Galois Theory, discriminants and torsion subgroups of elliptic curves

arXiv:0811.1463v3 [math.NT] 30 Jun 2009 GALOIS THEORY, DISCRIMINANTS AND TORSION SUBGROUP OF ELLIPTIC CURVES IRENE GARC´ IA–SELFA, ENRIQUE GONZ´ ALEZ–JIM´ ENEZ, AND JOS´ E M. TORNERO Abstract. We find a tight relationship between the torsion subgroup and the image of the mod 2 Galois representation associated to an elliptic curve defined over the rationals. This is shown using some characterizations for the squareness of the discriminant of the elliptic curve. 1. Introduction. In what follows we will denote by Cnand Snthe cyclic group of order nand the symmetric group acting on nelements, respectively. Let Ebe an elliptic curve defined over Q. Let pbe a prime number and let E[p] be the group of points of order pon E(Q), where Qdenotes an algebraic closure of Q. The action of the absolute Galois group GQ= Gal(Q/Q) on E[p] defines a mod pGalois representation ρE,p : GQ→Aut(E[p]) ∼ =GL2(Fp). Let Q(E[p]) be the number field generated by the coordinates of the points of E[p]. Therefore, the Galois extension Q(E[p])/Qhas Galois group Gal(Q(E[p])/Q)∼ =ρE,p(GQ) For p= 2 it is known that ρE,2(GQ) can be determined in terms of the discriminant ∆(E) and E(Q)[2], the points of order 2 defined over the rationals (cf. [20, 19, 21]): (1) ρE,2(GQ)∼ =       S3if p∆(E)6∈ Qand #E(Q)[2] = 1, C3if p∆(E)∈Qand #E(Q)[2] = 1, C2if p∆(E)6∈ Qand #E(Q)[2] >1, {id}if p∆(E)∈Qand #E(Q)[2] >1. Note that GL2(F2)∼ =S3, the non-split Cartan subgroup of GL2(F2) is isomorphic to C3and the conjugated Borel subgroup of GL2(F2) is isomorphic to C2. An elliptic curve Edefined over the rationals has always an integral short Weierstrass form. That is, Ehas a model of the form E:Y2=X3+AX +B, with A, B ∈Z. Then the discriminant of this model is ∆(E) = −24(4A3+ 27B2). Any change of variables over the rationals preserving this short form is of the type (x, y) = (u2x′, u3y′) with u∈Q, u 6= 0. Therefore, if E′is the curve obtained after such a change, we have u12∆(E′) = ∆(E). Then the squareness of the discriminant of E does not depend on the short model of Ebut on Eitself. Our aim is finding if there is a relationship between the torsion group of E(Q) (noted E(Q)tors in what follows), the discriminant of Eand ρE,2(GQ). Assume Eis an elliptic curve which has a non–trivial torsion subgroup. Taking into account Mazur’s exhaustive classification [16, 17], the possible structures of Date: June 30, 2009. 2000 Mathematics Subject Classification. 11G05 (primary); 11R16 (secondary). Key words and phrases. Elliptic curves, Cubic equations, Torsion subgroup, Galois theory. 1 2 IRENE GARC´ IA–SELFA, ENRIQUE GONZ´ ALEZ–JIM´ ENEZ, AND JOS´ E M. TORNERO E(Q)tors are Cnfor n= 2 ...10,12 and C2×C2nfor n= 1 ...4. The easiest cases are those in which the order of E(Q)tors is even and they will be treated at section 2. The four remaining cases, E(Q)tors =Cn, with n= 3,5,7,9 will be treated separately at sections 4, 5, 6, 7, respectively. In these cases, thanks to (1), the squareness of ∆(E) determines the image of ρE,2. We will prove that there are no elliptic curves over the rationals with square discriminant and points of order 5,7 and 9 respectively and we will give a parametrization of the elliptic curves with square discriminant and a point of order 3. Section 3 consists of the necessary background for elliptic curves with points of odd order. Before stating the main theorems at the last section, we will give a parametrization of all elliptic curves over the rationals having square discriminant at section 8. Some remarks on the case of trivial torsion will be given there too. At section 9, we will state the main theorems of this paper whose proofs will have been stablished by then. Finally, in an appendix, we give a complete parametrization of the integer solutions of the Diophantine equations x2+ 3y2= 4z3. These solutions will be needed at sections 4 and 8. Acknowledgements. We would like to thank to F. Beukers, H. Cohen, H. Darmon, L. Dieulefait, J. Gonz´alez and J. Fern´andez for useful comments during the preparation of this paper. Special thanks are due to F. Beukers for suggesting the complete parametrization of the Diophantine equation x2+ 3y2= 4z3. First and third authors were supported partially by FQM–218 and P08–FQM– 03894 (Junta de Andaluc´ıa) and MTM 2007–66929 (Ministerio de Educaci´on y Ciencia, Spain). Research of the second author was supported in part by grant MTM 2006–10548 (Ministerio de Educaci´on y Ciencia, Spain) and CCG07–UAM/ESP– 1814 (Universidad Aut´onoma de Madrid – Comunidad de Madrid, Spain). 2. The even case. Let E:Y2=F(X) = X3+AX +Bbe an elliptic curve over Qsuch that E(Q)[2] has an even positive number of points. Therefore, by (1), the squareness of ∆(E) determines the image of the mod 2 Galois representation attached to E. By definition ∆(E) = 24(α1−α2)2(α1−α3)2(α2−α3)2, where α1, α2, α3are the roots of F(X). Then if E(Q)tors is non–cyclic we have that α1, α2, α3∈Qand ∆(E) is a square over Q. Meanwhile, if E(Q)tors is cyclic then there is only a point of order 2 on E(Q) and therefore F(X) = (X−a)(X2+aX +b) where a, b ∈Qsatisfy A=b−a2,B=−ab and a2−4bis a non-square over Q. Since ∆(E) = 24∆(F) we have ∆(E) = 24(a2−4b)(2a2+b)2is not a square in Q. This proves the following: ρE,2(GQ)∼ =C2if #E(Q)[2] = 2, {id}if #E(Q)[2] = 4. 3. Families of elliptic curves with a torsion point of odd order. In this section we are going to introduce the necessary background related to elliptic curves defined over the rationals with a point of prescribed odd order. There are well-known rational parametrizations for the modular curve X1(N) with N∈ {3,5,7,9}(see e.g. Kubert [15]). Therefore these parametrizations give us families of elliptic curves defined over Qwith a point of order N∈ {3,5,7,9}. An old characterization of elliptic curves containing a rational point of order 3 is given by the Hessian form. Nevertheless we are going to use a new one (cf. [12]) since this will fit better our purposes. GALOIS THEORY, DISCRIMINANTS AND TORSION SUBGROUP OF ELLIPTIC CURVES 3 Let us introduce the construction given in [12]; every elliptic curve with a rational point of order 3 can be written in the following form: E3(α, β) : Y2=X3+ (27α4+ 6αβ)X+β2−27α6, α, β ∈Z. For the remaining cases that will be used below, an analogous expression can be achieved by means of the Tate normal form [15]: Tate(b, c) : Y2+ (1 −c)XY −bY =X3−bX2, b, c ∈Q∗. Denote by En(α) the one-parameter family of curves having a rational point of order n. Then E5(α) = Tate(α, α), E7(α) = Tate(α2(α−1), α(α−1))), E9(α) = Tate(α2(α−1)(α(α−1) + 1), α2(α−1)). Now we can take the above equations to a short Weierstrass form and find the parametric family containing all elliptic curves with points of order 5, 7 and 9. The actual families are: E5(α) : Y2=X3−27 α4−12α3+ 14α2+ 12α+ 1X + 54 α2+ 1α4−18α3+ 74α2+ 18α+ 1 E7(α) : Y2=X3−27 α8−12α7+ 42α6−56α5+ 35α4−14α2+ 4α+ 1X + 54 α12 −18α11 + 117α10 −354α9+ 570α8−486α7+ 273α6−222α5+ 174α4−46α3−15α2+ 6α+ 1 E9(α) : Y2=X3−27 α3−3α2+ 1 )( α9−9α8+ 27α7−48α6+ 54α5− 45α4+ 27α3−9α2+ 1X + 54 α18 −18α17 + 135α16 −570α15 + 1557α14 −2970α13+ 4128α12 −4230α11 + 3240α10 −2032α9+ 1359α8−1080α7+ 735α6−306α5+ 27α4+ 42α3−18α2+ 1 Therefore, an elliptic curve Edefined over Qwith a rational point of order n= 3 (resp. n= 5, 7 or 9) is Q–isomorphic to E3(α, β) (resp. En(α) for n= 5,7 or 9) for some α, β ∈Z(resp. α∈Q). These kind of arguments have proved fruitful in the last years, as a number of results have appeared based on them [1, 14, 11, 13]. Now we can write the discriminant ∆nfor the above elliptic curves En, to obtain ∆3(α, β) = −24·33·(5α3+β)(9α3+β)3 ∆5(α) = 212 ·312 ·α5(α2−11α−1) ∆7(α) = 212 ·312 ·α7(α−1)7(α3−8α2+ 5α+ 1) ∆9(α) = 212 ·312 ·α9(α−1)9(α2−α+ 1)3(α3−6α2+ 3α+ 1) In the following section we will study the rationality of the square root of the above discriminants to decide whether the corresponding Galois group is C3or S3. 4. The case n= 3 Elliptic curves with points of order three must yield a discriminant with the form ∆3(α, β) for some α, β ∈Z. So, in order to find an elliptic curve Ehaving square discriminant we are bound to find integral solutions to the equation ω2=−3·(5α3+β)(9α3+β). 4 IRENE GARC´ IA–SELFA, ENRIQUE GONZ´ ALEZ–JIM´ ENEZ, AND JOS´ E M. TORNERO Let us denote g= gcd(5α3+β, 9α3+β). This necessarily leads to 5α3+β=±gu2,±3gv2and 9α3+β=∓3gv2,∓gu2,respectively for some integers uand v. Solving the above Diophantine systems of equations is equivalent to finding the integer solutions to x2+ 3y2= 4z3, where (x, y, z) = (ug2, vg2,∓αg). Thus, for the first two systems we obtain that the elliptic curves E3(α, β) have square discriminant for: (2) (α, β) = ∓z g,±x2+ 5z3 g3, and for the last two systems we obtain (3) (α, β) = ∓z g,±3y2+ 5z3 g3. In all those cases x, y, z ∈Zsatisfies x2+ 3y2= 4z3. At the appendix, we give parametrizations of all integer solutions of the Diophantine equation x2+3y2= 4z3 at Lemma 1 in terms of parameters (a, b, c, d). Then we can clear denominators and obtain Q-isomorphic elliptic curves E(i)(a, b, c, d) : Y2=P(i)(a, b, c, d)(X) attached to the parametrization (i), for i= 1,2, where: P(1)(a, b, c, d)(X) = X3−9(c2+cd +d2)3(a2+ab +b2)(3a6c2+ 3a6cd +a6d2+ 9a5bc2−3a5bcd −3a5bd2−30a4b2cd −15a3b3c2−15a3b3cd + 25a3b3d2+ 30a2b4cd + 30a2b4d2+ 9ab5c2 +21ab5cd + 9ab5d2+ 3b6c2+ 3b6cd +b6d2)X +9(c2+cd +d2)4(6a12c4+ 12a12c3d+ 12a12c2d2+ 6a12cd3+a12d4+ 36a11bc4 +36a11bc3d+ 18a11bc2d2−6a11bcd3−6a11bd4+ 72a10b2c4−54a10b2c3d−72a10b2c2d2 −78a10b2cd3−18a10b2d4+ 30a9b3c4−318a9b3c3d−102a9b3c2d2−132a9b3cd3−4a9b3d4 −81a8b4c4−378a8b4c3d+ 162a8b4c2d2−252a8b4cd3+ 45a8b4d4−108a7b5c4−108a7b5c3d +108a7b5c2d2−576a7b5cd3+ 216a7b5d4−72a6b6c4−144a6b6c3d−468a6b6c2d2 −396a6b6cd3+ 600a6b6d4−108a5b7c4−324a5b7c3d−216a5b7c2d2+ 684a5b7cd3 +900a5b7d4−81a4b8c4+ 54a4b8c3d+ 810a4b8c2d2+ 1386a4b8cd3+ 756a4b8d4+ 30a3b9c4 +438a3b9c3d+ 1032a3b9c2d2+ 1002a3b9cd3+ 374a3b9d4+ 72a2b10c4+ 342a2b10c3d +522a2b10c2d2+ 384a2b10cd3+ 114a2b10d4+ 36ab11c4+ 108ab11c3d+ 126ab11c2d2 +78ab11cd3+ 18ab11d4+ 6b12c4+ 12b12c3d+ 12b12c2d2+ 6b12cd3+b12d4) P(2)(a, b, c, d)(X) = X3−3(c2+cd +d2)3(a2+ab +b2)(a6c2+a6cd + 7a6d2+ 3a5bc2+ 39a5bcd +39a5bd2+ 60a4b2c2+ 150a4b2cd + 60a4b2d2+ 115a3b3c2+ 115a3b3cd −5a3b3d2+ 60a2b4c2 −30a2b4cd −30a2b4d2+ 3ab5c2−33ab5cd + 3ab5d2+b6c2+b6cd + 7b6d2)X −(c2+cd +d2)4(2a12c4+ 4a12c3d−24a12c2d2−26a12cd3−37a12d4+ 12a11bc4−156a11bc3d −414a11bc2d2−534a11bcd3−366a11bd4−228a10b2c4−1446a10b2c3d−2880a10b2c2d2 −3246a10b2cd3−1434a10b2d4−1250a9b3c4−5902a9b3c3d−10758a9b3c2d2−9508a9b3cd3 −2876a9b3d4−4059a8b4c4−16002a8b4c3d−23382a8b4c2d2−14868a8b4cd3−2925a8b4d4 −8604a7b5c4−25740a7b5c3d−26676a7b5c2d2−10944a7b5cd3−936a7b5d4−11112a6b6c4 −22224a6b6c3d−13428a6b6c2d2−2316a6b6cd3+ 480a6b6d4−8604a5b7c4−8676a5b7c3d −1080a5b7c2d2+ 396a5b7cd3+ 468a5b7d4−4059a4b8c4−234a4b8c3d+ 270a4b8c2d2 −126a4b8cd3+ 504a4b8d4−1250a3b9c4+ 902a3b9c3d−552a3b9c2d2+ 698a3b9cd3+ 526a3b9d4 −228a2b10c4+ 534a2b10c3d+ 90a2b10c2d2+ 912a2b10cd3+ 150a2b10d4+ 12ab11c4+ 204ab11c3d +126ab11c2d2+ 222ab11cd3−78ab11d4+ 2b12c4+ 4b12c3d−24b12c2d2−26b12cd3−37b12d4) GALOIS THEORY, DISCRIMINANTS AND TORSION SUBGROUP OF ELLIPTIC CURVES 5 with the following discriminants ∆E(1)= 2436(c2+cd +d2)8(a3d+ 3a2bc + 3a2bd + 3ab2c−b3d)6 (2a3c+a3d+ 3a2bc −3a2bd −3ab2c−6ab2d−2b3c−b3d)2 ∆E(2)= 2434(c2+cd +d2)8(a3d+ 3a2bc + 3a2bd + 3ab2c−b3d)2 (2a3c+a3d+ 3a2bc −3a2bd −3ab2c−6ab2d−2b3c−b3d)6 Therefore we have proved the following result: Proposition 1. Let Ebe an elliptic curve defined over Qwith a rational point of order 3such that p∆(E)∈Q. Then there exist a, b, c, d ∈Zsuch that Eis Q-isomorphic to either E(1)(a, b, c, d)or E(2)(a, b, c, d). 5. The case n= 5 Let us have a look at the case n= 5. If we throw away quadratic factors in ∆5(α) we will find out that curves Ewith points of order 5, for which p∆(E)∈Q are parametrized by the affine rational points of the elliptic curve D5:z2=α(α2−11α−1), where the discriminant of the right–hand side polynomial is, remarkably, 53. This is a well–known elliptic curve, in fact is Q-isogenous to the modular curve X0(20). The elliptic curve D5is denoted by 20A4 in Cremona’s tables [7] or 20C in Antwerp tables [3]. Looking on that tables, or using a computer algebra package like SAGE or MAGMA ([24], [4] resp.), we check that D5(Q) = {(0,0)}∪{[0 : 1 : 0]}. Therefore the only affine rational point is (0,0), which implies α= 0. This precise value does not yield an elliptic curve, but a singular cubic on the family E5(α). This proves the following result: Proposition 2. Let Ebe an elliptic curve with C5⊂E(Q)tors. Then p∆(E)/∈Q. 6. The case n= 7 Move now to n= 7, where the analogous argument to the case n= 5 shows that curves with points of order seven for which p∆(E)∈Qare parametrized by the affine rational points of the hyperelliptic curve D7:z2=α(α−1)(α3−8α2+ 5α+ 1), where, by the way, we have that the discriminant for the right–hand side polynomial is 74. We have now a hyperelliptic curve of genus 2, a much harder nut to crack; but we are lucky. Using MAGMA we obtain that the rank of the Jacobian of this genus 2 curve is 0, which makes it perfect for Chabauty’s algorithm [5]. This method computes the full list of points in the jacobian, then all rational points in the curve, which turn out to be D7(Q) = {(0,0),(1,0)}∪{[0 : 1 : 0]}. Again the affine rational points annhilate the discriminant of E7(α) and hence we have proven the following result. Proposition 3. Let Ebe an elliptic curve with E(Q)tors ∼ =C7. Then p∆(E)/∈Q. 6 IRENE GARC´ IA–SELFA, ENRIQUE GONZ´ ALEZ–JIM´ ENEZ, AND JOS´ E M. TORNERO 7. The case n= 9 Finally, the case n= 9. This can also be dealt with in a similar way, but a little extra work is needed. Following the steps as the above sections, the hyperelliptic curve parametrizing curves with E(Q)tors ∼ =C9and square discriminant is D9:z2=α(α−1)(α2−α+ 1)(α3−6α2+ 3α+ 1) . Lemma. D9(Q) = {(0,0),(1,0)}∪{[0 : 1 : 0]}. Proof. Let u∈AutQ(D9) defined by u(X, Y ) = 1 1−X,Y (1 −X)4. We have that uhas order 3 and Riemann-Hurwitz formulae tell us that the quotient curve C/huihas genus 1. In fact this curve is an elliptic curve defined over Q, since (0,0) ∈C(Q) and uis defined over Q. We will denote this elliptic curve by E9. A Weierstrass equation for E9is given by v2=u3−27 and the quotient morphism is given by: π:D9−→ E9 (α, z)7→ (u, v) = α3−3α2+ 1 α(α−1) ,z(α2−α+ 1) α2(α−1)2 Using SAGE or MAGMA we compute that E9is Q-isogenous to the modular curve X0(36), and it is the elliptic curve denotes by 36A3 in Cremona’s tables or 36C in Antwerp tables. The Mordell-Weil group of this elliptic curve is: E9(Q) = {(3,0),[0 : 1 : 0]}∼ =Z/2Z. Now, to compute the set D9(Q) we just need to compute the preimages of the points of E9(Q) by the quotien morphism that are defined over Qand then we obtain the desired result.  Then we have proved the following result: Proposition 4. Let Ebe an elliptic curve with E(Q)tors ∼ =C9. Then p∆(E)/∈Q. 8. The generic elliptic curve with square discriminant. Let E:Y2=X3+AX +Bbe an elliptic curve with A, B ∈Z. Let us study when the discriminant ∆(E) is a square. This is equivalent to looking for integer solutions to the Diophantine equation 4A3+ 27B2=−C2. Making the change of variables (x, y, z) = (C, 3B, −A) we obtain that the integer solutions of the generalized Fermat equation x2+ 3y2= 4z3give us all the elliptic curves defined over the rationals with square discriminant. Lemma 1 from the appendix gives us a complete parametrization of the above Diophantine equation, which yields to the following elliptic curve: Ealt(a, b, c, d) : Y2=X3−34(c2+cd +d2)(a2+ab +b2)X +35(c2+cd +d2)(a3d+ 3a2bc + 3a2bd + 3ab2c−b3d) with discriminant ∆(Ealt) = 24312(c2+cd+d2)2(2a3c+a3d+3a2bc−3a2bd−3ab2c−6ab2d−2b3c−b3d)2 Propositions 2, 3 and 4, together with Section 2, tell us that the above elliptic curve has torsion subgroup either trivial, C3or non-cyclic. Then we have proved the following result: GALOIS THEORY, DISCRIMINANTS AND TORSION SUBGROUP OF ELLIPTIC CURVES 7 Proposition 5. Let Ebe an elliptic curve defined over Qsuch that p∆(E)∈ Q. Then there exist a, b, c, d ∈Zsuch that Eis Q-isomorphic to Ealt(a, b, c, d). Moreover, E(Q)tors is either trivial, non-cyclic or C3. Remark 1. Let P(X) be an irreducible polynomial with integer coefficients and degree 3 such that the cubic number field attached to P(X) is cyclic. Then the elliptic curve E:Y2=P(X) satisfies E(Q)tors is either trivial or C3. For example, let be Pm(X) = X3+mX2−(m+ 3)X+ 1 ∈Z[X], m∈Zand Em:Y2=Pm(X). The irreducible polynomial Pm(X) defines a cubic fields Kmthat has been studied by several authors. This family has been called the simplest cubic field ([22]). Its discriminant satisfies ∆(Pm) = (m2+ 3m+ 9)2, hence Km=Q(E[2]) is cyclic and therefore ρEm,2(GQ)∼ =Gal(Km/Q) = C3. Therefore Em(Q)tors is trivial or C3. Moreover, it has been proved [10] that if m2+ 3m+ 9 is square-free, then Em(Q)tors is trivial. Remark 2. We have checked on the extended Cremona’s tables [8] of elliptic curves with conductor less than 130.000. Among them, 452.724 curves have torsion subgroup either trivial or C3. At the table below appears the specific proportions of curves according to their torsion group and the squareness of their discriminant: E(Q)tors p∆(E)∈Qp∆(E)6∈ Q {O} 0.00383 0.9553 C30.00008 0.0408 9. Main theorems. To end up we will summarize our results in the following theorems. Theorem 1. Let Ebe an elliptic curve defined over Q. Then (1) If E(Q)tors is non–cyclic then p∆(E)∈Q. (2) If E(Q)tors ∼ =Cnfor n= 2,4,5...10,12 then p∆(E)6∈ Q. (3) E(Q)tors ∼ =C3and p∆(E)∈Qif and only if there exist a, b, c, d ∈Zsuch that Eis Q-isomorphic to either E(1)(a, b, c, d)or E(2)(a, b, c, d)and the corresponding polynomial P(i)(a, b, c, d)) is irreducible. (4) E(Q)tors is trivial and p∆(E)∈Qif and only if there exist a, b, c, d ∈Z such that Eis Q-isomorphic to Ealt(a, b, c, d)and a, b, c, d /∈ S2,S3where S2=((a, b, c, d)∈Z4Ψ2(a, b, c, d)(X) = 3 Y i=1 (X−αi)such that α1, α2, α3∈Z) S3=(a, b, c, d)∈Z4∃(α, β)∈Q2,such that Ψ3(a, b, c, d)(α) = 0 Ψ2(a, b, c, d)(α) = β2 and Ψn(a, b, c, d)(X)denotes the n-division polynomial attached to Ealt(a, b, c, d). Theorem 2. Let Ebe an elliptic curve defined over Q. Then (1) E(Q)tors is non–cyclic if and only if ρE,2(GQ) = {id}. (2) E(Q)tors ∼ =C2nif and only if ρE,2(GQ)∼ =C2. (3) If E(Q)tors ∼ =Cnfor n= 5,7,9, then ρE,2(GQ) = GL2(F2). (4) If E(Q)tors ∼ =C3then ρE,2(GQ)∼ =C3if and only if there exist a, b, c, d ∈ Zsuch that Eis Q-isomorphic to either E(1)(a, b, c, d)or E(2)(a, b, c, d). Otherwise, ρE,2(GQ) = GL2(F2). (5) If E(Q)tors is trivial then ρE,2(GQ)∼ =C3if and only if there exist a, b, c, d ∈ Zsuch that Eis Q-isomorphic to Ealt(a, b, c, d). Otherwise, ρE,2(GQ) = GL2(F2). 8 IRENE GARC´ IA–SELFA, ENRIQUE GONZ´ ALEZ–JIM´ ENEZ, AND JOS´ E M. TORNERO The next table summarizes part of the main results of this paper: E(Q)tors p∆(E)∈Q?ρE,2(GQ) {O} Yes / No C3/S3 C2No C2 C3Yes / No C3/S3 C4No C2 C5No S3 C6No C2 C7No S3 C8No C2 C9No S3 C10 No C2 C12 No C2 C2×C2Yes {id} C2×C4Yes {id} C2×C6Yes {id} C2×C8Yes {id} 10. Appendix: The generalized Fermat equation x2+ 3y2= 4z3 The generalized Fermat equation Axp+Byq=Czr, where A, B, C ∈Z∗and p, q, r ∈Z>0, have been studied by several people for the last decades. Starting with the huge work on Diophantine equations due to L.J. Mordell [18]. After that the main results are due to H. Darmon and A. Granville [9], who proved that the generalized Fermat equation has infinite primitive solutions (i.e. gcd(x, y, z) = 1) in the case 1/p + 1/q + 1/r > 1. Then F. Beukers [2] gave parametrizations for the solutions of this equation in the above case. H. Cohen in his huge new books about number theory calls this equation Super-Fermat equation and he also provides solutions for several cases. In particular, H. Cohen [6, Proposition 14.2.1(ii)] displays primitive solutions to the equation we are interested in. However, we would like to have all integer solutions, not just the primitive ones. The following parametrization was pointed out to us by F. Beukers: Lemma 1. The integer solutions of the equation x2+ 3y2= 4z3are parametrized by the following family of four variables: F:   x= (c2+cd +d2)(3a2b(c−d) + a3(2c+d)−b3(2c+d)−3ab2(c+ 2d)) y= (c2+cd +d2)(3ab2c+a3d−b3d+ 3a2b(c+d)) z= (c2+cd +d2)(a2+ab +b2) Proof. Let (x, y, z) be an integer solution of the equation x2+ 3y2= 4z3. Over Q(√−3)we have (x+√−3y)(x−√−3y) = 4z3. We are going to work over the ring of algebraic integers of Q(√−3), noted O=Z[ρ] where ρ= (1 + √−3)/2. This ring is a P.I.D., where it is easy to check that gcd(x+√−3y, x −√−3y) = r∈Z, for all x, y ∈Z. Therefore x+√−3y=r·µ,x−√−3y=r·¯µwith µ∈ O and gcd(µ, ¯µ) = 1. Let p6= 3 be a non inert prime dividing r. Thus p=α¯αfor some α∈ O. Suppose that αdivides µ, then we have r=r′pand µ=αµ′for some r′∈Zand µ′∈ O. In other words, x+√−3y= (r′α¯α)(αµ′). Therefore 4z3=r2µ¯µ= (r′)2p3µ′¯µ′, that is, pdivides z. Then we can remove all the primes as above obtaining 4w3= s2γ¯γsuch that s, γ, ¯γare pairwise coprimes. This yields s= 2t3and γ=β3for GALOIS THEORY, DISCRIMINANTS AND TORSION SUBGROUP OF ELLIPTIC CURVES 9 some t∈Zand β∈ O. Now collecting all the factors back we obtain x+√−3y= 2(a+bρ)3(c+dρ)(c2+cd +d2) for some a, b, c, d ∈Z. Its easy to check that z=NO((a+bρ)(c+dρ)), where NO denotes the norm on O. In order to attach the parametrization, we only have to expand the above expression and compute the coefficient of √−3 (corresponding to y) and the part in Z(corresponding to x).  References [1] M.A. Bennett; P. Ingram: Torsion subgroups of elliptic curves in short Weierstrass form. Trans. Amer. 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P.O. 1160. 41080 Sevilla (Spain). E-mail address:[email protected]