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,