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 :
+ 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
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 :
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.
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