52 2. Elliptic curves Theorem 2.9.3. Let E/Q be an elliptic curve y2 = x3 + Ax + B = (x e1)(x e2)(x e3) with distinct e1,e2,e3 Z and e1 + e2 + e3 = 0. There is a homomor- phism of groups δ : E(Q) Q×/(Q×)2 × Q×/(Q×)2 × Q×/(Q×)2 defined for P = (x0,y0) by δ(P ) = ⎪(1, ⎪(x0 ⎪(e2 1,1) if P = O e1,x0 e2,x0 e3) if y0 = 0 ((e1 e2)(e1 e3),e1 e2,e1 e3) if P = (e1,0) e1, (e2 e1)(e2 e3),e2 e3) if P = (e2,0) (e3 e1,e3 e2, (e3 e1)(e3 e2)) if P = (e3,0). If δ(P ) = (δ1,δ2,δ3), then δ1 ·δ2 ·δ3 = 1 in Q×/(Q×)2. Moreover, the kernel of δ is precisely 2E(Q) i.e., if δ(Q) = (1,1,1), then Q = 2P for some P E(Q). Proof. Let δ be the function defined in the statement of the theorem. Let us show that δ is a homomorphism of (abelian) groups i.e., we want to show that δ(P ) · δ(Q) = δ(P + Q). Notice first of all that δ(P ) = δ(x0,y0) = δ(x0, −y0) = δ(−P ), because the definition of δ does not depend on the sign of the y coordinate of P (in fact, it only depends on whether y(P ) = 0). Thus, it suffices to prove that δ(P ) · δ(Q) = δ(−(P + Q)) for all P, Q E(Q). Let P = (x0,y0), Q = (x1,y1) and R = −(P + Q) = (x2,y2), and let us assume, for simplicity, that yi = 0 for i = 1,2,3. By the definition of the addition rule on an elliptic curve (see Figure 1), the points P , Q and R are collinear. Let L = PQ be the line that goes through all three points, and suppose it has equation L : y = ax + b. Therefore, if we substitute y in the equation of E/Q, we obtain a polynomial p(x) = (ax + b)2 (x e1)(x e2)(x e3). The polynomial p(x) is cubic, its leading term is −1, and it has pre- cisely three rational roots, namely x0, x1 and x2. Hence, it factors: p(x) = (ax+b)2 −(x−e1)(x−e2)(x−e3) = −(x−x0)(x−x1)(x−x2).
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