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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 ≤ 2ν + 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,