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Chapter 9
Elliptic Curves over C

The goal of this chapter is to show that an elliptic curve over the complex
numbers is the same thing as a torus. First, we show that a torus is isomor-
phic to an elliptic curve. To do this, we need to study functions on a torus,
which amounts to studying doubly periodic functions on C, especially the
Weierstrass „˜-function. We then introduce the j-function and use its proper-
ties to show that every elliptic curve over C comes from a torus. Since most
of the ¬elds of characteristic 0 that we meet can be embedded in C, many
properties of elliptic curves over ¬elds of characteristic 0 can be deduced from
properties of a torus. For example, the n-torsion on a torus is easily seen to
be isomorphic to Zn • Zn , so we can deduce that this holds for all elliptic
curves over algebraically closed ¬elds of characteristic 0 (see Corollary 9.22).




9.1 Doubly Periodic Functions
Let ω1 , ω2 be complex numbers that are linearly independent over R. Then

L = Zω1 + Zω2 = {n1 ω1 + n2 ω2 | n1 , n2 ∈ Z}

is called a lattice. The main reason we are interested in lattices is that C/L
is a torus, and we want to show that a torus gives us an elliptic curve.
The set
F = {a1 ω1 + a2 ω2 | 0 ¤ ai < 1, i = 1, 2}
(see Figure 9.1) is called a fundamental parallelogram for L. A di¬er-
ent choice of basis ω1 , ω2 for L will of course give a di¬erent fundamental
parallelogram. Since it will occur several times, we denote

ω3 = ω1 + ω2 .

A function on C/L can be regarded as a function f on C such that f (z +
ω) = f (z) for all z ∈ C and all ω ∈ L. We are only interested in meromorphic


257

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



„¦3
„¦1




„¦2
0
Figure 9.1
The Fundamental Parallelogram


functions, so we de¬ne a doubly periodic function to be a meromorphic
function
f :C’C∪∞
such that
f (z + ω) = f (z)
for all z ∈ C and all ω ∈ L. Equivalently,
f (z + ωi ) = f (z), i = 1, 2.
The numbers ω ∈ L are called the periods of f .
If f is a (not identically 0) meromorphic function and w ∈ C, then we can
write
f (z) = ar (z ’ w)r + ar+1 (z ’ w)r+1 + · · · ,
with ar = 0. The integer r can be either positive, negative, or zero. De¬ne
the order and the residue of f at w to be
r = ordw f
a-1 = Resw f.
Therefore, ordw f is the order of vanishing of f at w, or negative the order of
a pole. The order is 0 if and only if the function is ¬nite and nonvanishing at
w. It is not hard to see that if f is doubly periodic, then ordw+ω f = ordw f
and Resw+ω f = Resw f for all ω ∈ L.
A divisor D is a formal sum of points:
D = n1 [w1 ] + n2 [w2 ] + · · · + nk [wk ],
where ni ∈ Z and wi ∈ F . In other words, we have a symbol [w] for each
w ∈ F , and the divisors are linear combinations with integer coe¬cients of
these symbols. The degree of a divisor is

deg(D) = ni .




© 2008 by Taylor & Francis Group, LLC
259
SECTION 9.1 DOUBLY PERIODIC FUNCTIONS

De¬ne the divisor of a function f to be
div(f ) = (ordw f )[w].
w∈F


THEOREM 9.1
Let f be a doubly periodic function for the lattice L and let F be a fundamental
parallelogram for L.
1. If f has no poles, then f is constant.
Resw f = 0.
2. w∈F

3. If f is not identically 0,
deg(div(f )) = ordw f = 0.
w∈F

4. If f is not identically 0,
w · ordw f ∈ L.
w∈F

5. If f is not constant, then f : C ’ C ∪ ∞ is surjective. If n is the sum
of the orders of the poles of f in F and z0 ∈ C, then f (z) = z0 has n
solutions (counting multiplicities).
6. If f has only one pole in F , then this pole cannot be a simple pole.
All of the above sums over w ∈ F have only ¬nitely many nonzero terms.


PROOF Because f is a meromorphic function, it can have only ¬nitely
many zeros and poles in any compact set, for example, the closure of F .
Therefore, the above sums have only ¬nitely many nonzero terms.
If f has no poles, then it is bounded in the closure of F , which is a compact
set. Therefore, f is bounded in all of C. Liouville™s theorem says that a
bounded entire function is constant. This proves (1).
Recall Cauchy™s theorem, which says that

f (z)dz = 2πi Resw f,
‚F w∈F

where ‚F is the boundary of F and the line integral is taken in the coun-
terclockwise direction. Write (assuming ω1 , ω2 are oriented as in Figure 9.1;
otherwise, switch them in the following)

f (z)dz =
‚F
ω2 ω2 +ω1 ω1 0
f (z)dz + f (z)dz + f (z)dz + f (z)dz.
0 ω2 ω1 +ω2 ω1




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

Since f (z + ω1 ) = f (z), we have
ω1 0 ω2
f (z)dz = ’
f (z)dz = f (z)dz.
ω1 +ω2 ω2 0

Similarly,
ω1 +ω2 0
f (z)dz = ’ f (z)dz.
ω2 ω1
Therefore, the sum of the four integrals is 0. There is a small technicality
that we have passed over. The function f is not allowed to have any poles on
the path of integration. If it does, adjust the path with a small detour around
such points as in Figure 9.2. The integrals cancel, just as in the above. This
proves (2).



„¦3
„¦1




„¦2
0
Figure 9.2




Suppose r = ordw f . Then f (z) = (z ’ w)r g(z), where g(w) is ¬nite and
nonzero. Then
r g (z)
f (z)
= + ,
z’w
f (z) g(z)
so
f
Resw = r.
f
If f is doubly periodic, then f is doubly periodic. Therefore, (2) applied to
f /f yields
f
2πi ordw f = 2πi Resw = 0.
f
w∈F w∈F

This proves (3).
For (4), we have
f f
w · ordw f = 2πi
2πi Resw z = z dz.
f f
‚F
w∈F w∈F




© 2008 by Taylor & Francis Group, LLC
261
SECTION 9.1 DOUBLY PERIODIC FUNCTIONS

However, in this case, the function zf /f is not doubly periodic. The integral
may be written as a sum of four integrals, as in the proof of (2). The double
periodicity of f and f yield
ω1 0
f (z) f (z)
z dz = (z + ω1 ) dz
ω1 +ω2 f (z) f (z)
ω2
ω2 ω2
f (z) f (z)
=’ dz ’ ω1
z dz.
f (z) f (z)
0 0

But ω2
1 f (z)
dz
2πi f (z)
0

is the winding number around 0 of the path

0 ¤ t ¤ 1.
z = f (tω2 ),

Since f (0) = f (ω2 ), this is a closed path. The winding number is an integer,
so
ω2 ω1
f (z) f (z)
z dz + z dz
f (z) ω1 +ω2 f (z)
0
ω2
f (z)
= ’ω1 dz ∈ 2πiZω1 .
f (z)
0

Similarly,
ω1 +ω2 0
f (z) f (z)
dz ∈ 2πiZω2 .
z dz + z
f (z) f (z)
ω2 ω1

Therefore,
w · ordw f ∈ 2πiL.
2πi
w∈F

This proves (4).
To prove (5), let z0 ∈ C. Then h(z) = f (z) ’ z0 is a doubly periodic
function whose poles are the same as the poles of f . By (3), the number
of zeros of h(z) in F (counting multiplicities) equals the number of poles
(counting multiplicities) of h, which is n. This proves (5).
For (6), suppose f has only a simple pole, say at w, and no others. Then
Resw f = 0 (otherwise, the pole doesn™t exist). The sum in (2) has only one
term, and it is nonzero. This is impossible, so we conclude that either the
pole cannot be simple or there must be other poles.

REMARK 9.2 As we saw in the proof of (5), part (3) says that the number
of zeros of a doubly periodic function equals the number of poles. This is a
general fact for compact Riemann surfaces (such as a torus) and for projective
algebraic curves (see [42, Ch. 8, Prop. 1] or [49, II, Cor. 6.10]).




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

If (6) were false for a function f , then f would give a bijective (by (5)) map
from the torus to the Riemann sphere (= C ∪ ∞). This is impossible for many
topological reasons (the torus has a hole but the sphere doesn™t).

So far, we do not have any examples of nonconstant doubly periodic func-
tions. This situation is remedied by the Weierstrass „˜-function.


THEOREM 9.3
Given a lattice L, de¬ne the Weierstrass „˜-function by
1 1 1
’2
„˜(z) = „˜(z; L) = + . (9.1)
(z ’ ω)2
z2 ω
ω∈L
ω=0


Then
1. The sum de¬ning „˜(z) converges absolutely and uniformly on compact
sets not containing elements of L.
2. „˜(z) is meromorphic in C and has a double pole at each ω ∈ L.
3. „˜(’z) = „˜(z) for all z ∈ C.
4. „˜(z + ω) = „˜(z) for all ω ∈ L.
5. The set of doubly periodic functions for L is C(„˜, „˜ ). In other words,
every doubly periodic function is a rational function of „˜ and its deriva-
tive „˜ .

PROOF Let C be a compact set, and let M = Max{|z| | z ∈ C}. If z ∈ C
and |ω| ≥ 2M , then |z ’ ω| ≥ |ω|/2 and |2ω ’ z| ¤ 5|ω|/2, so

z(2ω ’ z)
1
1
’2=
(z ’ ω)2 (z ’ ω)2 ω 2
ω
(9.2)
10M
M (5|ω|/2)
¤ = .
|ω|4 /4 |ω|3

The preceding calculation explains why the terms 1/ω 2 are included. With-
out them, the terms in the sum would be comparable to 1/ω 2 . Subtracting
this 1/ω 2 makes the terms comparable to 1/ω 3 . This causes the sum to con-
verge, as the following lemma shows.


LEMMA 9.4
If k > 2 then
1
|ω|k
ω∈L
ω=0




© 2008 by Taylor & Francis Group, LLC
263
SECTION 9.1 DOUBLY PERIODIC FUNCTIONS

converges.

PROOF Let F be a fundamental parallelogram for L and let D be the
length of the longer diagonal of F . Then |z| ¤ D for all z ∈ F . Let ω =
m1 ω1 + m2 ω2 ∈ L with |ω| ≥ 2D. If x1 , x2 are real numbers with mi ¤ xi <
mi + 1, then ω and x1 ω1 + x2 ω2 di¬er by an element of F , so
1
|m1 ω1 + m2 ω2 | ≥ |x1 ω1 + x2 ω2 | ’ D ≥ |x1 ω1 + x2 ω2 | ’ |m1 ω1 + m2 ω2 |,
2
since |m1 ω1 + m2 ω2 | ≥ 2D. Therefore,
2
|m1 ω1 + m2 ω2 | ≥ |x1 ω1 + x2 ω2 |.
3
Similarly,
|x1 ω1 + x2 ω2 | ≥ D.
Comparing the sum to an integral yields

(3/2)k
1
¤ (1/area of F ) dx1 dx2 .
|ω|k |x1 ω1 + x2 ω2 |k
|ω|≥2D |x1 ω1 +x2 ω2 |≥D

The change of variables de¬ned by u + iv = x1 ω1 + x2 ω2 changes the integral
to


1 1
r dr dθ < ∞,
C du dv = C k
(u2 + v 2 )k/2 θ=0 r=D r
|u+iv|≥D

where C = (3/2)k /(area of F ). Therefore, the sum for |ω| ≥ 2D converges.
Since there are only ¬nitely many ω with |ω| < 2D, we have shown that the
sum converges.

Lemma 9.4 and Equation 9.2 imply that the sum of the terms in Equa-
tion 9.1 with |ω| ≥ 2M converges absolutely and uniformly for z ∈ C. Since
only ¬nitely many terms have been omitted, we obtain (1). Since a uniform
limit of analytic functions is analytic, „˜(z) is analytic for z ∈ L. If z ∈ L,
then the sum of the terms for ω = z is analytic near z, so the term 1/(z ’ ω)2
causes „˜ to have a double pole at z. This proves (2).
To prove (3), note that ω ∈ L if and only if ’ω ∈ L. Therefore, in the sum
for „˜(’z), we can take the sum over ’ω ∈ L. The terms of this sum are of
the form
1 1 1
1
’ ’ 2.
=
(z ’ ω)
2 2 2
(’z + ω) (’ω) ω
Therefore the sum for „˜(’z) equals the sum for „˜(z).
The proof of (4) would be easy if we could ignore the terms 1/ω 2 , since
changing z to z + ω would simply shift the summands. However, these terms




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

are needed for convergence. With some care, one could justify rearranging the
sum, but it is easier to do the following. Di¬erentiating „˜(z) term by term
yields
1
„˜ (z) = ’2 .
(z ’ ω)3
ω∈L

Note that ω = 0 is included in the sum. This sum converges absolutely (by
comparison with the case k = 3 in Lemma 9.4) when z ∈ L, and changing z
to z + ω shifts the terms in the sum. Therefore,

„˜ (z + ω) = „˜ (z).

This implies that there is a constant cω such that

„˜(z + ω) ’ „˜(z) = cω ,

for all z ∈ L. Setting z = ω/2 yields

cω = „˜(’ω/2) ’ „˜(ω/2) = 0,

by (3). Therefore „˜(z + ω) = „˜(z). This proves (4).
Let f (z) be any doubly periodic function. Then

f (z) + f (’z) f (z) ’ f (’z)
f (z) = +
2 2
expresses f (z) as the sum of an even function and an odd function. Therefore,
it su¬ces to prove (5) for even functions and for odd functions. Since „˜(’z) =
„˜(z), it follows that „˜ (’z) = ’„˜ (z), so „˜ (z) is an odd function. If f (z)
is odd, then f (z)/„˜ (z) is even. Therefore, it su¬ces to show that an even
doubly periodic function is a rational function of „˜(z).
Let f (z) be an even doubly periodic function. We may assume that f is
not identically zero; otherwise, we™re done. By changing f , if necessary, to
af + b
cf + d

for suitable a, b, c, d with ad ’ bc = 0, we may arrange that f (z) does not have
a zero or a pole whenever 2z ∈ L (this means that we want f (0) = 0, ∞ and
f (ωi /2) = 0 for i = 1, 2, 3). If we prove (af + b)/(cf + d) is a rational function
of „˜, then we can solve for f and obtain the result for f .
Since f (z) is even and doubly periodic, f (ω3 ’ z) = f (z), so

ordw f = ordω3 ’w f.

We can therefore put the ¬nitely many elements in F where f (z) = 0 or
where f (z) has a pole into pairs (w, ω3 ’ w). Since we have arranged that
w = ω3 /2, the two elements of each pair are distinct. There is a slight




© 2008 by Taylor & Francis Group, LLC
265
SECTION 9.1 DOUBLY PERIODIC FUNCTIONS

problem if w lies on a side of F . Suppose w = xω1 with 0 < x < 1. Then
ω3 ’ w = (1 ’ x)ω1 + ω2 ∈ F . In this case, we translate by ω2 to get
(1 ’ x)ω1 ∈ F . Since w = ω1 /2, we have x = 1/2, hence xω1 = (1 ’ x)ω1 , and
again the two elements of the pair are distinct. The case w = xω2 is handled
similarly.
For a ¬xed w, the function „˜(z) ’ „˜(w) has zeros at z = w and z = ω3 ’ w.
By Theorem 9.1(5), these are the only two zeros in F and they are simple
zeros. Therefore, the function
ordw f
(„˜(z) ’ „˜(w))
h(z) =
(w, ω3 ’w)


(the product is over pairs (w, ω3 ’ w)) has a zero of order ordw f at w and
at ω3 ’ w when ordw f > 0 and has a pole of the same order as f when
ordw f = 0 by Theorem 9.1, the poles at z ∈ L of the
ordw f < 0. Since
factors in the product cancel. Therefore, f (z)/h(z) has no zeros or poles in F .
By Theorem 9.1(1), f (z)/h(z) is constant. Since h(z) is a rational function
of „˜(z), so is f (z). This completes the proof of Theorem 9.3.

In order to construct functions with prescribed properties, it is convenient
to introduce the Weierstrass σ-function. It is not doubly periodic, but it
satis¬es a simple transformation law for translation by elements of L.


PROPOSITION 9.5
Let
z (z/w)+ 1 (z/w)2
1’ e .
σ(z) = σ(z; L) = z 2
ω
ω∈L
ω=0


Then

1. σ(z) is analytic for all z ∈ C

2. σ(z) has simple zeros at each ω ∈ L and has no other zeros
d2
log σ(z) = ’„˜(z)
3. dz 2

4. given ω ∈ L, there exist a = aω and b = bω such that

σ(z + ω) = eaz+b σ(z)

for all z ∈ C.


PROOF The exponential factor is included to make the product converge.
A short calculation yields the power series expansion
2
1
(1 ’ u)eu+ 2 u = 1 + c3 u3 + c4 u4 + · · · .




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

Therefore, there is a constant C such that
2
1
|(1 ’ u)eu+ 2 u ’ 1| ¤ C|u|3

for u near 0. In particular, this inequality holds when u = z/ω for |ω| su¬-
|an | converges,
ciently large and z in a compact set. Recall that if a sum
then the product (1+an ) converges. Moreover, if (1+an ) = 0 for all n, then
|z/ω|3 converges by Lemma 9.4 with k = 3,
the product is nonzero. Since
the product de¬ning σ(z) converges uniformly on compact sets. Therefore,
σ(z) is analytic. This proves (1). Part (2) follows since the product of the
factors, omitting one ω, is nonzero at z = ω.
To prove (3), di¬erentiate the logarithm of the product for σ(z) to obtain
1 1 1 z
d
log σ(z) = + + +2 .
z’ω ω ω
dz z ω∈L
ω=0


Taking one more derivative yields the sum for ’„˜(z). This proves (3).
Let ω ∈ L. Since
d2 σ(z + ω)
log = 0,
2
dz σ(z)
there are constants a = aω and b = bω such that
σ(z + ω)
= az + b.
log
σ(z)
Exponentiating yields (4). We can restrict z in the above to lie in a small re-
gion in order to avoid potential complications with branches of the logarithm.
Then (4) holds in this small region, and therefore for all z ∈ C, by uniqueness
of analytic continuation.

We can now state exactly when a divisor is a divisor of a function. The
following is a special case of what is known as the Abel-Jacobi theorem,
which states when a divisor on a Riemann surface, or on an algebraic curve,
is the divisor of a function.

THEOREM 9.6
ni [wi ] be a divisor. Then D is the divisor of a function if and
Let D =
ni wi ∈ L.
only if deg(D) = 0 and

PROOF Parts (3) and (4) of Theorem 9.1 are precisely the statements
ni wi ∈ L.
that if D is the divisor of a function then deg(D) = 0 and
ni wi = ∈ L. Let
Conversely, suppose deg(D) = 0 and
σ(z)
σ(z ’ wi )ni .
f (z) =
σ(z ’ ) i




© 2008 by Taylor & Francis Group, LLC
267
SECTION 9.2 TORI ARE ELLIPTIC CURVES

If ω ∈ L, then

f (z + ω)
= eaω z+bω ’aω (z’ )’bω ni (aω (z’wi )+bω )
e = 1,
f (z)

since ni = 0 and ni wi = . Therefore, f (z) is doubly periodic. The
divisor of f is easily seen to be D, so D is the divisor of a function.

Doubly periodic functions can be regarded as functions on the torus C/L,
and divisors can be regarded as divisors for C/L. If we let C(L)— denote the
doubly periodic functions that do not vanish identically and let Div0 (C/L)
denote the divisors of degree 0, then much of the preceding discussion can be
expressed by the exactness of the sequence

div sum
0 ’’ C— ’’ C(L)— ’’ Div0 (C/L) ’’ C/L ’’ 0. (9.3)

The “sum” function adds up the complex numbers representing the points in
the divisor mod L. The exactness at C(L)— expresses the fact that a function
with no zeros and no poles, hence whose divisor is 0, is a constant. The
exactness at Div0 (C/L) is Theorem 9.6. The surjectivity of the sum function
is easy. If w ∈ C, then sum([w] ’ [0]) = w mod L.




9.2 Tori are Elliptic Curves
The goal of this section is to show that a complex torus C/L is naturally
isomorphic to the complex points on an elliptic curve.
Let L be a lattice, as in the previous section. For integers k ≥ 3, de¬ne the
Eisenstein series

ω ’k .
Gk = Gk (L) = (9.4)
ω∈L
ω=0



By Lemma 9.4, the sum converges. When k is odd, the terms for ω and ’ω
cancel, so Gk = 0.


PROPOSITION 9.7
For 0 < |z| < Min 0=ω∈L (|ω|),

1
(2j + 1)G2j+2 z 2j .
„˜(z) = 2 +
z j=1




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

When |z| < |ω|,
PROOF

1 1 1
’ 2 = ω ’2 ’1
(z ’ ω)2 (1 ’ (z/ω))2
ω

zn
’2
=ω (n + 1) .
ωn
n=1

Therefore,

zn
1
„˜(z) = 2 + (n + 1) n+2 .
z ω
n=1
ω=0

Summing over ω ¬rst, then over n, yields the result.


THEOREM 9.8
Let „˜(z) be the Weierstrass „˜-function for a lattice L. Then

„˜ (z)2 = 4„˜(z)3 ’ 60G4 „˜(z) ’ 140G6 .


PROOF From Proposition 9.7,

„˜(z) = z ’2 + 3G4 z 2 + 5G6 z 4 + · · ·
„˜ (z) = ’2z ’3 + 6G4 z + 20G6 z 3 + · · · .

Cubing and squaring these two relations yields

„˜(z)3 = z ’6 + 9G4 z ’2 + 15G6 + · · ·
„˜ (z)2 = 4z ’6 ’ 24G4 z ’2 ’ 80G6 + · · · .

Therefore,

f (z) = „˜ (z)2 ’ 4„˜(z)3 + 60G4 „˜(z) + 140G6 = c1 z + c2 z 2 + · · ·

is a power series with no constant term and with no negative powers of z.
But the only possible poles of f (z) are at the poles of „˜(z) and „˜ (z), namely,
the elements of L. Since f (z) is doubly periodic and, as we have just shown,
has no pole at 0, f (z) has no poles. By Theorem 9.1, f (z) is constant. Since
the power series for f (z) has no constant term, f (0) = 0. Therefore, f (z) is
identically 0.

It is customary to set

g2 = 60G4
g3 = 140G6 .




© 2008 by Taylor & Francis Group, LLC
269
SECTION 9.2 TORI ARE ELLIPTIC CURVES

The theorem then says that

„˜ (z)2 = 4„˜(z)3 ’ g2 „˜(z) ’ g3 . (9.5)

Therefore, the points („˜(z), „˜ (z)) lie on the curve

y 2 = 4x3 ’ g2 x ’ g3 .

It is traditional to leave the 4 as the coe¬cient of x3 , rather than performing a
change of variables to make the coe¬cient of x3 equal to 1. The discriminant
of the cubic polynomial is 16(g2 ’ 27g3 ).
3 2



PROPOSITION 9.9
∆ = g2 ’ 27g3 = 0.
3 2



PROOF Since „˜ (z) is doubly periodic, „˜ (ωi /2) = „˜ (’ωi /2). Since
„˜ (’z) = ’„˜ (z), it follows that

„˜ (ωi /2) = 0, i = 1, 2, 3. (9.6)

Therefore, each „˜(ωi /2) is a root of 4x3 ’ g2 x ’ g3 , by (9.5). If we can show
that these roots are distinct, then the cubic polynomial has three distinct
roots, which means that its discriminant is nonzero. Let

hi (z) = „˜(z) ’ „˜(ωi /2).

Then hi (ωi /2) = 0 = hi (ωi /2), so hi vanishes to order at least 2 at ωi /2. Since
hi (z) has only one pole in F , namely the double pole at z = 0, Theorem 9.1(5)
implies that ωi /2 is the only zero of hi (z). In particular,

hi (ωj /2) = 0, when j = i.

Therefore, the values „˜(ωi /2) are distinct.

The proposition implies that

E : y 2 = 4x3 ’ g2 x ’ g3

is the equation of an elliptic curve, so we have a map from z ∈ C to the
points with complex coordinates („˜(z), „˜ (z)) on an elliptic curve. Since „˜(z)
and „˜ (z) depend only on z mod L (that is, if we change z by an element of
L, the values of the functions do not change), we have a function from C/L
to E(C). The group C/L is a group, with the group law being addition of
complex numbers mod L. In concrete terms, we can regard elements of C/L
as elements of F . When we add two points, we move the result back into F by
subtracting a suitable element of L. For example, (.7ω1 + .8ω2 ) + (.4ω1 + .9ω2 )
yields .1ω1 + .7ω2 .




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

THEOREM 9.10
Let L be a lattice and let E be the elliptic curve y 2 = 4x3 ’ g2 x ’ g3 . The
map

¦ : C/L ’’ E(C)
z ’’ („˜(z), „˜ (z))
0 ’’ ∞

is an isomorphism of groups.

PROOF The surjectivity is easy. Let (x, y) ∈ E(C). Since the function
„˜(z) ’ x has a double pole, Theorem 9.1 implies that it has zeros, so there
exists z ∈ C such that „˜(z) = x. Theorem 9.8 implies that

„˜ (z)2 = y 2 ,

so „˜ (z) = ±y. If „˜ (z) = y, we™re done. If „˜ (z) = ’y, then „˜ (’z) = y and
„˜(’z) = x, so ’z ’ (x, y).
Suppose „˜(z1 ) = „˜(z2 ) and „˜ (z1 ) = „˜ (z2 ), and z1 ≡ z2 mod L. The only
poles of „˜(z) are for z ∈ L. Therefore, if z1 is a pole of „˜, then z1 ∈ L and
z2 ∈ L, so z1 ≡ z2 mod L. Now assume z1 is not a pole of „˜, so z1 is not in
L. The function
h(z) = „˜(z) ’ „˜(z1 )
has a double pole at z = 0 and no other poles in F . By Theorem 9.1, it has
exactly two zeros. Suppose z1 = ωi /2 for some i. From Equation 9.6, we
know that „˜ (ωi /2) = 0, so z1 is a double root of h(z), and hence is the only
root. Therefore z2 = z1 . Finally, suppose z1 is not of the form ωi /2. Since
h(’z1 ) = h(z1 ) = 0, and since z1 ≡ ’z1 mod L, the two zeros of h are z1
and ’z1 mod L. Therefore, z2 ≡ ’z1 mod L. But

y = „˜ (z2 ) = „˜ (’z1 ) = ’„˜ (z1 ) = ’y.

This means that „˜ (z1 ) = y = 0. But „˜ (z) has only a triple pole, so has only
three zeros in F . From Equation 9.6, we know that these zeros occur at ωi /2.
This is a contradiction, since z = ωi /2. Therefore, z1 ≡ z2 mod L, so ¦ is
injective.
Finally, we need to show that ¦ is a group homomorphism. Let z1 , z2 ∈ C
and let
¦(zi ) = Pi = (xi , yi ).
Assume that both P1 , P2 are ¬nite and that the line through P1 , P2 intersects
E in three distinct ¬nite points (this means that P1 = ±P2 , that 2P1 +P2 = ∞,
and that P1 + 2P2 = ∞). For a ¬xed z1 , this excludes ¬nitely many values of
z2 . There are two reasons for these exclusions. The ¬rst is that the addition
law on E has a di¬erent formula when the points are equal. The second is




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SECTION 9.2 TORI ARE ELLIPTIC CURVES

that we do not need to worry about the connection between double roots
in the algebraic calculations and double roots of the corresponding analytic
functions.
Let y = ax + b be the line through P1 , P2 . Let P3 = (x3 , y3 ) be the third
point of intersection of this line with E and let (x3 , y3 ) = P3 = ¦(z3 ) with
z3 ∈ C. The formulas for the group law on E show that
2
y2 ’ y 1
1
’ x1 ’ x2
x3 =
x2 ’ x1
4
2
„˜ (z2 ) ’ „˜ (z1 )
1
’ „˜(z1 ) ’ „˜(z2 ).
=
„˜(z2 ) ’ „˜(z1 )
4

The function
(z) = „˜ (z) ’ a„˜(z) ’ b
has zeros at z = z1 , z2 , z3 . Since (z) has a triple pole at 0, and no other
poles, it has three zeros in F . Therefore,

div( ) = [z1 ] + [z2 ] + [z3 ] ’ 3[0].

By Theorem 9.1(4), z1 + z2 + z3 ∈ L. Therefore,

„˜(z1 + z2 ) = „˜(’z3 ) = „˜(z3 ) = x3 .

We obtain
2
„˜ (z2 ) ’ „˜ (z1 )
1
’ „˜(z1 ) ’ „˜(z2 ).
„˜(z1 + z2 ) = (9.7)
„˜(z2 ) ’ „˜(z1 )
4

By continuity, this formula, which we proved with certain values of the zi
excluded, now holds for all zi for which it is de¬ned.
We now need to consider the y-coordinate. This means that we need to
compute „˜ (z1 + z2 ). We sketch the method (the interested and careful reader
may check the details). Di¬erentiating (9.7) with respect to z2 yields an
expression for „˜ (z1 + z2 ) in terms of x1 , x2 , y1 , y2 , and „˜ (z2 ). We need to
express „˜ in terms of „˜ and „˜ . Di¬erentiating (9.5) yields

2„˜ „˜ = (12„˜2 ’ g2 )„˜ .

Dividing by „˜ (z) (this is all right if „˜ (z) = 0; the other cases are ¬lled in by
continuity) yields

2„˜ (z2 ) = 12„˜(z2 )2 ’ g2 . (9.8)

Substituting this into the expression obtained for „˜ (z1 + z2 ) yields an expres-
sion for „˜ (z1 + z2 ) in terms of „˜(z1 ), „˜ (z1 ), „˜(z2 ), „˜ (z2 ). Some algebraic




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

manipulation shows that this equals the value for ’y3 obtained from the ad-
dition law for (x1 , y1 ) + (x2 , y2 ) = (x3 , ’y3 ). Therefore,

(„˜(z1 ), „˜ (z1 )) + („˜(z2 ), „˜ (z2 )) = („˜(z1 + z2 ), „˜ (z1 + z2 )).

This is exactly the statement that

¦(z1 ) + ¦(z2 ) = ¦(z1 + z2 ). (9.9)

It remains to check (9.9) in the cases where (9.7) is not de¬ned. The
cases where „˜(zi ) = ∞ and where z1 ≡ ’z2 mod L are easily checked. The
remaining case is when z1 = z2 . Let z2 ’ z1 in (9.7), use l™Hˆpital™s rule, and
o
use (9.8) to obtain
2
1 „˜ (z1 )
’ 2„˜(z1 )
„˜(2z1 ) =
4 „˜ (z1 )
2
6„˜(z1 )2 ’ 1 g2
1
’ 2„˜(z1 )
2
= (9.10)
4 „˜ (z1 )
2
6x2 ’ 1 g2
1 1
’ 2x1 .
2
=
4 y1
This is the formula for the coordinate x3 that is obtained from the addition
law on E. Di¬erentiating with respect to z1 yields the correct formula for the
y-coordinate, as above. Therefore,

¦(z1 ) + ¦(z1 ) = ¦(2z1 ).

This completes the proof of the theorem.

The theorem shows that the natural group law on the torus C/L matches
the group law on the elliptic curve, which perhaps looks a little unnatural.
Also, the classical formulas (9.7) and (9.10) for the Weierstrass „˜-function,
which look rather complicated, are now seen to be expressing the group law
for E.




9.3 Elliptic Curves over C
In the preceding section, we showed that a torus yields an elliptic curve. In
the present section, we™ll show the converse, namely, that every elliptic curve
over C comes from a torus.
Let L = Zω1 + Zω2 be a lattice and let

„ = ω1 /ω2 .




© 2008 by Taylor & Francis Group, LLC
273
SECTION 9.3 ELLIPTIC CURVES OVER C

Since ω1 and ω2 are linearly independent over R, the number „ cannot be
real. By switching ω1 and ω2 if necessary, we may assume that the imaginary
part of „ is positive:
(„ ) > 0.

In other words, we assume „ lies in the upper half plane

H = {x + iy ∈ C | y > 0}.

The lattice
L„ = Z„ + Z

is homothetic to L. This means that there exists a nonzero complex number
» such that L = »L„ . In our case, » = ω2 .
For integers k ≥ 3, de¬ne

1
Gk („ ) = Gk (L„ ) = . (9.11)
(m„ + n)k
(m,n)=(0,0)


We have
k
Gk („ ) = ω2 Gk (L),

where Gk (L) is the Eisenstein series de¬ned for L = Zω1 + Zω2 by (9.4). Let

q = e2πi„ .

It will be useful to express certain functions as sums of powers of q. If „ =
x + iy with y > 0, then |q| = e’2πy < 1. This implies that the expressions we
obtain will converge.


PROPOSITION 9.11

Let ζ(x) = n=1 n’x and let

σ (n) = d
d|n


be the sum of the th powers of the positive divisors of n. If k ≥ 2 is an
integer, then

(2πi)2k
σ2k’1 (n)q n
G2k („ ) = 2ζ(2k) + 2
(2k ’ 1)! n=1

(2πi)2k j 2k’1 q j
= 2ζ(2k) + 2 .
(2k ’ 1)! 1 ’ qj
j=1




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

PROOF We have
eπi„ + e’πi„
cos π„
π = πi πi„
e ’ e’πi„
sin π„
q+1 2πi
= πi = πi +
q’1 q’1

= πi ’ 2πi qj . (9.12)
j=0

Recall the product expansion

„ „
1’ 1+
sin π„ = π„
n n
n=1

(see [4]). Taking the logarithmic derivative yields

1 1 1
cos π„
=+ + . (9.13)
π
„ ’n „ +n
sin π„ „ n=1

Di¬erentiating (9.12) and (9.13) 2k ’ 1 times with respect to „ yields
∞ ∞
1
’ (2k ’ 1)!
2k 2k’1 j 2k’1
(2πi) j q = (’1) .
(„ + n)2k
n=’∞
j=1

Consider (9.11) with 2k in place of k. Since 2k is even, the terms for (m, n)
and (’m, ’n) are equal, so we only need to sum for m = 0, n > 0 and for
m > 0, n ∈ Z, then double the answer. We obtain
∞ ∞ ∞
1 1
G2k („ ) = 2 +2
2k (m„ + n)2k
n m=1 n=’∞
n=1
∞ ∞
(2πi)2k j 2k’1 mj
= 2ζ(2k) + 2 q
(2k ’ 1)!
m=1 j=1
∞ ∞
(2πi)2k
j 2k’1 q mj .
= 2ζ(2k) + 2
(2k ’ 1)! m=1 j=1

Let n = mj in the last expression. Then, for a given n, the sum over j can
be regarded as the sum over the positive divisors of n. This yields the ¬rst
expression in the statement of the proposition. The expansion m≥1 q mj =
q j /(1 ’ q j ) yields the second expression.

Recall that we de¬ned g2 = g2 (L) = 60G4 (L) and g3 = g3 (L) = 140G6 (L)
for arbitrary lattices L. Restricting to L„ , we de¬ne

g2 („ ) = g2 (L„ ), g3 („ ) = g3 (L„ ).




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SECTION 9.3 ELLIPTIC CURVES OVER C

Using the facts that

π4 π6
ζ(4) = and ζ(6) = ,
90 945
we obtain
⎛ ⎞

4π 4 4π 4 ⎝ j 3 qj ⎠
(1 + 240q + · · · ) =
g2 („ ) = 1 + 240
1 ’ qj
3 3 j=1
⎛ ⎞

8π 6 8π 6 ⎝ j 5 qj ⎠
(1 ’ 504q + · · · ) = 1 ’ 504
g3 („ ) = .
1 ’ qj
27 27 j=1


Since ∆ = g2 ’ 27g3 , a straightforward calculation shows that
3 2


∆(„ ) = (2π)12 (q + · · · ).

De¬ne
3
g2
j(„ ) = 1728 .

+ · · · . Including a few more terms in the above calculations
1
Then j(„ ) = q
yields
1
+ 744 + 196884q + 21493760q 2 + · · · .
j(„ ) =
q
For computational purposes, this series converges slowly since the coe¬cients
are large. It is usually better to use the following.


PROPOSITION 9.12

3
j 3 qj

1 + 240 j=1 1’q j
j(„ ) = 1728 2.
3
∞ ∞
j 3 qj j 5 qj
’ 1 ’ 504
1 + 240 j=1 1’q j j=1 1’q j



PROOF Substitute the above expressions for g2 , g3 into the de¬nition of
the j-function. The powers of π and other constants cancel to yield the present
expression.

It can be shown (see [70, p. 249]) that

(1 ’ q k )24 .
12
∆ = (2π) q
k=1




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

This yields the expression
3
j 3 qj

1 + 240
1 ’ qj
j=1
j= ,

q k=1 (1 ’ q k )24
which also works very well for computing j.
More generally, if L is a lattice, de¬ne
g2 (L)3
j(L) = 1728 .
g2 (L)3 ’ 27g3 (L)2
If » ∈ C— , then the de¬nitions of G4 and G6 easily imply that
g2 (»L) = »’4 g2 (L) g3 (»L) = »’6 g3 (L).
and (9.14)
Therefore
j(L) = j(»L).
’1
Letting L = Zω1 + Zω2 and » = ω2 , we obtain
j(Zω1 + Zω2 ) = j(„ ),
where „ = ω1 /ω2 .
Recall that
ab
a, b, c, d ∈ Z, ad ’ bc = 1
SL2 (Z) =
cd
acts on the upper half plane H by
a„ + b
ab
„=
cd c„ + d
for all „ ∈ H.


PROPOSITION 9.13
ab
Let „ ∈ H and let ∈ SL2 (Z). Then
cd
a„ + b
j = j(„ ).
c„ + d

PROOF We ¬rst compute what happens with Gk :
a„ + b 1
Gk =
(m a„+d + n)k
+b
c„ + d c„
(m,n)=(0,0)
1
= (c„ + d)k
(m(a„ + b) + n(c„ + d))k
(m,n)=(0,0)
1
= (c„ + d)k .
((ma + nc)„ + (mb + nd))k
(m,n)=(0,0)




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SECTION 9.3 ELLIPTIC CURVES OVER C

ab
Since has determinant 1, we have
cd
’1
d ’b
ab
= .
’c a
cd

Let
ab
(m , n ) = (m, n) = (ma + nc, mb + nd).
cd
Then
d ’b
(m, n) = (m , n ) ,
’c a
so there is a one-to-one correspondence between pairs of integers (m, n) and
pairs of integers (m , n ). Therefore,

a„ + b 1
= (c„ + d)k
Gk
(m „ + n )k
c„ + d
(m ,n )=(0,0)

= (c„ + d)k Gk („ ).

Since g2 and g3 are multiples of G4 and G6 , we have
a„ + b a„ + b
= (c„ + d)4 g2 („ ), = (c„ + d)6 g3 („ ).
g2 g3
c„ + d c„ + d
Therefore, when we substitute these expressions into the de¬nition of j, all
the factors (c„ + d) cancel.

Let F be the subset of z ∈ H such that
1 1 π π
|z| ≥ 1, ’ ¤ z = eiθ for
(z) < , <θ< .
2 2 3 2
Figure 9.3 is a picture of F . Since we will need to refer to it several times, we
let
ρ = e2πi/3 .


PROPOSITION 9.14
Given „ ∈ H, there exists

ab
∈ SL2 (Z)
cd

such that
a„ + b
= z ∈ F.
c„ + d
Moreover, z ∈ F is uniquely determined by „ .




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




i
Ρ



Figure 9.3
The Fundamental Domain for SL2 (Z)


The proposition says that F is a fundamental domain for the action of
SL2 (Z) on H. For a proof of the proposition, see [104] or [108].


COROLLARY 9.15
Let L ‚ C be a lattice. There exists a basis {ω1 , ω2 } of L with ω1 /ω2 ∈ F.
In other words,
L = (»)(Z„ + Z)
for some » ∈ C— and some uniquely determined „ ∈ F.


PROOF Let {±, β} be a basis for L and let „0 = ±/β. By changing the
sign of ± if necessary, we may assume that „0 ∈ H. Let

ab
∈ SL2 (Z)
cd

be such that
a„0 + b
= „ ∈ F.
c„0 + d
Let
ω1 = a± + bβ, ω2 = c± + dβ.
Since the matrix is in SL2 (Z),

L = Z± + Zβ = Zω1 + Zω2 = ω2 (Z„ + Z).

This proves the corollary.




© 2008 by Taylor & Francis Group, LLC
279
SECTION 9.3 ELLIPTIC CURVES OVER C

If z ∈ C, recall that ordz f is the order of f at z. That is,

f („ ) = („ ’ z)ordz (f ) g(„ ),

with g(z) = 0, ∞. We can also de¬ne the order of f at i∞. Suppose

f („ ) = an q n + an+1 q n+1 + · · · , (9.15)

with n ∈ Z and an = 0, and assume that this series converges for all q close
to 0 (with q = 0 when n < 0). Then

ordi∞ (f ) = n.

Note that q ’ 0 as „ ’ i∞, so ordi∞ (f ) expresses whether f vanishes (n > 0)
or blows up (n < 0) as „ ’ i∞.

PROPOSITION 9.16
Let f be a function meromorphic in H such that f is not identically zero and
such that
a„ + b ab
∈ SL2 (Z).
= f („ ) for all
f
cd
c„ + d
Then
1 1
ordi∞ (f ) + ordρ (f ) + ordi (f ) + ordz (f ) = 0.
3 2
z=i,ρ,i∞



REMARK 9.17 The function f can be regarded as a function on the

. 1
( 3)



>>