MINIMISATION AND REDUCTION OF 5-COVERINGS OF ELLIPTIC CURVES

TOM FISHER

Abstract. We consider models for genus one curves of degree 5, which arise in explicit 5-descent on elliptic curves. We prove a theorem on the existence of minimal models with the same invariants as the minimal model of the Jacobian elliptic curve and give an algorithm for computing such models. Finally we describe how to reduce genus one models of degree 5 defined over ℚ.

Introduction

Let E be an elliptic curve defined over a number field K. An n-covering of E is a morphism π : C → E where C is a smooth curve of genus one, and π = [n] ∘ ψ for some isomorphism ψ : C → E defined over \overline{K}. An n-descent on E computes the everywhere locally soluble n-coverings of E. For such n-coverings we have ψ^{∗}(n.0_{E}) ∼ D for some K-rational divisor D on C. The complete linear system |D| defines a morphism C → ℙ^{n−1}. Thus in the cases n = 2, 3, 4 we may represent C by a binary quartic, ternary cubic, or pair of quadrics in 4 variables. In the case n = 5 we obtain curves C ⊂ ℙ^{4} of degree 5 that are defined by the 4 × 4 Pfaffians of a 5 × 5 alternating matrix of linear forms.

The question naturally arises as to how we can choose co-ordinates on ℙ^{n−1} so that the equations for C have small coefficients. In the cases n = 2, 3, 4 this was answered in [CFS], using the combination of two techniques called minimisation and reduction. In this paper we extend to the case n = 5. We establish results on minimisation over an arbitrary local field (immediately implying results over any number field of class number 1), whereas those for reduction are specific to the case K = ℚ. Implementations of our algorithms in the case K = ℚ are available in MAGMA [BCP].

1. 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}(R) for the space of all genus one models with coefficients in a ring R. Models Φ and Φ′ are R-equivalent if Φ′ = [A, B]Φ for some A, B ∈ GL_{5}(R). Here the action of A is via Φ ↦ AΦA^{T}, and the action of B is via (Φ_{ij}(x_{1}, …, x_{5})) ↦ (Φ_{ij}(x′_{1}, …, x′_{5})) where x′_{j} = ∑_{i=1}^{5} B_{ij}x_{i}. The determinant of the transformation g = [A, B] is det g = (det A)^{2} det B.

Date: 21st December 2011.

We write Pf(Φ) for the row vector (p_{1}, …, p_{5}) where p_{i} is (−1)^{i−1} times the Pfaffian of the 4 × 4 submatrix obtained by deleting the ith row and column of Φ. This choice of signs is made so that Pf(Φ)Φ = 0. For A ∈ GL_{5}(R) we note that Pf(AΦA^{T}) = Pf(Φ) adj A.

A genus one model Φ ∈ X_{5}(K) over a field K is non-singular if the subscheme C_{Φ} = {rank Φ ≤ 2} ⊂ ℙ^{4} defined by the 4 × 4 Pfaffians of Φ is a smooth curve of genus one. We write K[X_{5}] for the polynomial ring in the 50 coefficients of a genus one model. A polynomial F ∈ K[X_{5}] is an invariant of weight k if F ∘g = (det g)^{k}F for all g = [A, B] with A, B ∈ GL_{5}(\overline{K}). Taking A and B to be scalar matrices shows that an invariant of weight k is a homogeneous polynomial of degree 5k.

Theorem 1.1. Let c_{4}, c_{6}, Δ ∈ ℤ[X_{5}] be the invariants of weights 4, 6, 12, satisfying c_{4}^{3} − c_{6}^{2} = 1728Δ, and scaled as specified in [F1].

(i) A model Φ ∈ X_{5}(K) is non-singular if and only if Δ(Φ) ≠ 0.

(ii) There exist a_{1}, a_{2}, a_{3}, a_{4}, a_{6} ∈ ℤ[X_{5}] and b_{2}, b_{4}, b_{6} ∈ ℤ[X_{5}] satisfying

(1)  \begin{aligned}b_{2} & = a_{1}^{2} + 4a_{2}, & b_{4} & = a_{1}a_{3} + 2a_{4},  b_{6} = a_{3}^{2} + 4a_{6}, \\ c_{4} & = b_{2}^{2} − 24b_{4}, & c_{6} & = −b_{2}^{3} + 36b_{2}b_{4} − 216b_{6}. \end{aligned}

(iii) If Φ ∈ X_{5}(K) is non-singular then C_{Φ} has Jacobian elliptic curve

y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}

where a_{i} = a_{i}(Φ).

For the proof of Theorem 1.1(ii) we use the following lemma.

Lemma 1.2. Let c_{4}, c_{6}, Δ ∈ R = ℤ[x_{1}, …, x_{N}] be primitive polynomials satisfying c_{4}^{3} −c_{6}^{2} = 1728Δ. If there exists a_{1} ∈ R satisfying a_{1}^{2}c_{4} +c_{6} ≡ 0 (mod 4) then there exist a_{2}, a_{3}, a_{4}, a_{6}, b_{2}, b_{4}, b_{6} ∈ R satisfying (1).

Proof: By unique factorisation in 𝔽_{3}[x_{1}, …, x_{N}] and the Chinese Remainder Theorem there exists b_{2} ∈ R with c_{4} ≡ b_{2}^{2} (mod 3), c_{6} ≡ −b_{2}^{3} (mod 3) and b_{2} ≡ a_{1}^{2} (mod 4). Then b_{2}c_{4} + c_{6} ≡ 0 (mod 12) and c_{4}^{3} ≡ c_{6}^{2} ≡ b_{2}^{2}c_{4}^{2} (mod 24). Since c_{4} is primitive it follows that c_{4} ≡ b_{2}^{2} (mod 24). Next putting x = b_{2} in an identity of Kraus [K],

(x^{2} − c_{4})^{3} = (x^{3} − 3xc_{4} − 2c_{6})(x^{3} + 2c_{6}) + 3(xc_{4} + c_{6})^{2} + c_{6}^{2} − c_{4}^{3},

we deduce b_{2}^{3} − 3b_{2}c_{4} − 2c_{6} ≡ 0 (mod 432). We put b_{4} = (b_{2}^{2} − c_{4})/24 and b_{6} = (b_{2}^{3} − 3b_{2}c_{4} − 2c_{6})/432. Then 0 ≡ c_{4}^{3} − c_{6}^{2} ≡ 16b_{2}^{2}(b_{2}b_{6} − b_{4}^{2}) (mod 64) and so b_{2}b_{6} ≡ b_{4}^{2} (mod 4). By unique factorisation in 𝔽_{2}[x_{1}, …, x_{N}] there exists a_{3} ∈ R with b_{4} ≡ a_{1}a_{3} (mod 2). Then b_{4}^{2} ≡ a_{1}^{2}a_{3}^{2} (mod 4) and b_{6} ≡ a_{3}^{2} (mod 4). We put a_{2} = (b_{2} − a_{1}^{2})/4, a_{4} = (b_{4} − a_{1}a_{3})/2 and a_{6} = (b_{6} − a_{3}^{2})/4. □

Proof of Theorem 1.1: (i) This is [F1, Theorem 4.4(ii)]. (ii) By Lemma 1.2 it suffices to construct a_{1} ∈ ℤ[X_{5}] with a_{1}^{2}c_{4} + c_{6} ≡ 0 (mod 4). In [F1, Section 10] we constructed an invariant a_{1} ∈ 𝔽_{2}[X_{5}] of weight 1 and showed that together with Δ it generates the ring of invariants in characteristic 2. In particular c_{4} ≡ a_{1}^{4} (mod 2) and c_{6} ≡ a_{1}^{6} (mod 2). So if we lift a_{1} to ℤ[X_{5}] then a_{1}^{2}c_{4} + c_{6} = 2f for some f ∈ ℤ[X_{5}]. Since a_{1} is an invariant mod 2, a_{1}^{2} is an invariant mod 4, and f is an invariant mod 2. Therefore f ≡ λa_{1}^{6} (mod 2) for some λ ∈ {0, 1}. Hence a_{1}^{2}c_{4} ± c_{6} ≡ 0 (mod 4). Specialising to one of the Weierstrass models in [F1, Section 6] shows that the sign is +. (iii) It is shown in [F1, Theorem 4.4(iii)] that if K is a perfect field with characteristic not 2 or 3 then C_{Φ} has Jacobian y^{2} = x^{3} − 27c_{4}(Φ)x − 54c_{6}(Φ). The proof is now identical to that of [CFS, Theorem 2.10]. This generalises a result of Artin, Rodriguez-Villegas and Tate [ARVT] in the case n = 3. □

2. Minimisation theorems

Let K be a discrete valuation field, with ring of integers O_{K}, and normalised valuation v : K^{×} → ℤ. We assume throughout that the residue field k is perfect. A genus one model Φ ∈ X_{5}(K) is integral if it has coefficient in O_{K}. If Φ is non-singular and integral then, by Theorem 1.1 and the standard formulae for transforming Weierstrass equations, we have v(Δ(Φ)) = v(Δ_{E}) + 12ℓ(Φ) where Δ_{E} is the minimal discriminant of E = Jac(C_{Φ}) and ℓ(Φ) is a non-negative integer we call the level. We say that Φ is minimal if v(Δ(Φ)), or equivalently the level, is minimal among all integral models K-equivalent to Φ. Notice that if Φ′ = gΦ for some g = [A, B] with A, B ∈ GL_{5}(K) then ℓ(Φ′) = ℓ(Φ) + v(det g).

Theorem 2.1. Let Φ ∈ X_{5}(K) be non-singular.

(i) (Weak minimisation theorem) If C_{Φ}(K) ≠ ∅ then Φ is K-equivalent to an integral model of level 0.

(ii) (Strong minimisation theorem) If C_{Φ}(L) ≠ ∅ where L is an unramified extension of K then Φ is K-equivalent to an integral model of level 0.

In this section we prove the weak minimisation theorem. In Section 3 we describe an explicit algorithm for minimising. Inspection of this algorithm shows that the minimal level is unchanged by an unramified extension. Theorem 2.1(ii) then follows from Theorem 2.1(i). In Section 7 we prove a converse to the strong minimisation theorem, thereby showing this result is best possible.

We refer to [CFS, Section 2] for notation and results analogous to those in Section 1 for genus one models of degree 4, i.e. quadric intersections. Let E be an elliptic curve over K, and D a K-rational divisor on E of degree n = 4 or 5. The complete linear system |D| defines an embedding E ⊂ ℙ^{n−1}. The image is defined by a genus one model Φ ∈ X_{n}(K), and this model is uniquely determined, up to K-equivalence, by the pair (E, [D]). Moreover every non-singular model Φ ∈ X_{n}(K) with C_{Φ}(K) ≠ ∅ arises in this way. Therefore Theorem 2.1(i) is an immediate consequence of the following.

Theorem 2.2. Let E/K be an elliptic curve, with integral Weierstrass equation

(2)  y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6},

and let D ∈ Div_{K}(E) be a divisor on E of degree n = 4 or 5. Then (E, [D]) can be represented by an integral genus one model with the same discriminant as (2).

The case n = 4 is proved in [CFS, Theorem 3.8]. To deduce the case n = 5 from the case n = 4 we use the following lemma.

Lemma 2.3. Let D ∈ Div_{K}(E) be a divisor of degree 4 and let P ∈ E(K). Let ℓ_{i}, α_{i}, β_{i} for i = 1, 2, 3 be linear forms in x_{1}, …, x_{4} over K. The following statements are equivalent.

(i) The pair (E, [D]) is represented by the quadric intersection

(3)  \begin{aligned}ℓ_{1}α_{1} + ℓ_{2}α_{2} + ℓ_{3}α_{3} & = 0 \\ ℓ_{1}β_{1} + ℓ_{2}β_{2} + ℓ_{3}β_{3} & = 0 \end{aligned}

and P is the point defined by ℓ_{1} = ℓ_{2} = ℓ_{3} = 0. (ii) The pair (E, [D + P]) is represented by the genus one model of degree 5

(4)  (\begin{matrix}0 & γ & α_{1} & α_{2} & α_{3} \\ & 0 & β_{1} & β_{2} & β_{3} \\ & & 0 & ℓ_{3} & −ℓ_{2} \\ & − & & 0 & ℓ_{1} \\ & & & & 0 \end{matrix})

where γ = x_{5} and P is the point (x_{1} : …: x_{5}) = (0 : …: 0 : 1).

Proof: An isomorphism ψ : C_{4} → C_{5}, between the curves C_{4} and C_{5} defined by (3) and (4), is given by

ψ : (x_{1} : x_{2} : x_{3} : x_{4}) ↦ (x_{1}ℓ_{i} : x_{2}ℓ_{i} : x_{3}ℓ_{i} : x_{4}ℓ_{i} : α_{j}β_{k} − α_{k}β_{j})

(where i, j, k are any cyclic permutation of 1, 2, 3) with inverse

ψ^{−1} : (x_{1} : x_{2} : x_{3} : x_{4} : x_{5}) ↦ (x_{1} : x_{2} : x_{3} : x_{4}).

This isomorphism identifies the points {ℓ_{1} = ℓ_{2} = ℓ_{3} = 0} ∈ C_{4}(K) and (0 : …: 0 : 1) ∈ C_{5}(K). To prove the equivalence of (i) and (ii) we note that if C_{4} ⊂ ℙ^{3} meets some plane in the divisor D = P_{1} + P_{2} + P_{3} + P_{4} then the points ψ(P_{i}) and (0 : …: 0 : 1) are a hyperplane section for C_{5} ⊂ ℙ^{4}. □

Lemma 2.4. The genus one models (3) and (4) have the same invariants.

Proof: Let Φ be the matrix (4) and write P = Pf(Φ) = (p_{1}, …, p_{5}). According to [F1, Section 5.4] there are invariant differentials ω_{4} on C_{4} = {p_{1} = p_{2} = 0} ⊂ ℙ^{3} and ω_{5} on C_{5} = {rank Φ ≤ 2} ⊂ ℙ^{4} given by

ω_{n} = \frac{x_{1}^{2}d(x_{2}/x_{1})}{Ω_{n}(x_{1}, …, x_{n})}

where

Ω_{4} = \frac{∂p_{1}}{∂x_{3}} \frac{∂p_{2}}{∂x_{4}} − \frac{∂p_{1}}{∂x_{4}} \frac{∂p_{2}}{∂x_{3}}  and  Ω_{5} = \frac{∂P}{∂x_{3}} \frac{∂Φ}{∂x_{5}} \frac{∂P^{T}}{∂x_{4}} .

In the expression for Ω_{5} we have written the partial derivative of a matrix as a short-hand for the matrix of partial derivatives. Since the only entries of Φ to involve x_{5} are in the top left 2×2 submatrix, it is clear that Ω_{4} = ±Ω_{5}. Hence the isomorphism ψ : C_{4} → C_{5} identifies the invariant differentials ω_{4} and ω_{5} (up to sign). It follows by [F1, Proposition 5.23] that (3) and (4) have the same invariants c_{4}, c_{6} and Δ. □

Proof of Theorem 2.2: Let D ∈ Div_{K}(E) be a divisor of degree 4, and let P ∈ E(K). We show that if the theorem holds for D then it holds for D + P. Suppose (E, [D]) is represented by an integral quadric intersection with discriminant Δ. Since O_{K} is a principal ideal domain, SL_{4}(O_{K}) acts transitively on ℙ^{3}(K). So we may assume P is the point (x_{1} : x_{2} : x_{3} : x_{4}) = (0 : 0 : 0 : 1). Our model is now of the form (3) with ℓ_{i} = x_{i} for i = 1, 2, 3. We may choose the linear forms α_{i} and β_{i} to have coefficients in O_{K}. Then the genus one model (4) is an integral model of discriminant Δ representing the pair (E, [D + P]). □

3. Minimisation algorithms

For Φ ∈ X_{5}(O_{K}) we write ϕ ∈ X_{5}(k) for its reduction mod π. The singular locus Sing C_{ϕ} is the set of points P ∈ C_{ϕ} with tangent space of dimension greater than 1. (We make this definition regardless of whether C_{ϕ} is a curve. In particular all points on components of dimension at least 2 are singular.) For example, if ϕ takes the form (4) with γ = x_{5} and ℓ_{i}, α_{i}, β_{i} linear forms in x_{1}, …, x_{4}, then P = (0 : …: 0 : 1) is singular if and only if ℓ_{1}, ℓ_{2}, ℓ_{3} are linearly dependent. An integral genus one model Φ ∈ X_{5}(O_{K}) is saturated if its 4 × 4 Pfaffians p_{1}, …, p_{5} are linearly independent mod π. We write I_{m} for the m × m identity matrix.

Our algorithm for minimising genus one models of degree 5 generalises the algorithm for models of degree 3 in [CFS, Section 4B].

Theorem 3.1. Let Φ ∈ X_{5}(O_{K}) be saturated and of positive level.

(i) The singular locus Sing C_{ϕ} does not span ℙ^{4}.

(ii) Let B ∈ GL_{5}(O_{K}) represent a change of co-ordinates on ℙ^{4} mapping the linear span of the singular locus in (i) to {x_{m+1} = … = x_{5} = 0}. Then there exist A ∈ GL_{5}(K) and μ ∈ K^{×} such that [A, μ Diag(I_{m}, πI_{5−m})B]Φ is an integral model of the same or smaller level.

(iii) If Φ is non-minimal then repeating the procedure in (ii) either gives a nonsaturated model or decreases the level after finitely many iterations.

As it stands Theorem 3.1 does not give an algorithm for minimising since we must show how to find A and μ in (ii), and show how to decrease the level of a non-saturated model. We do this in Theorem 3.2 below. Theorem 3.1 is proved in Sections 4 and 5. In Section 6 we bound the number of iterations required in (iii).

Theorem 3.2. Let Φ ∈ X_{5}(O_{K}) be non-singular. Let ℓ_{0} be the minimum of the levels of all integral models that are K-equivalent to Φ via a transformation of the form [A, μI_{5}] where A ∈ GL_{5}(K) and μ ∈ K^{×}.

(i) We may compute an integral model of the form [A, μI_{5}]Φ with level ℓ_{0} as

follows:

Step 1: Write Pf(Φ) = (p_{1}, …, p_{5}). Compute A = (a_{ij}) ∈ GL_{5}(K) and quadrics q_{1}, …, q_{5} ∈ O_{K}[x_{1}, …, x_{5}] such that p_{j} = ∑_{i=1}^{5} a_{ij}q_{i} and q_{1}, …, q_{5} are linearly independent mod π. Then replace Φ by [A, μI_{5}]Φ where μ ∈ K^{×} is chosen so that Φ has coefficients in O_{K} not all in πO_{K}.

Step 2: Replace Φ by [A, I_{5}]Φ where A ∈ GL_{5}(O_{K}) is chosen so that the first two rows of Φ are divisible by π^{e}, with e ≥ 0 as large as possible. Then divide the first row and column by π^{e}.

(ii) If the model computed in Step 1 is non-saturated, then we may compute an integral model of level smaller than ℓ_{0} by modifying Step 2 so that we divide the first two rows and columns by π^{e}, and then make a transformation of the form [I_{5}, B] to preserve integrality.

Proof: With the notation of Step 1 we have

Pf(AΦA^{T}) = Pf(Φ) adj A = (q_{1}, …, q_{5})A adj A = (det A)(q_{1}, …, q_{5}). So after Step 1 we have Pf(Φ) = (λq_{1}, …, λq_{5}) where λ := μ^{2} det A ∈ O_{K}. We split into the cases v(λ) = 0 and v(λ) ≥ 1. First we need two lemmas.

Lemma 3.3. Let Φ, Φ′ ∈ X_{5}(O_{K}) be non-singular models with Φ′ = [A, μI_{5}]Φ for some A ∈ GL_{5}(K) and μ ∈ K^{×}.

(i) If Φ is saturated then ℓ(Φ′) ≥ ℓ(Φ) with equality if and only if Φ and Φ′ are O_{K}-equivalent.

(ii) If Φ and Φ′ are of the form output by Step 1 then they are O_{K}-equivalent.

Proof: We have Pf(Φ′) = Pf(Φ)M where M := μ^{2} adj A.

(i) Since Φ is saturated, M has entries in O_{K}. Hence ℓ(Φ′)−ℓ(Φ) = \frac{1}{2}v(det M) ≥ 0 with equality if and only if M ∈ GL_{5}(O_{K}). If M ∈ GL_{5}(O_{K}) then replacing [A, μI_{5}] by [λA, λ^{−2}μI_{5}] for suitable λ ∈ K^{×} we may assume A ∈ GL_{5}(O_{K}). Since Φ and Φ′ have the same level they must therefore be O_{K}-equivalent. (ii) Since Pf(Φ) and Pf(Φ′) are scalar multiples of bases for the same O_{K}-module, some scalar multiple of M belongs to GL_{5}(O_{K}). Replacing [A, μI_{5}] by [λA, λ^{−2}μI_{5}] for suitable λ ∈ K^{×} we may assume A ∈ GL_{5}(O_{K}). Since Φ and Φ′ are primitive they must therefore be O_{K}-equivalent. □

Lemma 3.4. Let ϕ ∈ X_{5}(k) be a genus one model all of whose 4 × 4 Pfaffians are identically zero. Then ϕ is k-equivalent to either

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

where ℓ_{2}, …, ℓ_{5} are linear forms.

Proof: This is clear. □

We now complete the proof of Theorem 3.2. If v(λ) = 0 then Φ is saturated and we are done by Lemma 3.3(i). So suppose e := v(λ) ≥ 1. In Step 1 the matrix A has entries in O_{K}. So v(μ) ≤ 0 and the level is increased by

2v(det A) + 5v(μ) ≤ 2v(μ^{2} det A) = 2e.

Lemma 3.3(ii) shows that when we apply Step 1 to both Φ and the model implicit in the definition of ℓ_{0} then we obtain models that are O_{K}-equivalent. So it will suffice to show that Step 2 reduces the level by 2e, whereas the modified version in (ii) reduces the level by more than 2e.

Since Pf(Φ) = (λq_{1}, …, λq_{5}) we have (q_{1}, …, q_{5})Φ = 0. The reduction of Φ takes one of the forms specified in Lemma 3.4. In the first case we have q_{1}ℓ_{j} ≡ 0 (mod π) for j = 2, …, 5. This contradicts the choices of q_{1}, …, q_{5} and μ in Step 1. So we must be in the second case. Replacing Φ by an O_{K}-equivalent model we may assume it takes the form (4) with ℓ_{i} = x_{i} for i = 1, 2, 3 and α_{1}, α_{2}, α_{3}, β_{1}, β_{2}, β_{3}, γ linear forms that vanish mod π. By row and column operations we may assume α_{2} ∈ ⟨x_{2}, …, x_{5}⟩ and α_{3} ∈ ⟨x_{3}, …, x_{5}⟩. Then since π^{e} | (x_{1}α_{1} + x_{2}α_{2} + x_{3}α_{3}) we have π^{e} | α_{1}, α_{2}, α_{3}. Likewise we may assume π^{e} | β_{1}, β_{2}, β_{3}. The remaining Pfaffians show that π^{e} | γ. Steps 2 and its modified version in (ii) now reduce the level by 2e and 3e respectively. □

Corollary 3.5. For the proof of Theorem 3.1 we are free to replace Φ by an O_{K}-equivalent model, and to replace K by an unramified field extension.

Proof: Let Φ_{1}, Φ_{2} ∈ X_{5}(O_{K}) be O_{K}-equivalent models and Φ′_{1}, Φ′_{2} ∈ X_{5}(O_{K}) the models returned by Theorem 3.1(ii). Lemma 3.3(i) and [CFS, Lemma 4.1] together show that if Φ′_{1} is saturated and ℓ(Φ′_{1}) = ℓ(Φ′_{2}) then Φ′_{1} and Φ′_{2} are O_{K}-equivalent. Therefore the number of iterations required in Theorem 3.1(iii) depends only on the O_{K}-equivalence class of Φ.

For the final statement we note that the performance of the algorithms in Theorems 3.1 and 3.2 is unchanged by an unramified field extension. □

Replacing K by its strict Henselisation, we may assume in the next three sections that K is Henselian and its residue field k is algebraically closed.

4. The singular locus

In this section and the next we prove Theorem 3.1.

Lemma 4.1. Let ϕ ∈ X_{5}(k) be a genus one model. Suppose Γ ⊂ C_{ϕ} is either a line or a (non-singular) conic. Then either Γ ⊂ Sing C_{ϕ} or

#(Γ ∩ Sing C_{ϕ}) = \{ \begin{matrix}1 & if c_{4}(ϕ) = c_{6}(ϕ) = 0, \\ 2 & otherwise. \end{matrix}

Proof: (i) If C_{ϕ} contains the line Γ = {x_{3} = x_{4} = x_{5} = 0}, but not every point on Γ is singular, then (unless C_{ϕ} is a cone – which is an easy special case with c_{4}(ϕ) = c_{6}(ϕ) = 0) we may suppose ϕ is

(\begin{matrix}0 & x_{1} & x_{2} & ∗ & ∗ \\ & 0 & ∗ & α & β \\ & & 0 & γ & δ \\ & − & & 0 & x_{5} \\ & & & & 0 \end{matrix})

where α, β, γ, δ and the entries ∗ are linear forms in x_{3}, x_{4}, x_{5}. By row and column operations (and substitutions for x_{1} and x_{2}) we may suppose α, β, γ, δ do not involve x_{5}. We write α = α_{3}x_{3} + α_{4}x_{4}, …, δ = δ_{3}x_{3} + δ_{4}x_{4} and put

q(s, t) = det ((\begin{matrix}γ_{3} & γ_{4} \\ δ_{3} & δ_{4} \end{matrix}) s − (\begin{matrix}α_{3} & α_{4} \\ β_{3} & β_{4} \end{matrix}) t) .

By the Jacobian criterion we have

Γ ∩ Sing C_{ϕ} = {(s : t : 0 : 0 : 0) | q(s, t) = 0}.

A calculation using Lemma 2.4 shows that c_{4}(ϕ) = Δ(q)^{2} and c_{6}(ϕ) = −Δ(q)^{3} where Δ(q) is the discriminant of the binary quadratic form q.

(ii) Suppose C_{ϕ} contains the conic Γ = {f(x_{1}, x_{2}, x_{3}) = x_{4} = x_{5} = 0}, but not every point on Γ is singular. Let Pf(ϕ) = (p_{1}, …, p_{5}). Replacing ϕ by an equivalent model we may suppose p_{i}(x_{1}, x_{2}, x_{3}, 0, 0) = 0 for i = 1, 2, 3, 4 and p_{5}(x_{1}, x_{2}, x_{3}, 0, 0) = f. Since Pf(ϕ)ϕ = 0, and Γ is not contained in any component of C_{ϕ} of higher dimension, we may further suppose the last column of ϕ has entries x_{4}, x_{5}, 0, 0, 0. The monomials appearing in the invariants c_{4} and c_{6} are limited by the fact they are invariant under all pairs of diagonal matrices. These restrictions show that c_{4}(ϕ) and c_{6}(ϕ) are unchanged if we set x_{4} = x_{5} = 0 in all entries of ϕ outside the last row and column. Writing f = ∑_{i≤j} a_{ij}x_{i}x_{j} and ϕ_{34} = ∑ b_{i}x_{i} we put

δ = | \begin{matrix}2a_{11} & a_{12} & a_{13} & b_{1} \\ a_{12} & 2a_{22} & a_{23} & b_{2} \\ a_{13} & a_{23} & 2a_{33} & b_{3} \\ b_{1} & b_{2} & b_{3} & 0 \end{matrix} | .

A calculation using Lemma 2.4 shows that c_{4}(ϕ) = δ^{2} and c_{6}(ϕ) = −δ^{3}. By a change of co-ordinates we may suppose f = x_{1}x_{3}−x_{2}^{2}. Then δ is the discriminant of the binary quadratic form q(s, t) = ϕ_{34}(s^{2}, st, t^{2}, 0, 0) and by the Jacobian criterion

Γ ∩ Sing C_{ϕ} = {(s^{2} : st : t^{2} : 0 : 0) | q(s, t) = 0}. □

Lemma 4.2. Let ϕ ∈ X_{5}(k) be a genus one model. Suppose the 4 × 4 Pfaffians p_{1}, …, p_{5} are linearly independent and c_{4}(ϕ) = c_{6}(ϕ) = 0. Then either Sing C_{ϕ} is a linear subspace of ℙ^{4} or ϕ is equivalent to a model of the form

(5)  (\begin{matrix}0 & ξ & α & β & η \\ & 0 & γ & δ & x_{5} \\ & & 0 & x_{5} & 0 \\ & − & & 0 & 0 \\ & & & & 0 \end{matrix})

where ξ, η, α, β, γ, δ are linear forms in x_{1}, …, x_{5}.

Proof: If P_{1}, P_{2} ∈ Sing C_{ϕ} are distinct and the line ℓ between them is contained in C_{ϕ} then by Lemma 4.1 we have ℓ ⊂ Sing C_{ϕ}. So either Sing C_{ϕ} is a linear subspace of ℙ^{4} or there exist P_{1}, P_{2} ∈ Sing C_{ϕ} joined by a line not contained in C_{ϕ}. We move these points to (1 : 0 : …: 0) and (0 : 1 : …: 0). Writing ϕ = ∑ x_{i}M_{i}, the matrices M_{1} and M_{2} have rank 2 but their sum has rank 4. Therefore ϕ is equivalent to a model with ϕ_{12} = x_{1}, ϕ_{34} = x_{2} and all other ϕ_{ij} (for i < j) linear forms in x_{3}, x_{4}, x_{5}. Since P_{1} and P_{2} are singular, ϕ_{35} and ϕ_{45} are linearly dependent, and ϕ_{15} and ϕ_{25} are linearly dependent. So the space of linear forms spanned by the entries of the last column has dimension at most 2. In fact it has dimension exactly 2, since p_{1}, …, p_{5} are linearly independent.

Replacing ϕ by an equivalent model we may assume it has last column with entries x_{4}, x_{5}, 0, 0, 0. The transformation used here does not move P_{1} and P_{2} but may change the matrices M_{1} and M_{2}. Let Γ = {x_{4} = x_{5} = p_{5} = 0} ⊂ C_{ϕ}. Then P_{1} and P_{2} are contained in Γ but the line between them is not. It follows that Γ is either a non-singular conic or a pair of concurrent lines. In either case Lemma 4.1 shows that Γ ⊂ Sing C_{ϕ}. By the Jacobian criterion it follows that ϕ_{34} ∈ ⟨x_{4}, x_{5}⟩. However ϕ_{34} is non-zero since p_{1}, …, p_{5} are linearly independent. Therefore ϕ is equivalent to a model of the form (5). □

Lemma 4.3. Let Φ ∈ X_{5}(O_{K}) be a saturated non-singular model with reduction ϕ of the form (5). Suppose Sing C_{ϕ} has linear span {x_{m+1} = … = x_{5} = 0}.

(i) There exist A ∈ GL_{5}(K) and μ ∈ K^{×} such that [A, μ Diag(I_{m}, πI_{5−m})]Φ is an integral model of the same or smaller level.

(ii) Suppose that either δ = 0 and Φ_{45} ≡ 0 (mod π^{2}), or Φ_{35} ≡ Φ_{45} ≡ 0 (mod π^{2}). Then there is a transformation as in (i) that decreases the level.

Proof: Computing the 4 × 4 Pfaffians of (5) we find

(6)  C_{ϕ} = {η = x_{5} = αδ − βγ = 0} ∪ {γ = δ = x_{5} = 0}.

First suppose γ, δ, x_{5} are linearly dependent. By an O_{K}-equivalence we may assume δ = 0. Then {γ = x_{5} = 0} ⊂ Sing C_{ϕ} ⊂ {x_{5} = 0}. Therefore m = 3 or 4. The required transformations are as follows.

m = 3  m = 4
(i) A = Diag(π, 1, 1, 1, 1), μ = π^{−1} A = Diag(π, π, 1, 1, 1), μ = π^{−1}
(ii) A = Diag(π, 1, 1, 1, 1), μ = π^{−1} A = Diag(π, 1, 1, π^{−1}, π^{−1}), μ = 1

Now suppose γ, δ, x_{5} are linearly independent. Since Φ is saturated η, x_{5} are linearly independent. A calculation shows that Sing C_{ϕ} is the first of the two components in (6). Therefore m = 2 or 3. If m = 2 then we may assume β, γ, δ, η are linear combinations of x_{3}, x_{4}, x_{5}. The required transformations are as follows

m = 2  m = 3
(i) A = Diag(π, 1, 1, 1, 1), μ = π^{−1} A = Diag(1, 1, 1, 1, π^{−1}), μ = 1
(ii) A = Diag(1, 1, 1, π^{−1}, π^{−1}), μ = 1 A = Diag(π, π, 1, 1, π^{−1}), μ = π^{−1}

□

We now prove the first two parts of Theorem 3.1. Let Φ ∈ X_{5}(O_{K}) be saturated and of positive level. Lemma 4.2 shows that either Sing C_{ϕ} is a linear subspace or C_{ϕ} is contained in a hyperplane. Since C_{ϕ} is defined by 5 linearly independent quadrics it cannot be all of ℙ^{4}. This proves Theorem 3.1(i).

The proof of Theorem 3.1(ii) in the case ϕ takes the form (5) was already given in Lemma 4.3(i). So by Lemma 4.2 we may assume Sing C_{ϕ} = {x_{m+1} = … = x_{5} = 0}. We apply Lemma 3.4 to the reduction mod π of [I_{5}, Diag(I_{m}, πI_{5−m})]Φ. In the second case of that lemma we have m ≥ 3. We take A = Diag(1, 1, 1, 1, π^{−1}) and μ = 1. Otherwise we are in the first case. If m ≥ 2 then we take A = Diag(π, 1, 1, 1, 1) and μ = π^{−1}. It remains to treat the case m = 1, in other words the case Sing C_{ϕ} is a point.

By [F1, Lemma 5.8] every component of C_{ϕ} has dimension at least 1. So if Sing C_{ϕ} is just a point then there are also smooth points on C_{ϕ}. Since K is Henselian it follows that C_{Φ}(K) ≠ ∅ and so, by Theorem 2.1(i), Φ is non-minimal. With this extra hypothesis we show in the next section that the singular point on C_{ϕ} is non-regular (as a point on the O_{K}-scheme C_{Φ}).

We may suppose ϕ_{12} = x_{1} and all other ϕ_{ij} (for i < j) are linear forms in x_{2}, …, x_{5}. Since P = (1 : 0 : …: 0) is singular, ϕ_{34}, ϕ_{35}, ϕ_{45} are linearly dependent. So replacing Φ by an O_{K}-equivalent model we may assume ϕ_{45} = 0. In the presence of the stronger condition that P is non-regular we may further arrange that the coefficient of x_{1} in Φ_{45} is divisible by π^{2}. Taking A = Diag(1, 1, 1, π^{−1}, π^{−1}) and μ = 1 now preserves the level.

5. Weights and slopes

In this section we complete the proof of Theorem 3.1.

Definition 5.1. (i) The set of weights is

W = \{(r, s) ∈ ℤ^{5} × ℤ^{5}| \begin{matrix}r_{1} ≤ r_{2} ≤ … ≤ r_{5}, s_{1} ≤ s_{2} ≤ … ≤ s_{5}, \\ 2 ∑_{i=1}^{5} r_{i} = 1 + ∑_{i=1}^{5} s_{i} \end{matrix} \} .

(ii) A weight for Φ ∈ X_{5}(O_{K}) is (r, s) ∈ W such that the model

(7)  [Diag(π^{−r_{1}}, …, π^{−r_{5}}), Diag(π^{s_{1}}, …, π^{s_{5}})]Φ

has coefficients in O_{K}. (iii) Let w = (r, s) and w′ = (r′, s′) be weights. Then w dominates w′ if

max(r_{i} + r_{j} − s_{k}, 0) ≥ max(r′_{i} + r′_{j} − s′_{k}, 0)

for all 1 ≤ i < j ≤ 5 and 1 ≤ k ≤ 5.

Let 1 = (1, 1, …, 1). Then λ ∈ ℤ acts on W as (r, s) ↦ (r + λ1, s + 2λ1). Since weights in the same ℤ-orbit determine the same transformation (7) we may regard such weights as equivalent.

Lemma 5.2. Let Φ ∈ X_{5}(O_{K}) be an integral genus one model.

(i) If Φ is non-minimal then it is O_{K}-equivalent to a model with a weight. (ii) If Φ has weight w and w dominates w′ then Φ has weight w′.

Proof: (i) By hypothesis there exist A, B ∈ GL_{5}(K) with [A, B]Φ integral and 2v(det A) + v(det B) = −1. We put A and B in Smith normal form. (ii) Let Φ = (Φ_{ij}) with Φ_{ij} = ∑_{k} a_{ijk}x_{k}. Then Φ has weight (r, s) if and only if v(a_{ijk}) ≥ max(r_{i} + r_{j} − s_{k}, 0) for all 1 ≤ i < j ≤ 5 and 1 ≤ k ≤ 5. □

Lemma 5.3. Let Φ ∈ X_{5}(O_{K}) have weight (r, s) ∈ W with either r_{1} + r_{4} > s_{1} or r_{2} + r_{3} > s_{1}. Then P = (1 : 0 : …: 0) ∈ C_{ϕ} is a singular point. Moreover if s_{1} < s_{3} then P is non-regular (as a point on the O_{K}-scheme C_{Φ}).

Proof: We write ϕ = ∑ x_{i}M_{i}. If r_{1} + r_{4} > s_{1} then the only non-zero entries of M_{1} are in the top left 3 × 3 submatrix. If r_{2} + r_{3} > s_{1} then the only non-zero entries of M_{1} are in the first row and column. In both cases rank M_{1} ≤ 2 and so P ∈ C_{ϕ}. If M_{1} = 0 then P is singular (and non-regular). So we may assume M_{1} ≠ 0. We are free to multiply rows of Φ by units in O_{K} and to subtract O_{K}- multiples of later rows from earlier rows (it being understood that we also make the corresponding column operations). In particular these operations do not upset our hypothesis that Φ has weight (r, s). Let E_{ij} be the 5×5 matrix with a 1 in the (i, j)-place and zeros elsewhere. By row and column operations we reduce to the case M_{1} = E_{ij} − E_{ji} where (i, j) ∈ {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3)}. Let a < b < c be chosen such that {i, j, a, b, c} = {1, …, 5}. Since r_{i} + r_{j} ≤ s_{1} ≤ s_{2} it follows by the definition of W that

s_{3} + s_{4} + s_{5} < (r_{a} + r_{b}) + (r_{a} + r_{c}) + (r_{b} + r_{c}).

Therefore at least one of the following three inequalities holds:

\begin{aligned}s_{3} & < r_{a} + r_{b} & ⟹ & ϕ_{ab}, ϕ_{ac}, ϕ_{bc} & ∈ ⟨x_{4}, x_{5}⟩, \\ s_{4} & < r_{a} + r_{c} & ⟹ & ϕ_{ac}, ϕ_{bc} & ∈ ⟨x_{5}⟩, \\ s_{5} & < r_{b} + r_{c} & ⟹ & ϕ_{bc} & = 0. \end{aligned}

Since the tangent space at P is {ϕ_{ab} = ϕ_{ac} = ϕ_{bc} = 0} it follows that P ∈ C_{ϕ} is a singular point.

If s_{1} < s_{3} then the same argument shows there is some O_{K}-linear combination of Φ_{ab}, Φ_{ac}, Φ_{bc} (with not all coefficients in πO_{K}) that not only vanishes mod π, but whose coefficient of x_{1} vanishes mod π^{2}. Hence P is non-regular. □

Lemma 5.4. Let (r, s) ∈ W be a weight with r_{1} + r_{4} ≤ s_{1} and r_{2} + r_{3} ≤ s_{1}. Then (r, s) dominates one of the weights w_{1}, …, w_{7} in the following table.

r_{1} r_{2} r_{3} r_{4} r_{5} s_{1} s_{2} s_{3} s_{4} s_{5}
w_{1} 0 0 0 0 1 0 0 0 0 1
w_{2} 0 0 1 1 1 1 1 1 1 1
w_{3} 0 0 1 1 2 1 1 1 2 2
w_{4} 0 1 1 2 2 2 2 2 2 3
w_{5} 0 1 1 2 3 2 2 2 3 4
w_{6} 0 1 1 2 3 2 2 3 3 3
w_{7} 0 1 2 3 4 3 3 4 4 5

Proof: We checked the lemma by writing a computer program using the simplex algorithm. See the proof of Lemma 6.1 for details. □

Definition 5.5. The slope of Φ ∈ X_{5}(O_{K}) is the least possible value of v(det B) for B ∈ GL_{5}(K) a matrix with entries in O_{K} for which there exist A ∈ GL_{5}(K) and μ ∈ K^{×} such that [A, μB]Φ is an integral model of smaller level.

We now complete the proof of Theorem 3.1. Since Φ ∈ X_{5}(O_{K}) is non-minimal it has a slope σ, say. Lemma 3.3(i) shows that if σ = 0 then Φ is non-saturated. So we may assume σ > 0. By Lemma 5.2 (and Corollary 3.5) we may replace Φ by an O_{K}-equivalent model with a weight, say (r, s). Moreover we may assume the weight realises the slope, i.e. σ = ∑_{i=1}^{5}(s_{i} − s_{1}).

Suppose that either r_{1} + r_{4} > s_{1} or r_{2} + r_{3} > s_{1}. Since σ > 0 there exists 1 ≤ m ≤ 4 such that s_{1} = … = s_{m} < s_{m+1}. Lemma 5.3 shows (by first making unimodular transformations involving only x_{1}, …, x_{m}) that

(8)  {x_{m+1} = … = x_{5} = 0} ⊂ Sing C_{ϕ} .

Moreover if m = 1 then the point we have constructed is non-regular. (This is needed to complete the proof of Theorem 3.1(ii) at the end Section 4.)

Regardless of whether we have equality in (8) it follows that if the level is preserved then the slope is decreased. So after finitely many iterations Φ is either non-saturated or has weight (r, s) with r_{1} + r_{4} ≤ s_{1} and r_{2} + r_{3} ≤ s_{1}. In this last case Lemmas 5.2 and 5.4 show that Φ has weight w for some w ∈ {w_{1}, …, w_{7}}. If w ∈ {w_{1}, w_{2}, w_{6}} then Φ is non-saturated. If w ∈ {w_{5}, w_{7}} then Φ is O_{K}-equivalent to a model with weight w_{3}. (This is achieved by a unimodular transformation involving only the second and third rows and columns, respectively a unimodular transformation involving only x_{3} and x_{4}.) Finally if w ∈ {w_{3}, w_{4}} then Φ is O_{K}-equivalent to a model of the form considered in Lemma 4.3(ii)

6. The number of iterations

We have shown that if we start with a non-minimal model then iterating the procedure in Theorem 3.1(ii) eventually gives a non-saturated model or decreases the level. In this section we show that the maximum number of iterations required is 5. (In our MAGMA implementation we count the use of Theorem 3.2 to decrease the level of a non-saturated model as a further iteration. With this convention the maximum number of iterations is 6.)

Lemma 6.1. Let (r, s) ∈ W be a weight. Then (r, s) dominates one of the weights w_{1}, …, w_{29} in the following table. (The weights in Lemma 5.4 appear with new numberings. We have marked these weights in bold.)

r_{1} r_{2} r_{3} r_{4} r_{5} s_{1} s_{2} s_{3} s_{4} s_{5} λ_{ν}  r_{1} r_{2} r_{3} r_{4} r_{5} s_{1} s_{2} s_{3} s_{4} s_{5} λ_{ν}
w_{1} 0 0 0 0 0 −1 0 0 0 0 1 w_{16} 0 1 1 2 2 1 2 2 3 3 7
w_{2} 0 0 0 0 1 0 0 0 0 1 1 w_{17} 0 1 1 2 2 1 2 2 2 4 6
w_{3} 0 0 1 1 1 1 1 1 1 1 1 w_{18} 0 1 1 2 2 1 1 2 3 4 7
w_{4} 0 1 1 1 1 1 1 1 2 2 1 w_{19} 0 1 1 2 3 2 2 3 3 3 6
w_{5} 0 0 0 1 1 0 0 1 1 1 3 w_{20} 0 1 1 2 3 2 2 2 3 4 7
w_{6} 0 0 0 1 1 0 0 0 1 2 3 w_{21} 0 1 1 2 3 1 2 3 3 4 13
w_{7} 0 0 1 1 1 0 0 1 2 2 3 w_{22} 0 1 1 2 3 1 2 2 3 5 12
w_{8} 0 0 1 1 1 0 1 1 1 2 3 w_{23} 0 1 2 2 3 2 3 3 3 4 9
w_{9} 0 1 1 2 2 2 2 2 2 3 3 w_{24} 0 1 2 2 3 2 2 3 4 4 9
w_{10} 0 0 1 1 2 1 1 1 2 2 4 w_{25} 0 1 2 2 3 1 3 3 4 4 10
w_{11} 0 0 1 1 2 0 0 2 2 3 5 w_{26} 0 1 2 2 3 1 2 3 4 5 15
w_{12} 0 0 1 1 2 0 1 2 2 2 8 w_{27} 0 1 2 3 4 3 3 4 4 5 12
w_{13} 0 0 1 1 2 0 1 1 2 3 8 w_{28} 0 1 2 3 4 2 3 4 5 5 20
w_{14} 0 1 1 1 2 1 2 2 2 2 4 w_{29} 0 1 2 3 4 1 3 4 5 6 22
w_{15} 0 1 1 1 2 1 1 2 2 3 4

Proof: We define a standard inequality to be an inequality of the form r_{i} + r_{j} ≤ s_{k} + m where 1 ≤ i < j ≤ 5, 1 ≤ k ≤ 5 and m is a non-negative integer. The condition that (r, s) ∈ W does not dominate w_{ν} is equivalent to a list of λ_{ν} standard inequalities, at least one of which must hold, where λ_{ν} is as given in the table. For example, (r, s) ≱ w_{1} if and only if r_{1} + r_{2} ≤ s_{1}, whereas (r, s) ≱ w_{5} if and only if r_{1} + r_{4} ≤ s_{2} or r_{4} + r_{5} ≤ s_{2} + 1 or r_{4} + r_{5} ≤ s_{5}. (We have used the conditions r_{1} ≤ … ≤ r_{5} and s_{1} ≤ … ≤ s_{5} to remove redundant inequalities.)

We wrote a program using the simplex algorithm to maximise ∑(2r_{i} − s_{i}) for (r, s) ∈ ℝ^{10} subject to 0 ≤ r_{1} ≤ … ≤ r_{5}, 0 ≤ s_{1} ≤ … ≤ s_{5}, and a list of standard inequalities. Our program starts with the basic feasible solution (r, s) = (0, 0). If there is a finite maximum, and it is less than 1, then (by definition of W) there are no weights satisfying these inequalities. If the maximum is 1 then we add the constraint ∑(2r_{i} − s_{i}) = 1. We then use the simplex algorithm to maximise each of the functions r_{i} + r_{j} − s_{k} in turn. In the case of a finite maximum α we obtain an additional standard inequality r_{i} + r_{j} ≤ s_{k} + max(⌊α⌋, 0). Then running our original program on the enlarged set of standard inequalities we may still be able to show that ∑(2r_{i} − s_{i}) < 1.

After processing the inequalities coming from w_{1}, …, w_{ν} for ν = 1, …, 29 the number of cases remaining were as follows:

\begin{matrix}1, 1, 1, 1, 3, 5, 8, 13, 16, 30, 31, 49, 58, 47, 60, \\ 64, 58, 53, 45, 36, 39, 34, 25, 15, 14, 10, 3, 1, 0. \end{matrix}

The final zero indicates that no cases remain, and this proves the lemma. The proof of Lemma 5.4 is similar but easier. □

If Φ ∈ X_{5}(O_{K}) is non-minimal then by Lemmas 5.2 and 6.1 it has slope at most 14. This already shows that the algorithm in Theorem 3.1(iii) takes at most 14 iterations. The next lemma improves this bound to 7 iterations.

Lemma 6.2. If the procedure in Theorem 3.1(ii) returns a saturated model with the same level then the slope goes down by at least 2.

Proof: We revisit the proof of Theorem 3.1(iii) at the end of Section 5. If the slope goes down by only one then Sing C_{ϕ} spans a hyperplane. If Sing C_{ϕ} is a hyperplane then the proof of Theorem 3.1(ii) at the end of Section 4 shows that the level is decreased. Otherwise by Lemma 4.2 we may assume ϕ takes the form (5). We then follow the proof of Lemma 4.3(i) with m = 4. After applying the transformation suggested there, the second row of ϕ has at most one non-zero entry. This implies that Φ is non-saturated. □

The next lemma will be used to show that only 5 iterations are required.

Lemma 6.3. Let Φ ∈ X_{5}(O_{K}) be non-minimal and of slope greater than 10. Then replacing Φ by an O_{K}-equivalent model we may assume it has weight w_{29} and the coefficient of x_{k} in Φ_{ij} is a unit for

(i, j, k) ∈ {(1, 2, 1), (1, 4, 2), (1, 5, 3), (2, 3, 2), (2, 4, 3), (2, 5, 4), (3, 4, 4), (3, 5, 5)}.

Proof: By Lemma 5.2 we know that Φ is O_{K}-equivalent to a model with one of the 29 weights listed in Lemma 6.1. For all but one of these weights (r, s) we have ∑_{i=1}^{5}(s_{i} − s_{1}) ≤ 10. The remaining case is w_{29}. If one of the coefficients listed is not a unit then Φ has weight w_{ν} for some ν ∈ {1, 5, 13, 26, 16, 21, 8, 12}. □

We write [j, …, 5] for a linear combination of x_{j}, …, x_{5}, and underline in cases where we know the coefficient is non-zero. Lemma 6.3 shows that Φ ∈ X_{5}(O_{K}) has reduction ϕ ∈ X_{5}(k) of the form

(\begin{matrix}0 & [\underline{1}, 2, 3, 4, 5] & [2, 3, 4, 5] & [\underline{2}, 3, 4, 5] & [\underline{3}, 4, 5] \\ & 0 & [\underline{2}, 3, 4, 5] & [\underline{3}, 4, 5] & [\underline{4}, 5] \\ & & 0 & [\underline{4}, 5] & [\underline{5}] \\ & & & 0 & 0 \\ & & & & 0 \end{matrix}) .

Let Pf(ϕ) = (p_{1}, …, p_{5}). By considering the partial derivatives of p_{1}, p_{2}, p_{4} with respect to x_{1}, x_{2}, x_{3} we see that if P = (x_{1} : …: x_{5}) ∈ Sing C_{ϕ} then x_{5} = 0. Then since P ∈ C_{ϕ} we have x_{4} = x_{3} = x_{2} = 0. So (1 : 0 : …: 0) is the unique singular point.

Our algorithm applies the transformation

[Diag(1, 1, 1, π^{−1}, π^{−1}), Diag(1, π, π, π, π)].

The result is a model Φ with weight w_{26} = (0, 1, 2, 2, 3; 1, 2, 3, 4, 5) whose reduction ϕ takes the form

(\begin{matrix}0 & [\underline{1}] & 0 & [\underline{2}, 3, 4, 5] & [\underline{3}, 4, 5] \\ & 0 & 0 & [\underline{3}, 4, 5] & [\underline{4}, 5] \\ & & 0 & [\underline{4}, 5] & [\underline{5}] \\ & & & 0 & [5] \\ & & & & 0 \end{matrix}) .

A calculation similar to that above shows that Sing C_{ϕ} = {x_{3} = x_{4} = x_{5} = 0}. Our algorithm applies the transformation

[Diag(π, 1, 1, 1, 1), Diag(π^{−1}, π^{−1}, 1, 1, 1)]

The result is a model Φ with weight w_{13} = (0, 0, 1, 1, 2; 0, 1, 1, 2, 3) whose reduction ϕ takes the form

(\begin{matrix}0 & [\underline{1}] & 0 & [\underline{2}] & 0 \\ & 0 & [\underline{2}] & [2, \underline{3}, 4, 5] & [\underline{4}, 5] \\ & & 0 & [\underline{4}, 5] & [\underline{5}] \\ & & & 0 & [5] \\ & & & & 0 \end{matrix}) .

A calculation similar to that above shows that Sing C_{ϕ} = {x_{2} = x_{4} = x_{5} = 0}.

The next transformation [Diag(1, π, 1, 1, 1), Diag(π^{−1}, 1, π^{−1}, 1, 1)] gives a model with weight w_{15} = (0, 1, 1, 1, 2; 1, 1, 2, 2, 3). So after 3 iterations the slope is at most 4. It follows by Lemma 6.2 that at most 5 iterations are required.

Example 6.4. The simplest example of a genus one model satisfying the conditions of Lemma 6.3 is

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

We find that C_{Φ} is a rational curve with a cusp, parametrised by

(s : t) ↦ (−s^{5} : s^{3}t^{2} : s^{2}t^{3} : st^{4} : t^{5}).

In this case our algorithm takes the maximum of exactly 5 iterations to give a nonsaturated model. (The first 3 iterations are already described above.) Although the model in this example is singular, there are π-adically close non-singular models that are treated in the same way by our algorithm.

7. Insoluble models

In this section we prove a result converse to the strong minimisation theorem. This is analogous to the results for models of degrees n = 2, 3, 4 proved in [CFS, Section 5]. As in Section 2 we work over a discrete valuation field K. We write K^{sh} for the strict Henselisation of K. (If K is a p-adic field then this is the maximal unramified extension.)

Theorem 7.1. If Φ ∈ X_{5}(K) is non-singular and C_{Φ}(K^{sh}) = ∅ then the minimal level is at least 1, and is equal to 1 if char(k) ≠ 5.

As in Section 6 we write [j, …, 5] for a linear combination of x_{j}, …, x_{5}, and underline in cases where we require the coefficient is non-zero.

Definition 7.2. A genus one model Φ ∈ X_{5}(O_{K}) is critical if it has reduction mod π of the form

(\begin{matrix}0 & [\underline{1}, 2, 3, 4, 5] & [\underline{2}, 3, 4, 5] & [\underline{3}, 4, 5] & [\underline{4}, 5] \\ & 0 & [\underline{3}, 4, 5] & [\underline{4}, 5] & [\underline{5}] \\ & & 0 & [\underline{5}] & 0 \\ & & & 0 & 0 \\ & & & & 0 \end{matrix})

and π^{−1}Φ_{35}, π^{−1}Φ_{45} have reductions mod π of the form [\underline{1}, 2, 3, 4, 5], [\underline{2}, 3, 4, 5].

We show in the next three lemmas that critical models are insoluble, minimal and of positive level. We then show that every insoluble model Φ ∈ X_{5}(K) is K-equivalent to a critical model.

Lemma 7.3. Critical models are insoluble over K.

Proof: Suppose (x_{1}, …, x_{5}) ∈ K^{5} is a non-zero solution with min{v(x_{i})} = 0. By considering the 4 × 4 Pfaffians we successively deduce π | x_{5}, π | x_{4}, …, π | x_{1}. In particular min{v(x_{i})} > 0. This is the required contradiction. □

Since the definition of a critical model is unchanged by an unramified field extension, it follows immediately that critical models are insoluble over K^{sh}.

Lemma 7.4. Critical models are minimal.

Proof: It is easy to see that critical models are saturated. Moreover every point on C_{ϕ} = {x_{3} = x_{4} = x_{5} = 0} is singular. Our algorithm (see Theorem 3.1) makes the transformation [Diag(π, 1, 1, 1, 1), π^{−1} Diag(1, 1, π, π, π)]. This gives an integral model of the same level, that is O_{K}-equivalent (by a pair of cyclic permutation matrices) to a critical model.

If Φ were non-minimal then our algorithm would succeed in reducing the level. But on the contrary, when given a critical model, our algorithm endlessly cycles between five O_{K}-equivalence classes. □

The next lemma describes the possible levels of a critical model. To treat the cases char(k) = 2, 3 we need to work with the a-invariants defined in Section 1. Although these are not SL_{5} × SL_{5}-invariant, if we make our choices of a_{1}, b_{2}, a_{3} so as not to introduce any new monomials when we lift to characteristic 0, then they will be invariant under all pairs of diagonal matrices. It follows by the proof of Lemma 1.2 that a_{1}, …, a_{6} are isobaric, i.e.

a_{i} ∘ [Diag(λ_{1}, …, λ_{5}), Diag(μ_{1}, …, μ_{5})] = (∏ λ_{ν})^{2i}(∏ μ_{ν})^{i}a_{i}.

Lemma 7.5. The level of a critical model is at least 1 and equal to 1 if char(k) ≠ 5.

Proof: Applying [Diag(1, π^{−1/5}, π^{−2/5}, π^{−3/5}, π^{−4/5}), Diag(π^{1/5}, π^{2/5}, π^{3/5}, π^{4/5}, π)] to a critical model Φ gives a model with coefficients in O_{K}[π^{1/5}]. It follows by the isobaric property that π^{i} | a_{i}(Φ) for all i. Hence Φ has positive level.

The model with coefficients in O_{K}[π^{1/5}] has reduction

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

for some λ_{1}, …, λ_{5}, μ_{1} …, μ_{5} ∈ k^{×}. The invariants of this model are

c_{4}(λ, μ) = λ^{4} + 228λ^{3}μ + 494λ^{2}μ^{2} − 228λμ^{3} + μ^{4},
c_{6}(λ, μ) = −λ^{6} + 522λ^{5}μ + 10005λ^{4}μ^{2} + 10005λ^{2}μ^{4} − 522λμ^{5} − μ^{6},

and Δ(λ, μ) = λμ(λ^{2} − 11λμ − μ^{2})^{5}, where λ = ∏ λ_{i} and μ = ∏ μ_{i}. Computing a resultant shows that if char(k) ≠ 5 then c_{4}(λ, μ) and Δ(λ, μ) have no common roots. Therefore the critical model Φ we started with satisfies either v(c_{4}(Φ)) = 4 or v(Δ(Φ)) = 12. It follows that Φ has level at most 1. □

Remark 7.6. The following example of a critical model of level 2 over K = ℚ_{5} shows that the hypothesis char(k) ≠ 5 cannot be removed from Lemma 7.5.

(\begin{matrix}0 & x_{1} & x_{2} & −x_{3} & −x_{4} \\ & 0 & x_{3} & x_{4} & −x_{5} \\ & & 0 & x_{5} & 35x_{1} \\ & − & & 0 & 5x_{2} \\ & & & & 0 \end{matrix})

We recall that the minimal level is unchanged by an unramified field extension. Replacing K by K^{sh} we may assume for the rest of this section that K is Henselian and its residue field k is algebraically closed. To complete the proof of Theorem 7.1 we show

Theorem 7.7. If Φ ∈ X_{5}(O_{K}) is minimal and C_{Φ}(K) = ∅ then Φ is O_{K}-equivalent to a critical model.

We start the proof of Theorem 7.7 with the following lemma.

Lemma 7.8. If Φ ∈ X_{5}(O_{K}) is minimal then its reduction ϕ ∈ X_{5}(k) has the following properties.

(i) The 4 × 4 Pfaffians of ϕ are linearly independent.

(ii) The subscheme C_{ϕ} ⊂ ℙ^{4} does not contain a plane.

(iii) The entries of ϕ span the space of linear forms on ℙ^{4}.

Proof: (i) This follows by Theorem 3.2 and Lemma 3.3(i). (ii) Suppose C_{ϕ} contains the plane {x_{4} = x_{5} = 0}. By Lemma 3.4 we may assume the reduction mod π of [I_{5}, Diag(1, 1, 1, π, π)]Φ takes one of the two forms given in the lemma. We decrease the level by applying either [Diag(π, 1, 1, 1, 1), π^{−1}I_{5}] or [Diag(1, 1, 1, π^{−1}, π^{−1}), B] where B is chosen to preserve integrality. (iii) This is clear, as we could otherwise decrease the level by dividing one of the co-ordinates by π. □

Lemma 7.9. Let ϕ ∈ X_{5}(k) be a genus one model satisfying the conclusions of Lemma 7.8. Suppose that every point on C_{ϕ} is singular. Then ϕ is k-equivalent to

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

Our proof of Lemma 7.9 uses the following classification of degenerations of the twisted cubic. (Only the last sentence of the statement is needed.)

Lemma 7.10. Let ψ be a 3 × 2 matrix of linear forms in R = k[x_{1}, …, x_{4}]. Suppose the 2 × 2 minors of ψ are linearly independent and no linear combination of them has rank 1. Then ψ is GL_{2} × GL_{3} × GL_{4}-equivalent to one of the following:

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

In particular the locus of smooth points on Γ = {rank ψ ≤ 1} ⊂ ℙ^{3} spans ℙ^{3}.

Proof: We may realise Γ as the intersection of the image of the Segre embedding ℙ^{1} × ℙ^{2} → ℙ^{5} with a linear subspace ℙ^{3}. So every component of Γ has dimension at least 1. If every component has dimension 1 then by the Buchsbaum-Eisenbud acyclicity criterion there is a minimal free resolution

(10)  0⟶R(−3)^{2} ⟶^{ψ} R(−2)^{3} ⟶^{M} R

where M is the vector of 2 × 2 minors of ψ. If in addition dim T_{P} Γ = 1 for every P ∈ Γ then by an argument using Serre’s criterion (see [E, Section 18.3]) the ideal in R generated by the 2 × 2 minors of ψ is a prime ideal. By (10) the Hilbert polynomial is

h(t) = \binom{t + 3}{3} − 3\binom{t + 1}{3} + 2\binom{t}{3} = 3t + 1.

Therefore Γ is a twisted cubic and ψ is equivalent to the first of the matrices in (9).

In all other cases there exists P ∈ Γ with dim T_{P} Γ > 1. First suppose rank ψ(P) = 1. Moving P to (1 : 0 : 0 : 0) we may suppose

ψ = (\begin{matrix}x_{1} & α \\ δ & β \\ γ & 0 \end{matrix})

where α, β, γ, δ are linear forms in x_{2}, x_{3}, x_{4}. Our hypotheses on the 2 × 2 minors ensure that α, β, γ are linearly independent; say they are x_{2}, x_{3}, x_{4}. By row and column operations (and a substitution for x_{1}) we may assume δ is a multiple of x_{2}. This gives the second and third cases in (9).

Now suppose rank ψ(P) = 0. Let Q ∈ Γ be any other point. If rank ψ(Q) = 0 then the 2 × 2 minors are binary quadratic forms, and so some linear combination has rank 1. Therefore rank ψ(Q) = 1. If dim T_{Q} Γ > 1 then our earlier analysis applies (and in fact gives a contradiction). Otherwise we may assume

ψ = (\begin{matrix}x_{1} & 0 \\ α & x_{2} \\ β & x_{3} \end{matrix})

where α, β are linear forms in x_{2}, x_{3}. (The zero in the top right has been cleared by row operations.) Since αx_{3} − βx_{2} is a rank 2 quadratic form in x_{2}, x_{3} we can make a change co-ordinates so that Γ = {x_{1}x_{2} = x_{1}x_{3} = x_{2}x_{3} = 0}. Then ψ is equivalent to the last of the matrices in (9).

For the final statement, we note that the 4 cases correspond geometrically to (i) a twisted cubic, (ii) a conic and a line, (iii) three non-concurrent lines, and (iv) three concurrent lines. In each case Γ spans ℙ^{3} and the only singular points are the points where the components meet. □

Proof of Lemma 7.9: Let P ∈ C_{ϕ} be a singular point. Moving P to (1 : 0 : 0 : 0 : 0) we may assume ϕ takes the form

(\begin{matrix}0 & x_{1} & ℓ_{2} & α_{1} & β_{1} \\ & 0 & ℓ_{3} & α_{2} & β_{2} \\ & & 0 & α_{3} & β_{3} \\ & − & & 0 & 0 \\ & & & & 0 \end{matrix})

where ℓ_{i}, α_{i}, β_{i} are linear forms in x_{2}, …, x_{5}. Let ψ be the top right 3×2 submatrix and let Γ ⊂ ℙ^{3} be the curve defined by its 2 × 2 minors. Since the 2 × 2 minors of ψ are a subset of the 4 × 4 Pfaffians of ϕ, they are linearly independent. In particular α_{3} and β_{3} cannot both vanish identically. Without loss of generality α_{3} is non-zero.

Suppose no linear combination of the 2 × 2 minors of ψ has rank 1. Then by Lemma 7.10 there is a smooth point Q = (x_{2} : x_{3} : x_{4} : x_{5}) on Γ with α_{3}(Q) ≠ 0. Solving for x_{1} gives a smooth point (x_{1} : x_{2} : …: x_{5}) on C_{ϕ}. This is a contradiction. Therefore some linear combination of the 2 × 2 minors of ψ has rank 1. It is then easy to see that ϕ is k-equivalent to a model of the form (5).

By properties (i) and (ii), η, x_{5} are linearly independent and γ, δ, x_{5} are linearly independent. However if η, γ, δ, x_{5} were linearly independent then taking them to be x_{2}, …, x_{5} would give that (0 : 1 : 0 : 0 : 0) is a smooth point on C_{ϕ}. By row and column operations we may therefore suppose η = δ (= x_{4} say).

By property (ii), β, x_{4}, x_{5} are linearly independent and γ, x_{4}, x_{5} are linearly independent. By row and column operations (and substitutions for the x_{i}) we may suppose β = x_{3} and γ = x_{2} or x_{3}. If γ = x_{2} then by further row and column operations (and substitutions for the x_{i}) we may suppose α is a multiple of x_{1}. The lemma now follows using property (iii). □

Proof of Theorem 7.7: Since K is Henselian any smooth point on C_{ϕ} lifts to a K-point on C_{Φ}. So we may assume ϕ takes one of the three forms in Lemma 7.9. In the first two cases ϕ defines a pair of concurrent lines with multiplicities 2 and 3. (These cases may be distinguished by the dimension of the tangent space at the point of intersection). In the third case it defines a line with multiplicity 5.

We apply the transformation [Diag(1, 1, 1, 1, π^{−1}), Diag(1, 1, 1, π, π)]. This gives an integral model of the same level. So the reduction must again be k-equivalent to one of the three models in Lemma 7.9. We tidy up by an O_{K}-equivalence that cyclically permutes the rows and columns, and makes substitutions for x_{4} and x_{5}. The reduction ϕ ∈ X_{5}(k) now takes the form

(\begin{matrix}0 & x_{4} & x_{5} & α & β \\ & 0 & 0 & x_{1} & x_{3} \\ & & 0 & x_{2} & 0 \\ & − & & 0 & 0 \\ & & & & 0 \end{matrix}) or (\begin{matrix}0 & x_{4} & x_{5} & α & β \\ & 0 & x_{1} & 0 & x_{3} \\ & & 0 & x_{2} & 0 \\ & − & & 0 & 0 \\ & & & & 0 \end{matrix}) or (\begin{matrix}0 & x_{4} & x_{5} & α & β \\ & 0 & x_{1} & x_{2} & x_{3} \\ & & 0 & x_{3} & 0 \\ & − & & 0 & 0 \\ & & & & 0 \end{matrix})

where α and β are linear forms in x_{1}, x_{2}, x_{3}.

In the first case (0 : 0 : 0 : 1 : 0) is a point with tangent space of dimension 3 and C_{ϕ} contains points not on the line {x_{1} = x_{2} = x_{3} = 0}. So the transformation has moved us to the second case.

In second case we obtain a contradiction as follows. If α = x_{1} + λx_{2} + μx_{3} then adding μ times the fifth row/column to the third row/column, and making substitutions for x_{1} and x_{5} we may assume μ = 0. Then (0 : 0 : 1 : 0 : 0) is a smooth point on C_{ϕ}. Likewise if β = x_{1} + λx_{2} + μx_{3} then subtracting λ times the fourth row/column from the second row/column and making substitutions for x_{1} and x_{4} we may assume λ = 0. Then (0 : 1 : 0 : 0 : 0) is a smooth point on C_{ϕ}. We are forced to the conclusion that neither α nor β involves x_{1}. But then C_{ϕ} contains the plane {x_{2} = x_{3} = 0} and by Lemma 7.8 this contradicts that Φ is minimal.

In the third case we show that if the transformation above brings us back to the third case, then the original model is critical. If β = x_{1} + λx_{2} + μx_{3} then adding λ times the fourth row/column to the third row/column, and making substitutions for x_{1} and x_{5} we may assume λ = 0. Then C_{ϕ} contains the lines {x_{1} = x_{2} = x_{3} = 0} and {x_{1} = x_{3} = x_{5} = 0}. So if the transformation returns us to third case then β cannot involve x_{1}. Since C_{ϕ} does not contain a plane, and the 4 × 4 Pfaffians of ϕ are linearly independent, α must involve x_{1} and β must involve x_{2}. It follows by Definition 7.2 that the original model is O_{K}-equivalent to a critical model. □

8. Reduction

Let C ⊂ ℙ^{4} be a genus one normal curve of degree 5 defined over ℚ. We may represent it by a non-singular genus one model Φ ∈ X_{5}(ℤ). Running the algorithm in Section 3 locally at p, for all primes p dividing the discriminant Δ(Φ), we obtain a ℚ-equivalent model (still with coefficients in ℤ) whose discriminant is minimal in absolute value. If C is everywhere locally soluble then this discriminant is the minimal discriminant of E = Jac(C). It remains to make a GL_{5}(ℤ) change of co-ordinates on ℙ^{4} so that (after running the LLL algorithm on the space of 5 quadrics defining the curve) the coefficients (and not just the invariants) are small. The general method, described in [CFS, Section 6], is to run the LLL algorithm on the Gram matrix for the (unique) Heisenberg invariant inner product. In this section we outline how to compute this inner product in the case n = 5.

We recall that the Heisenberg group is the subgroup of SL_{5}(ℂ) consisting of matrices M_{T} that describe the action of T ∈ E[5] on C ⊂ ℙ^{4} by translation. For T ≠ 0_{E} we call the 5 points in ℙ^{4} fixed by M_{T} a syzygetic 5-tuple. It may be shown (for example by adapting the proof of [F2, Proposition 4.1] or using that H^{1}(ℝ, E[5]) is trivial) that Φ is SL_{5}(ℝ) × SL_{5}(ℝ)-equivalent to a model in Hesse form:

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

The invariants of this model are

c_{4} = a^{20} + 228a^{15}b^{5} + 494a^{10}b^{10} − 228a^{5}b^{15} + b^{20},
c_{6} = −a^{30} + 522a^{25}b^{5} + 10005a^{20}b^{10} + 10005a^{10}b^{20} − 522a^{5}b^{25} − b^{30},

and Δ = D^{5} where D = ab(a^{10} − 11a^{5}b^{5} − b^{10}). For a model in Hesse form the Heisenberg group is generated by Diag(1, ζ, …, ζ^{4}), where ζ is a primitive 5th root of unity, and a cyclic permutation matrix. Since these matrices are unitary, the Heisenberg invariant inner product is the standard inner product on ℝ^{5}.

The Hessian, introduced in [F2], is an SL_{5} × SL_{5}-equivariant polynomial map H : X_{5} → X_{5} with the property that the Hessian of (11) is of the same form with a and b replaced by −∂D/∂b and ∂D/∂a.

Theorem 8.1. Let Φ ∈ X_{5}(ℂ) be a non-singular genus one model with invariants c_{4} and c_{6}. Let A be the 3 × 5 matrix of quadrics such that λΦ + μH(Φ) has 4 × 4 Pfaffians

{λ^{2}A_{1i} + λμA_{2i} + μ^{2}A_{3i} : i = 1, …, 5}

Then X = {rank A ≤ 1} ⊂ ℙ^{4} consists of 30 points and the syzygetic 5-tuples for C_{Φ} are the fibres of the map α : X → ℙ^{2} given by the first (or indeed any) column of A. The image of α is the set of 6 points (x : y : z) ∈ ℙ^{2} satisfying

(12)  rank (\begin{matrix}0 & 5x & y & 6c_{4}x + z \\ x & y & 6c_{4}x − z & 8c_{6}x \\ y & −z & 8c_{6}x & 9c_{4}^{2}x \end{matrix}) ≤ 2.

Proof: It suffices to prove this for Φ in Hesse form. Then X is defined by

(13)  rank (\begin{matrix}x_{0}^{2} & x_{1}^{2} & x_{2}^{2} & x_{3}^{2} & x_{4}^{2} \\ x_{1}x_{4} & x_{0}x_{2} & x_{1}x_{3} & x_{2}x_{4} & x_{0}x_{3} \\ x_{2}x_{3} & x_{3}x_{4} & x_{0}x_{4} & x_{0}x_{1} & x_{1}x_{2} \end{matrix}) ≤ 1

and by [BHM, Proposition 1] is a set of 30 points. Evaluating the columns of (13) at these points we obtain (1 : 0 : 0) and (1 : ζ^{i} : ζ^{−i}) for i = 0, …, 4. These are the points (ξ : η : ν) ∈ ℙ^{2} satisfying

(14)  rank (\begin{matrix}ξ & η & ν & 0 \\ ν & ξ & 0 & −η \\ 0 & 0 & η & ν \end{matrix}) ≤ 2.

The remaining statements follow by direct calculation. In particular our description (12) of the image of α is checked by making the substitution

(\begin{matrix}x \\ y \\ z \end{matrix}) = (\begin{matrix}ab & b^{2} & −a^{2} \\ −a\frac{∂D}{∂a} + b\frac{∂D}{∂b} & −2b\frac{∂D}{∂a} & −2a\frac{∂D}{∂b} \\ −\frac{∂D}{∂b} \frac{∂D}{∂a} & (\frac{∂D}{∂a})^{2} & −(\frac{∂D}{∂b})^{2} \end{matrix}) (\begin{matrix}ξ \\ η \\ ν \end{matrix}) .

We note that this change of co-ordinates, and the matrix relating the 3 × 3 minors of (12) and (14), each have determinant a constant times a power of D. □

After computing the Hessian exactly (using the algorithm in [F2, Section 11]) we use Theorem 8.1 to compute the syzygetic 5-tuples numerically. We then compute a Gram matrix for the Heisenberg invariant inner product as follows.

Proposition 8.2. Let C ⊂ ℙ^{4} a genus one normal curve defined over ℝ. (i) Exactly two of the syzygetic 5-tuples for C are defined over ℝ, say

Y = {y_{i}y_{j} = 0 : i < j} ⊂ ℙ^{4},
Z = {z_{i}z_{j} = 0 : i < j} ⊂ ℙ^{4},

where y_{0}, …, y_{4} and z_{0}, …, z_{4} are linear forms in ℂ[x_{0}, …, x_{4}]. (ii) One of the 5-tuples in (i) has 5 real points and the other has 1 real point.

We may therefore arrange that y_{0}, …, y_{4} and z_{0} have real coefficients and that the pairs z_{1}, z_{4} and z_{2}, z_{3} are complex conjugates.

(iii) The Heisenberg invariant quadratic form spans the 1-dimensional real vec-

tor space

⟨y_{0}^{2}, …, y_{4}^{2}⟩ ∩ ⟨z_{0}^{2}, z_{1}z_{4}, z_{2}z_{3}⟩.

Proof: For C in Hesse form we may take y_{i} = x_{i} and z_{i} = ∑_{j=0}^{4} ζ^{ij}x_{j}. In this case the Heisenberg invariant quadratic form is x_{0}^{2} + … + x_{4}^{2}. □

9. Examples

Wuthrich [W] constructed an element of order 5 in the Tate-Shafarevich group of the elliptic curve E/ℚ with Weierstrass equation

y^{2} + xy + y = x^{3} + x^{2} − 3146x + 39049.

His example (see also [F1, Section 9]) is defined by the 4 × 4 Pfaffians of

(\begin{matrix}0 & 310x_{1} + 3x_{2} + 162x_{5} & −34x_{1} − 5x_{2} − 14x_{5} & 10x_{1} + 28x_{4} + 16x_{5} & 80x_{1} − 32x_{4} \\ & 0 & 6x_{1} + 3x_{2} + 2x_{5} & −6x_{1} + 7x_{3} − 4x_{4} & −14x_{2} − 8x_{3} \\ & & 0 & −x_{3} & 2x_{2} \\ & − & & 0 & −4x_{1} \\ & & & & 0 \end{matrix}) .

This model has discriminant 2^{132}Δ_{E} where Δ_{E} is the minimal discriminant of E. In other words, the model is minimal at all primes except p = 2, where the level is 11. Minimisation and reduction suggest the change of co-ordinates

(\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{matrix}) ← (\begin{matrix}0 & 4 & −8 & 4 & 8 \\ 0 & 0 & 0 & 0 & 16 \\ 0 & −4 & 4 & 0 & 12 \\ 4 & 5 & −15 & 2 & 7 \\ 4 & −12 & 20 & −12 & −8 \end{matrix}) (\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{matrix})

26 TOM FISHER

so that Wuthrich’s example simplifies to

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

Our MAGMA function DoubleGenusOneModel, described in [F3], computes a genus one model Φ′ that represents twice the class of Φ in the 5-Selmer group. This model has entries

Φ′_{12} = 3534132778x_{1} + 3583651940x_{2} − 881947110x_{3} − 323014538x_{4} + 3395115339x_{5},
Φ′_{13} = 5079379222x_{1} − 2965539950x_{2} + 11022202860x_{3} + 12821590868x_{4} + 640276471x_{5},
Φ′_{14} = −10098238458x_{1} − 1274966110x_{2} − 7873816170x_{3} − 3456923272x_{4} − 62353929x_{5},
Φ′_{15} = −12929747724x_{1} − 6790511810x_{2} − 11113305270x_{3} − 15161763156x_{4} + 3241937033x_{5},
Φ′_{23} = −3381247332x_{1} + 3810679160x_{2} + 5919634530x_{3} + 75326852x_{4} − 1245085426x_{5},
Φ′_{24} = −3572860258x_{1} − 5569480730x_{2} − 953739600x_{3} − 2138046812x_{4} − 858145244x_{5},
Φ′_{25} = −4674149266x_{1} − 943631490x_{2} − 6754488160x_{3} + 751535046x_{4} + 117685567x_{5},
Φ′_{34} = −1851228934x_{1} + 5238146110x_{2} − 165588410x_{3} − 2070411506x_{4} + 678105748x_{5},
Φ′_{35} = −6992835070x_{1} − 3744630360x_{2} + 3130208220x_{3} − 4523781310x_{4} + 433739425x_{5},
Φ′_{45} = 780078472x_{1} + 2039763820x_{2} − 450062790x_{3} − 7105731722x_{4} + 1625466111x_{5}.

The discriminant of Φ′ is Δ_{E}^{49}. In particular this model is non-minimal at all bad primes of E. Minimisation and reduction suggest the change of co-ordinates

(\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{matrix}) ← (\begin{matrix}92 & −36 & −153 & 129 & −131 \\ −54 & 84 & 5 & −206 & 139 \\ −63 & −174 & −60 & −79 & 53 \\ −111 & 106 & 206 & −115 & −162 \\ 314 & −466 & 158 & −328 & −12 \end{matrix}) (\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{matrix})

so that Φ′ simplifies to

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

References

[ARVT] M. Artin, F. Rodriguez-Villegas and J. Tate, On the Jacobians of plane cubics, Adv. Math. 198 (2005), no. 1, 366–382.

[BCP] W. Bosma, J. Cannon and C. Playoust, The MAGMA algebra system I: The user language,

     J. Symb. Comb. 24, (1997) 235–265. http://magma.maths.usyd.edu.au/magma/

[BHM] W. Barth, K. Hulek and R. Moore, Shioda’s modular surface S(5) and the Horrocks-Mumford bundle, Vector bundles on algebraic varieties (Bombay, 1984), 35–106, Tata Inst. Fund. Res. Stud. Math., 11, Tata Inst. Fund. Res., Bombay, 1987.

[CFS] J.E. Cremona, T.A. Fisher and M. Stoll, Minimisation and reduction of 2-, 3- and 4coverings of elliptic curves, Algebra & Number Theory 4 (2010), no. 6, 763-820.

[E] D. Eisenbud, Commutative algebra, with a view toward algebraic geometry, Graduate Texts in Mathematics 150, Springer-Verlag, New York, 1995.

[F1] T.A. Fisher, The invariants of a genus one curve, Proc. Lond. Math. Soc. (3) 97 (2008) 753-782.

[F2] T.A. Fisher, The Hessian of a genus one curve, to appear in Proc. Lond. Math. Soc.

[F3] T.A. Fisher, Invariant theory for the elliptic normal quintic, I. Twists of X(5), preprint,

     arXiv:1110.3520v1

[K] A. Kraus, Quelques remarques à propos des invariants c_{4}, c_{6} et Δ d’une courbe elliptique. Acta Arith. 54 (1989), no. 1, 75–80.

[W] C. Wuthrich, Une quintique de genre 1 qui contredit le principe de Hasse, Enseign. Math. (2) 47 (2001), no. 1-2, 161–172.

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

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