# 2 The rank of a surface and types of points

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faculteit Wiskunde en Natuurwetenschappen

## Notes on Elliptic Delsarte Surfaces

### Master's Thesis Mathematics

July 2014

Student: L.J. Disselhorst First Supervisor: Dr. J. Top

Second Supervisor: Dr. H.W. Broer

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Abstract

In this thesis, we try to find the generators of the Mordell-Weil group a particular elliptic K3 surface. We first determine the rank of the Mordell-Weil group and describe the relation between its genera- tors. Next we determine a point in a finite field, which we will lift to a field with characteristic zero.

### Contents

1 Introduction 3

2 The rank of a surface and types of points 3

2.1 The rank . . . 3

2.2 Torsion points . . . 6

3 Application of the theory 7 3.1 The rank . . . 7

3.2 Torsion Points . . . 10

3.3 The Galois action on the rational points . . . 11

4 Determining a point 15 4.1 Height Pairing . . . 15

4.2 A point over F41 . . . 18

4.3 Hensel lifting . . . 20

5 Conclusion 23 6 Additional proofs 24 7 Appendix Magma Code 26 7.1 Torsion Points . . . 26

7.2 Equations for the polynomial coefficients . . . 26

7.3 Points on a finite field . . . 27

7.4 Points of height pairing less than 4 . . . 27

7.5 Code to show that rank(E/F41(t)) = 19 . . . 28

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### 1 Introduction

In the PhD. thesis of Bas Heijne (), Heijne has described the generators of the Mordell-Weil groups of several Elliptic Delsarte surfaces. The only unsolved problem in the thesis, is to find the generators of the Mordell-Weil group of the surface defined by the equation y2 = x3 + t7x + 1. In this master’s thesis, I attempt to solve the problem. First, we discuss the method to determine the rank of a Mordell-Weil group, then we use Galois theory to describe the relation between the Mordell-Weil group’s generators. Finally, we determine a point on the elliptic surface in a finite field, which will be lifted to a field with characteristic zero. Throughout the thesis, the algebraic computing program magma is used.

### 2 The rank of a surface and types of points

In this section, we discuss a geometric approach to determine the group of rational points on an elliptic curve E defined over Q(t). Throughout, it is assumed that no elliptic curve E0 over Q exists such that E ∼= E0 over Q(t). Under this condition, we first observe that the Mordell-Weil group E(Q(t)) ∼= Zr× Z/aZ × Z/abZ for some integers r ≥ 0 and a > 0, b > 0 (See [9, p. 109] and [8, p. 242]). The number r is called the rank of E over Q(t) and the subgroup Etors(Q(t)) ∼= Z/aZ × Z/abZ, consisting of all points of finite order in E(Q(t)), is called the torsion subgroup.

### 2.1 The rank

An elliptic curve E over Q(t) is given by an equation y2 = x3 + ax + b, for certain a, b ∈ Q(t). Any u ∈ Q(t) gives rise to a coordinate change

η := u3y, ξ := u2x, (1)

which transforms the equation into

η2 = ξ3+ u4aξ + u6b.

In this way, one can clear the denominators of a, b, leading to an equation with a, b ∈ Q[t].

Now x3 + ax + b − y2 is a polynomial in Q[x, y, t], so its zeroes define a surface over Q. Moreover, this surface is equipped with a morphism to A1, namely (x, y, t) 7→ t. The elliptic surface E associated to E is a smooth, projective surface which is birational to the surface above, such that the

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given morphism defines a morphism π : E → P1, and E is “minimal” for these properties. For details, see section 8.8 of , which describes the more difficult situation where Q(t) and the principal ideal doman Q[t] are replaced by an arbitrary number field K and its ring of integers R.

The surface E has a “second Betti number”, b2 = dimCH2(E (C), C), which is a topological invariant of the space E (C).

Another invariant of E is its “Picard number”, ρ, which can be defined as the dimension over Q of the vector space H1,1(E (C), C) ∩ H2(E (C), Q).

A third invariant of E is its “Lefschetz number” λ, which is defined as:

λ := b2− ρ. (2)

There is a relation between the rank r of E(Q(t)) and these invariants.

To state this, we will introduce one more invariant, called ρtriv.

The morphism π : E → P1 is in fact an extention of the map (x, y, t) 7→ t.

For given t0 ∈ P1, the inverse image π−1(t0) ⊂ E is closely related to the curve with equation y2 = x3+ a(t0)x + b(t0). If this curve is an elliptic curve, then so is π−1(t0). Otherwise, π−1(t0) turns out to be a union of finitely many rational curves, which is called a singular fibre (for more information, see )

A method to determine the the rational curves, is explained in [6, p. 11] by Schuett and Shioda. We denote the surface’s equation by E : y2 = x3+ ax + b and determine its discriminant ∆(E (t0)) at t0. If ∆ = 0, then our cubic curve is singular at t0. Next, we verify the vanishing orders v of a and b. On [6, p. 16], there is a table which provides the types of fibres corresponding to v(a4) and v(a6). A comprehensive image of the types of curves can be found on [6, p. 13], so that we can determine the number of rational curves at t0. We denote this number by νt0, then:

ρtriv := 2 + X

t0∈P1 s.t.

π−1(t0) is not ellilptic

t0 − 1) (3)

We now state the following theorem:

Theorem 2.1. r = ρ − ρtriv.

This theorem is proven in e.g. 

We combine this theorem with equations 3 and 2, to obtain a formula for computing the rank:

Corollary 1. The rank of an elliptic surface is given by

r = b2− λ − ρtriv (4)

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We will now discuss the second betti number b2(S) of an elliptic surface, by combining some sections of .

On [6, p. 28], we obtain an equation for b2: b2(S) = e(S) − 2(1 − b1(C)),

where e(S) is the Euler number of an elliptic surface and b1(C) is the first Betti number of its base curve C. In our case C ∼= P1, hence b1 = 0.

Denote the elliptic surface in the usual Weierstrass form: y2 = x3+ax+b, with a, b ∈ Q[t] , as on [6, p. 16], then we define “minimal polynomials” as follows:

Definition 1. a(t) and b(t) are minimal polynomials if there is no c(t) ∈ Q[t] \ Q with the property that c4|a and c6|b.

Assuming we have an elliptic surface with minimal polynomials, we de- termine the smallest n ∈ N that satisfies the following three conditions (see [6, p. 37]:

1. deg(ai) ≤ ni for all i.

2. there is some i such that deg(ai) ≥ (n − 1)i.

3. For any finite place on P1 with valuation v, there is some i such that v(ai) < i.

In the minimal Weierstrass form, the latter two conditions are always met, the first reduces to finding the smallest n such that deg(a) ≤ 4n and deg(b) ≤ 6n.

Using [6, p. 38], we see that e(S) = 12χ(S) = 12n. We combine this with what we know about b2 to obtain:

b2 = 12n − 2. (5)

Finally, we discuss the algorithm for obtaining the Lefschetz number, which can also be found in . The Lefschetz number can not always be determined, but it is possible for Delsarte surfaces. These have the property that, when homogenized, they are of the form

3

X

i=0

Tai0Xai1ai2Zai3. (6)

With all exponents aij nonnegative. This homogeneous form is obtained from the standard form of an elliptic curve by the variable change y 7→ i · y. We

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place the exponents in a matrix as follows:

A =

a00 a01 a02 a03 a10 a11 a12 a13 a20 a21 a22 a23 a30 a31 a32 a33

Take the additive group (Q/Z)4 and construct a subgroup L by choosing the following generators:

l1 := (1, 0, 0, −1)A−1 l2 := (0, 1, 0, −1)A−1 l3 := (0, 0, 1, −1)A−1

Next, we define Λ ⊂ L. Λ is the set of all elements v = (b1, b2, b3, b4) ∈ L that satisfy:

1. v only has nonzero components.

2. There exists a “good” t ∈ Z such that ord(tv) = ord(v), where ord is the order in the additive group (Q/Z)4.

3. If such “good” t exist, we also need thatP3

i=0{t·bi} = 1 or 3 for at least one t. {bi} refers to the natural bijection between (Q/Z) and [0, 1).

The Lefschetz number is defined to be λ := #Λ.

### 2.2 Torsion points

To determine the torsion points of an elliptic curve, one can apply the Nagell- Lutz theorem:

Theorem 2.2. Let E (Q(t)) be an elliptic surface with Weierstrass equation y2 = x3 + a(t)x + b(t), with a(t), b(t) ∈ Q[t] and suppose that P = (α, β) ∈ E(Q(t)) is a nonzero torsion point. Then:

(either β = 0 P has order 2 and α ∈ Q[t] satisfies α3+ a(t)α + b(t) = 0, or β2|∆(E) order ≥ 3.

In , the proof has been given for elliptic curves of the form y2 = x3+ Ax + B, with A, B ∈ Z. This proof can be adapted to our case.

An alternative way to find torsion points, or prove in some cases that they do not exist, will be explained in section 3.2.

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### 3 Application of the theory

In this section, we apply the theory of section 1 to our surface.

### 3.1 The rank

We wish to determine the rank of the surface defined by:

y2 = x3+ t7x + 1, (7)

Figure 1: The surface E : y2 = x3+ t7x + 1

To determine the Lefschetz number of the our surface, we first need to transform it so that it can be homogenized into the form of equaiton 6, we first apply the variable change y 7→ (iy) = ˜y. Then our surface is defined by the equation x3+ t7x + ˜y2+ 1 = 0.

Theorem 3.1. The Lefschetz number of the surface defined by x3 + t7x +

˜

y2+ 1 = 0 equals 6.

Proof. We write the homogenized equation in the form as in equation 6, so P3

i=0Tai0Xai1ai2Zai3 = X3Z5+ T7X + ˜Y2Z6+ Z8 and put the exponents in a matrix A, so we get:

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A =

0 3 0 5 7 1 0 0 0 0 2 6 0 0 0 8

with inverse: A−1= 1 168

−8 24 0 5

56 0 0 −35

0 0 84 −63

0 0 0 21

 .

Then L ⊂ (Q/Z)4 is generated by:

l1 = (1, 0, 0, −1)A−1 = 1

42(−2, 6, 0, −4), l2 = (0, 1, 0, −1)A−1 = 1

42(14, 0, 0, −14), l3 = (0, 0, 1, −1)A−1 = 1

42(0, 0, 21, −21).

Claim: L ∼= Z/42Z, with generator l := l1 + l3.

In (Q/Z)4, ord(l1) = 21, ord(l3) = 2 and 14 · l1 = 421 (−28, 84, 0, −56) =

1

42(14, 0, 0, −14) = l2. Now let l := l1 + l3 = 421(−2, 6, 21, 17), which clearly has order 42. Then 22 · l = l1 and 21 · l = l3. The relation between l1 and l2 shows that 22 · 14 · l = l2. We conclude that L ∼= Z/42Z and that l := l1+ l3 is its generator, which proves the claim.

To describe all elements in L, let vn:= n · l, for n ∈ {0, . . . , 41}. We need to find all vn that are elements of Λ. First, observe that vn has only nonzero elemenhts if 2 - n and 7 - n. Next, for t ∈ Z, ord(t · vn) = ord(vn) if t is coprime with 42 and finally, we determine for which fully nonzero vn, there exists a t such that P4

i=1{t · ai} = 1 or 3. For six vn, this is the case (the smallest t is given):

n vn t ord(vn) P

3 (−213,37,32,5142) 5 42 1 9 (−219,97,92,15342) 11 42 1 15 (−1521,157,152,25542) 1 42 1 27 (−2721,277,272,45942) 1 42 3 33 (−3321,337,332,56142) 11 42 3 39 (−3921,397,392,66342) 5 42 3 So λ = #Λ = 6.

Next, we wish to determine the Betti number b2.

Theorem 3.2. the Betti number b2 of the surface defined by y2 = x3+t7x+1 equals 2.

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Proof. We recall the Weierstrass equation y2 = x3+ ax + b, with a, b ∈ Q[t].

With our equation (7), we have that a(t) = t7 and b(t) = 1. The polyno- mials are clearly minimal and not both are constant, so we can immediately compute n:

n = max(ddeg(a)

4 e, ddeg(b)

6 e) = 2, so b2 = 22.

Finally, we will prove the following about the number ρtriv of E : Theorem 3.3. The surface defined by y2 = x3+ t7x + 1 has ρtriv= 3.

Proof. As we have seen in section 2.1, ρtriv is the sum of the components of all rational fibres. We find the rational fibres by fixing t = t0 and determining for which t0our curve Et0/Q is not an elliptic curve. For finite t0, the singular curves can be found at the t0 for which ∆(Et0) = 0. For our curve, we have that

∆(Et0) = −4t21− 27. (8)

So ∆ has 21 simple roots, which means that E has 21 singular fibres of type I1, which have one component.

For infinite t, let t = 1s and apply the coordinate change as stated in equation 1, for u ∈ Q(s), to obtain E0 : η2 = ξ3+ sξ + s12. This equation is minimal at s = 0 and Es=00 is clearly not an elliptic curve. From [6, p. 14], we get that Et=∞ is a singular fibre of type III, which has ν= 2.

Recall equation 3:

ρtriv = 2 +X

t0 − 1) = 2 + 21 · (1 − 1) + (2 − 1) = 3

We can now determine the rank of our surface by equation 4:

r = b2− λ − ρtriv = 22 − 6 − 3 = 13

We conclude that there are thirteen solutions (xi(t), yi(t)), with i ≤ 13 of equation 7, that are independent in the group E(Q(t)).

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### 3.2 Torsion Points

We will now show that the elliptic curve E over Q(t) defined by equation 7 has no nontrivial torsion points.

Theorem 3.4. The elliptic curve E : y2 = x3+ t7x + 1 has no torsion points other than O

Proof. Our surface is defined by the Weierstrass equation y2 = x3+ t7x + 1.

Use theorem 2.2 and assume that P = (α, β) ∈ E(Q(t)) is a nonzero torsion point. ∆(E ) = −4t21 − 27, which has no double roots in Q, so β must be constant. First, assume β = 0, then α(α2 + t7) = −1, so α ∈ Q[t] = Q

. However, this contradicts that α must have a t component, because α3+ t7α + 1 has a degree of 7 in t. So β 6= 0 and therefore β ∈ Q. We have a β ∈ Q and we wish to determine if there is an α such that (α, β) ∈ E(Q(t)).

Assume there is, then α satisfies the equation α3+ t7α + 1 − β2 = 0.

Then α ∈ Q[t] and α|1 − β2. If β2 6= 1, then α is a nonzero constant.

However, α3 + t7α + 1 − β2 has degree 7 in t, so no such α can exist in this case. If β = ±1, then α3+ t7α = 0, so α = 0 or α2 = t7. In the latter case, we have that α /∈ Q[t], which contradicts an earlier finding. This means that α must be zero, so the only two candidate torsion points that we have left, are (0, ±1). Let P = (0, ±1), then 2 · P = (t144 , ∓t218 ∓ 1), which shows that β2 - ∆(E), so P is not a torsion point. We conclude that there are no torsion points.

We will now prove Theorem 3.4 in an alternative way, using reduction theory.

Proof. Use the fact that Q(t) = Q(1t) and let 1t = s. Then the elliptic curve is described as follows:

E : y2 = x3+ t7x + 1 (9)

= x3+ s−7x + 1

We multiply with s12 and use the standard coordinate change (see equation 1, with u = s2. Then

E : η2 = ξ3+ sξ + s12 (10)

which is another equation defining the curve E.

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Define E = E mod (s), then E : η2 = ξ3 over Q.

E(Q(s)) ⊂ E(Q((s)))−−−−→ E(Q)mod s (11)

E0(Q((s))) −φ:−−−−−mod s→ Ens(Q)

The map mod s is defined as follows. Since E ⊂ P2, every point P ∈ E(Q((s))) can be written as (a(s) : b(s) : c(s)) with a, b, c ∈ Q[[s]] and at least one of them in Q[[s]]. Now P mod s is defined as (a(0) : b(0) : c(0)) ∈ E(Q). This is analogous to section 7.2 of .

The set E0 consists of the points that are not singular when mapped by mod s, so in this case, (0, 0) is the only point that can be found in E(Q) but not in Ens(Q).

E clearly has a cusp at (0, 0), so we may apply proposition 2.5b of chapter III of , which tells us that Ens is isomorphic to an additive group, in our case with the structure (Q, +, 0).

Assume that P ∈ E(Q((s))) is a torsion point not equal to O. Because there is a type III fibre at s = 0, the table 15.1 at [8, p. 448] tells us that E0(Q((s))) is a subgroup of E(Q((s))) of index 2. So 2P = O, which we know to be impossible.

### 3.3 The Galois action on the rational points

From sections 3.1 and 3.2, we know that the Mordell-Weil group E(Q(t)) ∼= Z13.

E over Q(t) has j-invariant 17284t214t21+27, so there is a curve over Q(t21) with a j-invariant that has the same value. Some calculations, by the use of equation 1, show that E ∼= (η2 = ξ3 + t−21ξ + t−42).

Let t21 = T , then we have the following construction:

Z13∼=E(Q(t))  Gal(Q(t)/Q(t21)) ∼= Z/21Z

∪ Z∼=E(Q(T )).

The group Gal(Q(t)/Q(t21)) acts on E(Q(t)): for τ ∈Gal and P = (x0, y0) ∈ E(Q(t)) one takes τ (P ) := (τ (x0), τ (y0)). This action is linear in the sense that τ (P + Q) = τ (P ) + τ (Q) and τ (O) = O. Moreover, the Galois group is cyclic, by σ with σ(t) = ζt with ζ ∈ Q a primitive 21tst root of unity. If we fix a basis for E(Q(t)) ∼= Z13, then the action of σ on E(Q(t)) is given by an

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invertible matrix A ∈ Z13. As σ21= id, A21 = I, so A is diagonalizable over C. This tells that the eigenvalues of A form a satisfy λ21− 1 = 0, so they can only be 21st, 7th, 3rd roots of unity or 1.

Now let k be a field that contains the 21st roots of unity, then we have the following construction:

k(t)

Galois, degree 21



k(T )

Denote the Galois group by G = hσi, where σ has the property that σ(t) = ζt, with ζ a primitive 21st root of unity.

Non-trivial subgroups of G can be found by factoring 21 in its primes, so one can see that H3 = {id, σ7, σ14} = hσ7i and H7 = {id, σ3, . . . , σ18} = hσ3i are the only subgroups of G. Next, we determine k(t)H3:

k(t)H3 = {f ∈ k(t)|τ (f ) = f ∀τ ∈ H3}

= {f ∈ k(t)|σ7(f ) = f }

= {f ∈ k(t)|f (ζ7t) = f (t)}

= k(t3).

Similarly, we get that k(t)H7 = k(t7), so we can find subfields of k(t) as follows:

k(t)

k(t3)

3

;;

k(t7)

7

cc

k(T )

7

cc

3

;;

The Mordell-Weil groups of E over these function fields then have the same construction:

E(k(t)) ∼= Za+γ+β+δ

E(k(t3)) ∼= Za+γ

55

E(k(t7)) ∼= Za+β

ii

E(k(t21)) ∼= Za

ii 55

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To determine β, γ and δ, we note that the third roots of unity (which exist in k(t7)), come in pairs. β = 2b. Similarly, the seventh roots of unity come in sets of six, so γ = 6c. Finally, the primitive 21st roots of unity (in k(t)), show that δ = 12d. Then the rank of the Mordell-Weil groups of E over its respective function field, is determined to be the following:

a + 2b + 6c + 12d

a + 6c

66

a + 2b

hh

a

hh 66

If 1 is an eigenvalue of A, then there is a v ∈ Z13 that is a solution to (A − I)v = 0 for nonzero v. The point P = (0, 1) satisfies σ(P ) = P , so 1 is indeed an eigenvalue of A.

The third roots of unity satisfy σ3 = id, so we look at E(Q(t)3i) = E(Q(t7)). We determine the rank of E(Q(t7)) to see that it is isomorphic with Z (for the proof, see section 6), so the third roots of unity do not contribute to the set of eigenvalues or to the set of points (Means that we only have id as contribution, why?). Similarly, we can prove that E(Q(t)7i) = E(Q(t3)) ∼= Z, so the 7th roots of unity also do not contribute.

What remains are the 21st roots of unity: these are roots of the minimal cylcotomic polynomial Φ21(x) = x12− x11+ x9− x8+ x6− x4+ x3− x + 1.

So Q[σ] ∼= Q × Q(ζ21).

Matrix A works over Q and has its characteristic polynomial det(A − λI) in Z[λ], so the minimal polynomial of ζ21 divides the characteristic polyno- mial:

A ∼

 1

ζ21 ζ212

. ..

ζ21i . ..

ζ2120

 where i is coprime with 21. So

(A − I) · (A − ζ21I) · . . . · (A − ζ2120I) = (A − I) · (Φ21(A)) = 0. (12) This tells us that Q[A] ∼= Q × Q(ζ21) and even Z[A] ∼= Z[x]/(x · Φ21(x)) ∼= Z × Z[ζ21].

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In [10, p. 205], it is stated that Q(ζ21) has class number 1, so Z[ζ21] is a principal ideal ring. Let τ be the generator of Gal(C(t)/C(t21)) and Φ the minimal polynomial of the 21st roots of unity. Denote the Mordell- weil group of E by M . x − 1 and Φ(x) generate Z[x], so we have that φ(τ )M ⊕ (τ − 1)M = M .

Let P ∈ Φ(τ )M , then we have (τ − 1)P = O, or τ (P ) = P , so Φ(τ )M is the part of M that is invariant under τ , i.e. the group with generator P = (0, 1). (τ − 1)M is the kernel of Φ(τ ) and a module over Z[ζ21]. To show that his module is torsion free, let P ∈ (τ − 1)M and α ∈ Z[ζ21], such that αP = O. Then we also have that |α|P = O, but M is torsion free, so α = 1 for all cases. Since Z[ζ21] is a principal ideal ring, we have that (τ − 1)M is free over Z[ζ21]. Then we conclude that (τ − 1)M = Z[ζ21] · P for a P ∈ M . Recall the image of the subfields of the Mordell-Weil groups over k(t) and its subfields (12). The ranks are now provided.

E(C(t)) ∼= Z13

E(C(t3)) ∼= Z

66

E(C(t7)) ∼= Z

hh

E(C(t21)) ∼= Z

hh 66

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### 4 Determining a point

In this section, we will try to find a nontrivial point P ∈ E, that is a gen- erator of E(Q(t)). We let x(t) and y(t) be polynomials in Q(t). In , M.

Kuroda determined a form of x(t) and y(t), as well as the constraints of their coefficients, we will give a more detailled description. First, we will derive a relation between the degree of x at a point P and the height pairing hP, P i, which will help us choose the form of the polynomials. Then we use these polynomials to find a point solution to E over a finite field. Finally, we apply Hensel’s lemma (see ) to lift this solution to Q.

### 4.1 Height Pairing

Theorem 4.1. If P ∈ E(k(t)) satisfies hP, P i ≤ 4 (but nonzero), then x(P ) is a polynomial of degree ≤ 4. Moreover, deg(x(P )) = 4 if and only if hP, P i = 4 and deg(x(P )) ≤ 3 if and only if hP, P i = 72.

Proof. To show this, we turn to [7, p. 21], where the height paring is dis- cussed.

Let P ∈ E(Q), then hP, P i = 2χ + 2(P O) −P

v∈Rcontrv(P ). For us, χ = 2 and (P O) = k with k ∈ Z≥0. Moreover, contrv(P ) is the local contrivution at a bad fibre of E. We determine which componen ts P and O intersect at a bad fibre and what contribution they have. In section 2.1, we determined that E has 21 fibres of type I1 and one of type III (at ∞).

Fibres of type I1 have no contribution but:

contrv(P ) = (1

2 if P and O intersect different components at ∞

0 if P and O intersect the same components at ∞ (13) To look at the curve at t = ∞, recall equations 9 and 10:

E : y2 = x3+ t7x + 1

E0 : η2 = ξ3+ sξ + s12, so we determine its behaviour at s = 0.

A point P = (x(t), y(t)) in E is equivalent to P = (s4x(1s), s6y(1s)). Equiv- alently, the x-coordinate described by:

x(t) = adtd+ . . . + a0 bete+ . . . + b0,

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is the following in s:

s4· x(s−1) = s4ads−d+ . . . + a0

bes−e+ . . . + b0. (14) In the latter case, the degree of the numerator equals 4 − d and the degree of the denominator equals −e. We now have two cases.

Case 1: −e ≤ 4 − d. We multiply the rational function by se, to get s4· x(s−1) = ads4−d+e+ . . . + a0se

be+ h.o.t. . We substitute s = 0 to see that

s4· x(s−1) =

(0 if 4 − d + e > 0,

be if 4 − d + e = 0.

P (∞) = (0 : 0 : 1) if d < 4 + e and a smooth point if d = 4 + e.

Case 2: 4 − d < −e, then:

besd−4−e+ h.o.t..

At s = 0, the point is at ∞, and the curve is reduced to η2 = ξ3, which provides a smooth point. This means that equation 13 reduces to:

contrv(P ) =





1

2 if deg( numerator of x(t))

< 4 + deg(denominator of x(t)) 0 otherwise

The minimality of hP, P i is reached when (P O) = 0, so P (t0) 6= O =

∞ must hold for finite t0 and infinity. This means that x(P ) must be a polynomial, i.e. of the form adtd+ . . . a0, for finite t0. If t0 = ∞, or s0 = 0, we require s40· x(s−10 ) to be a polynomial (as shown by equation 14). So x(P ) must be a polynomial with deg(x(P )) ≤ 4.

The following table gives the degree in t of the right hand side of y2 = x3+ t7x + 1, depending on the degree of (x(t)), and what the resulting degree of y(t) is:

deg(x(t)) deg r.h.s. deg(y(t))

0 0 (if x = 0), 7 (if x 6= 0) 0 (cannot exist if x 6= 0)

1 8 4

2 9 cannot exist

3 10 5

4 12 6

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This concludes the proof that deg(x(t)) ≤ 4 and that deg(y(t)) ≤ 6.

In our search for points P = (x(t), y(t)) on E, we will first try to find points with the property hP, P i ≤ 4. The correspoding x(t) and y(t) are then denoted as follows:

x(t) =

4

X

m=0

amtm,

y(t) =

6

X

n=0

bmtm.

By comparing coefficients in the equality y(t))2 = (x(t))3+ x(t)t7+ 1, we obtain the following set of equations that (a0, . . . , a4, b1, . . . , b6) must satisfy for the pair (x(t), y(t)) to be a point P on E:

a34 = b26 3a3a24+ a4 = 2b5b6, 3a2a24+ 3a23a4+ a3 = 2b4b6+ b25, 3a1a24+ 6a2a3a4 + a2+ a33 = 2b3b6+ 2b4b5, 3a0a24+ 6a1a3a4+ a1+ 3a22a4+ 3a2a23 = 2b2b6+ 2b3b5+ b24, 6a0a3a4+ a0+ 6a1a2a4+ 3a1a32+ 3a22a3 = 2b1b6+ 2b2b5+ 2b3b4, 6a0a2a4+ 3a0a32+ 3a21a4+ 6a1a2a3+ a32 = 2b0b6+ 2b1b5+ 2b2b4+ b23,

6a0a1a4+ 6a0a2a3+ 3a21a3+ 3a1a22 = 2b0b5+ 2b1b4+ 2b2b3, 3a20a4+ 6a0a1a3+ 3a0a22+ 3a21a2 = 2b0b4+ 2b1b3+ b22,

3a20a3+ 6a0a1a2+ a31 = 2b0b3+ 2b1b2, 3a20a2+ 3a0a12 = 2b0b2+ b21,

3a20a1 = 2b0b1, a30+ 1 = b20,

The code to obtain this set can be found in 7.2. Should we try to find a solution depending only on (a0, . . . , a3, b0, . . . b5), i.e. a point P with height pairing 72, then magma is quick to find out that only the trivial soulutions can exist (see 7.4 for the code). A nontrivial solution depending on all twelve variables cannot be found by magma, but a different approach is discussed in the next sections.

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### 4.2 A point over F

41

Magma cannot find a set (a0, . . . , a4, b0, . . . , b6) that corresponds to a non- trivial point of E over Q(t), but when solving the problem over Fp, solutions can be found in a reasonable time. This is what we will now do, with p = 41.

When we have found points P ∈ E(F41), we will use Hensel’s lemma (as explained in section 4.3) to lift P to a point in Qp(t), the function field over the field of p-adic numbers.

The magma code which determines the equations in the previous section, is adapted to find solutions to this set, and can be found in 7.3. Using this code, we find 48 points (24 pairs of P and −P ) in E(F41(t)).

An example of such a solution is a = (0, −5, 4, 7, −8, 1, 0, 0, −1, 27, −1, 29).

This corresponds to P = (x(t), y(t)) = (−5t + 4t2+ 7t3− 8t4, 1 − t3+ 27t4− t5+ 29t6) ∈ E(F41(t)), with E : y2 + x3+ t7x + 1.

The field extension F41(t21) is not Galois, since the minimal polynomial of t over F41(t21) is x21 − t21, which does not split over F41(t). To obtain a splitting field, we need a primitive 21st root of unity. Since 21|412 − 1 = (41 − 1)(41 + 1), such a root ζ exists in F412, and we see that F412(t)/F412(t21) is Galois, with group Gal(F412(t)/F412(t21)) generated by σ, which is defined by σ(t) = ζt.

The curve ˜E : y2 = x3 + t−21x + t−42 is defined over F412(t21), so σ generates an automorphism of the group

E(F˜ 412(t)) ∼= E(F412(t)), with isomorphism:

P = x(t) t14 ,y(t)

t21  7→ (x(t), y(t)) (15) σ(P ) = x(ζt)

(ζt)14, y(ζt)

(ζt)21 7→ (ζ7x(ζt), y(ζt).

σ21= id in Gal(F41(t)/F412(t)), so we also have:

21− id) ∈ End(E(F412(t))) q

(σ − id)Φ3(σ)Φ7(σ)Φ21(σ)

Recall the image of the subfields of k(t), on page 12. With k = F41, we get the following construction of subfields:

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F412(t)

F412(t3)

3 99

F412(t7)

7

ee

F412(t21)

7

ee

3

99

Using magma (see 7.5 for the code), we determine (1−σ(P ))(Φ21(σ(P )) 6=

O. This is different from the situation over Q, where (1 − σ)(Φ21(σ)) = O in End(E(Q(t)).

Equation 15 shows that σ3(P ) = P , so

E(F412(t)) ⊃ E(F412(t))3=idi = E(F412(t7) = E(F412(t21)) = E(F412(t))hσ=idi, then E(F412(t7)) has rank 1. We deduce that (σ − id)Φ7(σ)Φ21(σ) = O in End(E(F412(t))). This means that E(F412(t))7=idi = E(F412(t3)) must have rank at least 7.

To check if (σ−id)Φ7(σ) = O, assume that there is a P ∈ E(F412(t))7=idi. Then σ7(P ) = P , so (ζ7x(ζ7t), y(ζ7t)) = P . However, take P = (−5t + 4t2+ 7t3− 8t4, 1 − t3+ 27t4− t5+ 29t6), then σ7(x(P )) has first coeficient equal to ζ14(−5) 6= −5, contradicting the assumption. So P is not exclusively found in E(F412(t))7=idi. Then (σ − id)Φ7(σ) 6= O.

As in the case of a field with characteristic 0, we have the following construction of the Mordell-Weil group of E over the different finite function fields:

E(F241(t)) ∼= Za+2b+6c+12d

E(F241(t3)) ∼= Za+6c

44

E(F241(t7)) ∼= Za+2b

jj

E(F241(t21)) ∼= Za

jj 44

We have established that rank(E(F412(t7))) = 1, so b = 0. Similarly, c ≥ 1 and d ≥ 1. E is a K3 surface, so it has rank at most 20. We conclude that c = d = 1. Since E(F412(t))hσ=idi has rank 1, rank(E(F412(t))) = 19.

As an example, let P = (−5t + 4t2 + 7t3 − 8t4, 1 − t3+ 27t4− t5 + 29t6) and apply the magma code from section 7.5. Then we indeed have:

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Φ21(σ(P ))Φ7(σ(P ))(1 − σ(P ))

= −P + σ(P ) − σ3(P ) + σ4(P ) − σ6(P ) + σ7(P ) − σ9(P ) + σ10(P )

− σ12(P ) + σ13(P ) − σ15(P ) + σ16(P ) − σ18(P ) + σ19(P )

= O.

### 4.3 Hensel lifting

In this section, we will lift the point P = (x(t), y(t)) from the previous section from Fp(t) to Qp(t), the function field over the field of p-adic numbers. To do so, we first discuss Hensel’s lemma:

Theorem 4.2. If f (X) ∈ Zp[X] and a a ∈ Zp satisfies

f (a) ≡ 0 mod p, f0(a) 6= 0 mod p, (16) then there is an α ∈ Zp such that f (α) = 0 ∈ Zp and α ≡ a mod p.

The theorem and proof can be found in . It is also the case n = 1 of theorem 4.3 below.

While this theorem helps us find a zero in Zp of a polynomial f (x), we need an analog for a system of polynomial equations in more variables. This is stated as follows:

Theorem 4.3. Let f1, . . . fn be a system of n polynomials in Zp[x1, . . . , xn] and let (a1, . . . , an) be a vector in Znp with:

fi(a) ≡ 0 mod p for all i ∈ {1, . . . , n}

and

det( ∂

∂xjfi)(a) 6= 0 mod p

Then α ∈ Znp exists such that fi(α) = 0 for all i and α ≡ a mod p.

Proof. We will construct a Cauchy sequence a(1) = a, a(2), . . . , a(m), . . . in Znp, such that fj(a(m)) ≡ 0 mod pm for all j and m. Then α := limm→∞a(m) is the desired solution.

Suppose m ≥ 1 and a(1). . . , a(m) have been constructed satisfying a(j) ≡ a(j−1) mod pj−1, for all j ≤ m, and fi(a(j)) ≡ 0 mod pj for all i. To determine a(m+1), try a(m+1) := a(m) + pm · λ for some λ ∈ Znp. Write fi(a(m)) = pm· bi with bi ∈ Zp, and i = 1, . . . , n.

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Then

fi(a(m+1)) mod pm+1

= fi(a(m)) mod pm+1+ pm(∇fi(a(m)))λ mod pm+1

= pm(bi+ (∇fi)λ) mod pm+1

This is 0 mod pm+1 for every i = 1, . . . , n, precisely when bi+ (∇fi)λ ≡ 0 mod p. Since a(m)≡ a(1) mod p, this means λ should satisfy:

∇f1(a)

∇f2(a) ...

∇fn(a)

 λ1 λ2

... λn

=

−b1

−b2

...

−bn

mod p.

By assumption, the matrix presented here is nonsingular modulo p, hence such a λ exists.

We will now use theorem 4.3 to show the following:

Proposition 1. There exist α = (α0, . . . , α4, β0, . . . , β6) ∈ Z1241, with α 6=

(0, . . . , 0, ±1, 0, . . . , 0) such that fi(a) = 0 for all 1 ≤ i ≤ 13, with:

f1 = a34− b26

f2 = 3a3a24+ a4− 2b5b6,

f3 = 3a2a24+ 3a23a4+ a3− 2b4b6+ b25,

f4 = 3a1a24+ 6a2a3a4+ a2+ a33− 2b3b6 + 2b4b5,

f5 = 3a0a24+ 6a1a3a4+ a1+ 3a22a4+ 3a2a23− 2b2b6+ 2b3b5+ b24, f6 = 6a0a3a4+ a0+ 6a1a2a4+ 3a1a23+ 3a22a3− 2b1b6+ 2b2b5 + 2b3b4, f7 = 6a0a2a4+ 3a0a23+ 3a21a4+ 6a1a2a3+ a32− 2b0b6+ 2b1b5 + 2b2b4+ b23, f8 = 6a0a1a4+ 6a0a2a3+ 3a21a3+ 3a1a22− 2b0b5+ 2b1b4 + 2b2b3,

f9 = 3a20a4+ 6a0a1a3+ 3a0a22+ 3a21a2− 2b0b4+ 2b1b3+ b22, f10= 3a20a3+ 6a0a1a2+ a31− 2b0b3 + 2b1b2,

f11= 3a20a2+ 3a0a21− 2b0b2+ b21, f12= 3a20a1− 2b0b1,

f13= a30− b20+ 1.

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Proof. For the moment, only consider the system f1 = f2 = . . . = f12 = 0, then a := (0, −5, 4, 7, −8, 1, 0, 0, −1, 27, −1, 29) satisfies

det( ∂

∂xjfi)(a) 6= 0 mod 41.

Namely,

det( ∂

∂xjfi)(a) =

0 0 0 0 28 0 0 0 0 0 0 24

0 0 0 28 −8 0 0 0 0 0 24 2

0 0 28 −8 −4 0 0 0 0 24 2 28

0 28 −8 −4 −2 0 0 0 24 2 28 2

28 −8 −4 −2 2 0 0 24 2 28 2 0

−8 −4 −2 2 3 0 24 2 28 2 0 0

−4 −2 2 3 −7 24 2 28 2 0 0 −2

−2 2 3 −7 0 2 28 2 0 0 −2 0

2 3 −7 0 0 28 2 0 0 −2 0 0

3 −7 0 0 0 2 0 0 −2 0 0 0

−7 0 0 0 0 0 0 −2 0 0 0 0

0 0 0 0 0 0 −2 0 0 0 0 0

0 0 0 0 0 −2 0 0 0 0 0 0

= 2 mod 41.

By theorem 4.3, we now know that α0 ∈ Z1241 exists such that fi0) = 0 for 1 ≤ i ≤ 12, and α0 ≡ a mod 41. Put

ξ(t) := α0+ α1t + . . . + α4t4, η(t) := α5+ α6t + . . . + α11t6.

By construction, ξ(t)3 + t7ξ(t) + 1 − η(t)2 is a constant c ∈ 41Z41, so η2 = ξ3+ t7ξ + 1 − c, with 1 − c 6= 0 and deg(ξ) = 4, deg(η) = 6. Applying Hensel’s lemma (see theorem 4.2) to the polynomials x2− 1 + c and x3− 1 + c, we obtain λ, µ ∈ Z41 with λ ≡ µ ≡ 1 mod 41 and λ2 = 1 − c = µ3. Now η1(t) := η(t)λ and ξ1(t) := ξ(t)µ satisfy η12 = ξ13+ µt72ξ1+ 1.

Apply Hensel’s lemma to x7 − µ2, which yields ν ∈ Z41 with ν ≡ 1 mod 41 and ν7 = µ2. Finally, write

s := t ν,

ξ2(s) := ξ1(sν) = ξ1(t), η2(s) := η1(sν) = η1(t).

Then ξ2, η2 are polyonomials of degree 4 resp. 6, and η22 = ξ23+ s7ξ2+ 1. The coefficients of ξ2, η2 provide the solution α ∈ Z1241.

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Fix a field homomorphism Qp → C, this embeds Z(Qp) → Z(C).

Claim: Z(C) is finite, hence Z(C) = Z(Q).

Proof. By construction, every point in Z(C) corresponds to a point P ∈ E(C)(t)) of height hP, P i ∈ {72, 4}. Since E(C(t)) contains only finitely many points of bounded height, this shows that Z(C) is finite. Suppose α ∈ Z(C) \ Z(Q). Then at least one of the coordinates of α is transcendental over Q. Specializing this coordinate to elements of Q then yields infinitely many points in Z(Q).

Since we showed #Z(Q41) ≥ 3, it follows that Z(C) = Z(Q) also has at least 3 elements. So indeed a point P ∈ E(Q(t)) exists with the x-coordinate a polynomial of degree 4.

### 5 Conclusion

In this thesis, we have tried to find the generators of the Mordell-Weil group of the elliptic curve E/Q(t) defined by the equation y2 = x3+ t7x + 1. We used a geometric approach to show that the Mordell-Weil group has rank 13, and showed that the group has no nontrivial torsion by using either the Nagell-Lutz theorem or reduction theory. By determing the Galois action on the rational points, we have shown that the thirteen generators are two sets: one only containing the trivial generator and one containing the twelve nontrivial generators. If we find one nontrivial generator, we can then apply the Galois action to find the other eleven.

So far, it is impossible to find a point on the curve over a function field with characteristic zero, but the computing software is able to determine a point on the curve over a finite field. We can then lift the point to a p-adic field using Hensel’s lemma. Using some transformations, we then proved that a solution α ∈ Z1241 exists. Finally, a homomorphism is fixed to embed Z(Qp) in Z(C) and the fact that Z(C) = Z(Q) tells us that there is a point P ∈ E(Q(t)) with the x-coordinate a polynomial of degree 4.

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In this section, we prove that rankE/Q(t3) = 1 and that rankE/Q(t7) = 1, facts we used in section 3.3.

Theorem 6.1. Let E : y2 = x3+ t−21x + t−42 be an elliptic curve over Q(t7).

Then this curve has rank equal to 1.

Proof. Let u = t−7, so Q(t7) = Q(u−1) = Q(u), the rank of E/Q(t7) is obviously the same as the rank of E/Q(u):

E : y2 = x3+ t−21x + t−42

E : η2 = ξ3+ u3ξ + u6

This curve is a rational elliptic curve, so equation 4 reduces to r = b2− ρtriv (cite?), with b2 = 10. To determine ρtriv, we apply the same method as we did in theorem 3.1.

∆(E/Q(u)) = 4u9+ 27u12 = u9(4 + 27u3), so E has three singular fibres of type I1 at finite nonzero u. For u = 0, we find a fibre of type III (which has eight components). In the case that u = ∞, let v = u1, to find that the curve defined by y2 = x3 + vx + 1 has a curve of type I0 (See [6, p. 14]).

Then ρtriv = 2 + 7 = 9 and r = 10 − 9 = 1.

Theorem 6.2. Let E : y2 = x3+ t−21x + t−42 be an elliptic curve over Q(t3).

Then this curve has rank equal to 1

Proof. Let s = t−3, so Q(t3) = Q(s−1) = Q(s). We transform E to a minimal curve in s:

E : y2 = x3+ t−21x + t−42

E : η2 = ξ3+ s3ξ + s8

This is a K3 surface, so we have to determine the Lefschetz number.

The homogenized form of the equation is X3Z5 + S3XZ5 − Y2Z6 + S8 and exponents form the following matrix:

A =

0 3 0 5 3 1 0 4 0 0 2 6 8 0 0 0

with inverse: A−1 = 1 56

0 0 0 7

32 −40 0 15

24 −72 28 27

−8 24 0 −9

 .

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