46 2. Elliptic curves

Example 2.7.6. Let E/Q be the elliptic curve y2 = x(x + 1)(x +

2), which already appeared in Example 1.1.1. Since the Weierstrass

equation of E is

y2

= x(x + 1)(x + 2) =

x3

+

3x2

+ 2x,

it follows from Theorem 2.7.4 that the rank RE satisfies

RE ≤

ν(A2

− 4B) + ν(B) − 1 = ν(1) + ν(2) − 1 = 0 + 1 − 1 = 0,

and therefore the rank is 0. The reader can check that

E(Q)torsion = {O, (0,0),(−1,0),(−2,0)}.

Since the rank is zero, the four torsion points on E/Q are the only

rational points on E.

Example 2.7.7. Let E :

y2

=

x3 +2308x2

+665858x. The primes 2

and 577 are the only prime divisors of (both) B and

A2

− 4B. Thus,

RE ≤

ν(A2

− 4B) + ν(B) − 1 = 2 + 2 − 1 = 3.

The points P1 = (−1681,25543), P2 = (−338,26), and P3 = (577/16,

332929/64) are of infinite order and the subgroup of E(Q) generated

by P1, P2 and P3 is isomorphic to

Z3.

Therefore, the rank of E is

equal to 3.

2.8. Linear independence of rational points

Let E/Q be the curve defined in Example 2.7.7. We claimed that

the subgroup generated by the points P1 = (−1681,25543), P2 =

(−338,26), and P3 = (577/16,332929/64) is isomorphic to

Z3.

But

how can we show that? In particular, why is P3 not a linear combi-

nation of P1 and P2? In other words, are there integers n1 and n2

such that P3 = n1P1 + n2P2? In fact, E/Q has a rational torsion

point T = (0,0) of order 2, so could some combination of P1, P2 and

P3 equal T ? This example motivates the need for a notion of linear

dependence and independence of points over Z.

Definition 2.8.1. Let E/Q be an elliptic curve. We say that the

rational points P1,...,Pm ∈ E(Q) are linearly dependent over Z if

there are integers n1,...,nm ∈ Z such that

n1P1 + n2P2 + · · · + nmPm = T,