Two Cases of Fermat's Last Theorem Using Descent on Elliptic Curves

A. M. Viergever Two cases of Fermat’s Last Theorem using descent on elliptic curves

Bachelor thesis 22 June 2018

Thesis supervisor: dr. R.M. van Luijk

Leiden University Mathematical Institute Contents

Introduction2

1 Some important theorems and deﬁnitions4 1.1 Often used notation, deﬁnitions and theorems...... 4 1.2 Descent by 2-isogeny...... 5 1.3 Descent by 3-isogeny...... 8

2 The case n = 4 10 2.1 The standard proof...... 10 2.2 Descent by 2-isogeny...... 11 2.2.1 C and C0 are elliptic curves...... 11 2.2.2 The descent-method applied to E ...... 13 2.2.3 Relating the descent method and the classical proof...... 14 2.2.4 A note on the automorphisms of C ...... 16 2.3 The method of descent by 2-isogeny for a special class of elliptic curves... 17 2.3.1 Addition formulas...... 17 2.3.2 The method of 2-descent on (Ct, O)...... 19 2.4 Twists and φ-coverings...... 22 2.4.1 Twists...... 22 2.4.2 A φ-covering...... 25

3 The case n = 3 26 3.1 The standard proof...... 26 3.2 Descent by 3-isogeny...... 28 3.2.1 The image of q ...... 28 3.2.2 The descent method applied to E ...... 30 3.2.3 Relating the descent method and the classical proof...... 32 3.2.4 A note on the automorphisms of C ...... 33 3.3 Twists and φ-coverings...... 34 3.3.1 The case d ∈ (Q∗)2 ...... 34 3.3.2 The case d∈ / (Q∗)2 ...... 38 A Appendix 41 A.1 An isomorphism of Ct to a curve in Weierstrass form...... 41 A.2 Smoothness of the Fermat curve...... 42

Bibliography 43 1 Introduction

In this thesis, we will consider two special cases of a very famous theorem which has a long history, and analyze them geometrically using elliptic curves. Let n ∈ Z≥1 and consider the equation xn + yn = zn (1) of which we are looking for integer solutions. For n = 1, this is trivial, and for n = 2, it turns out that there are inﬁnitely many solutions.

The case n = 2

An integer solution (x, y, z) ∈ Z3 to the equation x2 + y2 = z2 is called a Pythagorean triple. We say that it is primitive if gcd(x, y, z) = 1. Note that if (x, y, z) ∈ Z3 is a Pythagorean triple, we have for all λ ∈ Z that (λx, λy, λz) is also a Pythagorean triple. So once we know that, for example, (3, 4, 5) is an integer solution to equation (1), we have immediately proved that there are infinitely many Pythagorean triples. However, there is also a more geometric way to show this, and actually even more: we will show that there exist infinitely many primitive Pythagorean triples. Note that finding a solution in co-prime integers to the equation x2 + y2 = z2 is, since we can assume that z 6= 0 (for (0, 0, 0) is not primitive), equivalent to finding a rational solution to the equation

x2 + y2 = 1 (2)

i.e. to ﬁnding a rational point on the unit circle in R2. Now, let t ∈ Q be a rational number and consider the line given by y = tx + t, which intersects the unit circle in (−1, 0) and exactly one other point. This point is of the form (x, tx + t), for some x ∈ Q which we can ﬁnd by solving the equation

x2 + (tx + t)2 = 1 so (1 + t2)x2 + 2t2x + t2 − 1 = 0.

As we already knew, x = −1 is a solution, and we ﬁnd that the x-coordinate of the other 1−t2 point where the line intersects the unit circle is x = 1+t2 . This deﬁnes a map φ from Q to set of the rational solutions of (2), given by 1 − t2 2t φ: → {(x, y) ∈ 2 : x2 + y2 = 1}\{(−1, 0)}, t 7→ , . Q Q 1 + t2 1 + t2

On the other hand, if we have a rational solution (x0, y0) to (2), we take the line L from y0 y0 (−1, 0) to (x0, y0), which is given by y = x + and intersects the y-axis in the point x0+1 x0+1 (0, y0 ). This gives a map x0+1 y ψ : {(x, y) ∈ 2 : x2 + y2 = 1}\{(−1, 0)} → , (x, y) 7→ Q Q x + 1 and it can easily be checked that ψ provides an inverse for φ, which shows that φ is a bijection. We ﬁnd that the set of rational solutions of equation (2) can be parametrized by Q.

2 The case n ≥ 3 For n ≥ 3, we can still ﬁnd some “trivial solutions” to equation (1), for example (1, 0, 1) and (0, 1, 1). However, it is a very famous result that there are no nontrivial solutions.

Theorem 0.0.1 (Fermat’s Last Theorem). For n ≥ 3, equation (1) has no integer solutions (x, y, z) ∈ Z3 satisfying xyz 6= 0. This theorem has been named after Pierre de Fermat, a French nobleman born in 1601 with a lifelong interest in mathematics. He read a version of Diophantus’s “Arithmetica”, and wrote down some of his own conjectures in the margins of the book, including Theorem 0.0.1. Fermat claimed that he had a proof, but that the margins were, unfortunately, too small to write it down. Later on, all margin-conjectures of Fermat had been proven, but even after 350 years, nobody had managed to prove Theorem 0.0.1. It therefore became known as “Fermat’s Last Theorem”. Finally, in 1994, Theorem 0.0.1 has been proven by the British mathematician Andrew Wiles [Wil95], who — after working in isolation for seven years — found a brilliant proof which has been called “the proof of the century”1.

Structure and goal of this thesis Wiles’ proof is far too difficult to treat in this thesis, but for many special cases of The- orem 0.0.1, easier proofs had already been discovered before the general case had been established. The case n = 4, for example, is the only one of which we know a proof by Fermat himself, and the case n = 3 has been proven by Euler in 1770 (we will give versions of those proofs in Section 2.1 and Section 3.1, respectively). Both of those proofs come down to the principle of infinite descent. The key idea of this method is that whenever we have a solution (x, y, z) ∈ Z3 to equation (1) (for n = 3 or n = 4) satisfying xyz 6= 0, we can find another solution (u, v, w) ∈ Z3 satisfying uvw 6= 0 with |u| < |x|, |v| < |y| and |w| < |z|. Since this can never be the case when working with positive integers, we have a contradiction, so we conclude that there are no nontrivial solutions at all. This method has been the inspiration for the method of descent by an isogeny for elliptic curves, which can be used to prove special cases of the Weak Mordell–Weil Theorem. In this thesis, we will relate the classical proofs for the cases n = 3 and n = 4 of Fermat’s Last Theorem to a descent by 2- and 3-isogeny on elliptic curves. This will provide a geometric explanation for the fact that there are no nontrivial solutions. Chapter 1 will summarize some of the definitions and theorems which will be used. In Chapter 2, the case n = 4 is analyzed, whereas Chapter 3 is devoted to the case n = 3.

1By John Conway, who was interviewed about Wiles’ achievement in the documentary “The proof” produced by BBC in 1997.

3 Chapter 1

Some important theorems and deﬁnitions

This chapter will introduce some of the notation, deﬁnitions and theorems which will be used in the rest of this thesis. We will need some background knowledge about elliptic curves, which can be found in Chapter 1 till 3 from [Sil09] (of which we also adopt most of our notation). In Section 1.1, we will highlight some theorems, notations and deﬁnitions which we will use very often. Section 1.2 will introduce the method descent by 2-isogeny on elliptic curves, and Section 1.3 is devoted to the method of descent by 3-isogeny.

1.1 Often used notation, deﬁnitions and theorems

n Let n ∈ Z≥1 and choose an algebraic closure Q of Q. We will denote A for the aﬃne n-space over Q and Pn for the n-dimensional projective space over Q. Furthermore, we will need the following deﬁnition in Chapter 2.

Definition 1.1.1. Let w0, ..., wn ∈ Z≥1 be integers. We define the weighted projective space n+1 P(w0, ..., wn) over Q to be the set of equivalence classes of A \{(0, 0, ..., 0)} under the 0 0 ∗ relation ∼ defined by (a0, ..., an) ∼ (a0, ..., an) if and only if there exists some λ ∈ Q such wi 0 that λ ai = ai for all i ∈ {0, ..., n}. The notions of torsion points and dual isogenies will be of great interest to us, so we will 0 briefly recall them. Let (E1, O) and (E2, O ) be elliptic curves over Q. Then for m ∈ Z, we have the multiplication by m map [m]: E1 → E1 defined by

 m times z }| { P + P + ··· + P if m ≥ 1  [m]: E1 → E1,P 7→ − (P + P + ··· + P ) if m ≤ −1 .  | {z }  −m times O otherwise

Since [m](O) = O, we have that [m] is an endomorphism of E1. Identifying E1 with its set of points over Q, we denote the set of m-torsion points of E1 by E1[m] = {P ∈ E1 :[m]P = O}

4 1.2. Descent by 2-isogeny

S and its torsion subgroup by (E1)tors = m≥1 E1[m]. By Corollary 6.4 of [Sil09], if m 6= 0, ∼ 2 we have that E[m] = (Z/mZ) (we use here that we will only work over Q, which is of characteristic zero). In particular, the number of m-torsion points of E1 over Q is equal to m2. Now, let φ: E1 → E2 be a non-constant isogeny, then by Theorem 6.1 of [Sil09], there is a ˆ ˆ ˆ unique isogeny φ: E2 → E1 such that φ◦φ = [deg(φ)]. This isogeny φ is called the dual of φ. By Theorem 6.2 of [Sil09], we have that ˆ φ ◦ φˆ = [deg(φ)], deg(φˆ) = deg(φ) and φˆ = φ.

Further, denote the automorphism group of E1 over Q by Aut(E1). In Sections 2.2 and 3.2, we will need the following result, which is Corollary 10.2 of [Sil09].

Proposition 1.1.2. The automorphism group of E1 is a finite group, which has order 4 if the j-invariant of E1 equals 1728, order 6 if the j-invariant of E1 equals 0, and order 2 otherwise. In order to state another important result which we will use, we first need the following definition. Definition 1.1.3. Let C and C0 be curves. A birational map is a rational map C → C0 admitting an inverse which is also a rational map. If there exists a birational map C → C0, we say that C and C0 are birationally equivalent. In the final sections of both Chapter 2 and 3, we will need some advanced result, which is, for algebraically closed fields, Corollary I.6.11 of [Har77] (proof of which follows from Propositions I.6.7–I.6.9). The fact that this theorem is also true for general fields, follows from Proposition 2.6 of [Sta18, Algebraic Curves].

Theorem 1.1.4. Let C be a curve over Q, then there exists a smooth projective curve C over Q such that C and C are birationally equivalent, and C is unique up to isomorphism over Q. Furthermore, let D be a projective variety, and let φ: C → D be a rational map. Then φ induces a morphism φ: C → D.

1.2 Descent by 2-isogeny

We will now describe the procedure of descent by 2-isogeny on an elliptic curve, which will be one of the main subjects of this thesis. In our description of it, we will be following [Bri15], which is, in turn, following [Cas91]. We will not prove the lemmas stated in this section, but they can be found in those sources. The statement that for an elliptic curve E over a number field K we have that E(K)/2E(K) is finite is called the Weak Mordell–Weil Theorem, and descent by 2-isogeny is a way to prove it over Q, in case that the elliptic curve has a nontrivial rational 2-torsion point. So let (E, O) be an elliptic curve over Q which has a rational 2-torsion point other than O. Then, by making some translations over Q, we find that E can be given by a Weierstrass equation E : y2 = x(x2 + ax + b)

5 1.2. Descent by 2-isogeny

with a, b ∈ Z, which has the rational 2-torsion point (0, 0). Since E is smooth, the polynomial on the left hand side is separable, so we have that b 6= 0 and a2 − 4b 6= 0.

Lemma 1.2.1. Consider the second elliptic curve (E0, O0) over Q given by the equation E0 : v2 = u(u2 + a0u + b0), where a0 = −2a and b0 = a2 − 4b, and the isogeny ( (x + a + b , y − by ) if x 6= 0 φ: E → E0, (x, y) 7→ x x2 O0 otherwise then ker(φ) = {O, (0, 0)}.

Note that E0 is indeed an elliptic curve, since we have that b0 = a2 − 4b 6= 0 and also that a02 − 4b0 = 4a2 − 4a2 + 16b = 16b 6= 0, which implies that it is smooth. Lemma 1.2.1 also tells us that φ is of degree two. Using Lemma 1.2.1 again on E0 yields an elliptic curve E00 : y2 = x(x2 + 4ax + 16b) which is isomorphic to E (by E → E00, (x, y) → (4x, 8y)), and an isogeny

( 0 0 ( 1 (u + a0 + b ), 1 (y − b y )) if u 6= 0 φˆ: E0 → E, (u, v) 7→ 4 u 8 x2 O otherwise satisfying ker(φˆ) = {O0, (0, 0)}. The slight abuse of notation already hints towards the following lemma.

Lemma 1.2.2. The isogeny φˆ is the dual of φ.

In other words, φˆ ◦ φ = [2] on E.

Lemma 1.2.3. Deﬁne the map ( 0 ∗ ∗ 2 [u] if u 6= 0 q : E (Q) → Q /(Q ) , (u, v) 7→ [b0] if u = 0

and q(O) = [1]. Then q is a homomorphism of groups and the sequence

φ 0 q ∗ ∗ 2 E(Q) −→ E (Q) −→ Q /(Q )

is exact. Furthermore, let r ∈ Z be square-free, then we have that [r] ∈ im(q) if and only if the equation r2l4 + a0rl2m2 + b0m4 = rn2 has a solution (l, m, n) ∈ Z3 \{(0, 0, 0)}. This can only happen if r divides b0.

It follows that im(q) is finite, since there are only finitely many r ∈ Z dividing b0. Further- ∼ 0 0 more, by the Isomorphism Theorem we have that im(q) = E (Q)/ ker(q) = E (Q)/φ(E(Q)) is finite. Repeating the procedure on E0 yields that E(Q)/φˆ(E0(Q)) is also finite. We will now, and also often later, need the following tool from homological algebra.

6 1.2. Descent by 2-isogeny

Lemma 1.2.4 (Kernel-cokernel sequence). Let A, B and C be abelian groups and f : A → B and g : B → C group homomorphisms. Let i1 : ker(f) → ker(g ◦ f) be the natural inclusion map, let i2 : ker(g) → coker(f) be the restriction of the natural quotient map B → coker(f) and let i3 : coker(g ◦ f) → coker(g) be the natural quotient map. Then the sequence f g 0 → ker(f) −→i1 ker(g ◦ f) −→ ker(g) −→i2 coker(f) −→ coker(g ◦ f) −→i3 coker(g) → 0 is exact. So if we denote E(Q)[2] for the rational 2-torsion points of E, the sequence 0 0 0 → ker(φ) → E(Q)[2] → ker(φˆ) → E (Q)/φ(E(Q)) → E(Q)/2E(Q) → E(Q)/φˆ(E (Q)) → 0 is exact, from which it follows that E(Q)/2E(Q) is finite. This proves the Weak Mordell–Weil Theorem, which can be used to prove the “strong” Mordell–Weil Theorem. Theorem 1.2.5 (Mordell–Weil). Let E be an elliptic curve over a number field K. Then E(K) is a finitely generated abelian group. ∼ r By the Structure Theorem of abelian groups, we have that E(Q) = E(Q)tors ⊕ Z for some r ∈ Z≥0, which is called the rank of E. We will now show how the method of descent by 2-isogeny can be used to find this rank. We need the following two lemmas. Lemma 1.2.6. Let A be a finite abelian group and denote A[2] for its 2-torsion subgroup, then #(A/2A) = #A[2]. Proof. We have that A → A, x 7→ 2x is a group homomorphism with image 2A and kernel A[2], so by the Isomorphism Theorem, 2A ∼= A/A[2], which implies that #A #(2A) = #(A/A[2]) = #A[2] so #(A/2A) = #A[2]. Lemma 1.2.7. The cardinality of E(Q)/2E(Q) equals #coker(φ) · #coker(φˆ) if and only if 0 1 ˆ b is a square, and 2 · #coker(φ) · #coker(φ) otherwise. Proof. From the kernel-cokernel sequence in Lemma 1.2.4, we see that the sequence 0 0 → ker(φˆ)/φ(E(Q)[2]) → E (Q)/φ(E(Q)) → E(Q)/2E(Q) → E(Q)/φˆ(E(Q)) → 0 is also exact.This implies that

#(E0(Q)/φ(E(Q))) · #(E(Q)/φˆ(E(Q))) #(E(Q)/2E(Q)) = . #(ker(φˆ)/φ(E(Q)[2])) We know that E has four 2-torsion points over Q. The isogeny φ maps (0, 0) and O to O0, so it sends the two remaining 2-torsion points to (0, 0) (by the kernel of φˆ). So #(ker(φˆ)/φ(E(Q))[2]) equals 2 if E has no other rational 2-torsion points than (0, 0) and O, and 1 otherwise. Therefore, # (E(Q)/2E(Q)) equals #coker(φ) · #coker(φˆ) if and only if (0, 0) ∈ φ(E(Q)), so if and only if E has four rational 2-torsion points, which is the case if 0 1 ˆ and only if b is a square. In the same way, # (E(Q)/2E(Q)) equals 2 ·#coker(φ)·#coker(φ) if and only if (0, 0) ∈/ φ(E(Q)), so if and only if E has precisely two rational 2-torsion points, which is the case if and only if b0 is not a square.

7 1.3. Descent by 3-isogeny

The method of descent by 2-isogeny can be used to ﬁnd #coker(φ) and #coker(φˆ), from which we can compute the cardinality of E(Q)/2E(Q) using Lemma 1.2.7. Note that

r E(Q)/2E(Q) = (E(Q)tors/2E(Q)tors) ⊕ (Z/2Z) .

Since the torsion subgroup E(Q)tors is ﬁnite, we can apply Lemma 1.2.6 to it, then we have that # (E(Q)tors/2E(Q)tors) = #E(Q)[2]. We can therefore compute the rank r of E(Q) if we know the cardinality of the 2-torsion subgroup. This can be retrieved by observing that E has four rational 2-torsion points if and only if b0 is a square, and two otherwise.

1.3 Descent by 3-isogeny

Analogous to descent by 2-isogeny, there is also a method of descent by 3-isogeny, which we will summarize following [Coh07], to which we also redirect for proofs of the stated lemmas. Let E be an elliptic curve over Q given by the equation E : y2 = x3 + d(ax + b)2

where a, b, d ∈ are such that b 6= 0, d 6= 0 and 27b − 4a3d 6= 0 (this will ensure that E is Q ∗ smooth again, since the discriminant of E equals 16d2b3(27b − 4a3d)). Let δ ∈ Q be such that δ2 = d. Then one can check that (0, bδ) is a 3-torsion point, so it generates a subgroup of E of order 3 which is stable under Gal(Q/Q) (note that this need not be a group generated by a rational 3-torsion point).

Lemma 1.3.1. Consider the second elliptic curve E0 given by

v2 = u3 + d0(a0u + b0)2

0 0 27b−4a3d 0 0 where a = a, b = 9 and d = −3d. Then the isogeny φ: E → E given by

( 3 2 2 2 3 x +4d((a /3)x +abx+b ) , y(x −4bd(ax+2b)) if x 6= 0 (x, y) 7→ x2 x3 O0 otherwise

satisﬁes ker(φ) = {O, (0, bδ), (0, −bδ)}.

∗ It follows that deg(φ) = 3. Let δ0 ∈ Q be such that δ02 = d0. Using Lemma 1.3.1 again on E0, we get a third elliptic curve E00 with a00 = a, b00 = 3(27b − 4a3d) + 12a3d = 9b and d00 = 9d, which is isomorphic to E (by E → E00, (x, y) 7→ (9x, 27y)). Furthermore, we get a second isogeny φˆ: E → E0 given by

( 3 0 02 2 0 0 02 3 0 0 0 0 u +4d ((a /3)u +a b u+b ) , v(u −4b d (a u+2b )) if u 6= 0 (u, v) 7→ 9u2 27u3 O otherwise

and φˆ(O0) = O satisfying ker(φˆ) = {O0, (0, b0δ0), (0, −b0δ0)}. As the slight abuse of notation already indicates, these isogenies are dual.

8 1.3. Descent by 3-isogeny

Lemma 1.3.2. The isogeny φˆ ◦ φ is multiplication by 3 on E.

0 0 ∗ ∗ 3 We now set Kd0 = Q(δ ) and deﬁne the 3-descent map q : E (Q) → Kd0 /(Kd0 ) by ( [v − (a0u + b0)δ0] if u 6= 0 (u, v) 7→ [2b0δ0] otherwise

and q(O0) = [1].

Lemma 1.3.3. The map q is a homomorphism of groups, and the sequence

φ 0 q ∗ ∗ 3 E(Q) −→ E (Q) −→ Kd0 /(Kd0 ) is exact.

A proof that the image of q is finite is given in [vB10]. One can then use the kernel-cokernel sequence again to deduce that E(Q)/3E(Q) is finite. 0 ∗ 2 If d ∈/ (Q ) , an extra complexity arises in working with the field Kd0 instead of Q. The following proposition proved by Cohen will turn out to be useful.

0 2 ∗ ∗ 3 Proposition 1.3.4. If d ∈/ Q , the image of q is contained in the subset of Kd0 /(Kd0 ) consisting of all elements of which the norm is trivial in Q∗/(Q∗)3. In Chapter 3, we will need a bit of algebraic number theory. In particular, we will use the notion of the ring of integers O(Kd0 ) of Kd0 , which is deﬁned as

O(Kd0 ) = {x ∈ Kd0 : the minimum polynomial of x over Q has coeﬃcients in Z}.

0 1+δ0 0 0 By Exercise 1.15 of [Ste17], if d is square-free, we have that O(Kd ) = Z[ 2 ] if d ≡ 1 0 modulo 4, and O(Kd0 ) = Z[δ ] otherwise. By Exercise 1.23, the ring O(K−3) is a principal ideal domain, as we will need in Section 3.2.

9 Chapter 2

The case n = 4

For n = 4, a proof of Fermat’s Last Theorem was given by Fermat himself [Fer91]. We will give a version of it in Section 2.1. In Section 2.2, we will relate the standard proof to a descent by 2-isogeny. Then in Section 2.3, we will work out a version of the descent by 2-isogeny method on elliptic curves of a certain form, which “look like the Fermat curve”. Finally, in Section 2.4, we will view the method of descent by 2-isogeny from a diﬀerent perspective.

2.1 The standard proof

In giving this proof, we will follow [Bri15]. Theorem 2.1.1 (Fermat, n = 4). The equation x4 + y4 = z4 has no solutions (x, y, z) ∈ Z3 with xyz 6= 0. Proof. Suppose that (x, y, z) ∈ Z3 is a solution satisfying xyz 6= 0, then x4 + y4 = (z2)2 so (x, y, z2) is a nontrivial solution to the equation x4 + y4 = z2 (2.1) which implies that it suﬃces to prove that (2.1) has no nontrivial solutions. Suppose that (x, y, z) ∈ Z3 is a solution to (2.1) with xyz 6= 0 and max{|x|, |y|} minimal, then x, y, z may be assumed to be co-prime (for if a prime p would divide x, y and z, we can 2 x y z see from equation (2.1) that p divides z and then ( p , p , p2 ) is also a solution, contradicting the minimality of max{|x|, |y|}), and even pairwise co-prime (looking at (2.1), every prime dividing two of the variables must also divide the third). Considering the equation modulo 4 (where every square is equal to either 0 or 1) we see that x and y cannot both be odd, so without loss of generality, we may assume that x is even and y is odd, so z is also odd. We now factor the equation as x4 = (z + y2)(z − y2). Note that if a prime p divides both factors on the right hand side, it also divides (z + y2) + (z − y2) = 2z, and since it cannot divide z (then it would also divide (z + y2) − z = y2 and therefore z and y would not be co-prime, producing a contradiction) we have p = 2. So gcd(z + y2, z − y2) ∈ {1, 2}. Now, note that since y2 and z are both odd, we have that either z + y2 or z − y2 is a multiple of 4 and the other one is not (otherwise, we arrive at a contradiction modulo 4). Since x4 is a multiple of 16, there are two possibilities:

10 2.2. Descent by 2-isogeny

• There are positive integers u and v such that z + y2 = 8u4 and z − y2 = 2v4, where v is odd.

• There are positive integers u and v such that z + y2 = 2u4 and z − y2 = 8v4, where u is odd. However, if we look at the ﬁrst option and eliminate z we get y2 = 4u4 − v4, so y2 ≡ −1 modulo 4, which is impossible. So the second possibility holds. We can eliminate z again to get y2 = u4 − 4v4. We can factorize this equation as 4v4 = (u2 + y)(u2 − y), then since y and u are odd, we have again that both u2 + y and u2 − y are even. Since at least one of u2 + y and u2 − y is positive, this yields that there are positive integers s and r such that u2 + y = 2r4 and u2 − y = 2s4, so r4 + s4 = u2. Thus, we have found a new solution (r, s, u) ∈ Z3 to (2.1), where x4 = (z − y2)(z + y2) = 16u4v4 = 4u4(u2 + y)(u2 − y) = 16u4r4s4

which implies that r and s are nonzero and furthermore, that |r| < |x|, which is a contradiction. So (2.1) has no nontrivial solutions in integers, which implies that neither has the equation x4 + y4 = z4.

2.2 Descent by 2-isogeny

Looking more closely at the proof given above, we have two curves:

C : X4 + Y 4 = Z2 and C0 : U 4 − 4V 4 = W 2

over Q, in the weighted projective plane P(1, 1, 2) over Q, together with two rational maps: ψ : C → C0, [X : Y : Z] 7→ [Z : −XY : X4 − Y 4] and ψˆ: C0 → C, [U : V : W ] 7→ [2UV : W : U 4 + 4V 4]. In Section 2.2.1, we will show that C and C0 are elliptic curves, and that ψ and ψˆ are isogenies. (Note that apparently, we slightly abuse the notation for the dual, but later on, we will show that ψˆ is really the dual of ψ.) Then in Section 2.2.2, we can apply the method of descent by 2-isogeny described in Section 1.2 to C and C0. In Section 2.2.3, we will relate this to the classical proof in Section 2.1, to show that the inﬁnite descent in this proof actually comes down to a descent on elliptic curves. Finally in Section 2.2.4, we will say some words on the automorphisms of C.

2.2.1 C and C0 are elliptic curves We will show now that C and C0 are elliptic curves.

Proposition 2.2.1. Let t ∈ Z \{0} be a nonzero integer and let Ct be the curve over Q in the weighted projective plane P(1, 1, 2) given by the equation Z2 = X4 + tY 4, together with the rational point O = [1 : 0 : 1]. Then (Ct, O) is an elliptic curve.

11 2.2. Descent by 2-isogeny

Proof. First, we show that Ct is smooth. Let P = [a : b : c] ∈ Ct and suppose a 6= 0. 4 2 Y Z Then we consider the affine patch 1 + ty = z , where y = X and z = X2 . Defining 4 2 ∂g 4tb3 ∂g 2c g(x, y) = 1+ty −z , we have that ∂y (P ) = a3 and ∂z (P ) = − a so Ct can only be singular at P if b = c = 0, but the equation of Ct shows that such points are not on the curve. Now, 4 2 X suppose a = 0, then b 6= 0, so we consider the affine part u + t = w , where u = Y and Z 4 2 ∂h 4a3 ∂h 2c w = Y 2 . Denote h(u, w) = u + t − w then ∂u (P ) = b3 = 0 and ∂w (P ) = − b , and since the equation of Ct shows that c 6= 0, we conclude that Ct is smooth. 4 2 It remains to show that Ct has genus 1. Therefore, we go to the affine patch 1 + ty = z Y Z with coordinates y = X and z = X2 again and consider the maximal ideal

M = {f ∈ Q[Ct]: f(0, 1) = 0} ⊂ Q[Ct] for the point (0, 1), which is generated by y and z − 1. Localizing in this ideal, we ﬁnd the local ring f R = ∈ (C ): f, g ∈ [C ], g∈ / M O g Q t Q t

which has a unique maximal ideal MO generated by M. Note that in RO, we have that z2−1 ty4 z − 1 = z+1 = z+1 , so MO is generated by y alone. Similarly, we ﬁnd that the maximal ideal for the point (0, −1) is generated by y. So y is a uniformizer in both of those points. 1 Now, consider the projection map α: Ct → P , [X : Y : Z] 7→ [X : Y ] and for P ∈ Ct, denote eα(P ) for the ramiﬁcation index of α at P , then X deg(α) = eα(P ) = eα([1 : 0 : 1]) + eα([1 : 0 : −1]) = ord[1:0:1](y) + ord[1:0:−1](y) = 2. P ∈α−1([1:0])

It is clear that the ramification points of α are the four points on Ct satisfying Z = 0, and by the degree of φ, the ramification index in those points must be equal to 2. Since φ is non-constant and separable (since every field of characteristic 0 is perfect) we have by the Riemann–Hurwitz Theorem that the genus g of Ct satisfies X (2g − 2) = deg(φ)(2 · 0 − 2) + (eφ(P ) − 1) = −4 + 4 = 0

P ∈Ct

so g = 1. We conclude that (Ct, O) is an elliptic curve.

The elliptic curve (Ct, [0 : 1 : 0]) is isomorphic to the elliptic curve in Weierstrass form 2 3 1 Et : v = u − 4tu, an isomorphism given by

2 3 3 τt : Ct → Et, [X : Y : Z] 7→ 2Y (Z + X ) : 4(ZX + X ): Y .

In particular, we have shown that (C, [1 : 0 : 1]) and (C0, [1 : 0 : 1]) are both elliptic curves over Q (take t = 1 and t = −4). Now let O = [0 : 1 : 1] and O0 = [1 : 0 : −1], then we also have that (C, O) and (C0, O0) are elliptic curves, and it is easy to check that with those chosen points, ψ and ψˆ are isogenies. Therefore, in the remainder of this section we will use those chosen points, and denote C for the elliptic curve (C, O) and C0 for the elliptic

1 See Appendix A.1 for a proof that τt is indeed an isomorphism

12 2.2. Descent by 2-isogeny

curve (C0, O0). We have that C is isomorphic to the elliptic curve E : y2 = x3 − 4x in Weierstrass form, and that C0 is isomorphic to the elliptic curve E0 : v2 = u3 + 16u in Weierstrass form (these are precisely the elliptic curves E1 and E−4). By composing τ1 with the curve automorphism of C which switches X and Y and composing τ−4 with the curve automorphism of C0 taking V to −V and W to −W , we ﬁnd isomorphisms

τ : C → E, [X : Y : Z] 7→ 2(ZX + XY 2) : 4(ZY + Y 3): X3

and τ 0 : C0 → E0, [U : V : W ] 7→ −2(WV − VU 2) : 4(WU − U 3): V 3 .

2.2.2 The descent-method applied to E We will now apply the descent by 2-isogeny method in Section 1.2 to the elliptic curve E in Weierstrass form. This yields a second elliptic curve given by v2 = u(u2 + 16), which turns out to coincide with the curve E0 isomorphic to C0, and isogenies ( (x − 4 , y + 4y ) if x 6= 0 φ: E → E0, (x, y) 7→ x x2 O0 otherwise and ( 1 u + 16 , 1 v − 16v if u 6= 0 φˆ: E0 → E, (u, v) 7→ 4 u 8 u2 O otherwise

and by Lemma 1.2.2, we have that φ ◦ φˆ = [2]. Furthermore, by Lemma 1.2.3 the maps ( 0 ∗ ∗ 2 [u] if u 6= 0 q : E (Q) → Q /(Q ) , (u, v) 7→ [1] otherwise

(and q(O0) = [1]) and ( 0 ∗ ∗ 2 [x] if x 6= 0 q : E(Q) → Q /(Q ) , (x, y) 7→ [−1] otherwise

0 ∼ 0 (and q (O) = [1]) are group homomorphisms and we have that im(q) = E (Q)/φ(E(Q)) and 0 ∼ 0 im(q ) = E(Q)/φˆ(E (Q)). To ﬁnd E0(Q)/φ(E(Q)), we need to check whether the equation r2`4 + 16m4 = rn2 has a nonzero solution (l, m, n) ∈ Z3 for all square-free r ∈ Z such that r divides 16, so we need to check r ∈ {±1, ±2}. Trivially, we have that [1] ∈ im(q). Furthermore, it is clear from the equation that a negative r is never going to work. Thus, we need only to check whether the equation 2l4 + 8m4 = n2 (2.2) has a nonzero integer solution. Suppose that (l, m, n) ∈ Z3 \{(0, 0, 0)} is a solution, then we can assume that l, m and n are co-prime (if they had a common divisor a, equation (2.2)

13 2.2. Descent by 2-isogeny

shows that a2 would divide n, so we would obtain another solution (`/a, m/a, n/a2)). Note that n must be even, since the left hand side of equation (2.2) is even, which implies that all terms except possibly for 2l4 are divisible by 4, so l is also even. But then all terms on the left hand side are divisible by 8, so 4 divides n, which implies that 8m4 must be divisible by 16, so m is also even. This is a contradiction, so equation (2.2) has no nonzero integer solutions, which implies that im(q) = {[1]}, so E0(Q)/φ(E(Q)) is trivial. We will now repeat this procedure with q0 to ﬁnd E(Q)/φˆ(E0(Q)). We need to check whether the equation r2l4 − 4m4 = rn2 has a nonzero integer solution for all square-free integers r ∈ Z dividing −4, i.e. r ∈ {±1, ±2}. Note that [1] is always in im(q0), and that for r = 2, the equation 4(l4 − m4) = 2n2 has the solution (1, 1, 0), so [2] ∈ im(q0). Furthermore, we have that [−1] ∈ im(q0) since it is the image of the rational 2-torsion point (0, 0), so 0 0 ∼ 2 im(q ) = {±[1], ±[2]}, so E(Q)/φˆ(E (Q)) = (Z/2Z) . From the kernel-corkernel sequence (Lemma 1.2.4), we have that

0 0 E (Q)/φ(E(Q)) → E(Q)/2E(Q) → E(Q)/φˆ(E (Q)) → 0

is exact, from which we deduce that E(Q)/2E(Q) is isomorphic to (Z/2Z)2. Note that E has four rational 2-torsion points, which are (2,0), (-2,0), (0,0) and O (corresponding to [1 : 0 : 1], [1 : 0 : −1], [0 : 1 : 1] and [0 : 1 : −1] on C). By the Mordell–Weil Theorem (Theorem 1.2.5) we have that E(Q) and C(Q) are ﬁnitely generated, so we have that E(Q) is of rank zero, which implies that C(Q) is of rank zero. Let T denote the torsion subgroup ∼ of C(Q), then C(Q) = T so #(T/2T ) = #(C(Q)/2C(Q)) = 4. This is actually a special case of Lemma 1.2.6. By Lemma 1.2.4 (but now with φ and φˆ switched), we also have that the sequence

0 0 0 0 E(Q)/φˆ(E (Q)) → E (Q)/2E (Q) → E (Q)/φ(E(Q)) → 0

is exact, so E0(Q)/2E0(Q) is generated by φ(0, 0) = O0 (which does not contribute anything) and φ(2, 0) = (0, 0) and therefore isomorphic to Z/2Z. By the Mordell–Weil Theorem again, E0(Q) and C0(Q) are ﬁnitely generated, so since we know that E0 has the rational 2-torsion points (0, 0) and O0 (corresponding to [1 : 0 : 1] and [1 : 0 : −1] on C0), E0(Q) is of rank zero, and therefore C0(Q) is of rank zero2.

2.2.3 Relating the descent method and the classical proof The relation between the descent method in Section 2.2 and the classical proof in Section 2.1 will become more clear now.

Proposition 2.2.2. The diagram

ψ ψˆ C C0 C

τ τ 0 τ φ φˆ E E0 E

2In fact, one can prove that the two elliptic curves used in a descent by 2-isogeny always have the same rank.

14 2.2. Descent by 2-isogeny

commutes.

Proof. Let P = [X : Y : Z] ∈ C, then we have that

(τ 0 ◦ ψ)(P ) = τ 0([Z : −XY : X4 − Y 4]) = 2((X4 − Y 4)XY − XYZ2) : 4((X4 − Y 4)Z − Z3): −X3Y 3 = [−4XY 5 : −8ZY 4 : −X3Y 3] = 4XY 2 : 8ZY : X3

and if X 6= 0 (note that since X4 = (Z − Y 2)(Z + Y 2) we have that Z + Y 2 6= 0 too in that case), we have that

Z + Y 2 ZY + Y 3 (φ ◦ τ)(P ) = φ 2(ZX + XY 2) : 4(ZY + Y 3): X3 = φ 2 , 4 X2 X3 Z + Y 2 X2 ZY + Y 3 YX = 2 − , 4 + X2 Z + Y 2 X3 Z + Y 2 Z2 + 2ZY 2 + Y 4 − X4 Y (Z2 + 2Y 2Z + Y 4 + X4) = 2 , 4 X2(Z + Y 2) X3(Z + Y 2) 2ZY 2 + 2Y 4 2YZ(Z + Y 2) Y 2 ZY = 2 , 4 = 4 , 8 = [4XY 2 : 8ZY : X3] X2(Z + Y 2) X3(Z + Y 2) X2 X3 and if X = 0, it is easy to check that (φ ◦ τ)(O) = [0 : 1 : 0] = (φ ◦ τ)([0 : 1 : −1]). So the ﬁrst block of the diagram commutes. Furthermore, we have for all Q = [U : V : W ] ∈ C0 that:

(τ ◦ ψˆ)(Q) = τ([2UV : W : U 4 + 4V 4]) = 2(2(U 4 + 4V 4)UV + 2UVW 2) : 4((U 4 + 4V 4)W + W 3) : 8U 3V 3 = 8VU 5 : 8WU 4 : 8U 3V 3 = [VU 2 : WU : V 3]

and if V 6= 0 (which also implies that W − U 2 6= 0), we have that

W − U 2 WU − U 3 (φˆ ◦ τ 0)(Q) = φˆ−2(WV − VU 2) : 4(WU − U 3): V 3 = φˆ −2 , 4 V 2 V 3 1 W − U 2 V 2 1 WU − U 3 UV = −2 − 8 , 4 − 16 4 V 2 W − U 2 8 V 3 W − U 2 1 −2(W 2 − 2WU 2 + U 4) − 8V 4 1 4U(W 2 − 2WU 2 + U 4) − 16UV 4 = , 4 V 2(W − U 2) 8 V 3(W − U 2) 1 −4U 4 + 4U 2W 1 8UW (W − U 2) U 2 WU = , = , 4 V 2(W − U 2) 8 V 3(W − U 2) V 2 V 3 = [VU 2 : WU : V 3]

and for V = 0, we have that either Q = O, which implies that (φˆ ◦ τ 0)(Q) = [0 : 1 : 0], or Q = [1 : 0 : −1], which implies (φˆ ◦ τ 0)(Q) = ψˆ([0 : 1 : 0]) = [0 : 1 : 0], so the other half of the diagram also commutes.

15 2.2. Descent by 2-isogeny

2.2.4 A note on the automorphisms of C As we have seen in Section 2.2.2, ψ is surjective if we restrict it to rational points, however, ψˆ is not (of course, both maps are surjective if we do not restrict to rational points). One can, in fact, already see this at the beginning the standard proof of the case n = 4 of Fermat’s Last Theorem in Section 2.1, since two assumptions are made: we can choose Z to be positive, and we can assume that X is even and Y is odd. This corresponds to applying two curve automorphisms of C to our point: the one which sends [X : Y : Z] to [−X : Y : −Z] and the one which sends [X : Y : Z] to [Y : −X : Z]. Note that C has j-invariant 1728 (which we can see from the equation of E), and therefore, it has complex multiplication by Proposition 1.1.2: the map C → C, [X : Y : Z] → [iX : Y : Z] is an elliptic curve automorphism, which generates Aut(C). The automorphism [X : Y : Z] 7→ [−X : Y : Z] is therefore the only nontrivial elliptic curve automorphism of C which is defined over Q, so it must be the negation in the group law of C. Hence, every curve automorphism of C which is defined over Q is either the translation map by a point in C(Q), or such a translation map composed with [X : Y : Z] 7→ [−X : Y : Z]. Note that if a curve automorphism which is defined over Q is of order 2, it is either the translation map by a rational 2-torsion point, or the map C(Q) → C(Q),R 7→ P − R for a point P ∈ C(Q) (so a translation composed with negation). If in the latter case, the automorphism commutes with negation, we have that −P + R = P + R for all R ∈ C(Q), so P is a rational 2-torsion point. Therefore, since the curve automorphisms described above commute with negation, we have that they must correspond to the translation maps by points of order two, possibly composed with the negation map. We will now try to find out which points belong to which curve automorphisms. We have that [X : Y : Z] 7→ [−X : Y : −Z] is translation by a point, since it has no fixed points: suppose that there exists a point P = [X : Y : Z] ∈ C such that there exists some λ ∈ Q∗ such that [−X : Y : −Z] = [λX : λY : λ2Z]. Then we have that either λ2 = −1 or Z = 0. If λ2 = −1, we have that either −X 6= λX or λY 6= Y (since at least one of X and Y must be nonzero), so the first case is not possible. If Z = 0, we have that X,Y 6= 0, and since −X = λX implies that λ = −1 and Y = λY implies λ = 1, we also get a contradiction in this case. So [X : Y : Z] 7→ [−X : Y : −Z] is indeed translation by some point, which must be [0 : 1 : −1] (by the image of O). In a similar way, we have that the curve automorphism [X : Y : Z] 7→ [Y : −X : Z] is translation by [1 : 0 : 1]. It follows that the translation by [1 : 0 : −1] map must be the composition of the other two, and therefore, it corresponds to the automorphism [X : Y : Z] 7→ [Y : X : −Z]. So apparently, in the classical proof given in Section 2.1 it may be necessary to add something from the kernel of the multiplication by 2 map, i.e. translate by a rational 2-torsion point, in order to make sure that our point is in the right co-set (since we see that we do not need the negation map).

16 2.3. The method of descent by 2-isogeny for a special class of elliptic curves

2.3 The method of descent by 2-isogeny for a special class of elliptic curves

4 4 2 Let t ∈ Z \{0} be a nonzero integer and let Ct be the curve over Q given by X + tY = Z in the weighted projective plane P(1, 1, 2). In Section 2.2.1, we have shown that, together with the chosen point O = [1 : 0 : 1], (Ct, O) is an elliptic curve. We used this on C1 and ˆ C−4, and had to choose diﬀerent points on them, in order to make ψ and ψ into isogenies, which enabled us to compare the classical proof in Section 2.1 with a descent by 2-isogeny on elliptic curves in Weierstrass form. However, we can also develop a version of the method of descent by 2-isogeny for the elliptic curve (Ct, O), so that we do not need go to a Weierstrass curve ﬁrst. In this section, we will derive computational formulas for the addition on such elliptic curves, which will allow us to reformulate the lemmas in Section 1.2 for (Ct, O). Then, instead of using the isomorphism to a Weierstrass curve, we can also apply the descent by 2-isogeny method directly.

2.3.1 Addition formulas

Note that Ct has j-invariant 1728 (since we noted in section 2.2.1 that it is isomorphic to the elliptic curve in Weierstrass form Et), so by Proposition 1.1.2, it has complex multiplication: the map Ct → Ct, [X : Y : Z] 7→ [X : iY : Z] is an elliptic curve automorphism of order four, which generates the automorphism group of Ct. Therefore, we have that [X : Y : Z] 7→ [X : −Y : Z] must be the negation map (it is the only elliptic curve automorphism of Ct which is deﬁned over Q). So we get the following proposition.

Proposition 2.3.1 (Negation). Let P = [X : Y : Z] ∈ Ct then −P = [X : −Y : Z]. √ √ ∗ 2 It follows that, if we let t √∈ Q be such that√ ( t) = t, the 2-torsion points of Ct are [1 : 0 : 1], [1 : 0 : −1], [0 : 1 : t] and [0 : 1 : − t]. So Ct always has the rational 2-torsion point3 T = [1 : 0 : −1]. More geometrically, one can see that in the aﬃne patch z2 = 1 + ty2 Z Y (with coordinates z = X2 and y = X ), negation is just mirroring in the z-axis. Similar to Section 2.2.4, we have that the translation by T map is the curve√ automorphism Ct → Ct, [X : Y : Z] 7→ [X : −Y : −√Z], whereas√ the translation by [0 : 1 : t] map is√ the curve automorphism [X : Y : Z] 7→ [ tY : −X : tZ] and√ the 2-torsion√ point [0 : 1 : − t] corresponds to the curve automorphism [X : Y : Z] 7→ [ tY : X : − tZ]. We will now compute addition formulas for all other points of Ct. Considering the aﬃne 2 4 Z Y patch z = 1 + ty , where z = X2 and y = X , we have shown in the proof of Proposition z+1 2.2.1 that the maximal ideal of the local ring in O is generated by y. Therefore, y3 has a pole of order 3 at O, and since the other zero of y is the point [1 : 0 : −1] and z +1 has a zero of order 4 in this point, it has no other poles. So if we denote L(3O) for the Riemann-Roch z+1 z+1 space associated to 3O, we have that L(3O) is generated by 1, y2 and y3 .

3 One can check that the isomorphism τt which maps Ct to an elliptic curve in Weierstrass form sends T to (0, 0).

17 2.3. The method of descent by 2-isogeny for a special class of elliptic curves

Proposition 2.3.2 (Addition). Let P = (a, b),Q = (c, d) be two points in the affine patch z2 = 1 + ty4 such that a 6= 0 and c 6= 0. • If P 6= Q, define b + 1 (b + 1)c3 − a3(d + 1) 1 (b + 1)c3 − a3(d + 1) α = − + and β = − a2 (b + 1)ac3 − a3c(d + 1) a (b + 1)ac3 − a3c(d + 1) • If P = Q, define b(b + 1) 2ta4 + 3b(b + 1) α = and β = − 2a3(3b − 1) 2ta5 + 2b(b + 1)a

3 2tβ αy1 Let y1 = − 2 2 − a − c and z1 = −1 − . Then P + Q = (−y1, z1). tβ −α 1+βy1 Proof. By the Riemann-Roch Theorem, L(3O − P − Q) is a vector space of dimension 1 over Q, so there exist α, β, γ ∈ Q such that L(3O − P − Q) is generated by the function z+1 z+1 g = α + β y2 + γ y3 . Since P and Q are both not equal to O, we have that γ 6= 0. We will consider two cases: • If P 6= Q, we have – since g must have zero’s in P and Q – that b + 1 b + 1 d + 1 d + 1 α + β + γ = 0 and α + β + γ = 0 a2 a3 c2 c3 so b + 1 d + 1 b + 1 d + 1 β − + γ − = 0 a2 c2 a3 c3 b+1 d+1 from which we – scaling γ = 1 and denoting that since P 6= Q, we have that a2 6= c2 b+1 d+1 and a3 6= c3 – deduce that (b + 1)c3 − a3(d + 1) b + 1 (b + 1)c3 − a3(d + 1) 1 β = − so α = − + (b + 1)ac3 − a3c(d + 1) a2 (b + 1)ac3 − a3c(d + 1) a which gives us a generator for the Riemann-Roch space. • If P = Q, we still have that b + 1 b + 1 α + β + γ = 0 a2 a3 Furthermore, in order for P to be a double zero of g, we get the extra condition that the tangent spaces need to be equal at P . We scale γ = 1 again, then for z2−ty4−1, we have dz 4ty3 z+1 z+1 (z+)(βy+1) dz (z+1)(2βy+3) that dy = 2z and for α + β y2 + y3 = α + y3 , we get that dy = − y(βy+1) . So we have the following: 4ta3 (b + 1)(2βa + 3) 2ta4 + 3b(b + 1) = − so β = − 2b a(βa + 1) 2ta5 + 2b(b + 1)a from which we deduce that b + 1 1 b(b + 1)2 b(b + 1) α = − β + = = a2 a 2ta7 + 2b(b + 1)a3 2a3(3b − 1) which gives a generator for the Riemann-Roch space again.

18 2.3. The method of descent by 2-isogeny for a special class of elliptic curves

We can ﬁnd the third zero of g, by observing that for R = (y, z) ∈ Ct, we have that z+1 z+1 αy3 g(R) = α + β y2 + y3 = 0 if and only if z = −1 − 1+βy . Substituting this in the equation of Ct yields 2αy3 α2y6 0 = ty4 + 1 = z2 = 1 + + 1 + βy 1 + 2βy + β2y2 so ty(1 + βy)2 − 2α(1 + βy) − α2y3 = (tβ2 − α2)y3 + 2tβy2 + (t − 2αβ)y − 2α. We already know that a and c are roots of this polynomial, so looking at the quadratic term, 2tβ 2tβ the third zero must be y = − tβ2−α2 − a − c. Now if we deﬁne y1 = − tβ2−α2 − a − c and 3 αy1 z1 = −1 − and the point S = (y1, z1) ∈ Ct, we have in the zero part of the Piccard 1+βy1 0 group Pic (Ct) of Ct that [0] = [div(g)] = [3O − P − Q − S] so − [S − O] = [P − O] + [Q − O].

By Proposition 2.3.1, we conclude that P + Q = −S = (−y1, z1).

2.3.2 The method of 2-descent on (Ct, O)

We will now derive the descent by 2-isogeny procedure directly on Ct. To make the link with the original method clearer, we will refer to the corresponding lemmas in Section 1.2.

0 Lemma 2.3.3 (Lemma 1.2.1). Consider the second elliptic curve Ct over Q which is given by U 4 − 4tV 4 = W 2 and the isogeny

0 4 4 ψt : Ct → Ct, [X : Y : Z] 7→ [Z : XY : X − tY ] then ker(ψt) = {O,T }.

0 0 0 We will denote O = [1 : 0 : 1] for the chosen point of Ct = C−4t and T = [1 : 0 : −1] for its rational 2-torsion point.

0 Proof. First, note that Ct is indeed an elliptic curve over Q, as we already checked in Section 2.2.1. Furthermore, ψt is a rational map of smooth projective curves, and therefore a morphism, and since it sends O to O0, it is indeed an isogeny. What remains to check is that the kernel of ψt is indeed {O,T }. Therefore, note that ψt(T ) = O, so {O,T } ⊂ ker(ψt). On the ∗ other hand, if [X : Y : Z] ∈ ker(ψt), there exists some λ ∈ Q such that λZ = 1, −λXY = 0 2 4 4 and λ (X − tY ) = 1. It follows from the second equation√ that either Y = 0 or X = 0.√ In the latter case, we have that either [X : Y : Z] = [0 : 1 : t] or [X : Y : Z] = [0 : 1 : − t]. The ﬁrst equation then implies that λ = ± √1 , however, the third one gives that −tλ2 = 1, so t 2 1 λ = − t . This is a contradiction, so we have that Y = 0, and therefore, [X : Y : Z] ∈ {O,T }. We conclude that ker(ψt) = {O,T }.

00 4 4 2 We can use Lemma 2.3.3 again to ﬁnd a third elliptic curve Ct : X + 16tY = Z , which is 00 isomorphic to Ct (an isomorphism given by Ct → Ct, [X : Y : Z] 7→ [X : 2Y : Z]). So we get a second isogeny

ˆ 0 4 4 ψt : Ct → Ct, [U : V : W ] 7→ [W : 2UV : U + 64tV ]

19 2.3. The method of descent by 2-isogeny for a special class of elliptic curves

ˆ 0 0 with ker(ψt) = {O ,T }. This turns out to be the dual of ψt (which can be checked with the addition formulas in the previous sections), similar to Lemma 1.2.2.

Lemma 2.3.4 (First half of Lemma 1.2.3). Deﬁne the map

 2 [2 W +U ] if V 6= 0  V 2 0 ∗ ∗ 2 0 q : Ct(Q) → Q /(Q ) , [U : V : W ] 7→ [t] if [U : V : W ] = T . [1] if [U : V : W ] = O0

ψt 0 q ∗ ∗ 2 Then q is a homomorphism of groups, and the sequence Ct(Q) −→ Ct(Q) −→ Q /(Q ) is exact.

Proof. One can check that q is a homomorphism of groups using the formulas in the previous section4. To check that the sequence is exact, suppose that [U : V : W ] ∈ ker(q). If [U : V : W ] = O0, 0 it is clearly in the image√ of ψ, and if [U : V :√W ] = T√, we have that t is a square, which implies that [0 : 1 : t] ∈ Ct(Q) and ψt([0 : 1 : t]) = [ t : 0 : −t] = T . Otherwise, we have W +U 2 ∗ that 2 V 2 is a nonzero square in Q , so 4V 4 −W 2 + U 4 (W − U 2)(W + U 2) W − U 2 = = − = − 2V 2(W + U 2) 2tV 2(W + U 2) 2tV 2(W + U 2) 2tV 2

W −U 2 is also a square. Let α ∈ Q be a root of − 2tV 2 , and deﬁne P = [V : V α : UV α], then we have that W 2 − 2WU 2 + U 4 2U 4 − 2U 2W − 4tV 4 V 4 + tV 4α4 = V 4 + t = V 4 + t 4t2 4t2 W − U 2 = U 2 − = (UV α)2 2t

so P ∈ Ct(Q), and note that 2tV 4 W − U 2 ψ (P ) = UV α : V 2α : V 4 − tV 4α4 = U : V : − + t W − U 2 2 −4tV 4 + W 2 − 2WU 2 + U 4 2W 2 − 2WU 2 = U : V : = U : V : = [U : V : W ]. 2(W − U 2) 2(W − U 2)

So ker(q) ⊂ ψt(Ct(Q)). On the other hand, let [X : Y : Z] ∈ Ct(Q). If [X : Y : Z] ∈ {O,T }, it is in the kernel of q. Otherwise, we have that

X4 − tY 4 + Z2 X2 (q ◦ ψ )([X : Y : Z]) = q([Z : XY : X4 − tY 4]) = 2 = 4 = [1] t X2Y 2 Y 2

so [X : Y : Z] ∈ ker(q), which implies that ψt(Ct(Q)) ⊂ ker(q), so ψt(Ct(Q)) = ker(q), which completes the proof. 4However, the easiest way to prove this is to use the isomorphism to a Weierstrass equation

20 2.3. The method of descent by 2-isogeny for a special class of elliptic curves

Lemma 2.3.5 (Remainder of Lemma 1.2.3). Let r ∈ Z be a square-free integer, then we have that [r] ∈ im(q) if and only if the equation

r2l4 + tm4 = rn2 has a solution (l, m, n) ∈ Z3 \{(0, 0, 0)}. Furthermore, this can only happen if r divides t. Proof. The element [1] always lies in the image of q, and the corresponding equation also always has the solution (1, 0, 1). Also, [t] always lies in the image of q, and the equation t2l4 + tm4 = tn2 always has the solution (0, 1, 1). So now, suppose that [r] ∈/ {[1], [t]} and 0 0 0 h W +U 2 i [r] ∈ im(q). Then there exists a point [U : V : W ] ∈ Ct(Q)\{O ,T } such that 2 V 2 = [r], ∗ W +U 2 2 ra2V 2 2 so there is some a ∈ Q such that 2 V 2 = ra , which implies that W = 2 − U , so r2a4V 4 (aV )4 U 4 − 4tV 4 = W 2 = − ra2V 2U 2 + U 4 which implies that r2 + 4tV 4 = r(aV U)2. 4 4

v u 2 4 4 2 Writing V = w and U = w where u, v, w ∈ Z, this becomes r (av) + t(2v) = r(2auvw) . Substituting l = av, m = 2v and n = 2auvw, we obtain an equation of the desired form. On the other hand, substituting back we find that any nonzero solution to this equation gives 0 rise to a point in P ∈ Ct(Q) such that q(P ) = [r]. It remains to show that this can only happen if r divides t. Therefore, assume that there is a nonzero solution (l, m, n) ∈ Z3 to the equation r2l4 + tm4 = rn2, from which we may assume without loss of generality that l, m and n are co-prime, and that there exists a prime number p which divides r but does not divide t. Then p divides m (all terms except for tm4 are divisible by p and since t is not, m must be divisible by p), and therefore it also divides n (since r is square-free and the left hand side of the equation is divisible by p2). This implies that p also divides l (we have r2l4 = rn2 − tm4, where the right hand side is divisible by p3), which is a contradiction. As a corollary, we have that the image of q must be finite (there are only finitely many integers dividing t), so since

0 0 ∼ Ct(Q)/ψt(Ct(Q)) = Ct(Q)/ ker(q) = im(q)

0 we have shown that Ct(Q)/ψt(Ct(Q)) is finite. In the same way, we can find that ˆ 0 Ct(Q)/ψt(Ct(Q)) is finite. One can now use the kernel-cokernel sequence (Lemma 1.2.4) to prove that Ct(Q)/2Ct(Q) is also finite. Furthermore, we can use the isomorphism to im(q) to find generators for Ct(Q)/2Ct(Q). To illustrate this method, we go back to the original elliptic curve C : X4 + Y 4 = Z2 (so t = 1) and apply Lemma 2.3.1 to obtain the second elliptic curve C0 : U 4 − 4V 4 = W 2. Now let r ∈ Z be a square-free integer, then by Lemma 2.3.3, we have that [r] can only be in im(q) if r divides 1, i.e. r ∈ {−1, 1}. The class [1] is always in im(q) and it is clear that 0 r = −1 is not going to work, so C (Q)/ψ1(C(Q)) is trivial. In order to find the image of q0 : C(Q) → Q∗/(Q∗)2, we need to check for all square-free integers r dividing −4, so r ∈ {±1, ±2}, whether the equation r2`4 −4m4 = rn2 has solutions (l, m, n) ∈ Z3 \{(0, 0, 0)}. For r = 1, we know that this is always the case. For r = 2, the

21 2.4. Twists and φ-coverings

equation becomes 4(`4 −m4) = 2n2 which has (1, 1, 0) as a solution, and for r = −2, we have 4 4 2 ˆ 0 ∼ 2 that 4(` − m ) = −2n has the solution (1, 1, 0). So C(Q)/ψ1(C (Q)) = (Z/2Z) , which ∼ 2 implies that C(Q)/2C(Q) = (Z/2Z) . Note that this is the same result as we obtained in Section 2.2.

2.4 Twists and φ-coverings

A way to clarify the method of descent by 2-isogeny, is to view it from the perspective of twists and φ-coverings, which we will investigate in this section.

2.4.1 Twists

Let E : y2 = x(x2 + ax + b) be an elliptic curve over Q as in Section 1.2, so with a, b ∈ Z, where b 6= 0 and a2 − 4b 6= 0, and rational 2-torsion point (0, 0). Let r ∈ Q∗, then if [r] ∈/ {[1], [b]}, we have that [r] is in the image of the two descent map q0 : E(Q) → Q∗/(Q∗)2 if and only if there exists some s ∈ Q∗ and a point (x, y) ∈ E(Q) such that x = rs2. This motivates the deﬁnition of an algebraic set in P2 × A1 (where we take coordinates X,Y,Z in P2 and s in A1) deﬁned by the equations ( ZY 2 = X(X2 + aXZ + bZ2) . X = Zrs2

Proposition 2.4.1. This set is the union of the line {[0 : 1 : 0]} × A1 and some algebraic curve Fr which has ([0 : 0 : 1], 0) as its only singular point.

Proof. The subset of all elements satisfying Z = 0 is precisely the line {[0 : 1 : 0]} × A1. The Y X complement Fr is, by taking coordinates y = Z and x = Z , isomorphic to the algebraic set in A2 × A1 given by the equations ( y2 = x(x2 + ax + b) x = rs2

which, eliminating x, is the aﬃne algebraic set in A2 deﬁned as the zero set of the polynomial

2 2 2 4 2 F = y − rs (r s + ars + b) ∈ Q[y, s] √ 2 4 2 −a± a2−4b 2 2 Note that r s +ars +b has, by the quadratic formula, the zero’s 2r . Since a 6= a −4b (for b 6= 0), those roots are distinct, so in particular, r2s4 + ars2 + b is not a square in Q[s]. Since F is monic and of degree two, this implies that F is irreducible in (Q[s])[y]. Therefore, 2 2 4 2 Fr is an algebraic curve. Since rs (r s + ars + b) only has a double root at zero, it has only one singular point, which corresponds to ([0 : 0 : 1], 0).

0 By construction, Fr has a rational point precisely if [r] ∈ im(q ). By Theorem 1.1.4, we have that there exists a smooth projective curve Fr over Q (which is unique up to isomorphism over Q) such that Fr \{([0 : 0 : 1], 0)} is birationally equivalent to Fr. Furthermore, we get the following.

22 2.4. Twists and φ-coverings

Proposition 2.4.2. Fr is isomorphic to the smooth projective curve Er ⊂ P(1, 1, 2) given by the equation r2X4 + arX2Y 2 + bY 4 = rZ2.

Proof. Consider the map

ZY σ : F → E , ([X : Y : Z], s) 7→ sZ : Z : . r r rs

Then this is a well deﬁned rational map, since for all ([X : Y : Z], s) ∈ Fr we have that

Y 2Z = X(X2 + aXZ + bZ2) = rs2Z r2s4Z2 + ars2Z2 + bZ2

so ZY 2 r = r2(sZ)4 + ar(sZ)2Z2 + bZ4. sr Note that σ is regular everywhere, except for the singular point. Now we deﬁne the map

rX2 ZrX X σ0 : E → F , [X : Y : Z] 7→ : : Y , r r Y Y 2 Y

0 then σ is a well deﬁned rational map, since for all [X : Y : Z] ∈ Er satisfying Y 6= 0, we have that

Z2 X4 rZX 2 rX2 r2X4 r = r2 + arX2 + bY 2 so Y = + arX2 + bY 2 . Y 2 Y 2 Y 2 Y Y 2

We have that σ0 is also regular everywhere, except for the points satisfying Y = 0. Further- more, it is easy to check that σ0 provides a birational inverse for σ. By Theorem 1.1.4, σ determines an isomorphism Fr → Er.

X Y Now, we consider for the aﬃne part Z 6= 0 of Fr, with coordinates x = Z and y = Z , the projection map πr : Fr → E, (s, (x, y)) 7→ (x, y) which determines a unique map πr : Fr → E, and make the following observations:

1. First, note that over , we have that (F ) ∼ (F ) for all r ∈ ∗, an isomorphism Q r Q = 1 Q Q ∗ given by (F ) → (F ) , (s, (x, y)) 7→ (sρ, (x, y)), where ρ ∈ is such that ρ2 = r. r Q 1 Q Q These isomorphisms determine isomorphisms (F ) → (F ) . So we actually only need r Q 1 Q to consider F1 and the family of curves which are isomorphic to it over Q. Those are called the twists of F1.

2. Furthermore, if r0 = rγ2 for some γ ∈ Q∗, we can deﬁne the map s f : F → F 0 , (s, (x, y)) 7→ , (x, y) r r γ

which is an isomorphism over Q. Then we have that πr = πr0 ◦ f. So Fr and Fr0 are isomorphic over Q.

23 2.4. Twists and φ-coverings

3. Also, note that, up to isomorphism over Q, there are only finitely many r ∈ Q∗ for 0 which Fr has rational points, since we know that im(q ) is finite by Lemma 1.2.3. If Fr has got rational points which do not correspond to rational points of Fr, then either they correspond to the singular point with s = 0 or they are “at infinity”. In the first case, by the proof of Proposition 2.4.2, we have that those points correspond to the 4 2 points [X : Y : Z] ∈ Er(Q) satisfying X = 0, so bY = rZ . If this is a rational point, it follows that [r] = [b], so [r] ∈ im(q0). In the second case, we have that those points correspond to the points [X : Y : Z] ∈ Er(Q) satisfying Y = 0, so they satisfy 4 2 0 rX = Z , which implies that [r] = [1], so [r] ∈ im(q ). Therefore, Fr can only have a rational point if [r] ∈ im(q0). This implies that, up to isomorphism over Q, there are ∗ also only finitely many r ∈ Q for which Fr has rational points.

4. And finally, note that every point in E(Q) is the image under πr for precisely one of the curves Fr, up to isomorphism over Q. Equivalently, we have that [ E(Q) = πr(Fr(Q)) [r]∈Q∗/(Q∗)2 and this union is disjoint and contains only finitely many points. Applying the method of descent by 2-isogeny, we get a second elliptic curve given by E0 : v2 = u(u2 + a0u + b0) where a0 = −2a and b0 = a2 − 4b, for which we can consider the algebraic set in P2 × A1 (with coordinates U, V, W in P2 and s in A1) given by ( ( WV 2 = U(U 2 + a0UW + b0W 2) v2 = u(u2 + a0u + b0) or the affine equations U = W rs2 u = rs2

U V (where u = W and v = W ). Then by the results above, this is the union of a line and an 0 algebraic curve Fr which has ([0 : 1 : 0], 0) as its only singular point. By Theorem 1.1.4 0 0 again, there exists a smooth projective curve Fr which is birationally equivalent to Fr and unique up to isomorphism over Q. Proposition 2.4.3. There exists an isomorphism f :(F 0) → E over . r r Q Q Q Proof. Consider the map 1 v s v f : F 0 → E, ((u, v), s) 7→ u − a + , u − a + . 1 2 s 2 s 0 This is a well-deﬁned rational map, since we have for all ((u, v), s) ∈ F1 with s 6= 0 that the quadratic polynomial x2 + (a − u)x + b has zero’s r ! 1 √ 1 v2 1 v x = u − a ± u2 − 2au + a2 − 4b = u − a ± = u − a ± ± 2 2 u 2 s

b 2 2 2 so x± + a + = u, from which it follows that x+(x + ax+ + b) = ux = (sx+) , so x± + + f((u, v), s) ∈ E. Note that f is regular on all points except for the singular point. Further- more, we deﬁne the map y2 y2 y y E → F 0, (x, y) 7→ , 2x + a − , r x2 x2 x x

24 2.4. Twists and φ-coverings

then it is not difficult to check that it is a well defined rational map, which provides an inverse for f. It is regular for all (x, y) ∈ E with x 6= 0. Therefore, f determines a unique rational map f :(F 0) → E which is an isomorphism. Since F 0 is isomorphic to F 0 for all 1 Q Q r 1 r ∈ ∗, this implies that there exists an isomorphism f :(F 0) → E over . Q r r Q Q Q 0 So in fact, the twists Fr are isomorphic to E over Q. Combining this with Proposition 2.4.2, 0 ∗ we have that Er is isomorphic to E over Q for all r ∈ Q . So in fact, the method of descent by 2-isogeny is about replacing the original elliptic curve (of which rational points may be hard to find) by a set of twists of an isogenous curve, of which it may be easier to decide whether they have rational points. Moreover, it is already sufficient if we can decide whether 0 a twist Fr does have some rational point, to obtain information about E(Q)/2E(Q) and ultimately, about the rank of E(Q).

2.4.2 A φ-covering

0 0 2 1 2 Let πr : Fr → E denote the restriction of the natural projection map P × A → P as 0 described in the previous section. Then πr is a 2-to-1 map, since for almost all points in E , 0 0 there are two choices for s. Let πr : Fr → E denote the map determined by πr. Then we get the following.

Proposition 2.4.4. The diagram f 0 r Fr E π r φ

E0 commutes.

0 Proof. Let f : F1 → E be the map we used to prove Proposition 2.4.3. Using the notation 0 1 v of that proof again, for all ((u, v), s) ∈ F1 \{(0, 0), 0}, we deﬁne the zero x+ = 2 (u − a + s ) of the quadratic polynomial x2 + (a − u)x + b again, then we have that

b bs (φ ◦ f)(s, (u, v)) = φ(x+, sx+) = x+ + a + , sx+ − x+ x+ b = x+ + a + , s(2x+ − u + a) = (u, v) x+

Therefore, after composing with the isomorphism (F ) → (F ) from the ﬁrst remark of 1 Q r Q the previous section, the maps πr and f determine maps which make diagram commute.

Because of this interesting property, πr is called a φ-covering.

25 Chapter 3

The case n = 3

The case n = 3 of Fermat’s Last Theorem was first1 proven by Euler [Eul70] in 1770 using the method of infinite descent. We will give a version of this proof in Section 3.1. In Section 3.2, we will show that one can find a similar link with a descent by 3-isogeny procedure of elliptic curves as in the case n = 4. Finally, in Section 3.3 we will investigate the descent by 3-isogeny procedure from the viewpoint of twists and φ-coverings.

3.1 The standard proof

In order to give the classical proof of Theorem 0.0.1 for n = 3, we will need the following lemma, a proof2 of which can be found in, for example, [Rib99].

Lemma 3.1.1. Let s ∈ Z be odd and suppose there exist a, b ∈ Z with gcd(a, b) = 1 such that s3 = a2 + 3b2. Then there exist u, v ∈ Z such that s = u2 + 3v2, a = u(u2 − 9v2) and b = 3v(u2 − v2).

Following Ribenboim, we will give the standard proof of the case n = 3 now.

Theorem 3.1.2 (Fermat n = 3). The equation x3 + y3 = z3 has no solutions (x, y, z) ∈ Z3 satisfying xyz 6= 0.

Proof. Suppose that we have a solution (x, y, z) ∈ Z3 with xyz 6= 0. Since 2 is not a cube, we have that x, y and z are distinct. Furthermore, we can assume without loss of generality that x, y and z are pairwise co-prime, since any prime dividing two of the three variables divides the third. It follows that exactly one of x, y, z is even, say that x and y are odd and z is even (if z is odd, we have that either x or y is even, so say x is even, then (−x, z, y) is another solution). We will now assume that |z| is minimal under all solutions of which the last coordinate

1Actually, a full proof of Lemma 3.1.1 was missing, but Euler established this in an unpublished paper (of which an English translation and explanation can be found in [BL76]) which he wrote in 1760. However, there still turned out to be a missing link, which has been pointed out and solved by other mathematicians later. 2The elementary proof of Lemma 3.1.1 given by Ribenboim is quite long. However, with some more theory about unique factorization domains, there exist easier proofs.

26 3.1. The standard proof

is even, and derive a contradiction. Since x and y are odd, there are a, b ∈ Z such that x + y = 2a and x − y = 2b, and we have that a, b 6= 0, gcd(a, b) = 1 (for x = a + b and y = a−b) and precisely one of them is odd (for modulo 4, we have that either 2a ≡ x+y ≡ 0 or 2b ≡ x − y ≡ 0). Since

z3 = (a + b)3 + (a − b)3 = 2a(a2 + 3b2) (3.1) we have that b must be odd. k 2 2 Now let p be a prime and k ∈ Z≥1 and suppose that p divides both 2a and a + 3b , then since p 6= 2 (for a2 + 3b2 is odd) we have that pk divides a, and therefore also 3b2, and since it cannot divide b, we have that k = 1 and p = 3. So we consider two cases:

1. gcd(2a, a2 + 3b2) = 1. Then by equation (3.1), there exist r, s ∈ Z such that 2a = r3 and a2 + 3b2 = s3. By Lemma 3.1.1, there are u, v ∈ Z such that s = u2 + 3v2, a = u(u2 − 9v2) and b = 3v(u2 − v2). Then we have that v is odd and u is even (for b is odd), u, v 6= 0 and gcd(u, v) = 1 (any common divisor also divides a and b). Since u is not divisible by 3 (then a would also be divisible by 3), we have that 2u, u − 3v and u + 3v are pairwise co-prime, so from

r3 = 2u(u − 3v)(u + 3v) (3.2)

it follows that there are l, m, n ∈ Z \{(0, 0, 0)} such that 2u = n3, u − 3v = l3 and u + 3v = m3 and then we have that l3 + m3 = n3, so (l, m, n) is another nontrivial solution of the equation, and we have that n is even and

|z|3 = |2a(a2 + 3b2)| = |n3(u2 − 9v2)(a2 + 3b2)| > |n|3

so |n| < |z|, which contradicts the minimality of |z|.

2. gcd(2a, a2 + 3b2) = 3. In this case, there is some c ∈ Z such that a = 3c, so from (3.1), we have that z3 = 18c(3c2 + b2). Since gcd(b, c) = 1 (this follows from the fact that gcd(a, b) = 1), 3c2+b2 is odd and 3 does not divide b, we have that gcd(18c, 3c2+b2) = 1. It follows that there exist r, s ∈ Z such that 18c = r3 and 3c2+b2 = s3. By Lemma 3.1.1 again, there exist u, v ∈ Z such that s = u2 + 3v2, b = u(u2 − 9v2) and c = 3v(u2 − v2). Then we have that u is odd, v is even, u, v 6= 0 and gcd(u, v) = 1, so 2v, u − v and u + v are pairwise co-prime. Now from r 3 = 2v(u + v)(u − v) (3.3) 3

it follows that there are l, m, n ∈ Z \{(0, 0, 0)} such that 2v = n3, u + v = l3 and u − v = −m3 and l3 + m3 = n3. So (l, m, n) is another nontrivial solution and n is even, and since

|z|3 = 18|c(3c2 + b2)| = 27|n|3|u2 − v2|(3c2 + b2) > |n|3

we have that |n| < |z|, which is again a contradiction. We conclude that there are no nontrivial solutions to the equation x3 + y3 = z3.

27 3.2. Descent by 3-isogeny

3.2 Descent by 3-isogeny

Analyzing this proof, we start with the curve C : X3 + Y 3 = Z3 over Q in the projective plane P2. Together with the rational point O = [1 : −1 : 0], this is an elliptic curve, since C is smooth3 and has genus 1 by the genus-degree formula. In the proof, two cases are considered. In the first one, we get the curve given by equation (3.2), and in the second one, we get the curve given by equation (3.3). Both of these are isomorphic to the curve C0 : W 3 = 2U 3 − 2UV 2 over Q, which, together with the point O0 = [0 : 1 : 0], is also an elliptic curve (with similar reasoning). Furthermore, we find two rational maps (which are obtained from composing the rational maps we get in the first and the second case with isomorphisms to C and C0):

ψ : C → C0, [X : Y : Z] 7→ [Z3 : 2Y 3 − Z3 : 2XYZ] and ψˆ: C0 → C given by

V 3 V 3 V 2 [U : V : W ] 7→ U 3 − UV 2 − VU 2 + : U 3 − UV 2 + VU 2 − : W U 2 + 9 9 3

which are morphisms since they are maps between smooth projective curves. Furthermore, it is easy to check that ψ(O) = O0 and ψˆ(O0) = O, so ψ and ψˆ are isogenies. (Note that we apparently abuse the notation of the dual again, but later on, we will show that ψˆ is indeed the dual of ψ.) The curve C is isomorphic to the elliptic curve E : y2 = x3 − 432 in Weierstrass form. One isomorphism is given by τ : C → E, [X : Y : Z] 7→ [12Z : 36(Y −X): X +Y ]. Similarly, C0 is isomorphic to the elliptic curve in Weierstrass form E0 : v2 = u3 +11664, by the isomorphism τ 0 : C0 → E0, [U : V : W ] 7→ [−18W : 108V : U]. (These elliptic curves and isomorphisms have been obtained using SageMath [The18].)

3.2.1 The image of q

Let a, b, d ∈ Z be such that b, d 6= 0 and 27b − 4a3d 6= 0, then E : y2 = x3 + d(ax + b)2 defines an elliptic curve over Q, as in Section 1.3. Furthermore, assume that a3d is divisible by 9, then the second elliptic curve E0 : v2 = u3 + d0(a0u + b0)2 obtained from Lemma 1.3.1 satisfies ∗ a0, b0, d0 ∈ Z. Let δ0 ∈ Q be such that δ02 = d0. In order to perform a descent by 3-isogeny on E in the same way as we did it in the previous chapter, we need a way to explicitly construct the image of the three descent map defined by

( 0 0 0 0 ∗ 3 [v − δ (a u + b )] if u 6= 0 q : E (Q) → Kd0 /(Kd0 ) , (u, v) 7→ [2b0δ0] otherwise

0 0 0 and q(O ) = [1], as we will now do for the case that Kd0 = Q (i.e. the 3-torsion point (0, b δ ) is rational). Note that [r] ∈ im(q) holds if and only if [r2] ∈ im(q), which will be a helpful computational tool.

3See Appendix A.2

28 3.2. Descent by 3-isogeny

Proposition 3.2.1. Suppose that d0 ∈ (Q∗)2 and let r be a cube-free integer. Then [r] lies in the image of q if and only if the equation

2a0δ0rlmn + 2b0δ0n3 + rm3 = r2l3

has a solution (l, m, n) ∈ Z3 \{(0, 0, 0)} Furthermore, this can only happen if all primes dividing r also divide 2b0δ0.

Proof. Note that every class in Q∗/(Q∗)3 can be uniquely represented by a cube-free integer, and that if r0 = rs3 for some s ∈ Q∗, the equation has a solution for r if and only if it has a solution for r0: if (l, m, n) ∈ Z3 \{(0, 0, 0)} is a solution to the equation for r, we have that (l, sm, s2n) is a solution to the equation for r0 and the other way around. For r = 1, the equation has the solution (1, 1, 0), and [1] also always lies in the image of q. Furthermore, if [r] = [2b0δ0], we have that the equation for 2b0δ0 has the solution (0, 1, −1), and also, [2b0δ0] always lies in the image of q (it is the image of the rational 3-torsion point (0, b0δ0)). So we can assume without loss of generality that [r] ∈/ {[1], [2b0δ0], [4b02d0]}. Suppose that there exists some point (u, v) ∈ E0(Q) such that [r] = q(u, v) = [v − (a0u + b0)δ0].

Then there exists some t ∈ Q∗ such that v − (a0u + b0)δ0 = rt3. (3.4)

Since u3 = (v − (a0u + b0)δ0)(v + (a0u + b0)δ0) there also exists some s ∈ Q∗ such that v + (a0u + b0)δ0 = r2s3 (3.5)

and u = rst. Subtracting equations (3.4) and (3.5) yields

2δ0(a0u + b0) + rt3 = r2s3 so 2a0δ0rst + 2b0δ0 + rt3 = r2s3.

l m Write s = n and t = n , where l, m, n ∈ Z, then it follows that 2a0δ0rlmn + 2b0δ0n3 + rm3 = r2l3

as desired. On the other hand, going back through those substitutions, one can see that any nonzero solution to this equation gives rise to a rational point of E0 which maps to [r] under q. It remains to show that this can only happen if r divides 2b0δ0. Therefore, suppose that (l, m, n) ∈ Z \{(0, 0, 0)} is a solution to the equation, then we can assume without loss of generality that l, m and n are co-prime, and that there exists a prime number p dividing r but not dividing 2b0δ0. Looking at the equation, we then have that p must divide 2b0δ0n3, and therefore, it must divide n. Now, there are two possible situations:

29 3.2. Descent by 3-isogeny

1. p2 also divides r. In this case, p3 must divide rm3 (since it divides all other terms), so p divides m. Now, we have that all terms except possibly for 2b0δ0n3 are divisible by p4, so p2 also divides n. But then all terms on the left hand side are divisible by p5, so since r is cube-free, this implies that p also divides l. 2. If p2 does not divide r, we have that p must divide m, from which it follows that all terms on the left hand side are divisible by p3, so p also divides l. In all cases, p divides l, m and n, which is a contradiction. We conclude that all primes dividing r must also divide 2b0δ0, which completes the proof.

If the ring of integers of Kd0 is a principal ideal domain, which implies that it is also a unique factorization domain, one can also prove the following statement in a similar way.

Proposition 3.2.2. Suppose that O(Kd0 ) is a principal ideal domain and let r ∈ O(Kd0 ) be 0 cube-free. Then there exists a point (u, v) ∈ E (Kd0 ) for which we have that [v − δ0(a0u + b0)] = [r]

∗ ∗ 3 0 0 0 0 3 3 2 3 in Kd0 /(Kd0 ) if and only if the equation 2a δ rlmn + 2b δ n + rm = r l has a solution 0 0 (l, m, n) ∈ O(Kd0 )\{(0, 0, 0)}. This can only happen if all primes dividing r also divide 2b δ .

3.2.2 The descent method applied to E Applying the method of descent by 3-isogeny in Section 1.3 to the elliptic curve in Weierstrass form E : y2 = x3 − 432 which is isomorphic to C (so a = 0, b = 12 and d = −3), we obtain a second elliptic curve given by y2 = x3 + 11664, which turns out to be the Weierstrass curve E0 isomorphic to C0 again. Furthermore, we consider the isogenies

( 3 3 3 3 x −12 , y(x +2·12 ) if x 6= 0 φ: E → E0, (x, y) 7→ x2 x3 O0 otherwise √ √ satisfying ker(φ) = {O, (0, 12 −3), (0, −12 −3)} and

( 3 3 3 3 u +36 , v(u −2·36 ) if u 6= 0 φˆ: E0 → E, (u, v) 7→ 9u2 27u3 O otherwise

with ker(φˆ) = {O0, (0, 108), (0, −108}, which are dual (by Lemma 1.3.2). By Lemma 1.3.3, the maps q : E0(Q) → Q∗/(Q∗)3 deﬁned by ( [v − 108] if u 6= 0 (u, v) 7→ [1] otherwise √ √ (and q(O0) = [1]) and q0 : E(Q) → Q( −3)∗/(Q( −3)∗)3 deﬁned by ( √ [y − 12 −3] if x 6= 0 (x, y) 7→ [1] otherwise

30 3.2. Descent by 3-isogeny

∼ 0 (and q(O) = [1]) are group homomorphisms and we have that im(q) = E (Q)/φ(E(Q) and 0 ∼ 0 im(q ) = E(Q)/φˆ(E (Q)). We will now analyze the images of both maps using Proposition 3.2.1 and Proposition 3.2.2. Note that the 3-torsion point (0, 108) of E0 is rational, so in order to analyze the image of q, we need to check whether the equation

216n3 + rm3 = r2l3

has solutions in integers for all cube-free r ∈ Z dividing 23 · 33, i.e. for the following values of r:

• We know that [1] is always in the image of q, as it is the image of the point at inﬁnity.

• For r = 2, the equation becomes 22 · 33n3 + m3 = 2l3. Suppose that we have a solution (l, m, n) ∈ Z3 \{(0, 0, 0)}. Then we may assume that l, m and n are co-prime, since if a prime p divided all of them, we have that (l/p, m/p, n/p) is again a solution. Since all terms except possibly for m3 are even, we have that m must be even. But then 4 divides the left hand side of the equation, and therefore also the right hand side, so l must be even too. This, however, implies that all terms other than 22 · 33n3 are divisible by 8, so n is also even, which is a contradiction. So the equation has no solutions, which implies that [2] ∈/ im(q), and hence [4] ∈/ im(q).

• For r = 3, we have to ﬁnd out whether the equation 3l3 = m3 + 23 · 32n3 has solutions (l, m, n) ∈ Z3 \{(0, 0, 0)}. Suppose that (l, m, n) is such a solution and assume without loss of generality that l, m and n are co-prime. Then m must be divisible by 3. Similar to case r = 2, it follows that l is also divisible by 3, and therefore, 3 divides n too, which is a contradiction. We conclude that [3], [9] ∈/ im(q).

• For r = 6, the equation becomes 6l3 = m3 + 22 · 32n3. If (l, m, n) ∈ Z3 \{(0, 0, 0)} were a solution with again, l, m and n co-prime, we have that m must be even, so similar to the previous cases, we also have that l, and therefore n is even, which is a contradiction. So [6], [36] ∈/ im(q).

• For r = 12, we have that the equation 12l3 = m3 + 2 · 32n3 does not have solutions (l, m, n) ∈ Z3 \{(0, 0, 0)}. After all, if (l, m, n) is a solution, we can assume that l, m and n are co-prime again, and then it follows from the equation that m is even, so also n is even, which implies that l is also even, which is a contradiction. Hence, [12], [18] ∈/ im(q).

0 We conclude that E (Q)/φ(E(Q)) is trivial. √ Next, we analyze the image of q0. The 3-torsion point (0, 12 −3) of E is not rational, but √ √ we have that O( ( −3)) = [ζ], where ζ = −1+ −3 is a primitive third root of unity, is Q Z √ 2 √ √ a principal ideal domain. Note that 24 −3 = 23(− −3)3 and that 2 and −3 are prime elements of [ζ]. Therefore, by Proposition 3.2.2, if r ∈ [ζ] is cube-free, we have that Z √ Z [r] can only be in im(q0) if r is of the form r = ±ζk −3i2j, where i, j, k ∈ {0, 1, 2}. By

31 3.2. Descent by 3-isogeny

Proposition 1.3.4, we only need to check r = ζ. Note that (12, 36) ∈ E(Q) and √ √ √ −1 + −3 q0(12, −36) = [−36 − 12 −3] = [−4 + 4 −3] = 2 2 0 0 ∼ which implies that [ζ], [ζ ] ∈ im(q ). We conclude that E(Q)/φˆ(E (Q)) = Z/3Z. We can use Lemma 1.2.4 to find that the sequence 0 0 E (Q)/φ(E(Q)) → E(Q)/3E(Q) → E(Q)/φˆ(E (Q)) → 0 is exact, from which we deduce that E(Q)/3E(Q) is generated by (12, 36) (of which we also know now that it is a rational 3-torsion point4), and therefore, it is isomorphic to Z/3Z. By the Mordell–Weil Theorem, E(Q) and C(Q) are finitely generated, so we have that E(Q) is of rank zero, which implies that C(Q) has rank zero too. Similarly, we find that E0(Q)/3E0(Q) is generated by the rational 3-torsion point (0, 108), so it is also isomorphic to Z/3Z. Hence, E0(Q) is also of rank zero, from which it follows that C0(Q) has rank zero. We can also see from this that E and E0, and therefore C and C0, have precisely three rational 3-torsion points5.

3.2.3 Relating the descent method and the classical proof Similar to Section 2.2.3, we will now relate the classical proof in Section 3.1 and the method of descent by 3-isogeny in Section 3.2. Recall that we deﬁned the isogenies ψ : C → C0 and ψˆ: C0 → C in the introduction of this section, and also the isomorphisms τ : C → E and τ 0 : C0 → E0. Furthermore, in Section 3.2.2 we deﬁned the isogenies φ: E → E0 and φˆ: E0 → E. Proposition 3.2.3. The diagram

ψ ψˆ C C0 C

τ τ 0 τ φ φˆ E E0 E commutes. Proof. Let P = [X : Y : Z] ∈ C, then we have that (τ 0 ◦ ψ)(P ) = τ 0([Z3 : 2Y 3 − Z3 : 2XYZ]) = [−36XYZ : 216Y 3 − 108Z3 : Z3] and

(φ ◦ τ)(P ) = φ([12Z : 36(YX ): X + Y ]) = [124Z4 − 124Z(X + Y )3 : 36(Y − X)(123Z3 + 2 · 123(X + Y )3) : 123Z3(X + Y )] = [−12Z(3X2Y + 3Y 2X) : 36(3Z3(Y − X) + 6XY (Y + X)(Y − X)) : Z3(X + Y )] = [−12XYZ · 3(X + Y ) : 36(X + Y )(−3Z3 + 6Y 3): Z3(X + Y )] = [−36XYZ : 216Y 3 − 108Z3 : Z3]

4One can check that τ([0 : 1 : 1]) = (12, 36), so this should not be too surprising 5In fact, it can be shown that an elliptic curve over Q can have at most three rational 3-torsion points, as is done in [vB10] (following a lecture of Jaap Top).

32 3.2. Descent by 3-isogeny

so the ﬁrst half of the diagram commutes. Now let Q = [U : V : W ] ∈ C0, then

V 3 V 3 V 2 (τ ◦ ψˆ)(Q) = τ U 3 − UV 2 − VU 2 + : U 3 − UV 2 + VU 2 − : W U 2 + 9 9 3 V 2 V 3 = 12W U 2 + : 72 VU 2 − : 2(U 3 − UV 2) 3 9

and

(φˆ ◦ τ 0)(Q) = φˆ([−18W : 108V : U]) = [3 · (94 · 16W 4 − 2 · 9 · 363U 3W ) : 108V (−8 · 93W 3 − 2 · 363U 3): −8 · 27 · 93W 3U] = [54W (W 3 − 8U 3): −108V (W 3 + 16U 3): −27W 3U] V 2 V 3 = −12 · 27W U 2 + : −72 · 27U VU 2 − : −27W 3U 3 9 V 2 V 3 = 12W U 2 + : 72 VU 2 − : 2(U 3 − UV 2) 3 9

so the other half of the diagram also commutes.

3.2.4 A note on the automorphisms of C The descent by 3-isogeny in Section 3.2.2 shows that φ, and therefore ψ, is surjective if we restrict it to rational points. However, φˆ, so also ψˆ, is not surjective when restricted to rational points. Similar to Section 2.2.4, this can already be seen from the beginning of the classical proof in Section 3.1, since the assumption is made that for the nontrivial solution (x, y, z) of which we assume that it exists, we have that x and y are odd and z is even. This assumption comes down to applying the curve automorphisms C → C, [X : Y : Z] 7→ [Y : Z : X] and C → C, [X : Y : Z] 7→ [Z : X : Y ] to the point [x : y : z] ∈ C(Q). Note that since the j-invariant of C equals 0 (which can be seen from the equation of E), we have by Proposition 1.1.2 that C has complex multiplication: let ζ3 ∈ Q be a primitive third root of unity, then the automorphism group of C is of order 6, and it is generated by the automorphism [X : Y : Z] 7→ [Y : X : ζ3Z]. The automorphism [X : Y : Z] 7→ [Y : X : Z] is the only one which is deﬁned over Q, so this automorphism must be the negation6 in the group law of C. Therefore, every curve automorphism of C which is deﬁned over Q corresponds to translation by a rational point, possibly composed with this automorphism. So the curve automorphisms which are applied in the classical proof correspond to translations with rational 3-torsion points (since they are of order 3), possibly composed with the negation map. We will now identify which rational points correspond to which curve automorphisms. The automorphism C → C, [X : Y : Z] 7→ [Y : Z : X] is a translation by a rational point, since

6One can also compute the map τ −1 : E → C, [U : V : W ] 7→ [V − 36W : V + 36W : 6U] and then check directly that τ −1(−τ([X : Y : Z])) = −[X : Y : Z].

33 3.3. Twists and φ-coverings

otherwise, it would have ﬁxed points, which it doesn’t: if there exists some [X : Y : Z] ∈ C ∗ such that [X : Y : Z] = [Y : Z : X], there exists some λ ∈ Q such that λX = Y, λY = Z and λZ = X. It follows that X,Y,Z 6= 0 (if one of them is zero, also the other two are) and since λ3X = λ2Y = λZ = X, we have that λ3 = 1. But then, it follows that our point must be [λ : λ2 : 1]. However, clearly such points do not lie on C, so the automorphism is a translation by the rational point [1 : 0 : 1] (since it maps O to this point). Since −[1 : 0 : 1] = [0 : 1 : 1], it follows that the inverse of the translation by [1 : 0 : 1] map, i.e. the automorphism C → C, [X : Y : Z] 7→ [Z : X : Y ], is translation by [0 : 1 : 1]. So, in the classical proof, one may have to translate the point [x : y : z] by a rational 3-torsion point in order to get into the right co-set (for note that we do not need the negation map).

3.3 Twists and φ-coverings

Similarly to what we did in Section 2.4, we will now also investigate the method of descent by 3-isogeny from the viewpoint of twists and φ-coverings. Consider the elliptic curve E : y2 = x3 + d(ax + b)2 over Q, where a, b, d ∈ Q are such that b, d 6= 0 and 27b − 4a3d 6= 0. 2 Suppose that δ ∈ Q is such that δ = d, and denote Kd = Q(δ). Applying the descent by 0 ∗ ∗ 3 3-isogeny method in Section 1.3, we get a homomorphism of groups q : E(Q) → Kd /(Kd ) .

3.3.1 The case d ∈ (Q∗)2 Suppose that d ∈ (Q∗)2, i.e. δ ∈ Q∗, and let r ∈ Q∗. If [r] ∈/ {[1], [2bδ], [4b2d]} in Q∗/(Q∗)3, we have that [r] ∈ im(q0) if and only if there exists some (x, y) ∈ E(Q) and s ∈ Q∗ such that y − δ(ax + b) = rs3.

2 1 Proposition 3.3.1. Consider the algebraic set Fr ⊂ P × A (where we take coordinates X,Y,Z for P2 and s for A1) given by the equations: ( Y 2Z = X3 + dZ(aX + bZ)2 . Y − δ(aX + bZ) = rs3Z

Then this set is an algebraic curve, of which ([0 : bδ : 1], 0) is the only singular point.

Proof. Consider the part of Fr where Z 6= 0, which can be given by the aﬃne equations ( y2 = x3 + d(ax + b)2 y − δ(ax + b) = rs3

X Y where we take coordinates x = Z and y = Z . Eliminating y, this is isomorphic to the algebraic set in A2 which is the zero set of the polynomial F ∈ Q[x, s] given by F = x3 − 2δars3x − rs3(rs3 + 2δb).

Note that F is irreducible in (Q[s])[x], since we have that if f ∈ Q[s] is a zero, it needs to divide rs3(rs3 + 2δb), but it is easy to check that for all possible degrees of f, some terms

34 3.3. Twists and φ-coverings

remain which cannot vanish. Thus, F deﬁnes an algebraic curve, and since there are only ﬁnitely many points of Fr satisfying Z = 0, this implies that Fr is an algebraic curve. ∂F 2 3 ∂F 2 2 5 2 2 Note that ∂x = 3x − 2δars and ∂s = −6δars x − 6r s − 6r δbs . Clearly, both of those are zero if x = s = 0, which implies that Fr has a singular point at ([0 : bδ : 1], 0). On the other hand, if we require x 6= 0 (which is only possible if a 6= 0), we have that s 6= 0. Then δF δF rs3+rδb 2 2δars3 δx = δs = 0 can only hold if x = − δa and x = 3 , which is a contradiction. So Fr has no singular points except for ([0 : bδ : 1], 0).

0 By construction, we have that [r] ∈ im(q ) if and only if Fr has a rational point ([X : Y : Z], s) such that s 6= 0. By Theorem 1.1.4, there exists a smooth projective curve Fr which is birationally equivalent to Fr, and Fr is unique up to isomorphism over Q. In Proposition 2 3.2.1, we also considered the curve Er ⊂ P given by the equation 2arδUV W + 2bδW 3 + rV 3 = r2U 3.

One can check that this curve has singular points, but by Theorem 1.1.4 again, there exists some smooth projective curve Er over Q which is birationally equivalent to Er, and unique up to isomorphism over Q. Then we have the following.

Proposition 3.3.2. The curves Fr and Er are isomorphic over Q.

2 Proof. Consider the map σ : Fr → Er, ([X : Y : Z], s) 7→ [X : rZs : rZs]. Then this is a well deﬁned rational map, since we have for all ([X : Y : Z], s) ∈ Fr that

r2X3 = r2Z(Y − δ(aX + bZ))(Y + δ(aX + bZ)) = r3Z2s3(rs3Z + 2δ(aX + bZ)) = r(rs2Z)3 + 2aδr · rZs2 · rsZ · X + 2bδ(rZs)3.

Note that σ is regular everywhere, except for the singular point. Furthermore, deﬁne the 0 h rV 3 bW 2 W 2 i V map σ : Er → Fr, [U : V : W ] 7→ U : W 3 + δ aU + rV : rV , W then it is easy to check that σ0 is a well deﬁned rational map which provides an inverse for σ. It is regular everywhere, except for the points where V = 0 or W = 0. So σ admits a birational inverse, which implies that it determines a birational map σ : Fr → Er which, since both of those curves are smooth and projective, is an isomorphism.

∗ Note that for all r ∈ Q , we have that Fr is isomorphic to F1 over Q, since if we let ρ ∈ Q be such r = ρ3, the map (F ) → (F ) , ([X : Y : Z], s) 7→ ([X : Y : Z], ρs) is clearly a bijective r Q 1 Q rational map, and therefore, it determines an isomorphism (F ) → (F ) . Again, the F r Q 1 Q r 0 3 ∗ are called the twists of F1. Furthermore, suppose that r = rγ for some γ ∈ Q . Then we have that the map Fr0 → Fr, ([X : Y : Z], s) 7→ ([X : Y : Z], γs) is clearly a bijective rational map, which determines an isomorphism Fr0 → Fr over Q. Since im(q0) is finite by Proposition 3.2.1, this implies that up to isomorphism over Q, there are only finitely many of the curves Fr which have got rational points. Note that if Fr has a rational point and Fr has not, we have that those points either correspond to the singular point, so s = 0, or are “at infinity”. Using the proof of Proposition 3.3.2, we see that the rational points in the first case correspond to points [U : V : W ] ∈ Er(Q) satisfying V = 0,

35 3.3. Twists and φ-coverings

so 2bδW 3 = r2U 3, which implies that [r] = [2bδ], and this class is always in im(q0). In the second case, the rational points correspond to points [U : V : W ] ∈ Er(Q) satisfying 3 2 3 0 W = 0, so rV = r U , which implies that [r] = [1], which is also always in im(q ). So if Fr has a rational point, we always have that [r] ∈ im(q0), which implies that there are, up to isomorphism over Q, only ﬁnitely many twists of F1 which have rational points. 2 Now, consider the projection map πr : Fr → P , ([X : Y : Z], s) 7→ [X : Y : Z], which 2 determines a unique morphism πr : Fr → P . Then up to isomorphism over Q, every element in E(Q) is the image under πr for exactly one curve Fr. Summarizing the preceding discussion, we have that [ E(Q) = πr(Fr(Q)) [r]∈Q∗/(Q∗)3 where the union is disjoint and consists of ﬁnitely many points. Similarly, we suppose that d0 ∈ (Q∗)2, let δ0 be such that δ02 = d0 and consider the algebraic 0 2 1 set Fr ⊂ P × A given by ( V 2W = U 3 + d0W (a0U + b0W )2 V − δ0(a0U + b0W ) = rs3W

U V or, taking coordinates u = W and v = W , the aﬃne equations ( v2 = u3 + d0(a0u + b0)2 . v − δ0(a0u + b0) = rs3

0 Then Fr is a curve by Proposition 3.3.1. By Theorem 1.1.4, there exists some smooth 0 0 0 projective curve Fr such that Fr is birationally equivalent to Fr . The proof of the following proposition is due to [Coh07] (more speciﬁcally, Proposition 8.4.4 of his book). 0 Proposition 3.3.3. There exists an isomorphism f :(F ) → E over . r r Q Q Q u u 0 0 s+ s 0 s− s −b Proof. For ((u, v), s) ∈ F1 with s 6= 0, deﬁne u = and v = 0 , let x = 0 a and 2 2δ v + 3 consider the map 0 f : Fr → E, ((u, v), s) 7→ (x, u x) 0 0 0 u 0 0 0 02 0 02 02 02 Note that since s = u + v δ and s = u − v δ we have that u = u − d v = u + 3dv , and u3 s3 − = 2δ0(3u02v0 + v03d0) s while on the other hand, it follows from u3 s3 · = u3 = v2 − d0(a0u + b0)2 = (v − δ0(a0u + b0))(v + δ0(a0u + b0)) = s3(v + δ0(a0u + b0)) s u 3 0 0 0 that s = v + δ (a u + b ), so we have that u3 s3 − = v − δ0(a0u + b0) − (v + δ0(a0u + b0)) = −2δ0(a0u + b0) s 27b − 4a3d = −2δ0 a(u02 + 3dv02) + 9

36 3.3. Twists and φ-coverings

from which we see that ! a 4a3 a 2a2 b = −u02 v0 + + d v03 − av02 + = v0 + d v0 − − u02 . 3 27 3 3

It follows that a3 a a2 (x3 + d(ax + b)2) v0 + = −b3 + db2 v0 + −a + v0 + 3 3 3 ! ! a 2a2 2a2 = b2 v0 + − d v0 − − u02 + d v0 − 3 3 3 a a3 = b2 v0 + u02 = v0 + u02x2 3 3 so (x, u0x) ∈ E. Therefore, f is a well defined rational map. Furthermore, f has a birational 0 y 0 b a inverse, since for (x, y) ∈ E, we can define u = x and v = − x + 3 , then it is not difficult to check that the point ((u, v), s) with coordinates defined by u = u02 − d0v2, s = u0 + v0δ0 3 0 0 0 0 and v = s + δ (a u + b ) is in Fr, which defines an inverse for f. 0 Therefore, f determines an isomorphism f : F1 → E over . Since we already know that 0 0 Q 0 F is isomorphic to F over , this implies that there is an isomorphism f :(F ) → E 1 r Q r r Q Q over Q, as desired. It follows that there also exists an isomorphism (F ) → (E0) . So actually, in the method of r Q Q descent by 3-isogeny, we replace the problem of finding the rational points of E by checking whether another set of curves has a rational point, which may be less difficult. Furthermore, in order to obtain useful information about E(Q)/3E(Q) and ultimately, the rank of E(Q), it suffices to decide whether the twists do have rational points or not. The proof of the following is also due to [Coh07] (same proposition). Proposition 3.3.4. The following diagram commutes:

0 f r Fr E π r φ E0

0 Proof. Let ((u, v), s) ∈ F1 be such that s 6= 0, then we have that π1((u, v), s) = (u, v). On the other hand - using the notation of the previous proof again - we have that ! !!! a2 ab b 2 ab b 2 (φ ◦ f)(u, v) = φ(x, y) = x + 4d + + , u0 x − 4d + 2 3 x x x x

−b 02 0 2a 2 and note that x = 0 a = u − d v − so v + 3 3 ! a2 ab b 2 2a2 4a2 a a2 x + 4d + + = u02 − d v0 − + d − 4a v0 + + 4 v0 + 3 x x 3 3 3 3 = u02 + 3dv02 = u.

37 3.3. Twists and φ-coverings

1 3 u 3 03 0 02 We also have that v = 2 s + s = u − 9u v d which implies that !! ! ab b 2 2a2 a a2 u0 x − 4d + 2 = u0 u02 − d v0 − + d 4a v0 + − 8 v0 + x x 3 3 3 = u0(u02 − 9v02d) = v.

So (φ ◦ f)(u, v) = (u, v). Therefore, f and π1 make the diagram commute for r = 1. 0 0 Composing both maps with an isomorphism (F ) → (F ) , we see that the diagram r Q 1 Q commutes.

Again, πr is called a φ-covering.

3.3.2 The case d∈ / (Q∗)2 ∗ We want to develop a similar theory for the case that Kd 6= Q, i.e. δ∈ / Q . By Proposi- 0 ∗ ∗ 3 tion 1.3.4, the image of q is contained in the subﬁeld of Kd /(Kd ) consisting of all elements of which the norm is trivial in Q∗/(Q∗)3. ∗ ∗ Proposition 3.3.5. Let r = r1 − δr2 ∈ Kd , and let t ∈ Q be such that

3 N(r) = (r1 + δr2)(r1 − δr2) = t .

0 2 Then r ∈ im(q ) if and only if the smooth projective curve Fr ⊂ P given by the equation

2 2 3 3 2 2 3 atW (U − dV ) + bW = r2U + 3r1U V + 3dr2UV + dr1V has a rational point. Proof. For [r] = [1], the curve has the rational point [1 : 0 : 0], and [1] is also always in im(q0). (And note that we need not to check what happens for the 3-torsion point (0, b0δ0), since it is not rational.) Suppose that [r] 6= [1], and [r] ∈ im(q0), then there exists a point (x, y) ∈ E(Q) together with ∗ 3 some s = s1 − δs2 ∈ Kd such that y − δ(ax + b) = rs . Conjugating gives us the algebraic 4 set Er ⊂ A (with coordinates x, y, s1 and s2) deﬁned by:  y2 = x3 + d(ax + b)2  3 y − δ(ax + b) = (r1 − δr2)(s1 − δs2)  3 y + δ(ax + b) = (r1 + δr2)(s1 + δs2)

Combining the second and the last equation yields that x3 = y2 − d(ax + b)2 = t3N(s)3, so 2 2 since x ∈ Q, it follows that x = tN(s) = t(s1 − ds2). Combining this with the fact that (r + δr )(s + δs )3 − (r − δr )(s − δs )3 ax + b = 1 2 1 2 1 2 1 2 = r s3 + 3r s2s + 3dr s s2 + dr s3 2δ 2 1 1 1 2 2 1 2 1 2 we ﬁnd that 2 2 3 2 2 3 at(s1 − ds2) + b = r2s1 + 3r1s1s2 + 3dr2s1s2 + dr1s2 (3.6)

38 3.3. Twists and φ-coverings

0 so any [r] ∈ im(q ) gives rise to a rational point on Fr. 0 On the other hand, if Fr has a rational point [U : V : W ], this implies that [r] ∈ im(q ). If the rational point is not at inﬁnity (i.e. W 6= 0), this follows from the construction above. If 3 3 the rational point satisﬁes W = 0, it follows that (r1 + δr2)(U + δV ) = (r1 − δr2)(U − δV ) which implies that r(U − δV )3 ∈ Q. So t3N(U − δV )3 = r2(U − δV )6, which implies that r2 is a third power, so r itself is a third power, which implies that [r] = [1], which is always in im(q0).

4 Consider the aﬃne curve Kr ⊂ A given by the equations  x = t(s2 − ds2)  1 2 3 y − δ(ax + b) = (r1 − δr2)(s1 − δs2)  3 y + δ(ax + b) = (r1 + δr2)(s1 + δs2)

which is a subvariety of Er. By Theorem 1.1.4, there exists a smooth projective curve Kr which is birationally equivalent to Kr, and unique up to isomorphism over Q. It follows from the proof of Proposition 3.3.5 that Fr is isomorphic to Kr over Q, since if we consider the aﬃne part of Fr given by equation (3.6), the rational map

η : Kr → Fr, (x, y, s1, s2) 7→ (s1, s2)

is clearly well deﬁned, and it is easy to check that the map

0 2 2 3 2 2 η : Fr → Kr, (s1, s2) 7→ (t(s1 − ds2), r(s1 − δs2) + δ(at(s1 − ds2) + b), s1, s2) provides an inverse. Therefore, η determines an isomorphism (K ) → (F ) . r Q r Q Note that Fr is isomorphic to F1 over Q, since we can choose some ρ = ρ1 − δρ2 ∈ Q such 3 3 2 2 3 3 3 that r = ρ , i.e. r1 = ρ1 + 3dρ1ρ2 and r2 = 3ρ1ρ2 + dρ2, and from t = N(r) = N(ρ) we see that N(ρ) = t. Then we have that the map

σ :(F ) → (F ) , (u, v) 7→ (ρ u + dρ v, ρ u + ρ v) r Q 1 Q 1 2 2 1

0 0 is a well deﬁned rational map. In order to see this, let (u, v) ∈ Fr and denote σ(u, v) = (u , v ), then we have that

02 02 2 2 2 2 2 2 2 2 2 a(u − dv ) + b = a(ρ1u + 2dρ1ρ2uv + d ρ2v − d(ρ2u + 2ρ1ρ2uv + ρ1v )) + b 2 2 3 2 2 3 = at(u − dv ) + b = r2u + 3r1u v + 3dr2uv + dr1v 2 2 3 3 2 2 2 2 2 = (3ρ1ρ2 + dρ2)u + 3(ρ1 + 3dρ1ρ2)u v + 3d(3ρ1ρ2 + dρ2)uv 3 2 3 + d(ρ1 + 3dρ1ρ2)v 2 2 2 2 2 3 = 3(ρ1u + 2dρ1ρ2uv + d ρ2v )(ρ2u + ρ1v) + d(ρ2u + ρ1v) = 3u02v0 + dv03

so σ(u, v) ∈ F1. Since Fr is smooth, σ is a morphism. Furthermore, it is easy to check that the map σ0 :(F ) → (F ) , (u, v) 7→ ( ρ1u−ρ2dv , −ρ2u+ρ1v ) is well deﬁned and provides an 1 Q r Q t t inverse for σ, so σ is an isomorphism. Furthermore, we have that if r0 = γ3r for some γ ∈ Q∗, the curves Fr and Fr0 are isomorphic over Q. Indeed: the map Fr0 → F r, (u, v) 7→ (γu, γv)

39 3.3. Twists and φ-coverings

is an isomorphism (note that N(γ) = γ2, so N(r0) = t3γ6). Furthermore, up to isomorphism 0 over Q, there are only finitely many r for which Fr has a rational point (since im(q ) is finite by Proposition 3.2.2). Now, consider the projection map defined by

2 2 2 3 2 2 3 3 πr : Fr → P , [U : V : W ] 7→ [tW (U − dV ): r(U − V δ) + δW (at(U − dV ) + bW ): W ]. (Note that this is in fact the composition of the restriction of the natural projection map 2 0 Kr → P and the map determined by η .) Then we have that every element in E(Q) is the image under πr for exactly one of the curves Fr up to isomorphism over Q, so equivalently [ E(Q) = πr(Fr(Q)) ∗ ∗ 3 [r]∈Kd /(Kd )

(which is again a disjoint union consisting of only ﬁnitely many points). Analogously to Proposition 3.3.3, there exists an isomorphism f :(F ) → E over . This r r Q Q Q can be shown using the fact that over Q, Fr is isomorphic to F1, which is isomorphic to K1, then the proof is very similar to that of Proposition 3.3.3. In the same way, one can show that the analog of Proposition 3.3.4 still holds.

40 Appendix A

Appendix

A.1 An isomorphism of Ct to a curve in Weierstrass form

Let t ∈ Z \{0} be a nonzero integer and let Ct be the curve over Q in the weighted projective plane P(1, 1, 2) given by the equation Z2 = X4 + tY 4, together with the rational point O = [1 : 0 : 1]. In Section 2.1, we have shown that (Ct, O) is an elliptic curve, and 2 3 we claimed that it is isomorphic to the elliptic curve Et : v = u − 4tu in Weierstrass form in P2 (which is indeed an elliptic curve, since the discriminant of the Weierstrass equation equals −16 · 4 · 16t2 6= 0). We will now establish this.

Proposition A.1.1. Deﬁne the map

2 3 3 τt : Ct → Et, [X : Y : Z] 7→ 2Y (Z + X ) : 4(ZX + X ): Y

then τt is an isomorphism.

Proof. We have that τt is a well deﬁned, rational map because for all [X : Y : Z] ∈ Ct, we have that

8Y 3(Z + X2)3 − 4tY 6(2Y (Z + X2)) = 8(Z + X)Y 3(Z2 + 2ZX2 + X4 − tY 4) = 16(Z + X)2X2Y 3 = (4X(Z + X))2Y 3

so τt([X : Y : Z]) ∈ Et, and since Ct is smooth, τt is a morphism. Furthermore, note that τt(O) = [0 : 1 : 0] ∈ Et so it is an isogeny. We can deﬁne an inverse by

−1 3 2 2 τt : Et → Ct, [U : V : W ] 7→ WV : 2UW : 2U W − V W

This is again a well deﬁned rational map, since we have for all [U : V : W ] ∈ Et that

(2U 3W − V 2W 2)2 = 4U 6W 2 − 4U 3W 3V 2 + W 4V 4 = 4U 6W 2 − 4U 3W 2(U 3 − 4tUW 2) + W 4V 4 = 16tU 4W 4 + W 4V 4

41 A.2. Smoothness of the Fermat curve

−1 −1 so τt ([U : V : W ]) ∈ Ct, from which we conclude that τt is also a morphism. Furthermore, −1 −1 τt ([0 : 1 : 0]) = O so τt is an isogeny. −1 It remains to check that τt provides an inverse for τt. Therefore, note that we have for all [X : Y : Z] ∈ Ct that

−1 −1 2 3 3 (τt ◦ τt)[X : Y : Z] = τt 2Y (Z + X ) : 4(ZX + X ): Y = [4Y 3X(Z + X2) : 4Y 4(Z + X2) : 16Y 6((Z + X2)3 − X2(Z + X2)2)] = [X : Y : Z]

On the other hand, we have for all [U : V : W ] ∈ Et that

−1 3 2 2 (τt ◦ τt )([U : V : W ]) = τt WV : 2UW : 2U W − V W = 8U 4W 2 : 8U 3W 2V : 8U 3W 3 = [U : V : W ]

Hence, τt is an isomorphism, which maps Ct into a curve in Weierstrass form.

Note that once we know that τt is an isogeny, there is another way to show that it is an isomorphism: for [X : Y : Z] ∈ Ct, we have that [X : Y : Z] ∈ ker(τt) if and only if Y = 0 2 and Z + X 6= 0. It follows from the equation of Ct that [X : Y : Z] = O, so the kernel of τt is trivial. Since τt is separable, we have that deg(τt) = 1, so τt is an isomorphism.

A.2 Smoothness of the Fermat curve

Let n ≥ 3 be a positive integer and C be the curve over Q given by Xn + Y n = Zn in P2. Then C is a smooth curve. In order to see this, let P = [a : b : c] ∈ C be a point, n n Y Z and assume a 6= 0. Then we consider the aﬃne patch 1 + y = z , where y = X and z = X . Deﬁne f(x, y) = 1 + yn − zn, then we have that

∂f b n−1 ∂f c n−1 ∂f ∂f (P ) = n and (P ) = −n so if (P ) = (P ) = 0 ∂y a ∂z a ∂y ∂z

we must have bn−1 = cn−1 = 0, so b = c = 0. But since an + bn = cn, such P cannot exist. ∂f ∂f So the rank of ∂x (P ), ∂y (P ) is always equal to 1, which means that C is smooth at P . If a = 0, it follows from the equation of C that b 6= 0, and by symmetry, C is also smooth at P in that case. We conclude that C is smooth at all points P ∈ C.

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