68 2. Elliptic curves
Notice that, indeed, the elements of Sel2(E/Q) listed above form
a subgroup of Γ × Γ (Q×/(Q×)2)2. Since all the elements of
Sel2(E/Q) have rational points, we conclude that Sel2(E/Q) equals
E(Q)/2E(Q) and
X2(E/Q) = Sel2(E/Q)/(E(Q)/2E(Q)) = {C(1, 1)},
i.e., X2 is the trivial subgroup in this case.
Example 2.11.3. Let E :
y2
=
x3
6724x, as in Example 2.10.5.
The full group of homogeneous spaces H has 64 elements:
H = {C(δ1,δ2) : δi = ±1, ±2, ±41, ±82}.
The spaces in H with δ2 0 do not have points over R, so they
do not belong to Sel2(E/Q). Moreover, the spaces (δ1,δ2) = (2,2),
(41,2), (82,2), (−1,1), (−2,1), (−41,1), (−82,1), (−1,41), (−2,41),
(−41,41), (−82,41), (1,82), (2,82), (41,82), and (82,82) do not have
points over Q2. Therefore, they do not belong to Sel2(E/Q) either. It
turns out that the rest of the spaces (such as C(−1, 2)) are everywhere
locally solvable (we showed this for C(−1, 2)). Therefore they all
belong to Sel2(E/Q). Hence,
Sel2(E/Q) = {C(δ1,δ2) : (δ1,δ2) =
(1,1),(−1,82),(−82,2),(82,41),
(1,41),(82,1),(−82,82),(−2,2),
(41,1),(−41,82),(2,1),(−2,82),
(−41,2),(41,41),(−1,2),(2,41)}.
The spaces (1,1), (−1,82), (−82,2) and (82,41) have rational points
that correspond to (torsion) points on E(Q). However, none of the
other spaces have rational solutions! Thus, the rest are representative
of non-trivial elements of Sha, and we conclude that
E(Q)/2E(Q) = {C(1, 1),C(−1,82),C(−82,2),C(82,41)}
and X2(E/Q) = {C(δ1,δ2) : (δ1,δ2) = (1,1), (−1,2), (−2,2), (2,1)}.
Notice that the elements of X2 listed above are representatives
of all the classes in the quotient of Sel2(E/Q) by E(Q)/2E(Q). For
instance, (−1,2) · (1,41) = (−1,82) E(Q)/2E(Q). Thus, (−1,2) ·
(1,41) is trivial in X2.
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