34 2. Elliptic curves number of distinct rational points. Except for this criterion, Mazur’s theorem is not very helpful in effectively computing the torsion sub- group of a given elliptic curve. However, the following result, proven independently by E. Lutz and T. Nagell, provides a simple algorithm to determine E(Q)torsion: Theorem 2.5.5 (Nagell-Lutz, [Nag35], [Lut37]). Let E/Q be an elliptic curve with Weierstrass equation y2 = x3 + Ax + B, A, B Z. Then, every torsion point P = O of E satisfies: (1) The coordinates of P are integers, i.e., x(P ),y(P) Z. (2) If P is a point of order n 3, then 4A3 + 27B2 is divisible by y(P )2. (3) If P is of order 2, then y(P ) = 0 and x(P )3+Ax(P)+B = 0. For a proof, see [Sil86], Ch. VIII, Corollary 7.2, or [Mil06], Ch. II, Theorem 5.1. Example 2.5.6. Let E/Q : y2 = x3 2, so that A = 0 and B = −2. The polynomial x3 2 does not have any rational roots, so E(Q) does not contain any points of order 2. Also, 4A3 + 27B2 = 27 · 4. Thus, if (x(P),y(P)) are the coordinates of a torsion point in E(Q), then y(P ) is an integer and y(P )2 divides 27 · 4. This implies that y(P ) = ±1, ±2, ±3, or ±6. In turn, this implies that x(P )3 = 3, 6, 11 or 38, respectively. However, x(P ) is an integer, and none of 3, 6, 11 or 38 are a perfect cube. Thus, E(Q)torsion is trivial (i.e., the only torsion point is O). Example 2.5.7. Let p 2 be a prime number and let us define a curve Ep : y2 = x3 + p2. Since x3 + p2 = 0 does not have any rational roots, Ep(Q) does not contain points of order 2. Let P be a torsion point on Ep(Q). The list of all squares dividing 4A3 + 27B2 = 27p4 is short, and by the Nagell-Lutz theorem the possible values for y(P ) are: y = ±1, ±p, ±p2, ±3p, ±3p2, and ± 3. Clearly, (0, ±p) Ep(Q) and one can show that those two points and O are the only torsion points see Exercise 2.12.8. Thus, the torsion subgroup of Ep(Q) is isomorphic to Z/3Z for any prime p 2.
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