THERE ARE NO ELLIPTIC CURVES DEFINED OVER Q WITH POINTS OF ORDER 11

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1 THERE ARE NO ELLIPTIC CURVES DEFINED OVER Q WITH POINTS OF ORDER 11 ALLAN LACY 1. Introduction If E is an elliptic curve over Q, the set of rational points E(Q), form a group of finite type (Mordell-Weil Theorem): E(Q) = Z r E(Q) tors where E(Q) tors is finite. In 1977, Mazur gave complete list of the only subgroups that can appear as E(Q) tors as E varies. Mazur [6] proved that there are finitely many possibilities (all the possibilities occur infinitely often) Theorem 1 (Mazur, 1977). Let E/Q be an elliptic curve. Then the torsion subgroup of E(Q) is one of the following 15 groups: E(Q) tors = { Z/nZ n = 1, 2,..., 10, 12; Z/2Z Z/2nZ n = 1, 2, 3, 4 In particular, Mazur s Theorem implies that there is no elliptic curve over Q with a point of order 11. This particular result was proved originally in 1939 by Billing and Mahler [1]. In this pape we present a proof of this fact, following the main idea on the original proof. 2. The curve X 1 (11) We start with some general considerations about torsion points on elliptic curves. Let E/Q be a elliptic curve with Weierstrass model E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6 Lets assume that P E(Q) is point of exact order n 4. Since P has coordinates in Q, we can make a change of variables (over Q) to assume that P = (0, 0), and therefore that a 6 = 0. We have P = (0, a 3 ), and since we are assuming that P does not have order 2, it follows that a 3 0. Then the change of variable y y + a4 a 3 x allows us to write a WE for E with a 4 = 0. With a 4 = 0 we easily compute 2P = ( a 2, a 1 a 2 a 3 ), so in particular a 2 0, since otherwise 2P = P, and P would have order 3. Finally, if we set λ = a2 a 3, the change of variable x λ 2 x, y λ 3 y give us E : y 2 + a 1 λxy + λ 3 a 3 y = x 3 + λ 2 a 2 x 2 Note that λ 3 a 3 = λ 2 a 2, so setting b this common value, and c := 1 λa 1, we can write a model for E of the form 1

2 2 ALLAN LACY (1) E = E(b, c) : y 2 + (1 c)xy by = x 3 bx 2 The two-parameter Weierstrass equation (1) is called the Tate Normal Form for E with point P = (0, 0). Each point (b, c) Q Q for which (b, c), the discriminat of (1), is nonzero defines an elliptic curve over Q with P = (0, 0) E(Q) such that P, 2P, 3P. So far we have only used the fact that P is not a point of order 2 or 3. For any k Z, we can consider the point kp, which would have as coordinates rational functions of b and c. If we further assume that P has order n 4, then the equation np = defines a polynomial equation f n (b, c) = 0 over Z. Each point on the open affine curve f n (b, c) = 0, b 0, (b, c) 0 defines an elliptic curve E(b, c) together with a Q-rational point of order n. This is then an affine model of Y 1 (n), the parameter space (of isomorphism classes) of pairs (E, P ) of elliptic curves E with a point P of order n. The curve Y 1 (n) is not necessarily smooth, so we consider X 1 (n) to be its completion. The idea of the proof, roughly speaking, is to prove that the curve X 1 (11) does not contain points. Then, we start by computing an equation for X 1 (11). We follow the steps in [8] to find X 1 (11) : for n = 11, we compute X 1 (11) by looking at the condition imposed by 5P = 6P. We have 5P = (rs(s 1), r 2 s(s 1) 2 ); b = cr and c = s(r 1) 6P = ( mt, m 2 (m + 2t 1)); m(1 s) = s(1 r) and r s = t(1 s) so we get s(1 r) rs(s 1) = mt = (1 s) r s 1 s Clearing denominators we get r 2 4sr + 3s 2 r s 3 r + s = 0 If we further make the birational transformation r y + 1, s y x + 1 we get the following equation for X 1 (11) which is smooth (is a elliptic curve!): X 1 (11) : y 2 + y = x 3 x 2 We can trace back all the changes of variables and substitutions, to get b and c as a function of (x 0, y 0 ) X 1 (11): ( ) y0 (2) b = y 0 (y 0 + 1) + 1 x 0 and ( ) y0 c = y x 0 As we said roughly speakig, the idea is to prove that the set of points on the curve X 1 (11) is empty, since otherwise we can use (2) to define an elliptic curve E(b, c) with a point of order 11. Also, if we requiere that such E(b, c) is defined over Q, we have to restrict only to b, c Q, and therefore from ( ) 1 c b = c(y 0 + 1) and x 0 = y 0 1 y 0

3 THERE ARE NO ELLIPTIC CURVES DEFINED OVER Q WITH POINTS OF ORDER 11 3 we see that we also need make restrictions to the point (x 0, y 0 ) (in particular, it has to be Q-rational). Note that we can spot some rational points on X 1 (11): (0, 0), (1, 0), (0, 1) and (1, 1). Although, these 4 points produce b = 0 or b =, so have to be discarded (recall b = λ 3 a 3 0). We call these forbidden points cusps of X 1 (11). Now we can explain more rigorously the idea of the proof: we will prove that the only Q-rational points of X 1 (11) are these 4 cusps. We use the fact that X 1 (11) is an elliptic curve, and therefore X 1 (11)(Q) is a group. First of all, it not hard to check that these 4 points are torsion elements of X 1 (11). Moreover, these points (together with the point at infinity) form the whole torsion subgroup of X 1 (11)(Q). To see this, we use the following theorem, proved independently by Lutz and Nagell, to compute the torsion subgroup of an elliptic curve over Q: Theorem 2 (Lutz-Nagell). Let E/Q and elliptic curve with short Weierstrass equation y 2 = f(x) with f(x) Z[x]. If (x 0, y 0 ) E(Q) tors, then x 0, y 0 Z. Furthermore, either y 0 = 0 or y 2 0 D f, the discriminat of f. Then, in order to apply the Lutz-Nagell theorem to X 1 (11), we first need to write it in short form: completing the square for y and multiplying the resulting equation by 2 6 we get X 1 (11) : y 2 = x 3 4x = f(x); D f = First of all, the four (cusps) torsion points on y 2 + y = x 3 x 2 give us four torsion points on y 2 = f(x): (0, ±4) and (4, ±4). Now, we apply Lutz-Nagell theorem to prove these are the only torsion elements. Since the polynomial f(x) is irreducible over Q, we can discard the case y = 0. Then we have to look at the integers y such that y 2 2 8, that is y = ±2 i, i = 0, 1, 2, 3, 4. The case i = 2 give us the four points above. For i = 3 or i = 4, we need to solve x 2 (x 4) = 16(2 2i 4 1) which is not possible in Z: x must be divisible by 4, but if we write x = 4a, then 4a 2 (a 1) = 2 2i 4 1. The cases i = 0 and i = 1 can be ruled out easily too. We conclude that X 1 (11)(Q) tors = {, (0, 4), (0, 4), (4, 4), (4, 4)} = Z/5Z. We have proved that X 1 (11) has no more torsion points, so our next (and final) step is to prove that it has no points of infinite order. We make yet another change of variable to X 1 (11): completing the cube in f(x) and multiplying by 3 6 the resulting equation we get 1 X 1 (11) : y 2 = x 3 432x We will prove that the only rational points on this elliptic curve are ( 12, ±108) and (24, ±108) (which correspond to the cusps on y 2 + y = x 3 x 2 ). Assuming this result by the moment, we are ready: Lemma 1. The only rational points on y 2 + y = x 3 x 2 (0, 0), (0, 1), (1, ±1). are the four points 1 this is the curve appearing in [1], for which the authors prove to have only has 5 rational points (Lemma 2, [1])

4 4 ALLAN LACY Proof. The rational points on y 2 + y = x 3 x 2 are in bijection with the rational points of y 2 = x 3 432x , and there are only four such points. Theorem 3. There is no elliptic curve E/Q with a point of order 11. Proof. Suppose by contradiction that there is a elliptic curve E/Q with a point P E(Q) of order 11. We can assume that P = (0, 0) and that E is given by E = E(b, c) : y 2 + (1 c)xy by = x 3 bx 2 for some b, c Q, b 0. Since P has order 11, using formula (2) we define points x 0, y 0 Q satisfying y0 2 + y 0 = x 3 0 x 2 0. By construction (x 0, y 0 ) (0, 0), (0, 1), (1, ±1), since x 0 = 0, y 0 = 0, 1, x 0 are not allowed, according to (2). The previous Lemma gives us the contradiction. 3. The main tool The following is the key in [1], from where the authors conclude that there is no elliptic curve over Q with a point of order 11. Theorem 4. The elliptic curve E : y 2 = x 3 432x has no rational points of infinite order. Proof. The proof is by contradiction, so lets assume Q = (x 0, y 0 ) E(Q) is a point of infinite order. We start by recalling that the five points {, ( 12, ±108), (24, ±108)} = E(Q) tors form a cyclic group of order 5, generated say, by P. Then we have P = 6P and 3P = 8P, so every point of {, P, 2P, 3P, 4P } lie on 2E(Q). Therefore, if there is some Q E(Q) not in the subgroup 2E(Q), then Q is necessarily of infinite order. Conversely, if E(Q) contains a point of infinite order, we can find a point not in 2E(Q) (namely, a generator of E(Q)/E(Q) tors ). We are going to exploit this observation: we are going to prove that the quotient E(Q)/2E(Q) is trivial, by following the proof in [2] of the weak finite basis theorem, that in general, the quotient E(Q)/2E(Q) is finite. We start by factoring f(x) over Q: f(x) = x 3 432x := (x θ 1 )(x θ 2 )(x θ 3 ) Lets assume that θ = θ 1 R, and consider the number field K = Q(θ). We also consider a map given by { µ(q) = µ : E(Q) K /(K ) 2 1 if Q = (x θ) mod (K ) 2 if Q = (x, y) We use the following result, presented here without proof (for a proof, see [2]): Lemma 2. The map µ is a group homomorphism, and ker µ = 2E(Q).

5 THERE ARE NO ELLIPTIC CURVES DEFINED OVER Q WITH POINTS OF ORDER 11 5 Then our assumption that E(Q) contains a point Q of infinite order translates to Q / ker µ, that is, if Q = (x 0, y 0 ) then x 0 θ is not a square in K. If we write (x 0, y 0 ) = (r/t 2, s/t 3 ) Q 2 with (r, t) = (s, t) = 1, then y0 2 = f(x 0 ) is equivalent to s 2 = (r θt 2 )(r θ 2 t 2 )(r θ 3 t 2 ) Then µ(q) x 0 θ r θt 2 mod (K ) 2. We consider the principal ideal generated by r θt 2 in O K, the ring of integers of K. The bulk of the work is to prove that this ideal is a square. We also consider γ = (r θ 2 t 2 )(r θ 3 t 2 ) O K 2 and the factorization into primes of the ideals (r θt 2 ) and (γ): so (r θt 2 ) = p ai i and (γ) = p bi i (γ, r θt 2 ) = (γ) + (r θt 2 ) = p min(ai,bi) i Lets suppose that for p = p i in the above factorization we have a i = ord p (r θt 2 ) is odd. In particular a i > 0, and from s 2 = (r θt 2 )γ we have that b i = ord p (γ) > 0. Writing s 2 = (r θt 2 )γ = (r θt 2 ) (θ θ 2 )(θ θ 3 ) + (r θt 2 ) 2 (r + e 2 (θ θ 3 θ 3 )) we have = (r θt 2 ) f (θ) + (r θt 2 ) 2 (r + e 2 (θ θ 3 θ 3 )) ord p (f (θ)) = ord p ((r θt 2 )γ) min(a i, b i ) > 0 so we conclude that the ideal (γ, r θt 2 ) divides N K/Q (f (θ)) = (f) = , the discriminant of f. Therefore we can write (r θt 2 ) = IJ 2 for ideals I, J O K with I squarefree, divisible only by the primes above 2, 3 and 11. In O K we have the following factorization of these primes: so (2) = p 3 2, (3) = p 3, (11) = p 11 q 2 11 I = p a 2 p b 3 p c 11 q d 11, a, b, c, d {0, 1} We have prove that a = b = c = d = 0. Since s 2 = N K/Q (r θt 2 ) = N K/Q (I)N K/Q (J) 2 we see that N K/Q (I) = 2 a 3 b 11 c+d is a square. Then a = b = 0 and c = d = 0 or c = d = 1. Lets suppose that c = d = 1, so p 11 q 11 (r θt 2 ). Therefore p 2 11q 2 11 (r θt 2 ) 2 and since (11) = p 11 q 2 11 we see that 11 (r θt 2 ) 2, so (r θt2 ) 2 11 O K. But K has an integral basis { θ θ2 324, θ } 18 + θ2 54, θ2 36 so in particular, no element of O K has denominator 11. This ends the proof that the ideal (r θt 2 ) = J 2 is a square. 2 γ = r 2 rt 2 (θ 2 + θ 3 ) + θ 2 θ 3 t 4 = r 2 + rt 2 θ + t 4 bθ 1 O K

6 6 ALLAN LACY But K has class number 1, so there is some α O K such that J = (α), so (r θt 2 ) = (α 2 ). Therefore, there is a unit η O K such that (3) r θt 2 = ηα 2 Note that η cannot be the square of another unit, for otherwise, r θt 2 would be a square in K. The trick in [1] is to note that K = Q(θ) = Q(ϑ) where θ is the real root of x 3 4x 4 = 0. Specifically we are going to use the following information about K = Q(ϑ) θ = 24 9ϑ 2 = 12(1 + 3/ϑ) (so, in fact, θ and ϑ generate the same field extension) {1, ϑ, 1 2 ϑ2 } is an integral basis of K The unit group of K has rank 1, generated by the fundamental unit ϑ2 We have s 2 = N K/Q (r θt 2 ) = N K/Q (η)n K/Q (α) 2 so N K/Q (η) must be a square. This leave us with the only possibility η = ϑ2. Multiplying (3) by η we have η(r θt 2 ) = (ηα) 2 ( ϑ2 )(r (24 9ϑ 2 )t 2 ) = (ηα) 2 = (a + bϑ + c ϑ2 2 )2 for some a, b, c Z. Using ϑ 4 = 4ϑ + 4ϑ 2, equating coefficients above, we get the system of equations r + 24t 2 = a 2 + 4bc 18t 2 = c 2 + 2ab + 4bc 1 2 r 3t2 = b 2 + c 2 + ac From the third and first equations, r and a must be even. Since (r, t) = 1, it follows t is odd, hence the second equation leads to the impossible congruence c 2 2(mod 2). This contradiction proves that E(Q) does not contain any point of infinite order. 4. Remarks The result in Theorem 3 is no longer valid if we replace Q for arbitrary number field. For example, if we fix 0, 4 x 1 Q and let y 1 Q be a solution of y 2 = x 3 1 4x then x 0 = x 1 4, y 0 = y is a solution of y 2 + y = x 3 x 2. Then, if we define b, c Q(y 1 ) using (2), the elliptic curve E(b, c) has P = (0, 0) as a point of order 11. This procedure allow us to define quadratic (or cubic, if we fix y 1 first) extensions of Q and elliptic curves over these extensions having points of order 11. It is natural to ask for some kind of generalization of Mazur s Theorem to arbitrary numbers fields; that is, given an number field K, to characterize the groups that can

7 THERE ARE NO ELLIPTIC CURVES DEFINED OVER Q WITH POINTS OF ORDER 11 7 appear as the torsion part of E(K) for arbitrary elliptic curves E/K. Historically some partial results led to Conjecture 1 (Uniform Boundness Conjecture (UBC)). Let d 1. For any number field K/Q of degree d and any elliptic curve E/K, there is constant C = C(d) such that E(K) tors C. So in particular, the UBC implies that for d 1, the set T (d) = {E(K) tors : [K : Q] d and E/K is an elliptic curve} is finite. For quadratic extensions we have the following result, due to S. Kamienny [4] Theorem 5 (Kamienny, 1992). If E is an elliptic curve defined over a quadratic number field K, then E(K) tors is isomorphic to one of the following 26 groups: Z/nZ n = 1, 2,..., 16, 18; Z/2Z Z/2nZ n = 1, 2,..., 6 E(K) tors = Z/3Z Z/3nZ n = 1, 2 Z/4Z Z/4Z From Mazur s Theorem, we see that the only possible primes that appears as divisors of E(Q) tors are 2, 3, 5 and 7; while from Kamienny s Theorem, if K/Q is a quadratic extension, the only possible primes diving E(K) tors are 2, 3, 5, 7, 11 and 13. For d 1 lets consider S(d) = {primes p : there is K/Q of degree d and an elliptic curve E/K such that p E(K) tors } An obvious necessary condition for the UBC to hold is that for all d, the set S(d) is finite. It happens that this is also sufficient: Theorem 6 (Frey, Faltings). If S(d) is finite, then T (d) is finite. In particular, if S(d) is finite, the UBC holds in degree d. The following theorem, due to Merel [7], proves affirmatively the UBC: Theorem 7 (Merel, 1996). S(d) d 3d2 for all d 2 Shortly after Merel s result, his bound was improved by Osterlé: Theorem 8 (Oesterlé). S(d) (3 d/2 + 1) 2 for all d 2 Finally, we state some known results: Theorem 9.. (Kamienny and Parent, 1999) S(3) = {2, 3, 5, 7, 11, 13} (Kamienny, Stein and Stoll, 2010) S(4) = {2, 3, 5, 7, 11, 13, 17}

8 8 ALLAN LACY References [1] Billing G. and Mahler K., On exceptional points on cubic curves J. London Math. Soc. 15, (1940) [2] J.W.S. Cassels, Lectures on elliptic curves London Mathematical Society Students Texts 24, Cambridege University Press, 1991 [3] Husemöller, D., Elliptic Curves, Graduate Texts in Mathematics, Springer-Verlag, New York- Berlin, 1987 [4] Kamienny, S. Torsion points on elliptic curves and q-coefficients of modular forms Invent. Math. 109 (1992), no. 2, [5] Kamienny, S., Stein, W. and Stoll, M. Torsion points on elliptic curves over quadratic numbers fields, available online [6] Mazur, B. Modular curves and the Eisenstein ideal Inst. Hautes tudes Sci. Publ. Math. No. 47 (1977), (1978). [7] Merel, Loc. Bornes pour la torsion des courbes elliptiques sur les corps de nombres, Invent. Math. 124 (1996), no. 1-3, [8] Reichert, Markus A. Explicit determination of nontrivial torsion structures of elliptic curves over quadratic number fields Math. Comp. 46 (1986), no. 174,

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