52 2. Elliptic curves
Theorem 2.9.3. Let E/Q be an elliptic curve
y2
=
x3
+ Ax + B = (x − e1)(x − e2)(x − e3)
with distinct e1,e2,e3 ∈ Z and e1 + e2 + e3 = 0. There is a homomor-
phism of groups
δ : E(Q) →
Q×/(Q×)2
×
Q×/(Q×)2
×
Q×/(Q×)2
defined for P = (x0,y0) by
δ(P ) =
⎧
⎪
⎪
⎪(1,
⎪
⎪
⎪
⎪(x0
⎨
⎪
⎪
⎪
⎪
⎪(e2
⎪
⎪
⎩
1,1) if P = O;
− e1,x0 − e2,x0 − e3) if y0 = 0;
((e1 − e2)(e1 − e3),e1 − e2,e1 − e3) if P = (e1,0);
− e1, (e2 − e1)(e2 − e3),e2 − e3) if P = (e2,0);
(e3 − e1,e3 − e2, (e3 − e1)(e3 − e2)) if P = (e3,0).
If δ(P ) = (δ1,δ2,δ3), then δ1 ·δ2 ·δ3 = 1 in
Q×/(Q×)2.
Moreover, the
kernel of δ is precisely 2E(Q); i.e., if δ(Q) = (1,1,1), then Q = 2P
for some P ∈ E(Q).
Proof. Let δ be the function defined in the statement of the theorem.
Let us show that δ is a homomorphism of (abelian) groups; i.e., we
want to show that δ(P ) · δ(Q) = δ(P + Q). Notice first of all that
δ(P ) = δ(x0,y0) = δ(x0, −y0) = δ(−P ), because the definition of
δ does not depend on the sign of the y coordinate of P (in fact, it
only depends on whether y(P ) = 0). Thus, it suffices to prove that
δ(P ) · δ(Q) = δ(−(P + Q)) for all P, Q ∈ E(Q).
Let P = (x0,y0), Q = (x1,y1) and R = −(P + Q) = (x2,y2),
and let us assume, for simplicity, that yi = 0 for i = 1,2,3. By the
definition of the addition rule on an elliptic curve (see Figure 1), the
points P , Q and R are collinear. Let L = PQ be the line that goes
through all three points, and suppose it has equation L : y = ax + b.
Therefore, if we substitute y in the equation of E/Q, we obtain a
polynomial
p(x) = (ax +
b)2
− (x − e1)(x − e2)(x − e3).
The polynomial p(x) is cubic, its leading term is −1, and it has pre-
cisely three rational roots, namely x0, x1 and x2. Hence, it factors:
p(x) =
(ax+b)2
−(x−e1)(x−e2)(x−e3) = −(x−x0)(x−x1)(x−x2).
