56 2. Elliptic curves and, therefore, substituting a in Eq. (2.9) yields x(Q) = x1 = a2 2bcB = A x02 2y0 2 2bcB = (A2 2Ax2 0 + x4) 0 (2bcB)(4y2) 0 4y2 0 = (A2 2Ax2 0 + x4) 0 (2bcB)( 4 c2 ) 4y2 0 = (A2 2Ax2 0 + x4) 0 8Bx0 4y2 0 = x0 4 2Ax0 2 8Bx0 + A2 4y02 = x(2P ) as desired. In order to complete the proof of the fact that the kernel of δ is 2E(Q), we would need to consider the case when y(Q) = y1 = 0, but we leave this special case to the reader (Exercise 2.12.18). Thus, the previous proposition shows that there is a homomor- phism δ : E(Q) (Q×/(Q×)2)3 with kernel equal to 2E(Q). In fact, the theorem shows that there is a homomorphism from E(Q) into Γ = {(δ1,δ2,δ3) (Q×/(Q×)2)3 : δ1 · δ2 · δ3 = 1 Q×/(Q×)2}. Hence, δ induces an injection E(Q)/2E(Q) Γ (Q×/(Q×)2)3. The groups Q×/(Q×)2 and Γ are infinite, so such an injection does not tell us much about the size of E(Q)/2E(Q). However, the image of E(Q)/2E(Q) is much smaller than Γ. Example 2.9.4. Let E : y2 = x3 556x+3120 as in Example 2.9.2. It turns out that E(Q) Z/2Z Z/2Z Z2. The generators of the torsion part are T1 = (6,0) and T2 = (20,0), and the generators of the free part are P = (−8,84) and Q = (24,60). The image of the map δ in this case is, therefore, generated by the images of T1, T2, P and Q. δ(T1) = (−7, −14, 2), δ(T2) = (14,161,46), δ(P ) = (−14, −7, 2), δ(Q) = (2,1,2).
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