24 2. Elliptic curves
Obviously, Baker’s bound is not a very sharp bound, but it is theo-
retically interesting nonetheless.
2.4. The group structure on E(Q)
From now on, we will concentrate on trying to find all rational points
on a curve E : y2 = x3 + Ax + B. We will use the following notation
for the rational points on E:
E(Q) = {(x, y) ∈ E | x, y ∈ Q} ∪ {O}
where O = [0,1,0] is the point at infinity.
One of the aspects that makes the theory of elliptic curves so
rich is that the set E(Q) can be equipped with a group structure,
geometric in nature. The (addition) operation on E(Q) can be defined
as follows (see Figure 1). Let E be given by a Weierstrass equation
y2
=
x3
+Ax+B with A, B ∈ Q. Let P and Q be two rational points
in E(Q) and let L =
PQ be the line that goes through P and Q (if
P = Q, then we define L to be the tangent line to E at P ). Since
the curve E is defined by a cubic equation, and since we have defined
L so it already intersects E at two rational points, there must be a
third point of intersection R in L ∩ E, which is also defined over Q,
and
L ∩ E(Q) = {P, Q, R}.
The sum of P and Q, denoted by P + Q, is by definition the second
point of intersection with E of the vertical line that goes through R,
or in other words, the reflection of R across the x-axis.
It is easy to verify that the addition operation that we have de-
fined on points of E(Q) is commutative. The origin O is the zero
element, and for every P ∈ E(Q) there exists a point −P such that
P + (−P) = O. If E is given by
y2
=
x3
+ Ax + B and P = (x0,y0),
then −P = (x0, −y0). The addition is also associative (but this is
not obvious, and it is tedious to prove) and, therefore, (E,+) is an
abelian group.
Example 2.4.1. Let E be the elliptic curve
y2
=
x3
− 25x, as in
Example 1.1.2. The points P = (5,0) and Q = (−4,6) belong to
