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,
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