Download TORSION POINTS ON SOME SPECIAL ELLIPTIC CURVES. 1. Two

TORSION POINTS ON SOME SPECIAL ELLIPTIC CURVES.
IAN KIMING
1. Two special families of elliptic curves over Q.
Let A and B be non-zero integers. We will determine the groups of rational
torsion points for each of the curves:
y 2 = x3 + Ax ,
(A)
and
y 2 = x3 + B .
(B)
Also, we will draw from this analysis some interesting consequences concerning
rational solutions to an equation:
x3 + y 3 = d
where d is a non-zero integer.
Now notice first in connection with the curve (A) that we may assume that A
is fourth power free, i.e., is not divisible by any 4’th power of an integer other
than 1. For if n4 j A then rational solutions to (A) are in 1-1 correspondence with
rational solutions to:
A
y 2 = x3 + 4 · x
n
via (x, y) ↔ ( nx2 , ny3 ).
Similarly, for the curve (B) we may assume that B is sixth power free; for if
n6 j B, the rational solutions to (B) correspond to the rational solutions to:
B
y 2 = x3 + 6
n
again via (x, y) ↔ ( nx2 , ny3 ).
The following theorem is due to T. Nagell, cf. [2]. We can give a somewhat
shorter proof since we have more powerful tools at our disposal.
Theorem 1. Assume A ∈ Z, A 6= 0 and 4’th power free. For the elliptic curve
E : y 2 = x3 + Ax
we have:
E(Q)tors

 Z/Z4 generated by (2, 4) ,
∼
Z/Z2 × Z/Z2 generated by (0, 0) and (C, 0) ,
=

Z/Z2 generated by (0, 0) ,
1
if A = 4 ,
if A = −C 2 , C ∈ Z ,
otherwise.
2
IAN KIMING
Proof. We have seen in an exercise that #E(Q)tors is either 2 or 4.
The rational points of order 2 on E are those with y = 0. So we always have the
point (0, 0) of order 2. There is another point of order 2 if and only if A = −C 2
for some C ∈ Z; in that case we have additionally the points (±C, 0) of order 2,
and E(Q)tors is isomorphic to Klein’s 4-group, generated by the points (0, 0) and
(C, 0).
Since we easily verify that the curve y 2 = x3 + 4x has the point (2, 4) of order
4, the theorem now follows if we show that A must equal 4 in case E has a point
of order 4.
So assume that P = (a, b) is a rational point of order 4 on E. By Nagell-Lutz
the numbers a and b are integers, and by the above we deduce that E(Q)tors is
generated by P . Hence, E has a single rational point of order 2, namely (0, 0). It
follows that:
2 · P = (0, 0) .
Now, by the duplication formula we have (notice that b 6= 0 since P does not
have order 2):
2 · P = ((a2 − A)2 /4b2 , ·)
whence:
A = a2 .
Since A is 4’th power free, we deduce that a is square free. Since
b2 = a3 + Aa = 2a3
this implies that a is a power of 2 (any odd prime divisor of a must occur with
an even exponent in the prime factorization of a). Since now b2 = 2a3 and a is a
square free power of 2, we deduce a = 2 and hence A = 4.
The contents of the following theorem were first given by R. Fueter in [1]. Again
we can give a shorter argument because we have stronger tools.
Theorem 2. Assume B ∈ Z, B 6= 0 and 6’th power free. For the elliptic curve
E : y 2 = x3 + B
we have:
E(Q)tors

Z/Z6 generated




 Z/Z3 generated
∼
Z/Z3 generated
=



Z/Z2 generated


1,
by
by
by
by
(2, 3) ,
(0, U ) ,
(12, 36) ,
(−V, 0) ,
if B = 1 ,
if B = U 2 6= 1 , U ∈ Z ,
if B = −432 = −24 · 33 ,
if B = V 3 6= 1 , V ∈ Z ,
otherwise.
Proof. We showed in an exercise that #E(Q)tors is a divisor of 6. So our task is
to determine the conditions under which E has rational points of order 2 and 3,
respectively.
The curve has a rational point of order 2 if and only if x3 + B has a rational
root. This is the case exactly if B is the cube of an integer.
TORSION POINTS ON SOME SPECIAL ELLIPTIC CURVES.
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Suppose that P = (a, b) is an affine rational torsion point on E. According to
Nagell-Lutz the numbers a and b are then integers.
Now, P has order 3 if and only if 2 · P = −P = (a, −b). If P does not have order
2, i.e., if b 6= 0, the duplication formula for E implies that:
4
a − 8aB a6 + 20a3 B − 8B 2
2·P =
,
.
4b2
8b3
So we see that P has order 3 if and only if:
4
a − 8aB
= 4ab2 ,
6
3
2
a + 20a B − 8B = −8b4 .
Using b2 = a3 + B we see that these conditions reduce to the simple condition:
a4 + 4aB = 0 .
Hence there is a point P = (0, b) of order 3 if and only if B = b2 , i.e., if and only
if B is the square of an integer.
Suppose that P = (a, b) is of order 3 with a 6= 0. Then 4B = −a3 and so B must
have shape B = −2α3 for some integer α. But we also have −3B = B + a3 = b2 ;
hence B = −3β 2 for some integer β.
If p is a prime number 6= 2, 3, consider then s := ordp (B). Then s = 3 ordp (α) =
2 ordp (β), i.e., s is divisible by both 2 and 3, and hence by 6. Since B is 6’th power
free, we have s = 0. Consequently, B has form B = −2σ · 3τ . Then
6 > σ = ord2 (B) = 1 + 3 · ord2 (α) = 2 · ord2 (β) ,
which implies σ = 4, and
6 > τ = ord3 (B) = 3 · ord3 (α) = 1 + 2 · ord3 (β) ,
whence τ = 3. In other words, B = −24 · 33 = −432. Then we find a = 12 and
b = ±36.
Conversely, if B = −24 · 33 = −432 we confirm that (12, 36) is a rational point
of order 3.
The curve has a rational point of order 6 if and only if it has both a rational
point of order 2 and also a rational point of order 3. By what we have proved above,
we see that this is the case if and only if both B is the cube of an integer, and
either B = −432 or B is the square of an integer. Since −432 is not a cube, this
condition boils down to B being the 6’th power of an integer. Since B is assumed
6’th power free this is equivalent to B = 1.
If B = 1 then we have the points (−1, 0) and (0, −1) of order 2 and 3, respectively.
Then E(Q)tors is generated by the point
(−1, 0) + (0, −1) = (2, 3)
of order 6.
Piecing together all of the above we see that the proof of the theorem is complete.
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IAN KIMING
2. Applications.
Corollary 1. Suppose that c ∈ Z, c 6= 0, and 6’th power free.
If c 6= 1, −432 and if the equation
y 2 − x3 = c
has a solution in non-zero integers x, y then it has infinitely many rational
solutions. In fact, repeated application of the duplication formula
4
x − 8cx x6 + 20cx3 − 8c2
,
(x, y) 7→
4y 2
8y 3
necessarily gives infinitely many solutions.
Proof. Since xy 6= 0 and c 6= 1, −432 Theorem 2 implies that the rational point
P = (x, y) is a non-torsion point. Hence repeated duplication of the point leads to
infinitely many rational points.
Remark: Corollary 1 can not be improved: It can be shown that the curves
y 2 = x3 + 1 and y 2 = x3 − 432 both have rank 0 over Q, i.e., have only finitely
many rational points (which are then given by theorem 2).
Corollary 2. Let d ∈ N and assume that d is cube free.
If d 6= 1, 2 and if the equation
u3 + v 3 = d
has a rational solution then it has infinitely many rational solutions.
Proof. Assume d 6= 1, 2.
The statement clearly follows from the following claim:
Consider the cubic curve
(∗)
u3 + v 3 = dw3 .
Claim: If this curve has a rational point with w 6= 0 then it has infinitely many
rational points.
Now, one readily checks that if (u, v, w) is a rational solution to (∗) with w 6= 0,
then
(†)
(x, y, z) := (12dw, 36d(u − v), u + v)
is a rational point on the elliptic curve:
(∗∗)
y 2 z = x3 − 432d2 z 3 .
For this point (x, y, z) we clearly have x 6= 0. Conversely, if (x, y, z) is a rational
solution to (∗∗) with x 6= 0, we can solve (†) for u, v, w and thus obtain a rational
point (u, v, w) on (∗) with w 6= 0.
Suppose then that (u, v, w) is a rational point (u, v, w) on (∗) with w 6= 0. We
obtain a rational point (x, y) with x 6= 0 on the elliptic curve y 2 = x3 − 432d2 .
Write:
−432d2 = −24 · 33 · d2 = c · γ 6
TORSION POINTS ON SOME SPECIAL ELLIPTIC CURVES.
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where c, γ ∈ Z and γ largest possible.
Then c is not a cube: For otherwise −24 · d2 would be a cube, and this is seen
to imply d = 2, as d is cube free. But we excluded the case d = 2.
In particular, we have c 6= 1, and we must also have y 6= 0.
Furthermore, c 6= −432: Otherwise d would be a cube, and hence d = 1, contrary
to our assumptions.
Now corollary 1 implies that the curve y 2 = x3 −432d2 has a non-torsion rational
points, and hence infinitely many rational points. Hence (∗) also has infinitely many
rational points.
Remark: Corollary 2 is optimal: Each of the curves u3 +v 3 = w3 and u3 +v 3 = 2w3
has in fact only finitely many rational points: This follows because one can show
that the elliptic curves y 2 = x3 − 432 and y 2 = x3 − 27 both have rank 0 over Q.
References
[1] R. Fueter: ‘Ueber kubische diophantische Gleichungen.’ Comment. Math. Helv. 2 (1930),
69–89.
[2] T. Nagell: ‘Solution de quelques problemes dans la théorie arithmétique des cubiques planes
du premier genre.’ Skrifter utg. av Det Norske Videnskaps-Akademi i Oslo, I. Mat.-Naturv. Klasse
1 (1935), 1–25.
Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK2100 Copenhagen Ø, Denmark.
E-mail address: [email protected]