Descent Calculations for the Elliptic Curves of Conductor 11

Tom Fisher

April 18, 2001

Abstract Let A be one of the three elliptic curves over Q with conductor 11. We show that A has Mordell-Weil rank zero over its field of 5division points. In each case we also compute the 5-primary part of the Tate-Shafarevich group. Our calculations make use of the Galois equivariance of the Cassels-Tate pairing.

Introduction

Ever since the work of Mazur [Ma] the elliptic curves of conductor 11 have provided a testing ground for the Iwasawa theory of elliptic curves. We recall from [V1] that these curves form a single isogeny class, and have explicit Weierstrass equations

\begin{matrix}A_{0} = X_{0}(11) & y^{2} + y = x^{3} − x^{2} − 10x − 20 & 11A1 \\ A_{1} = X_{1}(11) & y^{2} + y = x^{3} − x^{2} & 11A3 \\ A_{2} & y^{2} + y = x^{3} − x^{2} − 7820x − 263580 & 11A2 \end{matrix}

Here the labels 11A1-3 are those used in [Cr], whereas the labels A_{0}, A_{1}, A_{2} are taken from [CS]. When there is no need to distinguish the three curves we shall simply write A to denote any one of them.

Coates and Howson [CH] have used the elliptic curves of conductor 11 to illustrate their work on non-abelian Iwasawa theory. A natural question to ask is

How does the Mordell-Weil rank behave as we pass up the tower of fields given by adjoining the 5-power division points?

Although we are still unable to answer this question, we prove that the rank is zero over the field of 5-division points for each of the three curves.

It seems that Mazur [Ma, Cor. 9.10] was the first to show rank A(Q) = 0 and � (A/Q)(5) = 0. An extension of this result to Q(μ_{5}), due to Greenberg, may be found in [CS]. In each case the authors put their classical descent calculations to good use in studying the behaviour of Selmer groups over the cyclotomic Z_{5}-extension. For instance in [CS] it is shown that rank A(Q(μ_{5^{∞}})) = 0. It is hoped that our results will have equally striking applications.

The curves of conductor 11 are chosen since they appear first in the list of modular curves, and they do not admit complex multiplication. The prime 5 is chosen to make the problem more tractable. Indeed there are isogenies of degree 5 defined over Q

A_{1} ⇄ A_{0} ⇄ A_{2}.  (1)

The curves A_{0} and A_{1} each have a rational point of order 5, whereas A_{2} does not. By properties of the Weil pairing we deduce A_{0}[5] ≃ μ_{5} ⊕ Z/5Z as a Galois module. Furthermore there are exact sequences

0 → Z/5Z → A_{1} → A_{0} → 0 0 → μ_{5} → A_{2} → A_{0} → 0.  (2)

The fields of 5-division points are k = Q(μ_{5}), K_{1} = Q(A_{1}[5]) and K_{2} = Q(A_{2}[5]). Since K_{1} and K_{2} are non-abelian and of degree 20, it should come as no surprise that our descent calculations are rather more involved than those cited above. Our conclusions are

Theorem 1 Let K_{1} = Q(A_{1}[5]). Then rank A(K_{1}) = 0 and

A_{0}(K_{1}) ≃ (Z/5Z)^{2} � (A_{0}/K_{1})(5) ≃ (Z/5Z)^{8}
A_{1}(K_{1}) ≃ (Z/5Z)^{2} � (A_{1}/K_{1})(5) ≃ (Z/5Z)^{2}
A_{2}(K_{1}) ≃ Z/5Z  � (A_{2}/K_{1})(5) ≃ (Z/5Z)^{4} ⊕ (Z/25Z)^{8}.

Theorem 2 Let K_{2} = Q(A_{2}[5]). Then rank A(K_{2}) = 0 and

A_{0}(K_{2}) ≃ (Z/5Z)^{2} � (A_{0}/K_{2})(5) ≃ (Z/5Z)^{8}
A_{1}(K_{2}) ≃ Z/5Z  � (A_{1}/K_{2})(5) = 0
A_{2}(K_{2}) ≃ (Z/5Z)^{2} � (A_{2}/K_{2})(5) ≃ (Z/5Z)^{6} ⊕ (Z/25Z)^{8}.

It is easy to check that these results are compatible with the isogeny invariance of the Birch Swinnerton-Dyer conjecture, as proved by Cassels [Ca3]. Let us note that for 𝔭|11, inspection of the j-invariants shows that the Tamagawa factors are c_{𝔭}(A_{0}) = 5 ord_{𝔭}(11) and c_{𝔭}(A_{1}) = c_{𝔭}(A_{2}) = ord_{𝔭}(11). At each infinite place, it follows by Vélu’s formulae [V2] that the periods Ω_{i} are related via Ω_{1}/Ω_{0} = Ω_{0}/Ω_{2} = 5.

In §1 we introduce some subfields of Q(A[5^{∞}]). In §2 we recall from [F0], [F1], a description of the Selmer groups attached to the 5-isogenies (1). The analogue of Theorems 1 and 2 for k = Q(μ_{5}) is an easy consequence. In §3 we give explicit Kummer generators for the fields introduced in §1. In §4 we recall the definition of the Cassels-Tate pairing. Following the work of McCallum [Mc] and Beaver [B] we give a formula for the pairing in the case we need. In §5 we discuss certain Galois modules, and the alternating pairings they admit. Finally in §6 and §7 we give the descent calculations proving Theorems 1 and 2.

We have made extensive use of the computer algebra package pari in the course of this work. However we have striven where possible to give arguments that may be checked by hand. For the proof of Theorem 1 this goal has largely been achieved. In contrast the proof of Theorem 2 relies on us exhibiting a “non-trivial” unit in K_{1}K_{2}. Our method here was to ask pari to find all units in a certain degree 25 subfield. (This took 1 hour and 20 minutes on a 800MHz Pentium-III with 128Mb RAM.)

In a separate note [F2] we prove an analogue of Theorems 1 and 2 for the field J_{1} = Q(μ_{5})Q(μ_{11})^{+}. Again the rank is zero. Curiously our argument in that case does not require any formula for the Cassels-Tate pairing.

Acknowledgements

I would like to thank John Coates, Ralph Greenberg, Karl Rubin and Ed Schaefer for a number of valuable suggestions.

Notation and Conventions

For F a perfect field we write G_{F} := Gal(\overline{F} /F) and H^{i}(F, −) = H^{i}(G_{F} , −). By Hilbert’s theorem 90 we identify H^{1}(F, μ_{5}) = F^{∗}/F^{∗5}. The number field F has ring of integers 𝔒_{F}, unit group 𝔒_{F}^{∗}, and class group ℭ𝔩_{F}. The local field F_{𝔭} has ring of integers 𝔒_{𝔭} and normalised valuation ord_{𝔭}. Since our interest is in descent via isogenies of odd degree we ignore the infinite places throughout.

Let C and D be elliptic curves defined over F, and let ψ : C → D be an isogeny of degree m. The Kummer exact sequence restricts to

0 ⟶ D(F)/ψC(F) ⟶^{δ} S^{(ψ)}(C/F) ⟶ � (C/F)[ψ] ⟶ 0.

We frequently avoid giving our isogeny a name by writing S(C → D/F) for S^{(ψ)}(C/F). The Weil pairing is denoted e_{ψ} : C[ψ] × D[ψ̂] → μ_{m}.

The following notation relating to the field k = Q(μ_{5}) is used throughout. We fix ζ a primitive 5th root of unity and write Ind_{ζ} : μ_{5} → Q/Z for the map ζ ↦ 1/5. Then k has fundamental unit ϕ = 1 + ζ + ζ^{−1}. (Taking ζ = exp(2πi/5) this is the golden ratio.) We write \overline{ϕ} = −1/ϕ = 1 − ϕ for its conjugate. In §3 we use ϕ to define involutions η and ε on P_{k}^{1}. The primes of k above 5 and 11 are 𝔩 = (1 − ζ) and 𝔭_{i} = (π_{i}) where π_{i} = 2 + ζ^{i}. We write ω : G_{Q} → (Z/5Z)^{∗} for the cyclotomic character.

1 A description of Gal(Q(A[5^{∞}])/Q)

Serre [Se2, §5.5] proved

Proposition 1.1 Gal(Q(A[p^{∞}])/Q) ≃ GL_{2}(Z_{p}) for all primes p ≠ 5.

In contrast for p = 5, Q(A[p^{∞}])/Q(μ_{p}) is a pro-p extension. A description of the Galois group in this case was given by Lang and Trotter [LT]. In this section we present an alternative proof of their result and go on to compute the torsion subgroups listed as part of Theorems 1 and 2.

Let C_{1} and C_{2} be the kernels of the degree 25 isogenies A_{1} → A_{2} and A_{2} → A_{1}. We shall be concerned with the fields J_{1} = k(C_{1}), J_{2} = k(C_{2}), K_{1} = Q(A_{1}[5]) and K_{2} = Q(A_{2}[5]).

Lemma 1.2 The fields J_{1}, J_{2}, K_{1}, K_{2} are degree 5 Kummer extensions of k.

Proof. All is clear, except perhaps that these extensions are non-trivial. In fact C_{1} is generated by the cusps on A_{1} = X_{1}(11) and these are defined over Q(μ_{11})^{+} = Q(μ_{11}) ∩ R. Thus J_{1} = Q(μ_{5})Q(μ_{11})^{+} and Q(μ_{25}) ⊂ J_{1}J_{2}. For K_{1} and K_{2} we must show that the exact sequences (2) do not split as G_{k}- modules. As explained in [CS, Chapter 4] an examination of the Tate periods shows that these exact sequences do not even split as G_{Q_{11}}-modules. □

We pick a basis P, Q for the Tate module T_{5}(A_{0}), such that the projections of P and Q in A_{0}[5] generate ker(A_{0} → A_{2}) ≃ Z/5Z and ker(A_{0} → A_{1}) ≃ μ_{5} respectively. Then the Galois representation ρ : G_{Q} → GL_{2}(Z_{5}) attached to A_{0} satisfies

ρ(σ) ≡ (\begin{matrix}1 & 0 \\ 0 & ω(σ) \end{matrix}) (mod 5).

In particular

ρ(G_{k}) ⊂ { M ∈ GL_{2}(Z_{5}) | M ≡ I (mod 5) }.  (3)

Lemma 1.3 For σ ∈ G_{k} let ρ(σ) = I + 5 (\begin{matrix}a & b \\ c & d \end{matrix}). There are isomorphisms

Gal(J_{1}/k) ≃ Z/5Z ; σ ↦ a  Gal(K_{1}/k) ≃ Z/5Z ; σ ↦ b
Gal(K_{2}/k) ≃ Z/5Z ; σ ↦ c  Gal(J_{2}/k) ≃ Z/5Z ; σ ↦ d.

Furthermore, the action of Gal(k/Q) on these Galois groups is described by ψ = 1, ω^{−1}, ω and 1 respectively.

Proof. We check the first of these isomorphisms, the other cases being similar. Let P_{r}, Q_{r} be the projections of P, Q in A_{0}[5^{r}]. The image of P_{2} under the 5-isogeny A_{0} → A_{1} is a generator for C_{1}. Thus for σ ∈ G_{k}

σ fixes J_{1} pointwise ⟺ σ(P_{2}) − P_{2} ∈ ker(A_{0} → A_{1})
⟺ aP_{1} + c Q_{1} ∈ ker(A_{0} → A_{1})
⟺ a ≡ 0 (mod 5).

It follows that the map σ ↦ a induces an isomorphism Gal(J_{1}/k) ≃ Z/5Z. Finally Gal(k/Q) acts on G_{k} by conjugation, and we compute

(\begin{matrix}1 & 0 \\ 0 & ω \end{matrix}) (\begin{matrix}a & b \\ c & d \end{matrix}) (\begin{matrix}1 & 0 \\ 0 & ω \end{matrix})^{−1} = ( \begin{matrix}a & b ω^{−1} \\ c ω & d \end{matrix}) .

□

Lemma 1.4 The fields J_{1}, J_{2}, K_{1}, K_{2} are independent degree 5 Kummer extensions of k.

Proof. Given the distinct actions of Gal(k/Q) it suffices to check that J_{1} and J_{2} are independent. But J_{1}J_{2} = Q(μ_{25})Q(μ_{11})^{+} so this is clear. □

The next proposition was originally proved by Lang and Trotter [LT, Part I, Theorem 8.1]. I am grateful to John Coates for pointing out to me the simpler proof presented here.

Proposition 1.5 The extension Q(A[5^{∞}])/Q has Galois group

ρ(G_{Q}) = \{ (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ GL_{2}(Z_{5}) | (\begin{matrix}a & b \\ c & d \end{matrix}) ≡ (\begin{matrix}1 & 0 \\ 0 & ∗ \end{matrix}) (mod 5)\}.

Proof. We prove by induction on r that the image of ρ(G_{k}) in GL_{2}(Z/5^{r}Z) is the kernel of the map GL_{2}(Z/5^{r}Z) → GL_{2}(Z/5Z). The case r = 2 follows from Lemmas 1.3 and 1.4. The induction step is well known, and may be found in [LT] or [Se1]. It makes use of the identity

(I + 5^{r−1}M)^{5} ≡ I + 5^{r}M (mod 5^{r+1}).

We deduce that equality holds in (3), and the proposition follows. □

In §3 we find explicit Kummer generators for the extensions K_{1}/k and K_{2}/k. From these we learn that the prime above 5 is split in K_{1}/k and is ramified in K_{2}/k. We prove the easy part of Theorems 1 and 2.

Corollary 1.6 The torsion subgroups for A(K_{1}) and A(K_{2}) are

A_{0}(K_{1})_{tors} ≃ (Z/5Z)^{2}  A_{0}(K_{2})_{tors} ≃ (Z/5Z)^{2}
A_{1}(K_{1})_{tors} ≃ (Z/5Z)^{2}  A_{1}(K_{2})_{tors} ≃ Z/5Z
A_{2}(K_{1})_{tors} ≃ Z/5Z  A_{2}(K_{2})_{tors} ≃ (Z/5Z)^{2}.

Proof. Since A has good reduction at 5 and Ã(F_{5}) ≃ Z/5Z, it suffices to check that the 5-power torsion is as claimed. We make the observation that A_{0} has no point of order 25 defined over K_{1}K_{2}. Indeed if σ ∈ G_{Q} satisfies ρ(σ) = 6I then σ fixes pointwise the fields K_{1} and K_{2}, but does not fix any point of order 25 on A_{0}. Thus A_{0}(K_{i})_{tors} ≃ (Z/5Z)^{2} for i = 1, 2. Again for i = 1, 2 the inverse image of A_{0}[5] under the 5-isogeny A_{i} → A_{0} has field of definition J_{i}K_{i}. The remaining statements now follow from Lemma 1.4. □

The Selmer groups used in our calculations are of the most concrete nature, namely those attached to isogenies. They therefore contain contributions from torsion in the Mordell-Weil group. For this reason we make frequent implicit use of Corollary 1.6. For future reference we give another result on torsion subgroups.

Lemma 1.7 Let [F : Q_{11}] < ∞. Then #A_{1}(F)(5) ≤ 5[F : Q_{11}].

Proof. We know that A_{1} had multiplicative reduction, with Tamagawa factor e = ord(11). The number of smooth points over the residue field is 11^{f} − 1, and the multiplication by 5 map on the formal group is an isomorphism. Hence

#A_{1}(F)(5) ≤ 5ef = 5[F : Q_{11}].
□

2 Explicit descent via 5-isogeny

The Selmer groups attached to the 5-isogenies (1) are defined as subgroups of H^{1}(F, μ_{5}) = F^{∗}/F^{∗5} and H^{1}(F, Z/5Z) = Hom(G_{F} , Z/5Z).

Proposition 2.1 Let F be a number field. Then

S(A_{0} → A_{1}/F) ≃ \{ θ ∈ F^{∗}/F^{∗5} | \begin{matrix}ord_{𝔭}(θ) ≡ 0 (mod 5) for all 𝔭 \\ and F (\sqrt[5]{θ})/F split at 𝔭|11 \end{matrix} \}
S(A_{1} → A_{0}/F) ≃ { χ ∈ Hom(G_{F} , Z/5Z) | χ unramified at all 𝔭 ∤ 11 }
S(A_{2} → A_{0}/F) ≃ { θ ∈ F^{∗}/F^{∗5} | ord_{𝔭}(θ) ≡ 0 (mod 5) for all 𝔭 ∤ 11 }
S(A_{0} → A_{2}/F) ≃ \{ χ ∈ Hom(G_{F} , Z/5Z) | \begin{matrix}χ unramified at all 𝔭 \\ and χ split at 𝔭|11 \end{matrix} \}

(Here χ split at 𝔭 means 𝔭 splits in the fixed field of the kernel of χ.)

Proof. More generally in [F1] we considered pairs of 5-isogenous elliptic curves C_{λ} and D_{λ} with ker(C_{λ} → D_{λ}) ≃ μ_{5} and ker(D_{λ} → C_{λ}) ≃ Z/5Z. Explicitly D_{λ} has Weierstrass equation

y^{2} + (1 − λ)xy − λy = x^{3} − λx^{2}  (4)

and Z/5Z ↪ D_{λ}(F) is generated by (x, y) = (0, 0). We see that A_{1} = D_{1} and A_{0} = D_{11}. For each prime 𝔭 there is an exact sequence

C_{λ}(F_{𝔭}) ⟶ D_{λ}(F_{𝔭}) ⟶^{δ_{𝔭}} F_{𝔭}^{∗}/F_{𝔭}^{∗5}.

We recall [F1, Propositions 2.15 and 2.16] that δ_{𝔭} has image

im δ_{𝔭} = \{ \begin{matrix}F_{𝔭}^{∗}/F_{𝔭}^{∗5} & if ord_{𝔭}(λ) ≠ 0 \\ 𝔒_{𝔭}^{∗}/𝔒_{𝔭}^{∗5} & if ord_{𝔭}(λ) = ord_{𝔭}(λ^{2} − 11λ − 1) = 0 \\ 1 & if ord_{𝔭}(λ^{2} − 11λ − 1) > 0 and 𝔭 ∤ 5. \end{matrix}

The descriptions of S(A_{0} → A_{1}/F) and S(A_{2} → A_{0}/F) now follow on taking λ = 1, respectively λ = 11. Tate local duality tells us that the images of the local connecting maps attached to an isogeny and its dual are exact annihilators with respect to the Tate pairing. The descriptions of S(A_{1} → A_{0}/F) and S(A_{0} → A_{2}/F) follow. □

Suppose F is number field for which we have a working knowledge of the unit group and the class group. It is now a straightforward exercise in Kummer theory to compute the Selmer groups S(A_{0} → A_{1}/F) and S(A_{2} → A_{0}/F). If μ_{5} ⊂ F, then the Selmer groups attached to the dual isogenies may be treated similarly. However there is a better way.

Proposition 2.2 Let F be a number field with r_{1} (resp. r_{2}), real (resp. pairs complex conjugate) embeddings and m primes above 11. Then

\frac{#S(A_{0} → A_{1}/F)}{#S(A_{1} → A_{0}/F)} = #μ_{5}(F) × 5^{r_{1}+r_{2}−1} × 5^{−m}

\frac{#S(A_{2} → A_{0}/F)}{#S(A_{0} → A_{2}/F)} = #μ_{5}(F) × 5^{r_{1}+r_{2}−1} × 5^{m}.

Proof. This is an application of Cassels’ formula [Ca3, Theorem 1.1]. The ratios of Tamagawa numbers are given in the introduction. □

Remark 2.3 In simple cases, for example if F has class number 1, it is a tolerable exercise in class field theory to deduce Proposition 2.2 directly from Proposition 2.1. The beauty of Cassels’ formula is that the class number of F does not appear.

Example 2.4 We use Propositions 2.1 and 2.2 to compute rank A(k). We recall that k has class number 1, and that 𝔬_{k}^{∗} is generated by ±ζ, ϕ, where ϕ = 1 + ζ + ζ^{−1}. Writing π_{i} = 2 + ζ^{i} we find

S(A_{0} → A_{1}/k) = 0  since ζ, ϕ ∉ (𝔬_{k}/π_{1})^{∗5}
S(A_{1} → A_{0}/k) ≃ (Z/5Z)^{2}  i.e. Hom(Gal(J_{1}K_{1}/k), Z/5Z)
S(A_{2} → A_{0}/k) ≃ (Z/5Z)^{6}  i.e. ⟨ζ, ϕ, π_{1}, π_{2}, π_{3}, π_{4}⟩ ⊂ k^{∗}/k^{∗5}
S(A_{0} → A_{2}/k) = 0  since h_{k} = 1.

We deduce rank A(k) = 0 and � (A_{i}/k)(5) = 0 for i = 0, 1. We further find � (A_{2}/k)(5) ≃ (Z/5Z)^{4}.

3 Torsion contributions and Kummer generators

Let C_{λ} and D_{λ} be as in the proof of Proposition 2.1. Then λ is a co-ordinate on X_{1}(5) ≃ P^{1} and this modular curve has cusps at λ = 0, ∞, ϕ^{5}, \overline{ϕ^{5}}. There is an involution η on X_{1}(5), permuting the cusps, such that μ_{5} ↪ C_{λ} is isomorphic to Z/5Z ↪ D_{η(λ)} over F (μ_{5}). We take

η : λ ↦ (ϕ^{5}λ + 1)/(λ − ϕ^{5}).

For λ ∈ F not a cusp of X_{1}(5) there is a Kummer exact sequence

0 ⟶ μ_{5}(F) ⟶ C_{λ}(F) ⟶ D_{λ}(F) ⟶^{δ} F^{∗}/F^{∗5}.  (5)

Lemma 3.1 The image of Z/5Z ↪ D_{λ}(F) under the connecting map δ is generated by λ.

Proof. In terms of the Weierstrass equation (4), the multiples of (0, 0) are (λ, λ^{2}), (λ, 0), and (0, λ). We recall from [F1] that if P = (x, y) ≠ (0, 0) then δ(P) = xy + y − x^{2}. The lemma follows. □

For λ ∈ F we deduce F (C_{λ}[5]) = F (μ_{5}, \sqrt[5]{λ}). In particular η(1) and 11 are Kummer generators for K_{1}/k and K_{2}/k. We also learn that X(5) ≃ P^{1} with forgetful map

X(5) → X_{1}(5) ; τ ↦ τ^{5}.

The cusps of X(5) are at τ = 0, ∞, ζ^{i}ϕ, ζ^{i}\overline{ϕ}. Under stereographic projection these points may be viewed as the vertices of an icosahedron. There is an action of PSL_{2}(Z/5Z) ≃ A_{5} on X(5) permuting the cusps, generated by τ ↦ ζτ and

ε : τ ↦ (ϕτ + 1)/(τ − ϕ).

Lemma 3.2 Suppose μ_{5} ⊂ F. Let τ ∈ F and put λ = η(τ^{5}). Then the image of (Z/5Z)^{2} ↪ D_{λ}(F) under δ is generated by

λ = ∏_{i=0}^{4} ε(ζ^{i}τ) and ∏_{i=0}^{4} ε(ζ^{i}τ)^{i}.

Proof. In terms of the Weierstrass equation (4), (Z/5Z)^{2} ↪ D_{λ}(F) is generated by (0, 0) and

( − λ \frac{(ζτ − ϕ)(ζ^{4}τ − ϕ)}{(ϕτ + 1)(τ − ϕ)}, −λ^{2} \frac{(ζτ − ϕ)(ζ^{2}τ − ϕ)(ζ^{4}τ − ϕ)^{2}}{(ϕτ + 1)^{2}(τ − ϕ)(ϕζτ + 1)} )  (6)

We conclude as in the proof of Lemma 3.1. □

Remark 3.3 One way to construct the point (6) is to observe that in the notation of [F0], [F1] the curve

T = T [λ; ε(τ), ε(ζτ), ε(ζ^{2}τ), ε(ζ^{3}τ), ε(ζ^{4}τ)] ⊂ P^{4}

has rational point

(τ − ϕ : ζτ − ϕ : ζ^{2}τ − ϕ : ζ^{3}τ − ϕ : ζ^{4}τ − ϕ).  (7)

There is a diagonal action of μ_{5} on T with quotient D_{λ}. In [F0, Appendix C] we give explicit equations for the map T → D_{λ} and this allows us to construct (6) from (7).

Applying Lemma 3.2 with τ = 1 gives Kummer generators for J_{1}/k and K_{1}/k. Applying Lemma 3.2 with τ = ε(1) = −ϕ^{3} gives Kummer generators for J_{2}/k and K_{2}/k. We re-write these Kummer generators in terms of ζ, ϕ, π_{1}, π_{2}, π_{3}, π_{4} and so obtain an alternative proof of Lemma 1.4.

L Kummer generator for L/k ψ^{−1}ω 𝔣(L/k)  d_{L}
J_{1}  ∏ ε(ζ^{i})^{i} ζ^{2}π_{1}π_{2}^{3}π_{3}^{2}π_{4}^{4} ω  𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4} 5^{15}11^{16}
K_{1}  η(1)  ϕ^{2}π_{1}π_{2}^{4}π_{3}^{4}π_{4} ω^{2} 𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4} 5^{15}11^{16}
K_{2}  η(−ϕ^{15})  π_{1}π_{2}π_{3}π_{4}  1 𝔩^{2}𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4} 5^{23}11^{16}
J_{2}  ∏ ε(−ζ^{i}ϕ^{3})^{i} π_{1}π_{2}^{3}π_{3}^{2}π_{4}^{4}  ω 𝔩^{5}𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4} 5^{35}11^{16}

We recall that if L/k is the Kummer extension corresponding to Δ ⊂ k^{∗}/k^{∗5}, then Gal(L/k) ≃ Hom(Δ, μ_{5}) as a Gal(k/Q)-module. Thus in the notation of Lemma 1.3, Δ is described by ψ^{−1}ω. This is born out in our table.

The final two columns of our table record the conductor 𝔣 = 𝔣(L/k) and the absolute discriminant d_{L}. They are related via d_{L} = (Norm 𝔣)^{4}d_{k}^{5}. It is clear that the primes above 11 ramify in each extension L/k. We determine the factorisation of the prime above 5.

(i) The extension J_{1}/k is a translate of Q(μ_{11})^{+}/Q, so 𝔩 is inert. (ii) The extension K_{1}/k has Kummer generator

η(1) = \frac{1 + ϕ^{5}}{1 − ϕ^{5}} = \overline{ϕ^{5}}(1 + \frac{ϕ^{5} − \overline{ϕ^{5}}}{1 + \overline{ϕ^{5}}} ).  (8)

Since (ϕ^{5} − \overline{ϕ^{5}})^{2} = 5^{3} the binomial theorem shows that η(1) is a 5th power in k_{𝔩}. Thus 𝔩 splits in K_{1}/k. (iii) The extension K_{2}/k has Kummer generator 11. The minimal polynomial for \sqrt[5]{11} − 1 is an Eisenstein polynomial. Thus 5 is totally ramified in K_{2}/Q. A useful intermediate step in computing d_{K_{2}} is to show that Q(\sqrt[5]{11})/Q has discriminant 5^{5}11^{4}. (iv) Since J_{1}J_{2} = Q(μ_{25})Q(μ_{11})^{+} it is clear that 𝔩 ramifies in J_{2}/k. We recall [W, Proposition 2.1] that Q(μ_{25}) has discriminant 5^{35}.

Remark 3.4 Another quick way to show that J_{1}/k and K_{1}/k are unramified above 5 is provided by Proposition 2.1 and the observation that our Kummer generators belong to S(A_{1} → A_{0}/k).

Lemma 3.5 (i) The 5-ray class field of k with conductor 𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4} is J_{1}K_{1}. (ii) The 5-ray class field of k with conductor 𝔩^{2}𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4} is J_{1}K_{1}K_{2}.

Proof. We recall from [Coh, §3.2] a well known formula of class field theory

[k(𝔪) : k] = \frac{h_{k}ϕ(𝔪)}{[𝔬_{k}^{∗} : 𝔬_{k}^{∗} ∩ k_{𝔪,1}]}.

In our case we know h_{k} = 1 and 𝔬_{k}^{∗} is generated by ±ζ, ϕ.

(i) For 𝔪 = 𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4} we have ϕ(𝔪) = 10^{4} and 𝔬_{k}^{∗} ∩ k_{𝔪,1} generated by ϕ^{10}. Thus the 5-ray class field has degree 5^{2}, and so must equal J_{1}K_{1}. (ii) For 𝔪 = 𝔩^{2}𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4} we have ϕ(𝔪) = 20.10^{4} and 𝔬_{k}^{∗} ∩ k_{𝔪,1} generated by ϕ^{20}. Thus the 5-ray class field has degree 5^{3}, and so must equal J_{1}K_{1}K_{2}. □

To end this section, we exhibit some (modular) units in the fields K_{1} and K_{2}. Our descent calculations in §6 and §7 shall require further units in addition to these.

Lemma 3.6 (i) Let α = \sqrt[5]{η(1)} and u_{i} = ε(ζ^{i}α). Then u_{i} is a unit in K_{1} = k(α). The extension J_{2}K_{1}/K_{1} has Kummer generator u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4}.

(i) Let β = \sqrt[5]{11} and u_{i} = ε(ζ^{i}β). Then u_{i} is a unit in K_{2} = k(β). The extension J_{1}K_{2}/K_{2} has Kummer generator u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4}.

Proof. The cusps ζ\overline{ϕ}, ζϕ have minimal polynomials

f(x) = x^{4} + 3x^{3} + 4x^{2} + 2x + 1, g(x) = x^{4} − 2x^{3} + 4x^{2} − 3x + 1.

We may check η(ε(x)^{5}) = xf(x)/g(x).

(i) Each u_{i} is a root of xf(x)−g(x) = 0 and so is a unit. We apply Lemma 3.2 with τ = α to give the stated Kummer generator. (ii) Since η(11) = −ϕ^{15}, each u_{i} is a root of xf(x) + ϕ^{15}g(x) = 0 and so is a unit. We apply Lemma 3.2 with τ = β to give the stated Kummer generator. □

4 The Cassels-Tate pairing

Let C, D, ψ be as in the Introduction. There is an exact sequence

0 ⟶ C[ψ] ⟶ C[m] ⟶^{ψ} D[ψ̂] ⟶ 0.  (9)

Taking Galois cohomology and restricting to Selmer groups we obtain

D[ψ̂](F) ⟶ S^{(ψ)}(C/F) ⟶ S^{(m)}(C/F) ⟶ S^{(ψ̂)}(D/K).

Proposition 4.1 There is an alternating pairing

S^{(ψ̂)}(D/K) × S^{(ψ̂)}(D/K) → Q/Z  (10)

whose kernel is the image of S^{(m)}(C/F).

Proof. See Cassels [Ca2] or Milne [Mi]. □

We recall the definition of the pairing in the case m is odd. Our treatment follows that of McCallum [Mc]. For (∗) a global element or map, we write (∗)_{𝔭} for the corresponding local object. The following commutative diagram is considered both in its own right and with F replaced by F_{𝔭} for each prime 𝔭.

C(F) ⟶^{ψ} D(F) ⟶^{δ_{ψ}} H^{1}(F, C[ψ])
|| ↓ψ̂ ↓^{ι}
C(F) ⟶^{×m} C(F) ⟶^{δ_{m}} H^{1}(F, C[m])
↓^{ψ}  ||  ↓^{ψ}
D(F) \underrightarrow{ψ̂} C(F) ⟶^{δψ̂} H^{1}(F, D[ψ̂])

To define the pairing we take x, y ∈ S^{(ψ̂)}(D/F) ⊂ H^{1}(K, D[ψ̂]) and suppose given x_{1} ∈ H^{1}(F, C[m]) lifting x. At each prime 𝔭 we choose x_{𝔭,1} ∈ im δ_{m,𝔭} such that ψ(x_{𝔭,1}) = x_{𝔭}. Then ψ(x_{𝔭,1} − x_{1,𝔭}) = 0 and so there exists ξ_{𝔭} ∈ H^{1}(F_{𝔭}, C[ψ]) with ι(ξ_{𝔭}) = x_{𝔭,1} − x_{1,𝔭}. We define

⟨x, y⟩ = ∑_{𝔭}(ξ_{𝔭}, y_{𝔭})_{𝔭}  (11)

where (·, ·)_{𝔭} is the Tate pairing H^{1}(F_{𝔭}, C[ψ]) × H^{1}(F_{𝔭}, D[ψ̂]) → Q/Z. Using Tate local duality and the product formula for the Tate pairing we may check that the definition is independent of all choices. It is clear that ⟨x, y⟩ = 0 whenever x is in the image of S^{(m)}(C/F).

Remark 4.2 More generally, the Cassels-Tate pairing is defined on the Tate-Shafarevich group � (D/F). The restriction to � (D/F)[ψ̂] is the pairing induced by (10). We may implicit use of this fact in due course.

Remark 4.3 Suppose C, D, ψ are defined over F_{0} ⊂ F. If F is a normal extension of F_{0} then (10) is Gal(F/F_{0})-equivariant. It is to be understood that the Galois action on Q/Z is trivial.

We give a formula for (10) in the case where ψ is the 5-isogeny A_{1} → A_{0} and F contains K_{1} = Q(A_{1}[5]). The exact sequence (9) becomes

0 ⟶ A_{1}[ψ] ⟶ A_{1}[5] ⟶^{ψ} A_{0}[ψ̂] ⟶ 0.  (12)

We choose a section for the map ψ in (12) and use this to construct x_{1} from x. As McCallum [Mc] observes we may now express (11) as a sum of local pairings

⟨x, y⟩ = ∑_{𝔭}⟨x_{𝔭}, y_{𝔭}⟩_{𝔭}.

Furthermore the local pairing is trivial outside the usual set of bad primes, in our case those above 5 and 11. The description of S(A_{0} → A_{1}/F) given in Proposition 2.1 tells us that x, y are already trivial at 𝔭|11. So it only remains to compute the local pairing at 𝔭|5.

In §3 we saw that K_{1}/k has Kummer generator η(1) = (1 + ϕ^{5})/(1 − ϕ^{5}). We put α = \sqrt[5]{η(1)}. By (8) the primes above 5 split in K_{1}/k. We label them 𝔏_{0}, 𝔏_{1}, …, 𝔏_{4} such that α ≡ ζ^{i}ϕ (mod 𝔏_{i}^{2}).

Lemma 4.4 Let F be a number field with F ⊃ K_{1}. Let 𝔭|5 be a prime and let e = e(𝔭/5). Then there exists i = i(𝔭) in Z/5Z such that α ≡ ζ^{i}ϕ (mod 𝔭^{e/2}).

Proof. We have i = i(𝔭) if and only if 𝔭|𝔏_{i}. □

Proposition 4.5 Let F be a number field with F ⊃ K_{1}. Then the Cassels-Tate pairing on S(A_{0} → A_{1}/F) ⊂ F^{∗}/F^{∗5} is given, up to scalars, by

⟨θ, θ′⟩ = ∑_{𝔭|5} Ind_{ζ}(θ, θ′)_{𝔭}^{i(𝔭)}

where (·, ·)_{𝔭} is the Hilbert norm residue symbol.

Remark 4.6 To remove the qualifier “up to scalars” we must specify the isomorphism A_{0}[ψ̂] ≃ μ_{5} used to embed S(A_{0} → A_{1}/F) inside F^{∗}/F^{∗5}. For the proof of Theorems 1 and 2, a formula “up to scalars” is good enough.

Lemma 4.7 Let P ∈ A_{1}[ψ] and Q ∈ A_{0}[ψ̂] with e_{ψ}(P, Q) = ζ. Then we may label the inverse image ψ^{−1}(Q) = {Q_{0}, Q_{1}, …, Q_{4}} such that Q_{i} belongs to the kernel of reduction mod 𝔏_{i}.

Proof. Let denote reduction mod 𝔏_{i}. By inspection of the Weierstrass equation (4) the reduction Ã_{1}(F_{5}) ≃ Z/5Z is generated by P̃. The kernel of the reduction map A_{1}[5] → Ã_{1}(F_{5}) is cyclic of order 5. We choose a generator Q_{i} with ψ(Q_{i}) = Q. Then Gal(K_{1}/k) permutes both the 𝔏_{i} and the Q_{i}. □

The Weil pairing and Hilbert’s theorem 90 allow us to identify

H^{1}(F, A_{1}[ψ]) = Hom(A_{0}[ψ̂], F^{∗}/F^{∗5})
H^{1}(F, A_{1}[5]) = Hom(A_{1}[5], F^{∗}/F^{∗5})
H^{1}(F, A_{0}[ψ̂]) = Hom(A_{1}[ψ], F^{∗}/F^{∗5}).

(13)

We give a more precise version of Proposition 4.5.

Lemma 4.8 Let F be a number field with F ⊃ K_{1}. Let P, Q_{i} ∈ A_{1}[5] be chosen as in Lemma 4.7. Then the Cassels-Tate pairing on S(A_{0} → A_{1}/F) ⊂ Hom(A_{1}[ψ], F^{∗}/F^{∗5}) is given by

⟨x, y⟩ = ∑_{𝔭|5} Ind_{ζ}(x(Q_{1} − Q_{0}), y(P))_{𝔭}^{i(𝔭)}.

Proposition 4.5 follows immediately from Lemma 4.8, since P and Q_{1} − Q_{0} are both generators for A_{1}[ψ].

Lemma 4.9 Let 𝔭|5 with i = i(𝔭). Then the local connecting map δ_{5,𝔭} : A_{1}(F_{𝔭}) → Hom(A_{1}[5], F_{𝔭}^{∗}/F_{𝔭}^{∗5}) has image

\{ x ∈ Hom(A_{1}[5], 𝔒_{𝔭}^{∗}/𝔒_{𝔭}^{∗5}) | F_{𝔭}(\sqrt[5]{x(Q_{i})})/F_{𝔭} is unramified \}. (14)

Proof. Let x = δ_{5,𝔭}(T) for some T ∈ A_{1}(F_{𝔭}). The description of im δ_{ψ̂,𝔭} used in the proof of Proposition 2.1 shows that x(P) is a unit. Let T′ ∈ A_{1}(\overline{F}_{𝔭}) with 5T′ = T. Then x is represented by the cocycle σ(T′) − T′ in H^{1}(F_{𝔭}, A_{1}[5]). But if σ belongs to the inertia subgroup, then σ(T′) − T′ belongs to the kernel of reduction mod 𝔭 and e_{5}(Q_{i}, σ(T′) − T′) = 1. Hence x(Q_{i}) is unramified.

We have shown that im δ_{5,𝔭} belongs to (14). But im δ_{5,𝔭} ⊂ H^{1}(F_{𝔭}, A_{1}[5]) is a maximal isotropic subspace with respect to the Tate pairing. A counting argument completes the proof of the lemma. □

The identifications (13) allow us to express the Tate pairing in terms of the Hilbert norm residue symbol.

Lemma 4.10 The Tate pairing H^{1}(F_{𝔭}, A_{1}[ψ]) × H^{1}(F_{𝔭}, A_{0}[ψ̂]) → Q/Z is given by

(x, y)_{𝔭} = Ind_{ζ}(x(Q), y(P))_{𝔭}^{−1}

where (·, ·)_{𝔭} on the right is the Hilbert norm residue symbol.

Proof. We recall e_{ψ}(P, Q) = ζ. The lemma follows by a standard cup product calculation. □

Proof of Lemma 4.8. The map ψ in (12) has section Q ↦ Q_{0}. Let x, y belong to S(A_{0} → A_{1}/F). By (13) we view x, y as maps A_{1}[ψ] → F^{∗}/F^{∗5}. Then x_{1} : A_{1}[5] → F^{∗}/F^{∗5} extends x via Q_{0} ↦ 1. For each 𝔭|5 we extend x_{𝔭} to x_{𝔭,1} : A_{1}[5] → F_{𝔭}^{∗}/F_{𝔭}^{∗5} via Q_{i(𝔭)} ↦ 1. Then x_{𝔭,1} ∈ im δ_{5,𝔭} and ξ_{𝔭}(Q) = x(Q_{0} − Q_{i(𝔭)}) = x(Q_{1} − Q_{0})^{−i(𝔭)}. The local pairing is

⟨x_{𝔭}, y_{𝔭}⟩_{𝔭} = (ξ_{𝔭}, y_{𝔭})_{𝔭}
= Ind_{ζ}(ξ_{𝔭}(Q), y_{𝔭}(P))_{𝔭}^{−1}
= Ind_{ζ}(x_{𝔭}(Q_{1} − Q_{0}), y_{𝔭}(P))_{𝔭}^{i(𝔭)}.

This completes the proof of Lemma 4.8 and so of Proposition 4.5. □

5 Some Galois modules

The polynomials x^{5} + 2x^{4} + 6x^{3} − 2x^{2} + 4x − 1 and x^{5} − 11 have splitting fields K_{1} = Q(A_{1}[5]) and K_{2} = Q(A_{2}[5]). In each case the Galois group is

G := ⟨ σ, τ | σ^{4} = τ^{5} = 1, στσ^{−1} = τ^{2} ⟩.

In preparation for the proof of Theorems 1 and 2, we give some preliminaries on Z/5Z[G]-modules. We define ψ : G → (Z/5Z)^{∗} via σ ↦ 2 and τ ↦ 1. Any Z/5Z[G]-module M may be decomposed into σ-eigenspaces

M = M^{1} ⊕ M^{ψ} ⊕ M^{ψ^{2}} ⊕ M^{ψ^{3}}  (15)

where M^{χ} = {x ∈ M|σx = χ(σ)x}. If M = M^{χ} we say that M is described by χ. In particular ψ describes the action of G on ⟨τ⟩ via conjugation.

Lemma 5.1 Let M be a Z/5Z[G]-module with M/(τ − 1)M ≃ Z/5Z as an abelian group. Then

(i) M/(τ − 1)M is described by some character χ : G → (Z/5Z)^{∗}. (ii) M has dimension d := dim_{Z/5Z} M with d ≤ 5. (iii) The pair (χ, d) uniquely determines M as a G-module. (iv) If d ≤ 4 then End_{G}(M) ≃ Z/5Z.

Proof. (i) This is clear. (ii) Let M_{i} = (τ − 1)^{i}M. The decreasing filtration of Z/5Z[G]-modules

M = M_{0} ⊃ M_{1} ⊃ M_{2} ⊃ …  (16)

satisfies dim_{Z/5Z} M_{i}/M_{i+1} ≥ dim_{Z/5Z} M_{i+1}/M_{i+2}. But dim_{Z/5Z} M_{0}/M_{1} = 1, so d = min{i | M_{i} = 0}. Since (τ − 1)^{5} ≡ 0 (mod 5) we must have d ≤ 5. (iii) We pick x ∈ M^{χ} a generator for M/(τ − 1)M. Then M has basis x, (τ − 1)x, …, (τ − 1)^{d−1}x as a Z/5Z-vector space. The actions of σ and τ on this basis are uniquely determined. (iv) The quotient M_{i}/M_{i+1} is described by χψ^{i}. Thus for d ≤ 4 the decomposition (15) is into 1-dimensional spaces, and the element x in the proof of (iii) is uniquely determined up to scalars. □

We write M(χ, d) for the Z/5Z[G]-module described in Lemma 5.1. We abbreviate M(χ) = M(χ, 1). The filtration (16) becomes

M(χ, d) ⊃ M(χψ, d − 1) ⊃ … ⊃ M(χψ^{d−1}, 1) ⊃ 0.

For M a G-module we recall that M^{∗} := Hom(M, Z/5Z) is a G-module via gθ = θ.g^{−1}.

Lemma 5.2 M(χ, d)^{∗} ≃ M(χ^{−1}ψ^{1−d}, d).

Proof. The case d = 1 is clear. The general case follows from Lemma 5.1 and the observation coker(τ − 1|M^{∗}) ≃ ker(τ − 1|M)^{∗}. □

With properties of the Cassels-Tate pairing in mind, we say that a bilinear form ⟨ , ⟩ : M × N → Q/Z is G-equivariant if ⟨gx, gy⟩ = ⟨x, y⟩ for all g ∈ G. Equivalently M → N^{∗} is a G-module homomorphism.

Lemma 5.3 Any non-zero G-equivariant pairing on M(χ, d) has odd rank. In particular there are no non-zero alternating G-equivariant pairings.

Proof. Suppose f : M(χ, d) → M(χ^{−1}ψ^{1−d}, d) is a G-module map of rank r. Then im f = M(χ^{−1}ψ^{1−r}, r). We deduce χ = χ^{−1}ψ^{1−r} and so r is odd. □

Lemma 5.4 Assume d ≤ 4. Then any non-zero alternating G-equivariant pairing on M := M(χ, d) ⊕ M(χ^{−1}ψ^{1−d}, d) is non-degenerate.

Proof. We claim that, up to scalars, M admits a unique alternating Gequivariant pairing. By Lemma 5.3 any such pairing is trivial when restricted to either summand. We are reduced to showing that there is a unique Gequivariant bilinear form

M(χ, d) × M(χ^{−1}ψ^{1−d}, d) → Z/5Z.

Lemma 5.2 gives the existence. Lemma 5.1(iv) gives the uniqueness up to scalars. Finally we observe that the pairing constructed is non-degenerate. □

We give an example typical of the G-modules we encounter. The group G acts on the affine line Z/5Z via σ : x ↦ 2x and τ : x ↦ x + 1. The corresponding permutation representation, with coefficients in Z/5Z, is M(1, 5). We construct further G-modules using M(χ_{1}χ_{2}, d) = M(χ_{1}) ⊗ M(χ_{2}, d).

6 Descent calculations over K_{1} = Q(A_{1}[5])

We recall that K_{1}/k has Kummer generator η(1) = (1 + ϕ^{5})/(1 − ϕ^{5}). We put α = \sqrt[5]{η(1)}. Then Gal(K_{1}/Q) ≃ G via

\begin{matrix}σ(ζ) = ζ^{3} & σ(α) = −1/α \\ τ(ζ) = ζ & τ(α) = ζα. \end{matrix}

The cyclotomic character ω, and the character ψ of §5 are related via ψ = ω^{−1}. The primes 5 and 11 factor in K_{1} as

(5) = 𝔏_{0}^{4}𝔏_{1}^{4}𝔏_{2}^{4}𝔏_{3}^{4}𝔏_{4}^{4} (11) = 𝔓_{1}^{5}𝔓_{2}^{5}𝔓_{3}^{5}𝔓_{4}^{5}

with α ≡ ζ^{i}ϕ (mod 𝔏_{i}^{2}) and 𝔓_{i}|𝔭_{i}. In §3 we saw that H_{1} := J_{1}K_{1} is the 5-ray class field of k for conductor (11) = 𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4}.

Lemma 6.1 The extension H_{1}/K_{1} is unramified at all primes. Furthermore the primes above 5 and 11 are inert in this extension.

Proof. Only primes above 11 ramify in H_{1}/k. By considering suitable ratios of our Kummer generators for J_{1}/k and K_{1}/k we see that 𝔭_{i} cannot be totally ramified in H_{1}/k. Hence H_{1}/K_{1} is unramified as claimed. Now H_{1}/K_{1} is a translate of Q(μ_{11})^{+}/Q. Since 5 is inert in Q(μ_{11})^{+}/Q and 𝔏_{i} has residue field F_{5}, it follows that 𝔏_{i} is inert. Finally the definitions of J_{1} and K_{1} give #A_{1}(H_{1})(5) ≥ 5^{3}, so by Lemma 1.7 the 𝔓_{i} are inert. □

Proposition 6.2 The 5-class group of K_{1} is ℭ𝔩_{K_{1}}(5) ≃ Z/5Z. It is generated by any prime above 5 or 11.

Proof. Let B = ℭ𝔩_{K_{1}}(5). By Lemma 3.5(i) we know that H_{1} is the maximal unramified 5-extension of K_{1} which is abelian over k. Thus

B/(τ − 1)B ≃ Gal(H_{1}/K_{1}) ≃ Z/5Z

as abelian groups. By Lemma 6.1, 𝔓_{1} is inert in H_{1}/K_{1} and so generates B/(τ − 1)B. Since (τ − 1)^{5} ⊂ 5Z_{5}[τ] it follows that 𝔓_{1} generates B as a Z_{5}[τ]-module. But τ(𝔓_{1}) = 𝔓_{1} and 𝔓_{1}^{5} = 𝔭_{1} is principal. Thus B ≃ Z/5Z as claimed. By Lemma 6.1, B is generated by any prime above 5 or 11. □

Remark 6.3 Since σ(𝔏_{0}) = 𝔏_{0} and τ(𝔓_{1}) = 𝔓_{1} the action of G on ℭ𝔩_{K_{1}}(5) is trivial. In particular 𝔓_{1} ∼ 𝔓_{2} ∼ 𝔓_{3} ∼ 𝔓_{4}.

We turn our attention to the units in K_{1}. We write

u_{i} = \frac{ϕζ^{i}α + 1}{ζ^{i}α − ϕ} v_{i} = \frac{ζ^{i}α + 1}{ζ^{i}α − 1}.

The u_{i} are units by Lemma 3.6. For the v_{i} we have (v_{i} + 1)^{5}/(v_{i} − 1)^{5} = (1 + ϕ^{5})/(1 − ϕ^{5}). Thus each v_{i} is a root of

ϕ^{5}(x^{5} + 10x^{3} + 5x) − (5x^{4} + 10x^{2} + 1) = 0

and so is a unit. It is easy to check

\begin{matrix}σ(u_{i}) = u_{2i} & σ(v_{i}) = −1/v_{2i} \\ τ(u_{i}) = u_{i+1} & τ(v_{i}) = v_{i+1} \end{matrix}

and we have relations ∏ u_{i} = 1, ∏ v_{i} = ϕ^{−5}. Thus the subgroups of K_{1}^{∗}/K_{1}^{∗5} generated by u_{1}, u_{2}, u_{3}, u_{4} and v_{1}, v_{2}, v_{3}, v_{4} are quotients of the G-modules M(1, 4) and M(ω^{2}, 4).

Proposition 6.4 The units ζ, ϕ, u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, v_{4} generate a subgroup of 𝔒_{K_{1}}^{∗} of index prime to 5.

Proof. By Dirichlet it suffices to check that the elements listed are independent in K_{1}^{∗}/K_{1}^{∗5}. There is a G-module homomorphism

M := M(ω) ⊕ M(ω^{2}) ⊕ M(1, 4) ⊕ M(ω^{2}, 4) → K_{1}^{∗}/K_{1}^{∗5}

where the summands correspond to ζ, ϕ, ⟨u_{1}, u_{2}, u_{3}, u_{4}⟩ and ⟨v_{1}, v_{2}, v_{3}, v_{4}⟩. We suppose for a contradiction that this map has non-trivial kernel. Then this kernel meets

\begin{aligned}ker(τ − 1|M) & = & M(ω) ⊕ M(ω^{2}) ⊕ M(ω) ⊕ M(ω^{3}) \\ & = & ⟨ ζ, ϕ, u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4}, v_{1}v_{2}^{2}v_{3}^{3}v_{4}^{4} ⟩. \end{aligned}

Dividing into σ-eigenspaces we learn that one of the elements

ζ^{i}(u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4})^{j}, ϕ, v_{1}v_{2}^{2}v_{3}^{3}v_{4}^{4}

is a 5th power. For ζ and ϕ this is clearly false, since no unit can be a Kummer generator for K_{1}/k. The smallest prime to split completely in K_{1}/Q is p = 101. Reducing modulo primes above p we obtain a contradiction. □

Remark 6.5 Further to the proof Proposition 6.4, a brutal computer calculation shows

\begin{aligned}π_{1}π_{2}^{3}π_{3}^{2}π_{4}^{4} & ≡ & (u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4})^{3} & (mod K_{1}^{∗5}) \\ π_{1}π_{2}^{2}π_{3}^{3}π_{4}^{4} & ≡ & (v_{1}v_{2}^{2}v_{3}^{3}v_{4}^{4})^{3} & (mod K_{1}^{∗5}). \end{aligned}

Remark 6.6 According to pari the field K_{1} has class number 5, and

ϕ, u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, v_{4}

is a set of fundamental units. However pari assumes the Generalised Riemann Hypothesis, whereas our results are unconditional.

We apply Propositions 2.1 and 2.2 in the case F = K_{1}.

Proposition 6.7 The Selmer groups attached to the 5-isogenies (1) are

\begin{aligned}S(A_{0} → A_{1}/K_{1}) & ≃ & (Z/5Z)^{8} & S(A_{2} → A_{0}/K_{1}) & ≃ & (Z/5Z)^{14} \\ S(A_{1} → A_{0}/K_{1}) & ≃ & (Z/5Z)^{2} & S(A_{0} → A_{2}/K_{1}) & = & 0. \end{aligned}

Proof. By Propositions 6.2 and 6.4 the space

{ θ ∈ K_{1}^{∗}/K_{1}^{∗5} | ord_{𝔭}(θ) ≡ 0 (mod 5) for all 𝔭 }

has basis ζ, ϕ, u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, v_{4}, 11. Here 11 is a contribution from the class group or “virtual unit”. We choose characters (𝔒_{K_{1}}/𝔓_{i})^{∗} → Z/5Z and compute these characters on our basis. Notice that by inspection of our Kummer generator for K_{1}/k we have 11 ≡ (1 + ϕ^{5})^{2} ≡ (1 − ϕ^{5})^{2} (mod K_{1}^{∗5}). Thus our table is easily computed by hand.

𝔓_{1} 𝔓_{2} 𝔓_{3} 𝔓_{4}
ζ 1 3 2 4
ϕ 2 3 3 2
u_{i} 2 2 2 2
v_{i} 0 0 0 0
11 2 2 2 2

By Proposition 2.1 we deduce

S(A_{0} → A_{1}/K_{1}) ≃ ⟨u′_{1}, u′_{2}, u′_{3}, u′_{4}, v_{1}, v_{2}, v_{3}, v_{4}⟩ ⊂ K_{1}^{∗}/K_{1}^{∗5}

where u′_{i} = u_{i}/11. Propositions 2.1 and 6.2 show that S(A_{0} → A_{2}/K_{1}) is trivial. The remaining statements follow by Proposition 2.2. □

Proposition 6.7 furnishes the estimate rank A(K_{1}) ≤ 8. We improve on this by computing the Cassels-Tate pairing

S(A_{0} → A_{1}/K_{1}) × S(A_{0} → A_{1}/K_{1}) → Q/Z.  (17)

As a G-module we have

S(A_{0} → A_{1}/K_{1}) ≃ M(1, 4) ⊕ M(ω^{2}, 4)

where the summands correspond to ⟨u′_{1}, u′_{2}, u′_{3}, u′_{4}⟩ and ⟨v_{1}, v_{2}, v_{3}, v_{4}⟩. By Lemma 5.3 the pairing (17) is trivial when restricted to either summand. It therefore suffices for us to compute the entries ⟨u′_{r}, v_{s}⟩. By Proposition 4.5 and the action of Gal(K_{1}/k) we have

\begin{aligned}⟨u′_{r}, v_{s}⟩ & = & ∑_{i=0}^{4} Ind_{ζ}(u′_{r}, v_{s})_{𝔏_{i}}^{i} \\ & = & ∑_{i=0}^{4} Ind_{ζ}(u′_{r−i}, v_{s−i})_{𝔏_{0}}^{i} \end{aligned}

We recall that 𝔏_{0} is the prime of K_{1} above 5 such that α ≡ ϕ (mod 𝔏_{0}^{2}). But α is a 5th root of

η(1) = \frac{1 + ϕ^{5}}{1 − ϕ^{5}} = −ϕ^{15}(1 + \frac{10(\overline{ϕ^{5}} − ϕ^{5})}{1 + 10ϕ^{5}}).  (18)

The binomial theorem gives α ≡ −ϕ^{3} (mod 𝔏_{0}^{6}) and for r ≠ 0 it follows

\begin{aligned}u_{r} & ≡ & (ζ^{r}ϕ^{4} − 1)/(ζ^{r}ϕ^{3} + ϕ) & (mod 𝔏_{0}^{6}) \\ v_{s} & ≡ & (ζ^{s}ϕ^{3} − 1)/(ζ^{s}ϕ^{3} + 1) & (mod 𝔏_{0}^{6}). \end{aligned}

Using these approximations we are reduced to computing the Hilbert norm residue symbol at the prime 𝔩 = (1−ζ) of k = Q(μ_{5}). This is straightforward, if tedious, to do by hand. See [CF, Exercises 1 and 2]. We find

\begin{matrix}( , )_{𝔏_{0}} & v_{0} & v_{1} & v_{2} & v_{3} & v_{4} & 5⟨ , ⟩ & v_{0} & v_{1} & v_{2} & v_{3} & v_{4} \\ u′_{0} & 1 & ζ^{2} & ζ^{4} & ζ & ζ^{3} & u′_{0} & 0 & 4 & 1 & 1 & 4 \\ u′_{1} & ζ^{3} & ζ^{2} & ζ^{4} & 1 & ζ & u′_{1} & 4 & 0 & 4 & 1 & 1 \\ u′_{2} & ζ & 1 & ζ^{4} & ζ^{2} & ζ^{3} & u′_{2} & 1 & 4 & 0 & 4 & 1 \\ u′_{3} & ζ^{4} & ζ^{2} & ζ^{3} & ζ & 1 & u′_{3} & 1 & 1 & 4 & 0 & 4 \\ u′_{4} & ζ^{2} & ζ^{4} & 1 & ζ & ζ^{3} & u′_{4} & 4 & 1 & 1 & 4 & 0 \end{matrix}

The matrix on the right has rank 3, and so the pairing (17) has rank 6. The kernel is generated by u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4} and v_{1}v_{2}^{2}v_{3}^{3}v_{4}^{4}. By Corollary 1.6 and Lemma 3.6 the first of these elements is accounted for by torsion, whereas the second is not.

By Propositions 4.1 and 6.7 we deduce

S(A_{1} →^{×5} A_{1}/K_{1}) ≃ (Z/5Z)^{4}.

This furnishes the estimate rank A(K_{1}) ≤ 2. Since rank A(k) = 0 the action of Gal(K_{1}/k) on A(K_{1}) ⊗ Q forces rank A(K_{1}) ≡ 0 (mod 4). Thus rank A(K_{1}) = 0 and � (A_{1}/K_{1})[5] ≃ (Z/5Z)^{2}. However, to identify the 5primary part of the Tate-Shafarevich group we must work harder.

We aim to compute S(A_{0} →^{×5} A_{0}/K_{1}) as a G-module. To identify it as an abelian group we make use of the exact sequences

\begin{matrix}0 ⟶ S(A_{0} → A_{1}/K_{1}) ⟶ S(A_{0} →^{×5} A_{0}/K_{1}) ⟶^{ψ_{1}} S(A_{1} → A_{0}/K_{1}) \\ 0 ⟶ S(A_{0} → A_{2}/K_{1}) ⟶ S(A_{0} →^{×5} A_{0}/K_{1}) ⟶^{ψ_{2}} S(A_{2} → A_{0}/K_{1}) \end{matrix} (19)

The images of the maps ψ_{1} and ψ_{2} are the kernels of the Cassels-Tate pairings

\begin{matrix}Ψ_{1} : S(A_{1} → A_{0}/K_{1}) × S(A_{1} → A_{0}/K_{1}) → Z/5Z \\ Ψ_{2} : S(A_{2} → A_{0}/K_{1}) × S(A_{2} → A_{0}/K_{1}) → Z/5Z \end{matrix}  (20)

Lemma 6.8 The pairings Ψ_{1} and Ψ_{2} have ranks 0 and 4 respectively.

Proof. The alternating pairing Ψ_{1} is defined on S(A_{1} → A_{0}/K_{1}) ≃ (Z/5Z)^{2}. By Corollary 1.6 this Selmer group contains a contribution from torsion. Hence Ψ_{1} is trivial. The exact sequences (19), together with Propositions 4.1 and 6.7, now tell us that S(A_{0} →^{×5} A_{0}/K_{1}) ≃ (Z/5Z)^{10} and that the pairing Ψ_{2} has rank 4. □

The exact sequences (19) provide inclusions

S(A_{0} → A_{1}/K_{1}) ⊂ S(A_{0} →^{×5} A_{0}/K_{1}) ⊂ S(A_{2} → A_{0}/K_{1}) ⊂ K_{1}^{∗}/K_{1}^{∗5} (21)

By Remark 6.5 there exist w_{1}, w_{2} in K_{1} with

\begin{aligned}w_{1}^{5} & = & π_{1}π_{2}^{−2}π_{3}^{2}π_{4}^{−1}(u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4})^{2} \\ w_{2}^{5} & = & π_{1}π_{2}^{2}π_{3}^{−2}π_{4}^{−1}(v_{1}v_{2}^{2}v_{3}^{3}v_{4}^{4})^{2}. \end{aligned}

A rather tedious calculation suggests we write x_{1} = w_{1}u_{2}^{2}u_{3}u_{4}^{3}(v_{1}v_{2}^{2}v_{3}^{3}v_{4}^{4})^{3} and x_{2} = w_{2}v_{1}^{3}v_{2}^{3}(u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4})^{3}, whereupon, multiplying x_{2} by a 5th root of unity if necessary

σ(x_{1}) ≡ x_{1}^{3}  (mod K_{1}^{∗5}) σ(x_{2}) ≡ x_{2}^{2}  (mod K_{1}^{∗5})
τ(x_{1}) ≡ x_{1}u_{2}^{3}u_{3}u_{4}^{3} (mod K_{1}^{∗5}) τ(x_{2}) ≡ x_{2}ϕ^{2}v_{2}^{3}v_{3}v_{4}^{3} (mod K_{1}^{∗5}) (22)

Proposition 6.9 Multiplying x_{1} by a 5th root of unity if necessary, the Selmer groups (21) are

\begin{aligned}S(A_{0} → A_{1}/K_{1}) & ≃ & ⟨u′_{1}, u′_{2}, u′_{3}, u′_{4}, v_{1}, v_{2}, v_{3}, v_{4}⟩ \\ S(A_{0} →^{×5} A_{0}/K_{1}) & ≃ & ⟨u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, v_{4}, 11, x_{1}⟩ \\ S(A_{2} → A_{0}/K_{1}) & ≃ & ⟨ζ, ϕ, u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, v_{4}, 11, α, x_{1}, x_{2}⟩. \end{aligned}

Proof. The descriptions of S(A_{0} → A_{1}/K_{1}) and S(A_{2} → A_{0}/K_{1}) follow from Proposition 2.1. Now S(A_{0} →^{×5} A_{0}/K_{1}) is the kernel of the pairing Ψ_{2} and this pairing induces a pairing on the quotient

\frac{S(A_{2} → A_{0}/K_{1})}{S(A_{0} → A_{1}/K_{1})} ≃ ⟨ζ, ϕ, 11, α, x_{1}, x_{2}⟩.  (23)

Decomposing into σ-eigenspaces, we learn that S(A_{0} →^{×5} A_{0}/K_{1}) has basis

u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, v_{4}, 11, z

for some z = ζ^{i}x_{1}^{j}. Since we consider ourselves free to multiply x_{1} by a 5th root of unity, it only remains to show ζ ∉ S(A_{0} →^{×5} A_{0}/K_{1}). To do this we identify (23) as a G-module. By (22)

\begin{aligned}τ(x_{1}) & ≡ & x_{1}11^{2} & mod ⟨u′_{1}, u′_{2}, u′_{3}, u′_{4}, K_{1}^{∗5}⟩ \\ τ(x_{2}) & ≡ & x_{2}ϕ^{2} & mod ⟨v_{1}, v_{2}, v_{3}, v_{4}, K_{1}^{∗5}⟩ \end{aligned}

whereas τ(α) = ζα and ζ, ϕ, 11 ∈ k are fixed by τ. Thus

\frac{S(A_{2} → A_{0}/K_{1})}{S(A_{0} → A_{1}/K_{1})} ≃ M(ω, 2) ⊕ M(ω^{2}, 2) ⊕ M(ω^{3}, 2).

By Lemma 5.4 the pairing Ψ_{2} restricted to

M(ω^{2}, 2) ⊕ M(ω^{3}, 2) ≃ ⟨ζ, ϕ, α, x_{2}⟩

is either zero or non-degenerate. But it cannot be zero by our earlier consideration of σ-eigenspaces. Thus ζ ∉ S(A_{0} →^{×5} A_{0}/K_{1}) and we are done. □

Remark 6.10 An alternative proof of Proposition 6.9 is given by computing Ψ_{2} via the formula of [F0, §7.3].

S(A_{0} →^{×5} A_{0}/K_{1}) ≃ M(1) ⊕ M(ω, 5) ⊕ M(ω^{2}, 4)

where the summands correspond to 11, ⟨u_{1}, u_{2}, u_{3}, u_{4}, x_{1}⟩ and ⟨v_{1}, v_{2}, v_{3}, v_{4}⟩. The contributions from torsion are 11 and u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4}. Thus

\frac{S(A_{0} →^{×5} A_{0}/K_{1})}{im(δ|A_{0}[5])} ≃ M(ω, 4) ⊕ M(ω^{2}, 4).  (24)

Our earlier calculation of the pairing (17) shows that the Cassels-Tate pairing on (24) is non-zero. By Lemma 5.4 the Cassels-Tate pairing on (24) is nondegenerate. It follows that � (A_{0}/K_{1})(5) ≃ (Z/5Z)^{8}.

Earlier we saw � (A_{1}/K_{1})[5] ≃ (Z/5Z)^{2}. We claim that � (A_{1}/K_{1}) contains no element of order 25. Indeed, from the exact sequences

A_{1}(K_{1}) ⟶^{ψ} A_{0}(K_{1}) ⟶ S(A_{1} → A_{0}/K_{1}) ⟶ � (A_{1}/K_{1})[ψ] ⟶ 0

0 ⟶ � (A_{1}/K_{1})[ψ] ⟶ � (A_{1}/K_{1})(5) ⟶^{ψ} � (A_{0}/K_{1})(5)

we learn that � (A_{1}/K_{1})(5) is finite and contains no copy of (Z/25Z)^{2}. Our claim is now a well known consequence of the Cassels-Tate pairing.

Finally the exact sequence

S(A_{2} → A_{0}/K_{1}) ⟶ S(A_{2} →^{×5} A_{2}/K_{1}) ⟶ S(A_{0} → A_{2}/K_{1}) = 0

shows that S(A_{2} →^{×5} A_{2}/K_{1}) ≃ (Z/5Z)^{13}. By Lemma 6.8 the Cassels-Tate pairing on this Selmer group has rank 4. Since the multiplication by 5 map on A_{2} factors through A_{0} and � (A_{0}/K_{1})(5) is 5-torsion, we deduce

� (A_{2}/K_{1})(5) ≃ (Z/5Z)^{4} ⊕ (Z/25Z)^{8}.

This completes the proof of Theorem 1.

7 Descent calculations over K_{2} = Q(A_{2}[5])

We recall that K_{2}/k has Kummer generator 11. We put β = \sqrt[5]{11}. Then Gal(K_{2}/Q) ≃ G via

\begin{matrix}σ(ζ) = ζ^{2} & σ(β) = β \\ τ(ζ) = ζ & τ(β) = ζβ. \end{matrix}

The cyclotomic character ω and the character ψ of §5 are equal. The primes above 5 and 11 ramify in K_{2}/k. We write

(5) = 𝔏^{20} (11) = 𝔓_{1}^{5}𝔓_{2}^{5}𝔓_{3}^{5}𝔓_{4}^{5}

with 𝔓_{i}|𝔭_{i}. In §3 we saw that H_{2} := J_{1}K_{1}K_{2} is the 5-ray class field of k for conductor 𝔩^{2}𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4}.

Lemma 7.1 The extension H_{2}/K_{2} is unramified at all primes.

(i) The primes above 5 are inert in J_{1}K_{2}/K_{2} and split in K_{1}K_{2}/K_{2}. (ii) The primes above 11 are split in J_{1}K_{2}/K_{2} and inert in K_{1}K_{2}/K_{2}.

Proof. Only primes above 11 ramify in J_{1}/k and K_{1}/k. By considering ratios of Kummer generators we see that the extensions J_{1}K_{2}/K_{2} and K_{1}K_{2}/K_{2} are unramified. Hence the composite H_{2}/K_{2} is unramified.

(i) The extensions J_{1}K_{2}/K_{2} and K_{1}K_{2}/K_{2} are translates of Q(μ_{11})^{+}/Q and K_{1}/k. Since 5 is totally ramified in K_{2}/Q our claims follow. (ii) The Kummer generator for J_{1}K_{2}/K_{2} belongs to S(A_{0} → A_{2}/K_{2}). By Proposition 2.1 the primes above 11 split in J_{1}K_{2}/K_{2}. By Lemma 6.1 the primes above 11 cannot split completely in H_{2}/K_{2}. They are therefore inert in K_{1}K_{2}/K_{2}. □

Proposition 7.2 The 5-class group of K_{2} is ℭ𝔩_{K_{2}}(5) ≃ (Z/5Z)^{2}. It is generated by the primes above 5 and 11.

Proof. Let B = ℭ𝔩_{K_{2}}(5). By Lemma 3.5(ii) we know that H_{2} is the maximal unramified 5-extension of K_{2} which is abelian over k. Thus

B/(τ − 1)B ≃ Gal(H_{2}/K_{2}) ≃ (Z/5Z)^{2}

as abelian groups. By Lemma 7.1, the primes 𝔓_{1} and 𝔏 generate B/(τ −1)B. Since (τ − 1)^{5} ⊂ 5Z_{5}[τ] it follows that 𝔓_{1} and 𝔏 generate B as a Z_{5}[τ]- module. But τ(𝔓_{1}) = 𝔓_{1}, τ(𝔏) = 𝔏 and 𝔓_{1}^{5} = 𝔭_{1}, 𝔏^{5} = 𝔩 are principal. Thus B ≃ (Z/5Z)^{2} as claimed. □

Remark 7.3 The action of τ on ℭ𝔩_{K_{2}}(5) is trivial. By Lemma 1.3 we have ℭ𝔩_{K_{2}}(5) ≃ Gal(H_{2}/K_{2}) ≃ M(1) ⊕ M(ω^{3}). In particular 𝔓_{1} ∼ 𝔓_{2}^{2} ∼ 𝔓_{3}^{3} ∼ 𝔓_{4}^{4}.

We now give a description of the units in K_{2}. Substituting β^{5} for λ in the identity (λ − ϕ^{5})(λ − \overline{ϕ^{5}}) = λ^{2} − 11λ − 1 we learn Norm_{K_{2}|Q}(β − ϕ) = 1. Thus β − ϕ is a unit. We choose to work with the units

u_{i} = \frac{ϕζ^{i}β + 1}{ζ^{i}β − ϕ} v_{i} = (ζ^{i}β)^{2} − ζ^{i}β − 1.

It is easy to check

\begin{matrix}σ(u_{i}) = −1/u_{2i} & σ(v_{i}) = v_{2i} \\ τ(u_{i}) = u_{i+1} & τ(v_{i}) = v_{i+1} \end{matrix}

and we have relations ∏ u_{i} = −ϕ^{15}, ∏ v_{i} = −1. Thus the subgroups of K_{2}^{∗}/K_{2}^{∗5} generated by u_{1}, u_{2}, u_{3}, u_{4} and v_{1}, v_{2}, v_{3}, v_{4} are quotients of the Gmodules M(ω^{2}, 4) and M(1, 4).

Lemma 7.4 The extension K_{1}K_{2}/K_{2} has Kummer generator ϕv_{1}v_{2}^{4}v_{3}^{4}v_{4}.

Proof. By Lemma 7.1 and Proposition 7.2 it is sufficient to show that the stated element is (i) a 5th power locally at the prime 𝔏 above 5, and (ii) not a 5th power locally at the primes 𝔓_{i} above 11. A suitable version of Hensel’s lemma shows that a ∈ 𝔒_{K_{2}} is a 5th power locally at 𝔏 if and only if x^{5} ≡ a (mod 𝔏^{26}) is soluble. We prove (i) by using pari to perform calculations in the group (𝔒_{K_{2}}/𝔏^{26})^{∗}. Since ϕv_{1}v_{2}^{4}v_{3}^{4}v_{4} ≡ ϕ (mod 𝔓_{i}), claim (ii) is clear without computer calculation. This completes the proof of the lemma. Of course more brutal computer calculations are possible, showing

ϕ^{2}π_{1}π_{2}^{4}π_{3}^{4}π_{4} ≡ (ϕv_{1}v_{2}^{4}v_{3}^{4}v_{4})^{−1} (mod K_{2}^{∗5}).
□

A consequence of Lemma 7.4 is that K_{2}(\sqrt[5]{v_{1}v_{2}^{2}v_{3}^{3}v_{4}^{4}})/K_{2} is unramified at all primes. Furthermore this extension is split at the primes above 5 and 11. By Proposition 7.2 we deduce

v_{1}v_{2}^{2}v_{3}^{3}v_{4}^{4} = w^{5}

for some w ∈ K_{2}. Multiplying w by a 5th root of unity if necessary

\begin{aligned}σ(w) & ≡ & w^{3} & mod ⟨v_{1}, v_{2}, v_{3}, v_{4}⟩ \\ τ(w) & ≡ & wv_{0} & (mod K_{2}^{∗5}). \end{aligned}

Proposition 7.5 The units ζ, ϕ, u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, w generate a subgroup of 𝔒_{K_{2}}^{∗} of index prime to 5.

Proof. By Dirichlet it suffices to check that the elements listed are independent in K_{2}^{∗}/K_{2}^{∗5}. There is a G-module homomorphism

M := M(ω) ⊕ M(ω^{2}) ⊕ M(ω^{2}, 4) ⊕ M(ω^{3}, 4) → K_{2}^{∗}/K_{2}^{∗5}

where the summands correspond to ζ, ϕ, ⟨u_{1}, u_{2}, u_{3}, u_{4}⟩ and ⟨v_{1}, v_{2}, v_{3}, w⟩. We suppose for a contradiction that this map has non-trivial kernel. Then this kernel meets

\begin{aligned}ker(τ − 1|M) & = & M(ω) ⊕ M(ω^{2}) ⊕ M(ω) ⊕ M(ω^{2}) \\ & = & ⟨ ζ, ϕ, u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4}, v_{1}v_{2}^{4}v_{3}^{4}v_{4} ⟩. \end{aligned}

Dividing into σ-eigenspaces we learn that one of the elements

ζ^{i}(u_{1}u_{2}^{2}u_{3}^{3}u_{4}^{4})^{j}, ϕ^{i}(v_{1}v_{2}^{4}v_{3}^{4}v_{4})^{j}

is a 5th power. By Lemmas 3.6 and 7.4 these elements are Kummer generators for J_{1}J_{2}K_{2}/K_{2} and for K_{1}K_{2}(\sqrt[5]{ϕ})/K_{2}. This gives the required contradiction. □

Remark 7.6 According to pari the field K_{2} has class number 25, whereas ϕ, β −ϕ and its conjugates, generate a subgroup of index 5 in 𝔒_{K_{2}}^{∗}/(torsion). Again this is conditional on the Generalised Riemann Hypothesis.

We apply Propositions 2.1 and 2.2 in the case F = K_{2}.

Proposition 7.7 The Selmer groups attached to the 5-isogenies (1) are

S(A_{0} → A_{1}/K_{2}) ≃ (Z/5Z)^{8}  S(A_{2} → A_{0}/K_{2}) ≃ (Z/5Z)^{15}
S(A_{1} → A_{0}/K_{2}) ≃ (Z/5Z)^{2}  S(A_{0} → A_{2}/K_{2}) ≃ Z/5Z.

Proof. By Propositions 7.2 and 7.5 the space

{ θ ∈ K_{2}^{∗}/K_{2}^{∗5} | ord_{𝔭}(θ) ≡ 0 (mod 5) for all 𝔭 }  (25)

has basis ζ, ϕ, u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, w, π_{1}, 1 − ζ. Here π_{1} and 1 − ζ are contributions from the class group or “virtual units”. We choose characters (𝔒_{K_{2}}/𝔓_{i})^{∗} → Z/5Z and compute these characters on a basis for (25)

𝔓_{1} 𝔓_{2} 𝔓_{3} 𝔓_{4}
ζ 1 3 2 4
ϕ 2 3 3 2
u_{i} 3 2 2 3
v_{i} 0 0 0 0
w 2 4 1 3
π_{1}π_{2}^{2}π_{3}^{3}π_{4}^{4} 3 1 4 2
2ϕ − 1 2 2 2 2

By Proposition 2.1 we deduce

S(A_{0} → A_{1}/K_{2}) ≃ ⟨u′_{1}, u′_{2}, u′_{3}, u′_{4}, v_{1}, v_{2}, v_{3}, w′⟩ ⊂ K_{2}^{∗}/K_{2}^{∗5}

where u′_{i} = u_{i}ϕ and w′ = wπ_{1}π_{2}^{2}π_{3}^{3}π_{4}^{4}. Propositions 2.1 and 7.2 give S(A_{0} → A_{2}/K_{2}) ≃ Z/5Z. The remaining statements follow by Proposition 2.2. □

Proposition 7.7 furnishes the estimate rank A(K_{2}) ≤ 8. We improve on this by computing the Cassels-Tate pairing on S(A_{0} → A_{1}/K_{2}). As a Gmodule we have

S(A_{0} → A_{1}/K_{2}) ≃ M(ω^{2}, 4) ⊕ M(ω^{3}, 4).  (26)

By [Ca1, §7 (5)] there is a commutative diagram

⟨·, ·⟩_{K_{2}} : S(A_{0} → A_{1}/K_{2}) × S(A_{0} → A_{1}/K_{2}) → Q/Z
↑ cores ↓ res ||
⟨·, ·⟩_{K_{1}K_{2}} : S(A_{0} → A_{1}/K_{1}K_{2}) × S(A_{0} → A_{1}/K_{1}K_{2}) → Q/Z

where res and cores are the maps K_{2}^{∗}/K_{2}^{∗5} ⇄ (K_{1}K_{2})^{∗}/(K_{1}K_{2})^{∗5} induced by the natural inclusion and norm map respectively. We compute ⟨c, d⟩_{K_{2}} where

c = β^{4} + 3β^{3} + 4β^{2} + 2β + 1, d = β^{2} − β − 1.

A computer search yields a unit γ ∈ K_{1}K_{2} with Norm_{K_{1}K_{2}|K_{2}}(γ) = c^{2}. An explicit expression for 55γ in terms of u = ε(α) and β is

(271858β^{4} − 724855β^{3} + 1158870β^{2} − 663207β − 928521)u^{4} +(942679β^{4} − 1424893β^{3} + 1970490β^{2} − 2288047β − 1777424)u^{3} +(2185832β^{4} − 3700207β^{3} + 8323117β^{2} − 6646061β − 10380876)u^{2} +(−25520β^{4} + 52792β^{3} − 893918β^{2} + 423925β + 1617231)u +(1351406β^{4} − 2180822β^{3} + 4850841β^{2} − 4066025β − 6097410)

By Lemma 7.1, the primes above 11 are inert in K_{1}K_{2}/K_{2}. Any non-trivial character F_{11^{5}}^{∗} → Z/5Z factors via the norm map F_{11^{5}}^{∗} → F_{11}^{∗}. So by Proposition 2.1, γ belongs to S(A_{0} → A_{1}/K_{1}K_{2}). The prime 5 factors as

(5) = 𝔏_{0}^{4}𝔏_{1}^{4}𝔏_{2}^{4}𝔏_{3}^{4}𝔏_{4}^{4}  in K_{1}
(5) = 𝔏^{20}  in K_{2}
(5) = L_{0}^{20}L_{1}^{20}L_{2}^{20}L_{3}^{20}L_{4}^{20} in K_{1}K_{2}

with L_{i}|𝔏_{i}. Proposition 4.5 and the above diagram give

⟨c^{2}, d⟩_{K_{2}} = ⟨γ, d⟩_{K_{1}K_{2}}
= ∑_{i=0}^{4} Ind_{ζ}(γ, d)_{L_{i}}^{i}
= ∑_{i=0}^{4} Ind_{ζ}(γ_{i}, d)_{𝔏}^{i}

where γ_{i} ∈ K_{2} is chosen L_{i}-adically close to γ. To do this we use (18) and the binomial theorem to choose α_{i} ∈ k 𝔏_{i}-adically close to α. Then we substitute ε(α_{i}) for u = ε(α) in our expression for γ.

We compute the Hilbert norm residue symbol (·, ·)_{𝔏} using the product formula and Euler’s criterion. Finally a computer calculation shows

⟨c, d⟩_{K_{2}} ≠ 0.

This calculation, together with Lemma 5.4, shows that the Cassels-Tate pairing on (26) is non-degenerate. By Proposition 4.1 it follows that S(A_{1} →^{×5} A_{1}/K_{2}) ≃ Z/5Z. Thus rank A(K_{2}) = 0 and � (A_{1}/K_{2})(5) = 0.

From the exact sequences

A_{0}(K_{2}) \underrightarrow{ψ̂} A_{1}(K_{2}) ⟶ S(A_{0} → A_{1}/K_{2}) ⟶ � (A_{0}/K_{2})[ψ̂] ⟶ 0

0 ⟶ � (A_{0}/K_{2})[ψ̂] ⟶ � (A_{0}/K_{2})(5) \underrightarrow{ψ̂} � (A_{1}/K_{2})(5)

we learn � (A_{0}/K_{2})(5) ≃ (Z/5Z)^{8}. Proposition 4.1 and the exact sequences

\begin{matrix}0 ⟶ S(A_{0} → A_{2}/K_{2}) ⟶ S(A_{0} →^{×5} A_{0}/K_{2}) ⟶ S(A_{2} → A_{0}/K_{2}) \\ 0 ⟶ S(A_{2} → A_{0}/K_{2}) ⟶ S(A_{2} →^{×5} A_{2}/K_{2}) ⟶ S(A_{0} → A_{2}/K_{2}) \end{matrix}

show that S(A_{2} →^{×5} A_{2}/K_{2}) ≃ (Z/5Z)^{16} and that the Cassels-Tate pairing on this Selmer group has rank 6. Since the multiplication by 25 map on A_{2} factors through A_{1} and � (A_{1}/K_{2})(5) = 0, we deduce

� (A_{2}/K_{2})(5) ≃ (Z/5Z)^{6} ⊕ (Z/25Z)^{8}.

This completes the proof of Theorem 2.

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