62 2. Elliptic curves
If C(δ1,δ2) has a rational point but C(δ1,δ2) does not
have a rational point, then C(δ1 · δ1, δ2 · δ2) does not
have a rational point.
Example 2.10.4. Let E :
y2
=
x3
1156x = x(x 34)(x+34). The
only divisors of ΔE are 2 and 17. Thus, Γ = {±1, ±2, ±17, ±34}. Let
us choose e1 = 0, e2 = −34 and e3 = 34. Therefore, the homogeneous
spaces for this curve are all of the form
C(δ1,δ2) :
δ2Y 2 δ1X2 = 34,
δ2Y
2

δ1δ2Z2
= 68
with δ1,δ2 Γ . We analyze these spaces, case by case. There are 64
pairs (δ1,δ2) to take care of:
(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
= 68
cannot have solutions (in Q or R) because the left-hand side
is always negative for any X, Z Q.
(3) (δ1 0 and δ2 0). The equation δ2Y
2

δ1X2
= 34
cannot have solutions (in Q or R), because the left-hand
side is always negative.
(4) (δ1 = −1, δ2 = 34). The space C(−1, 34) has a rational
point (X, Y, Z) = (0,1,1), which maps to T1 = (0,0) on
E(Q) via Eq. (2.12).
(5) (δ1 = −34, δ2 = 2). The space C(−34, 2) has the rational
point (X, Y, Z) = (1,0,1), which maps to T2 = (−34,0) on
E(Q) via Eq. (2.12).
(6) (δ1 = 34, δ2 = 17). If δ(T1) = δ((0, 0)) equals (−1,34, −34),
and δ(T2) = (−34,2, −17), then
δ(T1 + T2) = δ(T1) · δ(T2) = (−1,34, −34) · (−34,2, −17) = (34,17,2).
Thus, the space C(34, 17) must have a point that maps
back to T1 + T2 = (34,0). Indeed, C(34, 17) has a point
(X, Y, Z) = (1,2,0) that maps to (34,0) via Eq. (2.12).
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