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:
⎧
⎪y0
⎪
⎪
⎪
⎨
⎪
⎪
⎪x0
⎪
⎩
2
= (x0 − e1)(x0 − e2)(x0 − e3),
x0 − e1 =
δ1u2,
− 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,
