56 2. Elliptic curves
and, therefore, substituting a in Eq. (2.9) yields
x(Q) = x1 =
a2
− 2bcB =
A − x0
2
2y0
2
− 2bcB
=
(A2 − 2Ax0 2 + x0) 4 − (2bcB)(4y0)2
4y02
=
(A2 − 2Ax0 2 + x0) 4 − (2bcB)(
4
c2
)
4y02
=
(A2 − 2Ax0 2 + x0) 4 − 8Bx0
4y02
=
x0
4
− 2Ax0
2
− 8Bx0 +
A2
4y02
= x(2P )
as desired. In order to complete the proof of the fact that the kernel of
δ is 2E(Q), we would need to consider the case when y(Q) = y1 = 0,
but we leave this special case to the reader (Exercise 2.12.18).
Thus, the previous proposition shows that there is a homomor-
phism δ : E(Q) →
(Q×/(Q×)2)3
with kernel equal to 2E(Q). In fact,
the theorem shows that there is a homomorphism from E(Q) into
Γ = {(δ1,δ2,δ3) ∈
(Q×/(Q×)2)3
: δ1 · δ2 · δ3 = 1 ∈
Q×/(Q×)2}.
Hence, δ induces an injection
E(Q)/2E(Q) → Γ ⊂
(Q×/(Q×)2)3.
The groups Q×/(Q×)2 and Γ are infinite, so such an injection does
not tell us much about the size of E(Q)/2E(Q). However, the image
of E(Q)/2E(Q) is much smaller than Γ.
Example 2.9.4. Let E :
y2
=
x3
− 556x+3120 as in Example 2.9.2.
It turns out that E(Q)
∼
=
Z/2Z ⊕ Z/2Z ⊕
Z2.
The generators of the
torsion part are T1 = (6,0) and T2 = (20,0), and the generators of
the free part are P = (−8,84) and Q = (24,60). The image of the
map δ in this case is, therefore, generated by the images of T1, T2, P
and Q.
δ(T1) = (−7, −14, 2), δ(T2) = (14,161,46),
δ(P ) = (−14, −7, 2), δ(Q) = (2,1,2).
