2.7. The rank and the free part of E(Q) 45

(2) For all P ∈ E(Q) and m ∈ Z, h(mP ) = m2 ·h(P ). (Note: in

particular, the height of mP is much larger than the height

of P , for any m = 0,1.)

(3) Let P ∈ E(Q). Then h(P ) ≥ 0, and h(P ) = 0 if and only if

P is a torsion point.

For the proofs of these properties, see [Sil86], Ch. VIII, Thm.

9.3, or [Mil06], Ch. IV, Prop. 4.5 and Thm. 4.7.

As we mentioned at the beginning of this section, we can calculate

upper bounds on the rank of a given elliptic curve (see [Sil86], p. 235,

exercises 8.1, 8.2). Here is an example:

Theorem 2.7.4 ([Loz08], Prop. 1.1). Let E/Q be an elliptic curve

given by a Weierstrass equation of the form

E :

y2

=

x3

+

Ax2

+ Bx, with A, B ∈ Z.

Let RE be the rank of E(Q). For an integer N ≥ 1, let ν(N) be the

number of distinct positive prime divisors of N . Then

RE ≤

ν(A2

− 4B) + ν(B) − 1.

More generally, let E/Q be any elliptic curve with a non-trivial point

of 2-torsion and let a (resp. m) be the number of primes of additive

(resp. multiplicative) bad reduction of E/Q. Then

RE ≤ m + 2a − 1.

Example 2.7.5. Pierre de Fermat proved that n = 1 is not a con-

gruent number (see Example 1.1.2) by showing that x4 + y4 = z2 has

no rational solutions. As an application of the previous theorem, let

us show that the curve

E1 :

y2

=

x3

− x = x(x − 1)(x + 1)

only has the trivial solutions (0,0), (±1,0), which are torsion points

of order 2. Indeed, the minimal discriminant of E1 is ΔE1 = 64.

Therefore p = 2 is the unique prime of bad reduction. Moreover, the

reader can check that the reduction at p = 2 is multiplicative. Now

thanks to Theorem 2.7.4 we conclude that RE1 = 0 and E1 only has

torsion points. Finally, using Proposition 2.6.15 or Theorem 2.5.5,

we can show that the only torsion points are the three trivial points

named above.