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50 2. Elliptic curves The current implementation of the algorithm is more fully explained in Cremona’s book [Cre97]. Let E/Q be a curve given by y2 = x3 + Ax + B, with A, B ∈ Z. The most general case of the method of descent is quite involved, so we will concentrate on a particular case where the calculations are much easier: we will assume that E(Q) has 4 distinct rational points of 2-torsion (including O). As we saw before (Theorem 2.5.5, or Exercise 2.12.6), a point P = (x, y) ∈ E(Q) is of 2-torsion if and only if y = 0 and x3 + Ax + B = 0 (or P = O). Thus, if E(Q) has 4 distinct rational points of order 2, that means that x3 + Ax + B has three (integral) roots and it factors completely over Z: x3 + Ax + B = (x − e1)(x − e2)(x − e3) with ei ∈ Z. Since x3 +Ax+B does not have an x2 term, we conclude that e1 + e2 + e3 = 0. Suppose, then, that E : y2 = (x − e1)(x − e2)(x − e3), where the roots satisfy ei ∈ Z and e1 + e2 + e3 = 0. We would like to find a solution (x0,y0) ∈ E with x0,y0 ∈ Q, i.e., y0 2 = (x0 − e1)(x0 − e2)(x0 − e3). Thus, each term (x0 − ei) must be almost a square, and we can make this precise by writing (x0 − e1) = au2, (x0 − e2) = bv2, (x0 − e3) = cw2, y0 2 = abc(uvw)2, where a, b, c, u, v, w ∈ Q, the numbers a, b, c ∈ Q are square-free, and abc is a square (in Q). Example 2.9.1. Let E : y2 = x3 − 556x + 3120 = (x − 6)(x − 20)(x + 26) so that e1 = 6, e2 = 20 and e3 = −26. The point (x0,y0) = ( 164184 289 , 66469980 4913 ) is rational and on E. We can write x0 − e1 = 164184 289 − 6 = 2 · 285 17 2 and, similarly, x0 − e2 = ( 398 17 )2 and x0 − e3 = 2 · ( 293 17 )2. Thus, following the notation of the preceeding paragraphs a = 2, b = 1, c = 2, u = 285 17 , v = 398 17 , w = 293 17 .