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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 iв€ћ 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 ПЃ includes only
1/3 of a disc from F (namely, 1/6 near ПЃ and 1/6 near 1 + ПЃ, which is folded
over to meet ПЃ). 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 ПЃ, 1 + ПЃ,
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 ПЃ, 1 + ПЃ, and i have radius , where is
chosen small enough that there are no zeros or poles of f inside the circles,
except possibly at ПЃ, 1 + ПЃ, or i. Similarly, the top part of C is chosen to have

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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 iв€ћ. This is possible
since
f (z) = q n (an + an+1 q + В· В· В· ).
Since the series an + an+1 q + В· В· В· is assumed to converge for q small, it is п¬Ѓnite
and is approximately equal to an = 0 for suп¬ѓciently small q.
As in the proof of Theorem 9.1, we have

1 f (z)
dz = ordz (f ).
2ПЂi f (z)
C zв€€F
z=i,ПЃ

11
в€€ SL2 (Z) gives the map z в†’ 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 weвЂ™ll show that the integral over the part of the unit circle to the left of
0 в€’1
в€€
i cancels the part to the right. This is proved by using the fact that
10
SL2 (Z) gives the map z в†’ в€’1/z, which interchanges the left and right arcs of
the unit circle. In addition, diп¬Ђerentiating the relation f (в€’1/z) = f (z) yields

в€’1 в€’1
f f
d = (z) dz.
f z z f

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SECTION 9.3 ELLIPTIC CURVES OVER C

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

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

The integral over the small semicircle near i is

f (i + eiОё ) iОё
1
ie dОё, (9.18)
f (i + eiОё )
2ПЂi Оё

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

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

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

2ПЂian+1 q + В· В· В·
f (П„ )
= 2ПЂin + .
an + В· В· В·
f (П„ )

The second term goes to 0 as q в†’ 0, hence as N в†’ в€ћ. The limit of the
integral as N в†’ в€ћ is therefore
в€’1
1 2
2ПЂin dt = в€’n = в€’ordiв€ћ (f ).
2ПЂi t= 1
2

Combining all of the above calculations yields the theorem.

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

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282 CHAPTER 9 ELLIPTIC CURVES OVER C

PROOF First, we need to calculate j(ПЃ) and j(i). Recall that П„ corresponds
to the lattice LП„ = ZП„ + Z. Since ПЃ2 = в€’1 в€’ ПЃ, it follows easily that ПЃLПЃ вЉ† LПЃ .
Therefore,
LПЃ = ПЃ3 LПЃ вЉ† ПЃ2 LПЃ вЉ† ПЃLПЃ вЉ† LПЃ ,
so ПЃLПЃ = LПЃ . It follows from (9.14) that

g2 (LПЃ ) = g2 (ПЃLПЃ ) = ПЃв€’4 g2 (LПЃ ) = ПЃв€’1 g2 (LПЃ ).

Since ПЃ = 1, we have g2 (ПЃ) = g2 (LПЃ ) = 0. Therefore,

g2 (LПЃ )3
j(ПЃ) = 1728 =0
g2 (LПЃ )3 в€’ 27g3 (LПЃ )2

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

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

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

If z = 0, 1728, then h has order 0 at ПЃ and at i. Therefore, h has a unique zero
in F, so j(П„ ) = 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, ПЃ; 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(П„ ) в€’ 1728 has a double zero at i and no other zero in F . Similarly, j(П„ ) has
a triple zero at ПЃ and no other zero in F .

COROLLARY 9.19
ab
Let П„1 , П„2 в€€ H. Then j(П„1 ) = j(П„2 ) if and only if there exists в€€ SL2 (Z)
cd
such that
aП„1 + b
= П„2 .
cП„1 + d

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

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SECTION 9.3 ELLIPTIC CURVES OVER C

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

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

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

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

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

THEOREM 9.21
Let y 2 = 4x3 в€’ Ax в€’ B deп¬Ѓne 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 в€’ 27B 2
By Corollary 9.18, there exists a lattice L = ZП„ +Z such that j(П„ ) = j(L) = j.
Assume п¬Ѓrst that g2 (L) = 0. Then j = j(L) = 0, so A = 0. Choose О» в€€ CГ—
such that
g2 (О»L) = О»в€’4 g2 (L) = A.
The equality j = j(L) implies that
g3 (О»L)2 = B 2 ,
so g3 (О»L) = В±B. If g3 (О»L) = B, weвЂ™re done. If g3 (О»L) = в€’B, then
g3 (iО»L) = iв€’6 g3 (О»L) = B g2 (iО»L) = i4 g2 (О»L) = A.
and

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284 CHAPTER 9 ELLIPTIC CURVES OVER C

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

g3 (ВµL) = Вµв€’6 g3 (L) = B.

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

C/L в€’в†’ 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 = ZП‰1 + ZП‰2 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
П‰1 + П‰2 , 0 в‰¤ j, k в‰¤ n в€’ 1.
n n
It follows that
Zn вЉ• Zn .
E[n]
In fact, we can use this observation to give a proof of Theorem 3.2 for all п¬Ѓelds
of characteristic 0.

COROLLARY 9.22
Let K be a п¬Ѓeld of characteristic 0, and let E be an elliptic curve over K.
Then
E[n] = {P в€€ E(K) | nP = в€ћ} Zn вЉ• Zn .

PROOF Let L be the п¬Ѓeld generated by Q and the coeп¬ѓcients of the
equation of E. Then L has п¬Ѓnite 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 вЉ• Zn .
There is a technical point to worry about. The deп¬Ѓnition of E[n] that
we have used requires the coordinates of the n-torsion to lie in the algebraic
closure of the base п¬Ѓeld. How can we be sure that the п¬Ѓeld K isnвЂ™t so large
that it allows more torsion points than C? Suppose that E[n] вЉ‚ 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 п¬Ѓnite transcendence degree over Q,
hence can be embedded into C. The coordinates of the n2 + 1 points will yield

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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 subп¬Ѓeld 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 deп¬Ѓning E, they are both algebraic or both
transcendental over K). Let Пѓ be an automorphism of C such that Пѓ(x) =
x+1, and such that Пѓ is the identity on L. Such an automorphism exists: take
Пѓ to be the desired automorphism of K(x), then use ZornвЂ™s Lemma to extend
Пѓ to all of C (see Appendix C). The points Пѓ 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 вЉ† K. Therefore, the
passage from K to C does not aп¬Ђect E[n].

Suppose we have an elliptic curve E deп¬Ѓned 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 п¬Ѓrst that the lattice is rectangular: L = ZП‰1 + ZП‰2 with П‰1 в€€ iR
and П‰2 в€€ R. Then
(в„˜(z), в„˜ (z)) в€€ E(R)
when
with 0 в‰¤ t < 1,
(I) z = tП‰2
and also when
with 0 в‰¤ t < 1.
(II) z = (1/2)П‰1 + tП‰2
The п¬Ѓrst of these is easy to see: if z is real and the lattice L is preserved by
complex conjugation, then conjugating the deп¬Ѓning expression for в„˜(z) leaves
it unchanged, so в„˜ maps reals to reals. The second is a little more subtle:
conjugating z = (1/2)П‰1 + tП‰2 yields z = в€’(1/2)П‰1 + tП‰2 , which is equivalent
to z mod L. Therefore, the deп¬Ѓning expression for в„˜(z) is again unchanged
by complex conjugation, so в„˜ 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 diп¬Ђer by П‰2 . 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 в€ћ
in the Weierstrass form), we obtain a slightly deformed version of the graph
of Figure 2.1(a) on page 10.

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286 CHAPTER 9 ELLIPTIC CURVES OVER C

в„¦1

1 II
в„¦
21

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

In the case of a skewed parallelogram, that is, when П‰2 is real and the
imaginary part of П‰1 is half of П‰2 , 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 = ZП‰1 + ZП‰2 via the doubly periodic functions
в„˜ and в„˜ , but how do we п¬Ѓnd the periods П‰1 and П‰2 ?
For simplicity, letвЂ™s consider the case where E is deп¬Ѓned over R and E вЉ‚
E(R). Then the equation for E can be put in the form

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

We may assume П‰2 в€€ R with П‰2 > 0 and П‰1 в€€ iR with (П‰1 ) > 0, as in
Figure 9.5. The graph of E is as in Figure 2.1(a) on page 10. The Weierstrass
в„˜-function and its derivative map C/L to E via

(x, y) = (в„˜(z), в„˜ (z)).

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

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SECTION 9.4 COMPUTING PERIODS

Consider now the integral
в€ћ
dx
.
4(x в€’ e1 )(x в€’ e2 )(x в€’ e3 )
e3

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

Therefore,
в€ћ
dx
П‰2 = .
(x в€’ e1 )(x в€’ e2 )(x в€’ e3 )
e3

The change of variables

e3 в€’ (e3 в€’ e1 )(e3 в€’ e2 ) t + e3 + (e3 в€’ e1 )(e3 в€’ e2 )
x=
t+1
(plus a lot of algebraic manipulation) changes the integral to
1
2 dt
П‰2 = в€љ в€љ ,
e3 в€’ e1 + e3 в€’ e2 (1 в€’ t2 )(1 в€’ k 2 t2 )
в€’1

where
в€љ в€љ
e3 в€’ e1 в€’ e3 в€’ e2
k=в€љ в€љ . (9.19)
e3 в€’ e1 + e3 в€’ 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
П‰2 = в€љ в€љ .
e3 в€’ e1 + e3 в€’ e2 (1 в€’ t2 )(1 в€’ k 2 t2 )
0

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

In the following, weвЂ™ll see how to compute K(k) numerically very accurately
and quickly, but п¬Ѓrst letвЂ™s п¬Ѓnd an expression for П‰1 .
When z runs along the vertical line from П‰2 /2 to П‰2 /2 + П‰1 /2, the function
в„˜(z) takes on real values (see Exercise 9.6) from e3 to e2 , and its derivative

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288 CHAPTER 9 ELLIPTIC CURVES OVER C

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

1/k
2i dt
П‰1 = в€љ в€љ .
e3 в€’ e1 + e3 в€’ e2 (t2 в€’ 1)(1 в€’ k 2 t2 )
1

в€љ
1 в€’ k 2 and make the substitution
Let k =

t = (1 в€’ k u2 )в€’1/2 .
2

The integral becomes

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

Therefore,

2i
П‰1 = в€љ в€љ K( 1 в€’ k 2 ).
e3 в€’ e1 + e3 в€’ e2

Therefore, both П‰1 and П‰2 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. Deп¬Ѓne an and bn by

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

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

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SECTION 9.4 COMPUTING PERIODS

Example 9.1
в€љ
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
aв€ћ = bв€ћ = 1.198140234735592207439922492 . . . .

The rapid convergence is explained by the following.

PROPOSITION 9.23
Suppose a в‰Ґ b > 0. Then
bnв€’1 в‰¤ bn в‰¤ an в‰¤ anв€’1
and
1
0 в‰¤ an в€’ bn в‰¤ (anв€’1 в€’ bnв€’1 ). (9.21)
2
Therefore
M (a, b) = lim an = lim bn
nв†’в€ћ nв†’в€ћ
exists. Moreover, if b в‰Ґ 1 then
2m
an в€’ bn
an+m в€’ bn+m в‰¤ 8 (9.22)
8
for all m, n в‰Ґ 0.

PROOF The fact that an в‰Ґ bn for all n is the arithmetic-geometric mean
inequality, or the fact that
1в€љ
an в€’ bn = ( anв€’1 в€’ bnв€’1 )2 в‰Ґ 0.
2

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290 CHAPTER 9 ELLIPTIC CURVES OVER C

Therefore, since anв€’1 в‰Ґ bnв€’1 , it follows immediately from (9.20) that
1
an в‰¤ bn в‰Ґ
(anв€’1 + anв€’1 ) = anв€’1 and bnв€’1 bnв€’1 = bnв€’1 .
2
Also,
1в€љ 2
an в€’ b n = anв€’1 в€’ bnв€’1
2
1в€љ в€љ
в‰¤ anв€’1 в€’ bnв€’1 anв€’1 + bnв€’1
2
1
= (anв€’1 в€’ bnв€’1 ).
2
Therefore, an в€’ bn в‰¤ (1/2)n (a в€’ b), so an в€’ bn в†’ 0. Since the an вЂ™s are
a decreasing sequence bounded below by the increasing sequence of bn вЂ™s, it
follows immediately that the two sequences converge to the same limit, so
в€љ
M (a, b) exists. If bnв€’1 в‰Ґ 1, then anв€’1 + bnв€’1 в‰Ґ 2, so

an в€’ bn 1в€љ 2
anв€’1 в€’
= bnв€’1
8 16
в€љ 2
1в€љ anв€’1 + bnв€’1
2
в‰¤ anв€’1 в€’ bnв€’1
16 4
2
anв€’1 в€’ bnв€’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 в‰Ґ 1. Also, since M (b, a) = M (a, b)
(because a1 and b1 are symmetric in a, b), we may always arrange that a в‰Ґ b.
By Inequality (9.21), an в€’ bn < 1 for suп¬ѓciently 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. Deп¬Ѓne
ПЂ/2
dОё
.
I(a, b) =
a2 cos2 Оё + b2 sin2 Оё
0

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SECTION 9.4 COMPUTING PERIODS

Then
a+b в€љ
I , ab = I(a, b).
2
Moreover,
ПЂ/2
.
I(a, b) =
M (a, b)

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

Therefore,
a+b в€љ в€ћ
1 du
, ab
I = .
2 2 в€’в€ћ ( a+b )2 )(u2
(u2 + + ab)
2

Let
1 ab
vв€’ 0 < v < в€ћ.
u= ,
2 v
в€љ
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 в€љ в€ћ
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 ) = В· В· В· .
Let
aв€ћ = bв€ћ = M (a, b) = lim an = lim bn .
nв†’в€ћ nв†’в€ћ
It is fairly easy to justify taking the limit inside the integral sign to obtain
I(a, b) = lim I(an , bn )
nв†’в€ћ
= I(aв€ћ , bв€ћ )
ПЂ/2
dОё
=
a2 cos2 Оё + b2 sin2 Оё
0
в€ћ в€ћ
ПЂ/2
1 dОё
=
M (a, b) cos2 Оё + sin2 Оё
0
ПЂ/2
= .
M (a, b)

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292 CHAPTER 9 ELLIPTIC CURVES OVER C

PROPOSITION 9.25
If 0 < k < 1, then

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

PROOF
1
dt
K(k) =
(1 в€’ t2 )(1 в€’ k 2 t2 )
0
ПЂ/2
dОё
= (let t = sin Оё)
1в€’ 2
k2 sin Оё
0
ПЂ/2
dОё
=
cos2 Оё + (1 в€’ k 2 ) sin2 Оё
0

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

Putting everything together, we can now express the periods П‰1 and П‰2 in
terms of arithmetic-geometric means.

THEOREM 9.26
Suppose E is given by
y 2 = 4x3 в€’ g2 x в€’ g3 = 4(x в€’ e1 )(x в€’ e2 )(x в€’ e3 )
with real numbers e1 < e2 < e3 . Then ZП‰1 + ZП‰2 is a lattice for E, where
ПЂi
в€љ в€љ
П‰1 =
M ( e3 в€’ e1 , e2 в€’ e1 )
ПЂ
в€љ в€љ
П‰2 = .
M ( e3 в€’ e1 , e3 в€’ e2 )

PROOF We have, with k as in (9.19),
4
П‰2 = в€љ в€љ K(k)
e3 в€’ e1 + e3 в€’ e2
4
=в€љ в€љ I(1 + k, 1 в€’ k).
e3 в€’ e1 + e3 в€’ e2
Use the deп¬Ѓnition (9.19) of k and the relation cI(ca, cb) = I(a, b) with
в€љ в€љ
e3 в€’ e1 + e3 в€’ e2
c=
2

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SECTION 9.4 COMPUTING PERIODS

to obtain
в€љ в€љ
П‰2 = 2I( e3 в€’ e1 , e3 в€’ e2 )
ПЂ
в€љ в€љ
= .
M ( e3 в€’ e1 , e3 в€’ e2 )
The proof of the formula for П‰1 uses similar reasoning to obtain
2i
П‰1 = в€љ в€љ K( 1 в€’ k 2 )
e3 в€’ e1 + e3 в€’ e2
2i
=в€љ в€љ I(1, k)
e3 в€’ e1 + e3 в€’ e2
в€љ в€љ в€љ в€љ
= 2iI( e3 в€’ e1 + e3 в€’ e2 , e3 в€’ e1 в€’ e3 в€’ e2 ).

If we let
в€љ в€љ в€љ в€љ
e3 в€’ e1 + e3 в€’ e2 , e3 в€’ e1 в€’ e3 в€’ e2 ,
a= b=

then (9.20) yields в€љ в€љ
e3 в€’ e1 , e2 в€’ e1 .
a1 = b1 =
Proposition 9.24 therefore implies that
в€љ в€љ
П‰1 = 2iI( e3 в€’ e1 , e2 в€’ e1 )
ПЂi
в€љ в€љ
= .
M ( e3 в€’ e1 , e2 в€’ e1 )

Example 9.2
Consider the elliptic curve E given by

y 2 = 4x3 в€’ 4x.

Then e1 = в€’1, e2 = 0, e3 = 1, so
ПЂi
в€љ
П‰1 = = i2.62205755429211981046483959 . . .
M ( 2, 1)
ПЂ
в€љ
П‰2 = = 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 О“(1/4)О“(1/2)
в€љ = ,
1 в€’ x4 2 О“(3/4)
в€’1

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

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294 CHAPTER 9 ELLIPTIC CURVES OVER C

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

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

2ПЂ
в€љ в€љ
П‰1 = (9.23)
M ( 4e , 2e + 3e1 )
П‰1 ПЂi
в€љ
П‰2 = в€’ в€љ
+ . (9.24)
M ( 4e , 2e в€’ 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 ПЂ very accurately and how it behaves for complex arguments, see
 and .

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. WeвЂ™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 п¬Ѓeld of characteristic 0, given by an
equation y 2 = x3 + Ax + B. All of the equations describing the group law
are deп¬Ѓned over Q(A, B). Since C is algebraically closed and has inп¬Ѓnite
transcendence degree over Q, it is easy to see that Q(A, B) may be considered
as a subп¬Ѓeld of C. Therefore, we regard E as an elliptic curve deп¬Ѓned over
C. By Theorem 9.21, there is a lattice L corresponding to E. Let в„˜(z) be the
associated Weierstrass в„˜-function, which satisп¬Ѓes the relation
(в„˜ )2 = 4в„˜3 в€’ g2 в„˜ в€’ g3 ,
with g2 = в€’4A, g3 = в€’4B. WeвЂ™ll derive formulas for в„˜(nz) and в„˜ (nz), then
use x = в„˜(z) and y = в„˜ (z)/2 to obtain the desired formulas for n(x, y).

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

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

The sign of fn can be chosen so that

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SECTION 9.5 DIVISION POLYNOMIALS

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

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

these factors contribute в„˜ /2 to fn . The remaining factors can be paired up,
as in the case when n is odd, to obtain a polynomial in в„˜ of degree (n2 в€’ 4)/2
and leading coeп¬ѓcient n. Therefore, fn has the desired form.
Since в„˜(z) = z в€’2 + В· В· В· and в„˜ (z) = в€’2z в€’3 + В· В· В· , we immediately obtain
the expansion of fn at 0.
Clearly fn has a zero at each nonzero u в€€ (C/L)[n]. There are n2 в€’ 1 such
points. Since the only pole mod L of fn is one of order n2 в€’ 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 в‰Ґ 2. Then
fnв€’1 (z)fn+1 (z)
в„˜(nz) = в„˜(z) в€’ .
fn (z)2

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

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

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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 в‰Ў В±w в‰Ў 0 (mod L). For such
w,

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

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

(n в€’ 1)2 в€’ 1 в€’ 3Оґ + (n + 1)2 в€’ 1 в€’ 3Оґ + 6Оґ = 2n2

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

fnв€’1 fn+1
=
2
fn
2
(в€’1)n (n в€’ 1) (в€’1)n+2 (n + 1) (в€’1)n+1 n
+ В·В·В· + В·В·В· + В·В·В·
z n2 в€’1
z (nв€’1)2 в€’1 z (n+1)2 в€’1
1
= 1 в€’ 2 z в€’2 + В· В· В· ,
n

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

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

LEMMA 9.29
f2n+1 = fn+2 fn в€’ fn+1 fnв€’1 .
3 3

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SECTION 9.5 DIVISION POLYNOMIALS

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

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

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

the result follows.

LEMMA 9.31
For all n в‰Ґ 1,
1
fn (z) = П€n в„˜(z), в„˜ (z)
2
where П€n is deп¬Ѓned in Section 3.2.

PROOF Since П€1 = 1 and П€2 = 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
в€’ = в€’ 2 = в„˜(2z) в€’ в„˜(z) (9.25)
(в„˜ )2 f2
2
1 в„˜ (z)
в€’ 2в„˜(z) в€’ в„˜(z)
=
4 в„˜ (z)
3в„˜4 в€’ 3 g2 в„˜2 в€’ 3g3 в„˜ в€’ 12
16 g2
=в€’ 2
.
(в„˜ )2

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298 CHAPTER 9 ELLIPTIC CURVES OVER C

Therefore,
3 12
f3 = 3в„˜4 в€’ g2 в„˜2 в€’ 3g3 в„˜ в€’ g2
2 16
= 3в„˜ + 6Aв„˜ + 12Bв„˜ в€’ A2 = П€3 (в„˜).
4 2

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

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

f4 f2
в€’ 2 = в„˜(3z) в€’ в„˜(z)
f3
2 2
в„˜ (2z) в€’ в„˜ (z)
1 1 в„˜ (2z) + в„˜ (z)
в€’
=
в„˜(2z) в€’ в„˜(z) в„˜(2z) в€’ в„˜(z)
4 4
в„˜ (2z)в„˜ (z)
=в€’
(в„˜(2z) в€’ в„˜(z))2
в„˜ (2z)в„˜ (z)
=в€’ (by Equation 9.25)
(в€’f3 /в„˜ (z)2 )2
в„˜ (2z)в„˜ (z)5
=в€’ .
2
f3

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

1
f4 = П€4 в„˜, в„˜ ,
2

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

LEMMA 9.32

f2n
в„˜ (nz) = .
4
fn

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SECTION 9.5 DIVISION POLYNOMIALS

PROOF The function в„˜ (nz) has triple poles at all points of (C/L)[n].
Therefore, there are 3n2 poles. Since the zeros of в„˜ are at the points in
(C/L) other than 0, the zeros of в„˜ (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 в„˜ (nz) at z = 0 is
в€’2
+ В·В·В· .
в„˜ (nz) =
n3 z 3
4
The function f2n /fn is easily seen to have the same divisor as в„˜ (nz) and
their expansions at z = 0 have the same leading coeп¬ѓcients. 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 п¬Ѓeld of characteristic not 2, let n be a
positive integer, and let (x, y) be a point on E. Then
П†n П‰n
n(x, y) = ,3 ,
2
П€n П€n
where П†n , П€n , and П‰n are deп¬Ѓned in Section 3.2.

PROOF First, assume E is deп¬Ѓned over a п¬Ѓeld of characteristic 0. As
above, we regard E as being deп¬Ѓned over C. We have
1 1
(x, y) = в„˜(z), в„˜ (z) , n(x, y) = в„˜(nz), в„˜ (nz)
2 2
for some z. Therefore,
fnв€’1 fn+1
в„˜(nz) = в„˜(z) в€’ 2
fn
в„˜fn в€’ fnв€’1 fn+1
2
= 2
fn
xП€n в€’ П€nв€’1 П€n+1
2
= (by Lemma 9.31)
2
П€n
П†n
= 2.
П€n
This proves the formula for the x-coordinate.
For the y-coordinate, observe that the deп¬Ѓnition of П‰n can be rewritten as
1 П€2n
П‰n = .
2 П€n

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300 CHAPTER 9 ELLIPTIC CURVES OVER C

Therefore, by Lemmas 9.32 and 9.31,
1 П€2n П‰n
1
в„˜ (nz) = = 3.
4
2 2 П€n П€n
This completes the proof of the theorem when the characteristic of the п¬Ѓeld
is 0.
Suppose now that E is deп¬Ѓned over a п¬Ѓeld K of arbitrary characteristic
(not 2) by y 2 = x3 + Ax + B. Let (x, y) в€€ E(K). Let О±, ОІ, and X be three
independent transcendental elements of C and let Y satisfy Y 2 = X 3 +О±X +ОІ.
There is a ring homomorphism
ПЃ : Z[О±, ОІ, X, Y ] в€’в†’ K(x, y)

such that

g(О±, ОІ, X, Y ) в†’ g(A, B, x, y)
Лњ
for all polynomials g. Let R = Z[О±, ОІ, X, Y ] and let E be the elliptic curve
over R deп¬Ѓned by y 2 = x3 + О±x + ОІ. We want to say that by Corollary 2.33,
ПЃ induces a homomorphism
Лњ
ПЃ : E(R) в€’в†’ 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 в†’
K(x, y). Since K(x, y) is a п¬Ѓeld, 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 вЉ† RM and the map ПЃ extends to a map

ПЃ : RM в€’в†’ K(x, y).
Since RM is a local ring, and projective modules over local rings are free, it
can be shown that RM satisп¬Ѓes 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 satisп¬Ѓes Condition (1). Now we can apply
Corollary 2.33.
Лњ
The point n(X, Y ) in E(RM ) is described by the polynomials П€j , П†j , and П‰j ,
which are polynomials in X, Y with coeп¬ѓcients in Z[О±, ОІ]. Applying ПЃ shows
that these polynomials, regarded as polynomials in x, y with coeп¬ѓcients 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 п¬Ѓeld K. Let f (x, y) be a function from E to
K в€Є {в€ћ} and let n в‰Ґ 1 be an integer not divisible by the characteristic of K.

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SECTION 9.5 DIVISION POLYNOMIALS

Suppose f (P + T ) = f (P ) for all P в€€ E(K) and all T в€€ 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 в€€ 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
в€љ
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

ПѓT : K(x, y) в†’ K(x, y)
f (x, y) в†’ f (R, S)

is a homomorphism from K(x, y) to itself. Since Пѓв€’T is the inverse of ПѓT , the
map ПѓT is an automorphism. Because (x, y) + T = (x, y) + T when T = T ,
we have ПѓT (x, y) = ПѓT (x, y) when T = T . Therefore, we have a group of n2
distinct automorphisms ПѓT , 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 п¬Ѓeld L, then the п¬Ѓxed п¬Ѓeld F of G satisп¬Ѓes [L : F ] = |G|. Therefore, the
п¬Ѓeld F of functions f satisfying the conditions of the proposition satisп¬Ѓes

[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)) вЉ† F. (9.27)

Moreover,

[ K(gn (x), y hn (x)) : K(gn (x))] в‰Ґ 2 (9.28)

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

[K(x, y) : K(gn (x))] в‰Ґ 2n2 .

From Theorem 3.6,
П†n
gn (x) = ,
2
П€n
2
and П†n and П€n are polynomials in x. Therefore, X = x is a root of the
polynomial

P (X) = П†n (X) в€’ gn (x)П€n (X) в€€ K[gn (x)][X].
2

By Lemma 3.5,
2
П†n (X) = X n + В· В· В·

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

and П€n (X) has degree n2 в€’ 1. Therefore,
2

2
P (X) = X n + В· В· В· ,

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))] в‰¤ 2n2 .
Combined with the previous inequality from above, we obtain equality, which
means that we had equality in all of our calculations. In particular,
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