2.9. Descent and the weak Mordell-Weil theorem 55 where all the congruences are modulo f(x) = x3 + Ax + B. The congruences in the previous equation say that a polynomial of degree 1, call it g(x) = x1 x, is congruent to a polynomial of degree 2, call the last line h(x), modulo a polynomial of degree 3, namely f(x). Then h(x) g(x) is a polynomial of degree 2, divisible by a polynomial of degree 3. This implies that h(x) g(x) must be zero and h(x) = g(x), i.e., x1 x = (2ac + b2 Ac2)x2 + (2ab Bc2 2Abc)x + (a2 2bcB). If we match coefficients, we obtain the following equalities: 2ac + b2 Ac2 = 0, (2.7) 2ab Bc2 2Abc = −1, (2.8) a2 2bcB = x1. (2.9) If c = 0, then b = 0 by Eq. (2.7) therefore, p(x) = a +bx+cx2 = a is a constant function, and so t1 = t2 = t3. By Eq. (2.6), it follows that e1 = e2 = e3, which is a contradiction with our assumptions. Hence, c must be non-zero. We multiply Eq. (2.8) by 1 c2 and Eq. (2.7) by b c3 to obtain 2ab c2 B 2Ab c = 1 c2 , (2.10) 2ab c2 + b3 c3 Ab c = 0. (2.11) We subtract Eq. (2.10) from Eq. (2.11) to get: b c 3 + A b c + B = 1 c 2 . Hence, the point P = (x0,y0) = ( b c , 1 c ) is a rational point on E(Q). It remains to show that x(2P ) = x(Q). From Eq. (2.11) we deduce that a = Ab c b3 c3 2b c2 = A ( b c )2 2 · 1 c = A x2 0 2y0 ,
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