SOME MINIMISATION ALGORITHMS IN ARITHMETIC INVARIANT THEORY

TOM FISHER AND LAZAR RADIČEVIĆ

Abstract. We extend the work of Cremona, Fisher and Stoll on minimising genus one curves of degrees 2, 3, 4, 5, to some of the other representations associated to genus one curves, as studied by Bhargava and Ho. Specifically we describe algorithms for minimising bidegree (2, 2)-forms, 3 × 3 × 3 cubes and 2 × 2 × 2 × 2 hypercubes. We also prove a theorem relating the minimal discriminant to that of the Jacobian elliptic curve.

1. Introduction

Let F be a homogeneous polynomial in several variables with rational coefficients. Then making a linear change of variables and rescaling the polynomial by a rational number does not change the isomorphism class of the hypersurface defined by F. Thus a natural question is to find a change of variables and a rescaling of the polynomial so that its coefficients are small integers.

More generally we may consider the following situation. Let G be a product of general linear groups, acting linearly on a ℚ-vector space W. We fix a basis for W, and represent a vector w ∈ W by its vector of co-ordinates (w_{1}, …, w_{N}) relative to this basis. We refer to these co-ordinates as the coefficients. Then given w ∈ W we seek to find g ∈ G(ℚ) such that g · w has small integer coefficients.

An invariant is a polynomial I ∈ ℤ[w_{1}, …, w_{N}] such that:

I(g · w) = χ(g)I(w)

for all g ∈ G(ℂ) and w ∈ W, where χ is a rational character on G (i.e. a product of determinants). In practice there will be an invariant Δ, which we call the discriminant, and the elements w ∈ W of interest will be those with Δ(w) ≠ 0. We note that if w has integer coefficients then Δ(w) is an integer. Our strategy is to first find g ∈ G(ℚ) making this discriminant as small as possible (in absolute value). This is known as minimisation. This is a local problem, in that for each prime p dividing Δ(w) we seek to minimise the p-adic valuation v_{p}(Δ(w)), without changing the valuations at the other primes. Once we’ve minimised the discriminant, the next step is to find a transformation in G(ℤ), making the coefficients as small as possible. This is known as reduction.

Date: 6th March 2017.

This strategy has been carried out in [3] and [4], for the models (i.e. collections of polynomials) defining genus one curves of degrees 2, 3, 4 and 5. In these cases the invariants give a Weierstrass equation for the Jacobian of the genus one curve. In this article, we extend these techniques to some of the other representation associated to genus one curves, as studied in [1]. Specifically we describe algorithms for minimising bidegree (2, 2)-forms, 3 × 3 × 3 cubes and 2 × 2 × 2 × 2 hypercubes. In each of these cases the invariants define not only the Jacobian elliptic curve E, but also one or two marked points on E. One possible application of these algorithms is in computing the Cassels-Tate pairing (see [5]).

As explained below, each (2, 2)-form F determines a pair of binary quartics G_{1}, G_{2}, each 3 × 3 × 3 cube S determines a triple of ternary cubics F_{1}, F_{2}, F_{3}, and each 2 × 2 × 2 × 2 hypercube H determines a quadruple of binary quartics G_{1}, …, G_{4}. Therefore a natural approach would be to minimise and reduce the corresponding binary quartics and ternary cubics, using the algorithms in [3], and then apply the transformations that arise in this way to F, S or H. This strategy works for reduction (which we therefore do not study further in this article), but not for minimisation. For example if F ∈ ℤ[x_{1}, x_{2}; y_{1}, y_{2}] is a (2, 2)-form with F ≡ x_{2}^{2}y_{2}^{2} (mod p^{2}) then the binary quartics G_{1} and G_{2} vanish mod p^{2}. The algorithm for minimising binary quartics says that we should divide each G_{i} by p^{2}. However this information on its own does not tell us how to minimise F.

Since minimisation is a local problem, we work in the following setting. Let K be a field with a discrete valuation v : K^{×} → ℤ. We write O_{K} for the valuation ring, and π for a uniformiser, i.e. an element π ∈ K with v(π) = 1. The residue field is k = O_{K}/πO_{K}. For example we could take K = ℚ or ℚ_{p}, and v = v_{p} the p-adic valuation. In these cases O_{K} = ℤ_{(p)} or ℤ_{p}. We make no restrictions on the characteristics of K and k.

Since it serves as a prototype for our work, we briefly recall the algorithm for minimising binary quartics. See [3] for further details. A binary quartic is a homogeneous polynomial of degree 4 in two variables:

G(x_{1}, x_{2}) = ax_{1}^{4} + bx_{1}^{3}x_{2} + cx_{1}^{2}x_{2}^{2} + dx_{1}x_{2}^{3} + ex_{2}^{4}.

If R is any ring then there is an action of G(R) = R^{×} × GL_{2}(R) on the space of binary quartics over R via

(1)  [λ, (\begin{matrix}r & s \\ t & u \end{matrix})] : G(x_{1}, x_{2}) ↦ λ^{2}G(rx_{1} + tx_{2}, sx_{1} + ux_{2}).

We say that binary quartics are R-equivalent if they belong to the same orbit for this action. A polynomial I ∈ ℤ[a, b, c, d, e] is an invariant of weight p if

I([λ, A] · G) = (λ det A)^{p}I(G)

for all [λ, A] ∈ G(ℂ). The invariants of a binary quartic are

I = 12ae − 3bd + c^{2}
J = 72ace − 27ad^{2} − 27b^{2}e + 9bcd − 2c^{3}

of weights 4 and 6, and Δ = (4I^{3} − J^{2})/27 of weight 12.

A binary quartic G is integral if it has coefficients in O_{K}, and non-singular if Δ(G) ≠ 0. We write v(G) for the minimum of the valuations of the coefficients of G. Given a non-singular binary quartic, we seek to find a K-equivalent integral binary quartic G with v(Δ(G)) as small as possible.

We write G̃ for the reduction of π^{−v(G)}G mod π. If a binary quartic G(x_{1}, x_{2}) is non-minimal, then it is O_{K}-equivalent to a binary quartic with

G(x_{1}, π^{s}x_{2}) ≡ 0 (mod π^{2s+2})

for some integer s ≥ 0. The least such integer s is called the slope, and can only take values 0, 1 and 2. If v(G) ≤ 1 (i.e. the slope is positive) then G̃ has a unique multiple root, and if we move this root to (1 : 0) then π^{−2}G(x_{1}, πx_{2}) is an integral binary quartic with the same invariants, but with smaller slope. After at most two iterations we reach a form of slope 0. We can then divide through by π^{2}, and repeat the process until a minimal binary quartic is obtained.

Our algorithms for minimising (2, 2)-forms, 3 × 3 × 3 cubes and 2 × 2 × 2 × 2 hypercubes are described in Sections 2, 3 and 4. We also give formulae for the Jacobian elliptic curve and the marked points that work in all characteristics. (In [1] the authors worked over a field of characteristic not 2 or 3, and the formulae were not always given explicitly.) In Section 5 we prove a theorem about the minimal discriminant, and describe how it is improved by our minimisation algorithms.

2. Bidegree (2,2)-forms

A (2, 2)-form is a polynomial in x_{1}, x_{2}, y_{1}, y_{2}, that is homogeneous of degree 2 in both sets of variables. We can view a (2, 2)-form F as a binary quadratic form in y_{1}, y_{2} whose coefficients are binary quadratic forms in x_{1}, x_{2}:

F (x_{1}, x_{2}; y_{1}, y_{2}) = F_{1}(x_{1}, x_{2})y_{1}^{2} + F_{2}(x_{1}, x_{2})y_{1}y_{2} + F_{3}(x_{1}, x_{2})y_{2}^{2}.

The discriminant G_{1} = F_{2}^{2} − 4F_{1}F_{3} is then a binary quartic in x_{1}, x_{2}. Switching the two sets of variables we may likewise define a binary quartic G_{2} in y_{1}, y_{2}. It may be checked that G_{1} and G_{2} have the same invariants I and J. We define c_{4}(F) = I and c_{6}(F) = J/2. The discriminant is Δ(F) = (c_{4}^{3} − c_{6}^{2})/1728.

A non-zero (2, 2)-form F over a field defines a curve in ℙ^{1} × ℙ^{1}. If Δ(F) ≠ 0 then this curve C_{F} is a smooth curve of genus one. It may be written as a double cover of ℙ^{1} (ramified over the roots of G_{1} or G_{2}) by projecting to either factor.

Let R be a ring. There is an action of G(R) = R^{×} × GL_{2}(R) × GL_{2}(R) on the space of (2,2)-forms over R given by

[λ, A, B] : F (x_{1}, x_{2}; y_{1}, y_{2}) ↦ λF ((x_{1}, x_{2})A; (y_{1}, y_{2})B).

We say that (2, 2)-forms are R-equivalent if they belong to the same orbit for this action. If [λ, A, B] · F = F′ then the binary quartics G_{1} and G_{2} determined by F, and the binary quartics G′_{1} and G′_{2} determined by F′, are related by

(2)  \begin{aligned}G′_{1} & = [λ det B, A] · G_{1} \\ G′_{2} & = [λ det A, B] · G_{2} \end{aligned}

where the action on binary quartics is that defined in (1).

We may represent F by a 3 × 3 matrix via:

(3)  F (x_{1}, x_{2}; y_{1}, y_{2}) = (x_{1}^{2} x_{1}x_{2} x_{2}^{2}) (\begin{matrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}) (\begin{matrix}y_{1}^{2} \\ y_{1}y_{2} \\ y_{2}^{2} \end{matrix}) .

A polynomial I ∈ ℤ[a_{ij}] is an invariant of weight p if

I([λ, A, B] · F) = (λ det A det B)^{p}I(F)

for all [λ, A, B] ∈ G(ℂ). In particular the polynomials c_{4}, c_{6} and Δ are invariants of weights 4, 6 and 12. Over a field of characteristic not 2 or 3, the invariants determine a pair (E, P) where E is an elliptic curve (the Jacobian of C_{F}) and P is a marked point on E. The next lemma gives formulae for

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

and P = (ξ, η) that work in all characteristics.

Lemma 2.1. There exist ξ, η, a_{1}, a_{2}, a_{3}, a_{4}, a_{6} ∈ ℤ[a_{ij}] such that

(i) We have c_{4} = b_{2}^{2} −24b_{4} and c_{6} = −b_{2}^{3} +36b_{2}b_{4} −216b_{6}, where b_{2} = a_{1}^{2} +4a_{2}, b_{4} = a_{1}a_{3} + 2a_{4} and b_{6} = a_{3}^{2} + 4a_{6},

(ii) The polynomials u = 12ξ + a_{1}^{2} + 4a_{2} and v = 2η + a_{1}ξ + a_{3} are invariants

of weights 2 and 3 satisfying (108v)^{2} = (3u)^{3} − 27c_{4}(3u) − 54c_{6}. (iii) We have η^{2} + a_{1}ξη + a_{3}η = ξ^{3} + a_{2}ξ^{2} + a_{4}ξ + a_{6}.

Proof. We put ξ = a_{11}a_{33} + a_{13}a_{31} and η = a_{11}a_{22}a_{33}.

(i) We put

a_{1} = −a_{22},
a_{2} = −(a_{11}a_{33} + a_{12}a_{32} + a_{13}a_{31} + a_{21}a_{23}),
a_{3} = a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} − a_{11}a_{23}a_{32} − a_{12}a_{21}a_{33}.

Since we already defined c_{4} and c_{6}, we may solve for a_{4} and a_{6}. We find that these too are polynomials in the a_{ij} with integer coefficients. (ii) The invariants u and v were denoted δ_{2} and δ_{3} in [1, Section 6.1.2]. In fact we have v = det(a_{ij}). (iii) This follows from (i) and (ii), exactly as in [6, Chapter III]. □

Let (E, P) be a pair consisting of an elliptic curve E/K and a point 0_{E} ≠ P ∈ E(K). On a minimal Weierstrass equation for E, the point P has co-ordinates (x_{P} , y_{P}), where either x_{P} , y_{P} ∈ O_{K} or v(x_{P}) = −2r, v(y_{P}) = −3r for some integer r ≥ 1. We define κ(P) = 0 in the first case, and κ(P) = r in the second. We write Δ_{E} for the minimal discriminant of E.

We say that a (2, 2)-form F is integral if it has coefficients in O_{K}, and nonsingular if Δ(F) ≠ 0.

Lemma 2.2. Let F be a non-singular integral (2, 2)-form. Let (E, P) be the pair specified in Lemma 2.1. Then

v(Δ(F)) = v(Δ_{E}) + 12κ(P) + 12ℓ(F)

where ℓ(F) ≥ 0 is an integer we call the level.

Proof. The formulae in Lemma 2.1 give an integral Weierstrass equation W for E, upon which P is a point with integral coordinates. The smallest possible discriminant of such an equation is v(Δ_{E}) + 12κ(P). Since the discriminant of F is equal to the discriminant of W, the result follows. □

In this section we give an algorithm for minimising (2, 2)-forms. That is, given a non-singular (2, 2)-form F over K, we explain how to find a K-equivalent integral (2, 2)-form with level (equivalently, valuation of the discriminant) as small as possible. In Section 5 we show that if C_{F} (K) ≠ ∅ then the minimal level is zero.

By clearing denominators, we may start with an integral (2, 2)-form. If this form is K-equivalent to an integral form of smaller level, then our task is to find such a form explicitly. Define v(F) to be the minimum of the valuations of the coefficients of F. If v(F) ≥ 1 then we can divide through by π, reducing the level of F. We may therefore assume v(F) = 0.

Our algorithm for minimising (2, 2)-forms is described by the following theorem.

Theorem 2.3. Let F be a non-minimal (2, 2)-form with v(F) = 0. Let f be reduction of F mod π. Then we are in one of the following three situations.

(i) The form f factors as a product of binary quadratic forms, both of which have a repeated root. By an O_{K}-equivalence we may assume f = x_{2}^{2}y_{2}^{2}.

Then at least one of the forms

π^{−2}F (x_{1}, πx_{2}; y_{1}, y_{2})
π^{−2}F (x_{1}, x_{2}; y_{1}, πy_{2})
π^{−3}F (x_{1}, πx_{2}; y_{1}, πy_{2})

is an integral (2, 2)-form of smaller level. (ii) The form f factors as a product of binary quadratic forms, exactly one of

which has a repeated root. By an O_{K}-equivalence, and switching the two sets of variables if necessary, we may assume that f = x_{2}^{2}h(y_{1}, y_{2}). Then π^{−1}F (x_{1}, πx_{2}; y_{1}, y_{2}) is an integral (2, 2)-form of the same level.

(iii) The curve C_{f} ⊂ ℙ^{1}×ℙ^{1} has a unique singular point. By an O_{K}-equivalence, this is the point ((1 : 0), (1 : 0)). Then π^{−2}F (x_{1}, πx_{2}; y_{1}, πy_{2}) is an integral (2, 2)-form of the same level.

Moreover the (2, 2)-form F computed in (ii) or (iii) either has v(F) ≥ 1 or has reduction mod π of the form specified in (i).

Remark 2.4. Let F be an integral (2, 2)-form, with associated binary quartics G_{1} and G_{2}. It is clear by (2) that if either G_{1} or G_{2} is minimal then F is minimal. However the converse is not true. For example if F ≡ (x_{1}y_{1} + x_{2}y_{2})^{2} (mod π^{2}), then F is minimal by Theorem 2.3, yet we have G_{1} ≡ G_{2} ≡ 0 (mod π^{2}).

Exactly as in the case of binary quartics, any non-minimal (2, 2)-form F is O_{K}- equivalent to a form whose level can be reduced using diagonal transformations. Indeed, suppose that [λ, A_{1}, A_{2}] ∈ G(K) is a transformation reducing the level. By clearing denominators, we may assume that the A_{i} have entries in O_{K}, not all in πO_{K}. Then writing these matrices in Smith normal form we have A_{i} = Q_{i}D_{i}P_{i} where P_{i}, Q_{i} ∈ GL_{2}(O_{K}) and

D_{1} = (\begin{matrix}1 & 0 \\ 0 & π^{a} \end{matrix}) , D_{2} = (\begin{matrix}1 & 0 \\ 0 & π^{b} \end{matrix}) ,

for some integers a, b ≥ 0. Replacing F by an O_{K}-equivalent form, it follows that

π^{−a−b−1}F (x_{1}, π^{a}x_{2}; y_{1}, π^{b}y_{2})

is an integral (2, 2)-form. We say that the pair (a, b) is admissible for F.

Lemma 2.5. Let F be an integral (2, 2)-form. If some pair (a, b) is admissible for F then at least one of the following pairs is admissible:

(0, 0), (1, 0), (0, 1), (1, 1), (2, 1), (1, 2).

Proof. The coefficients of F, arranged as in (3), have valuations satisfying

≥ a + b + 1  ≥ a + 1  ≥ a − b + 1
≥ b + 1  ≥ 1  ≥ −b + 1
≥ −a + b + 1  ≥ −a + 1  ≥ −a − b + 1.

Conversely, if the valuations satisfy these inequalities then the pair (a, b) is admissible. If a = b = 0 then we are done as (0, 0) is on the list. If a ≥ 1, b = 0 or a = 0, b ≥ 1, then (1, 0) or (0, 1) is admissible. If a = b > 0, then (1, 1) is admissible. If a > b > 0 or b > a > 0, then (2, 1) or (1, 2) is admissible. □

Proof of Theorem 2.3. For the proof we are free to replace the (2, 2)-form F by an O_{K}-equivalent form. Indeed the transformations specified in the statement of the theorem induce well-defined maps on O_{K}-equivalence classes, as may be verified using [3, Lemma 4.1]. We may therefore assume that one of the pairs (a, b) listed in Lemma 2.5 is admissible for F. Since v(F) = 0 we cannot have a = b = 0. By switching the two sets of variables, we may assume a ≥ b. This leaves us with three cases.

Case 1. We assume (1, 0) is admissible for F. The coefficients of F have valuations satisfying

\begin{aligned}≥ 2 & ≥ 2 & ≥ 2 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned}

We have f = x_{2}^{2}h(y_{1}, y_{2}) where h is a binary quadratic form. If h has a repeated root, then the first transformation in (i) decreases the level. Otherwise the transformation in (ii) gives a (2, 2)-form F with v(F) ≥ 1.

Case 2. We assume (1, 1) is admissible for F. The coefficients of F have valuations satisfying

\begin{aligned}≥ 3 & ≥ 2 & ≥ 1 \\ ≥ 2 & ≥ 1 & ≥ 0 \\ ≥ 1 & ≥ 0 & ≥ 0 \end{aligned}

We have

f = x_{2}y_{2}(αx_{1}y_{2} + βx_{2}y_{1} + γx_{2}y_{2})

for some α, β, γ ∈ k. If α = β = 0 then the third transformation in (i) decreases the level. If exactly one of the coefficients α and β is zero then the transformation in (ii) gives a (2, 2)-form whose reduction mod π is either zero, or of the form specified in (i). If α and β are both non-zero then C_{f} ⊂ ℙ^{1} × ℙ^{1} has a unique singular point at ((1 : 0), (1 : 0)). The transformation in (iii) gives a (2, 2)-form F with v(F) ≥ 1.

Case 3. We assume (2, 1) is admissible for F. The coefficients of F have valuations satisfying

\begin{aligned}≥ 4 & ≥ 3 & ≥ 2 \\ ≥ 2 & ≥ 1 & = 0 \\ = 0 & ≥ 0 & ≥ 0 \end{aligned}

The two valuations indicated are zero, as we would otherwise be in Case 1 or Case 2. A calculation shows that C_{f} ⊂ ℙ^{1} × ℙ^{1} has a unique singular point at ((1 : 0), (1 : 0)). The transformation in (iii) gives a (2, 2)-form whose reduction mod π is of the form specified in (i). □

The following lemma will be needed in Section 4, in connection with our study of 2 × 2 × 2 × 2 hypercubes.

Lemma 2.6. Let F be a non-minimal (2, 2)-form, and let f = F mod π.

(i) If C_{f} ⊂ ℙ^{1} × ℙ^{1} is singular at ((1 : 0), (1 : 0)), then the coefficients of F have valuations satisfying

\begin{aligned}≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned} or \begin{aligned}≥ 1 & ≥ 1 & ≥ 0 \\ ≥ 1 & ≥ 1 & ≥ 0 \\ ≥ 1 & ≥ 0 & ≥ 0 \end{aligned}

(ii) If f = x_{2}^{2}y_{2}^{2} then the coefficients of F have valuations satisfying

\begin{aligned}≥ 2 & ≥ 2 & ≥ 2 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & = 0 \end{aligned} or \begin{aligned}≥ 2 & ≥ 1 & ≥ 1 \\ ≥ 2 & ≥ 1 & ≥ 1 \\ ≥ 2 & ≥ 1 & = 0 \end{aligned} or \begin{aligned}≥ 3 & ≥ 2 & ≥ 1 \\ ≥ 2 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & = 0 \end{aligned}

Proof. (i) The singular point forces a_{11} ≡ a_{12} ≡ a_{21} ≡ 0 (mod π). The vanishing of the invariants u and v in Lemma 2.1 gives

8a_{13}a_{31} + a_{22}^{2} ≡ a_{13}a_{22}a_{31} ≡ 0 (mod π).

It follows that a_{22} ≡ 0 (mod π). The same lemma shows that (ξ, η) = (a_{13}a_{31}, 0) is a singular point on the curve with Weierstrass equation y^{2} ≡ x^{2}(x − a_{13}a_{31}) (mod π). Therefore a_{13}a_{31} ≡ 0 (mod π).

(ii) The proof of Theorem 2.3 shows that F is O_{K}-equivalent to a (2, 2)-form F_{1} with

(4)  F_{1}(x_{1}, π^{a}x_{2}; y_{1}π^{b}y_{2}) ≡ 0 (mod π^{a+b+1})

for some (a, b) = (1, 0), (0, 1) or (1, 1). Working mod π we have F_{1} ≡ x_{2}^{2}h(y_{1}, y_{2}), g(x_{1}, x_{2})y_{2}^{2} or x_{2}y_{2}(αx_{1}y_{2} + βx_{2}y_{1} + γx_{2}y_{2}). In the last case it follows from our assumption F ≡ x_{2}^{2}y_{2}^{2} (mod π) that α = β = 0. The equivalence relating F and F_{1} must now fix the points (x_{1} : x_{2}) = (1 : 0) mod π, (y_{1} : y_{2}) = (1 : 0) mod π, or both. It follows that F also satisfies (4). □

3. 3 × 3 × 3 Rubik’s cubes

We consider polynomials in x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}, z_{1}, z_{2}, z_{3} that are linear in each of the three sets of variables. Such a form may be represented as

∑_{1≤i,j,k≤3} s_{ijk}x_{i}y_{j}z_{k}

where S = (s_{ijk}) is a 3×3×3 cubical matrix. A Rubik’s cube S may be partitioned into three 3 × 3 matrices in three distinct ways:

(i) M^{1} = (s_{1jk}) is the front face, N^{1} = (s_{2jk}) is the middle slice and P^{1} = (s_{3jk}) is the back face.

(ii) M^{2} = (s_{i1k}) is the top face, N^{2} = (s_{i2k}) is the middle slice and P^{2} = (s_{i3k})

is the bottom face. (iii) M^{3} = (s_{ij1}) is the left face, N^{3} = (s_{ij2}) is the middle slice and P^{3} = (s_{ij3})

is the right face. To each slicing (M^{i}, N^{i}, P^{i}), we may associate a ternary cubic form

F_{i}(x, y, z) = det(M^{i}x + N^{i}y + P^{i}z).

Following [3, Section 2] we scale the invariants c_{4}, c_{6}, Δ of a ternary cubic so that c_{4}(xyz) = 1, c_{6}(xyz) = −1 and c_{4}^{3} − c_{6}^{2} = 1728Δ. It may be checked that the F_{i} have the same invariants. We define c_{4}(S) = c_{4}(F_{i}), c_{6}(S) = c_{6}(F_{i}) and Δ(S) = Δ(F_{i}).

If S is defined over a field and Δ(S) ≠ 0 then each of the F_{i} defines a smooth curve of genus 1 in ℙ^{2}. These curves are isomorphic, although not in a canonical way. (See [1, Section 3.2] for further details.) We write C_{S} to denote any one of them.

Let R be a ring. For each 1 ≤ i ≤ 3 there is an action of GL_{3}(R) on the space of Rubik’s cubes over R given by

A = (a_{ij}) : (M^{i}, N^{i}, P^{i}) ↦ (a_{11}M^{i} + a_{12}N^{i} + a_{13}P^{i},

a_{21}M^{i} + a_{22}N^{i} + a_{23}P^{i}, a_{31}M^{i} + a_{32}N^{i} + a_{33}P^{i}).

These actions commute, and so give an action of G(R) = GL_{3}(R)^{3}. We say that 3 × 3 × 3 cubes are R-equivalent if they belong to the same orbit for this action. If [A_{1}, A_{2}, A_{3}] · S = S′ then the associated ternary cubics are related by

(5)  F′_{i}(x, y, z) = det(A_{j}A_{k})F_{i}((x, y, z)A_{i})

where {i, j, k} = {1, 2, 3}.

A polynomial I ∈ ℤ[s_{ijk}] is an invariant of weight p if

I([A_{1}, A_{2}, A_{3}] · S) = (det A_{1} det A_{2} det A_{3})^{p}I(S)

for all [A_{1}, A_{2}, A_{3}] ∈ G(ℂ). In particular the polynomials c_{4}, c_{6} and Δ are invariants of weights 4, 6 and 12. Over a field of characteristic not 2 or 3, the invariants determine a pair (E, P) where E is an elliptic curve (the Jacobian of C_{S}) and P is a marked point on E. The next lemma gives formulae for E and P that work in all characteristics.

Lemma 3.1. There exist ξ, η, a_{1}, a_{2}, a_{3}, a_{4}, a_{6} ∈ ℤ[s_{ijk}] such that

(i) We have c_{4} = b_{2}^{2} −24b_{4} and c_{6} = −b_{2}^{3} +36b_{2}b_{4} −216b_{6}, where b_{2} = a_{1}^{2} +4a_{2}, b_{4} = a_{1}a_{3} + 2a_{4} and b_{6} = a_{3}^{2} + 4a_{6},

(ii) The polynomials u = 12ξ + a_{1}^{2} + 4a_{2} and v = 2η + a_{1}ξ + a_{3} are invariants

of weights 2 and 3 satisfying (108v)^{2} = (3u)^{3} − 27c_{4}(3u) − 54c_{6}. (iii) We have η^{2} + a_{1}ξη + a_{3}η = ξ^{3} + a_{2}ξ^{2} + a_{4}ξ + a_{6}.

Proof. We define matrices A, B, C by the rule

(adj(λN^{1} + μP^{1}))M^{1} = λ^{2}A + λμB + μ^{2}C.

We put ξ = −tr(AC) and η = −tr(CBA).

(i) We put

a_{1} = tr(B),
a_{2} = tr(AC) + tr(A)tr(C) − tr(adj(B)),
a_{3} = tr(ABC) + tr(CBA) + tr(AC)tr(B).

Since we already defined c_{4} and c_{6}, we could now in principle solve for a_{4} and a_{6}. However it is simpler to argue as follows. Let a′_{1}, …, a′_{6} be the a-invariants (as defined in [3, Lemma 2.9]) of the ternary cubic F_{1}. We checked by computer algebra that there exist r, s, t ∈ ℤ[s_{ijk}] satisfying

a′_{1} = a_{1} + 2s,
a′_{2} = a_{2} − sa_{1} + 3r − s^{2},
a′_{3} = a_{3} + ra_{1} + 2t.

It follows by the transformation formulae for Weierstrass equations (see [6]) that a_{4}, a_{6} ∈ ℤ[s_{ijk}]. Note that our reason for working with a_{1}, …, a_{6}, in preference to a′_{1}, …, a′_{6}, is that this helped us find particularly simple expressions for ξ and η. (ii) The invariants u and v were denoted 4c_{6} and c_{9} in [1, Section 5.1.3]. In fact we have v = tr(ABC) − tr(CBA). (iii) This follows from (i) and (ii) exactly as in [6, Chapter III]. □

A Rubik’s cube S is integral if it has coefficients in O_{K}, and non-singular if Δ(S) ≠ 0.

Lemma 3.2. Let S be a non-singular integral Rubik’s cube. Let (E, P) be the pair specified in Lemma 3.1. Then

v(Δ(S)) = v(Δ_{E}) + 12κ(P) + 12ℓ(S)

where ℓ(S) ≥ 0 is an integer we call the level.

Proof. The proof is identical to that of Lemma 2.2. □

In this section we give an algorithm for minimising Rubik’s cubes. In Section 5 we show that if C_{S}(K) ≠ ∅ then the minimal level is zero.

We say that an integral cube S is saturated if for each i = 1, 2, 3 the matrices M^{i}, N^{i}, P^{i} ∈ Mat_{3}(O_{K}) are linearly independent mod π. If an integral cube is not saturated, then it is obvious how we may decrease the level.

Our algorithm for minimising 3 × 3 × 3 cubes is described by the following theorem.

Theorem 3.3. Let S be a non-minimal saturated Rubik’s cube. Let F_{1}, F_{2}, F_{3} be the associated ternary cubics, and f_{1}, f_{2}, f_{3} their reductions mod π. Then we are in one of the following two situations.

(i) Two or more of the f_{i} are non-zero and have a repeated linear factor, say f_{1} and f_{2} are divisible by z^{2}. We apply a transformation

[(\begin{matrix}1 \\ & 1 \\ & & π \end{matrix}) , (\begin{matrix}1 \\ & 1 \\ & & π \end{matrix}) , A_{3}]

where A_{3} ∈ GL_{3}(K) is chosen such that M^{3}, N^{3}, P^{3} ∈ Mat_{3}(O_{K}) are linearly independent mod π.

(ii) Two or more of the f_{i} define a curve with a unique singular point, say f_{1} and f_{2} define curves with singular points at (1 : 0 : 0). We apply a transformation

[(\begin{matrix}1 \\ & π \\ & & π \end{matrix}) , (\begin{matrix}1 \\ & π \\ & & π \end{matrix}) , A_{3}]

where A_{3} ∈ GL_{3}(K) is chosen such that M^{3}, N^{3}, P^{3} ∈ Mat_{3}(O_{K}) are linearly independent mod π.

The procedures in (i) and (ii) give an integral cube of the same or smaller level. Repeating these procedures either gives a non-saturated cube or decreases the level after at most three iterations.

Remark 3.4. Let S be an integral Rubik’s cube, with associated ternary cubics F_{1}, F_{2}, F_{3}. It is clear by (5) that if any of the F_{i} are minimal then S is minimal. However the converse is not true. For example if S ≡ (ε_{ijk}) (mod π), where ε_{ijk} is the Levi-Civita symbol (as appears in the definition of the cross product), then S is minimal by Theorem 3.3, yet we have F_{1} ≡ F_{2} ≡ F_{3} ≡ 0 (mod π).

Exactly as in the case of (2, 2)-forms, any non-minimal Rubik’s cube S is O_{K}- equivalent to a cube whose level can be reduced using diagonal transformations. Indeed, suppose that [π^{−s}A_{1}, A_{2}, A_{3}] ∈ G(K) is a transformation reducing the level. By clearing denominators, we may assume that the A_{i} have entries in O_{K}, not all in πO_{K}. Then writing these matrices in Smith normal form we have A_{i} = Q_{i}D_{i}P_{i} where P_{i}, Q_{i} ∈ GL_{3}(O_{K}) and

D_{i} = (\begin{matrix}π^{a_{1i}} & 0 & 0 \\ 0 & π^{a_{2i}} & 0 \\ 0 & 0 & π^{a_{3i}} \end{matrix} )

with min(a_{1i}, a_{2i}, a_{3i}) = 0. If this transformation reduces the level then ∑ a_{ij} < 3s. In fact, by increasing one of the a_{ij}, we may assume ∑ a_{ij} = 3s − 1. We will from now on assume a_{11} = a_{12} = a_{13} = 0. If the new cube has coefficients in O_{K} then we say that the tuple (a_{21}, a_{31}; a_{22}, a_{32}; a_{23}, a_{33}) is admissible for S.

Lemma 3.5. Let S be a non-minimal Rubik’s cube. Then after permuting the three slicings, and replacing S by an O_{K}-equivalent cube, at least one of the following tuples is admissible.

τ_{1} = (1, 1; 0, 0; 0, 0),  τ_{2} = (0, 1; 0, 1; 0, 0),  τ_{3} = (1, 2; 0, 1; 0, 1),
τ_{4} = (1, 1; 1, 1; 0, 1),  τ_{5} = (1, 2; 1, 2; 1, 1),  τ_{6} = (2, 3; 1, 2; 1, 2).

Proof. We define the set of weights

W = \{(A, s) ∈ Mat_{3}(ℤ) × ℤ | \begin{matrix}a_{11} = a_{12} = a_{13} = 0, \\ a_{ij} ≥ 0 for all i, j, \\ ∑ a_{ij} = 3s − 1 \end{matrix} \} .

If (A, s) ∈ W then (a_{21}, a_{31}; a_{22}, a_{32}; a_{23}, a_{33}) is admissible for S if and only if

v(s_{ijk}) ≥ max(s − a_{i1} − a_{j2} − a_{k3}, 0)

for all i, j, k ∈ {1, 2, 3}. We define a partial order on W by (A, s) ≤ (A′, s′) if

max(s − a_{i1} − a_{j2} − a_{k3}, 0) ≤ max(s′ − a′_{i1} − a′_{j2} − a′_{k3}, 0)

for all i, j, k ∈ {1, 2, 3}. A computer calculation, using Lemma 3.6 below, shows that (W, ≤) has exactly 81 minimal elements. By an O_{K}-equivalence we may assume a_{2i} ≤ a_{3i} for i = 1, 2, 3, and by permuting the three slicings of S we may assume a_{31} ≥ a_{32} ≥ a_{33}. Only 8 of the 81 minimal elements satisfy these additional conditions. These are the 6 elements listed in the statement of the lemma, together with two more that are the same as τ_{4} up to permuting the slicings. □

Lemma 3.6. If (A, s) ∈ W is minimal then s ≤ 10.

Proof. Suppose for a contradiction that (A, s) is minimal with s > 10. Without loss of generality we have

(6)  a_{21} ≤ a_{31}, a_{22} ≤ a_{32}, a_{23} ≤ a_{33} and a_{31} ≥ a_{32} ≥ a_{33}.

Since 6a_{31} ≥ ∑ a_{ij} = 3s − 1 we certainly have a_{31} > 3. Let A′ be the matrix obtained from A by replacing a_{31} by a_{31} − 3. Then (A′, s − 1) ∈ W, and by our minimality assumption (A′, s − 1) ≰ (A, s). Therefore

max(s − 1 − a′_{i1} − a′_{j2} − a′_{k3}, 0) > max(s − a_{i1} − a_{j2} − a_{k3}, 0)

for some i, j, k ∈ {1, 2, 3}. Since we only changed the entry a_{31} we must have i = 3 and s − 1 − (a_{31} − 3) > 0. Therefore

(7)  s + 1 ≥ a_{31}.

The following inequalities are obtained in an entirely analogous way:

(i) If a_{33} > 0 then by considering (a_{21}, a_{31} − 1; a_{22}, a_{32} − 1; a_{23}, a_{33} − 1), we have s ≥ a_{32} + a_{33}.

(ii) If a_{21}, a_{22}, a_{23} > 0 then by considering (a_{21} −1, a_{31} −1, a_{22} −1, a_{32} −1, a_{23} −

1, a_{33} − 1), we have s ≥ a_{21} + a_{22} + a_{23}. (iii) If a_{22} > 0 then by considering (a_{21}, a_{31} − 1; a_{22} − 1, a_{32} − 1; a_{23}, a_{33}), we

have s ≥ a_{31} + a_{22}. (iv) If a_{21}, a_{32} > 0 then by considering (a_{21} − 1, a_{31} − 1; a_{22}, a_{32} − 1; a_{23}, a_{33}), we

have s ≥ a_{21} + a_{32}.

(v) If a_{23} > 0 then by considering (a_{21}, a_{31} − 1; a_{22}, a_{32}; a_{23} − 1, a_{33} − 1), we have s ≥ a_{31} + a_{23}.

We now claim that if a_{33} > 0 then s ≥ a_{21} + a_{22} + a_{23}. Indeed if a_{21}, a_{22}, a_{23} > 0 then this is (ii). If a_{21} = 0 then we instead use (i). If a_{21} > 0 and a_{23} = 0 then (noting that a_{32} ≥ a_{33} > 0) we instead use (iv). If a_{23} > 0 and a_{22} = 0 then we instead use (v).

To complete the proof of the lemma, we first suppose a_{33} > 0. Then the inequalities in (i) and (ii) hold without further hypothesis. We weaken the inequalities (iii), (iv) and (v) to

(8)  s + 1 ≥ a_{31} + a_{22}
(9)  s + 1 ≥ a_{21} + a_{32}
(10)  s + 1 ≥ a_{31} + a_{23}

so that in cases where some of the a_{ij} are zero, these still hold by (6) and (7). Adding together all five inequalities gives

5s + 3 + a_{33} ≥ 2 ∑ a_{ij} = 2(3s − 1)

and hence a_{33} ≥ s − 5. Using (i) again gives

s ≥ a_{32} + a_{33} ≥ 2a_{33} ≥ 2(s − 5)

and hence s ≤ 10, as required.

If a_{33} = 0 then we still have (8) and (9) giving 2(s + 1) ≥ ∑ a_{ij} = 3s − 1, and hence s ≤ 3. □

Proof of Theorem 3.3. We represent S as a triple of matrices A, B, C, say.

\begin{matrix}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{matrix}  \begin{matrix}B_{11} & B_{12} & B_{13} \\ B_{21} & B_{22} & B_{23} \\ B_{31} & B_{32} & B_{33} \end{matrix}  \begin{matrix}C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{matrix}

The action of G(K) = GL_{3}(K)^{3} may be described as follows. The first factor replaces A, B, C by linear combinations of these matrices. The second factor acts by row operations (applied to A, B, C simultaneously), and the third factor acts by column operations.

We may assume one of the tuples τ_{1}, …, τ_{6} in Lemma 3.5 is admissible for S. We therefore split into these 6 cases.

Case 1. We assume (1, 1; 0, 0; 0, 0) is admissible for S. Then the entries of A have valuation at least one, and so the cube S is not saturated.

Case 2. We assume (0, 1; 0, 1; 0, 0) is admissible for S. The entries of A, B and C have valuations satisfying

\begin{aligned}≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned} \begin{aligned}≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned} \begin{aligned}≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned}

Since S is saturated we may assume by column operations that v(C_{11}) = 0, v(C_{12}) ≥ 1 and v(C_{13}) ≥ 1. Subtracting a multiple of the first row from the second row gives v(C_{21}) ≥ 1, and again by column operations v(C_{22}) = 0 and v(C_{23}) ≥ 1. Subtracting multiples of the first two rows from the third, the valuations now satisfy

\begin{aligned}≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned} \begin{aligned}≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned} \begin{aligned}= 0 & ≥ 1 & ≥ 1 \\ ≥ 1 & = 0 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 0 \end{aligned}

We compute f_{1} = C_{11}C_{22}z^{2}(A_{33}x + B_{33}y + C_{33}z) mod π. Since S is saturated it follows that f_{1} is nonzero. The same argument shows that f_{2} has a repeated factor and is nonzero. On the other hand we have f_{3} = 0. The procedure in (i) multiplies C and the third row by π, and then divides the cube by π. This transformation decreases the level.

Case 3. We assume (1, 2; 0, 1; 0, 1) is admissible for S. The entries of A, B and C have valuations satisfying

\begin{aligned}≥ 2 & ≥ 2 & ≥ 1 \\ ≥ 2 & ≥ 2 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 0 \end{aligned} \begin{aligned}≥ 1 & ≥ 1 & ≥ 0 \\ ≥ 1 & ≥ 1 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned} \begin{aligned}≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned}

Since S is saturated we have v(A_{33}) = 0. If B_{13} ≡ B_{23} ≡ 0 (mod π) then we are in Case 2, and likewise if B_{31} ≡ B_{32} ≡ 0 (mod π). By operating on the first two rows and columns, and then subtracting a multiple of A from B, the valuations now satisfy

\begin{aligned}≥ 2 & ≥ 2 & ≥ 1 \\ ≥ 2 & ≥ 2 & ≥ 1 \\ ≥ 1 & ≥ 1 & = 0 \end{aligned} \begin{aligned}≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & = 0 \\ ≥ 1 & = 0 & ≥ 1 \end{aligned} \begin{aligned}≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned}

Working mod π we compute

f_{1} = −B_{23}B_{32}C_{11}y^{2}z + z^{2}( ⋯)
f_{2} = −A_{33}B_{32}z^{2}(C_{11}x + C_{21}y + C_{31}z)
f_{3} = −A_{33}B_{23}z^{2}(C_{11}x + C_{12}y + C_{13}z)

Since S is saturated, it is clear that f_{2} and f_{3} are nonzero.

We note that multiplying C, the last row and the last column by π, and then dividing the whole cube by π, gives an integral model of the same level which is not saturated. These transformations are carried out by the procedure in (i), except possibly in the case where f_{1} has a repeated factor, and this factor is not z^{2}. In this remaining case v(C_{11}) = 0. We may assume by row and column operations that C_{12} ≡ C_{13} ≡ C_{21} ≡ C_{31} ≡ 0 (mod π). Subtracting multiples of A and B from C gives C_{32} = C_{33} = 0 (mod π). Now f_{1} = C_{11}z(A_{33}C_{22}xz − B_{23}B_{32}y^{2} − B_{32}C_{23}yz), and so C_{22} ≡ C_{23} ≡ 0 (mod π).

If the procedure in (i) picks f_{1} and f_{2} then we multiply B and the last row by π. Dividing the last two columns by π gives a model of the same level with valuations satisfying

\begin{aligned}≥ 2 & ≥ 1 & ≥ 0 \\ ≥ 2 & ≥ 1 & ≥ 0 \\ ≥ 2 & ≥ 1 & = 0 \end{aligned} \begin{aligned}≥ 2 & ≥ 1 & ≥ 1 \\ ≥ 2 & ≥ 1 & = 0 \\ ≥ 3 & = 1 & ≥ 2 \end{aligned} \begin{aligned}= 0 & ≥ 0 & ≥ 0 \\ ≥ 1 & ≥ 0 & ≥ 0 \\ ≥ 2 & ≥ 1 & ≥ 1 \end{aligned}

Since the first two columns of A and B are divisible by π, we are now in Case 2. The case where the procedure in (i) picks f_{1} and f_{3} works in the same way.

Case 4. We assume (1, 1; 1, 1; 0, 1) is admissible for S. The entries of A, B and C have valuations satisfying

\begin{aligned}≥ 2 & ≥ 2 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 0 \\ ≥ 1 & ≥ 1 & ≥ 0 \end{aligned} \begin{aligned}≥ 1 & ≥ 1 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned} \begin{aligned}≥ 1 & ≥ 1 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned}

Working mod π we compute

f_{1} = (B_{13}y + C_{13}z) |(\begin{matrix}B_{21} & B_{22} \\ B_{31} & B_{32} \end{matrix}) y + (\begin{matrix}C_{21} & C_{22} \\ C_{31} & C_{32} \end{matrix}) z| ,

and

f_{2} = (A_{23}y + A_{33}z) |(\begin{matrix}B_{21} & B_{22} \\ C_{21} & C_{22} \end{matrix}) y + (\begin{matrix}B_{31} & B_{32} \\ C_{31} & C_{32} \end{matrix}) z| .

Since S is saturated, the linear factors ℓ_{1} = B_{13}y + C_{13}z and ℓ_{2} = A_{23}y + A_{33}z cannot be identically zero. Let q_{1} and q_{2} be the quadratic factors. These are binary quadratic forms associated to the same 2 × 2 × 2 cube. In particular q_{1} and q_{2} have the same discriminant, say δ. If this cube is not saturated, it is easy to see we are in Case 1 or Case 2. Therefore f_{1} and f_{2} are nonzero.

Replacing B and C by suitable linear combinations, and likewise the last two rows, we may suppose that the linear factors ℓ_{1} and ℓ_{2} are multiples of z, i.e.

(11)  B_{13} ≡ A_{23} ≡ 0 (mod π)

Under this assumption f_{3} = −A_{33}C_{13}z^{2}(B_{21}x + B_{22}y + B_{23}z), and this is nonzero as we would otherwise be in Case 2.

If f_{1} and f_{2} don’t have repeated factors, then each defines a curve with a unique singular point at (1 : 0 : 0). The procedure in (ii) multiplies B, C and the last two rows by π. The level is then reduced using columns operations, in exactly the way suggested by the definition of Case 4.

Now suppose that at least one of the forms f_{1} and f_{2} has a repeated factor. Then the procedure in (i) is applied. We say we are in the good situation if the two of the f_{i} chosen are multiples of z^{2} and B_{21} ≡ B_{22} ≡ 0 (mod π). Indeed in the good situation, the procedure in (i) reduces us to Case 1 or Case 2.

Suppose that f_{1} and f_{3} are chosen. Dropping the assumption (11) we may assume that f_{1} has repeated factor z^{2}. Then q_{1} has no y^{2} term and by row operations we reach the good situation. The case where f_{2} and f_{3} are chosen is similar. Finally we suppose that f_{1} and f_{2} are chosen. If q_{1} has a factor z, we may assume as above that B_{21} ≡ B_{22} ≡ 0 (mod π). But then q_{2} has a factor z. So if δ = 0, i.e. q_{1} and q_{2} each have a repeated factor, then we reach the good situation. Otherwise we make the assumption (11), and deduce that f_{1} and f_{2} are now multiples of z^{2}. The procedure in (i) multiplies C and the last row by π. The only coefficients not to vanish mod π are now those in the second row of B. It follows that after suitable column operations the level is preserved and we are reduced to Case 2 or Case 3.

Case 5. We assume (1, 2; 1, 2; 1, 1) is admissible for S. The entries of A, B and C have valuations satisfying

\begin{aligned}≥ 3 & ≥ 2 & ≥ 2 \\ ≥ 2 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 0 & ≥ 0 \end{aligned} \begin{aligned}≥ 2 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned} \begin{aligned}≥ 1 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned}

Since S is saturated, we may assume by column operations that v(A_{32}) ≥ 1 and v(A_{33}) = 0. Then v(B_{31}) = v(C_{12}) = v(C_{21}) = 0, otherwise we would be in Case 4. By row and column operations, and subtracting multiples of A from B and C we reduce to the case

\begin{aligned}≥ 3 & ≥ 2 & ≥ 2 \\ ≥ 2 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & = 0 \end{aligned} \begin{aligned}≥ 2 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 0 & ≥ 0 \\ = 0 & ≥ 1 & ≥ 1 \end{aligned} \begin{aligned}≥ 1 & = 0 & ≥ 1 \\ = 0 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1 \end{aligned}

Working mod π we compute

f_{1} = C_{12}z(B_{31}B_{23}y^{2} − A_{33}C_{21}xz)
f_{2} = −A_{33}z(B_{22}C_{21}y^{2} − B_{31}C_{12}xz)
f_{3} = −A_{33}C_{12}yz(B_{22}y + B_{23}z)

If B_{22} ≢ 0 (mod π) and B_{23} ≢ 0 (mod π) then f_{1}, f_{2}, f_{3} each define a curve with a unique singular point at (1 : 0 : 0). If we multiply B, C, the last two rows and the last two columns by π, then the cube is divisible by π^{2}. From this we see that whichever two of the f_{i} are chosen by the procedure in (ii), the level is preserved and we are reduced to Case 2.

If B_{22} ≢ 0 (mod π) and B_{23} ≡ 0 (mod π) then f_{1} and f_{3} have repeated factors but f_{2} does not. The procedure in (i) multiplies C and the middle column by π. Then dividing the first two rows by π preserves the level and reduces us to Case 4 with δ = 0. The observation that δ = 0 is needed to show that at most three iterations are required, as claimed in the statement of the theorem.

If B_{22} ≡ 0 (mod π) and B_{23} ≢ 0 (mod π) then we switch the first two slicings (i.e. A, B, C are replaced by the matrices formed from the first, second, third rows). Then switching the last two columns brings us to the situation considered in the previous paragraph.

Finally, if B_{22} ≡ B_{23} ≡ 0 (mod π) then we are already in Case 2.

Case 6. We assume (2, 3; 1, 2; 1, 2) is admissible for S. The entries of A, B and C have valuations satisfying

\begin{aligned}≥ 4 & ≥ 3 & ≥ 2 \\ ≥ 3 & ≥ 2 & ≥ 1 \\ ≥ 2 & ≥ 1 & ≥ 0 \end{aligned} \begin{aligned}≥ 2 & ≥ 1 & ≥ 0 \\ ≥ 1 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned} \begin{aligned}≥ 1 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \\ ≥ 0 & ≥ 0 & ≥ 0 \end{aligned}

Since S is saturated, we have v(A_{33}) = 0. Then v(B_{22}) = 0, otherwise we would be in Case 3. We also have v(C_{12}) = v(C_{21}) = 0, otherwise we would be in Case 4, and v(B_{13}) = v(B_{31}) = 0 otherwise we would be in Case 5. By row and column operations, and subtracting multiples of A from B and C we reduce to the case

\begin{aligned}≥ 4 & ≥ 3 & ≥ 2 \\ ≥ 3 & ≥ 2 & ≥ 1 \\ ≥ 2 & ≥ 1 & = 0 \end{aligned} \begin{aligned}≥ 2 & ≥ 1 & = 0 \\ ≥ 1 & = 0 & ≥ 0 \\ = 0 & ≥ 0 & ≥ 1 \end{aligned} \begin{aligned}≥ 1 & = 0 & ≥ 1 \\ = 0 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1 \end{aligned}

Working mod π we compute

f_{1} = −B_{31}B_{22}B_{13}y^{3} − C_{12}C_{21}A_{33}xz^{2} + ( ⋯)y^{2}z
f_{2} = A_{33}z(B_{31}C_{12}xz − C_{21}y(B_{22}y + B_{32}z))
f_{3} = A_{33}z(B_{13}C_{21}xz − C_{12}y(B_{22}y + B_{23}z))

We see that f_{1}, f_{2}, f_{3} each define a curve with a unique singular point at (1 : 0 : 0). If we multiply B, C, the last two rows and the last two columns by π, then the cube is divisible by π^{2}. From this we see that whichever two of the f_{i} are chosen by the procedure in (ii), the level is preserved and we are reduced to Case 3. □

4. 2 × 2 × 2 × 2 hypercubes

We consider polynomials in x_{1}, x_{2}, y_{1}, y_{2}, z_{1}, z_{2}, t_{1}, t_{2} that are linear in each of the four sets of variables. Such a polynomial may be represented as

(12)  ∑_{1≤i,j,k,l≤2} H_{ijkl}x_{i}y_{j}z_{k}t_{l}

where H = (H_{ijkl}) is a 2×2×2×2 hypercube. A hypercube H may be partitioned into two 2 × 2 × 2 cubes in four distinct ways:

(i) A_{1} = (H_{1jkl}) and B_{1} = (H_{2jkl})

(ii) A_{2} = (H_{i1kl}) and B_{2} = (H_{i2kl})

(iii) A_{3} = (H_{ij1l}) and B_{3} = (H_{ij2l})

(iv) A_{4} = (H_{ijk1}) and B_{4} = (H_{ijk2}) Let R be a ring. For each 1 ≤ i ≤ 4 there is an action of GL_{2}(R) on the space of hypercubes over R via

(\begin{matrix}r & s \\ t & u \end{matrix}) : (A_{i}, B_{i}) ↦ (rA_{i} + sB_{i}, tA_{i} + uB_{i}).

These actions commute, and so give an action of GL_{2}(R)^{4}. We say that hypercubes are R-equivalent if they belong to the same orbit for this action.

For each 1 ≤ i < j ≤ 4 there is an associated (2, 2)-form F_{ij}. Indeed if we view (12) as a bilinear form in z_{k} and t_{l}, then the determinant of this form is a (2, 2)-form in x_{i} and y_{j}:

F_{12} = ( ∑_{1≤i,j≤2} H_{ij11}x_{i}y_{j})( ∑_{1≤i,j≤2} H_{ij22}x_{i}y_{j}) − ( ∑_{1≤i,j≤2} H_{ij12}x_{i}y_{j})( ∑_{1≤i,j≤2} H_{ij21}x_{i}y_{j}).

The other F_{ij} are defined similarly. If [M_{1}, M_{2}, M_{3}, M_{4}] · H = H′ then the (2, 2)- forms are related by

[det(M_{3}) det(M_{4}), M_{1}, M_{2}] · F_{12} = F′_{12}.

As seen in Section 2, each (2, 2)-form determines a pair of binary quartics. It turns out that the binary quartics in x_{1}, x_{2} associated to F_{12}, F_{13}, F_{14} are all equal. Thus a hypercube H determines four binary quartics G_{1}, …, G_{4}, one in each of the four sets of variables. Each of these binary quartics has the same invariants I and J. Therefore the six (2, 2)-forms F_{ij} all have the same invariants c_{4}, c_{6} and Δ. We define c_{4}(H) = c_{4}(F_{ij}), c_{6}(H) = c_{6}(F_{ij}) and Δ(H) = Δ(F_{ij}).

If H is defined over a field and Δ(H) ≠ 0 then each of the F_{ij} defines a genus one curve in ℙ^{1} × ℙ^{1}. These curves are isomorphic, although not in a canonical way. (See [1, Section 2.3] for further details.) We write C_{H} to denote any one of them.

Ley u and v be the invariants in Lemma 2.1. We find that u(F_{12}) = u(F_{34}) and v(F_{12}) = v(F_{34}). Therefore F_{12} and F_{34} determine isomorphic pairs (E, P). (A further calculation is needed to check this in characteristics 2 and 3, but we omit the details.) Repeating for the other F_{ij} gives a tuple (E, P_{1}, P_{2}, P_{3}) where E is an elliptic curve and 0_{E} ≠ P_{1}, P_{2}, P_{3} ∈ E with P_{1} + P_{2} + P_{3} = 0_{E}.

We say that a hypercube H is integral if it has coefficients in O_{K}, and nonsingular if Δ(H) ≠ 0.

Lemma 4.1. Let H be a non-singular integral hypercube. Let (E, P_{1}, P_{2}, P_{3}) be the tuple determined by H. Then

v(Δ(H)) = v(Δ_{E}) + 12 max(κ(P_{1}), κ(P_{2}), κ(P_{3})) + 12ℓ(H)

where ℓ(H) ≥ 0 is an integer we call the level.

Proof. This is immediate from Lemma 2.2. □

An integral hypercube is saturated if for all 1 ≤ i ≤ 4 the cubes A_{i} and B_{i} are linearly independent mod π. If an integral hypercube is not saturated, then it is obvious how we may decrease the level.

Our algorithm for minimising hypercubes is described by the following theorem.

Theorem 4.2. Let H be a saturated hypercube with associated (2, 2)-forms F_{ij}. Suppose that all of the F_{ij} are non-minimal. Then by an O_{K}-equivalence, and permuting the sets of variables, we are in one of the following two situations:

(i) The reduction of F_{12} mod π defines a curve in ℙ^{1}×ℙ^{1} with a unique singular point at ((1 : 0), (1 : 0)), and the transformation

(13)  [\frac{1}{π} (\begin{matrix}1 & 0 \\ 0 & π \end{matrix}) , (\begin{matrix}1 & 0 \\ 0 & π \end{matrix}) , (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) , (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix})]

gives an integral hypercube of the same level. (ii) We have F_{12} ≡ x_{2}^{2}y_{2}^{2} (mod π) and the transformation (13) gives a non-

saturated hypercube of the same level. Moreover, at most two iterations of the procedure in (i) are needed to give a nonsaturated hypercube, or to reach the situation in (ii).

We initially used the methods in Sections 2 and 3 to prove Theorem 4.2 under the hypothesis that H is non-minimal. The advantage of the theorem as stated here is that it has the following consequence.

Corollary 4.3. Let H be a integral hypercube with associated (2, 2)-forms F_{ij}. Then H is minimal if and only if some F_{ij} is minimal.

Remark 4.4. We may represent H = (H_{ijkl}) as a 4 × 4 matrix:

(14)  ( \begin{matrix}H_{1111} & H_{1211} & H_{1112} & H_{1212} \\ H_{2111} & H_{2211} & H_{2112} & H_{2212} \\ H_{1121} & H_{1221} & H_{1122} & H_{1222} \\ H_{2121} & H_{2221} & H_{2122} & H_{2222} \end{matrix} ) .

If we write r_{1}, r_{2}, r_{3}, r_{4} for the rows, then the first copy of GL_{2} acts by row operations simultaneously on {r_{1}, r_{2}} and {r_{3}, r_{4}}, the third copy of GL_{2} acts by row operations on {r_{1}, r_{3}} and {r_{2}, r_{4}}, and the other two copies of GL_{2} act by column operations.

Remark 4.5. Let H be an integral hypercube with associated binary quartics G_{1}, …, G_{4}. It is clear that if any of the G_{i} are minimal then H is minimal. However the converse is not true. For example if

H ≡ ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} ) (mod π^{2})

then H is minimal (since F_{12} ≡ (x_{1}y_{1}+x_{2}y_{2})^{2} (mod π^{2}) and we saw in Remark 2.4 that this is minimal), yet we have G_{1} ≡ … ≡ G_{4} ≡ 0 (mod π^{2}).

For the proof of Theorem 4.2 we need the following lemma.

Lemma 4.6. Let H be an integral hypercube. Suppose that at least one of the associated (2, 2)-forms F_{ij} is non-minimal. Then by an O_{K}-equivalence, and permuting the sets of variables, we may assume H_{11kl} ≡ 0 (mod π) for all 1 ≤ k, l ≤ 2.

Proof. We suppose that F_{12} is non-minimal. If the reduction of F_{12} mod π is non-zero, then by Theorem 2.3 it defines a curve in ℙ^{1} × ℙ^{1} with singular locus a point, a line or a pair of lines. We may assume by an O_{K}-equivalence that the curve is singular at ((1 : 0), (1 : 0)). If H_{11kl} ≢ 0 (mod π) for some 1 ≤ k, l ≤ 2 then we may assume by an O_{K}-equivalence that H_{1111} ≢ 0 (mod π). A further O_{K}-equivalence gives

H_{2111} ≡ H_{1211} ≡ H_{1121} ≡ H_{1112} ≡ 0 (mod π).

Since the coefficients of x_{1}^{2}y_{1}^{2}, x_{1}^{2}y_{1}y_{2} and x_{1}x_{2}y_{1}^{2} in F_{12} vanish mod π, we have

H_{1122} ≡ H_{1222} ≡ H_{2122} ≡ 0 (mod π).

Lemma 2.6(i) now shows that either

H_{1221}H_{1212} ≡ 0 (mod π) or H_{2121}H_{2112} ≡ 0 (mod π).

By switching the first two sets of variables and switching the last two sets of variables, as necessary, we may assume H_{1212} ≡ 0 (mod π). Now H_{1jk2} ≡ 0 (mod π) for all 1 ≤ j, k ≤ 2, and this proves the lemma. □

Proof of Theorem 4.2. By Lemma 4.6 we may assume H_{11kl} ≡ 0 (mod π) for all 1 ≤ k, l ≤ 2. Applying Lemma 2.6(i) to F_{12}, and switching the first two sets of variables if necessary, we have

H_{1211}H_{1222} − H_{1212}H_{1221} ≡ 0 (mod π).

By an O_{K}-equivalence we may assume H_{1jkl} ≡ 0 (mod π) for all 1 ≤ j, k, l ≤ 2, except (j, k, l) = (2, 2, 2). Since H is saturated we have H_{1222} ≢ 0 (mod π). Again by Lemma 2.6(i) we have H_{2111} ≡ 0 (mod π).

We now split into cases, according as to whether

(15)  H_{2211} ≡ H_{2121} ≡ H_{2112} ≡ 0 (mod π).

If this condition is not satisfied, then by permuting the last three sets of variables, we may suppose H_{2211} ≢ 0 (mod π). By an O_{K}-equivalence we have

(16)  H ≡ ( \begin{matrix}0 & 0 & 0 & 0 \\ 0 & 1 & β & 0 \\ 0 & 0 & 0 & 1 \\ α & 0 & γ & 0 \end{matrix} ) (mod π)

for some α, β, γ ∈ k. We compute F_{12} ≡ x_{1}x_{2}y_{2}^{2} + x_{2}^{2}(αβy_{1}^{2} + γy_{1}y_{2}) (mod π). The conclusions in (i) are satisfied unless αβ = γ = 0. In the remaining case we may assume, by switching the last two sets of variables if necessary, that α = 0. Now switching the first and last sets of variables, and swapping over the third set of variables (i.e. z_{1} ↔ z_{2}), we may swap over β and γ. Therefore β = γ = 0, and this contradicts that H is saturated.

Now suppose the condition (15) is satisfied. Then by an O_{K}-equivalence (and our assumption that H is saturated) we have

(17)  H ≡ ( \begin{matrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & −1 & −1 & 0 \end{matrix} ) (mod π).

We compute F_{12} ≡ x_{2}^{2}y_{2}^{2} (mod π). Let F_{12} have coefficients a_{ij} as labelled in (3). Lemma 2.6(ii) shows that either v(a_{12}) ≥ 2 or v(a_{21}) ≥ 2. Therefore v(H_{1111}) ≥ 2. Again by Lemma 2.6(ii) we have either v(a_{11}) ≥ 3, v(a_{13}) ≥ 2 or v(a_{31}) ≥ 2. Therefore at least one of the coefficients H_{2111}, H_{1211}, H_{1121}, H_{1112} has valuation at least two. By permuting the sets of variables we may suppose v(H_{1112}) ≥ 2. The conclusions in (ii) are now satisfied.

To prove the last part of the theorem, we need the following lemma.

Lemma 4.7. Let H be a hypercube over a field k with associated (2, 2)-forms F_{ij}. We write

F_{12} = f_{1}(x_{1}, x_{2})y_{1}^{2} + f_{2}(x_{1}, x_{2})y_{1}y_{2} + f_{3}(x_{1}, x_{2})y_{2}^{2}
F_{13} = g_{1}(x_{1}, x_{2})z_{1}^{2} + g_{2}(x_{1}, x_{2})z_{1}z_{2} + g_{3}(x_{1}, x_{2})z_{2}^{2}

(i) We have g_{2} = f_{2} + 2h and g_{1}g_{3} = f_{1}f_{3} + f_{2}h + h^{2} for some h ∈ k[x_{1}, x_{2}].

(ii) If f_{1} = f_{2} = 0 and g_{1}, g_{2} are multiples of x_{2}^{2}, then F_{13} is either zero or factors as a product of binary quadratic forms.

Proof. (i) We have already remarked that f_{2}^{2} − 4f_{1}f_{3} = g_{2}^{2} − 4g_{1}g_{3}. The result follows by considering the f_{i} and g_{i} as polynomials in ℤ[H_{ijkl}][x_{1}, x_{2}].

(ii) By (i) we have g_{1} = αx_{2}^{2}, g_{2} = 2βx_{2}^{2} and αx_{2}^{2}g_{3}(x_{1}, x_{2}) = βx_{2}^{4}. If α = 0 then g_{1} = g_{2} = 0, whereas if α ≠ 0 then g_{1}, g_{2}, g_{3} are multiples of x_{2}^{2}. □

We say that a (2, 2)-form F is slender if F mod π is either zero, or factors as a product of binary quadratic forms. Theorem 2.3 shows that if F is non-minimal then either F mod π defines a curve with a unique singular point, or F is slender. These possibilities are mutually exclusive.

We now complete the proof of Theorem 4.2. Applying the transformation in (i) to H has the effect of applying the transformation in Theorem 2.3(iii) to F_{12}. The last sentence of Theorem 2.3 tells us that, after applying this transformation, F_{12} mod π is either zero, or factors as a product of binary quadratic forms both of which have a repeated root. In particular F_{12} is slender.

We claim that F_{13} is slender. If not then F_{13} mod π defines a curve with a unique singular point. By an O_{K}-equivalence we may assume that this point is ((1 : 0), (1 : 0)), and that F_{12} ≡ f_{3}(x_{1}, x_{2})y_{2}^{2} (mod π) for some binary quadratic form f_{3}. Lemmas 2.6(i) and 4.7(ii) now show that F_{13} is slender.

The same argument shows that all of the F_{ij} are slender, except possibly F_{34}. Since F_{34} was unchanged by the transformation (13), it follows that after at most two iterations, all of the F_{ij} are slender. In particular we cannot return to the situation in (i), and this completes the proof. □

5. Minimisation Theorems

The algorithms in [3] and [4] for minimising genus one curves of degree 2, 3, 4, 5 were complemented by a more theoretical result. This stated that if a genus one curve is soluble over K (or more generally over an unramified extension) then the discriminant of a minimal model is the same as that for the Jacobian elliptic curve. In this section we prove the analogue of this result for (2, 2)-forms, 3 × 3 × 3 cubes and 2 × 2 × 2 × 2 hypercubes.

In earlier papers, most notably [2, Lemmas 3,4,5], the minimisation algorithms and minimisation theorems were treated together. Following [3] we separate these out, and this leads to clean results that work the same in all residue characteristics. We phrase our result in terms of the level, as defined in Lemmas 2.2, 3.2 and 4.1.

Theorem 5.1. Let Φ be a nonsingular (2, 2)-form, 3 × 3 × 3 cube, or 2 × 2 × 2 × 2 hypercube defined over K. If C_{Φ}(K) ≠ ∅ then Φ has minimal level 0.

Remark 5.2. The algorithms in Sections 2, 3 and 4 show that the minimal level is unchanged by an unramified field extension. The hypothesis in Theorem 5.1 may therefore be weakened to solubility over an unramified field extension. We give examples below to show that this hypothesis cannot be removed entirely.

Let E/K be an elliptic curve and n ∈ {2, 3}. Let D and D′ be K-rational divisors on E of degree n. The image of E in ℙ^{n−1} × ℙ^{n−1} via |D| × |D′| is defined by a (2, 2)-form in the case n = 2, and three bilinear forms in the case n = 3. The coefficients of the latter give a 3 × 3 × 3 cube. We note that the (2, 2)-form, respectively 3 × 3 × 3 cube, is uniquely determined up to K-equivalence by the triple (E, [D], [D′]), where [D] denotes the linear equivalence class of D. Moreover every (2, 2)-form, respectively 3 × 3 × 3 cube, defining a non-singular genus one curve with a K-rational point, arises in this way. Therefore the first two cases of Theorem 5.1 are immediate from the following theorem.

We write sum : Div_{K}(E) → E(K) for the map that sends a formal sum of points to its sum using the group law on E.

Theorem 5.3. Let E/K be an elliptic curve with integral Weierstrass equation

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

and let P = (0, 0) ∈ E(K). Let D, D′ ∈ Div_{K}(E) be divisors of degree n ∈ {2, 3} with sum(D′ − D) = P. Then (E, [D], [D′]) may be represented by an integral (2, 2)-form, or 3 × 3 × 3 cube, with the same discriminant as (18).

We start by proving Theorem 5.3 in the case D ∼ n.0_{E}. Since sum(D′ −D) = P we have D′ ∼ (n − 1).0_{E} + P. We put

f = \frac{y + a_{1}x + a_{3}}{x} = \frac{x^{2} + a_{2}x + a_{4}}{y}

and split into the cases n = 2 and n = 3.

Case n = 2. The embedding E → ℙ^{1} × ℙ^{1} via |D| × |D′| is given by

(x, y) ↦ ((1 : x), (1 : f)).

The image is defined by the (2, 2)-form

F (x_{1}, x_{2}; y_{1}, y_{2}) = x_{2}^{2}y_{1}^{2} − x_{1}x_{2}y_{2}^{2} + x_{1}y_{1}(a_{1}x_{2}y_{2} + a_{2}x_{2}y_{1} + a_{3}x_{1}y_{2} + a_{4}x_{1}y_{1}),

with the same discriminant as (18).

Case n = 3. The embedding E → ℙ^{2} × ℙ^{2} via |D| × |D′| is given by

(x, y) ↦ ((1 : x : y), (1 : x : f)).

The image is defined by bilinear forms

B_{1}(x_{1}, x_{2}, x_{3}; y_{1}, y_{2}, y_{3}) = x_{2}y_{1} − x_{1}y_{2},
B_{2}(x_{1}, x_{2}, x_{3}; y_{1}, y_{2}, y_{3}) = x_{3}y_{1} + a_{1}x_{2}y_{1} + a_{3}x_{1}y_{1} − x_{2}y_{3},
B_{3}(x_{1}, x_{2}, x_{3}; y_{1}, y_{2}, y_{3}) = x_{2}y_{2} + a_{2}x_{2}y_{1} + a_{4}x_{1}y_{1} − x_{3}y_{3}.

The coefficients of B_{1}, B_{2}, B_{3} give a 3 × 3 × 3 cube, and this has the same discriminant as (18).

Lemma 5.4. Let S be a 3 × 3 × 3 cube corresponding to bilinear forms B_{1}, B_{2}, B_{3}, defining C ⊂ ℙ^{2} × ℙ^{2} a smooth curve of genus one, embedded via |D| × |D′|.

(i) If Q = ((0 : 0 : 1), (0 : 0 : 1)) ∈ C(K) then for i = 1, 2, 3 we can write

B_{i} = L_{i}(y_{1}, y_{2})x_{3} + M_{i}(x_{1}, x_{2})y_{3} + N_{i}(x_{1}, x_{2}; y_{1}, y_{2}).

(ii) The image of C in ℙ^{1}×ℙ^{1} via |D−Q|×|D′−Q| is defined by the (2, 2)-form

F (x_{1}, x_{2}; y_{1}, y_{2}) = |\begin{matrix}L_{1} & M_{1} & N_{1} \\ L_{2} & M_{2} & N_{2} \\ L_{3} & M_{3} & N_{3} \end{matrix}| .

(iii) We have Δ(F) = Δ(S).

Proof. We map C → ℙ^{1} × ℙ^{1} via ((x_{1} : x_{2}), (y_{1} : y_{2})). The first two statements are clear. For (iii) we checked by a generic calculation that F and S have the same invariants c_{4} and c_{6}. □

Lemma 5.5. Let F be a (2, 2)-form defining C ⊂ ℙ^{1} ×ℙ^{1} a smooth curve of genus one, embedded via |D| × |D′|.

(i) If Q = ((1 : 0), (1 : 0)) ∈ C(K) then we can write

F (x_{1}, x_{2}; y_{1}, y_{2}) = (x_{1}^{2} x_{1}x_{2} x_{2}^{2}) (\begin{matrix}0 & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}) (\begin{matrix}y_{1}^{2} \\ y_{1}y_{2} \\ y_{2}^{2} \end{matrix}) .

(ii) The image of C in ℙ^{2} ×ℙ^{2} via |D +Q|×|D′ +Q| is defined by the 3×3×3 cube S with entries

(\begin{matrix}0 & 1 & 0 \\ 1 & a_{22} & a_{23} \\ 0 & a_{32} & a_{33} \end{matrix})  (\begin{matrix}0 & 0 & 0 \\ 0 & a_{12} & a_{13} \\ −1 & 0 & 0 \end{matrix})  (\begin{matrix}0 & 0 & −1 \\ 0 & a_{21} & 0 \\ 0 & a_{31} & 0 \end{matrix}) .

(iii) We have Δ(S) = Δ(F).

Proof. We have D ∼ Q + R and D′ ∼ Q + R′ where R = ((1 : 0), (−a_{13} : a_{12})) and R′ = ((−a_{31} : a_{21}), (1 : 0)). Choosing bases for the space of bilinear forms vanishing at R′, and the space of bilinear forms vanishing at R, we find that the map C → ℙ^{2} × ℙ^{2} via |D + Q| × |D′ + Q| is given by

((x_{1} : x_{2}), (y_{1} : y_{2})) ↦ (((a_{21}x_{1} + a_{31}x_{2})y_{1} : x_{1}y_{2} : x_{2}y_{2}),
(x_{1}(a_{12}y_{1} + a_{13}y_{2}) : x_{2}y_{1} : x_{2}y_{2})).

The image is defined by

B_{1} = x_{2}y_{1} + x_{1}y_{2} + a_{22}x_{2}y_{2} + a_{32}x_{3}y_{2} + a_{23}x_{2}y_{3} + a_{33}x_{3}y_{3}
B_{2} = −x_{3}y_{1} + a_{12}x_{2}y_{2} + a_{13}x_{2}y_{3},
B_{3} = −x_{1}y_{3} + a_{21}x_{2}y_{2} + a_{31}x_{3}y_{2}.

The coefficients of these forms give the cube S in the statement of the lemma. Again we prove (iii) by a generic calculation. □

Proof of Theorem 5.3. We split into the cases n = 2 and n = 3.

Case n = 2. We have D ∼ 3.0_{E} − Q for some Q ∈ E(K). By the special case of the theorem already established, there is an integral 3 × 3 × 3 cube representing (E, [D + Q], [D′ + Q]), with the same discriminant as (18). We have E ⊂ ℙ^{2} × ℙ^{2}. Since SL_{3}(O_{K}) acts transitively on ℙ^{2}(K) we may assume Q = ((0 : 0 : 1), (0 : 0 : 1)). Then Lemma 5.4 give an integral (2, 2)-form representing (E, [D], [D′]), with the same discriminant as (18).

Case n = 3. We have D ∼ 2.0_{E} + Q for some Q ∈ E(K). By the special case of the theorem already established, there is an integral (2, 2)-form representing (E, [D − Q], [D′ − Q]), with the same discriminant as (18). We have E ⊂ ℙ^{1} × ℙ^{1}. Since SL_{2}(O_{K}) acts transitively on ℙ^{1}(K) we may assume Q = ((1 : 0), (1 : 0)). Then Lemma 5.5 give an integral 3 × 3 × 3 cube representing (E, [D], [D′]), with the same discriminant as (18). □

This completes the proof of Theorem 5.1 for (2, 2)-forms and 3 × 3 × 3 cubes. We now deduce the result for hypercubes from the result for (2, 2)-forms. Let H be a non-singular hypercube over K, with associated (2, 2)-forms F_{ij}. The genus one curve C_{H} is that defined by any of the F_{ij}. So if C_{H}(K) ≠ 0 then the result for (2, 2)-forms shows that each F_{ij} has minimal level 0. By the definitions in Lemmas 2.2 and 4.1, we have ℓ(H) = min ℓ(F_{ij}). It follows by Corollary 4.3 that H has minimal level 0.

Remark 5.6. We give some examples to show that the minimal level can be positive. We assume for convenience that char(k) ≠ 2, 3. A binary quartic, or ternary cubic is called critical (see [3, Section 5]) if the valuations of its coefficients satisfy

= 1 ≥ 2 ≥ 2 ≥ 3 = 3  or  \begin{aligned}= 2 \\ ≥ 2 & ≥ 2 \\ ≥ 1 & ≥ 1 & ≥ 2 \\ = 0 & ≥ 1 & ≥ 1 & = 1 \end{aligned}

We now define a critical (2, 2)-form, 3 × 3 × 3 cube or 2 × 2 × 2 × 2 hypercube, to be one whose coefficients have valuations satisfying

\begin{aligned}= 2 & ≥ 2 & = 1 \\ ≥ 2 & ≥ 1 & ≥ 1 \\ = 1 & ≥ 1 & = 0 \end{aligned}

or

\begin{aligned}≥ 2 & = 1 & ≥ 1 \\ = 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & = 0 \end{aligned} \begin{aligned}= 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & = 0 \\ ≥ 1 & = 0 & ≥ 0 \end{aligned} \begin{aligned}≥ 1 & ≥ 1 & = 0 \\ ≥ 1 & = 0 & ≥ 0 \\ = 0 & ≥ 0 & ≥ 0 \end{aligned}

or

\begin{aligned}≥ 2 & = 1 & = 1 & ≥ 1 \\ = 1 & ≥ 1 & ≥ 1 & = 0 \\ = 1 & ≥ 1 & ≥ 1 & = 0 \\ ≥ 1 & = 0 & = 0 & ≥ 0 \end{aligned}

Either by using our algorithms, or observing that the corresponding binary quartics and ternary cubics are critical, we see that any such model Φ is minimal. However by applying the transformation

[π^{−2}, A_{2}, A_{2}], [π^{−4/3}A_{3}, A_{3}, A_{3}] or [π^{−3/2}A_{2}, A_{2}, A_{2}, A_{2}],

where

A_{2} = (\begin{matrix}1 \\ & π^{1/2} \end{matrix}) and A_{3} = (\begin{matrix}1 \\ & π^{1/3} \\ & & π^{2/3} \end{matrix}) ,

we see that I(Φ) ≡ 0 (mod π^{p}) for any invariant I of weight p. Therefore Φ has positive level.

References

[1] M. Bhargava and W. Ho, Coregular spaces and genus one curves, Camb. J. Math. 4 (2016), no. 1, 1–119.

[2] B.J. Birch and H.P.F. Swinnerton-Dyer, Notes on elliptic curves I, J. reine angew. Math. 212 (1963) 7–25.

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

[4] T.A. Fisher, Minimisation and reduction of 5-coverings of elliptic curves, Algebra & Number Theory 7 (2013), no. 5, 1179–1205.

[5] T.A. Fisher and R.D. Newton, Computing the Cassels-Tate pairing on the 3-Selmer group of an elliptic curve, Int. J. Number Theory 10 (2014), no. 7, 1881–1907.

[6] J.H. Silverman, The arithmetic of elliptic curves, Graduate Texts in Mathematics 106, Springer-Verlag, New York, 1986.

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

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