<< стр. 3(всего 3)СОДЕРЖАНИЕ

F = K(gn (x), y hn (x)).

The functions in F are those that are invariant under translation by elements
of E[n]. Those on the right are those that are of the form h(n(x, y)). There-
fore, we have proved the proposition.

9.6 The Torsion Subgroup: DoudвЂ™s Method
Let E : y 2 = x3 + Ax + B be an elliptic curve deп¬Ѓned over Z. The Lutz-
Nagell Theorem (Section 8.1) says that if (x, y) в€€ E(Q) is a torsion point,
then either y = 0 or y 2 |4A3 + 27B 2 . This allows us to determine the torsion,
as long as we can factor 4A3 + 27B 2 , and as long as it does not have many
square factors. In this section, we present an algorithm due to Doud  that
avoids these diп¬ѓculties and is usually much faster in practice.
Let p в‰Ґ 11 be a prime not dividing 4A3 +27B 2 . By Theorem 8.9, the kernel
of the map from the torsion of E(Q) to E(Fp ) is trivial. Therefore, the order
of the torsion subgroup of E(Q) divides #E(Fp ). If we use a few values of
p and take the greatest common divisor of the values of #E(Fp ), then we
obtain a value b that is a multiple of the order of the torsion subgroup of
E(Q). We consider divisors n of b, running from largest divisor to smallest,
and look for a point of order n on E (of course, we should look at only the
values of n allowed by MazurвЂ™s theorem).
In order to work analytically, we multiply the equation for E by 4 to obtain
E1 : y1 = 4x3 + 4Ax + 4B, with y1 = 2y.
2

The period lattice for E1 is generated by П‰1 and П‰2 , with П‰2 в€€ R. The
points in the fundamental parallelogram corresponding to real x, y under the
map of Theorem 9.10 lie on the line П‰2 R, and also on the line 1 П‰1 + П‰2 R
2
when the cubic polynomial 4x3 + 4Ax + 4B has 3 real roots. Doubling a point
on the second line yields a point on the п¬Ѓrst line. Therefore, if n is odd, all

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SECTION 9.6 THE TORSION SUBGROUP: DOUDвЂ™S METHOD

n-torsion points come from the line П‰2 R, hence lie in the subgroup generated
by n П‰2 , so в„˜( n П‰2 ) must be an integer. If n is even and z в€€ C/(ZП‰1 + ZП‰2 )
1 1

has order n, then z generates the same subgroup of C/(ZП‰1 + ZП‰2 ) as one of
1 1 1 1 1 1
n П‰2 or n П‰2 + 2 П‰1 or n П‰2 + 2 П‰1 + 2 П‰2 . Therefore, if there is a torsion point
of order n, then at least one of these three values of z must yield an integral
value of x = в„˜(z).
The strategy is therefore to evaluate
1
if n is odd or if 4x3 + 4Ax + 4B has only one real root
в„˜( П‰2 )
n
1 1 1 1 1 1
в„˜( П‰2 ), в„˜( П‰2 + П‰1 ), в„˜( П‰2 + П‰1 + П‰2 )
n n 2 n 2 2
3
if n is even and 4x + 4Ax + 4B has 3 real roots
for each divisor of b, starting with the largest n. If we п¬Ѓnd a numerical value
of x that is close to an integer, we test whether y 2 = x3 + Ax + B yields
an integral value of y. It can be checked whether or not (x, y) has order n
by computing n(x, y). If so, then (since n is the largest divisor of b not yet
excluded), we have the largest cyclic subgroup of the torsion group. Since
only the 2-torsion can be noncyclic (Corollary 3.13), we need to see only if
there is a point of order 2 not already in the subgroup generated by (x, y).
If n(x, y) = в€ћ, we continue with n and smaller divisors that are still allowed
by MazurвЂ™s theorem and the value of b. We thus obtain all torsion points in
E(Q).
The AGM method (Theorem 9.24) calculates П‰1 and П‰2 quickly. The fol-
lowing allows us to compute в„˜.

PROPOSITION 9.35
Let z в€€ C and let u = e2ПЂiz/П‰2 . Let П„ = П‰1 /П‰2 (with the requirement that П„
is in the upper half plane) and let q = e2ПЂiП„ . Then в„˜(z) =
в€ћ
2
2ПЂi 1 u u u 2
в€’
qn
+ + +n .
12 (1 в€’ u) (1 в€’ q n u)2 (q в€’ u)2 (1 в€’ q n )2
2
П‰2 n=1

PROOF Let f (z) denote the right-hand side of the equation. Since |q| < 1,
it is easy to see that the series deп¬Ѓning f (z) converges uniformly on compact
subsets of C that do not contain points in the lattice П‰1 Z + П‰2 Z. Therefore,
f (z) is analytic away from these lattice points. Moreover, it has a double pole
at each lattice point. Using the fact that u = 1 + (2ПЂi/П‰2 )z + В· В· В· , we п¬Ѓnd
that the Laurent expansion of f (z) around z = 0 starts (1/z 2 ) + В· В· В· .
Since u is invariant under z в†’ z + П‰2 , so is f (z). Changing z to z + П‰1
multiplies u by q. A straightforward calculation shows that f is invariant
under u в†’ qu. Therefore f is doubly periodic.
The diп¬Ђerence f (z)в€’в„˜(z) is a doubly periodic function with no poles except
possibly simple poles at the lattice points. By Theorem 9.1, this implies that

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304 CHAPTER 9 ELLIPTIC CURVES OVER C

the diп¬Ђerence is a constant; call it C. The roots of the cubic polynomial
4T 3 в€’ g2 T в€’ g3 are the x-coordinates of the points of order 2, namely в„˜( 1 П‰1 ),
2
в„˜( 2 П‰2 ), and в„˜( 1 (П‰1 + П‰2 )). Since there is no T 2 term, the sum of these three
1
2
roots is 0. Therefore,
1 1 1
f ( П‰1 ) + f ( П‰2 ) + f ( (П‰1 + П‰2 )) = 3C.
2 2 2
The following lemma shows that C = 0, which yields the proposition.

LEMMA 9.36
1 1 1
f ( П‰1 ) + f ( П‰2 ) + f ( (П‰1 + П‰2 )) = 0.
2 2 2

PROOF The values of u corresponding to the three values of z are

u = q в€’1/2 .
u = в€’1, u = q 1/2 ,

We may divide by the factor (2ПЂi/П‰2 )2 , hence we ignore it. The sum of
the three terms 1/12 yields 1/4, which cancels the value of u/(1 в€’ u)2 when
u = в€’1. The п¬Ѓnal term inside the sum deп¬Ѓning f (z) is independent of u and
thus yields
в€ћ
qn
в€’6 . (9.29)
(1 в€’ q n )2
n=1

We now consider the remaining terms.
The value u = в€’1 (substituted into the sum in f ) yields
в€ћ
qn
в€’2 . (9.30)
(1 + q n )2
n=1

Combining (9.29) and (9.30) yields
в€ћ
q n + q 2n + q 3n
в€’8 . (9.31)
(1 в€’ q 2n )2
n=1

The value u = q 1/2 (substituted into the sum in f ) yields (the value of
u/(1 в€’ u)2 at u = q 1/2 is included in the п¬Ѓrst summation)
в€ћ в€ћ
1 1
q n+ 2 q nв€’ 2
+
1 1
n=0 (1 в€’ q (q nв€’ 2 в€’ 1)2
n+ 2 )2
n=1
(9.32)
в€ћ 1
q nв€’ 2
=2 .
1
n=1 (1 в€’ q
nв€’ 2 )2

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SECTION 9.6 THE TORSION SUBGROUP: DOUDвЂ™S METHOD

Similarly, the value u = в€’q 1/2 (substituted into the sum in f ) yields
в€ћ 1
q nв€’ 2
в€’2 . (9.33)
1
n=1 (1 в€’ q
nв€’ 2 )2

Since 1 1
q nв€’ 2 q nв€’ 2 4q 2nв€’1
в€’ = ,
(1 в€’ q 2nв€’1 )2
1 1
(1 в€’ q nв€’ 2 )2 (1 + q nв€’ 2 )2
the sum of (9.31), (9.32), (9.33) is
в€ћ
q n + q 2n + q 3n q 2nв€’1
в€’
8 + . (9.34)
(1 в€’ q 2n )2 (1 в€’ q 2nв€’1 )2
n=1

Diп¬Ђerentiating the series for 1/(1 в€’ X) yields the identity
в€ћ
1
mX mв€’1 .
=
(1 в€’ X)2 m=1

Substituting X = q 2nв€’1 , multiplying by q 2nв€’1 , and summing over n yields
в€ћ в€ћ в€ћ
q 2nв€’1
mq (2nв€’1)m
= (9.35)
(1 в€’ q 2nв€’1 )2
n=1 m=0 n=1
вЋ› вЋћ
в€ћ
вЋќ dвЋ  q N .
= (9.36)
N =1 d|N, N/d odd

Similarly, we obtain
в€ћ в€ћ в€ћ
qn
mq n(2mв€’1)
=
(1 в€’ q 2n )2
n=1 m=0 n=1
вЋ› вЋћ (9.37)
в€ћ
d + 1вЋ  N
вЋќ
= q
2
N =1 d|N, d odd

and
в€ћ в€ћ в€ћ
q 3n
mq n(2m+1)
=
(1 в€’ q 2n )2
n=1 m=0 n=1
вЋ› вЋћ (9.38)
в€ћ
d в€’ 1вЋ  N
вЋќ
= q.
2
N =1 d|N, d odd

Also, the method yields
вЋ› вЋћ
в€ћ в€ћ
2n
q вЋќ dвЋ  q N .
= (9.39)
(1 в€’ q 2n )2
n=1 N =1 d|N, N/d even

В© 2008 by Taylor & Francis Group, LLC
306 CHAPTER 9 ELLIPTIC CURVES OVER C

Using (9.35), (9.37), (9.38), and (9.39), we п¬Ѓnd that (9.34) equals
вЋ› вЋћ
в€ћ
вЋќ dвЋ  q N .
в€’ в€’
8 d d (9.40)
N =1 d|N, N/d odd d|N, d odd d|N, N/d even

We claim that for all N в‰Ґ 1,

в€’ в€’
d d d = 0.
d|N, N/d odd d|N, d odd d|N, N/d even

Write N = 2a u with a в‰Ґ 0 and u odd. Then

2a d1
d=
d1 |u
d|N, N/d odd

d= d
d|N, d odd d|u

d= d2 .
d2 |2aв€’1 u
d|N, N/d even

If a = 0, the last sum is interpreted to be 0. In this case, the claim is easily
seen to be true. If a в‰Ґ 1, then the divisors of 2aв€’1 u are of the form 2j d3 with
0 в‰¤ j в‰¤ a в€’ 1 and d3 |u. Therefore,

(1 + 2 + 22 + В· В· В· + 2aв€’1 )d3 = (2a в€’ 1)
d2 = d3 .
d2 |2aв€’1 u d3 |u d3 |u

The claim follows easily. This completes the proof of the lemma.

Since C = 0, the proof of the proposition is complete.

Example 9.3
Consider the curve

E : y 2 = x3 в€’ 58347x + 3954150.

We have 4A2 + 27B 2 = в€’372386507784192, which factors as 218 317 11, al-
though we do not need this factorization. Since 11 divides this number, we
skip 11 and start with p1 = 13. The number of points in E(F13 ) is 10. The
number of points in E(F17 ) is also 10. Either of these facts implies that the
number of torsion points in E(Q) divides 10. Using the AGM, we calculate

П‰2 = 0.198602 В· В· В· .
П‰1 = i0.156713 . . . ,

This yields П„ = i0.78908 В· В· В· and q = 0.0070274 В· В· В· . We calculate

в„˜(П‰2 /10) = 2539.82553 . . . ,

В© 2008 by Taylor & Francis Group, LLC
307
EXERCISES

which is not close to an integer. However,
1
в„˜(П‰2 /10 + П‰1 ) = в€’213.00000 . . . .
2
This yields the point
(x, y) = (в€’213, 2592)
on E. An easy check shows that this is a point of order 10. Since the order of
the torsion subgroup divides 10, we have determined that the torsion in E(Q)
consists of the multiples of (в€’213, 2592).

Exercises
9.1 (a) Show that d3 в‰Ў d5 (mod 12) for all integers d.
(b) Show that
d5 в‰Ў 0 (mod 12)
d3 + 7
5
d|n d|n

for all positive integers n.
(c) Show that
(2ПЂ)4
g2 = (1 + 240X),
12
в€ћ n
where X = n=1 Пѓ3 (n)q .
(d) Show that
(2ПЂ)6
(1 в€’ 504Y ),
g3 =
216
в€ћ
Пѓ5 (n)q n .
where Y = n=1
(e) Show that

1728(2ПЂ)в€’12 в€† = (1 + 240X)3 в€’ (1 в€’ 504Y )2 . (9.41)

(f) Show that the right side of (9.41) is congruent to 144(5X + 7Y )
mod 1728.
в€ћ
(g) Conclude that (2ПЂ)в€’12 в€† = dn q n , with dn в€€ Z.
n=0
(h) Compute enough coeп¬ѓcients to obtain that
в€ћ
в€’12
en q n )
(2ПЂ) в€† = q(1 +
n=1

with en в€€ Z.

В© 2008 by Taylor & Francis Group, LLC
308 CHAPTER 9 ELLIPTIC CURVES OVER C

в€ћ
(i) Show that (2ПЂ)12 /в€† = q в€’1 fn q n with fn в€€ Z.
n=0
(j) Show that
в€ћ
1
cn q n
j= +
q n=0
with cn в€€ Z.
ai bi
в€€ SL2 (Z) for i = 1, 2, 3 with M2 M1 = M3 . Let
9.2 Let Mi =
ci di
П„1 в€€ F. Let
a1 П„1 + b1 a2 П„2 + b2
П„2 = , П„3 = .
c1 П„1 + d1 c2 П„2 + d2
Show that
a3 П„1 + b3
П„3 = .
c3 П„1 + d3
9.3 Let k в‰Ґ 0 be an integer. Let f be a meromorphic function on the upper
half plane such that f has a q-expansion at iв€ћ (as in Equation (9.15))
and such that
aП„ + b
= (cП„ + d)k f (П„ )
f
cП„ + d
ab
for all П„ в€€ H and for all в€€ SL2 (Z). Show that
cd

1 1 k
ordiв€ћ (f ) + ordПЃ (f ) + ordi (f ) + ordz (f ) = .
3 2 12
z=i,ПЃ,iв€ћ

ab
9.4 The stabilizer in SL2 (Z) of a point z в€€ H is the set of matrices
cd
such that (az + b)/(cz + d) = z.
(a) Show that the stabilizer of i has order 4.
(b) Show that the stabilizer of ПЃ has order 6.
(c) Show that the stabilizer of iв€ћ consists of the matrices of the form
1b
В± with b в€€ Z.
01
(d) Show that the stabilizer of each z в€€ H has order at least 2.
It can be shown that the stabilizer of each element in the fundamental
domain F has order 2 except for i and ПЃ.
9.5 Let E : y 2 = 4x3 + Ax + B, with A, B в€€ R be an elliptic curve deп¬Ѓned
over R. We know by Theorem 9.21 that E(C) C/L for some lattice
L. The goal of this exercise is to show that L has one of the two shapes
given in part (i) below.

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309
EXERCISES

(a) Let П„ в€€ F. Show that j(П„ ) = j(в€’П„ ).
(b) Show that if П„ is in the fundamental domain F , then either в€’П„ в€€ F,
or (П„ ) = в€’1/2, or |П„ | = 1 with в€’1/2 в‰¤ (П„ ) в‰¤ 0.
(c) Suppose that П„ в€€ F and that j(П„ ) в€€ R. Show that (П„ ) = 0,
or (П„ ) = в€’1/2, or |П„ | = 1 with в€’1/2 в‰¤ (П„ ) в‰¤ 0 (Hint: Use
Corollary 9.18.)
(d) Let П„ в€€ H. Show that if |П„ | = 1 then (в€’1/(П„ + 1)) = в€’1/2.
(e) Let L be a lattice with g2 (L) = в€’A and g3 (L) = в€’B. Show
that there exists П„ в€€ H such that (П„ ) = 0 or в€’1/2 and j(L) =
j(ZП„ + z).
(f) Show that if П„ в€€ H is such that (П„ ) = 0 or в€’1/2, then we have
g2 (П„ ), g3 (П„ ) в€€ R.
(g) By Corollary 9.20, there exists О» в€€ C such that L = (О»)(ZП„ + Z).
Show that if j = 0, 1728 then О»2 в€€ R. (Hint: Use Equations
(9.14).)
This shows that L is obtained from the lattice ZП„ + Z by an ex-
pansion by |О»| and a rotation by 0, 90в—¦ , 180в—¦ , or 270в—¦ .
(h) Let 0 = y в€€ R. Let M be the lattice ( 1 + iy)Z + Z. Show that iM
2
has {y + 1 i, 2y} as a basis.
2
(i) Assume that j = 0, 1728. Show that L has a basis {П‰1 , П‰2 } with
П‰2 в€€ R and (П‰1 ) = 0 or 1 П‰2 . Therefore, the lattice L is either
2
rectangular or a special shape of parallelogram.
(j) Use the facts that j(ПЃ) = 0 and j(i) = 1728 to prove (i) in the
cases that j(E) = 0 and j(E) = 1728. (The condition that О»2 в€€ R
gets replaced by О»6 в€€ R and О»4 в€€ R, respectively. However, the
lattices for П„ = ПЃ and П„ = i have extra symmetries.)

9.6 Let L be a lattice that is stable under complex conjugation (that is, if
П‰ в€€ L then П‰ в€€ L). This is the same as requiring that the elliptic curve
associated to L is deп¬Ѓned over R (see Exercise 9.5).

(a) Show that в„˜(z) = в„˜(z).
(b) Show that if t в€€ R and if П‰2 в€€ R is a real period, then

1
в€€ R.
в„˜ П‰2 + it
2

(Hint: Use (a), the periodicity of в„˜, and the fact that в„˜(в€’z) =
в„˜(z).)
(c) Diп¬Ђerentiate the result of (b) to show that в„˜ (z) в€€ iR for the
points 1 П‰2 + it in (b). This path, for 0 в‰¤ t в‰¤ П‰1 , corresponds to
2
x moving along the x-axis between the two parts of the graph in

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310 CHAPTER 9 ELLIPTIC CURVES OVER C

Figure 2.1(a) on page 10. The points donвЂ™t appear on the graph
because y is imaginary. For the curve in Figure 2.1(b) on page 10, x
moves to the left along the x-axis, from the point on the x-axis back
to the point at inп¬Ѓnity, corresponding to the fact that П‰1 = 1 П‰2 +it
2
for appropriate t (see Exercise 9.5).
9.7 Deп¬Ѓne the elliptic integral of the second kind to be
в€љ
1 в€’ k 2 x2
1
в€љ dx, в€’1 < k < 1.
E(k) =
1 в€’ x2
0

(a) Show that
ПЂ/2
(1 в€’ k 2 sin2 Оё)1/2 dОё.
E(k) =
0

(b) Show that the arc length of the ellipse

x2 y2
+ 2 =1
a2 b

with b в‰Ґ a > 0 equals 4bE( 1 в€’ (a/b)2 ).
This connection with ellipses is the origin of the name вЂњelliptic inte-
gral.вЂќ The relation between elliptic integrals and elliptic curves, as in
Section 9.4, is the origin of the name вЂњelliptic curve.вЂќ For more on
elliptic integrals, see .
9.8 Let E be the elliptic curve y 2 = 4x3 в€’ 4x. Show that
в€ћ 1
dx 1
tв€’3/4 (1 в€’ t)в€’1/2 dt = ОІ(1/4, 1/2),
П‰2 = =
x(x2 в€’ 1) 2
1 0

1 pв€’1
в€’ t)qв€’1 dt is the beta function. A classical
where ОІ(p, q) = t (1
0
result says that
О“(p)О“(q)
.
ОІ(p, q) =
О“(p + q)
Therefore,
1 О“(1/4)О“(1/2)
П‰2 = .
2 О“(3/4)

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