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).