2.9. Descent and the weak Mordell-Weil theorem 53 If we evaluate p(x) at x = ei, we obtain p(ei) = (aei + b)2 = −(ei x0)(ei x1)(ei x2) or, equivalently, (x0 ei)(x1 ei)(x2 ei) = (aei + b)2. Thus, the product δ(P ) · δ(Q) · δ(R) equals δ(P ) · δ(Q) · δ(R) = (x0 e1,x0 e2,x0 e3) ·(x1 e1,x1 e2,x1 e3) ·(x2 e1,x2 e2,x2 e3) = ((x0 e1)(x1 e1)(x2 e1), (x0 e2)(x1 e2)(x2 e2), (x0 e3)(x1 e3)(x2 e3)) = ((ae1 + b)2, (ae2 + b)2, (ae3 + b)2) = (1,1,1) (Q×/(Q×)2)3. Hence, δ(P ) · δ(Q) · δ(R) = 1. If we multiply both sides by δ(R) and notice that a2 = 1 for any a Q×/(Q×)2, we conclude that δ(P ) · δ(Q) = δ(R) = δ(−(P + Q)) = δ(P + Q), as desired. In order to completely prove that δ is a homomorphism, we would need to check the cases when P , Q or R is one of the points (ei,0) or O, but we leave those special cases for the reader to check (Exercise 2.12.15). If δ(P ) = (δ1,δ2,δ3), then it follows directly from the definition of δ that δ1 · δ2 · δ3 = 1 in Q×/(Q×)2. Indeed, this is clear for P = O or P = (ei,0), and if P = (x0,y0) with y0 = 0, then (x0 e1)(x0 e2)(x0 e3) = y0, 2 which is a square, and is therefore trivial in Q×/(Q×)2. Next, let us show that the kernel of δ is 2E(Q). Clearly, 2E(Q) is in the kernel of δ, because δ is a homomorphism with image in (Q×/(Q×)2)3, as we just proved. Indeed, if P E(Q), then δ(2P ) = δ(P ) · δ(P ) = δ(P )2 = (δ2,δ2,δ2) 1 2 3 = (1,1,1), because squares are trivial in Q×/(Q×)2. Now let us show the reverse inclusion, i.e., that the kernel of δ is contained in 2E(Q). Let Q = (x1,y1) E(Q) such that δ(Q) = (1,1,1). We want to find P = (x0,y0) such that 2P = Q. Notice that
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