<<

. 2
( 3)



>>

surface obtained as follows. Identify the left and right sides on F to get a
tube, then fold the part with |z| = 1 at i. Then pinch the open end at i∞ to
a point. This gives a surface that is topologically a sphere. The proposition
expresses the fact that the number of poles of f equals the number of zeros on
such a surface, just as occurred for doubly periodic functions in Theorem 9.1.
The point i is special since a small neighborhood around i contains only half
of a disc inside F . Similarly, a small neighborhood around ρ includes only
1/3 of a disc from F (namely, 1/6 near ρ and 1/6 near 1 + ρ, which is folded
over to meet ρ). This explains the factors 1/2 and 1/3 in the proposition. For
a related phenomenon, see Exercise 9.3.

PROOF Let C be the path shown in Figure 9.4. Essentially, C goes around
the edge of F. However, it consists of a small circular arc past each of ρ, 1 + ρ,
and i. If there is a pole or zero of f at a point on the path, we make a small
detour around it and a corresponding detour at the corresponding point on
the other side of F. The arcs near ρ, 1 + ρ, and i have radius , where is
chosen small enough that there are no zeros or poles of f inside the circles,
except possibly at ρ, 1 + ρ, or i. Similarly, the top part of C is chosen to have




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




Figure 9.4




imaginary part N , where N is large enough that f (z) has no zeros or poles
with imaginary part greater than N , except perhaps at i∞. This is possible
since
f (z) = q n (an + an+1 q + · · · ).
Since the series an + an+1 q + · · · is assumed to converge for q small, it is ¬nite
and is approximately equal to an = 0 for su¬ciently small q.
As in the proof of Theorem 9.1, we have

1 f (z)
dz = ordz (f ).
2πi f (z)
C z∈F
z=i,ρ



11
∈ SL2 (Z) gives the map z ’ z + 1, we have
Since
01

f (z) = f (z + 1). (9.16)

Therefore, the integrals over the left and right vertical parts of C are the
same, except that they are in opposite directions, so they cancel each other.
Now we™ll show that the integral over the part of the unit circle to the left of
0 ’1

i cancels the part to the right. This is proved by using the fact that
10
SL2 (Z) gives the map z ’ ’1/z, which interchanges the left and right arcs of
the unit circle. In addition, di¬erentiating the relation f (’1/z) = f (z) yields

’1 ’1
f f
d = (z) dz.
f z z f




© 2008 by Taylor & Francis Group, LLC
281
SECTION 9.3 ELLIPTIC CURVES OVER C

Therefore, the integral over C from ρ to i equals the integral from ’1/ρ = 1+ρ
to ’1/i = i, which is the negative of the integral from i to 1 + ρ. Therefore,
the two parts cancel.
All that remains are the parts of C near ρ, 1 + ρ, i, and i∞. Near i, we
have f (z) = (z ’ i)k g(z) for some k, with g(i) = 0, ∞. Therefore,

f (z) k g (z)
= + . (9.17)
z’i
f (z) g(z)

The integral over the small semicircle near i is

f (i + eiθ ) iθ
1
ie dθ, (9.18)
f (i + eiθ )
2πi θ

where θ ranges from slightly more than π to slightly less than 0. (Note that
C is traveled clockwise. Because of the curvature of the unit circle, the limits
are 0 and π only in the limit as ’ 0.) Substitute (9.17) into (9.18) and
let ’ 0. Since g /g is continuous at i, the integral of g /g goes to 0. The
integral of k/(z ’ i) yields
0
1 1
1
ki dθ = ’ k = ’ ordi (f ).
2πi 2 2
θ=π

Similarly, the contributions from the parts of C near ρ and 1 + ρ add up to
’(1/3)ordρ (f ) (we are using the fact that f (ρ) = f (ρ + 1), by (9.16)).
Finally, the integral along the top part of C is
’1
1 f (t + iN )
2
dt.
2πi f (t + iN )
t= 1
2


Since f („ ) = q n (an + an+1 q + · · · ), we have

2πian+1 q + · · ·
f („ )
= 2πin + .
an + · · ·
f („ )

The second term goes to 0 as q ’ 0, hence as N ’ ∞. The limit of the
integral as N ’ ∞ is therefore
’1
1 2
2πin dt = ’n = ’ordi∞ (f ).
2πi t= 1
2



Combining all of the above calculations yields the theorem.


COROLLARY 9.18
If z ∈ C, then there is exactly one „ ∈ F such that j(„ ) = z.




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

PROOF First, we need to calculate j(ρ) and j(i). Recall that „ corresponds
to the lattice L„ = Z„ + Z. Since ρ2 = ’1 ’ ρ, it follows easily that ρLρ ⊆ Lρ .
Therefore,
Lρ = ρ3 Lρ ⊆ ρ2 Lρ ⊆ ρLρ ⊆ Lρ ,
so ρLρ = Lρ . It follows from (9.14) that

g2 (Lρ ) = g2 (ρLρ ) = ρ’4 g2 (Lρ ) = ρ’1 g2 (Lρ ).

Since ρ = 1, we have g2 (ρ) = g2 (Lρ ) = 0. Therefore,

g2 (Lρ )3
j(ρ) = 1728 =0
g2 (Lρ )3 ’ 27g3 (Lρ )2

(note that the denominator is nonzero, by Proposition 9.9).
Similarly, „ = i corresponds to the lattice Li = Zi + Z, and iLi = Li .
Therefore,
g3 (Li ) = g3 (iLi ) = i’6 g3 (Li ) = ’g3 (Li ),
so g3 (i) = g3 (Li ) = 0. Therefore,

g2 (Li )3
j(i) = 1728 = 1728.
g2 (Li )3 ’ 27g3 (Li )2

We now look at the other values of „ . Consider the function h(„ ) = j(„ )’z.
Then h has a pole of order 1 at i∞ and no other poles. By Proposition 9.16,
we have
1
1
ordρ (h) + ordi (h) + ordz (h) = 1.
3 2
z=i,ρ,∞

If z = 0, 1728, then h has order 0 at ρ and at i. Therefore, h has a unique zero
in F, so j(„ ) = z has a unique solution in F. If z = 1728, then (1/2)ordi (h) >
0. Since the order of h at a point is an integer, the order must be 0 when z =
i, ρ; otherwise, the sum would be larger than 1. Also, there is no combination
of m/2 + n/3 that equals 1 except when either m = 0 or n = 0. Therefore,
j(„ ) ’ 1728 has a double zero at i and no other zero in F . Similarly, j(„ ) has
a triple zero at ρ and no other zero in F .

COROLLARY 9.19
ab
Let „1 , „2 ∈ H. Then j(„1 ) = j(„2 ) if and only if there exists ∈ SL2 (Z)
cd
such that
a„1 + b
= „2 .
c„1 + d

PROOF Proposition 9.13 gives one direction of the statement. Assume
conversely that j(„1 ) = j(„2 ). Let „1 , „2 ∈ F map to „1 , „2 via the action of




© 2008 by Taylor & Francis Group, LLC
283
SECTION 9.3 ELLIPTIC CURVES OVER C

SL2 (Z), as in Proposition 9.14. Then, by Proposition 9.13,
j(„1 ) = j(„1 ) = j(„2 ) = j(„2 ).
By Corollary 9.18, „1 = „2 . Since an element of SL2 (Z) maps „1 to „1 , and
an element of SL2 (Z) maps „1 = „2 to „2 , the product of these two matrices
(see Exercise 9.2) maps „1 to „2 , as desired.

There is also a version of Corollary 9.19 for lattices (the j-invariant of a
lattice is de¬ned on page 276).


COROLLARY 9.20
Let L1 , L2 ‚ C be lattices. Then j(L1 ) = j(L2 ) if and only if there exists
0 = » ∈ C such that »L1 = L2 .

PROOF One direction was proved on page 276. Conversely, suppose
j(L1 ) = j(L2 ). Write Li = (»i )(Z„i + Z) with „i ∈ F, as in Corollary 9.15.
Then j(„1 ) = j(L1 ) = j(L2 ) = j(„2 ), so Corollary 9.18 implies that „1 = „2 .
Let » = »2 /»1 . Then »L1 = L2 .

We can now show that every elliptic curve over C corresponds to a torus.


THEOREM 9.21
Let y 2 = 4x3 ’ Ax ’ B de¬ne an elliptic curve E over C. Then there is a
lattice L such that
g2 (L) = A and g3 (L) = B.
There is an isomorphism of groups
C/L E(C).

PROOF Let
A3
j = 1728 .
A3 ’ 27B 2
By Corollary 9.18, there exists a lattice L = Z„ +Z such that j(„ ) = j(L) = j.
Assume ¬rst that g2 (L) = 0. Then j = j(L) = 0, so A = 0. Choose » ∈ C—
such that
g2 (»L) = »’4 g2 (L) = A.
The equality j = j(L) implies that
g3 (»L)2 = B 2 ,
so g3 (»L) = ±B. If g3 (»L) = B, we™re done. If g3 (»L) = ’B, then
g3 (i»L) = i’6 g3 (»L) = B g2 (i»L) = i4 g2 (»L) = A.
and




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

Therefore, either »L or i»L the desired lattice.
If g2 (L) = 0, then j = j(L) = 0, so A = 0. Since A3 ’ 27B 2 = 0 by
assumption and since g2 (L)3 ’ 27g3 (L)2 = 0 by Proposition 9.9, we have
B = 0 and g3 (L) = 0. Choose µ ∈ C— such that

g3 (µL) = µ’6 g3 (L) = B.

Then g2 (µL) = µ’4 g2 (L) = 0 = A, so µL is the desired lattice.
By Theorem 9.10, the map

C/L ’’ E(C)

is an isomorphism.

The elements of L are called the periods of L.
Theorem 9.21 gives us a good way to work with elliptic curves over C. For
example, let n be a positive integer and let E be an elliptic curve over C.
By Theorem 9.21, there exists a lattice L = Zω1 + Zω2 such that C/L is
isomorphic to E(C). It is easy to see that the n-torsion on C/L is given by
the points
k
j
ω1 + ω2 , 0 ¤ j, k ¤ n ’ 1.
n n
It follows that
Zn • Zn .
E[n]
In fact, we can use this observation to give a proof of Theorem 3.2 for all ¬elds
of characteristic 0.


COROLLARY 9.22
Let K be a ¬eld of characteristic 0, and let E be an elliptic curve over K.
Then
E[n] = {P ∈ E(K) | nP = ∞} Zn • Zn .


PROOF Let L be the ¬eld generated by Q and the coe¬cients of the
equation of E. Then L has ¬nite transcendence degree over Q, hence can be
embedded into C (see Appendix C). Therefore, we can regard E as an elliptic
curve over C. Therefore, the n-torsion is Zn • Zn .
There is a technical point to worry about. The de¬nition of E[n] that
we have used requires the coordinates of the n-torsion to lie in the algebraic
closure of the base ¬eld. How can we be sure that the ¬eld K isn™t so large
that it allows more torsion points than C? Suppose that E[n] ‚ E(K) has
order larger than n2 . Then we can choose n2 + 1 of these points and adjoin
their coordinates to L. Then L still has ¬nite transcendence degree over Q,
hence can be embedded into C. The coordinates of the n2 + 1 points will yield




© 2008 by Taylor & Francis Group, LLC
285
SECTION 9.3 ELLIPTIC CURVES OVER C

n2 +1 points in E(C) that are n-torsion points. This is impossible. Therefore,
E[n] is no larger than it should be.
There is also the reverse possibility. How do we know that K is large enough
to account for all the n-torsion points that we found in E(C)? We need to
show that the n-torsion points in E(C) have coordinates that are algebraic
over L (where L is regarded as a sub¬eld of C). Let P = (x, y) be an n-torsion
point in E(C), and suppose that x and y are transcendental over L (since x
and y satisfy the polynomial de¬ning E, they are both algebraic or both
transcendental over K). Let σ be an automorphism of C such that σ(x) =
x+1, and such that σ is the identity on L. Such an automorphism exists: take
σ to be the desired automorphism of K(x), then use Zorn™s Lemma to extend
σ to all of C (see Appendix C). The points σ m (P ) for m = 1, 2, 3, . . . , have
distinct x-coordinates x + 1, x + 2, x + 3, . . . , hence are distinct points. Each
must be an n-torsion point of E in E(C). But there are only n2 such points,
so we have a contradiction. Therefore, the coordinates of the n-torsion points
are algebraic over L, hence are algebraic over K, since L ⊆ K. Therefore, the
passage from K to C does not a¬ect E[n].

Suppose we have an elliptic curve E de¬ned over the real numbers R.
Usually, it is represented by a graph, as in Chapter 2 (see Figure 2.1 on
page 10). It is interesting to see how the torus we obtain relates to this graph.
It can be shown (Exercise 9.5) that the lattice L for E has one of two shapes.
Suppose ¬rst that the lattice is rectangular: L = Zω1 + Zω2 with ω1 ∈ iR
and ω2 ∈ R. Then
(„˜(z), „˜ (z)) ∈ E(R)
when
with 0 ¤ t < 1,
(I) z = tω2
and also when
with 0 ¤ t < 1.
(II) z = (1/2)ω1 + tω2
The ¬rst of these is easy to see: if z is real and the lattice L is preserved by
complex conjugation, then conjugating the de¬ning expression for „˜(z) leaves
it unchanged, so „˜ maps reals to reals. The second is a little more subtle:
conjugating z = (1/2)ω1 + tω2 yields z = ’(1/2)ω1 + tω2 , which is equivalent
to z mod L. Therefore, the de¬ning expression for „˜(z) is again unchanged
by complex conjugation, so „˜ maps reals to reals.
Fold the parallelogram into a torus by connecting the right and left sides to
form a tube, then connecting the ends. The paths (I) (see Figure 9.5) starts
and ends at points that di¬er by ω2 . Therefore the endpoints are equivalent
mod L, so (I) yields a circle on the torus. Similarly, (I) yields a circle on the
torus.
When the ends of path (I) are disconnected at 0 (which corresponds to ∞
in the Weierstrass form), we obtain a slightly deformed version of the graph
of Figure 2.1(a) on page 10.




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



„¦1

1 II
„¦
21

I „¦2
0
Figure 9.5
The Real Points on C/L


In the case of a skewed parallelogram, that is, when ω2 is real and the
imaginary part of ω1 is half of ω2 , the real axis is mapped to the reals, but
the analogue of path (II) is not mapped to the reals. This corresponds to the
situation of Figure 2.1(b) on page 10.




9.4 Computing Periods
Suppose E is an elliptic curve over C. From Theorem 9.21, we know that
E corresponds to a lattice L = Zω1 + Zω2 via the doubly periodic functions
„˜ and „˜ , but how do we ¬nd the periods ω1 and ω2 ?
For simplicity, let™s consider the case where E is de¬ned over R and E[2] ‚
E(R). Then the equation for E can be put in the form

y 2 = 4x3 ’ g2 x ’ g3 = 4(x ’ e1 )(x ’ e2 )(x ’ e3 ) with e1 < e2 < e3 .

We may assume ω2 ∈ R with ω2 > 0 and ω1 ∈ iR with (ω1 ) > 0, as in
Figure 9.5. The graph of E is as in Figure 2.1(a) on page 10. The Weierstrass
„˜-function and its derivative map C/L to E via

(x, y) = („˜(z), „˜ (z)).

As z goes from 0 to ω2 /2, the function „˜(z) takes on real values, starting with
x = ∞. The ¬rst point of order two is encountered when z = ω2 /2. Which
point (ei , 0) is it? The graph of the real points of E has two components. The
one connected to ∞ contains the point (e3 , 0) of order two, so x = „˜(z) must
run from ∞ to e3 as z goes from 0 to ω2 /2. The expansion of „˜ (z) starts
with the term ’2/z 3 , from which it follows that y = „˜ (z) < 0 near z = 0,
hence „˜ (z) < 0 for 0 < z < ω2 /2.




© 2008 by Taylor & Francis Group, LLC
287
SECTION 9.4 COMPUTING PERIODS

Consider now the integral

dx
.
4(x ’ e1 )(x ’ e2 )(x ’ e3 )
e3


Substitute x = „˜(z). The denominator becomes „˜ (z)2 = ’„˜ (z) (recall
that „˜ (z) < 0) and the limits of integration are from z = ω2 /2 to 0. Adjusting
the direction of integration and the sign yields
ω2 /2
ω2
dz = .
2
0

Therefore,

dx
ω2 = .
(x ’ e1 )(x ’ e2 )(x ’ e3 )
e3

The change of variables

e3 ’ (e3 ’ e1 )(e3 ’ e2 ) t + e3 + (e3 ’ e1 )(e3 ’ e2 )
x=
t+1
(plus a lot of algebraic manipulation) changes the integral to
1
2 dt
ω2 = √ √ ,
e3 ’ e1 + e3 ’ e2 (1 ’ t2 )(1 ’ k 2 t2 )
’1

where
√ √
e3 ’ e1 ’ e3 ’ e2
k=√ √ . (9.19)
e3 ’ e1 + e3 ’ e2

Since the integrand is an even function, we can take twice the integral over
the interval from 0 to 1 and obtain
1
4 dt
ω2 = √ √ .
e3 ’ e1 + e3 ’ e2 (1 ’ t2 )(1 ’ k 2 t2 )
0

This integral is called an elliptic integral (more precisely, an elliptic integral
of the ¬rst kind). It is usually denoted by
1
dt
K(k) = .
(1 ’ t2 )(1 ’ k 2 t2 )
0

In the following, we™ll see how to compute K(k) numerically very accurately
and quickly, but ¬rst let™s ¬nd an expression for ω1 .
When z runs along the vertical line from ω2 /2 to ω2 /2 + ω1 /2, the function
„˜(z) takes on real values (see Exercise 9.6) from e3 to e2 , and its derivative




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

„˜ (z) takes on purely imaginary values. Reasoning similar to that above
(including the same change of variables) yields

1/k
2i dt
ω1 = √ √ .
e3 ’ e1 + e3 ’ e2 (t2 ’ 1)(1 ’ k 2 t2 )
1



1 ’ k 2 and make the substitution
Let k =


t = (1 ’ k u2 )’1/2 .
2




The integral becomes

1
dt
= K(k ) = K( 1 ’ k 2 ).
(1 ’ t2 )(1 ’ k 22
0 t)


Therefore,

2i
ω1 = √ √ K( 1 ’ k 2 ).
e3 ’ e1 + e3 ’ e2


Therefore, both ω1 and ω2 can be expressed in terms of elliptic integrals.




9.4.1 The Arithmetic-Geometric Mean

In this subsection, we introduce the arithmetic-geometric mean. It yields a
very fast and ingenious method, due to Gauss, for computing elliptic integrals.
Start with two positive real numbers a, b. De¬ne an and bn by


a0 = a, b0 = b
1
an = (an’1 + bn’1 ) (9.20)
2
bn = an’1 bn’1 .


Then an is the arithmetic mean (=average) of an’1 and bn’1 , and bn is their
geometric mean.




© 2008 by Taylor & Francis Group, LLC
289
SECTION 9.4 COMPUTING PERIODS

Example 9.1

Let a = 2 and b = 1. Then
a1 = 1.207106781186547524400844362 . . .
b1 = 1.189207115002721066717499970 . . .

a2 = 1.198156948094634295559172166 . . .
b2 = 1.198123521493120122606585571 . . .


a3 = 1.198140234793877209082878869 . . .
b3 = 1.198140234677307205798383788 . . .

a4 = 1.198140234735592207440631328 . . .
b4 = 1.198140234735592207439213655 . . . .
The sequences are converging very quickly to the limit
a∞ = b∞ = 1.198140234735592207439922492 . . . .



The rapid convergence is explained by the following.


PROPOSITION 9.23
Suppose a ≥ b > 0. Then
bn’1 ¤ bn ¤ an ¤ an’1
and
1
0 ¤ an ’ bn ¤ (an’1 ’ bn’1 ). (9.21)
2
Therefore
M (a, b) = lim an = lim bn
n’∞ n’∞
exists. Moreover, if b ≥ 1 then
2m
an ’ bn
an+m ’ bn+m ¤ 8 (9.22)
8
for all m, n ≥ 0.

PROOF The fact that an ≥ bn for all n is the arithmetic-geometric mean
inequality, or the fact that
1√
an ’ bn = ( an’1 ’ bn’1 )2 ≥ 0.
2




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

Therefore, since an’1 ≥ bn’1 , it follows immediately from (9.20) that
1
an ¤ bn ≥
(an’1 + an’1 ) = an’1 and bn’1 bn’1 = bn’1 .
2
Also,
1√ 2
an ’ b n = an’1 ’ bn’1
2
1√ √
¤ an’1 ’ bn’1 an’1 + bn’1
2
1
= (an’1 ’ bn’1 ).
2
Therefore, an ’ bn ¤ (1/2)n (a ’ b), so an ’ bn ’ 0. Since the an ™s are
a decreasing sequence bounded below by the increasing sequence of bn ™s, it
follows immediately that the two sequences converge to the same limit, so

M (a, b) exists. If bn’1 ≥ 1, then an’1 + bn’1 ≥ 2, so

an ’ bn 1√ 2
an’1 ’
= bn’1
8 16
√ 2
1√ an’1 + bn’1
2
¤ an’1 ’ bn’1
16 4
2
an’1 ’ bn’1
= .
8

Inequality 9.22 follows easily by induction.

The limit M (a, b) is called the arithmetic-geometric mean of a and b.
Since
M (ca, cb) = cM (a, b),
we can always rescale a and b to make b ≥ 1. Also, since M (b, a) = M (a, b)
(because a1 and b1 are symmetric in a, b), we may always arrange that a ≥ b.
By Inequality (9.21), an ’ bn < 1 for su¬ciently large n. The numbers an+m
and bn+m give approximations to M (a, b). Inequality (9.22) predicts that
the number of decimal places of accuracy doubles with each iteration. This
phenomenon occurs in the above example.
The reasons we are interested in the arithmetic-geometric mean are the
following two propositions.


PROPOSITION 9.24
Let a, b be positive real numbers. De¬ne
π/2

.
I(a, b) =
a2 cos2 θ + b2 sin2 θ
0




© 2008 by Taylor & Francis Group, LLC
291
SECTION 9.4 COMPUTING PERIODS

Then
a+b √
I , ab = I(a, b).
2
Moreover,
π/2
.
I(a, b) =
M (a, b)

PROOF Let u = b tan θ. The integral becomes
∞ ∞
du 1 du
I(a, b) = = .
2
(u2 + a2 )(u2 + b2 ) (u2 + a2 )(u2 + b2 )
’∞
0

Therefore,
a+b √ ∞
1 du
, ab
I = .
2 2 ’∞ ( a+b )2 )(u2
(u2 + + ab)
2

Let
1 ab
v’ 0 < v < ∞.
u= ,
2 v

u2 + ab. Since
Then v = u +
2
a+b 1
2
(v 2 + a2 )(v 2 + b2 ),
u+ = 2
2 4v
it is straightforward to obtain
a+b √ ∞
dv
, ab =
I = I(a, b).
2 (v 2 + a2 )(v 2 + b2 )
0

By induction, we obtain
I(a, b) = I(a1 , b1 ) = I(a2 , b2 ) = · · · .
Let
a∞ = b∞ = M (a, b) = lim an = lim bn .
n’∞ n’∞
It is fairly easy to justify taking the limit inside the integral sign to obtain
I(a, b) = lim I(an , bn )
n’∞
= I(a∞ , b∞ )
π/2

=
a2 cos2 θ + b2 sin2 θ
0
∞ ∞
π/2
1 dθ
=
M (a, b) cos2 θ + sin2 θ
0
π/2
= .
M (a, b)




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

PROPOSITION 9.25
If 0 < k < 1, then

1 ’ k 2 = I(1 + k, 1 ’ k).
K(k) = I 1,


PROOF
1
dt
K(k) =
(1 ’ t2 )(1 ’ k 2 t2 )
0
π/2

= (let t = sin θ)
1’ 2
k2 sin θ
0
π/2

=
cos2 θ + (1 ’ k 2 ) sin2 θ
0


1 ’ k2 )
= I(1,
= I(1 + k, 1 ’ k).
The last equation follows from Proposition 9.24, with a = 1 + k and b = 1 ’ k.


Putting everything together, we can now express the periods ω1 and ω2 in
terms of arithmetic-geometric means.

THEOREM 9.26
Suppose E is given by
y 2 = 4x3 ’ g2 x ’ g3 = 4(x ’ e1 )(x ’ e2 )(x ’ e3 )
with real numbers e1 < e2 < e3 . Then Zω1 + Zω2 is a lattice for E, where
πi
√ √
ω1 =
M ( e3 ’ e1 , e2 ’ e1 )
π
√ √
ω2 = .
M ( e3 ’ e1 , e3 ’ e2 )

PROOF We have, with k as in (9.19),
4
ω2 = √ √ K(k)
e3 ’ e1 + e3 ’ e2
4
=√ √ I(1 + k, 1 ’ k).
e3 ’ e1 + e3 ’ e2
Use the de¬nition (9.19) of k and the relation cI(ca, cb) = I(a, b) with
√ √
e3 ’ e1 + e3 ’ e2
c=
2




© 2008 by Taylor & Francis Group, LLC
293
SECTION 9.4 COMPUTING PERIODS

to obtain
√ √
ω2 = 2I( e3 ’ e1 , e3 ’ e2 )
π
√ √
= .
M ( e3 ’ e1 , e3 ’ e2 )
The proof of the formula for ω1 uses similar reasoning to obtain
2i
ω1 = √ √ K( 1 ’ k 2 )
e3 ’ e1 + e3 ’ e2
2i
=√ √ I(1, k)
e3 ’ e1 + e3 ’ e2
√ √ √ √
= 2iI( e3 ’ e1 + e3 ’ e2 , e3 ’ e1 ’ e3 ’ e2 ).

If we let
√ √ √ √
e3 ’ e1 + e3 ’ e2 , e3 ’ e1 ’ e3 ’ e2 ,
a= b=

then (9.20) yields √ √
e3 ’ e1 , e2 ’ e1 .
a1 = b1 =
Proposition 9.24 therefore implies that
√ √
ω1 = 2iI( e3 ’ e1 , e2 ’ e1 )
πi
√ √
= .
M ( e3 ’ e1 , e2 ’ e1 )

Example 9.2
Consider the elliptic curve E given by

y 2 = 4x3 ’ 4x.

Then e1 = ’1, e2 = 0, e3 = 1, so
πi

ω1 = = i2.62205755429211981046483959 . . .
M ( 2, 1)
π

ω2 = = 2.62205755429211981046483959 . . . .
M ( 2, 1)
Therefore, the fundamental parallelogram for the lattice is a square. This also
follows from the fact that E has complex multiplication by Z[i]. See Chapter
10. The number 2.622 . . . can be shown (see Exercise 9.8) to equal
1
dx “(1/4)“(1/2)
√ = ,
1 ’ x4 2 “(3/4)
’1

where “ is the gamma function (for its de¬nition, see Section 14.2). This is
a special case of the Chowla-Selberg formula, which expresses the periods of




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

elliptic curves with complex multiplication in terms of values of the gamma
function (see [101]).

There are also formulas similar to those of Theorem 9.26 for the case where
g2 , g3 ∈ R but 4x3 ’ g 2 x ’ g3 has only one real root. Let e1 be the unique
real root of 4x3 ’ g2 x ’ g3 and let e = 3e2 ’ (1/4)g2 . Then
1


√ √
ω1 = (9.23)
M ( 4e , 2e + 3e1 )
ω1 πi

ω2 = ’ √
+ . (9.24)
M ( 4e , 2e ’ 3e1 )
2
The proof is similar to the one when there are three real roots.
For more on the arithmetic-geometric mean, including how it has been used
to compute π very accurately and how it behaves for complex arguments, see
[17] and [30].




9.5 Division Polynomials
In this section, we prove Theorem 3.6, which gives a formula for n(x, y),
where n > 1 is an integer and (x, y) is a point on an elliptic curve. We™ll
start with the case of an elliptic curve in characteristic zero, then use this to
deduce the case of positive characteristic.
Let E be an elliptic curve over a ¬eld of characteristic 0, given by an
equation y 2 = x3 + Ax + B. All of the equations describing the group law
are de¬ned over Q(A, B). Since C is algebraically closed and has in¬nite
transcendence degree over Q, it is easy to see that Q(A, B) may be considered
as a sub¬eld of C. Therefore, we regard E as an elliptic curve de¬ned over
C. By Theorem 9.21, there is a lattice L corresponding to E. Let „˜(z) be the
associated Weierstrass „˜-function, which satis¬es the relation
(„˜ )2 = 4„˜3 ’ g2 „˜ ’ g3 ,
with g2 = ’4A, g3 = ’4B. We™ll derive formulas for „˜(nz) and „˜ (nz), then
use x = „˜(z) and y = „˜ (z)/2 to obtain the desired formulas for n(x, y).

LEMMA 9.27
There is a doubly periodic function fn (z) such that

(„˜(z) ’ „˜(u)).
fn (z)2 = n2
0=u∈(C/L)[n]

The sign of fn can be chosen so that




© 2008 by Taylor & Francis Group, LLC
295
SECTION 9.5 DIVISION POLYNOMIALS

1. if n is odd, fn = Pn („˜), where Pn (X) is a polynomial of degree (n2 ’1)/2
with leading coe¬cient n,
2. if n is even, fn = „˜ Pn („˜), where Pn (X) is a polynomial of degree
(n2 ’ 4)/2 with leading coe¬cient n/2.
The expansion of fn at 0 is
(’1)n+1 n
+ ··· .
fn (z) =
z n2 ’1
The zeros of fn are at the points 0 = u ∈ (C/L)[n], and these are simple
zeros.

PROOF The product is over the nonzero n-torsion in C/L. Since „˜(u) =
„˜(’u), the factors for u and ’u are equal. Suppose n is odd. Then u is
never congruent to ’u mod L, so every factor in the product occurs twice.
Therefore, fn can be taken to be n („˜(z) ’ „˜(u)), where we use only one
member of each pair (u, ’u). This is clearly a polynomial in „˜(z) of degree
(n2 ’ 1)/2 and leading coe¬cient n. When n is even, there are three values
of u that are congruent to their negatives mod L, namely, ωj /2 for j = 1, 2, 3.
Since
(„˜ )2 = 4 („˜ ’ „˜(ωj /2)),
j

these factors contribute „˜ /2 to fn . The remaining factors can be paired up,
as in the case when n is odd, to obtain a polynomial in „˜ of degree (n2 ’ 4)/2
and leading coe¬cient n. Therefore, fn has the desired form.
Since „˜(z) = z ’2 + · · · and „˜ (z) = ’2z ’3 + · · · , we immediately obtain
the expansion of fn at 0.
Clearly fn has a zero at each nonzero u ∈ (C/L)[n]. There are n2 ’ 1 such
points. Since the only pole mod L of fn is one of order n2 ’ 1 at z = 0, and
since the number of zeros equals the number of poles (counting multiplicities),
these zeros must all be simple.

LEMMA 9.28
Let n ≥ 2. Then
fn’1 (z)fn+1 (z)
„˜(nz) = „˜(z) ’ .
fn (z)2

PROOF Let g(z) = „˜(nz) ’ „˜(z). We™ll show that g and fn’1 fn+1 /fn 2

have the same divisors.
The function g(z) has a double pole at each u ∈ (C/L)[n] with u = 0. At
z = 0, it has the expansion
1 1
’ 2 + ··· ,
g(z) =
n2 z 2 z




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

so g also has a double pole at 0. Therefore, g has a total of 2n2 poles, counting
multiplicities.
The function g has a zero at z = w when nw ≡ ±w ≡ 0 (mod L). For such
w,

d
= n„˜ (nw) ’ „˜ (w) = ±n„˜ (w) ’ „˜ (w) = (±n ’ 1)„˜ (w).
g(z)
dz z=w

Since the zeros of „˜ (z) occur when z = ωj /2, we have g (w) = 0 when
w = ωj /2, so such w are simple zeros of g. Moreover, when n is odd, n(ωj /2) ≡
ωj /2, so the points ωj /2 are at least double zeros of g in this case.
If nw ≡ w (mod L), then (n ’ 1)w ≡ 0. Let δ = 0 if n is even and δ = 1
if n is odd. There are (n ’ 1)2 ’ 1 ’ 3δ points w with (n ’ 1)w = 0 and
w = 0, ωj /2. Similarly, there are (n + 1)2 ’ 1 ’ 3δ points w with (n + 1)w = 0
and w = 0, ωj /2. There are at least 6δ zeros (counting multiplicities) at the
points ωj /2. Therefore, we have accounted for at least

(n ’ 1)2 ’ 1 ’ 3δ + (n + 1)2 ’ 1 ’ 3δ + 6δ = 2n2

zeros. Since g(z) has exactly 2n2 poles, we have found all the zeros and their
multiplicities.
The function
fn’1 fn+1
2
fn
has a double pole at each of the zeros of fn . If w ≡ 0 and (n ± 1)w ≡ 0 then
fn±1 has a simple zero at w. If both (n + 1)w ≡ 0 and (n ’ 1)w ≡ 0, then
2w ≡ 0, so w ≡ ωj /2 for some j. Therefore, fn’1 fn+1 has a simple zero at
each w with (n ± 1)w ≡ 0, except for those where w ≡ ωj /2, at which points
it has a double zero. At z = 0, the expansions of the functions yield

fn’1 fn+1
=
2
fn
2
(’1)n (n ’ 1) (’1)n+2 (n + 1) (’1)n+1 n
+ ··· + ··· + ···
z n2 ’1
z (n’1)2 ’1 z (n+1)2 ’1
1
= 1 ’ 2 z ’2 + · · · ,
n

so there is a double zero at z = 0. Therefore, ’fn’1 fn+1 /fn has the same
2

divisor as „˜(nz) ’ „˜(z), so the two functions are constant multiples of each
other. Since their expansions at 0 have the same leading coe¬cient, they must
be equal. This proves the lemma.


LEMMA 9.29
f2n+1 = fn+2 fn ’ fn+1 fn’1 .
3 3




© 2008 by Taylor & Francis Group, LLC
297
SECTION 9.5 DIVISION POLYNOMIALS

PROOF As in the proof of Lemma 9.28, we see that
f2n+1
„˜((n + 1)z) ’ „˜(nz) = ’ 2 2
fn+1 fn
since the two sides have the same divisors and their expansions at 0 have the
same leading coe¬cient. Since
„˜((n + 1)z) ’ „˜(nz) = („˜((n + 1)z) ’ „˜(z)) ’ („˜(nz) ’ „˜(z))
fn+2 fn fn+1 fn’1
=’ 2 + ,
2
fn+1 fn
the result follows by equating the two expressions for „˜((n + 1)z) ’ „˜(nz).



LEMMA 9.30
„˜ f2n = (fn )(fn+2 fn’1 ’ fn’2 fn+1 ).
2 2



PROOF As in the proofs of the previous two lemmas, we have
„˜ f2n
„˜((n + 1)z) ’ „˜((n ’ 1)z) = ’ .
2 2
fn’1 fn+1
(A little care is needed to handle the points ωj /2.) Since
„˜((n + 1)z) ’ „˜((n ’ 1)z) = („˜((n + 1)z) ’ „˜(z)) ’ („˜((n ’ 1)z) ’ „˜(z))
fn+2 fn fn fn’2
=’ 2 +2 ,
fn+1 fn’1

the result follows.


LEMMA 9.31
For all n ≥ 1,
1
fn (z) = ψn „˜(z), „˜ (z)
2
where ψn is de¬ned in Section 3.2.

PROOF Since ψ1 = 1 and ψ2 = 2y, the lemma is easily seen to be true for
n = 1, 2. From Equations (9.10) and (9.8) in Section 9.2, we have
f3 f3
’ = ’ 2 = „˜(2z) ’ „˜(z) (9.25)
(„˜ )2 f2
2
1 „˜ (z)
’ 2„˜(z) ’ „˜(z)
=
4 „˜ (z)
3„˜4 ’ 3 g2 „˜2 ’ 3g3 „˜ ’ 12
16 g2
=’ 2
.
(„˜ )2




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

Therefore,
3 12
f3 = 3„˜4 ’ g2 „˜2 ’ 3g3 „˜ ’ g2
2 16
= 3„˜ + 6A„˜ + 12B„˜ ’ A2 = ψ3 („˜).
4 2


This proves the lemma for n = 3.
By Equation (9.7) in Section 9.2, we have
2
„˜ (2z) ’ „˜ (z)
1
’ „˜(2z) ’ „˜(z)
„˜(2z + z) =
„˜(2z) ’ „˜(z)
4

and
2
„˜ (2z) + „˜ (z)
1
„˜(2z ’ z) = ’ „˜(2z) ’ „˜(z).
„˜(2z) ’ „˜(z)
4
Therefore,

f4 f2
’ 2 = „˜(3z) ’ „˜(z)
f3
2 2
„˜ (2z) ’ „˜ (z)
1 1 „˜ (2z) + „˜ (z)

=
„˜(2z) ’ „˜(z) „˜(2z) ’ „˜(z)
4 4
„˜ (2z)„˜ (z)
=’
(„˜(2z) ’ „˜(z))2
„˜ (2z)„˜ (z)
=’ (by Equation 9.25)
(’f3 /„˜ (z)2 )2
„˜ (2z)„˜ (z)5
=’ .
2
f3

This yields f4 f2 = „˜ (2z)„˜ (z)5 . We know that 1 „˜ (2z) is the y-coordinate of
2
2(„˜(z), 1 „˜ (z)), which means that „˜ (2z) can be expressed in terms of „˜(z)
2
and „˜ (z), using the formulas for the group law. When this is done, we obtain

1
f4 = ψ4 „˜, „˜ ,
2

so the lemma is true for n = 4.
Since the fn ™s satisfy the same recurrence relations as the ψn ™s (see Lem-
mas 9.29 and 9.30 and the de¬nition of the ψn ™s), and since the lemma holds
for enough small values of n, the lemma now follows for all n.


LEMMA 9.32

f2n
„˜ (nz) = .
4
fn




© 2008 by Taylor & Francis Group, LLC
299
SECTION 9.5 DIVISION POLYNOMIALS

PROOF The function „˜ (nz) has triple poles at all points of (C/L)[n].
Therefore, there are 3n2 poles. Since the zeros of „˜ are at the points in
(C/L)[2] other than 0, the zeros of „˜ (nz) are at the points that are in
(C/L)[2n] but not in (C/L)[n]. There are 3n2 such points. Since the num-
ber of zeros equals the number of poles, all of these zeros are simple. The
expansion of „˜ (nz) at z = 0 is
’2
+ ··· .
„˜ (nz) =
n3 z 3
4
The function f2n /fn is easily seen to have the same divisor as „˜ (nz) and
their expansions at z = 0 have the same leading coe¬cients. Therefore, the
functions are equal.

Finally, we can prove the main result of this section.


THEOREM 9.33
Let E be an elliptic curve over a ¬eld of characteristic not 2, let n be a
positive integer, and let (x, y) be a point on E. Then
φn ωn
n(x, y) = ,3 ,
2
ψn ψn
where φn , ψn , and ωn are de¬ned in Section 3.2.


PROOF First, assume E is de¬ned over a ¬eld of characteristic 0. As
above, we regard E as being de¬ned over C. We have
1 1
(x, y) = „˜(z), „˜ (z) , n(x, y) = „˜(nz), „˜ (nz)
2 2
for some z. Therefore,
fn’1 fn+1
„˜(nz) = „˜(z) ’ 2
fn
„˜fn ’ fn’1 fn+1
2
= 2
fn
xψn ’ ψn’1 ψn+1
2
= (by Lemma 9.31)
2
ψn
φn
= 2.
ψn
This proves the formula for the x-coordinate.
For the y-coordinate, observe that the de¬nition of ωn can be rewritten as
1 ψ2n
ωn = .
2 ψn




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

Therefore, by Lemmas 9.32 and 9.31,
1 ψ2n ωn
1
„˜ (nz) = = 3.
4
2 2 ψn ψn
This completes the proof of the theorem when the characteristic of the ¬eld
is 0.
Suppose now that E is de¬ned over a ¬eld K of arbitrary characteristic
(not 2) by y 2 = x3 + Ax + B. Let (x, y) ∈ E(K). Let ±, β, and X be three
independent transcendental elements of C and let Y satisfy Y 2 = X 3 +±X +β.
There is a ring homomorphism
ρ : Z[±, β, X, Y ] ’’ K(x, y)

such that

g(±, β, X, Y ) ’ g(A, B, x, y)
˜
for all polynomials g. Let R = Z[±, β, X, Y ] and let E be the elliptic curve
over R de¬ned by y 2 = x3 + ±x + β. We want to say that by Corollary 2.33,
ρ induces a homomorphism
˜
ρ : E(R) ’’ E(K(x, y)).

But we need to have R satisfy Conditions (1) and (2) of Section 2.11. The
easiest way to accomplish this is to let M be the kernel of the map R ’
K(x, y). Since K(x, y) is a ¬eld, M is a maximal ideal of R. Let RM be the
localization of R at M (this means, we invert all elements of R not in M).
Then R ⊆ RM and the map ρ extends to a map

ρ : RM ’’ K(x, y).
Since RM is a local ring, and projective modules over local rings are free, it
can be shown that RM satis¬es Condition (2). Since we are assuming that
K(x, y) has characteristic not equal to 2, it follows that 2 is not in M, hence
is invertible in RM . Therefore, RM satis¬es Condition (1). Now we can apply
Corollary 2.33.
˜
The point n(X, Y ) in E(RM ) is described by the polynomials ψj , φj , and ωj ,
which are polynomials in X, Y with coe¬cients in Z[±, β]. Applying ρ shows
that these polynomials, regarded as polynomials in x, y with coe¬cients in K,
describe n(x, y) on E. Therefore, the theorem holds for E.

As an application of the division polynomials, we prove the following result,
which will be used in Chapter 11.


PROPOSITION 9.34
Let E be an elliptic curve over a ¬eld K. Let f (x, y) be a function from E to
K ∪ {∞} and let n ≥ 1 be an integer not divisible by the characteristic of K.




© 2008 by Taylor & Francis Group, LLC
301
SECTION 9.5 DIVISION POLYNOMIALS

Suppose f (P + T ) = f (P ) for all P ∈ E(K) and all T ∈ E[n]. Then there is
a function h on E such that f (P ) = h(nP ) for all P .


PROOF The case n = 1 is trivial, so we assume n > 1. Let T ∈ E[n].
There are rational functions R(x, y), S(x, y) depending on T such that

(x, y) + T = (R(x, y), S(x, y)).

Let y 2 = x3 +Ax+B be the equation of E and regard K(x, y) as the quadratic

extension of K(x) given by adjoining x3 + Ax + B. Since (R, S) lies on E,
we have S 2 = R3 + AR + B. The map

σT : K(x, y) ’ K(x, y)
f (x, y) ’ f (R, S)

is a homomorphism from K(x, y) to itself. Since σ’T is the inverse of σT , the
map σT is an automorphism. Because (x, y) + T = (x, y) + T when T = T ,
we have σT (x, y) = σT (x, y) when T = T . Therefore, we have a group of n2
distinct automorphisms σT , where T runs through E[n], acting on K(x, y). A
basic result in Galois theory says that if G is a group of distinct automorphisms
of a ¬eld L, then the ¬xed ¬eld F of G satis¬es [L : F ] = |G|. Therefore, the
¬eld F of functions f satisfying the conditions of the proposition satis¬es

[K(x, y) : F ] = n2 . (9.26)

Let n(x, y) = (gn (x), y hn (x)) for rational functions gn , hn . Then

K(gn (x), y hn (x)) ⊆ F. (9.27)

Moreover,

[ K(gn (x), y hn (x)) : K(gn (x))] ≥ 2 (9.28)

since clearly y hn (x) ∈ K(gn (x)). Therefore, by (9.26), (9.27), and (9.28),

[K(x, y) : K(gn (x))] ≥ 2n2 .

From Theorem 3.6,
φn
gn (x) = ,
2
ψn
2
and φn and ψn are polynomials in x. Therefore, X = x is a root of the
polynomial

P (X) = φn (X) ’ gn (x)ψn (X) ∈ K[gn (x)][X].
2


By Lemma 3.5,
2
φn (X) = X n + · · ·




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

and ψn (X) has degree n2 ’ 1. Therefore,
2

2
P (X) = X n + · · · ,

so x is of degree at most n2 over K(gn (x)). Since

[ K(x, y) : K(x)] = 2,

we obtain
[ K(x, y) : K(gn (x))] ¤ 2n2 .
Combined with the previous inequality from above, we obtain equality, which
means that we had equality in all of our calculations. In particular,

<<

. 2
( 3)



>>