64 2. Elliptic curves

we have reached a contradiction and C(2, 2) cannot have a

rational point. One can rule out all the other (δ1,δ2) in the

list similarly.

We have analyzed all 64 possible pairs (δ1,δ2) and have found that

the image of E(Q)/2E(Q) via δ has order

24.

Therefore,

2RE+2

=

24

and RE = 2. The rank of the curve is exactly 2 and T1,T2,P and Q

(as found above) are generators of E(Q)/2E(Q). (In fact, they are

generators of E(Q) as well.)

Example 2.10.5. Let E :

y2

=

x3

− 6724x = x(x − 82)(x +82). Let

e1 = 0, e2 = −82 and e3 = 82. The only divisors of ΔE are 2 and

41, hence Γ = {±1, ±2, ±41, ±82}. Let us analyze the homogeneous

spaces

C(δ1,δ2) :

δ2Y

2

−

δ1X2

= 82,

δ2Y 2 − δ1δ2Z2 = 164

as we did in the previous example. Once again, there are 64 pairs to

check:

(1) ((δ1, δ2,δ3) = (1,1,1)). The point at infinity (i.e., the origin)

is sent to (1,1,1) via δ, i.e., δ(O) = (1,1,1).

(2) (δ1 0 and δ2 0). The equation δ2Y 2 − δ1δ2Z2 = 164

cannot have rational solutions because the left-hand side is

always negative for any X, Z ∈ Q.

(3) (δ1 0 and δ2 0). The equation δ2Y

2 −δ1X2

= 82 cannot

have rational solutions, because the left-hand side is always

negative.

(4) ((δ1, δ2) = (−1,82), (−82,2), (82,41)). The corresponding

spaces have (trivial) rational points that map, respectively,

to T1 = (0,0), T2 = (−82,0) and T3 = T1 + T2 = (82,0) via

Eq. (2.12).

(5) ((δ1, δ2) = (1,2)). The space C(1, 2) does not have rational

points (same reason as for Exercise 2.12.21). In fact, it does

not have any solutions over Q2.

(6) ((δ1, δ2) = (−1,41), (−82,1), (82,82)). The correspond-

ing spaces cannot have rational points, because these ele-

ments of ΓΔ are the product of (1,2,2), with no points,