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

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