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.