34 2. Elliptic curves

number of distinct rational points. Except for this criterion, Mazur’s

theorem is not very helpful in effectively computing the torsion sub-

group of a given elliptic curve. However, the following result, proven

independently by E. Lutz and T. Nagell, provides a simple algorithm

to determine E(Q)torsion:

Theorem 2.5.5 (Nagell-Lutz, [Nag35], [Lut37]). Let E/Q be an

elliptic curve with Weierstrass equation

y2

=

x3

+ Ax + B, A, B ∈ Z.

Then, every torsion point P = O of E satisfies:

(1) The coordinates of P are integers, i.e., x(P ),y(P) ∈ Z.

(2) If P is a point of order n ≥ 3, then 4A3 + 27B2 is divisible

by y(P )2.

(3) If P is of order 2, then y(P ) = 0 and x(P )3+Ax(P)+B = 0.

For a proof, see [Sil86], Ch. VIII, Corollary 7.2, or [Mil06], Ch.

II, Theorem 5.1.

Example 2.5.6. Let E/Q :

y2

=

x3

− 2, so that A = 0 and B = −2.

The polynomial

x3

− 2 does not have any rational roots, so E(Q)

does not contain any points of order 2. Also,

4A3

+

27B2

= 27 · 4.

Thus, if (x(P),y(P)) are the coordinates of a torsion point in E(Q),

then y(P ) is an integer and y(P

)2

divides 27 · 4. This implies that

y(P ) = ±1, ±2, ±3, or ±6. In turn, this implies that x(P

)3

= 3, 6,

11 or 38, respectively. However, x(P ) is an integer, and none of 3, 6,

11 or 38 are a perfect cube. Thus, E(Q)torsion is trivial (i.e., the only

torsion point is O).

Example 2.5.7. Let p ≥ 2 be a prime number and let us define a

curve Ep :

y2

=

x3

+

p2.

Since

x3

+

p2

= 0 does not have any rational

roots, Ep(Q) does not contain points of order 2. Let P be a torsion

point on Ep(Q). The list of all squares dividing

4A3

+

27B2

=

27p4

is short, and by the Nagell-Lutz theorem the possible values for y(P )

are:

y = ±1, ±p,

±p2,

±3p,

±3p2,

and ± 3.

Clearly, (0, ±p) ∈ Ep(Q) and one can show that those two points and

O are the only torsion points; see Exercise 2.12.8. Thus, the torsion

subgroup of Ep(Q) is isomorphic to Z/3Z for any prime p ≥ 2.