On ternary cubic forms that determine central simple algebras of degree 3

Mélanie Raczek

November 13, 2008

Abstract Fixing a field F of characteristic different from 2 and 3, we consider pairs (A, V) consisting of a degree 3 central simple F -algebra A and a 3-dimensional vector subspace V of the reduced trace zero elements of A which is totally isotropic for the trace quadratic form. Each such pair gives rise to a cubic form mapping an element of V to its cube; therefore we call it a cubic pair over F. Using the Okubo product in the case where F contains a primitive cube root of unity, and extending scalars otherwise, we give an explicit description of all isomorphism classes of such pairs over F. We deduce that a cubic form associated with an algebra in this manner determines the algebra up to (anti-)isomorphism.

Introduction

Consider a field F of characteristic different from 2. Let A be a quaternion algebra over F and let A^{0} denote the subspace of reduced trace zero elements of A. Then for all x ∈ A^{0} we have x^{2} ∈ F. We thus obtain a quadratic form on A^{0} mapping x to x^{2}. Up to the sign, this quadratic form is the norm form of the quaternion algebra restricted to A^{0}. By Theorem 2.5, p. 57, in [Lam, 2005], this quadratic form determines the quaternion algebra up to isomorphism.

In this paper we shall generalize this construction for algebras of degree 3. Consider a field F of characteristic different from 2 and 3 and let A be a degree 3 central simple algebra over F. Again let A^{0} denote the subspace of reduced trace zero elements of A. Then the cube of an arbitrary element x ∈ A^{0} need not be in F in general. In fact, it is in F if and only if the reduced trace of x^{2} is equal to zero. Let q: A^{0} → F be the trace quadratic form on A^{0} (mapping x to the reduced trace of x^{2}). Then the Witt index of q is equal to 4 if F contains a primitive cube root of unity and is equal to 3 otherwise (see Lemma 0.1). In both cases there exist 3-dimensional subspaces of A^{0} which are totally isotropic for the trace quadratic form. Each such vector subspace V ⊂ A^{0} gives rise to a cubic form. In this paper we shall prove that this cubic form determines the algebra up to isomorphism or anti-isomorphism.

In the first two sections we shall give an explicit description of the pairs (A, V) where A and V are as above. In the first section we assume that the field F contains a primitive cube root of unity and we use the fact that we may write V in terms of the Okubo product. In the second section we assume that F does not contain a primitive cube root of unity, and we shall minimally extend the field F to use the results of the previous case. In the last section we use these descriptions to prove that a cubic form associated with a pair (A, V) determines A up to (anti-)isomorphism.

Throughout the paper, we denote by F a field of characteristic different from 2 and 3, by F_{s} a separable closure of F, and by Γ the absolute Galois group Gal(F_{s}/F). We fix ω ∈ F_{s} a primitive cube root of unity. We say that a pair (A, V) is a cubic pair over F if A is a degree 3 central simple F -algebra and V is a 3-dimensional subspace of A^{0} (= the subspace of reduced trace zero elements of A) which is totally isotropic for the trace quadratic form. For a cubic pair (A, V) over F we define a cubic form

f_{A,V} : V → F: x ↦ x^{3}.

We say that Θ: (A, V) → (B, W) is an isomorphism of cubic pairs over F if Θ: A → B is an F -algebra isomorphism such that Θ(V) = W. Note that if (A, V) and (B, W) are isomorphic then f_{A,V} and f_{B,W} are isometric (i.e. there exists an F -vector space isomorphism Θ: V → W such that f_{A,V} = f_{B,W} ∘ Θ). For a field extension L over F we write A_{L} (resp. V_{L}) for A⊗_{F} L (resp. V ⊗_{F} L). Further we let Trd_{A} denote the reduced trace of A, and for an F -algebra K, we denote by Tr_{K} (resp. N_{K}) the trace (resp. the norm) of K.

Acknowledgement. Most results in this paper were already found in my PhD thesis (see [Raczek, 2007]). I would like to thank Jean-Pierre Tignol, my thesis supervisor, for his help during this work.

0. Some results on quadratic forms

Before we start the classification of cubic pairs, we need preliminary results on quadratic forms.

Let A be a degree 3 central simple algebra over F. First we shall compute the Witt index of the trace quadratic form of A.

Lemma 0.1 Let q be the trace quadratic form on A^{0}. Then the Witt index of q is equal to 4 if F contains a primitive cube root of unity and is equal to 3 otherwise.

Proof : There exists a splitting field L of A of odd degree over F. Indeed, we may choose L := F if A is split and we choose a splitting field of degree 3 over F otherwise. Then straightforward computations show that the class of the form q_{L}: A_{L}^{0} → L is equal to the class of 2⟨1, 3⟩ in the Witt ring of L. Hence, by Springer’s Theorem about odd degree extensions (see Theorem 2.7, p. 194, in [Lam, 2005]), the Witt index of q is greater than or equal to 3 and it is 4 if and only if F contains a primitive cube root of unity. □

Suppose that F contains a primitive cube root of unity.

Lemma 0.2 Let V be a 3-dimensional totally isotropic subspace of A^{0}. Then there exist exactly two maximal totally isotropic subspaces W_{1}, W_{2} of A^{0} containing V; thus V = W_{1} ∩ W_{2}.

Proof : A more general fact is proved in III.1.11 of [Chevalley, 1997]: in a quadratic space with a Witt index equal to n, all (n − 1)-dimensional totally isotropic subspaces are contained in exactly two maximal totally isotropic subspaces. □

1. Classification of cubic pairs over a field with a primitive cube root of unity

We assume that F contains a primitive cube root of unity.

1.1. Okubo product

Let A be a degree 3 central simple F -algebra. In [Knus et al., 1998] the Okubo product over A^{0} is defined as follows:

x ⋆ y := μxy + (1 − μ)yx − \frac{1}{3}Trd_{A}(xy) = \frac{1}{1 − ω} (yx − ωxy) − \frac{1}{3}Trd_{A}(xy)

where μ := \frac{1−ω}{3} . Let q denote the trace quadratic form on A^{0}. Because F contains a primitive cube root of unity, the Witt index of q is equal to 4. In [2008], Matzri interprets the results of van der Blij and Springer [1960] on triality, in the language of the Okubo product: he gives a description of the 4-dimensional subspaces of A which are totally isotropic for q, in terms of the Okubo product.

Theorem 1.1 (Matzri) Let u ∈ A^{0} \ {0} be such that Trd_{A}(u^{2}) = 0. Then u ⋆ A^{0} and A^{0} ⋆ u are 4-dimensional totally isotropic subspaces of A^{0}. Moreover any 4-dimensional totally isotropic subspace is of this form.

We may also write the 3-dimensional totally isotropic subspaces of A^{0} in terms of the Okubo product. By Lemma 0.2, the 3-dimensional totally isotropic subspaces of A^{0} are the intersections of two subspaces of the form u ⋆ A^{0} or A^{0} ⋆ u. We can be more precise using the following:

Theorem 1.2 (Matzri) Let u, v ∈ A^{0} \ {0} be such that Trd_{A}(u^{2}) = 0 and Trd_{A}(v^{2}) = 0. Then

1. the dimension of u ⋆ A^{0} ∩ v ⋆ A^{0} is even;

2. if u ⋆ v ≠ 0, then dim(u ⋆ A^{0} ∩ A^{0} ⋆ v) = 1;

3. if u ⋆ v = 0, then dim(u ⋆ A^{0} ∩ A^{0} ⋆ v) = 3.

Note that the Okubo product depends on the choice of the primitive cube root of unity. If we set, for a primitive cube root of unity ρ,

x ⋆_{ρ} y := μ_{ρ}xy + (1 − μ_{ρ})yx − \frac{1}{3}Trd_{A}(xy)

where μ_{ρ} := \frac{1−ρ}{3}, then x ⋆_{ω} y = y ⋆_{ω^{2}} x. So by Theorem 1.2, the dimension of A^{0} ⋆ u ∩ A^{0} ⋆ v is also even.

Corollary 1.3 Let (A, V) be a cubic pair over F. Then there exist nonzero u, v ∈ A^{0} with Trd_{A}(u^{2}) = 0, Trd_{A}(v^{2}) = 0 and u ⋆ v = 0 such that V = u ⋆ A^{0} ∩ A^{0} ⋆ v. □

The vectors u and v are in fact uniquely determined up to scalars. We shall prove this as a part of a more general statement:

Lemma 1.4 If (B, W) is another cubic pair over F with W = r ⋆ B^{0} ∩ B^{0} ⋆ s as in Corollary 1.3, then an F -algebra isomorphism Θ: A → B induces an isomorphism Θ: (A, V) → (B, W) of cubic pairs if and only if Θ(u)F = rF and Θ(v)F = sF.

Proof : Assume that Θ: A → B is an F -algebra isomorphism. If Θ(V) = W, then W = Θ(u) ⋆ B^{0} ∩ B^{0} ⋆ Θ(v) = r ⋆ B^{0} ∩ B^{0} ⋆ s. By Lemma 0.2, we have

{Θ(u) ⋆ B^{0}, B^{0} ⋆ Θ(v)} = {r ⋆ B^{0}, B^{0} ⋆ s}.

If r ⋆ B^{0} = B^{0} ⋆ Θ(v), then W = Θ(u) ⋆ B^{0} ∩ r ⋆ B^{0} and by Theorem 1.2, the dimension of W is even. Hence Θ(u) ⋆ B^{0} = r ⋆ B^{0} and B^{0} ⋆ Θ(v) = B^{0} ⋆ s. By Theorem 2.10 in [Matzri, 2008], we then have Θ(u)F = rF and Θ(v)F = sF. The converse is obvious. □

1.2. Classification

We shall describe a cubic pair (A, V) over F up to isomorphism. By Corollary 1.3, there exist nonzero u, v ∈ A^{0} such that Trd_{A}(u^{2}) = 0, Trd_{A}(v^{2}) = 0, u ⋆ v = 0 and V = u ⋆ A^{0} ∩ A^{0} ⋆ v. Since Trd_{A}(v) = Trd_{A}(v^{2}) = 0 we have v^{3} ∈ F, and similarly u^{3} ∈ F. Note that u ⋆ v = 0 implies that

vu = \frac{1 − ω}{3} Trd_{A}(uv) + ωuv.

Set t := uv − \frac{1}{3}Trd_{A}(uv). Then

tu = uvu − \frac{1}{3}Trd_{A}(uv)u = \frac{1 − ω}{3} Trd_{A}(uv)u + ωu^{2}v − \frac{1}{3}Trd_{A}(uv)u = ωut

and similarly vt = ωtv. We deduce that t^{3} ∈ F. Indeed,

t^{2} = t(uv − \frac{1}{3}Trd_{A}(uv))
= ωutv − \frac{1}{3}Trd_{A}(uv)t
= ωu^{2}v^{2} + \frac{ω^{2}}{3} Trd_{A}(uv)uv + \frac{1}{9}Trd_{A}(uv)^{2}

and thus

t^{3} = t^{2}(uv − \frac{1}{3}Trd_{A}(uv))
= ω^{2}ut^{2}v − \frac{1}{3}Trd_{A}(uv)t^{2}
= u^{3}v^{3} − \frac{1}{27}Trd_{A}(uv)^{3} ∈ F.  (1)

This implies in particular that Trd_{A}(t^{2}) = 0, so Trd_{A}(u^{2}v^{2}) = \frac{1}{3}Trd_{A}(uv)^{2}. We shall prove that t^{2}, t^{2}u, t^{2}v ∈ V. First we observe that

u ⋆ A^{0} = {x ∈ A^{0} | x ⋆ u = 0}.

Indeed, by Proposition (34.19) in [Knus et al., 1998] we have (u ⋆ x) ⋆ u = \frac{1}{6}Trd_{A}(u^{2})x = 0 for all x ∈ A^{0}. Hence

u ⋆ A^{0} ⊂ {x ∈ A^{0} | x ⋆ u = 0} = ker(R_{u})

where R_{u}: A^{0} → A^{0} : x ↦ x ⋆ u. But dim ker(R_{u}) + dim im(R_{u}) = 8, thus dim ker(R_{u}) = dim(u ⋆ A^{0}) = 4. Similarly,

A^{0} ⋆ v = {x ∈ A^{0} | v ⋆ x = 0}.

One can see that t^{2} ∈ V since ut^{2} = ωt^{2}u and t^{2}v = ωvt^{2}. Also u(ut^{2}) = ω(ut^{2})u and (vt^{2})v = ωv(vt^{2}) imply (ut^{2}) ⋆ u = 0 and v ⋆ (vt^{2}) = 0. Now

v ⋆ (ut^{2}) = \frac{1}{1 − ω} (ut^{2}v − ωvut^{2}) − \frac{1}{3}Trd_{A}(ut^{2}v)
= \frac{1}{1 − ω} (ut^{2}v − ω \frac{1 − ω}{3} Trd_{A}(uv)t^{2} − ω^{2}uvt^{2}) − \frac{1}{3}Trd_{A}(ut^{2}v)
= ut^{2}v − \frac{ω}{3} Trd_{A}(uv)t^{2} − \frac{1}{3}Trd_{A}(ut^{2}v)
= \frac{ω^{2}}{27} Trd_{A}(uv)^{3} − \frac{ω^{2}}{9} Trd_{A}(uv)Trd_{A}(u^{2}v^{2})
= 0.

Since (vt^{2})⋆u = ω^{2}v ⋆(ut^{2}) = 0, we obtain that ut^{2}, vt^{2} ∈ V. So t^{2}, t^{2}u, t^{2}v ∈ V because vt = ωtv and tu = ωut.

To work out the classification of cubic pairs we shall distinguish different situations:

First case: We assume that u, v, t ∈ A^{×}. Observe that t = \frac{1}{1−ω} (uv − vu) ≠ 0, hence u and v are linearly independent. Since Trd_{A}(u) = Trd_{A}(v) = 0, the vectors 1, u, v are also linearly independent. Therefore t^{2}, t^{2}u, t^{2}v span V. Set ξ := t^{2}, η := t^{2}v and λ := \frac{1}{3t^{3}} Trd_{A}(uv). Because t = uv − \frac{1}{3}Trd_{A}(uv), we have u = tv^{−1} + \frac{1}{3}Trd_{A}(uv)v^{−1}. One can check that

t^{2}u = \frac{ω}{v^{3}t^{3}} (ξη^{2} + λξ^{2}η^{2}).

Finally A is the symbol algebra (a, b)_{ω,F} generated by ξ and η such that ξ^{3} = a, η^{3} = b, ξη = ωηξ, and V is the vector subspace spanned by ξ, η and ξη^{2} +λξ^{2}η^{2} where 1 + λ^{3}a ≠ 0 since u^{3} ≠ 0. In this basis of V, the cubic form f_{A,V} takes the generalized Hesse normal form:

f_{A,V} (xξ + yη + z(ξη^{2} + λξ^{2}η^{2})) = ax^{3} + by^{3} + ab^{2}(1 + λ^{3}a)z^{3} − 3ω^{2}abλxyz.

The form f_{A,V} is nonsingular.

Conversely, suppose that B is the symbol algebra (a, b)_{ω,F} generated by ξ, η such that ξ^{3} = a, η^{3} = b, ξη = ωηξ, and W the vector subspace spanned by ξ, η, ξη^{2} + λξ^{2}η^{2}, for some a, b ∈ F^{×} and λ ∈ F such that 1 + λ^{3}a ≠ 0. Then one can check that (B, W) is a cubic pair over F such that f_{B,W} is nonsingular.

Second case: We suppose that u, v ∈ A^{×} and t = 0. Then uv = \frac{1}{3}Trd_{A}(uv) ∈ F^{×}. Thus we may assume that u = v^{2}. We need the following:

Lemma 1.5 Let ξ ∈ A^{0} be such that ξ^{3} ∈ F^{×}. Then there exists η ∈ A such that η^{3} ∈ F^{×} and ξη = ωηξ.

Proof : Assume that ξ^{3} ∉ F^{×3}, then F (ξ) is a subfield of A. Let σ: F (ξ) → F (ξ) be the F -automorphism defined by σ(ξ) = ω^{2}ξ. By the Skolem-Noether Theorem, there exists η ∈ A^{×} such that ηxη^{−1} = σ(x) for all x ∈ F (ξ). In particular ξη = ωηξ. Because η = ωξ^{−1}ηξ and η^{2} = ω^{2}ξ^{−1}η^{2}ξ, we have Trd_{A}(η) = 0, Trd_{A}(η^{2}) = 0; so η^{3} ∈ F^{×}.

Now we suppose that ξ^{3} ∈ F^{×3}. Then we may assume that ξ^{3} = 1 and A = M_{3}(F). The minimal polynomial of ξ divides t^{3} − 1, so ξ is diagonalizable and its eigen values are cube roots of unity. Hence we may assume that

ξ = ( \begin{matrix}λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix} )

with λ_{i} ∈ {1, ω, ω^{2}}. Since tr(ξ) = 0 we have {λ_{1}, λ_{2}, λ_{3}} = {1, ω, ω^{2}}. Conjugating by

( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} )

if necessary, we may assume that λ_{1} = 1, λ_{2} = ω, λ_{3} = ω^{2}. Then

η := ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} )

is such that η^{3} ∈ F^{×} and ξη = ωηξ. □

Let w ∈ A be such that w^{3} ∈ F^{×} and vw = ωwv. Then the subspace of the elements x in A such that vx = ω^{2}xv is spanned by w^{2}, vw^{2}, v^{2}w^{2}, and it is contained in V; therefore

V = {x ∈ A | vx = ω^{2}xv}.

Set ξ := w^{2} and η := vw^{2}, then A is the symbol algebra (a, b)_{ω,F} generated by ξ and η such that ξ^{3} = a, η^{3} = b and ξη = ωηξ, and V is the vector subspace spanned by ξ, η and ξ^{2}η^{2}. In this basis of V, the form f_{A,V} takes the generalized Hesse normal form:

f_{A,V} (xξ + yη + zξ^{2}η^{2}) = ax^{3} + by^{3} + a^{2}b^{2}z^{3} − 3ω^{2}abxyz.

The form f_{A,V} is singular; more precisely, it is triangular, i.e., there exist linearly independent forms φ_{1}, φ_{2}, φ_{3} ∈ (V ⊗_{F} F_{s})^{⋆} such that f_{A,V} = φ_{1}φ_{2}φ_{3} as a cubic form over V ⊗_{F} F_{s}.

Conversely, suppose that B is the symbol algebra (a, b)_{ω,F} generated by ξ, η such that ξ^{3} = a, η^{3} = b, ξη = ωηξ, and W is the vector subspace spanned by ξ, η, ξ^{2}η^{2}, for some a, b ∈ F^{×}. Then (B, W) is a cubic pair over F and f_{B,W} is triangular.

Third case: We suppose that either u ∉ A^{×}, or v ∉ A^{×}, or t ∉ A^{×} and t ≠ 0. Then the algebra A is split, so we may assume that A = M_{3}(F). This case is less interesting and thus we shall not give an explicit description of the pair (A, V), but we shall only prove that the cubic form f_{A,V} is singular and not triangular (it is possible to describe V by matrix computations distinguishing several cases; details can be found in [Raczek, 2007]).

To show that f_{A,V} is not triangular, we first prove a more general Lemma on triangular forms.

Lemma 1.6 Suppose that (B, W) is any cubic pair over F such that f_{B,W} is triangular. Then W = s^{2} ⋆ B^{0} ∩ B^{0} ⋆ s for some s ∈ B^{0} such that s is invertible.

Proof : We may assume that F = F_{s} and B = M_{3}(F). Let e_{1}, e_{2}, e_{3} ∈ W be such that

f_{B,W} (x_{1}e_{1} + x_{2}e_{2} + x_{3}e_{3}) = x_{1}x_{2}x_{3}

for all x_{1}, x_{2}, x_{3} ∈ F. Observe that x^{3} = \frac{1}{3}tr(x^{3}) for all x ∈ W, hence f_{B,W} (x_{1}e_{1} + x_{2}e_{2} + x_{3}e_{3}) is equal to

∑_{i=1}^{3} e_{i}^{3}x_{i}^{3} + ∑_{i≠j} tr(e_{i}^{2}e_{j})x_{i}^{2}x_{j} + tr(e_{1}e_{2}e_{3} + e_{2}e_{1}e_{3})x_{1}x_{2}x_{3}

for all x_{1}, x_{2}, x_{3} ∈ F. We deduce that tr(e_{i}^{2}e_{j}) = 0 for all i, j. Set e_{2} = (x_{ij}) and e_{3} = (y_{ij}). Suppose that e_{1}^{2} ≠ 0. Since e_{1}^{3} = 0, we may assume that

e_{1} = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} ) .

Because tr(e_{2}) = 0, tr(e_{1}e_{2}) = 0 and tr(e_{1}^{2}e_{2}) = 0, we have

x_{33} = −x_{11} − x_{22}, x_{32} = −x_{21}, x_{31} = 0.

From tr(e_{1}e_{2}^{2}) = 0 we deduce that x_{21}(2x_{11}+x_{22}) = 0. If x_{21} = 0 then tr(e_{2}^{2}) = 0 and e_{2}^{3} = 0 imply x_{11} = x_{22} = 0. Then e_{1}e_{2} + e_{2}e_{1} = (x_{12} + x_{23})e_{1}^{2} and it contradicts the fact that tr(e_{1}^{2}e_{3}) = 0 and tr(e_{1}e_{2}e_{3}+e_{2}e_{1}e_{3}) = 1. If x_{22} = −2x_{11} then

e_{1}e_{2} + e_{2}e_{1} = ( \begin{matrix}x_{21} & −x_{11} & x_{12} + x_{23} \\ 0 & 0 & −x_{11} \\ 0 & 0 & −x_{21} \end{matrix} ) .

By symmetry we know that

e_{3} = ( \begin{matrix}y_{11} & y_{12} & y_{13} \\ y_{21} & −2y_{11} & y_{23} \\ 0 & −y_{21} & y_{11} \end{matrix} ) ,

thus tr(e_{1}e_{2}e_{3} + e_{2}e_{1}e_{3}) = 1 is impossible. Therefore e_{1}^{2} = 0 (by symmetry we also have e_{2}^{2} = 0, e_{3}^{2} = 0) and we may assume that

e_{1} = ( \begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) .

Since tr(e_{2}) = 0 and tr(e_{1}e_{2}) = 0, we have x_{33} = −x_{11} − x_{22} and x_{31} = 0. Then e_{2}^{2} = 0 implies x_{21}x_{32} = 0. Observe that

e_{1}e_{2} + e_{2}e_{1} = ( \begin{matrix}0 & x_{32} & −x_{22} \\ 0 & 0 & x_{21} \\ 0 & 0 & 0 \end{matrix} ) .

Because tr(e_{1}e_{2}e_{3} + e_{2}e_{1}e_{3}) = 1 we have either x_{21} = 0 = y_{32} and x_{32}, y_{21} ≠ 0 or x_{21}, y_{23} ≠ 0 and x_{32} = 0 = y_{21}. Thus we may assume that x_{21} = 0 and x_{32} = 1. From e_{2}^{2} = 0 we deduce that

e_{2} = ( \begin{matrix}0 & α & −αβ \\ 0 & β & −β^{2} \\ 0 & 1 & −β \end{matrix} )

for some α, β ∈ F. We may assume that α = β = 0 since the invertible matrix

m = ( \begin{matrix}1 & 0 & −α \\ 0 & 1 & −β \\ 0 & 0 & 1 \end{matrix} )

is such that me_{1}m^{−1} = e_{1} and

me_{2}m^{−1} = ( \begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) .

Similarly we see that

e_{3} = ( \begin{matrix}α & −α^{2} & 0 \\ 1 & −α & 0 \\ 0 & 0 & 0 \end{matrix} )

for some α ∈ F. Again we may assume that α = 0 conjugating by

( \begin{matrix}1 & −α & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} )

if necessary. Then one can check that W = s^{2} ⋆ B^{0} ∩ B^{0} ⋆ s with

s = ( \begin{matrix}1 & 0 & 0 \\ 0 & ω^{2} & 0 \\ 0 & 0 & ω \end{matrix} ) .

□

In our case, the subspace V is equal to u ⋆ A^{0} ∩ A^{0} ⋆ v where either u ∉ A^{×}, or v ∉ A^{×}, or t ∉ A^{×} and t ≠ 0. Observe that if u, v ∈ A^{×}, then uF = v^{2}F if and only if t = 0. Thus, by the previous Lemma, the form f_{A,V} is not triangular.

To prove that f_{A,V} is singular we shall distinguish different cases.

1. Suppose that v ∉ A^{×}. If tr(uv) ≠ 0 then, by the relation (1), t is invertible; hence V is spanned by t^{2}, t^{2}u and t^{2}v. Since tr(x(t^{2}v)^{2}) = 0 for all x ∈ V, the point t^{2}vF_{s} of the projective plane ℙ_{V} (F_{s}) is a singular zero of f_{A,V}. If v^{2} ≠ 0 and tr(uv) = 0, then v^{2} ∈ V and v^{2}F_{s} is a singular zero of f_{A,V}. If v^{2} = 0, then we may assume that

v = ( \begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) , so u = ( \begin{matrix}ω^{2}α_{1} & α_{2} & α_{3} \\ 0 & ωα_{1} & α_{4} \\ 0 & 0 & α_{1} \end{matrix} )

for some α_{i} ∈ F. If α_{1} = 0 then vF_{s} is a singular zero of f_{A,V}. If α_{1} ≠ 0 then V is spanned by

( \begin{matrix}α_{1} & 0 & −α_{3} \\ 0 & ωα_{1} & 0 \\ 0 & 0 & ω^{2}α_{1} \end{matrix} ) , ( \begin{matrix}0 & (ω − 1)α_{1} & α_{4} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) , ( \begin{matrix}0 & 0 & α_{2} \\ 0 & 0 & (ω − ω^{2})α_{1} \\ 0 & 0 & 0 \end{matrix} )

and the cubic curve associated with f_{A,V} is a triple line.

2. Suppose that u ∉ A^{×}, then, by symmetry, we deduce that f_{A,V} is also singular.

3. Suppose that u, v ∈ A^{×}, t ∉ A^{×} and t ≠ 0. If t^{2} ≠ 0 then t^{2}F_{s} is a singular zero of f_{A,V}. If t^{2} = 0, we shall prove that there exists a nonzero s ∈ A such that s^{2} = 0, vs = ω^{2}sv, s(tv^{−1}) = (tv^{−1})s = 0. Since u = tv^{−1} + \frac{1}{3}tr(uv)v^{−1}, we then have s ∈ V and so sF_{s} is a singular zero of f_{A,V}. Let w ∈ A be such that w^{3} ∈ F^{×} and vw = ωwv. Since v(tv^{−1}) = ω(tv^{−1})v and tv^{−1} ≠ 0, there exist α_{i} ∈ F not all zero such that tv^{−1} = α_{0}w + α_{1}vw + α_{2}v^{2}w. But (tv^{−1})^{2} = ω^{2}t^{2}v^{−2} = 0, so α_{0} ≠ 0, α_{2} = α_{1}^{2}α_{0}^{−1} and α_{0}^{3} = v^{3}α_{1}^{3}. Hence v^{3} ∈ F^{×3} and we may assume that v^{3} = 1. Replacing w by α_{0}^{−1}w if necessary, we may assume that α_{0} = 1. Then α_{1} = 1, ω or ω^{2}. Conjugating by w or w^{−1} if necessary, we may assume that tv^{−1} = w + vw + v^{2}w. Then we may choose s = w^{2} + vw^{2} + v^{2}w^{2}.

We summarize the above classification in the following Theorem.

Theorem 1.7 Suppose that F contains a primitive cube root of unity. Let (A, V) be a cubic pair over F.

1. If f_{A,V} is nonsingular, then

(A, V) ≅ ((a, b)_{ω,F} , span_{F} ⟨ξ, η, ξη^{2} + λξ^{2}η^{2}⟩)

for some a, b ∈ F^{×}, λ ∈ F such that 1+λ^{3}a ≠ 0, where ξ, η are generators of the symbol algebra such that ξ^{3} = a, η^{3} = b, ξη = ωηξ. Conversely, let a, b ∈ F^{×}, λ ∈ F be such that 1 + λ^{3}a ≠ 0. Let B be the symbol algebra (a, b)_{ω,F} generated by ξ, η such that ξ^{3} = a, η^{3} = b, ξη = ωηξ, and W the subspace spanned by ξ, η, ξη^{2} + λξ^{2}η^{2}. Then (B, W) is a cubic pair over F and f_{B,W} is nonsingular. In the basis (ξ, η, ξη^{2} + λξ^{2}η^{2}), the form f_{B,W} takes the generalized Hesse normal form:

(xξ + yη + z(ξη^{2} + λξ^{2}η^{2}))^{3} = ax^{3} + by^{3} + ab^{2}(1 + λ^{3}a)z^{3} − 3ω^{2}abλxyz.

2. If f_{A,V} is triangular, then

(A, V) ≅ ((a, b)_{ω,F} , span_{F} ⟨ξ, η, ξ^{2}η^{2}⟩)

for some a, b ∈ F^{×}, where ξ, η are generators of the symbol algebra such that ξ^{3} = a, η^{3} = b and ξη = ωηξ. Conversely, let a, b ∈ F^{×}, let B be the symbol algebra (a, b)_{ω,F} generated by ξ, η such that ξ^{3} = a, η^{3} = b, ξη = ωηξ, and W the subspace spanned by ξ, η, ξ^{2}η^{2}. Then (B, W) is a cubic pair over F and f_{B,W} is triangular. In the basis (ξ, η, ξ^{2}η^{2}), the form f_{B,W} takes the generalized Hesse normal form:

(xξ + yη + zξ^{2}η^{2})^{3} = ax^{3} + by^{3} + a^{2}b^{2}z^{3} − 3ω^{2}abxyz.

3. If f_{A,V} is singular and not triangular, then A is split.

2. Classification of cubic pairs over a field without primitive cube root of unity

Suppose that F does not contain a primitive cube root of unity. We shall give the classification of cubic pairs over F in the case where the associated cubic form is nonsingular or triangular. For the remaining cases we know by Theorem 1.7 that the algebra is split and it is just a matter of matrix computations to describe all the possible subspaces up to conjugacy (see [Raczek, 2007] for details). We shall extend the scalars to F (ω) to use the previous classification. To simplify notations, let T denote the reduced trace of A_{F (ω)}. Throughout this section, we denote by σ the F -automorphism of F (ω) such that σ(ω) = ω^{2}.

2.1. Nonsingular form

Let (A, V) be a cubic pair over F such that f_{A,V} is nonsingular. By Subsection 1.2, there exist nonzero u, v ∈ A_{F (ω)}^{0} such that T(u^{2}) = 0, T(v^{2}) = 0, u ⋆ v = 0 and

V_{F (ω)} = (u ⋆ A_{F (ω)}^{0}) ∩ (A_{F (ω)}^{0} ⋆ v)

where u, v, t := uv − \frac{1}{3}T(uv) ∈ A_{F (ω)}^{×}. Then V_{F (ω)} is spanned by t^{2}, t^{2}v and t^{2}u.

We extend σ to an F -automorphism of A_{F (ω)}: for x ∈ A_{F (ω)} and λ ∈ F (ω), define σ(x ⊗ λ) = x ⊗ σ(λ). Then A (resp. V) consists of the elements of A_{F (ω)} (resp. V_{F (ω)}) which are fixed under σ. Note that σ(x ⋆ y) = σ(y) ⋆ σ(x) for all x, y ∈ A_{F (ω)}. Since σ(V_{F (ω)}) = V_{F (ω)}, we have

(σ(v) ⋆ A_{F (ω)}^{0}) ∩ (A_{F (ω)}^{0} ⋆ σ(u)) = (u ⋆ A_{F (ω)}^{0}) ∩ (A_{F (ω)}^{0} ⋆ v).

Hence there exists λ ∈ F (ω)^{×} such that σ(u) = λv. Replacing v by λv if necessary we may assume that σ(u) = v and thus σ(v) = u. Recall that

vu = \frac{1 − ω}{3} T(uv) + ωuv,

hence we have

σ(t) = σ(uv) − \frac{1}{3}σ(T(uv))
= σ(u)σ(v) − \frac{1}{3}T(σ(uv))
= vu − \frac{1}{3}T(vu)
= ωt.

Therefore ω^{2}t ∈ A and ωt^{2} ∈ V. Set e := ωt^{2}, then V is spanned by e, e(v + u) and e(ωv + ω^{2}u).

We shall find a Galois ℤ/3ℤ-algebra (L, ρ) such that L ⊂ A, the vector e and L generate A, and ex = ρ(x)e for all x ∈ L. To do this we first construct an element η ∈ A_{F (ω)} such that eη = ωηe, η^{3} ∈ F (ω)^{×} and σ(η)η ∈ F^{×}. Recall that t^{3} = u^{3}v^{3} − \frac{1}{27}T(uv)^{3}, vt = ωtv, tu = ωut and

t^{2} = ωu^{2}v^{2} + \frac{ω^{2}}{3} T(uv)uv + \frac{1}{9}T(uv)^{2}.

Set

η := \frac{T(uv)}{3}v + \frac{1}{t^{3}} e^{2}v,

then eη = ωηe, η^{3} = u^{3}v^{6} and

σ(η)η = \frac{T(uv)^{2}}{9} uv − \frac{ωT(uv)}{3}utv + ω^{2}ut^{2}v
= u^{3}v^{3}.

We set λ := u^{3}v^{3}, so η + λη^{−1} is fixed under σ. If η^{3} ∉ F (ω)^{×3}, then set L := F (η + λη^{−1}) and let ρ: L → L be the F - automorphism defined by ρ(η + λη^{−1}) = ωη + ω^{2}λη^{−1}. Then L is a cyclic extension of degree 3 over F which is contained in A and with a Galois group generated by ρ. Moreover ex = ρ(x)e for all x ∈ L. If η^{3} ∈ F (ω)^{×3}, then η^{3} = ν^{3} for some ν ∈ F (ω)^{×}. Replacing η by ν^{−1}η if necessary, we may assume that η^{3} = 1 and σ(η)η = 1. Set

L := F · 1 + F · (η + η^{−1}) + F · (ωη + ω^{2}η^{−1})

and define ρ as the F -automorphism of L such that ρ(η + η^{−1}) = ωη + ω^{2}η^{−1} and ρ(ωη + ω^{2}η^{−1}) = ω^{2}η + ωη^{−1}. Since (η + η^{−1})^{2} = η + η^{−1} + 2, the algebra L is isomorphic to F × F × F. Again (L, ρ) is a Galois ℤ/3ℤ-algebra such that L ⊂ A and ex = ρ(x)e for all x ∈ L. In both cases we obtain that A = ⨁_{i=0}^{2} Le^{i} where the multiplication in A is determined by e^{3} = a ∈ F^{×} and ex = ρ(x)e for all x ∈ L.

To finish the description of the pair (A, V), we shall write ev in function of e and η:

ev = e(\frac{T(uv)}{3} + \frac{1}{t^{3}} e^{2})^{−1}η
= \frac{1}{u^{3}v^{3}} (α + a^{−1}α^{2}e + e^{2})η

where α := −T(uv)t^{3}/3. Hence V_{F (ω)} is spanned by e, (α + a^{−1}α^{2}e + e^{2})η and (α + a^{−1}α^{2}e + e^{2})λη^{−1}. Note that α^{3} ≠ a^{2} since ev is invertible. We obtain that

V = span_{F} ⟨e, (α + a^{−1}α^{2}e + e^{2})θ, (α + a^{−1}α^{2}e + e^{2})ρ(θ)⟩

where θ = η + λη^{−1} ∈ L \ {0} is such that θ + ρ(θ) + ρ^{2}(θ) = 0.

We shall prove that the cubic form f_{A,V} is isometric to the form

Tr_{K,β} − 3bN_{K}: K → F: x ↦ Tr_{K}(βx^{3}) − 3bN_{K}(x)

for K = F × F (ω) and, for some β ∈ K^{×} and b ∈ F. We set

v_{1} := e, v_{2} := (α + a^{−1}α^{2}e + e^{2})θ, v_{3} := (α + a^{−1}α^{2}e + e^{2})ρ(θ)

so that V is the vector space spanned by v_{1}, v_{2}, v_{3}. Then f_{A,V} (xv_{1} + yv_{2} + zv_{3}) is equal to

(xu_{1} + (y + ωz)u_{2} + (y + ω^{2}z)u_{3})^{3}

where u_{1} := e, u_{2} := (α + a^{−1}α^{2}e + e^{2})η, u_{3} := (α + a^{−1}α^{2}e + e^{2})λη^{−1}. But (xu_{1} + yu_{2} + zu_{3})^{3} is equal to

ax^{3} + η^{3}(a^{−1}α^{3} − a)^{2}y^{3} + σ(η^{3})(a^{−1}α^{3} − a)^{2}z^{3} − 3λα(a^{−1}α^{3} − a)xyz.

Therefore f_{A,V} is isometric to the form Tr_{K,β} − 3bN_{K} where K := F × F (ω), β := (a, η^{3}(a^{−1}α^{3} − a)^{2}) and b := λα(a^{−1}α^{3} − a).

Observe that if (L, ρ) is a Galois ℤ/3ℤ-algebra, then there exist λ ∈ F^{×} and ϕ ∈ L ⊗_{F} F (ω) such that ϕ^{3} ∈ F (ω)^{×}, ϕ ∉ F (ω), σ(ϕ^{3}) = λ^{3}ϕ^{−3} and

L = F · 1 + F · (ϕ + λϕ^{−1}) + F · (ωϕ + ω^{2}λϕ^{−1}).

Indeed, suppose that (L, ρ) is not split, then L(ω) ≅ L ⊗_{F} F (ω) is a Galois extension of F with a Galois group isomorphic to ℤ/2ℤ × ℤ/3ℤ. Let σ̃, τ be F -automorphisms of L(ω) such that L (resp. F (ω)) is the subfield of L(ω) which is fixed under σ̃ (resp. τ). Let ϕ ∈ L(ω) be such that ϕ^{3} ∈ F (ω)^{×} and ϕ ∉ F (ω). Replacing ϕ by ϕ^{−1} if necessary, we may assume that τ(ϕ) = ωϕ. Hence

τσ̃(ϕ) = σ̃τ(ϕ) = σ̃(ωϕ) = ω^{2}σ̃(ϕ).

Thus σ̃(ϕ) = λϕ^{−1} for some λ ∈ F (ω)^{×}. But

ϕ = σ̃(λϕ^{−1}) = σ̃(λ)λ^{−1}ϕ,

so σ̃(λ) = λ and λ ∈ F^{×}. Then we have L = F (ϕ + λϕ^{−1}) with

σ(ϕ^{3}) = (σ̃(ϕ))^{3} = (λϕ^{−1})^{3} = λ^{3}ϕ^{−3}.

Suppose that L = F × F × F, then we may choose λ = 1 and ϕ = (1, ω, ω^{2}).

Now let (L, ρ) be a Galois ℤ/3ℤ-algebra, a ∈ F^{×} and α ∈ F such that α^{3} ≠ a^{2}. Set

(B, W) := ( ⨁_{i=0}^{2} Le^{i}, span_{F} ⟨e, (α + a^{−1}α^{2}e + e^{2})θ, (α + a^{−1}α^{2}e + e^{2})ρ(θ)⟩),

where the multiplication in B is defined by e^{3} = a, ex = ρ(x)e for all x ∈ L, and θ ∈ L \ {0} is such that θ + ρ(θ) + ρ^{2}(θ) = 0. Then one can check that (B, W) is a cubic pair over F and f_{B,W} is nonsingular.

We summarize this subsection by the following Theorem:

Theorem 2.1 Suppose that F does not contain a primitive cube root of unity. Let (A, V) be a cubic pair over F such that f_{A,V} is nonsingular. Then (A, V) is isomorphic to

( ⨁_{i=0}^{2} Le^{i}, span_{F} ⟨e, (α + a^{−1}α^{2}e + e^{2})θ, (α + a^{−1}α^{2}e + e^{2})ρ(θ)⟩)

for some Galois ℤ/3ℤ-algebra (L, ρ), a ∈ F^{×}, α ∈ F such that α^{3} ≠ a^{2}, where e^{3} = a, ex = ρ(x)e for all x ∈ L and θ ∈ L \ {0} is such that θ + ρ(θ) + ρ^{2}(θ) = 0. Conversely, let (L, ρ) be a Galois ℤ/3ℤ-algebra, a ∈ F^{×}, α ∈ F such that α^{3} ≠ a^{2}. Let B = ⨁_{i=0}^{2} Le^{i} be the algebra with multiplication defined by e^{3} = a, ex = ρ(x)e for all x ∈ L, and let W be the subspace spanned by e, (α + a^{−1}α^{2}e + e^{2})θ, (α + a^{−1}α^{2}e + e^{2})ρ(θ), where θ ∈ L \ {0} is such that θ + ρ(θ) + ρ^{2}(θ) = 0. Then (B, W) is a cubic pair over F and f_{B,W} is nonsingular. Let ϕ ∈ L ⊗_{F} F (ω) and λ ∈ F^{×} be such that ϕ^{3} ∈ F (ω)^{×}, ϕ ∉ F (ω), σ(ϕ^{3}) = λϕ^{−3} and 1, ϕ + λϕ^{−1}, ωϕ + ω^{2}λϕ^{−1} span L, where σ is the nontrivial F -automorphism of F (ω). Then f_{B,W} is isometric to Tr_{K,β} − 3bN_{K} where K = F × F (ω), β = (a, ϕ^{3}(a^{−1}α^{3} − a)^{2}) and b = λα(a^{−1}α^{3} − a).

2.2. Triangular form

Let (A, V) be a cubic pair over F such that f_{A,V} is triangular. By Subsection 1.2, there exists v ∈ A_{F (ω)}^{0} such that T(v^{2}) = 0, v ∈ A_{F (ω)}^{×} and

V = (v^{2} ⋆ A_{F (ω)}^{0}) ∩ (A_{F (ω)}^{0} ⋆ v);

then V = {x ∈ A_{F (ω)} | vx = ω^{2}xv}. Fix e ∈ V, then ev = ωve. We extend the F -automorphism σ of F (ω) to A_{F (ω)}. Then σ(v) = λv^{2} for some λ ∈ F (ω)^{×}. Since

v = σ(λv^{2}) = σ(λ)λ^{2}v^{4},

we deduce that σ(λ)λ^{2}v^{3} = 1. Hence

σ(λv) = σ(λ)σ(v) = λ^{−2}v^{−3}λv^{2} = (λv)^{−1};

so we may assume that σ(v) = v^{−1}. Set

L := F · 1 + F · (v + v^{−1}) + F · (ωv + ω^{2}v^{−1})

and define ρ as the F -automorphism of L such that ρ(v + v^{−1}) = ωv + ω^{2}v^{−1} and ρ(ωv + ω^{2}v^{−1}) = ω^{2}v + ωv^{−1}. Then (L, ρ) is a Galois ℤ/3ℤ-algebra (note that (L, ρ) is split if and only if v^{3} ∈ F (ω)^{×3}). Moreover L ⊂ A, ex = ρ(x)e for all x ∈ L, A = ⨁_{i=0}^{2} Le^{i} and V = eL. It is easy to check that f_{A,V} is isometric to aN_{L}, where a := e^{3}.

Conversely, let (L, ρ) be a Galois ℤ/3ℤ-algebra and a ∈ F^{×}. Set

(B, W) := ( ⨁_{i=0}^{2} Le^{i}, eL)

where the multiplication in B is defined by e^{3} = a and ex = ρ(x)e for all x ∈ L. Then (B, W) is a cubic pair over F and f_{B,W} is triangular.

Thus we obtain:

Theorem 2.2 Suppose that F does not contain a primitive cube root of unity. Let (A, V) be a cubic pair over F such that f_{A,V} is triangular. Then (A, V) is isomorphic to

( ⨁_{i=0}^{2} Le^{i}, eL)

for some Galois ℤ/3ℤ-algebra (L, ρ) and a ∈ F^{×}, where e^{3} = a and ex = ρ(x)e for all x ∈ L. Conversely, let (L, ρ) be a Galois ℤ/3ℤ-algebra and a ∈ F^{×}. Let B = ⨁_{i=0}^{2} Le^{i} be the algebra with multiplication defined by e^{3} = a, ex = ρ(x)e for all x ∈ L, and set W := eL. Then (B, W) is a cubic pair over F and f_{B,W} is triangular. Moreover f_{B,W} is isometric to aN_{L}.

3. The form determines the algebra

Let (A, V) and (A′, V′) be cubic pairs over F and suppose that f_{A,V} and f_{A′,V′} are isometric. In this section we shall prove that A and A′ are either isomorphic or anti-isomorphic.

We may assume that F contains a primitive cube root of unity. Indeed, if A ⊗_{F} F (ω) ≅ A′ ⊗_{F} F (ω), then A ≅ A′ since A and A′ are central simple algebras of degree 3 and F (ω)/F is an extension of degree at most 2. We may also assume that A is division, because there is nothing to prove if A and A′ are split. Therefore, by Theorem 1.7, the cubic form f_{A,V} is either nonsingular or triangular.

First case: Suppose that f_{A,V} is nonsingular, then so is f_{A′,V′}. By Theorem 1.7, there exist a_{i}, a′_{i} ∈ F^{×}, λ, λ′ ∈ F such that 1 + λ^{3}a_{1}, 1 + λ^{′3}a′_{1} ≠ 0 and

A = (a_{1}, a_{2})_{ω,F} , V = span_{F} ⟨ξ_{1}, ξ_{2}, ξ_{1}ξ_{2}^{2} + λξ_{1}^{2}ξ_{2}^{2}⟩
A′ = (a′_{1}, a′_{2})_{ω,F} , V′ = span_{F} ⟨ξ′_{1}, ξ′_{2}, ξ′_{1}ξ_{2}^{′2} + λ′ξ_{1}^{′2}ξ_{2}^{′2}⟩

where A (resp. A′) is generated by ξ_{1}, ξ_{2} such that ξ_{i}^{3} = a_{i} and ξ_{1}ξ_{2} = ωξ_{2}ξ_{1} (resp. ξ′_{1}, ξ′_{2} such that ξ_{i}^{′3} = a′_{i} and ξ′_{1}ξ′_{2} = ωξ′_{2}ξ′_{1}). Set

ξ_{3} := ξ_{1}ξ_{2}^{2} + λξ_{1}^{2}ξ_{2}^{2}, a_{3} := ξ_{3}^{3}, ξ′_{3} := ξ′_{1}ξ_{2}^{′2} + λ′ξ_{1}^{′2}ξ_{2}^{′2}, a′_{3} := ξ_{3}^{′3}.

We recall properties of nonsingular cubic forms and we refer to [Brieskorn and Knörrer, 1986] or [Hirschfeld, 1979] for more details. A nonsingular cubic form f on a 3-dimensional vector space V has 9 inflexion points in the projective plane ℙ_{V} (F_{s}). There are four triangles (i.e. cubic curves associated with triangular cubic forms) in ℙ_{V} (F_{s}) with the property that each inflexion point is incident with one and only one line of the triangle and each line of the triangle passes through exactly 3 inflexion points. These triangles are called inflexional triangles of f. For a triangular cubic form g = φ_{1}φ_{2}φ_{3} over V, we denote by g = 0 the triangle formed by the zeros of the linear forms φ_{i} in ℙ_{V} (F_{s}).

The map V → K := F × F × F which sends ξ_{1}, ξ_{2}, ξ_{3} on the canonical basis of K is an F -vector space isomorphism. Under this isomorphism, f_{A,V} is isometric to the form

Tr_{K,α} − 3bN_{K}: K → F: x ↦ Tr_{K}(αx^{3}) − 3bN_{K}(x)

where α = (a_{1}, a_{2}, a_{3}) and b = ω^{2}a_{1}a_{2}λ. The inflexional triangles of the form Tr_{K,α} − 3bN_{K} are N_{K} = 0 and Tr_{K,α} − 3θN_{K} = 0 for all θ ∈ F_{s} such that θ^{3} = N_{K}(α). Let (φ_{1}, φ_{2}, φ_{3}) denote the dual basis of (ξ_{1}, ξ_{2}, ξ_{3}), then under the previous isomorphism V → K, the form φ_{1}φ_{2}φ_{3} is isometric to N_{K}. Hence Γ acts trivially on the lines of the corresponding inflexional triangle. In fact, we have the following:

Lemma 3.1 There exists a unique inflexional triangle of f_{A,V} whose lines are defined over F.

Proof : Suppose that Γ acts trivially on the lines of Tr_{K,α} − 3θN_{K} = 0, for some θ ∈ F_{s} such that θ^{3} = N_{K}(α). For x = (x_{1}, x_{2}, x_{3}) ∈ K, Tr_{K}(αx^{3}) − 3θN_{K}(x) is equal to

(θ_{1}x_{1} + θ_{2}x_{2} + θ_{3}x_{3})(θ_{1}x_{1} + ωθ_{2}x_{2} + ω^{2}θ_{3}x_{3})(θ_{1}x_{1} + ω^{2}θ_{2}x_{2} + ωθ_{3}x_{3})

for some θ_{i} ∈ F_{s} such that θ_{i}^{3} = a_{i} and θ_{1}θ_{2}θ_{3} = θ. Since Γ acts trivially on the line θ_{1}x_{1} + θ_{2}x_{2} + θ_{3}x_{3} = 0, there exists a nonzero u ∈ F such that θ_{2} = uθ_{1}. This implies that a_{2} = u^{3}a_{1}, which contradicts the assumption that A is division. □

Let (φ′_{1}, φ′_{2}, φ′_{3}) be the dual basis of (ξ′_{1}, ξ′_{2}, ξ′_{3}). Then φ′_{1}φ′_{2}φ′_{3} = 0 is an inflexional triangle of f_{A′,V′} whose lines are defined over F. Let Θ: V → V′ be an F -vector space isomorphism such that f_{A,V} = f_{A′,V′} ∘ Θ, then

φ_{1}φ_{2}φ_{3}F = (φ′_{1}φ′_{2}φ′_{3} ∘ Θ)F.

Thus there exist λ_{i} ∈ F^{×} and a permutation π of {1, 2, 3} such that

φ′_{i} ∘ Θ = λ_{π(i)}φ_{π(i)} for all i ∈ {1, 2, 3}.

For all i, j ∈ {1, 2, 3}, we have

φ′_{i} ∘ Θ(ξ_{π(j)}) = λ_{π(i)}φ_{π(i)}(ξ_{π(j)}) = δ_{ij}λ_{π(i)},

hence Θ(ξ_{π(j)}) = λ_{π(j)}ξ′_{j}. We obtain that a_{π(j)} = λ_{π(j)}^{3}a′_{j} and b = λ_{1}λ_{2}λ_{3}b′ where b′ = ω^{2}a′_{1}a′_{2}λ′. But a_{1} is the only scalar among the a_{i}’s such that

\frac{a_{1}a_{2}a_{3} − b^{3}}{a_{i}^{2}} ∈ F^{×3}.

Indeed a_{1}a_{2}a_{3} − b^{3} = a_{1}^{2}a_{2}^{3}, thus a_{1}^{−2}(a_{1}a_{2}a_{3} − b^{3}) ∈ F^{×3}; if a_{2}^{−2}(a_{1}a_{2}a_{3} − b^{3}) ∈ F^{×3}, then a_{1}F^{×3} = a_{2}F^{×3} and it contradicts the assumption that A is division; similarly, a_{3}^{−2}(a_{1}a_{2}a_{3} − b^{3}) ∉ F^{×3}. On the other hand, we have a_{1}^{′−2}(a′_{1}a′_{2}a′_{3} − b^{′3}) ∈ F^{×3} and

\frac{a_{1}a_{2}a_{3} − b^{3}}{a_{π(1)^{2}}} = (\frac{λ_{1}λ_{2}λ_{3}}{λ_{π(1)}^{2}} )^{3} \frac{a′_{1}a′_{2}a′_{3} − b^{′3}}{a_{1}^{′2}} ∈ F^{×3};

therefore π(1) = 1. If π(2) = 2, then

(λ_{i}ξ′_{i})^{3} = a_{i} and (λ_{1}ξ′_{1})(λ_{2}ξ′_{2}) = ω(λ_{2}ξ′_{2})(λ_{1}ξ′_{1}),

thus A′ ≅ A. If π(2) = 3, then

(λ_{1}ξ′_{1})^{3} = a_{1}, (λ_{2}ξ′_{3})^{3} = a_{2} and (λ_{1}ξ′_{1})(λ_{2}ξ′_{3}) = ω^{2}(λ_{2}ξ′_{3})(λ_{1}ξ′_{1}),

thus A′ ≅ A^{op}.

Second case: Suppose that f_{A,V} is triangular. Then there exist a_{i}, a′_{i} ∈ F^{×} such that

A = (a_{1}, a_{2})_{ω,F} , V = span_{F} ⟨ξ_{1}, ξ_{2}, ξ_{1}^{2}ξ_{2}^{2}⟩
A′ = (a′_{1}, a′_{2})_{ω,F} , V′ = span_{F} ⟨ξ′_{1}, ξ′_{2}, ξ_{1}^{′2}ξ_{2}^{′2}⟩

where A (resp. A′) is generated by ξ_{1}, ξ_{2} such that ξ_{i}^{3} = a_{i} and ξ_{1}ξ_{2} = ωξ_{2}ξ_{1} (resp. ξ′_{1}, ξ′_{2} such that ξ_{i}^{′3} = a′_{i} and ξ′_{1}ξ′_{2} = ωξ′_{2}ξ′_{1}). Let θ ∈ F_{s} be a cube root of a_{1}^{−1}a_{2}. Since A is division, θ ∉ F. We have

f_{A,V} (x_{1}ξ_{1} + x_{2}ξ_{2} + x_{3}ξ_{1}^{2}ξ_{2}^{2}) = a_{1}N_{F (θ)}(x_{1} + x_{2}θ + ω^{2}a_{1}x_{3}θ^{2}).

Let θ′ ∈ F_{s} be a cube root of a_{1}^{′−1}a′_{2}. Similarly, f_{A′,V′}(x_{1}ξ′_{1} + x_{2}ξ′_{2} + x_{3}ξ_{1}^{′2}ξ_{2}^{′2}) is equal to

a′_{1}(x_{1} + x_{2}θ′ + ω^{2}a′_{1}x_{3}θ^{′2})(x_{1} + ωx_{2}θ′ + ωa′_{1}x_{3}θ^{′2})(x_{1} + ω^{2}x_{2}θ′ + a′_{1}x_{3}θ^{′2}).

Since f_{A,V} ≅ f_{A′,V′}, we have θ′ ∉ F and, by Proposition 8 in [Raczek and Tignol, 2008], the fields F (θ) and F (θ′) are isomorphic. We deduce that either θ′F = θF or θ′F = θ^{2}F. Identifying θ with ξ_{1}^{−1}ξ_{2} (resp. θ′ with ξ_{1}^{′−1}ξ′_{2}), we have

A = ⨁_{i=0}^{2} F (θ)ξ_{1}^{i} and A′ = ⨁_{i=0}^{2} F (θ′)ξ_{1}^{′i}

where ξ_{1}θ = ωθξ_{1} and ξ′_{1}θ′ = ωθ′ξ′_{1}. Because f_{A,V} is isometric to f_{A′,V′}, there exist u_{i} ∈ F such that

a_{1}N_{F (θ)}(u_{1} + u_{2}θ + u_{3}θ^{2}) = a′_{1}.

Set η_{1} := ξ_{1}(u_{1} + u_{2}θ + u_{3}θ^{2}), then

η_{1}^{3} = a_{1}N_{F (θ)}(u_{1} + u_{2}θ + u_{3}θ^{2}) = a′_{1} and η_{1}θ = ωθη_{1}.

Hence A = ⨁_{i=0}^{2} F (θ)η_{1}^{i} with η_{1}^{3} = a′_{1} and η_{1}θ = ωθη_{1}. So

A ≅ \{ \begin{matrix}A′ & if θ′F = θF, \\ A^{′op} & if θ′F = θ^{2}F. \end{matrix}

We thus obtain the following:

Theorem 3.2 Let (A, V) and (A′, V′) be cubic pairs over F. Suppose that f_{A,V} and f_{A′,V′} are isometric, then the algebras A and A′ are either isomorphic or anti-isomorphic.

References

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[2] [Claude Chevalley, 1997] The algebraic theory of spinors and Clifford algebras, Springer-Verlag, Berlin.

[3] [James William Peter Hirschfeld, 1979] Projective geometries over finite fields, The Clarendon Press Oxford University Press, New York.

[4] [Tsit Yuen Lam, 2005], Introduction to quadratic forms over fields, Graduate Studies in Mathematics, vol. 67, American Mathematical Society, Providence, RI.

[5] [Eli Matzri, 2008] Totally isotropic subspaces in degree 3 symbol algebras, preprint.

[6] [Mélanie Raczek, 2007] Ternary cubic forms and central simple algebras of degree 3, PhD thesis from Université Catholique de Louvain, supervised by Jean-Pierre Tignol.

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