EXPLICIT 5-DESCENT ON ELLIPTIC CURVES

TOM FISHER

Abstract. We compute equations for genus one curves representing non-trivial elements of order 5 in the Tate-Shafarevich group of an elliptic curve. We explain how to write the equations in terms of Pfaffians and give examples for elliptic curves over the rationals both with and without a rational 5-isogeny.

1. Introduction

An explicit descent calculation on an elliptic curve E over a number field K computes the Selmer group (attached to some isogeny) and represents its elements by giving equations for the corresponding covering curves. These may be used to help search for generators of the Mordell-Weil group E(K) or to exhibit non-trivial elements of the Tate-Shafarevich group � (E/K).

Let C be a smooth curve of genus one representing an element of order n in � (E/K). Cassels [7] showed that C admits a K-rational divisor D of degree n. So for n ≥ 3 we may embed C ⊂ ℙ^{n−1} by the complete linear system |D|. The result is called a genus one normal curve of degree n. For n ≥ 4 it is well known (see for example [15], [24]) that the homogeneous ideal of such a curve is generated by a vector space of quadrics of dimension n(n − 3)/2.

The equations for a genus one normal curve of degree 5 may conveniently be written as the 4 × 4 Pfaffians of a 5 × 5 alternating matrix of linear forms. Over the complex numbers this is a classical fact. In general it is a consequence of the Buchsbaum-Eisenbud structure theorem [5], [6] for Gorenstein ideals of codimension 3. In Section 4 we explain how to compute these matrices of linear forms.

The author has been compiling [18] a list of explicit elements of � (E/ℚ)[5] for elliptic curves E/ℚ of small conductor (taken from the Cremona database [9]). The equations are computed using either descent by 5-isogeny, full 5-descent, or visibility. We give details of the first two of these methods in Sections 5 and 6, expanding on the treatments in [13] and [10]. Our use of visibility is described in [16].

Date: 6th February 2012.

2. Background on descent

Let ϕ : E → E′ be an isogeny of elliptic curves over K. A ϕ-covering of E′ is a pair (C, π) where C is a smooth curve of genus one and π : C → E′ is a morphism (both defined over K) such that

\xymatrix{C \ar[d]_{ψ} \ar[dr]^{π} \\ E \ar[r]_{ϕ} & E′}

commutes for some isomorphism ψ : C → E defined over \overline{K}.

We write H^{i}(K, −) as a shorthand for H^{i}(Gal(\overline{K}/K), −). Taking Galois cohomology of the short exact sequence of Gal(\overline{K}/K)-modules

0 ⟶ E[ϕ] ⟶ E ⟶^{ϕ} E′ ⟶ 0

gives a long exact sequence of abelian groups

(1) … → E(K) →^{ϕ} E′(K) →^{δ} H^{1}(K, E[ϕ]) → H^{1}(K, E) → …

The group H^{1}(K, E[ϕ]) parametrises the ϕ-coverings of E′, up to isomorphism over K. The subgroup of everywhere locally soluble coverings is the ϕ-Selmer group S^{(ϕ)}(E/K). Likewise the group H^{1}(K, E) parametrises the torsors (or principal homogeneous spaces) under E, up to isomorphism over K. The subgroup of everywhere locally soluble torsors is the Tate-Shafarevich group � (E/K). There is then an exact sequence

0 → E′(K)/ϕE(K) →^{δ} S^{(ϕ)}(E/K) → � (E/K)[ϕ_{∗}] → 0

where ϕ_{∗} : � (E/K) → � (E′/K) is the map induced by ϕ.

There are two natural ways to construct a rational divisor class on C. Let m be the least positive integer such that E[ϕ] ⊂ E[m]. Then D = ψ^{∗}(m.0_{E}) and D′ = π^{∗}(0_{E′}) are divisors on C of degrees m and n = deg ϕ. A calculation shows that D is linearly equivalent to all its Galois conjugates, whereas D′ is already defined over K. For each σ ∈ Gal(\overline{K}/K) we pick h_{σ} ∈ \overline{K}(C)^{×} with div(h_{σ}) = σD − D. There is then an obstruction map (see [22], [25], [10, Paper I])

Ob : H^{1}(K, E[ϕ]) → Br(K) = H^{2}(K, \overline{K}^{×})

that sends (C, π) to the class of the 2-cocycle (σ, τ) ↦ σ(h_{τ})h_{σ}/h_{στ}. Since H^{1}(K, \overline{K}(C)^{×}) = 0 it follows that D is linearly equivalent to a K-rational divisor if and only if (C, π) has trivial obstruction.

If #(E[ϕ] ∩ E[2]) = 1 or 4 then the elements of E[ϕ] sum to 0_{E}, in which case ϕ^{∗}(0_{E′}) ∼ n.0_{E} and D′ ∼ (n/m)D.

In this paper we are interested in the following two cases where we may write C as a genus one normal curve of degree 5 with hyperplane section D.

(i) ϕ is an isogeny of degree 5 and (C, π) ∈ H^{1}(K, E[ϕ]).

(ii) ϕ is multiplication-by-5 on E and (C, π) ∈ S^{(5)}(E/K). The obstruction is trivial in both cases. In the first case this is because D ∼ D′, whereas in the second case the proof (which we follow in our calculations) uses the local-to-global principle for the Brauer group.

3. Pfaffians

We recall some basic facts about Pfaffians. Let A = (a_{ij}) be an n×n alternating matrix. If n = 2m is even then the Pfaffian of A is

(2)  pf(A) = \frac{1}{2^{m}m!} ∑_{σ∈S_{n}} sign(σ) ∏_{i=1}^{m} a_{σ(2i−1)σ(2i)}.

Standard calculations (see [2, Chap. IX, §5]) show that pf(P AP^{T}) = det(P) pf(A) and det(A) = pf(A)^{2}. Since det(A) in an integer coefficient polynomial in the entries of A, the same must be true of pf(A). This is used to define the Pfaffian over an arbitrary ring.

Pfaffians, just like determinants, may be expanded along a row. We write A^{{i,j}} for the matrix obtained from A by deleting the ith and jth rows and columns. It may be shown using (2) that

(3)  pf(A) = ∑_{j=2}^{n}(−1)^{j}a_{1j} pf(A^{{1,j}}).

For example in the 4 × 4 case we have

pf (\begin{matrix}0 & a_{12} & a_{13} & a_{14} \\ & 0 & a_{23} & a_{24} \\ & & 0 & a_{34} \\ & & & 0 \end{matrix}) = a_{12}a_{34} − a_{13}a_{24} + a_{14}a_{23}.

Definition 3.1. Let A be an n×n alternating matrix with n odd. The row vector of submaximal Pfaffians of A is Pf(A) = (p_{1}, …, p_{n}) where p_{i} = (−1)^{i} pf(A^{{i}}) and A^{{i}} is the matrix obtained by deleting the ith row and column of A.

Lemma 3.2. If A is an n × n alternating matrix with n odd then (i) Pf(A)A = 0, (ii) Pf(P AP^{T}) = Pf(A) adj(P), (iii) adj(A) = Pf(A)^{T} Pf(A).

Proof: Since we only need the case n = 5 (which may be checked by a generic computation) we omit the proof. □

4. Computing genus one models

A genus one model (of degree 5) is a 5 × 5 alternating matrix of linear forms in variables x_{1}, …, x_{5}. We write X_{5}(K) for the space of all genus one models with coefficients in a field K, and C_{Φ} ⊂ ℙ^{4} for the subscheme defined by the 4 × 4 Pfaffians of Φ ∈ X_{5}(K).

Theorem 4.1. Let C ⊂ ℙ^{4} be a genus one normal curve of degree 5 defined over a field K.

(i) There exists Φ ∈ X_{5}(K) such that C = C_{Φ}.

(ii) If Φ_{1}, Φ_{2} ∈ X_{5}(K) with C = C_{Φ_{1}} = C_{Φ_{2}} then there exist A ∈ GL_{5}(K) and μ ∈ K^{×} such that Φ_{2} = μAΦ_{1}A^{T}.

Theorem 4.1 is a consequence of the Buchsbaum-Eisenbud structure theorem [5], [6] for Gorenstein ideals of codimension 3. We give a simplified form of the proof and use it to give explicit algorithms for computing Φ and A. These algorithms are needed in our work [15],

[16] on the invariant theory of genus one models.

Example 4.2. Let E be the elliptic curve y^{2} = x^{3} + ax + b. For any n ≥ 3 we may embed E ⊂ ℙ^{n−1} via the complete linear system |n.0_{E}| to give a genus one normal curve of degree n. If n = 5 then the embedding is given by

(x_{1} : …: x_{5}) = (1 : x : y : x^{2} : xy)

and the image is defined by the 4 × 4 Pfaffians of

(\begin{matrix}0 & bx_{1} & x_{5} & x_{4} + ax_{1} & −x_{3} \\ & 0 & −x_{4} & −x_{3} & x_{2} \\ & & 0 & −x_{2} & 0 \\ & − & & 0 & −x_{1} \\ & & & & 0 \end{matrix}) .

(Since the homogeneous ideal is generated by a 5-dimensional space of quadrics, it suffices to check that the 4 × 4 Pfaffians are linearly independent and that they vanish on E.)

Let R = K[x_{1}, …, x_{n}] = ⊕_{d≥0}R_{d} be the polynomial ring with its usual grading by degree. Let R_{+} = ⊕_{d≥1}R_{d} be the irrelevant ideal.

Definition 4.3. Let M be a finitely generated graded R-module. A graded free resolution of M is a complex of graded free R-modules

F_{•} : 0 ⟶ F_{s} ⟶^{φ_{s}} F_{s−1} ⟶ … ⟶ F_{2} ⟶^{φ_{2}} F_{1} ⟶^{φ_{1}} F_{0} ⟶ 0

that is exact at all terms except F_{0} where we have F_{0}/ im(φ_{1}) ≅ M. The resolution F_{•} is minimal if ϕ_{i}(F_{i}) ⊂ R_{+}F_{i−1} for all i.

We shall need the following two facts.

Lemma 4.4. Let F_{•} be a minimal graded free resolution of M. Then any graded free resolution of M is a direct sum of F_{•} and a trivial complex. In particular F_{•} is unique up to isomorphism.

Proof: See [12, Section 20.1]. □

Lemma 4.5. (Buchsbaum-Eisenbud acyclicity criterion) The complex F_{•} is acyclic (i.e. exact at all terms except F_{0}) if and only if for all 1 ≤ i ≤ s,

rank F_{i} = rank φ_{i} + rank φ_{i+1}

and the ideal generated by the r_{i} × r_{i} minors of φ_{i} (where r_{i} = rank φ_{i}) has codimension at least i.

Proof: See [4, Theorem 1.4.13] or [12, Theorem 20.9]. We use here that R is Cohen-Macaulay, so that the codimension (also called height) of an ideal is the same as the grade (also called depth). □

We follow the convention that maps of graded R-modules preserve the degree. Let R(d) be R as a graded module over itself with degrees shifted by d, i.e. R(d)_{e} = R_{d+e}. We use the same notation for maps of R-modules and the matrices that represent them (with respect to the standard bases).

Theorem 4.6. Let C ⊂ ℙ^{4} be a genus one normal curve of degree 5 with homogeneous ideal I = I(C) ⊂ R = K[x_{1}, …, x_{5}].

(i) The minimal graded free resolution of R/I takes the form

(4)  0 ⟶ R(−5) ⟶^{Q^{T}} R(−3)^{5} ⟶^{Φ} R(−2)^{5} ⟶^{P} R ⟶ 0

(This means that P = (p_{1}, …, p_{5}) and Q = (q_{1}, …, q_{5}) are vectors of quadrics and Φ is a 5 × 5 matrix of linear forms.)

(ii) The K-vector space

{B ∈ Mat_{5}(K) : ΦB is alternating}

is 1-dimensional and contains a non-singular matrix. (iii) If Φ is alternating then P and Q are scalar multiples of Pf(Φ).

Proof: The conclusions of the theorem are unchanged if we extend our field K. So we may assume K is algebraically closed. Then C is an elliptic curve, and up to translation any two divisors on C of the same degree are linearly equivalent. So we may change co-ordinates on ℙ^{4} so that C = C_{Φ} where Φ is as given in Example 4.2. (If char(K) = 2 or 3 then we use the more general formula in [15, Section 6].) By Lemma 3.2(i) there is a complex

(5)  0 ⟶ R(−5) ⟶^{P^{T}} R(−3)^{5} ⟶^{Φ} R(−2)^{5} ⟶^{P} R ⟶ 0

with P = Pf(Φ). Since P is not identically zero we have rank(Φ) = 4 and rank(P) = 1. By Lemma 3.2(iii) the ideals generated by the 4 × 4 Pfaffians of Φ and the 4 × 4 minors of Φ have the same radical. Since C ⊂ ℙ^{4} has codimension 3, the conditions of Lemma 4.5 are satisfied and so (5) is the minimal graded free resolution of R/I. This proves (i) and shows by Lemma 4.4 that for any resolution (4) there exist A_{1}, A_{2} ∈ GL_{5}(K) such that A_{1}ΦA_{2} is alternating. Replacing Φ by ΦA_{2}A_{1}^{−T} we may assume for the proof of (ii) that Φ is alternating.

Suppose that both Φ and ΦB are alternating for some B ∈ Mat_{5}(K). Then P Φ = P ΦB = 0 and ΦP^{T} = ΦBP^{T} = 0. Since (4) is exact it follows that P^{T} and BP^{T} are scalar multiples of Q^{T}. Therefore B is a scalar matrix. This proves (ii). To prove (iii) we apply the same argument starting with the identity Pf(Φ)Φ = 0. □

Theorem 4.6 not only proves Theorem 4.1(i) but gives the following algorithm for computing a genus one model Φ with C = C_{Φ}. We start with a basis p_{1}, …, p_{5} for the space of quadrics vanishing on C. We then solve by linear algebra for a matrix Ψ whose columns are a basis for the space of all 5-tuples of linear forms (ℓ_{1}, …, ℓ_{5}) ∈ R^{5} satisfying ∑_{i=1}^{5} ℓ_{i}p_{i} = 0. Finally we take Φ = ΨB where B ∈ Mat_{5}(K) is any non-zero matrix satisfying ΨB = −B^{T} Ψ^{T}.

To prove Theorem 4.1(ii) we put P_{1} = Pf(Φ_{1}), P_{2} = Pf(Φ_{2}) and note that by Lemma 4.4 there is an isomorphism of complexes

[figure: 0; R(−5); P_{1}^{T}; μ; R(−3)^{5}; Φ_{1}; A^{−T}; R(−2)^{5}; P_{1}; B^{T}; R; 0; 0; R(−5); P_{2}^{T}; R(−3)^{5}; Φ_{2}; R(−2)^{5}; P_{2}; R; 0]

for some A, B ∈ GL_{5}(K) and μ ∈ K^{×}. Commutativity of this diagram gives P_{1}^{T} = μA^{T} P_{2}^{T} = BP_{2}^{T} and Φ_{2} = B^{T} Φ_{1}A^{T}. Since the entries of P_{2} are linearly independent it follows that B = μA^{T} and so Φ_{2} = μAΦ_{1}A^{T} as required. The proof shows that A ∈ GL_{5}(K) is uniquely determined up to scalars by the condition Pf(Φ_{1}) ∝ Pf(Φ_{2})A. This observation (which also follows by Lemma 3.2(ii)) gives a convenient way to compute A. If K is algebraically closed then we may scale A so that μ = 1. With this convention A is unique up to sign.

5. Descent by isogeny

We return to working over a number field K. Let ϕ : E → E′ be a cyclic isogeny of degree n and ϕ̂: E′ → E its dual isogeny. If (C, π) is a ϕ-covering of E′ then (C, ϕ̂∘ π) is an n-covering of E. In general not all n-coverings of E arise in this way. Instead an upper bound for the rank is obtained by computing both S^{(ϕ)}(E/K) and S^{(ϕ̂)}(E′/K).

Since the Weil pairing E[ϕ] × E′[ϕ̂] → μ_{n} is non-degenerate, the action of Galois on E[ϕ], E′[ϕ̂] and μ_{n} is described by characters

χ^{−1}ω, χ, ω : Gal(\overline{K}/K) → (ℤ/nℤ)^{×}.

Let L = K(E′[ϕ̂]) be the fixed field of the kernel of χ, and let G = Gal(L/K). If n is prime then [L : K] divides n − 1 and so is coprime to n. By the inflation-restriction exact sequence we have

H^{1}(K, E[ϕ]) ≅ H^{1}(L, E[ϕ])^{G}.

Since H^{1}(L, E[ϕ]) ≅ H^{1}(L, μ_{n}) ≅ L^{×}/(L^{×})^{n} it follows (by keeping track of the G-actions) that H^{1}(K, E[ϕ]) ≅ (L^{×}/(L^{×})^{n})^{χ} where for A a G-module A^{χ} = {a ∈ A | σ(a) = a^{χ(σ)} for all σ ∈ G}.

There is an analogue of the exact sequence (1) obtained by replacing K by its completion K_{v}. Let δ_{v} be the connecting map in this exact sequence. The Selmer group attached to ϕ is

S^{(ϕ)}(E/K) = {θ ∈ H^{1}(K, E[ϕ]) | res_{v}(θ) ∈ im δ_{v} for all places v}

where res_{v} : H^{1}(K, E[ϕ]) → H^{1}(K_{v}, E[ϕ]) is the restriction map. Assuming we can compute the groups

L(S, n) = {θ ∈ L^{×}/(L^{×})^{n} | v_{𝔭}(θ) ≡ 0 (mod n) for all 𝔭 ∉ S}

for S a finite set of primes, the problem of computing the Selmer group reduces to that of computing the images of the local connecting maps δ_{v}. Since we give equations for the covering curves, the im δ_{v} may be computed by working out conditions for these curves to be locally soluble. See for example [13], [14], [8]. Alternatively, as described for example in [19], [23], the im δ_{v} may be computed as the cokernel of the map ϕ : E(K_{v}) → E′(K_{v}).

We take n = 5 and split into the cases where χ has order 1, 2 or 4. If χ is trivial then E[ϕ] ≅ μ_{5} and E′[ϕ̂] ≅ ℤ/5ℤ as Galois modules. We recall from [13] that E ≅ C_{λ} and E′ ≅ D_{λ} for some λ ∈ K where

(6)  \begin{aligned}C_{λ} : & y^{2} + (1 − λ)xy − λy & = x^{3} − λx^{2} + a_{4}x + a_{6}, \\ D_{λ} : & y^{2} + (1 − λ)xy − λy & = x^{3} − λx^{2}, \end{aligned}

and a_{4} = −5λ(λ^{2} + 2λ − 1), a_{6} = −λ(λ^{4} + 10λ^{3} − 5λ^{2} + 15λ − 1).

Theorem 5.1. If λ_{0}, …, λ_{4} ∈ K with

(7)  λ = ∏_{i=0}^{4} λ_{i} and θ ≡ ∏_{i=0}^{4} λ_{i}^{i} mod (K^{×})^{5}

then the ϕ-covering of D_{λ} corresponding to θ ∈ K^{×}/(K^{×})^{5} is defined by the 4 × 4 Pfaffians of

(\begin{matrix}0 & λ_{1}x_{1} & x_{2} & −x_{3} & −λ_{4}x_{4} \\ & 0 & λ_{3}x_{3} & x_{4} & −x_{0} \\ & & 0 & λ_{0}x_{0} & x_{1} \\ & − & & 0 & λ_{2}x_{2} \\ & & & & 0 \end{matrix}) .

Proof: See [13, Proposition 2.12]. The analogue of this result for cyclic isogenies of degrees n = 3, 4 is given in [14, Section 1.2]. □

Example 5.2. Taking K = ℚ and (λ_{0}, …, λ_{4}) = (1, 1, 2, 3, 5) gives an element of order 5 in � (C_{30}/ℚ).

If χ is a quadratic character then E and E′ are the quadratic twists by χ of C_{λ} and D_{λ} for some λ ∈ K. We write L = K(\sqrt{d}).

Theorem 5.3. If r, s ∈ K (not both zero) then the ϕ-covering of E′ corresponding to θ = (r + s\sqrt{d})/(r − s\sqrt{d}) ∈ (L^{×}/(L^{×})^{5})^{χ} is defined by the 4 × 4 Pfaffians of

(\begin{matrix}0 & λx_{0} & d(x_{2} − x_{4}) & −x_{1} + x_{3} & −x_{3} \\ & 0 & −x_{1} − x_{3} & x_{2} + x_{4} & x_{4} \\ & & 0 & (r^{2} − s^{2}d)x_{0} & rx_{1} + sdx_{2} \\ & − & & 0 & sx_{1} + rx_{2} \\ & & & & 0 \end{matrix}) .

Proof: Let α = r + s\sqrt{d} and α′ = r − s\sqrt{d}. We apply Theorem 5.1 over L with

(λ_{0}, …, λ_{4}) = (λ/(αα′), α, 1, 1, α′).

We then substitute x_{0} ← −(r^{2} − s^{2}d)x_{0} and

(x_{1}, …, x_{4}) ← (x_{1} + \sqrt{d}x_{2}, x_{3} + \sqrt{d}x_{4}, x_{3} − \sqrt{d}x_{4}, x_{1} − \sqrt{d}x_{2})

to give a curve defined over K. Since [L : K] and deg ϕ are coprime the restriction map H^{1}(K, E[ϕ]) → H^{1}(L, E[ϕ]) is injective. Since the curve we have found and the curve we are looking for are isomorphic over L, they must therefore be isomorphic over K. □

Example 5.4. Taking K = ℚ and λ = 11, d = 5, r = s = 1 gives an element of order 5 in � (E/ℚ) where E is the elliptic curve 275b3 in Cremona’s tables [9].

Remark 5.5. The curve in Theorem 5.1 is defined by the 5 quadrics

λ_{i}x_{i}^{2} + x_{i−1}x_{i+1} − λ_{i−2}λ_{i+2}x_{i−2}x_{i+2} = 0

where the subscripts are read modulo 5. If λ′_{i} = −λ_{2i}/(λ_{2i−2}λ_{2i+2}) then the curves defined by λ_{0}, …, λ_{4} and λ′_{0}, …, λ′_{4} are isomorphic via

(x_{0} : …: x_{4}) ↦ (x_{0} : x_{2} : x_{4} : x_{1} : x_{3}).

Taking Jacobians it follows that C_{λ} ≅ C_{−1/λ}. Alternatively this last statement may be checked using the Weierstrass equations (6).

Now suppose χ has order 4. Let σ be the generator of Gal(L/K) with χ(σ) = 2. Then E and E′ are isomorphic over L to C_{λ} and D_{λ} for some λ ∈ L satisfying σ(λ) = −1/λ.

Theorem 5.6. If α ∈ L^{×} then the ϕ-covering of E′ corresponding to

(8)  θ = α^{4}σ(α)^{2}σ^{2}(α)σ^{3}(α)^{3} ∈ (L^{×}/(L^{×})^{5})^{χ}

is isomorphic over L to the curve in Theorem 5.1 with

(λ_{0}, …, λ_{4}) = (λσ(α)σ^{3}(α), \frac{α}{λσ(α)σ^{3}(α)}, \frac{λσ(α)}{α}, \frac{λσ^{3}(α)}{σ^{2}(α)}, \frac{σ^{2}(α)}{λσ(α)σ^{3}(α)}) .

Moreover a model for this curve over K is obtained by substituting

(\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}) ← (\begin{matrix}β_{1} & β_{2} & β_{3} & β_{4} \\ σ(β_{1}) & σ(β_{2}) & σ(β_{3}) & σ(β_{4}) \\ σ^{3}(β_{1}) & σ^{3}(β_{2}) & σ^{3}(β_{3}) & σ^{3}(β_{4}) \\ σ^{2}(β_{1}) & σ^{2}(β_{2}) & σ^{2}(β_{3}) & σ^{2}(β_{4}) \end{matrix}) (\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix})

where β_{1}, …, β_{4} is a basis for L over K.

Proof: The first part is clear since we have chosen λ_{0}, …, λ_{4} to satisfy (7). We have also arranged that σ(λ_{i}) = −λ_{2i}/(λ_{2i−2}λ_{2i+2}). The second part then follows by Remark 5.5. □

Remark 5.7. Since ρ = 4 + 2σ + σ^{2} + 3σ^{3} ∈ 𝔽_{5}[G] is an idempotent satisfying σρ = 2ρ, every element of (L^{×}/(L^{×})^{5})^{χ} is of the form (8).

Example 5.8. Let E and E′ be the 5-isogenous elliptic curves

E = 23808c3  y^{2} = x^{3} − x^{2} − 785949x − 271615419,
E′ = 23808c2  y^{2} = x^{3} − x^{2} + 7651x + 676677.

Then L = ℚ(ε) where ε = \sqrt{2 + \sqrt{2}}. Moreover λ = (49 + 41\sqrt{2})/31 and σ : ε ↦ ε^{3} − 3ε. We take α = 1 + ε and β_{j} = ε^{j−1} for j = 1, …, 4.

After following the construction in Theorem 5.6, the algorithms for minimisation and reduction in [17] suggest the change of co-ordinates

(\begin{matrix}x_{0} \\ x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}) ← (\begin{matrix}0 & 0 & 0 & 0 & 62 \\ 0 & 6 & −6 & 14 & 0 \\ 13 & −13 & −7 & −7 & 0 \\ 0 & 1 & −1 & −8 & 0 \\ −3 & 3 & 4 & 4 & 0 \end{matrix}) (\begin{matrix}x_{0} \\ x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}) .

The result is C ⊂ ℙ^{4} defined by the 4 × 4 Pfaffians of

(\begin{matrix}0 & x_{0} − x_{1} + x_{3} + 4x_{4} & x_{1} − x_{2} − x_{4} & −x_{2} − 2x_{3} + 4x_{4} & x_{1} \\ & 0 & −x_{2} − 4x_{4} & x_{1} − x_{2} + x_{4} & x_{3} \\ & & 0 & x_{0} − x_{1} − x_{3} − 4x_{4} & x_{2} \\ & − & & 0 & x_{0} \\ & & & & 0 \end{matrix}) .

Computing the invariants, as described in [15], and using Bruin’s programs [3] to check local solubility, we find that C represents an element of � (E/ℚ)[5]. It is non-trivial since E′(ℚ) = 0 and θ ∉ (L^{×})^{5}.

6. An example of full 5-descent

In this section we compute equations for an element of order 5 in the Tate-Shafarevich group of the elliptic curve E/ℚ:

6727a1 y^{2} + xy = x^{3} − x^{2} − 202951x − 34841040.

Since E has no rational 5-isogenies, our method is to use full 5-descent, that is, descent with respect to the multiplication-by-5 map on E. Further details of the calculation are given in a MAGMA [1] file that is available on the author’s webpage.

Let T = (x_{T} , y_{T}) be a non-trivial 5-torsion point on E. Then L = ℚ(T) is a number field of degree 24. Let σ_{2} be the automorphism of L with σ_{2}(T) = 2T. We shall write elements of L in terms of u and v where v = −31(2y_{T} + x_{T})/(x_{T}^{2} + 480x_{T} + 87391) and u = v/σ_{2}(v). Explicitly u has minimal polynomial

X^{12} + 4X^{11} − 6X^{10} − 20X^{9} + 15X^{8} − 303X^{7}
+323X^{6}+303X^{5} + 15X^{4} + 20X^{3} − 6X^{2} − 4X + 1

and v is a square root of \frac{1}{3}(2u^{11} + 9u^{10} − 8u^{9} − 46u^{8} + 10u^{7} − 591u^{6} +343u^{5} + 928u^{4} + 331u^{3} + 60u^{2} + 8u − 9).

We recall from [11], [27] that there is an injective map

H^{1}(ℚ, E[5]) → L^{×}/(L^{×})^{5}

whose image is contained in the σ_{2}-eigenspace

(9)  {x ∈ L^{×}/(L^{×})^{5} : σ_{2}(x) ≡ x^{2} mod (L^{×})^{5}}.

The primes of bad reduction for E are 7 and 31, with Tamagawa numbers c_{7} = 1 and c_{31} = 2. Since the Tamagawa numbers are coprime to 5, we have S^{(5)}(E/ℚ) ⊂ L(S, 5) where S = {𝔭_{1}, 𝔭_{2}} is the set of primes of L above 5.

The number field L is too large for an unconditional computation of its class group and units. However according to PARI/GP [26] (which by default makes heuristic assumptions) the class number is 2. We also used PARI/GP to compute a set of fundamental units, and generators for the prime ideals 𝔭_{1} and 𝔭_{2}. This gives a basis for L(S, 5) ≅ (ℤ/5ℤ)^{15}. The intersection of L(S, 5) with the σ_{2}-eigenspace (9) is 3-dimensional. One of the non-trivial elements is

a = \frac{1}{294}(4600u^{11} + 8325u^{10} − 72155u^{9} − 50035u^{8} + 289975u^{7} − 1450795u^{6}

+ 4510595u^{5} − 592350u^{4} − 3962957u^{3} − 1755928u^{2} − 811953u − 191035)v

+ \frac{1}{294}(158985u^{11} + 661975u^{10} − 836070u^{9} − 3280275u^{8} + 1784950u^{7} − 48064875u^{6}

+ 43645605u^{5} + 52498690u^{4} + 14516335u^{3} + 7628705u^{2} + 310520u − 311257).

We have (a) = 𝔠^{5} for some integral ideal 𝔠, and aσ_{2}(a)^{2} = b^{5} where

b = \frac{1}{294}(452u^{11} + 1935u^{10} − 2186u^{9} − 9743u^{8} + 4070u^{7} − 135379u^{6}

+ 108106u^{5} + 172665u^{4} + 54912u^{3} + 14840u^{2} − 4879u − 12762)v

+ \frac{1}{294}(−1983u^{11} − 9082u^{10} + 7240u^{9} + 46137u^{8} − 7149u^{7} + 585937u^{6}

− 289205u^{5} − 957562u^{4} − 338134u^{3} − 139997u^{2} − 62943u + 7646).

We recall some of the theory from [10]. Let E be an elliptic curve over a field K of characteristic 0. Let R be the K-algebra of all Galois equivariant maps E[n] → \overline{K} and let w : E[n] → \overline{R}^{×} = Map(E[n], \overline{K}^{×}) be the map induced by the Weil pairing e_{n}. If σ ↦ ξ_{σ} is a cocycle representing ξ ∈ H^{1}(K, E[n]) then by Hilbert’s theorem 90 there exists γ ∈ \overline{R}^{×} with σ(γ)/γ = w(ξ_{σ}) for all σ ∈ Gal(\overline{K}/K). We put α = γ^{n} and ρ = ∂γ where

∂ : \overline{R}^{×} → (\overline{R} ⊗ \overline{R})^{×} = Map(E[n] × E[n], \overline{K}^{×})

is given by (∂z)(T_{1}, T_{2}) = z(T_{1})z(T_{2})/z(T_{1} + T_{2}). Then according to [10, Paper I, Section 3] there are group homomorphisms

w_{1} : H^{1}(K, E[n]) → R^{×}/(R^{×})^{n};  ξ ↦ α,
w_{2} : H^{1}(K, E[n]) → (R ⊗ R)^{×}/∂R^{×}; ξ ↦ ρ.

The map w_{1} is injective for n prime, whereas w_{2} is always injective.

Let Ob : H^{1}(K, E[n]) → Br(K) be the obstruction map as defined in Section 2.

Theorem 6.1. Assume n is odd. Let ξ ∈ H^{1}(K, E[n]) and ρ ∈ (R ⊗ R)^{×} with w_{2}(ξ) = ρ∂R^{×}. Let A = (R, +, ∗_{ρ}) where the new multiplication ∗_{ρ} is defined by

z_{1} ∗_{ρ} z_{2} : T ↦ ∑_{T_{1}+T_{2}=T} e_{n}(T_{1}, T_{2})^{(n+1)/2}ρ(T_{1}, T_{2})z_{1}(T_{1})z_{2}(T_{2}).

Then A is a central simple algebra over K of dimension n^{2} representing the class of Ob(ξ) in Br(K).

Proof: See [10, Paper I, Lemma 3.11 and Section 4]. □

Returning to our numerical example we write α and β for the elements (1, a) and (1, b) in the étale algebra R = ℚ × L. To compute ρ exactly (using ∂α = ρ^{5}) we must extract a 5th root in a number field of degree \frac{1}{2}# GL_{2}(ℤ/5ℤ) = 240. This would be the direct analogue of what we do for 3-descent (see [10, Paper III, Section 8]), but is clearly not very promising. So instead we write ρ = ∂γ and (fixing an embedding \overline{ℚ} ⊂ ℂ) represent γ ∈ \overline{R} = Map(E[5], \overline{ℚ}) numerically. Since γ^{5} = α there are at first sight 5^{25} possibilities for γ. We cut down to just 5^{3} choices by requiring that

(i) γ(T)γ(2T)^{2} = β(T) for all T ∈ E[5], and (ii) γ : E(ℂ)[5] → ℂ is Gal(ℂ/ℝ)-equivariant.

To explain these conditions we recall that σ(γ)/γ = w(ξ_{σ}) for all σ ∈ Gal(\overline{ℚ}/ℚ). From this it is easy to see that T ↦ γ(T)γ(2T)^{2} is Galois equivariant. Since α(T)α(2T)^{2} = β(T)^{5}, and there are no non-trivial 5th roots of unity in R, this proves (i). Let τ ∈ Gal(\overline{ℚ}/ℚ) be complex conjugation. (Recall that we fixed an embedding \overline{ℚ} ⊂ ℂ.) Since H^{1}(ℝ, E[5]) = 0 we have τ(γ)/γ = w(ξ_{τ}) = w(τ(S) − S) for some S ∈ E(ℂ)[5]. Dividing γ by w(S) now gives (ii). Multiplying γ by w(T) for T ∈ E(ℝ)[5] does not change ρ = ∂γ, so in fact we only need to loop over 5^{2} choices for γ.

Let T_{1}, T_{2} be a basis for E[5](ℂ) with \overline{T}_{1} = T_{1}, \overline{T}_{2} = −T_{2}. Then ζ = e_{5}(T_{1}, T_{2}) is a primitive 5th root of unity. We define

h(T_{1}) = (\begin{matrix}0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \end{matrix}) , h(T_{2}) = (\begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & ζ & 0 & 0 & 0 \\ 0 & 0 & ζ^{2} & 0 & 0 \\ 0 & 0 & 0 & ζ^{3} & 0 \\ 0 & 0 & 0 & 0 & ζ^{4} \end{matrix}) ,

and h : E[5](ℂ) → Mat_{5}(ℂ) ; rT_{1} + sT_{2} ↦ ζ^{−rs/2}h(T_{1})^{r}h(T_{2})^{s}, where the exponent of ζ is read as an element of ℤ/5ℤ.

We compute the structure constants for A_{ρ} from the real trivialisation given in [10, Paper III, Section 5], i.e.

(10)  A_{ρ} ⊗ ℝ ≅ Mat_{5}(ℝ) ; z ↦ ∑_{T ∈E[5]} γ(T)z(T)h(T).

As recommended there we choose as our ℚ-basis for L, a ℤ-basis for 𝔠^{−1} that is LLL reduced with respect to the inner product

⟨z_{1}, z_{2}⟩ = ∑_{0≠T ∈E[5]} |α(T)|^{2/5}z_{1}(T)\overline{z_{2}(T)}.

This makes the structure constants small integers, and so easy to recognise from floating point approximations. The incorrect choices of γ are quickly discarded since the structure constants do not in general turn out to be integers.

To record our final choice of γ we let T_{1}, T_{2} be the basis for E(ℂ)[5] given (approximately) by

T_{1} = (1996.32, −87675.66), T_{2} = (−643.55, 321.77 − 13079.33i).

Then γ is the 5th root of α given (approximately) by the following matrix, with entries γ(rT_{1} + sT_{2}) for r, s = 0, …, 4.

(\begin{matrix}1.00 & −3.96 + 0.90i & 1.39 − 4.05i & 1.39 + 4.05i & −3.96 − 0.90i \\ −0.92 & 5.87 − 2.18i & 2.39 + 1.96i & 2.39 − 1.96i & 5.87 + 2.18i \\ −2.20 & 4.41 + 3.00i & −3.56 − 4.19i & −3.56 + 4.19i & 4.41 − 3.00i \\ −2.12 & −7.13 − 4.33i & −0.29 + 3.75i & −0.29 − 3.75i & −7.13 + 4.33i \\ 4.44 & 0.14 − 0.12i & −0.96 − 0.48i & −0.96 + 0.48i & 0.14 + 0.12i \end{matrix})

Although our method for choosing a basis for L as a ℚ-vector space works well on a computer, the basis vectors (which are elements of 𝔠^{−1}) are extremely messy to write down. To assist in recording some details of the calculation, we replace α and γ by their inverses. Our ℚ-basis u_{1}, …, u_{24} for L is now a ℤ-basis for 𝔠. Its first two elements are

u_{1} =\frac{1}{147}(906u^{11} + 3697u^{10} − 5099u^{9} − 18382u^{8} + 11847u^{7} − 274284u^{6}

+ 271264u^{5} + 284304u^{4} + 51522u^{3} + 31261u^{2} − 4247u − 3174),

u_{2} =\frac{1}{294}(640u^{11} + 2621u^{10} − 3565u^{9} − 13051u^{8} + 8154u^{7} − 193589u^{6}

+ 188894u^{5} + 204155u^{4} + 40745u^{3} + 21338u^{2} − 5548u − 2903)v

+ \frac{1}{294}(−221u^{11} − 943u^{10} + 1135u^{9} + 4972u^{8} − 2330u^{7} + 65086u^{6}
− 53197u^{5} − 99488u^{4} − 12061u^{3} + 13094u^{2} + 5473u + 4980).

Then R has basis r_{1}, …, r_{25} where r_{1} = (1, 0) and r_{i+1} = (0, u_{i}). Let A_{ρ} = (R, +, ∗_{ρ}) with basis a_{1}, …, a_{25} corresponding to r_{1}, …, r_{25}. Note that a_{1} is the identity. The structure constants turn out to be integers with maximum absolute value 448 and mean absolute value 22.65. As predicted by [10, Paper III, Lemma 5.2] the order with basis the a_{i} has discriminant 5^{48} · 7^{16} · 31^{18} = 5^{25} · Disc(L). The basis vectors a_{i} have minimal polynomials

X − 1, X^{5} + 435X^{3} + 7315X^{2} + 835X + 32172,
X^{5} − 390X^{3} − 4885X^{2} + 17560X + 1407822, …

If α ∈ R^{×}/(R^{×})^{5} corresponds to a Selmer group element then by the local-to-global principle for the Brauer group we have A_{ρ} ≅ Mat_{5}(ℚ). The problem of finding such an isomorphism (called a trivialisation) is addressed in [10, Paper III], [20], [21]. By using MAGMA to compute a maximal order (and running LLL on the change of basis matrix) we found a basis with minimal polynomials

X^{2}, X^{2}, X^{2}, X^{4}, X^{2}, X^{3}, X^{2}, X^{3} − X, X^{5} − 2X^{3} + X,

X^{4} − X^{2}, X^{4} − X^{2}, X^{5} + X^{3}, X^{2}, X^{4} − X^{2}, X^{3}, X^{4} − 2X^{2},

X^{4} − X^{2}, X^{5} − X^{3} + X^{2} + X, X^{5} − X^{3} − 4X^{2} + 4X,

X^{4} − 2X^{2} − X, X^{4} + X^{3} − X^{2} − X, X^{5} − X^{3}, X^{5} − 2X^{3},

X^{5} − 5X^{2} + X, X^{3} + X^{2}.

Any reducible minimal polynomial gives a zero-divisor in A_{ρ}, and once we know a zero-divisor it is easy to find a trivialisation. In this way we found a trivialisation τ that maps a_{1} ↦ I_{5} and

a_{2} ↦ (\begin{matrix}13 & −5 & −20 & −20 & −15 \\ −40 & −22 & 40 & 20 & 10 \\ −20 & −35 & 3 & −15 & −15 \\ −15 & 0 & 30 & 13 & 0 \\ 15 & 15 & −10 & −5 & −7 \end{matrix}) , a_{3} ↦ (\begin{matrix}−12 & 5 & 0 & 10 & 5 \\ −30 & −42 & 35 & 5 & 0 \\ −50 & −35 & 38 & 20 & 5 \\ −110 & −50 & 65 & 8 & 5 \\ 45 & 45 & −30 & 0 & 8 \end{matrix}) .

These calculations show that α ∈ R^{×}/(R^{×})^{5} corresponds to an element of H^{1}(ℚ, E[5]) with trivial obstruction. It may therefore be represented by a genus one normal curve C ⊂ ℙ^{4}.

We compute equations for C using the “Hesse pencil method”, as described in [10, Paper I, Section 5.1]. Let r_{1}^{∗}, …, r_{25}^{∗} be the basis for R with Tr_{R/ℚ}(r_{i}r_{j}^{∗}) = δ_{ij}. It is shown that

M = ∑_{i=1}^{25} r_{i}^{∗}τ(a_{i}) ∈ GL_{5}(R) = Map_{ℚ}(E[5], GL_{5}(\overline{ℚ}))

describes the action of E[5] on C ⊂ ℙ^{4}. In [16, Section 12] we gave a practical method for computing all genus one normal curves C ⊂ ℙ^{4} that have Jacobian E and are invariant under the matrices M_{T} for T ∈ E[5]. As predicted by [10, Paper I, Proposition 5.5] there is only one such curve defined over ℚ. We use the algorithms for minimisation and reduction in [17] to make a final change of co-ordinates. In this example the model obtained is already minimal, whereas reduction suggests the change of co-ordinates

(\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{matrix}) ← (\begin{matrix}−1 & 2 & 1 & −2 & 1 \\ −1 & 1 & 1 & −1 & 1 \\ 1 & −1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \end{matrix}) (\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{matrix}) .

The result is C ⊂ ℙ^{4} defined by the 4 × 4 Pfaffians of

(\begin{matrix}0 & −x_{1} + x_{2} + x_{3} & x_{1} + 3x_{2} + x_{4} & −2x_{2} + x_{3} + x_{5} & 2x_{2} − 2x_{3} + x_{5} \\ & 0 & −x_{1} − x_{2} − x_{3} + x_{5} & x_{2} − x_{4} + x_{5} & −x_{2} + x_{3} + x_{4} \\ & & 0 & −x_{3} + x_{5} & −x_{1} + x_{3} − x_{5} \\ & − & & 0 & x_{4} \\ & & & & 0 \end{matrix}) .

Computing the invariants, as described in [15], and using Bruin’s programs [3] to check local solubility, we find that C represents an element of � (E/ℚ)[5]. It is non-trivial since E(ℚ)/5E(ℚ) = 0 and α ∉ (R^{×})^{5}.

The theory in [10, Paper I, Section 3] shows that if M^{5} = α′I_{5} then α′/α ∈ (R^{×})^{5}. This is a condition we can check exactly. So even though we made use of floating point approximations (and did not check at the outset that α is in the image of w_{1}; methods for doing this are however described in [11],[27]), we can be sure that C corresponds to our original choice of α.

Repeating for other choices of α, we found a subgroup of � (E/ℚ) isomorphic to (ℤ/5ℤ)^{2}. For these, and examples for other elliptic curves E/ℚ of small conductor, see [18]. The main difficulty in computing further examples is that the computation of class group and units is often prohibitively expensive.

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University of Cambridge, DPMMS, Centre for Mathematical Sciences, Wilberforce Road, Cambridge CB3 0WB, UK

  E-mail address: T.A.Fisher@dpmms.cam.ac.uk