# The rank of elliptic curves of the form E

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## The rank of elliptic curves of the form E A,B : y 2 = x 3 + A (x − B) 2

### Lianne van Timmeren

Master thesis mathematics July 2015

Supervisor: Prof. dr. J. Top

Second reader: Prof. dr. ir. R.W.C.P. Verstappen

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Abstract

The rank of a finitely generated abelian group A is by definition the maximal number of independent points in A. Here we are especially interested in the group, EA,B(Q), of rational points on the elliptic curve EA,B. Under the special condition that the mul- tiplication by 3 : EA,B → EA,B factors as a product 3 = φ ◦ ψ, we compute the rank of EA,B(Q). We need to know something about the torsion points of order 3 that are rational. After introducing a map α we can compute the rank. A computer program will be explained for computing the rank of EA,B. We will try to make families of elliptic curves with a higher rank.

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## Contents

1 Introduction 4

2 Elliptic curves 6

2.1 Definition of elliptic curves . . . . 6

2.2 Rank of an elliptic curve . . . . 12

2.3 Points of order 3 . . . . 13

2.4 The multiplication map . . . . 16

2.5 The map α . . . . 17

3 Computing the rank of the elliptic curve EA,B: y2= x3+ A (x − B)2 19 3.1 Algebraic number theory . . . . 19

3.2 Formula for the rank . . . . 20

3.3 Image of α . . . . 23

3.4 Computing the rank . . . . 27

4 Explanation of the code for computing the rank 29 5 Examples 36 5.1 Example E8,1: y2= x3+ 8 (x − 1)2 . . . . 36

5.2 Example E79,4: y2= x3+ 79 (x − 4)2 . . . . 37

5.3 Example E−388728,5184: y2= x3− 388728 (x − 5184)2 . . . . 39

6 Families of elliptic curves of higher rank. 41 6.1 Elliptic curves with at least 2 rational points . . . . 41

6.1.1 The case B = 4 . . . . 43

6.2 Elliptic curves of the form y2= x3+ a (t) (x − b)2. . . . 45

6.2.1 The case β = 3, t = 2 . . . . 46

6.2.2 The case β = 9 . . . . 47

6.3 Elliptic curves of the form y2= x3+ at2+ b x − ct2+ dt + e2 . . . . 48

7 Conclusion 50

A Code in magma for computing the rank 53

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## Introduction

In this thesis we are interested in elliptic curves over Q of the form EA,B: y2 = x3+ A (x − B)2.

The group of rational points on this elliptic curve will be denoted by EA,B(Q). The main goal of this thesis is computing the rank of the group EA,B(Q).

In [vB10] there was already a good start with the subject of this thesis. Monique van Beek did a lot of research about the necessary theory on computing the rank of the group EA,B(Q). When reading the master thesis [vB10] several questions remained about how this exactly works and how to compute the rank of EA,B(Q).

The research of this master thesis started with reading thesis [vB10] and understand- ing the theory explained there. We started with making a few examples computing the rank. It became clear that some aspects of the theory explained in [vB10] could be stated more precisely and computing the rank by hand was really not doable. Moreover, we wanted to make families of elliptic curves that have higher rank. So we had three goals during this thesis:

1. Making the theory of computing the rank of EA,B(Q) more precise and more prac- tical.

2. Making a computer program in magma to compute the rank with help of this theory.

3. Finding families of elliptic curves that have higher rank.

In chapter 2 we start with general theory about elliptic curves. First we define elliptic curves in Weierstrass normal form. We recall the group law and explicit formulas for the group law of elliptic curves. After that we will prove an elliptic curve is an abelian group,

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then we explain where the special form

EA,B : y2= x3+ A (x − B)2 (1.1) comes from. After that more theory about rational points of the elliptic curve EA,B will be discussed. In section 2.2 we will state the definition of rank and some theorems about the rank of elliptic curves. To compute the rank of the group EA,B(Q) we need some theory about the points of order dividing 3, this will be explained in section 2.3.

In section 2.4 we define two isogenies φ and ψ such that φ ◦ ψ = [3] where [3] is the mul- tiplication by 3 map. We use this multiplication map to compute the rank of EA,B(Q).

In the next section we will define a homomorphism α : EA,B(Q) → Q√

A

/Q√

A∗3

which is probably the most important homomorphism in this thesis. With help of α we get an explicit formula for the rank.

Chapter 3 starts with some general algebraic number theory. This is needed to prove that the image of α is contained in a certain set Λ. Then we will derive a formula to compute the rank with help of the image of α. After this we will define the set Λ. We will conclude the chapter with a roadmap to compute the rank.

In chapter 4 the computer program made for computing the rank will be explained.

The code can be found in appendix A. In chapter 5 some examples will be given. In chapter 6 we will try to make some families of elliptic curves with higher rank.

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## Elliptic curves

In this chapter some general theory about elliptic curves will be explained. In the first section the definition of elliptic curves will be treated. Also the group law will be ex- plained. After that we tell something about our choice of working with elliptic curves of the special form EA,B : y2 = x3 + A (x − B)2. We treat some theorems about rational points on this curve. In the next section we tell something about the rank of elliptic curves. Before we can do that, we need to know what the order of a point means, this will also be explained. Section 2.3 of this chapter contains theory about points of order 3 on our elliptic curve. The last two sections contain theory about two different maps, the multiplication map and the map α, respectively.

### 2.1 Definition of elliptic curves

A general cubic equation is given by

ax3+ bx2y + cxy + dy3+ ex2+ f xy + gy2+ hx + iy + j = 0.

A point (x, y) satisfying this cubic equation is rational if x ∈ Q and y ∈ Q. Any cubic equation which contains at least one rational point and which has the property that its homogenization

ax3+ bx2y + cxyz + dy3+ ex2z + f xyz + gy2z + hxz2+ iyz2+ jz3

defines a smooth cubic curve in P2 can be transformed into the Weierstrass normal form.

In general this equation is given by

y2 = 4x3− g2x − g3.

The above equation can be transformed into a more general form namely y2= f (x) = x3+ ax2+ bx + c.

We refer to this equation if we are talking about the Weierstrass normal form.

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Definition 1. Elliptic curves are non-singular curves of the form

F : y2 = f (x) = x3+ ax2+ bx + c (2.1) together with a point at infinity O.

The discriminant of this curve is given by

∆ = −4a3c + a2b2+ 18abc − 4b3− 27c2.

The roots on the right hand side of equation 2.1 must be distinct, which means the discriminant cannot be zero. Elliptic curves can be defined over any field K, however F in the form of equation 2.1 does not define a non-singular curve in case char(K) = 2.

Definition 2. The group law of elliptic curves is defined as follows: Take two points P and Q on the elliptic curve, draw a line through P and Q, this line will intersect with the curve on a third point −R. Draw a vertical line through −R, this line will again intersect with the curve. The point where it intersects is the point R = P + Q. See figure 2.1. Be aware that if P = (x, y) then −P = (x, −y).

Figure 2.1: Group law elliptic curves

For the next theorem we will use explicit formulas for computing the coordinates of the point P + Q on an elliptic curve. We start with the points P1 = (x1, y1), P2 = (x2, y2) and P1+ P2 = (x3, y3), where x1 6= x2. From the group law described above we see we first have to draw a line through P1and P2. So let the equation of this line be y = αx + β.

Then we have α = xy2−y1

2−x1 and β = y1− αx1= y2− αx2. Substituting this in the equation of F , using y2 = (αx + β)2, we get

(αx + β)2= x3+ ax2+ bx + c

0 = x3+ a − α2 x2+ (b − 2αβ) x + c − β2 .

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Since this equation has three roots, namely x1, x2 and x3, we have x3+ a − α2 x2+ (b − 2αβ) x + c − β2

= (x − x1) (x − x2) (x − x3)

= x3− (x1+ x2+ x3) x2+ (x1x2+ x1x3+ x2x3) − x1x2x3. So we have α2− a = x1+ x2+ x3. Summarized we get the equations

α = y1− y2 x1− x2 β = y1− αx1

x3 = α2− a − x1− x2 y3 = αx3+ β.

Using this set of equations we can compute the coordinates of P1+P2 = (x3, y3) explicitly.

If x1= x2 and y1 = y2 6= 0, the only difference is that α = f02y(x).

Theorem 1. An elliptic curve is an abelian group, consisting of the points given by equation 2.1 and a point O at ”infinity”.

Proof. To prove that an elliptic curve is a commutative group, we have to prove associativ- ity, that a point at infinity exists, that every point has an inverse and that commutativity holds:

• G1, associativity: For all P, Q, R ∈ F we have (P + Q) + R = P + (Q + R). Since this proof is quite lengthy we will not do it here. The proof can be found in [Fri].

• G2, point at infinity: For all P ∈ F we have P + O = P = O + P . This follows immediately from the definition of the group law on elliptic curves.

• G3, inverse: For all P ∈ F exist a Q ∈ F such that P + Q = O = Q + P . Since we have to find a point Q ∈ F such that P and Q lie on a vertical line, we see immediately that this must be the point Q = −P . It is easy to see that every point on an elliptic curve has such a point.

• G4, commutativity: For all P, Q ∈ F we have P + Q = Q + P . With the for- mulas given above this property can be checked easily. This is also easy to check geometrically, as can be seen in figure 2.1.

This thesis is about the rank of elliptic curves EA,B of the form

EA,B: y2= x3+ A (x − B)2 (2.2)

y2= x3+ Ax2− 2ABx + AB2,

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where A and B are integers. The discriminant then becomes ∆ = A2B3(−4A − 27B).

Be aware that if A and/or B are not integer, then we can multiply the equation by their denominator(s) and replace the variables x, y to get a new elliptic curve where A, B are integers.

To explain where the form of EA,B comes form, we first need the definition of an isogeny.

Definition 3. An isogeny between two elliptic curves E1 and E2 is a morphism that maps OE1 to OE2, so

φ : E1→ E2 with φ (OE1) = OE2.

Two elliptic curves are isogenous if there exists an isogeny between E1 and E2 such that φ (E1) 6= {∅}.

It is known, see [Sil08], that isogenies are group homomorphism. An important isogeny for this thesis is the multiplication by m map. This will be denoted by [m]. In this thesis we are interested in the map [3]. Our goal is to find two isogenies φ and ψ that together make the map [3]: φ ◦ ψ = [3].

We start with the elliptic curve F defined over Q. Now we want to define the map [3] and the isogenies φ and ψ such that we get figure 2.2.

F

F F

φ ψ

[3]

Figure 2.2: The isogenies φ and ψ and the map [3].

The isogeny φ maps to F , where F is also an elliptic curve defined over Q. Then we have that

degree (φ) · degree (ψ) = 9 = degree ([3]) .

If the degree of φ (or ψ) is 1 this isogeny is not interesting. Instead we want both isogenies to have degree 3. We know that the degree(φ) = ker (φ) where the ker(φ) is given by

ker (φ) = {P ∈ F (C) | φ (P ) = O} .

This gives us a subgroup of order 3, namely {P, 2P = −P, O}. Because φ is defined over Q we must have that ker (φ) is stable under automorphism. Take the point P in the subgroup as (α, β). We have for all σ ∈ Aut (C) that

{σ (P ) , σ (−P ) = −σ (P ) , σ (O)} = {P, −P, O} .

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So we have σ (P ) = ±P , and it follows that σ (α) = α for all σ. Therefore, we find that α ∈ Q.

So starting with the equation of F : y2 = x3 + ax2 + bx + c and translating it with α ∈ Q, so replace x by x − α we get

y2 = (x − α)3+ a (x − α)2+ b (x − α) + c

= x3− 3αx3+ 3α2x − α3+ ax2− 2aαx + aα2+ bx − bα + c

= x3+ (a − 3α) x2+ 3α2− 2aα + b x + aα2− α3− bα + c

= x3+ ˜ax2+ ˜bx + ˜c with point

 0,√

˜ c



. Now we want that this point has order 3. We construct a tangent line on F in point P . The equation of the line will be y = ax +√

˜ c with a = dy

dx P

= 3x2+ 2˜ax + ˜b 2y

P

= ˜b 2√

˜ c. Since the order of P is not equal to 2 we have that√

˜

c 6= 0, so we can divide by√

˜ c. Then the tangent line is given by y = ˜b

2

˜ cx +√

˜

c. Intersecting the tangent line with the elliptic curve gives

˜b

4˜cx2+ ˜bx + ˜c = x3+ ˜ax2+ ˜bx + ˜c which has a triple intersection point in x = 0 if and only if

˜ a = ˜b

4˜c. This gives two solutions:

1. ˜a = ˜b = 0, so y2 = x3+ ˜c. Indeed

 0,√

˜ c



has order 3, but this is not the curve we are interested in in this thesis.

2. ˜a 6= 0, then ˜c = ˜b2a. The resulting elliptic curve is y2 = x3+ ˜ax2+ ˜bx + ˜b2

4˜a

= x3+ ˜a x2+

˜b

˜ a+

˜b2 4˜a2

!

= x3+ ˜a x + ˜b 2˜a

! .

A change of variables A := ˜a and B := −˜ab gives us the result we wanted: the curve EA,B: y2 = x3+ Ax2− 2ABx + AB2= x3+ A (x − B)2.

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The reader should take note that while in this thesis we derive everything for the elliptic curve EA,B specifically, most of the theorems and definitions used in this thesis are also correct for a general elliptic curve in Weierstrass normal form.

Definition 4. The set of rational points on EA,B: y2= x3+ A (x − B)2 is given by EA,B(Q) :=

n

(x, y) ∈ Q × Q | y2 = x3+ A (x − B)2 o

.

In the next section we will see that the group of rational points is a finitely generated abelian group.

Theorem 2. Any rational point (x, y) 6= (0, 0) on the curve 2.1 can be written in the form

(x, y) =

m e2, n

e3

 . where m, n, e ∈ Z and gcd(m, e) = gcd(n, e) = 1.

Proof. Since x, y are rational, we can write them as x = Mm and y = Nn where M, N > 0 and gcd(N, n) = gcd(M, m) = 1. Substituting this in the equation of EA,B we get

y2= x3+ A (x − B)2 n2

N2 = m3

M3 + Am

M − B2

n2

N2 = m3

M3 + Am2

M2 − 2ABm

M + AB2

M3n2= m3N2+ Am2M N2− 2ABmM2N2+ AB2M3N2 (∗) . Now we want to show that M3 = N2, so we have to show M3|N2 and N2|M3.

1. The right hand side of the last equation (∗) contains in each term N2 so we have:

N2|M3n2, and since we have gcd(N, n) = 1 we have N2|M3.

2. From (∗) we have that M |m3N2, since gcd(M, m) = 1 we have M |N2. Using this back in (∗) we have M2|N2m3 so M2|N2 and M |N and again using this back in (∗) we have M3|N2m3 so M3|N2.

So now we have proven that M3 = N2 and shown M |N , let e = MN. Then we have:

e2 = N2

M2 = M3

M2 = M, e3 = N3 M3 = N3

N2 = N, from which we have the required result

(x, y) =m e2, n

e3

 , where m, n, e ∈ Z and gcd(m, e) = gcd(n, e) = 1.

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### 2.2 Rank of an elliptic curve

Definition 5. The order of a point P of an elliptic curve is the smallest positive integer m, satisfying:

mP = P + P + · · · + P

| {z }

m times

= O.

If no such m exists, we say P has order infinity.

Definition 6. A torsion point is a point with finite order.

In this thesis we are especially interested in the rank of the elliptic curve EA,B.

Definition 7. The rank of an elliptic curve is the maximal number of independent ra- tional points on EA,B.

So for our research about the rank of the elliptic curve EA,Bwe first need to know all the rational points of EA,B. Then we need to check whether they have infinite order or not.

To say something about the rank of EA,B we make use of the Mordell’s theorem.

Theorem 3. Mordell’s theorem The group of rational points EA,B(Q) on the elliptic curve EA,B : y2 = x3 + A (x − B)2 is a finitely generated abelian group isomorphic to Zr⊕ EA,B(Q)tors, where EA,B(Q)tors is finite.

• r is the rank of the elliptic curve.

• EA,B(Q)tors = {P ∈ EA,B(Q) | P has finite order}.

Proof. We refer to [vB10] for the proof, since it is derived there explicitly for the same elliptic curve we are discussing.

There is one more important theorem regarding to points of finite order. We will not use it directly in this thesis, but since we want to set out the complete theory about the order of a point on the elliptic curve EA,B we do mention this theorem. One can use it to compute EA,B(Q)tors.

Theorem 4. Nagell-Lutz theorem If P = (x, y) ∈ EA,B(Q) is a torsion point, then x, y ∈ Z and either y = 0 (then P has order 2) or y2|∆.

Proof. The proof can be found in [ST92].

This theorem can sometimes be used to prove that a point P has infinite order. Namely compute 2P, 3P, · · · until you get a point nP with coordinates that are not integers. Then you know that nP , and hence P , are of infinite order.

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### 2.3 Points of order 3

Furthermore, we need some theory about the points of order dividing 3 on the elliptic curve 2.2. This can also be found in [ST92] and in [vB10]. Since this is also very impor- tant for computing the rank of the elliptic curve EA,B, we will set out the theory in this thesis.

We will first derive the theory for the elliptic curve EA,B over a general field. After we have derived the possible points, we will look if they can be rational.

For points to have order 3 we need to have points P on EA,B such that 3P = O. To make it slightly easier we look at the points that satisfy 2P = −P . With use of the group law of elliptic curves we immediately see that x (2P ) = x (−P ) = x (P ). So the points of order 3 satisfy x (2P ) = ±x (P ). If we use the group law of elliptic curves in the Weierstrass form y2 = x3+ ax2+ bx + c we can find an explicit formula for the x−coordinate of 2P , namely:

x4− 2bx2− 8cx + b2− 4ac 4x3+ 4ax2+ 4bx + 4c .

If we set this equal to x and do some algebra then we get the following theorem.

Theorem 5. A point P = (x, y) of the elliptic curve EA,B : y2 = x3+ A (x − B)2 has order three if and only if x is a root of the polynomial

g (x) = 3x4+ 4Ax3− 12ABx2+ 12AB2x.

Proof. The proof is almost worked out in the text above, and can be found by doing some algebra.

To continue our question: ”How many points of order dividing 3 are on the elliptic curve 2.2?” we see from [ST92] that the polynomial of theorem 5 has 4 distinct roots. So we get 8 different points of order 3 on our elliptic curve EA,B, namely the 4 roots with positive and the 4 roots with negative y−value. Together with the point infinity, which is the only point with order 1 and so the only point with order dividing 3, this gives us in total 9 points of order dividing 3.

It is important for us to know how many of the points of order dividing 3 can be rational and therefore are in EA,B(Q). We notice that there is only one abelian group with nine elements that all have order dividing 3: the product of two cyclic groups of order three, so we see that only 1, 3 or 9 points can be rational.

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We immediately see that 1 and 3 points can be rational and order dividing 3, for example E2,1 : y2= x3+ 2 (x − 1)2

E2,1(Q)tors= {O} , E1,1 : y2= x3+ (x − 1)2

E1,1(Q)tors= {O, (0, 1) , (0, −1)} .

These are the only possibilities as we shall prove now. To prove this, we first need the definition of an inflection point on an elliptic curve.

Definition 8. A nonsingular point P on an elliptic curve is a point of inflection if the tangent line through P has an intersection multiplicity ≥ 3 with EA,B at P .

Lemma 1. A point P 6= O at EA,B is an inflection point if and only if it is of order 3.

Proof. Let P 6= O be a point of order 3 on the elliptic curve EA,B. Now take the tangent line L through P and call the point of intersection with the elliptic curve Q. Now P + P = 2P = −Q. Now take the line through 2P = −Q and P . Since P is a third order point we must have that the line through P and Q intersects at O, meaning the line has to be vertical. This means that we have P = Q. Now we see that the line L has an intersection multiplicity of 3 with EA,B at P . Therefore, P must be a point of inflection on the elliptic curve EA,B.

Theorem 6. M¨obius theorem. The graph y = √

x3+ Ax2− 2ABx + AB2 contains exactly one point of inflection if x3+ Ax2− 2ABx + AB2 has only simple zeros.

Proof. Recall that for a twice differentiable function, an inflection point is a point where the graph of the second derivative changes sign. To make the proof slightly easier, we will proof it in general and look at the equation y =√

x3+ ax2+ bx + c =pf (x). If we can prove it for this case, then substitute a = A, b = −2AB and c = AB2, we have the result in this special case. The first derivative is

dy dx = 1

2f12 · 3x2+ 2ax + b . And the second derivative

d2y dx2 = −1

4f32 · 3x2+ 2ax + b2

+ 1

2f12 · (6x + 2a)

= −1

4f32 · f02

+1

2f12 · (6x + 2a)

= −1 4f32

 f02

− 2 · f · f00



=

2f · f00− f02

4y · f = F (x) 4y · f.

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If we differentiate F (x) = 2f · f00− f0

2

we get F0(x) = 2f0· f00+ 12f − 2f0· f00 = 12f . So we have F (x) = 3x4+ lower order terms and limx→±∞ = +∞. The local extrema of F (x) are at point(s) α with f (α) = 0. At these points we have F (α) = −f (α)2. Suppose f (x) only has simple zeros, then f (x) and f0(x) have no zero in common, so F (α) < 0.

So it follows that F (x) has exactly two zeros. In our case we have f (x) > 0, so we have only one zero and therefore one point of inflection.

Recalling the equation of EA,B to be:

EA,B: y2 = x3+ A (x − B)2

= x3+ Ax2− 2ABx + AB2 we see that EA,B consist of two parts,

y =p

x3+ Ax2− 2ABx + AB2 y = −p

x3+ Ax2− 2ABx + AB2.

By M¨obius’ theorem we have that each of these two parts has exactly one point of inflec- tion and by lemma 1 we see we have at most two real points of order 3 on EA,B. Since a rational point is real, we have at most two rational points of order 3. With O included we have at most three rational points of order dividing 3.

For now it is important to determine when EA,B(Q) can contain those two points of order 3. By theorem 5 we have that the x−coordinate of a point of order 3 must be a zero of g (x) = 3x4+ 4Ax3− 12ABx2+ 12AB2x.

One obvious solution for this equation is of course x = 0. Therefore we have that the point (0, y) is a point on EA,B

y2 = 03+ A (0 − B)2= AB2.

In order to obtain a rational solution we must have that A = a2, for some a ∈ Q. If A is of this form then we always have 2 points of order 3 in EA,B(Q).

Now we have to look for solutions of g (x) = 3x4 + 4Ax3− 12ABx2 + 12AB2x where x 6= 0. So we have to solve 3x3 + 4Ax2 − 12ABx + 12AB2 = 0. If we multiply the equation of EA,B by 3 we get: 3y2 = 3x3+ 3A (x − B)2. Subtracting these two equations from each other results in:

3y2 = 3x3+ 3A (x − B)2− 3x3− 4Ax2+ 12ABx − 12AB2

= 3x3+ 3Ax2− 6ABx + 3AB2− 3x3− 4Ax2+ 12ABx − 12AB2

= −Ax2+ 6ABx − 9AB2

= −A (x − 3B)2.

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Since y 6= 0, this implies A = −3a2 for some a ∈ Q. If we put A = −3a2 in the equation of g (x) and solve it algebraically (with help of a computer program) for x we get two solutions that can be rational, one is the obvious soltuion x = 0 and the other is

x = 2

3 8a6− 18Ba413

−3

2 · 4Ba2169a4 (8a6− 18Ba4)13

+4 3a2.

From the above solution for x we see that 8a6− 18Ba4 has to be a third power to let x be a rational solution. So 8a6− 18Ba4= a3 8a3− 18Ba = m3 for a certain m ∈ Q, m > 0.

Therefore, we see that we can only have rational points of order dividing 3 if A = a2, a ∈ Q or A = −3a2 and 8a3− 18Ba = m3 for a, m ∈ Q, m > 0.

Definition 9. EA,B(Q) [3] = {points of order dividing 3 in EA,B(Q)}

We have:

#EA,B(Q) [3] =









3 if A = a2 or A = −3a2 and 8a3− 18Ba = m3 for some a, m ∈ Q, m > 0

1 otherwise.

### 2.4 The multiplication map

In section 2.1 we saw that we used two isogenies and the multiplication by 3 map to get the special form of EA,B. Since this is the way EA,B is constructed, we know those isogenies exists. So again we have the isogenies φ, ψ and the multiplication by 3 map as in figure 2.3.

EA,B

EA,B EA,B

φ ψ

[3]

Figure 2.3: The isogenies φ and ψ and the map [3].

Here φ and ψ are given by:

φ : EA,B→ EA,B ψ : EA,B→ EA,B.

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From [Top91] we see the elliptic curves EA,B and EA,B are given by the equations EA,B : y2= x3+ A (x − B)2

EA,B : η2= ξ3+ A ξ − B2

,

where A = −27A and B = 4A + 27B. The two isogenies are defined as follows φ : EA,B→ EA,B, φ (x, y) = (ξ, η)

where

ξ = 9 x2



2y2+ 2AB2− x3− 2 3Ax2



η = 27y

x3 −4ABx + 8AB2− x3

and ψ we split by two maps, namely σ and τ , such that ψ = τ ◦ σ, where σ is given by σ : EA,B → E

A,B, σ (ξ, η) = (x, y) where

x = 9 ξ2



2+ 2AB2− ξ3−2 3Aξ2



y = 27η ξ3



−4ABξ + 8AB2− ξ3

and τ : E

A,B→ EA,B is given by replacing y by 39y and x by 36x and divide the equation by 318.

If we start with the elliptic curve EA,B and first apply φ followed by σ, then we get the resulting elliptic curve E

A,B : y2 = x3 + 36A x − 36B2

. If we replace y by 39y and x by 36x and divide the equation by 318 then we get our original elliptic curve EA,B

back. Thus we see that the group of rational points of E

A,B is isomorphic to the group of rational points of EA,B.

### 2.5 The map α

To determine the rank of the elliptic curve EA,B : y2 = x3+ A (x − B)2 we need to make use of a homomorphism α. Therefore we need to rewrite the equation of EA,B:

x3 = y2− A (x − B)2

=

y + (x − B)√ A 

y − (x − B)√ A

.

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The map α is defined as follows

α : EA,B(Q) → Q

√

A



/Q

√

A

∗3

(2.3)

α (P ) =

1 · Q√

A∗3

if P = O



y + (x − B)√ A



· Q√

A

∗3

if P = (x, y) ∈ EA,B(Q) This map α is well-defined if

y + (x − B)√ A

6= 0. This is not the case when A = a2. There are two special points in this case, namely P = (0, ±aB). We define the image of α for these points separately:

α (0, aB) = 1

2aB · Q∗3 α (0, −aB) = 2aB · Q∗3.

Since proving that α is a homomorphism is quite lengthy and does not contain any useful theory for this thesis we will not discuss it here. For the interested reader, the complete proof can be found in [vB10] and in a more abstract way in [Top91].

(22)

## Computing the rank of the elliptic curve E A,B : y 2 = x 3 + A (x − B) 2

Although most of this chapters content can be found in [vB10] we will repeat it here for three reasons. First, this chapter is probably the most important for determining the rank of an elliptic curve EA,B of the form: EA,B: y2 = x3+ A (x − B)2. Secondly the notation will be slightly different from [vB10], hoping to clarify the theory. At last we will make a roadmap to explain how exactly the rank can be computed, this was not done in [vB10].

In chapter 4 a computer program is made with theory of this chapter.

### 3.1 Algebraic number theory

Before we can compute the rank of the elliptic curve EA,B : y2 = x3+ A (x − B)2 we recall some algebraic number theory. This will be stated in this section. The reader who is familiar with algebraic number theory can skip this section.

Write A = D · n2, with D ∈ Z square free. In this thesis we work with the quadratic extension of the number field Q, the number field Q√

D := n

a + b√

D | a, b ∈ Qo . Inside the number field Q√

D



we havethe ring of integers O

Q(D), given by

OQ(D) =

 Z

h√

Di

if D ≡ 2, 3 mod 4 Z

h1+ D 2

i

if D ≡ 1 mod 4.

The following important theorem holds.

Theorem 7. The ring of integers O

Q(D) is a Dedekind domain.

Proof. Since the proof doesn’t add anything to this thesis, it will not be included here.

The proof can be found in [Ste04].

(23)

An important property of a Dedekind domain is that every ideal can be written as a unique product of prime ideals. So in the ring of integers of Q√

D

we have unique ideal factorization.

Furthermore we need the definition of a prime ideal to be inert, split or ramified. This theory can also be found in [Cha00]. Since we are only working in the ring of integers OQ(D) we will define this only for this number field. So start with Q

√

D

 where D 6≡ 1 mod 4. Then the ring of integers O

Q(D) = Z h√

Di

, where √

D has minimum polynomial X2− D. Given a prime number p ∈ Z, we have three different cases regarding to the prime ideal factorization of (p):

• split primes: The factorization of pO

Q(D) is P1· P2 with P1, P2 different.

• inert primes: pO

Q(D) is prime by itself.

• ramified primes: The factorization of pO

Q(D) is P12.

For the units of the ring of integers we make use of the Dirichlet unit theorem. We will only use the Dirichlet unit theorem in the case Zh√

D i

, D nonsquare. The complete theoreom can be found in [Ste].

Theorem 8. Dirichlet unit theorem, special case.

Z h√

Di

, D nonsquare:

• D < 0: two units, namely {±1}, D 6= −1, D 6= −3.

• D > 1: the units are generated by -1 and a fundamental unit, which has infinity order.

Finally we recall the definition of the class group.

Definition 10. The class group, Cl, of O

Q(D) exists of the equivalence classes of ideals not equal to zero, for the equivalence relation we have: J1 ∼ J2 if ∃a, b 6= 0 ∈ O

Q(D), such that aJ1 = bJ2. This is a group with multiplication: [J1] · [J2] = [J1J2] .

### 3.2 Formula for the rank

In chapter 2 we derived the homomorphism α and explained about the isogeny [3]. Recall the isogeny φ : EA,B → EA,B.

φ : EA,B → EA,B, φ (x, y) = (ξ, η) ,

(24)

where

ξ = 9 x2



2y2+ 2AB2− x3− 2 3Ax2



η = 27y

x3 −4ABx + 8AB2− x3 .

Together with the isogeny ψ : EA,B → EA,B, which is defined in a similar way, we have φ ◦ ψ = [3].

With some knowledge we can derive a formula to compute the rank of the elliptic curve EA,B : y2 = x3 + A (x − B)2. Since this theory is exactly the same as in [vB10] we will not derive it again completely, but we will state the most important formulas.

From [vB10] we see that

EA,B(Q) ∼= Zr⊕ (Z/pν11Z) ⊕ . . . ⊕ (Z/pνssZ) , where r is the rank of the elliptic curve. Now we have

3EA,B(Q) ∼= 3Zr⊕ 3 (Z/pν11Z) ⊕ . . . ⊕ 3 (Z/pνssZ) . Now it follows

EA,B(Q) 3EA,B(Q)

∼= Z 3Z

r

⊕ (Z/pν11Z) 3 (Z/pν11Z)

⊕ . . . ⊕ (Z/pνssZ) 3 (Z/pνssZ), where



Z/pνjjZ

 3



Z/pνjjZ

 ∼=

Z/3Z if pj = 3 0 if pj 6= 3.

So we have

[EA,B(Q) : 3EA,B(Q)] = 3r+number of j with pj=3.

Recall E (Q) [3] = {points of order dividing 3 in EA,B(Q)}. From [vB10] we see

#E (Q) [3] = 3number of j with pj=3.

Now we find an important formula for determining the rank of the curve EA,B: 3r= [EA,B(Q) : 3EA,B(Q)]

#EA,B(Q) [3] .

Here comes the point where we will use the isogenies φ and ψ. We rewrite [EA,B(Q) : 3EA,B(Q)] = [EA,B(Q) : ψ ◦ φ (EA,B(Q))] .

(25)

Denote EA,B(Q) as the group of rational points on EA,B. We have 3EA,B(Q) ⊆ ψ

EA,B(Q)

⊆ EA,B(Q) . So we find

[EA,B(Q) : 3EA,B(Q)] = h

EA,B(Q) : ψ



EA,B(Q)

i

·h ψ



EA,B(Q)



: ψ ◦ φ (EA,B(Q)) i

. From elementary group theory we find (the derivation can be found in [vB10]):

[EA,B(Q) : 3EA,B(Q)] =h

EA,B(Q) : ψ

EA,B(Q)i

· h

EA,B(Q) : φ (EA,B(Q))i [ker (ψ) : ker (ψ) ∩ φ (EA,B(Q))]. So in total we get:

3r= h

EA,B(Q) : ψ

EA,B(Q)i

·h

EA,B(Q) : φ (EA,B(Q))i [ker (ψ) : ker (ψ) ∩ φ (EA,B(Q))] · #EA,B(Q) [3] .

Now it is time to simplify the numerator and denominator. We start with the numerator.

Recall the map α to be:

α : EA,B(Q) → Q

√

A



/Q

√

A

∗3

α (P ) =

1 · Q√

A∗3

if P = O



y + (x − B)√ A



· Q√

A

∗3

if P = (x, y) ∈ EA,B(Q) . We have a simular map for EA,B, called α:

EA,B : y2 = x3+ A x − B2

α (x, y) =

y + x − Bp A

· Qp A∗3

with A = −27A and B = 4A + 27B.

Lemma 2. ker(α) = φ (EA,B(Q)) . Proof. The proof can be found in [vB10].

From this it follows that α

EA,B(Q) ∼= EA,B(Q) ker (α)

∼= EA,B(Q) φ (EA,B(Q)) and now we have the result

h

EA,B(Q) : φ (EA,B(Q))i

= #α

EA,B(Q) .

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In the simular way we have ker(α) = ψ

EA,B(Q)

, so we findh

EA,B(Q) : ψ

EA,B(Q)i

=

#α (EA,B(Q)). Hence we have simplified the numerator.

3r=

#α (EA,B(Q)) · #α



EA,B(Q)



[ker (ψ) : ker (ψ) ∩ φ (EA,B(Q))] · #EA,B(Q) [3].

Lets start now with the denominator. From chapter 2 we know everything about the number of points of order dividing 3, namely:

#EA,B(Q) [3] =

3 if A = a2 or A = −3a2 and 8a3− 18Ba = m3 for some a, m ∈ Q 1 otherwise.

So the term we are left with is [ker (ψ) : ker (ψ) ∩ φ (EA,B(Q))]. So our question is of what elements does ker(ψ) consists. So which elements of EA,B(Q) are mapped to O by ψ. For certain we have ψ O = O. So we only have to look at non-trivial points in ker(ψ). From the map ψ we see that if ξ 6= 0 then (ξ, η) 6∈ ker (ψ). Furthermore the point (0, 0) is never an element of EA,B(Q). So we are left with the points (0, η) , η 6= 0. Such a point is only a point of EA,B(Q) if it satisfies η2 = ξ3+ A ξ − B2

. Hence we must have η2 = A · B2, so A has to be a perfect square. Since A = −27A we must have A = −3a2 for some a ∈ Z. So

#ker (ψ) =

3 if A = −3a2 1 otherwise.

So the term [ker (ψ) : ker (ψ) ∩ φ (EA,B(Q))] disappears if A 6= −3a2. Because φ ◦ ψ = [3], the only elements of order dividing 3 are sent by φ to ker(ψ). Following [vB10] we see in total:

[ker (ψ) : ker (ψ) ∩ φ (EA,B(Q))] =









3 if A = −3a2

and EA,B(Q) contains no points of order 3 1 otherwise.

Combining all the formulas we derived in this section we have a formula for the rank as follows:

3r =

1

3 · #α (EA,B(Q)) · #α

EA,B(Q)

if A = −3a2 or A = a2

#α (EA,B(Q)) · #α

EA,B(Q)

otherwise.

(3.1)

### 3.3 Image of α

From the previous chapter we have derived a formula for the rank, namely

3r =

1

3 · #α (EA,B(Q)) · #α

EA,B(Q)

if A = −3a2 or A = a2

#α (EA,B(Q)) · #α

EA,B(Q)

otherwise.

(27)

So for now it is important to determine #α (EA,B(Q)) and #α

EA,B(Q)

. We will do this for the cases where A 6= −3a2 or A 6= a2. Remember the image of α to be

α : EA,B(Q) → Q

√

A



/Q

√

A

∗3

. Define three sets as follows, these sets are fixed for fixed A and B:

QA,B:= {Q1, . . . , Qm ⊂ OK | Qi prime ideals, Qi nonprincipal ramified prime, A ∈ Qi} , RA,B:= {R1, . . . , Rt⊂ OK | Ri prime ideals, Ri split prime, AB ∈ Ri} ,

PA,B:= {P1, . . . , Pn⊂ OK | Pi together with the primes in QA,B generate Cl} .

Theorem 9.

α (EA,B(Q)) ⊂

 βQ√

A∗3

| β 6= 0, β ∈ OK and βOK= P131· · · Pn3n· Q11· · · Qmm· Rγ11· · · Rtγto

where Pj ∈ PA,B, Qi ∈ QA,B and Rk ∈ RA,B, j, δi, γk∈ Z≥0, j < the order of Pj, δi <

the order of Qi and γk< 3 times the order of Rk . Proof. We start with the elliptic curve:

EA,B: y2 = x3+ A (x − B)2. (3.2) Like we stated before any rational point of on the curve 3.2 can be written in the form

(x, y) =

m e2, n

e3

 ,

where m, n, e ∈ Z and gcd(m, e) = gcd(n, e) = 1. We substitute this into the equation 3.2 and find:

n e3

2

=m e2

3

+ Am

e2 − B2

=m e2

3

+ Am e2

2

− 2Am

e2B + AB2 n2 = m3+ Am2e2− 2AmBe4+ AB2e6. This can be factorized into:

m3 =

n + me − Be3√ A 

n − me − Be3√ A

= β · β and the map α becomes

αm e2, n

e3



=n e3 +m

e2 − B√ A

· Q√

A∗3

=

n + me − Be3√ A

· Q√

A∗3

= β · Q√

A∗3

. Now we have three important facts about β:

(28)

1. β · β = m3 6= 0 2. β ∈ OK

3. gcd(m, e) = gcd(n, e) = 1.

Since the ring of integers of Q√

A

is a Dedekind domain we have unique prime ideal factorization.

So we can make a prime ideal factorization of β. This can exsist of 3 different kind of ideals, namely: inert ideals, ramified ideals, and split ideals. So we have:

βOK= V1ζ1· · · Vuζu

= V1ζ1· · · Viζi· Vi+1ζi+1· · · Vjζj · Vj+1ζj+1· · · Vuζu,

where V1· · · Vi the inert ideals, Vi+1· · · Vj the ramified ideals and Vj+1· · · Vu the split ideals.

The norm of β must be divisible by 3, so we must have:

|N (β) | = v11· · · vii· vi+1i+1· · · vjj· vj+1ζj+1· · · vuζu. By the property of an inert ideal we have that p3 = 1 · Q√

A

for a inert prime p. The same result we have for principal ramified primes. So we are left with:

βOK = Vi+1i+1· · · Vjj· Vj+1ζj+1· · · Vuζu,

where Vi+1· · · Vj the non-principal ramified ideals and Vj+1· · · Vu the split ideals.

Let’s have a look at a non-principal ramified prime V that can occur in Λ. From β =



n + me − Be3√ A

we see that

n + me − Be3√ A

∈ V , from this it follows that A ∈ V .

A split prime is a prime that satisfies (V ) = P ·Q, where P 6= Q. Here we have to consider two possibilities:

1. P and Q appear both in the prime ideal factorization of β. Here we have



n + me − Be3√ A



∈ V . This means that V divides n and me − be3√ A = e m − be2√

A. One possibility is that V divides both n and e, but since gcd(n, e) = 1 this is not possible. So we must have V |A or V | m − Be2. If V | m − Be2 we have that V |m and V |B, since V |e is not possible as we saw before. So if P and Q appear both, we have that AB ∈ V .

(29)

2. Only P (or Q) appear in the prime ideal factorization of β. This case needs a little algebraic number theory. The class group Cl of the number field Q√

A

is a finite abelian group. Since only P occurs in the prime ideal factorization of β, and because the norm of β is a cube we must have that P occurs to some power 3 with  ∈ N.

We have P ∼ Q

iPiαi ·Q

jQβjj since PA,B ∪ QA,B generate the class group. So P = Q

iPiαi ·Q

jQβjj · (λP) where λP ∈ Q√

A



. So since we have that P must occur to some power 3i this λP will disappear modulo third powers. So we only need to consider the Pi that generators the class group together with the elements in the set QA,B.

So we are left with:

βOK = Vi+1i+1· · · Vjj· Vj+1ζj+1· · · Vuζu,

where Vi+1· · · Vj are the non-principal ramified ideals which divide A and Vj+1· · · Vu are the split primes which divide AB or the ideals that generate the class group. So rewriting the prime ideal factorization of β we get the set:

 βQ

√

A

∗3

| β 6= 0, β ∈ OK and βOK = P131· · · Pn3n· Q11· · · Qmm· Rγ11· · · Rγtt

 . We still need to prove: j < the order of Pj, δi < the order of Qi and γk < 3 times the order of Rk.

So start with βOK = P131· · · Pn3n · Q11· · · Qmm · Rγ11· · · Rtγt in the set above and δ = 3 · order [Pj]. Write 3i= q · δ + r where r < δ. Since 3|δ it follows that r ≡ 0 mod 3.

(Pj)3i =

(Pj)δq

· Pjr = (λ)q· Pjr= Pjr· something in Q√

A∗3

.

Since r < δ this proves the claim about the power of Pi. We have the same arguments for the maximum of δi and γk.

The large set of the last theorem we give the name Λ, so:

Λ :=

 βQ√

A∗3

| β 6= 0, β ∈ OK and βOK = P131· · · Pn3n· Q11· · · Qmm· Rγ11· · · Rγtt

 . The following two theorems tells us more about the set Λ.

Theorem 10. The set Λ is finite.

Proof. This proof is totally given in the proof of theorem 9.

Theorem 11. The set Λ is a group.

Proof.

(30)

• G1, associativity: For all β1, β2, β3 ∈ Λ we have (β1∗ β2) ∗ β3 = β1∗ (β2∗ β3). This follows from the definition of the image α en the set Λ.

• G2, there exist a unit: For all β ∈ Λ we have β ∗ e = β = e ∗ β. The unit in Λ is 1, since β · 1 · Q√

A∗3

= β · Q√

A∗3

= 1 · β · Q√

A∗3

.

• G3, inverse: For all β ∈ Λ exist a β ∈ Λ such that β ∗ β = e = β ∗ β. This follows directly from: m3 =



n + me − Be3√ A

 

n − me − Be3√ A



= β · β and m3 = 1 · Q√

A

∗3

.

Compute the rank

To compute the rank of the elliptic curve EA,B: y2 = x3+ A (x − B)2, in the case A and

−3A not a square, we first compute the sets Λ and ¯Λ for EA,B as EA,B. This gives an upper bound rup for the rank, namely:

rup= #Λ · #Λ. (3.3)

After this we search for rational points on the curve EA,B and calculate their image of α in Λ. Doing the same for EA,B and combining the results gives a lower bound rlow for the rank, in terms of the subgroup of Λ generated by the images.

We know the rank of the elliptic curve EA,B if rup = rlow. In all other cases we have only an upper bound and lower bound for the rank given by rlow≤ r ≤ rup.

### 3.4 Computing the rank

Since there is a lot of theory in the text above, the procedure to compute the upper and lower bound of the elliptic curve EA,B : y2= x3+ A (x − B)2 is summarized here.

1. Compute the fundamental unit of the ring of integers O of the quadratic field Q

√A .

2. Compute the ideals Pj, that are the generators of the class group of the ring of integers O.

3. Compute the non-principal ramified primes Qi such that A ∈ Qi. 4. Compute the split primes Rk such that AB ∈ Rk.

5. Compute with help of the order of the ideals (see theorem 9) the maximum power to which the ideals Pj, Qi and Rk can occur.

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