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

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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 .

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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

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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

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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]).

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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

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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

∞

2π

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

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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

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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 + · · · .

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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

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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

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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 .

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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 .

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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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271

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

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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

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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„ ).

© 2008 by Taylor & Francis Group, LLC

275

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

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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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277

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 „ .

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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