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.
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