60 2. Elliptic curves Thus, Γ has as many elements as {(t0,...,tν) : ti = 0 or 1}, which clearly has 2ν+1 elements. Moreover, the set ΓΔ, as defined in Corol- lary 2.9.6, has as many elements as Γ ×Γ , i.e., 22ν+2 elements. Since E(Q)/2E(Q) injects into ΓΔ, we conclude that it also has at most 22ν+2 elements. Since the size of E(Q)/2E(Q) is 2RE+2, we conclude that RE + 2 + 2 and RE 2ν, as claimed. Example 2.10.2. Let E : y2 = x3 1156x = x(x 34)(x + 34). The discriminant of E/Q is ΔE = 98867482624 = 212 · 176. Hence, ν(ΔE) = 2 and the rank of E is at most 4. (The rank is in fact 2 see Example 2.10.4 below.) The bound RE 2ν(ΔE) is, in general, not very sharp (The- orem 2.7.4 is an improvement). However, the method we followed to come up with the bound yields a strategy to find generators for E(Q)/2E(Q) as follows. Recall that E(Q)/2E(Q) embeds into ΓΔ via the map δ, so we want to identify which elements of ΓΔ may belong to the image of δ. Suppose (δ1,δ2,δ3) ΓΔ belongs to the image of δ and it is not the image of a torsion point. Then there exists P = (x0,y0) E(Q) such that: ⎪y2 ⎪x 0 = (x0 e1)(x0 e2)(x0 e3), x0 e1 = δ1u2, 0 e2 = δ2v2, x0 e3 = δ3w2 for some rational numbers u, v, w. We may substitute the last equa- tion in the previous two, and obtain e3 e1 = δ1u2 δ3w2, e3 e2 = δ2v2 δ3w2. Recall that the elements (δ1,δ2,δ3) that are in the image of δ satisfy δ1 · δ2 · δ3 = 1 modulo squares. Thus, δ3 = δ1 · δ2 · λ2 and if we do a change of variables (u, v, w) (X, Y, Z λ ), we obtain a system C(δ1,δ2) : e3 e1 = δ1X2 δ1δ2Z2, e3 e2 = δ2Y 2 δ1δ2Z2,
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