponding to the first coordinates of points of order nine, since an elliptic curve can only have six such different points, and, in Weierstrass short form, three of them should be symmetric to the other three w.r.t. the x–axis. 3 Points of order 5and 7 Orders five and seven share an ad–hoc strategy for obtaining the respective systems Σ5and Σ7. Assume (x1, w1)∈E(Q) has order five. Then the first coordinate of [4](x1, w1) must be again x1, and if we write [2](x1, w1) = (x2, w2), then it must hold x16=x2. We will have, from duplication formula, x4 1−2Ax2 1−8Bx1+A2 4(x3 1+Ax1+B)=x2;x4 2−2Ax2 2−8Bx2+A2 4(x3 2+Ax2+B)=x1. This equations yield a system whose (naive) general solution for B is given by x1=x1, x2=x2, A =A, B =1 4(x1+x2)(x2 1−4x1x2+x2 2−2A), where x1, x2, A ∈Zand Amust verify the second degree polynomial A2+ (2x2+ 2xv + 2v2)A−x4+x3v+ 9x2v2+xv3−v4= 0. 6 The discriminant of this equation is 4(x1−x2)2(2x1+x2)(x1+ 2x2), hence there must exist x3∈Zsuch that x2 3= (2x1+x2)(x1+ 2x2). With these notations we can finally write A=−x2 1−x1x2−x2 2+ (x1−x2)x3, B=−1 4(x1+x2)(−3x2 1+ 2x1x2−3x2 2+ 2(x1−x2)x3). Now, since w2 1=(x1−x2)2(3x1+ 2x3+ 3x2) 4, there must be x4∈Zverifying x2 4= 3x1+ 2x3+ 3x2. This completes a system Σ′ 5, given by A=−x2 1−x1x2−x2 2+ (x1−x2)x3 4B=−(x1+x2)(−3x2 1+ 2x1x2−3x2 2+ 2(x1−x2)x3) 0 = x2 3−(2x1+x2)(x1+ 2x2) 0 = x2 4−(3x1+ 2x3+ 3x2). Note that the polynomials obtained in this case are actually homogeneous, except for S5. We can also get the torsion points with a little extra work. With our previous notations it must hold w2 2= (x1−x2)2(3x1−2x3+ 3x2)/4, hence we can find t∈Zsuch that with t2= 3x1−2x3+ 3x2. Then the full list of torsion points of order five is x1,±(x1−x2)x4 2,x2,±(x1−x2)t 2. 7 By plugging these points into the curve we see that conditions given by Σ′ 5are sufficient to guarantee the existence of torsion points of order five. We then eliminate x3from the last equation and rename z1=x1, z2=x2, z3=x4 to finally get the equivalent system Σ5: Σ5:       62·A=P5= 18 ·h(z1−z2)2−6z2 1+ (z1−z2)z2 3i 63·B=Q5= 54 ·(z1+z2)h6z2 1−2z1z2+ (z2−z1)z2 3i 0 = R5=z4 3−6z2 3(z1+z2) + (z1−z2)2 Let us do then the case n= 7. Assume (x1, w1)∈E(Q) is a point of order seven, and denote [2](x1, w1) = (x2, w2) and [3](x1, w1) = (x3, w3), being x1, x2and x3different integers. From the duplication formula, as above, we have F1(x1, x2) = x4 1−2Ax2 1−8Bx1+A2−4(x3 1+Ax1+B)x2= 0. From the expression of x3used in the order nine case we can write F2(x1, x3) = x9 1−12Ax7 1−96Bx6 1+ 30A2x5 1−24ABx4 1+ +(36A3+ 48B2)x3 1+ 48A2Bx2 1+ (96AB2+ 9A4)x1+ 8B(A3+ 8B2)− −x3(3x4 1+ 6Ax2 1+ 12Bx1−A2)2= 0. Also, using the addition formula for (x1, w1) and (x2, w2), we can also get x3as an expression in x1and x2. Now, the difference between our two expressions for x3(one in x1and another in x1and x2) must vanish, and we get F3(x1, x2) = 0, where F3is a polynomial in x1and x2with 139 terms whose degree is 20. We will get still another relationship between our parameters, using now two different ways to express x3, as it is also the first coordinate of [4](x1, w1). The first one comes from [2](x2, w2) and depends only on x2, and the second one from the addition formula for (x1, w1) and (x3, w3) which depends on x1and x3. Thus we get F4(x1, x2, x3) = 0, 8 where F4is a polynomial in x1, x2and x3with 321 terms whose degree is 18. Now if we consider the ideal hF1, F2, F3, F4i ⊂ Q[x1, x2, x3] and calculate a Gr¨obner basis with respect to the graded reverse lexicographic order we get a basis consisting of five polynomials. The general solution for {A, B}of this system is nx1=x1, x2=x2, x3=x3, A =A, B=1 4x3x2 1−2x1x2x3+x2 2x3+x3 1−4x2x2 1−x2 2x1−2(x2+x1)Ao, where x1, x2, x3, A ∈Zand Averifies A2+ (2x2 1+ 2x2 2+ 2x2x1)A− x4 1+ 3x2x3 1−2x3 1x3+x3 2x1+ 3x2x3x2 1+ 6x2 1x2 2−x3 2x3= 0. As a side remark, it was not easy to find this relation, which parallels case n= 5. In fact, for most methods (including most orders considered for Gr¨obner bases computations), Aended up verifying a fourth degree equation, which became then a dead end for our hopes (see below). The discriminant of the previous second degree equation verified by Ais 4(x2+ 2x1)(x1−x2)2(x1+x3+x2) hence there must be x4∈Zwith x2 4= (x2+ 2x1)(x1+x3+x2). Therefore A=−x2 1−x2 2−x1x2+x4(x1−x2), 4B= (3x3 1+x3x2 1+ 3x2 2x1+x2 2x3−2x1x2x3+ 2x3 2+ 2(x2 2−x2 1)x4). Now again, as (x1, w1) has order 7, we can write x3as the first coordinate of [3](x1, w1) and [4](x1, w1), to get Aas a root of the polynomial Z4+ 6x2(x1+x2)Z3+ (−8x4 1+ 15x2 1x2 2+ 11x2x3 1+ 23x3 2x1+ 13x4 2)Z2 +(−24x2x5 1+ 58x3 2x3 1+ 42x4 2x2 1−6x2 2x4 1+ 4x6 1+ 24x5 2x1+ 10x6 2)Z 9 Then z1= 1, z2= 2, z3=−2, z4=−8; is a solution for the first three equations of Σ7, but S7(1,2,−2,−8) = −443. Again, the torsion subgroup of Eis trivial. Example.– The curve E:Y2=X3−4X has T(E(Q)) = (Z/2Z)×(Z/2Z). If we only onsider the first, second and fourth equation from Σ7we find the solution: z1= 2, z2=−2, z3=z4= 0; but R7(2,−2,0,0) = 4. Order eight Example.– Let us take E:Y2=X3−92X+ 480. The first two equations from Σ8admit the common solution z1= 3, z2= 6. However that leads to R8(3,6, z3) = z2 3−54, which has no integral solutions. As expected, T(E(Q)) = Z/4Z. 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