CA3. Elliptic Curves over Complex Plane- Ryan Lok-Wing Pang

Elliptic Curves over C

Ryan Lok-Wing Pang

Department of MathematicsHong Kong University of Science and Technology

Topics in Complex Analysis, 2014

Ryan Lok-Wing Pang (HKUST) Elliptic Curves over CTopics in Complex Analysis, 2014 1 /

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Outline

1 Introduction to Elliptic Curves

2 The Group Law

3 Elliptic Curves over C

4 Application to Number Theory


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Introduction to Elliptic Cueves

An elliptic curve over a field F is a curve given by the equation

E : y2 + a1xy + a3 = x3 + a2x2 + a4x + a6,

where ai ∈ F satisfying one extra condition that the curve isnon-singular. i.e. The partial derivatives cannot both vanishes at thesame point. This means a1y = 3x3 + 2a2 + a4, 2y + a1x = 0 cannotbe solved simultaneously over the algebraic closure of F.

In the case char(F) 6= 2, 3, we may make a change of variables andwrite:

E : y2 = x3 + Ax + B,A,B ∈ F.

The non singular condition is equivalent to the cubic on the righthaving three distinct roots. From now on, we assume thatchar(F) 6= 2, 3. Then the nonsingularity condition becomes4A3 + 27B2 6= 0.


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Introduction to Elliptic Cueves

An elliptic curve over a field F is a curve given by the equation

E : y2 + a1xy + a3 = x3 + a2x2 + a4x + a6,

where ai ∈ F satisfying one extra condition that the curve isnon-singular. i.e. The partial derivatives cannot both vanishes at thesame point. This means a1y = 3x3 + 2a2 + a4, 2y + a1x = 0 cannotbe solved simultaneously over the algebraic closure of F.

In the case char(F) 6= 2, 3, we may make a change of variables andwrite:

E : y2 = x3 + Ax + B,A,B ∈ F.

The non singular condition is equivalent to the cubic on the righthaving three distinct roots. From now on, we assume thatchar(F) 6= 2, 3. Then the nonsingularity condition becomes4A3 + 27B2 6= 0.


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The Group Law

We let E (F) = {(x , y) ∈ F2|y2 = x3 + Ax + B} ∪ {O}, here the pointO is the point at ∞.

Elliptic Curves have a group structure.

Let E : y2 = f (x), where f (x) = x3 + ax + b, a, b ∈ F. LetP = (x1, y1),Q = (x2, y2) ∈ E (F). Connect the two points by astraight line. Since f (x) is a cubic, this line will cross at a third point,and we call it P ∗ Q. Then we reflect the point P ∗ Q about thex-axis, and call this new point on the curve P + Q.

E (F) is an abelian group with identity O under this operation.

Checking Closure and Associativity are simple but tedious exercise inhigh school algebra. Commutativity is obvious. For inverses, one caneasily check that if P = (x , y), then −P = (x ,−y).


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The Group Law







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The Group Law







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The Group Law







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The Group Law







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Recall

Defintion (Eisenstein Series)

Let τ ∈ H and Λ = 〈1, τ〉 is the corresponding lattice. For each integerk > 2, denote (called Eisenstein Series)

Gk(τ) =∑

ω∈Λ\{0}

1

ω2k.

Defintion (Weiestrass ℘-function)

The Weiestrass ℘-function is given by

℘(z) =1

z2+

∑ω∈Λ,ω 6=0

(1

(z − ω)2− 1

ω2),

which is meromorphic in C with double poles precisely at the points in Λ.


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Recall

Defintion (Eisenstein Series)

Let τ ∈ H and Λ = 〈1, τ〉 is the corresponding lattice. For each integerk > 2, denote (called Eisenstein Series)

Gk(τ) =∑

ω∈Λ\{0}

1

ω2k.

Defintion (Weiestrass ℘-function)

The Weiestrass ℘-function is given by

℘(z) =1

z2+

∑ω∈Λ,ω 6=0

(1

(z − ω)2− 1

ω2),

which is meromorphic in C with double poles precisely at the points in Λ.


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Structure of E (C)

Let E : y2 = 4x3 − g2x − g3, g2, g3 ∈ C. We now study the structureof the group E (C).

Let g2 = 60G2, g3 = 140G3.

Then we have g2(∞) = 120ζ(4) = 43π

4, g3(∞) = 280ζ(6) = 827π

6.


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Structure of E (C)


Let g2 = 60G2, g3 = 140G3.

Then we have g2(∞) = 120ζ(4) = 43π

4, g3(∞) = 280ζ(6) = 827π

6.


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Structure of E (C)


Let g2 = 60G2, g3 = 140G3.

Then we have g2(∞) = 120ζ(4) = 43π

4, g3(∞) = 280ζ(6) = 827π

6.


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Structure of E (C)


Let g2 = 60G2, g3 = 140G3.

Then we have g2(∞) = 120ζ(4) = 43π

4, g3(∞) = 280ζ(6) = 827π

6.


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Some Useful Theorems

Theorem

℘′2 = 4℘3 − g2℘− g3.

Theorem (Addition Formula)

℘(z + w) = 14 (℘

′(z)−℘′(w)℘(z)−℘(w) )2 − ℘(z)− ℘(w)

Theorem

A holomorphic elliptic function is constant.

Corollary

An elliptic function with no zeros is constant.

Proof.

If f has no zeros, then 1/f is holomorphic, hence constant.


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Theorem

℘′2 = 4℘3 − g2℘− g3.

Theorem (Addition Formula)

℘(z + w) = 14 (℘

′(z)−℘′(w)℘(z)−℘(w) )2 − ℘(z)− ℘(w)

Theorem

A holomorphic elliptic function is constant.

Corollary

An elliptic function with no zeros is constant.

Proof.

If f has no zeros, then 1/f is holomorphic, hence constant.


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Theorem

If f : C/Λ→ C is meromorphic, then f has the same number of poles(counting multiplicities) as the number of zeros (counting multiplicities),denote this number the order of f .

Theorem (Viete’s Theorem)

Let F be a field. If p(x) = anxn + an−1x

n−1 + · · ·+ a1x + a0 ∈ F [x ], thenp(x) have n (not necessarily distinct) roots x1, x2, · · · , xn in an algebraicclosure F of F and

x1 + x2 + · · ·+ xn =−an−1

an.

Theorem

℘(z) is an even elliptic function, ℘′(z) is an odd elliptic function.


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Theorem

Let g2 = g2(Λ) and g3 = g3(Λ) be the quantities associated to a latticeΛ ∈ C, then y2 = 4x3 − g2x − g3 is an elliptic curve over C. i.e.f (x) = 4x3 − g2x − g3 has distinct roots.

Proof.

Let {ω1 = 1, ω2 = τ} be a basis for Λ and let ω3 = ω1 + ω2.

Then since ℘′(z) is an odd elliptic function, we have℘′(ωi

2 ) = −℘′(−ωi2 ) = −℘′(ωi

2 ).

So ℘′(ωi/2) = 0. Hence f (x) = 0 for x = ℘(ωi/2). So it suffices toshow that these three values are distinct.

The function ℘(z)− ℘(ωi/2) has a double zero at z = ωi/2.However, it is an elliptic function of order 2, so it has exactly twozeros or a single double zero. Hence ℘(ωi/2) 6= ℘(ωj/2) for i 6= j .


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Theorem


Proof.



2 ) = −℘′(−ωi2 ) = −℘′(ωi

2 ).




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Theorem


Proof.



2 ) = −℘′(−ωi2 ) = −℘′(ωi

2 ).




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Theorem


Proof.



2 ) = −℘′(−ωi2 ) = −℘′(ωi

2 ).




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Theorem


Proof.



2 ) = −℘′(−ωi2 ) = −℘′(ωi

2 ).




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E (C) is a torus

Theorem

Using the above notation: E (C) is isomorphic to C/Λ via the mapφ : C/Λ→ E (C), z + Λ 7→ (℘(z), ℘′(z)).

Proof.

The image of any nonzero point z of C/Λ is a point in E (C) because℘′2 = 4℘3 − g2℘− g3.

the map is a group homomorphism: Let y = mx + c be the linethrough the points P = (x , y) and P ′ = (x ′, y ′) on the curvey2 = 4x3 − g2x − g3, then the x , x ′ and x(P + P ′) are the roots ofthe polynomials (mx + c)2 − 4x3 + g2x + g3. So

x(P + P ′) + x + x ′ = 14m

2 = 14 ( y−y

′

x−x ′ )2, which agrees with theaddition formula of ℘, hence this is a group homomorphism.


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E (C) is a torus

Theorem


Proof.



x(P + P ′) + x + x ′ = 14m

2 = 14 ( y−y

′



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E (C) is a torus

Theorem


Proof.



x(P + P ′) + x + x ′ = 14m

2 = 14 ( y−y

′



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The Proof

Proof.

To see φ is surjective, let (x , y) ∈ E (C), then ℘(z)− x is anon-constant elliptic function, so it has a zero, say z = a. Hence℘(a)′2 = y2. Replacing a by −a if necessary, we obtain ℘(a)′ = y .Then φ(a) = (x , y).

To see φ is injective, it is equivalent to prove that ker(φ) = {0}. NowConsider the embedding E (C) ↪→ P2(C), (x , y) 7→ (x , y , 1). Thenz + Λ ∈ ker(φ) iff ℘(z + Λ) = ℘′(z + Λ) = 0 iff z is a double zero of ℘iff z = 0, done.


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The Proof

Proof.

To see φ is surjective, let (x , y) ∈ E (C), then ℘(z)− x is anon-constant elliptic function, so it has a zero, say z = a. Hence℘(a)′2 = y2. Replacing a by −a if necessary, we obtain ℘(a)′ = y .Then φ(a) = (x , y).

To see φ is injective, it is equivalent to prove that ker(φ) = {0}. NowConsider the embedding E (C) ↪→ P2(C), (x , y) 7→ (x , y , 1). Thenz + Λ ∈ ker(φ) iff ℘(z + Λ) = ℘′(z + Λ) = 0 iff z is a double zero of ℘iff z = 0, done.


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Application to FLT for Entire Functions

There is an interesting application of the isomorphism to the so calledFermat’s Last Theorem for entire functions.

The classical Fermat’s Last Theorem says if n ≥ 3, then the equationan + bn = cn admits no positive integer solutions. We want to knowwhether the same holds if a, b, c are allowed to be entire functions.

we change notation a little bit and we write f = a/c , and g = b/c ,where f , g are now meromorphic functions in C. We wish to knowwether f n + gn = 1 forces f and g to be constant functions, wheref , g are meromorphic functions in C.

The answer is NO if n = 1, 2, 3 and YES if n ≥ 4.


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The cases n = 1, 2 are easy. We prove the case for n = 3.

For n = 3, we can transform the equation f 3 + g3 = 1 into anequation of the form: Y 2 = X 3 − 432, via settingf = (36− Y )/6X , g = (36 + Y )/6X .

E : Y 2 = X 3 − 432 can be realized as an elliptic curve.

Now, via this isomorphism, we will get non-constant f , g in terms of℘(z), and ℘′(z).

For n ≥ 4, the proof uses theory of compact Riemann surface.


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References

[1] L. Ahlfors, Complex Analysis, Mcgraw Hill, 1979.[2] J. Silverman, The Arithmetic of Elliptic Curves, Springer GTM, 2008.[3] A. Knapp, Elliptic Curves, Princeton U Press, 1992.


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Thanks

Thanks!


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Documents

CA3. Elliptic Curves over Complex Plane- Ryan Lok-Wing Pang