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:

⎪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,
Previous Page Next Page