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).
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