On the 3-Pfister number of quadratic forms

Mélanie Raczek^{∗}

November 2, 2010

Abstract

For a field F of characteristic different from 2, containing a square root of -1, endowed with an F^{×2}-compatible valuation v such that the residue field has at most two square classes, we use a combinatorial analogue of the Witt ring of F to prove that an anisotropic quadratic form over F with even dimension d, trivial discriminant and Hasse-Witt invariant can be written in the Witt ring as the sum of at most (d^{2})/8 3-fold Pfister forms.

Introduction

Let F be a field of characteristic different from 2 which contains a square root of −1. Let W (F) be the Witt ring of F, I(F) the fundamental ideal of W (F) and I^{m}(F) the m-th power of I(F). Since −1 is a square in F, the Witt ring W (F) is an algebra over the field 𝔽_{2} with two elements.

The quadratic form a_{1}X_{1}^{2} + … + a_{d}X_{d}^{2} with a_{i} ∈ F^{×} is denoted by ⟨a_{1}, …, a_{d}⟩. Each [q] ∈ I^{m}(F) is the sum of Witt classes of m-fold Pfister forms, i.e. quadratic forms of the type

⟨⟨a_{1}, …, a_{m}⟩⟩ = ⟨1, a_{1}⟩ ⊗ … ⊗ ⟨1, a_{m}⟩.

Brosnan, Reichstein and Vistoli [2010] define the m-Pfister number Pf_{m}(q) of a quadratic form [q] ∈ I^{m}(F) to be the least number of terms needed to write the Witt class of q as a sum of Witt classes of m-fold Pfister forms. They prove the existence of quadratic forms whose 3-Pfister number increases exponentially with their dimension. In general, no upper bound is known for the 3-Pfister number.

In this paper, we shall prove that for a valued field (F, v) such that 1 + 𝔪_{v} ⊂ F^{×2} and (\overline{F} _{v}^{×} : \overline{F} _{v}^{×2}) ≤ 2, the 3-Pfister number of a quadratic form in I^{3}(F) of dimension d is less than or equal to (d^{2})/8. This result on 3-Pfister numbers extends those on the 1- and 2-Pfister numbers in [Parimala et al., 2009]. Indeed, our proof technique is similar: we use the group algebra 𝔽_{2}[F^{×}/F^{×2}] as a combinatorial analogue of the Witt ring of F, allowing us to make explicit computations.

However, our result does not hold for general fields: the combinatorial analogue – which works for the computation of 1- and 2-Pfister numbers over general fields, see [Parimala et al., 2009] – is not powerful enough to deal with 3-Pfister numbers in general. We shall indicate the obstruction.

I would like to thank Ido Efrat, Karim Becher and Jean-Pierre Tignol for their helpful comments.

1. A combinatorial analogue of the Witt ring

For F a field of characteristic different from 2 which contains a square root of −1, we set V_{F} := F^{×}/F^{×2}; it is a vector space over 𝔽_{2}, so let 𝔽_{2}[V_{F}] denote its group algebra. This algebra can be seen as a combinatorial analogue of the Witt ring for the following reason: it is a discrete object which only depends on the square classes of F such that the map

Ψ: 𝔽_{2}[V_{F}] → W (F): (a_{1}F^{×2}) + … + (a_{d}F^{×2}) ↦ [⟨a_{1}, …, a_{d}⟩]

∗Postdoctoral Fellow of the FNRS, Université Catholique de Louvain, Belgium, melanie.raczek@uclouvain.be

is a well-defined surjective 𝔽_{2}-algebra homomorphism. Moreover, in the case where F is the field of iterated Laurent series ℂ((x_{1})) … ((x_{n})), this homomorphism is in fact an isomorphism [Parimala et al., 2009, p. 333].

It is natural to study a more general situation: the group algebra 𝔽_{2}[V] with V any vector space over 𝔽_{2}. Since the addition in V induces the multiplication in 𝔽_{2}[V], we write X^{v} for the image of v ∈ V in 𝔽_{2}[V], so that

𝔽_{2}[V] = { ∑_{v∈V} α_{v}X^{v} | α_{v} ∈ 𝔽_{2} and {v ∈ V | α_{v} ≠ 0} is finite}

where X^{0} = 1 and X^{u} · X^{v} = X^{u+v}.

We now define an ideal of 𝔽_{2}[V], which is an analogue of the fundamental ideal of W (F). Let ϵ_{0} : 𝔽_{2}[V] → 𝔽_{2} be the 𝔽_{2}-algebra homomorphism defined by

ϵ_{0}(∑_{v∈V} α_{v}X^{v}) = ∑_{v∈V} α_{v}

and let I[V] be its kernel. It has the property that for all ξ ∈ 𝔽_{2}[V_{F}],

ξ ∈ I[V_{F}] ⟺ Ψ(ξ) ∈ I(F).

For m ≥ 1, let I^{m}[V] denote the m-th power of I[V]; it is generated as a group by elements of the type

[[v_{1}, …, v_{m}]] := (1 + X^{v_{1}}) ⋯ (1 + X^{v_{m}}),

which we call m-fold Pfister elements. The m-Pfister number Pf_{m}(ξ) of an element ξ ∈ I^{m}[V] is the least number of terms needed to write ξ as a sum of m-fold Pfister elements. It is related to the m-Pfister number of a quadratic form in the following way: for all ξ ∈ I^{m}[V_{F}],

Pf_{m}(Ψ(ξ)) ≤ Pf_{m}(ξ).

This is so because Ψ carries m-fold Pfister elements in 𝔽_{2}[V_{F}] to m-fold Pfister forms in W (F). Furthermore, we define the m-scaled Pfister number of an element ξ ∈ I^{m}[V], and we write GPf_{m}(ξ), as the least number of terms needed to write ξ as a sum of scaled m-fold Pfister elements, i.e. elements of the form X^{a}[[b_{1}, …, b_{m}]]. Clearly GPf_{m}(ξ) ≤ Pf_{m}(ξ) and since X^{a}(1+X^{b}) = (1+X^{a})+(1+X^{a+b}), we have Pf_{m}(ξ) ≤ 2GPf_{m}(ξ).

In this section, we shall first study the structure of I^{3}[V] and, by way of example, compute the 3-Pfister number of certain of its elements. Next we exhibit an algorithm which allows us to classify the elements in I^{3}[V]. Finally, we use the scaled Pfister numbers to find an upper bound for the 3-Pfister number of an element in I^{3}[V].

1.1. Structure of I^{3}[V]

We shall prove that I^{3}[V] is the kernel of some group homomorphism. By [Parimala et al., 2009, Corollary 1.2],

I^{2}[V] = {ξ ∈ I[V] | ϵ_{1}(ξ) = 0},

where ϵ_{1} : 𝔽_{2}[V] → V is the group homomorphism defined by ϵ_{1}(∑_{v∈V} α_{v}X^{v}) = ∑_{v∈V} α_{v}v. We define ϵ_{2} : 𝔽_{2}[V] → V ∧ V by

ϵ_{2}(X^{v_{1}} + … + X^{v_{r}}) = ∑_{1≤i<j≤r} v_{i} ∧ v_{j}.

To prove that ϵ_{2} is well-defined it is sufficient to check the obvious fact that if v_{r} = v_{r−1}, then

∑_{1≤i<j≤r−2} v_{i} ∧ v_{j} = ∑_{1≤i<j≤r} v_{i} ∧ v_{j}.

Even though the map ϵ_{2} is not a group homomorphism, the following property holds:

Lemma 1.1 Let ξ_{1}, ξ_{2} ∈ 𝔽_{2}[V] be such that either ϵ_{1}(ξ_{1} + ξ_{2}) = 0 or ϵ_{1}(ξ_{1}) = 0 or ϵ_{1}(ξ_{2}) = 0. Then ϵ_{2}(ξ_{1} + ξ_{2}) = ϵ_{2}(ξ_{1}) + ϵ_{2}(ξ_{2}). In particular, ϵ_{2}|_{I^{2}[V ]} is a group homomorphism.

Proof : Write ξ_{1} = X^{v_{1}} + … + X^{v_{r}} and ξ_{2} = X^{w_{1}} + … + X^{w_{s}}. Then

ϵ_{2}(ξ_{1} + ξ_{2}) = ∑_{1≤i<j≤r} v_{i} ∧ v_{j} + ∑_{i=1}^{r} ∑_{j=1}^{s} v_{i} ∧ w_{j} + ∑_{1≤i<j≤s} w_{i} ∧ w_{j}
= ϵ_{2}(ξ_{1}) + ϵ_{1}(ξ_{1}) ∧ ϵ_{1}(ξ_{2}) + ϵ_{2}(ξ_{2}).

By assumption, ϵ_{1}(ξ_{1}) ∧ ϵ_{1}(ξ_{2}) = 0, hence ϵ_{2}(ξ_{1} + ξ_{2}) = ϵ_{2}(ξ_{1}) + ϵ_{2}(ξ_{2}). □

Observe that the image of a 3-fold Pfister element by ϵ_{2} is zero, hence

I^{3}[V] ⊂ {ξ ∈ I^{2}[V] | ϵ_{2}(ξ) = 0}.

We shall prove that the inclusion is in fact an equality. First we define the support of an element ξ = ∑_{v∈V} α_{v}X^{v} as the set

D(ξ) := {v ∈ V | α_{v} = 1}.

Let ξ ∈ I^{2}[V] be a nonzero element such that ϵ_{2}(ξ) = 0. Let a ∈ V \ {0} and W a subspace of V be such that span_{𝔽_{2}}⟨D(ξ)⟩ = W ⊕ 𝔽_{2} · a (where span_{𝔽_{2}}⟨D(ξ)⟩ is the subspace of V spanned by D(ξ)). Then there exist unique η_{1}, η_{2} ∈ 𝔽_{2}[W] such that ξ = η_{1} + X^{a} · η_{2}.

Lemma 1.2 We have η_{2}, η_{1} + η_{2} ∈ I^{2}[W] and ϵ_{2}(η_{1} + η_{2}) = 0.

Proof : Since ξ ∈ I^{2}[W], we have ϵ_{0}(ξ) = 0 and ϵ_{1}(ξ) = 0. The map ϵ_{0} is a ring homomorphism, so

0 = ϵ_{0}(ξ) = ϵ_{0}(η_{1}) + ϵ_{0}(η_{2})

and thus ϵ_{0}(η_{1}) = ϵ_{0}(η_{2}). Because ϵ_{1} is a group homomorphism, we have

0 = ϵ_{1}(ξ) = ϵ_{1}(η_{1}) + ϵ_{1}(X^{a} · η_{2}).

Observe that

ϵ_{1}(X^{a} · η_{2}) = ∑_{v∈D(η_{2})} a + v = |D(η_{2})|a + ϵ_{1}(η_{2}) = ϵ_{0}(η_{2})a + ϵ_{1}(η_{2});

hence ϵ_{1}(η_{1}) + ϵ_{1}(η_{2}) + ϵ_{0}(η_{2})a = 0 where ϵ_{1}(η_{1}) + ϵ_{1}(η_{2}) ∈ W. It implies that ϵ_{1}(η_{1}) + ϵ_{1}(η_{2}) = 0 and ϵ_{0}(η_{2}) = 0. Therefore η_{1} + η_{2} ∈ I^{2}[W] and η_{2} ∈ I[W]. Write

η_{1} = X^{v_{1}} + … + X^{v_{r}} and η_{2} = X^{w_{1}} + … + X^{w_{s}}.

By Lemma 1.1, ϵ_{2}(ξ) = ϵ_{2}(η_{1}) + ϵ_{2}(X^{a} · η_{2}). But

ϵ_{2}(X^{a} · η_{2}) = ∑_{1≤i<j≤s}(a + w_{i}) ∧ (a + w_{j})

= ∑_{j=2}^{s}(j − 1)a ∧ w_{j} + ∑_{i=1}^{s−1} w_{i} ∧ (s − i)a + ∑_{1≤i<j≤s} w_{i} ∧ w_{j}

= ∑_{i=1}^{s}(s − 1)a ∧ w_{i} + ϵ_{2}(η_{2})
= a ∧ ϵ_{1}(η_{2}) + ϵ_{2}(η_{2}).

Thus ϵ_{2}(ξ) = 0 implies ϵ_{2}(η_{1})+ϵ_{2}(η_{2})+a∧ϵ_{1}(η_{2}) = 0 where ϵ_{2}(η_{1})+ϵ_{2}(η_{2}) ∈ W ∧W and ϵ_{1}(η_{2}) ∈ W. Hence ϵ_{2}(η_{1})+ϵ_{2}(η_{2}) = 0 and a∧ϵ_{1}(η_{2}) = 0. It implies that ϵ_{2}(η_{1} +η_{2}) = 0 and ϵ_{1}(η_{2}) = 0. Therefore η_{2}, η_{1} + η_{2} ∈ I^{2}[W] and ϵ_{2}(η_{1} + η_{2}) = 0. □

Corollary 1.3 We have I^{3}[V] = {ξ ∈ I^{2}[V] | ϵ_{2}(ξ) = 0}.

Proof : Let ξ ∈ I^{2}[V] be such that ϵ_{2}(ξ) = 0. We shall prove that ξ ∈ I^{3}[V] by induction on m := dim span_{𝔽_{2}}⟨D(ξ)⟩, the dimension of span_{𝔽_{2}}⟨D(ξ)⟩. If m = 0, then ξ = 0 (by [Parimala et al., 2009, Lemma 1.1], ξ ∈ I^{2}[V] \ {0} implies |D(ξ)| ≥ 4). Suppose that m > 0, then m ≥ 2 because |D(ξ)| ≥ 4. Let a ∈ V \ {0} and W be a subspace of V such that span_{𝔽_{2}}⟨D(ξ)⟩ = W ⊕ 𝔽_{2} · a. Let η_{1}, η_{2} ∈ 𝔽_{2}[W] be such that ξ = η_{1} + X^{a} · η_{2}. Then ξ = (η_{1} + η_{2}) + (1 + X^{a}) · η_{2} and by Lemma 1.2, η_{2}, η_{1} + η_{2} ∈ I^{2}[W] and ϵ_{2}(η_{1} + η_{2}) = 0. Since dim⟨D(η_{1} + η_{2})⟩ ≤ dim W < m, by induction assumption, η_{1} + η_{2} ∈ I^{3}[W]. But (1 + X^{a}) · η_{2} ∈ I^{3}[V] because η_{2} ∈ I^{2}[W], so ξ ∈ I^{3}[V]. □

1.2. Computation of 3-Pfister numbers

Let e = (e_{1}, …, e_{n}) be a family of linearly independent vectors of V with n ≥ 3. Set e_{0} := e_{1} +…+e_{n} and

η_{e} := n + X^{e_{1}} + … + X^{e_{n}} + nX^{e_{0}} + X^{e_{0}+e_{1}} + … + X^{e_{0}+e_{n}}.

Then |D(η_{e})| = 2n if n is even and |D(η_{e})| = 2(n + 1) otherwise. Set

ξ = n + X^{e_{1}} + … + X^{e_{n−1}} + X^{e_{1}+...+e_{n−1}},

then ξ ∈ I^{2}[V] and η_{e} = (1 + X^{e_{0}}) · ξ, hence η_{e} ∈ I^{3}[V]. By [Parimala et al., 2009, Proposition 1.4], Pf_{2}(ξ) = n − 2, thus Pf_{3}(η_{e}) ≤ n − 2. The following proposition proves that we have in fact an equality.

Proposition 1.4 Pf_{3}(η_{e}) = n − 2.

Proof : We prove the equality by induction on n. Suppose that n = 3, then η_{e} = [[e_{1}, e_{2}, e_{3}]] and Pf_{3}(η_{e}) = 1. Suppose now that n > 3. Let W be a subspace of V containing e_{1}, …, e_{n−1} such that V = W ⊕ 𝔽_{2} · e_{n}. Let φ: V → V be the linear map defined by φ(x) = x for all x ∈ W and φ(e_{n}) = 0. Let φ_{∗}: 𝔽_{2}[V] → 𝔽_{2}[V] be the 𝔽_{2}-algebra homomorphism induced by φ. Then φ_{∗}(η_{e}) = η_{e′}, where e′ = (e_{1}, …, e_{n−1}). Suppose that η_{e} = π_{1} + … + π_{p} where the π_{i}’s are 3fold Pfister elements and p = Pf_{3}(η_{e}). Renumbering if necessary, we may assume that e_{n} ∈ D(π_{p}) and thus π_{p} = [[e_{n}, a, b]] for some a, b ∈ V. Observe that φ_{∗}(π_{i}) is a 3-fold Pfister element, for all i ∈ {1, …, p} and φ_{∗}(η_{e}) = φ_{∗}(π_{1}) + … + φ_{∗}(π_{p−1}). Therefore Pf_{3}(η_{e′}) ≤ p − 1. By induction assumption, Pf_{3}(η_{e′}) = n−3, hence p ≥ n−2. Since the other inequality holds, the proof is complete.

□

1.3. Algorithm and classification

We shall give an algorithm to classify elements in I^{3}[V] which is based on the following lemma.

Lemma 1.5 Let ξ ∈ I^{3}[V] be nonzero and let a ∈ D(ξ) \ {0} and W be a subspace of V such that span_{𝔽_{2}}⟨D(ξ)⟩ = W ⊕ 𝔽_{2} · a. Then there exist unique ξ_{1} ∈ I^{3}[W] and ξ_{2} ∈ I^{2}[W] such that ξ = ξ_{1} + (1 + X^{a})ξ_{2}. Moreover, the elements ξ_{1} and ξ_{2} satisfy the following properties:

1. 0 ∈ D(ξ_{2});

2. |D(ξ_{1})| = |D(ξ)| − 2|D(ξ_{1} + ξ_{2}) ∩ D(ξ_{2})|;

3. 0 ∈ D(ξ) ⟺ 0 ∈ D(ξ_{1} + ξ_{2}) ⟺ 0 ∉ D(ξ_{1}).

In particular, |D(ξ_{1})| < |D(ξ)| if 0 ∈ D(ξ).

Proof : By Lemma 1.2, the elements η_{1}, η_{2} ∈ 𝔽_{2}[W] such that ξ = η_{1} + X^{a} · η_{2} satisfy η_{1} + η_{2} ∈ I^{3}[W] and η_{2} ∈ I^{2}[W]. Hence ξ = ξ_{1} + (1 + X^{a})ξ_{2} with ξ_{1} = η_{1} + η_{2} ∈ I^{3}[W] and ξ_{2} = η_{2} ∈ I^{2}[W]. □

These properties are the key to the construction of an algorithm to classify the elements in I^{3}[V], which goes as follows. Assuming we have already classified all the elements ξ ∈ I^{3}[V] such that |D(ξ)| < d, where d > 0, we classify first all the ξ ∈ I^{3}[V] such that |D(ξ)| = d and 0 ∈ D(ξ). To do so, we fix a nonzero a ∈ V and a vector space W such that V = W ⊕ 𝔽_{2} · a. Then we choose ξ_{1} ∈ I^{3}[W] such that |D(ξ_{1})| < d and 0 ∉ D(ξ_{1}), and we set

r := \frac{d − |D(ξ_{1})|}{2} − 1

(r ∈ ℕ because d and |D(ξ_{1})| are even). We construct now an element ξ_{2} ∈ I^{2}[W] such that |D(ξ_{1} + ξ_{2}) ∩ D(ξ_{2})| = r + 1. Thereto, we pick distinct nonzero a_{1}, …, a_{r} ∈ W \ D(ξ_{1}) and distinct b_{1}, …, b_{s} ∈ D(ξ_{1}) such that r + s is odd and a_{1} + … + a_{r} + b_{1} + … + b_{s} = 0. Set

ξ_{2} := X^{0} + X^{a_{1}} + … + X^{a_{r}} + X^{b_{1}} + … + X^{b_{s}}

and ξ := ξ_{1} + (1 + X^{a})ξ_{2}. Then ξ_{2} ∈ I^{2}[W], so ξ ∈ I^{3}[V] and

|D(ξ)| = |D(ξ_{1} + ξ_{2})| + |D(ξ_{2})| = (|D(ξ_{1})| − s + r + 1) + (r + 1 + s) = |D(ξ_{1})| + 2(r + 1) = d.

Moreover all the ξ ∈ I^{3}[V] such that |D(ξ)| = d and 0 ∈ D(ξ) are constructed in this way. The second step is to classify all the elements ξ ∈ I^{3}[V] such that |D(ξ)| = d and 0 ∉ D(ξ). We fix a nonzero a ∈ V and a vector space W such that V = W ⊕ 𝔽_{2} · a. Choose ξ_{1} ∈ I^{3}[W] such that |D(ξ_{1})| ≤ d and 0 ∈ D(ξ_{1}), and set

r := \frac{d − |D(ξ_{1})|}{2} .

To construct an element ξ_{2} ∈ I^{2}[W] such that |D(ξ_{1} + ξ_{2}) ∩ D(ξ_{2})| = r, we pick distinct nonzero a_{1}, …, a_{r} ∈ W \ D(ξ_{1}) and distinct nonzero b_{1}, …, b_{s} ∈ D(ξ_{1}) such that r + s is odd and a_{1} + … + a_{r} + b_{1} + … + b_{s} = 0. Set

ξ_{2} := X^{0} + X^{a_{1}} + … + X^{a_{r}} + X^{b_{1}} + … + X^{b_{s}}

and ξ := ξ_{1} + (1 + X^{a})ξ_{2}. Then ξ_{2} ∈ I^{2}[W], so ξ ∈ I^{3}[V] and

|D(ξ)| = |D(ξ_{1} + ξ_{2})| + |D(ξ_{2})| = (|D(ξ_{1})| − (s + 1) + r) + (r + s + 1) = |D(ξ_{1})| + 2r = d.

Again, all the ξ ∈ I^{3}[V] such that |D(ξ)| = d and 0 ∉ D(ξ) are constructed in this way.

Below we shall classify the elements ξ ∈ I^{3}[V] with |D(ξ)| ≤ 14. (As |D(ξ)| grows, the number of different types of elements ξ ∈ I^{3}[V] becomes huge.) Observe first that |D(ξ)| is even for all ξ ∈ I^{3}[V] because I^{3}[V] ⊂ I[V]. In the next proposition, we start with the elements ξ ∈ I^{3}[V] such that |D(ξ)| ≤ 12. Here we do not need our algorithm to classify those elements (even if it works perfectly) because we can use well-known results on quadratic forms.

Proposition 1.6 Let ξ ∈ I^{3}[V].

1. If |D(ξ)| < 8, then ξ = 0.

2. If |D(ξ)| = 8, then ξ = X^{a}[[e_{1}, e_{2}, e_{3}]] for some vectors e_{1}, e_{2}, e_{3}, a ∈ V where e_{1}, e_{2}, e_{3} are linearly independent. In particular,

GPf_{3}(ξ) = 1 and Pf_{3}(ξ) = \{ \begin{matrix}1 & if 0 ∈ D(ξ), \\ 2 & if 0 ∉ D(ξ). \end{matrix}

3. There is no η ∈ I^{3}[V] such that |D(η)| = 10.

4. If |D(ξ)| = 12, then ξ = ξ_{1} · ξ_{2} for some ξ_{1}, ξ_{2} ∈ 𝔽_{2}[V] such that ξ_{2} ∈ I^{2}[V], |D(ξ_{1})| = 2 and |D(ξ_{2})| = 6. In particular,

GPf_{3}(ξ) = 2 and Pf_{3}(ξ) = \{ \begin{matrix}3 & if 0 ∈ D(ξ), \\ 2 or 4 & if 0 ∉ D(ξ). \end{matrix}

Proof : Replacing V by span_{𝔽_{2}}⟨D(ξ)⟩ if necessary, we may assume that V is finite dimensional. By the proof of Theorem 2.2 in [Parimala et al., 2009], the vector space V may be identified with V_{E} = E^{×}/E^{×2} for some field E of iterated Laurent series so that the map Ψ: 𝔽_{2}[V_{E}] → W (E) is an isomorphism. Let q be a quadratic form over E such that Ψ(ξ) = [q] ∈ I^{3}(E) and the dimension of q is equal to |D(ξ)|.

(1) If |D(ξ)| < 8, then q is isotropic by [Lam, 2005, Hauptsatz 5.1, page 352], hence ξ = 0.

(2) If |D(ξ)| = 8, then q is a scalar multiple of a 3-fold Pfister form (see [Lam, 2005, Theorem 5.6, page 355]), hence ξ = X^{a}[[e_{1}, e_{2}, e_{3}]] for some e_{1}, e_{2}, e_{3}, a ∈ V and GPf_{3}(ξ) = 1. The vectors e_{1}, e_{2}, e_{3} ∈ V are linearly independent because otherwise ξ = 0. If 0 ∈ D(ξ), then ξ = [[e_{1}, e_{2}, e_{3}]] and Pf_{3}(ξ) = 1; otherwise ξ = η_{e} where e = (a, a + e_{1}, a + e_{2}, a + e_{3}) is a family of 4 linearly independent vectors and in particular Pf_{3}(ξ) = 2.

(3) There is no η ∈ I^{3}[V] such that |D(η)| = 10, because if |D(ξ)| = 10, then q is isotropic (see [Pfister, 1966, page 123]); hence ξ = 0 which leads to a contradiction.

(4) Suppose that |D(ξ)| = 12, then by [Pfister, 1966, page 123-124], q can be written as ϕ_{1} ⊥ ϕ_{2} ⊥ ϕ_{3} where the ϕ_{i}’s are 4-dimensional, [ϕ_{i}] ∈ I^{2}(E) and the product of the Clifford invariants C(ϕ_{1})C(ϕ_{2})C(ϕ_{3}) is trivial in the Brauer group of E. Hence there exists a quadratic extension E(\sqrt{δ}) of E which splits all C(ϕ_{i}). Then q = q_{1} ⊗ q_{2} where q_{1} = ⟨1, −δ⟩ and for some 6-dimensional form q_{2} such that [q_{2}] ∈ I^{2}(E). Hence ξ = ξ_{1} · ξ_{2} for some ξ_{1}, ξ_{2} ∈ 𝔽_{2}[V] with ξ_{2} ∈ I^{2}[V], |D(ξ_{1})| = 2 and |D(ξ_{2})| = 6. We may assume that ξ_{1} = X^{0} + X^{a} for some a ∈ V. Let e_{1}, …, e_{5} ∈ V be such that

ξ_{2} = X^{e_{1}} + … + X^{e_{5}} + X^{e_{1}+...+e_{5}}.

Then ξ = X^{e_{1}}[[a, e_{1}+e_{2}, e_{1}+e_{3}]]+X^{e_{4}}[[a, e_{4}+e_{5}, e_{1}+e_{2}+e_{3}+e_{4}]] and GPf_{3}(ξ) ≠ 1 because |D(ξ)| ≠ 8; hence GPf_{3}(ξ) = 2. If 0 ∈ D(ξ), then we may assume that e_{5} = 0. Since |D(ξ)| = 12, the vectors e_{1}, …, e_{4}, a are linearly independent. Then ξ = η_{e} where e = (e_{1}, e_{2}, e_{3}, e_{4}, e_{1} + e_{2} + e_{3} + e_{4} + a) is a family of 5 linearly independent vectors and in particular Pf_{3}(ξ) = 3. If 0 ∉ D(ξ) and e_{1}, …, e_{5}, a are linearly independent, then ξ = η_{e} where e = (e_{1}, …, e_{5}, e_{1} + … + e_{5} + a) is a family of 6 linearly independent vectors, so Pf_{3}(ξ) = 4. Suppose that 0 ∉ D(ξ) and e_{1}, …, e_{5}, a are linearly dependent. If e_{1}, …, e_{5} are linearly independent, then there exists a permutation σ of {1, …, 5} such that e_{σ(5)} = e_{σ(1)} + e_{σ(2)} because |D(ξ)| = 12. Renumbering the e_{1}, …, e_{5} if necessary, we may assume that e_{5} = e_{1} + e_{2}. Then

ξ = (X^{0} + X^{a})(X^{e_{1}} + X^{e_{2}} + X^{e_{1}+e_{2}} + X^{e_{3}} + X^{e_{4}} + X^{e_{3}+e_{4}}) = [[a, e_{1}, e_{2}]] + [[a, e_{3}, e_{4}]]

where e_{1}, …, e_{4}, a are linearly independent. If a ∈ span_{𝔽_{2}}⟨e_{1}, …, e_{5}⟩, then there exists a permutation σ of {1, …, 5} such that a = e_{σ(1)} + e_{σ(2)} + e_{σ(3)}. Renumbering the e_{1}, …, e_{5} if necessary, we may assume that a = e_{1} + e_{2} + e_{3}. Then

ξ = (X^{0} + X^{e_{1}+e_{2}+e_{3}})(X^{e_{1}} + X^{e_{2}} + X^{e_{1}+e_{2}} + X^{e_{4}} + X^{e_{5}} + X^{e_{4}+e_{5}})

and we are in the situation of the previous case. Hence, if 0 ∈ D(ξ) and e_{1}, …, e_{5}, a are linearly dependent, then Pf_{3}(ξ) = 2. □

In the next proposition, we use the above algorithm to classify the elements ξ ∈ I^{3}[V] such that |D(ξ)| = 14.

Proposition 1.7 Suppose that ξ ∈ I^{3}[V] is such that |D(ξ)| = 14. Then there exist e_{1}, …, e_{7} ∈ V such that

ξ = [[e_{1}, e_{2}, e_{3}]] + [[e_{4}, e_{5}, e_{6}]] + [[e_{1}, e_{2}, e_{6}]] + [[e_{4}, e_{5}, e_{7}]] = X^{e_{6}}([[e_{1}, e_{2}, e_{3} + e_{6}]] + [[e_{4}, e_{5}, e_{6} + e_{7}]])

and in particular, GPf_{3}(ξ) = 2. More precisely:

1. If 0 ∈ D(ξ), then ξ = [[e_{1}, e_{2}, e_{3}]] + [[e_{4}, e_{5}, e_{6}]] + [[e_{1}, e_{2}, e_{6}]] for some linearly independent vectors e_{1}, …, e_{6} ∈ V and Pf_{3}(ξ) = 3.

2. If 0 ∉ D(ξ), then there exist linearly independent vectors e_{1}, …, e_{6} ∈ V such that • either ξ = [[e_{1}, e_{2}, e_{3}]] + [[e_{4}, e_{5}, e_{6}]] and Pf_{3}(ξ) = 2;

• or ξ = [[e_{1}, e_{2}, e_{3}]] + [[e_{4}, e_{5}, e_{6}]] + [[e_{1}, e_{2}, e_{4}]] + [[e_{1}, e_{4}, e_{5}]] and Pf_{3}(ξ) = 4;

• or ξ = [[e_{1}, e_{2}, e_{3}]]+[[e_{4}, e_{5}, e_{6}]]+[[e_{1}, e_{2}, e_{6}]]+[[e_{4}, e_{5}, e_{7}]] for some e_{7} ∈ V \span_{F_{2}}⟨e_{1}, …, e_{6}⟩, and Pf_{3}(ξ) = 4.

Proof : Let a, W, ξ_{1}, ξ_{2} be as in Lemma 1.5.

(1) Suppose that 0 ∈ D(ξ). Then |D(ξ_{1})| < 14 and 0 ∉ D(ξ_{1}). Observe that ξ_{1} ≠ 0 because otherwise |D(ξ_{2})| = 7 which is impossible. Hence |D(ξ_{1})| is equal to 8 or 12.

a) If |D(ξ_{1})| = 12, then D(ξ_{1} + ξ_{2}) ∩ D(ξ_{2}) = {0}. Suppose that ξ = η_{e} where e is a family of 6 linearly independent vectors, then ξ_{2} = X^{0} + X^{v_{1}} + … + X^{v_{r}} where r is odd, v_{1}, …, v_{r} ∈ D(η_{e}) are distinct pairwise and v_{1} + … + v_{r} = 0; but such vectors v_{1}, …, v_{r} do not exist in D(η_{e}). Hence, by the proof of Proposition 1.6, there exist linearly independent vectors e_{1}, …, e_{5} ∈ V such that ξ_{1} = [[e_{1}, e_{2}, e_{3}]] + [[e_{3}, e_{4}, e_{5}]]. Then

ξ_{2} = X^{0} + X^{v_{1}} + … + X^{v_{r}} ,
ξ_{1} = X^{v_{1}} + … + X^{v_{12}},
ξ = X^{0} + X^{v_{r+1}} + … + X^{v_{12}} + X^{a}(X^{0} + X^{v_{1}} + … + X^{v_{r}})

for some vectors v_{1}, …, v_{r} ∈ D(ξ_{1}). Replacing W by span_{𝔽_{2}}⟨e_{1}, e_{2}, e_{3}, e_{4}, a⟩ if necessary, we may assume that |D(ξ_{2})| = 4. Choosing another basis of span_{𝔽_{2}}⟨D(ξ_{1})⟩ if necessary, we may assume that ξ_{2} = X^{0} + X^{e_{1}} + X^{e_{2}} + X^{e_{1}+e_{2}} and then

ξ = [[e_{1}, e_{2}, a]] + [[e_{4}, e_{5}, e_{3}]] + [[e_{1}, e_{2}, e_{3}]].

b) If |D(ξ_{1})| = 8, then D(ξ_{1} + ξ_{2}) ∩ D(ξ_{2}) = {0, b, c} and ξ_{1} = η_{e} where e is a family of 4 linearly independent vectors. Therefore

ξ_{2} = X^{0} + X^{b} + X^{c} + X^{v_{1}} + … + X^{v_{r}}

for some pairwise distinct vectors v_{1}, …, v_{r} ∈ D(ξ_{1}) such that r is odd and v_{1} + … + v_{r} = b + c. Observe that b + c ∈ D(η_{e}). Since b, c ∉ D(η_{e}) and there do not exist vectors u, v ∈ span_{𝔽_{2}}⟨D(η_{e})⟩ \ D(η_{e}) such that u + v ∈ D(η_{e}), we have b, c ∉ span_{𝔽_{2}}⟨D(η_{e})⟩. Replacing W by span_{𝔽_{2}}⟨e_{1}, e_{2}, e_{3}, e_{4}, a⟩ if necessary, we may assume that |D(ξ_{2})| = 4 and choosing another basis of span_{𝔽_{2}}⟨D(η_{e})⟩ we may assume that ξ_{2} = X^{0} + X^{e_{1}} + X^{b} + X^{e_{1}+b}. Then

ξ = [[e_{0}, e_{1} + e_{2}, e_{3}]] + [[a, b, e_{1}]] + [[e_{0}, e_{1} + e_{2}, e_{1}]]

where e_{0} := e_{1} + … + e_{4}.

In both the cases,

ξ = [[f_{1}, f_{2}, f_{3}]] + [[f_{4}, f_{5}, f_{6}]] + [[f_{1}, f_{2}, f_{6}]] + [[f_{4}, f_{5}, f_{7}]]

for some linearly independent vectors f_{1}, …, f_{6} ∈ V and with f_{7} = 0. Since 0 ∈ D(ξ), the 3-Pfister number of ξ is odd. Hence Pf_{3}(ξ) = 3 because if Pf_{3}(ξ) = 1, then |D(ξ)| = 8.

(2) Suppose that 0 ∉ D(ξ). Then 0 ∈ D(ξ_{1}) and |D(ξ_{1})| ∈ {8, 12, 14}.

a) If |D(ξ_{1})| = 14, then ξ_{1} = [[e_{1}, e_{2}, e_{3}]] + [[e_{4}, e_{5}, e_{6}]] + [[e_{1}, e_{2}, e_{6}]] for some linearly independent vectors e_{1}, …, e_{6} ∈ V and D(ξ_{2}) ⊂ D(ξ_{1}). Replacing W by span_{𝔽_{2}}⟨e_{2}, …, e_{6}, a⟩ and choosing another basis of span_{𝔽_{2}}⟨D(ξ_{1})⟩, we may assume that ξ_{2} = X^{0} + X^{e_{4}} + X^{e_{5}} + X^{e_{4}+e_{5}}. Hence

ξ = [[e_{1}, e_{2}, e_{3}]] + [[e_{4}, e_{5}, e_{6}]] + [[e_{1}, e_{2}, e_{6}]] + [[e_{4}, e_{5}, a]]

where e_{1}, …, e_{6}, a are linearly independent. Since 0 ∉ D(ξ), the 3-Pfister number of ξ is even. If Pf_{3}(ξ) = 2, then the dimension of span_{𝔽_{2}}⟨D(ξ)⟩ is less than or equal to 6 and we get a contradiction; so Pf_{3}(ξ) = 4. b) If |D(ξ_{1})| = 12, then D(ξ_{1} + ξ_{2}) ∩ D(ξ_{2}) = {b} and ξ_{1} = η_{e} for some family e of 5 linearly

independent vectors. Since b ∈ span_{𝔽_{2}}⟨D(ξ_{1})⟩ \ D(ξ_{1}), there exists a permutation σ of {1, …, 5} such that either b = e_{σ(1)} + e_{σ(2)} or b = e_{σ(1)} + e_{σ(2)} + e_{σ(3)}. Renumbering the e_{i}’s, we may assume that b = e_{1} +e_{2} or b = e_{1} +e_{2} +e_{3}. Then replacing W by span_{𝔽_{2}}⟨e_{1} +e_{2}, e_{3}, e_{4}, e_{5}, a⟩ and choosing another basis of span_{𝔽_{2}}⟨D(ξ_{1})⟩, we may assume that ξ_{2} = X^{0} + X^{e_{1}} + X^{e_{2}} + X^{e_{1}+e_{2}}. Thus

ξ = [[e_{0}, e_{1} + e_{2} + e_{3}, e_{4}]] + [[e_{1} + e_{2}, e_{1}, e_{0}]] + [[e_{0}, e_{1} + e_{2} + e_{3}, e_{1} + e_{2}]] + [[e_{1} + e_{2}, e_{1}, a]]

where e_{0} := e_{1} + … + e_{5}. So ξ can be written as

[[f_{1}, f_{2}, f_{3}]] + [[f_{4}, f_{5}, f_{6}]] + [[f_{1}, f_{2}, f_{6}]] + [[f_{4}, f_{5}, f_{7}]]

for some linearly independent vectors f_{1}, …, f_{5}, f_{7} ∈ V with f_{6} = f_{1} + f_{4}. We know that Pf_{3}(ξ) is even and Pf_{3}(ξ) ≤ 4. To prove that Pf_{3}(ξ) = 4, we consider a linear map φ: V → V such that φ|_{W} is the identity and φ(a) = 0. The map φ induces an algebra homomorphism φ_{∗}: 𝔽_{2}[V] → 𝔽_{2}[V] such that φ_{∗}(ξ) = η_{e}. Then 3 = Pf_{3}(η_{e}) ≤ Pf_{3}(ξ), so Pf_{3}(ξ) = 4.

c) If |D(ξ_{1})| = 8, then D(ξ_{1} + ξ_{2}) ∩ D(ξ_{2}) = {b, c, d} and ξ_{1} = [[e_{1}, e_{2}, e_{3}]] for some linearly independent vectors e_{1}, e_{2}, e_{3} ∈ W. Therefore

ξ_{2} = X^{v_{1}} + … + X^{v_{r}} + X^{b} + X^{c} + X^{d}

for some vectors v_{1}, …, v_{r} ∈ D(ξ_{1}) such that v_{1} + … + v_{r} = b + c + d. The vectors e_{1}, e_{2}, e_{3}, b are linearly independent because b ∉ D(ξ_{1}). Observe that e_{1}, e_{2}, e_{3}, b, c are linearly independent because otherwise either c ∈ D(ξ_{1}) or d ∈ D(ξ_{1}) and we get a contradiction. Replacing W by span_{𝔽_{2}}⟨e_{1}, e_{2}, e_{3}, a, b⟩ if necessary, we may assume that |D(ξ_{2})| = 4. Then ξ_{2} = X^{0} + X^{b} + X^{c} + X^{b+c},

ξ = [[e_{1}, e_{2}, e_{3}]] + [[a, b, c]]

and Pf_{3}(ξ) = 2. □

Remark 1.8 Proposition 1.7 shows that, already for 14-dimensional quadratic forms, the combinatorial analogue 𝔽_{2}[V] diverges from the Witt ring of a general field. Indeed, Hoffmann and Tignol [1998, Example 6.3] constructed a field over which there exist 14-dimensional quadratic forms in the third power of the fundamental ideal which are not a scalar multiple of a sum of two 3-fold Pfister forms. However, Proposition 1.7 shows that, in the combinatorial analogue, all the elements ξ ∈ I^{3}[V] such that |D(ξ)| = 14 can be written as a scaled sum of two 3-fold Pfister elements.

1.4. Bounds for the 3-Pfister number

We shall compute an upper bound for the 3-Pfister number of ξ ∈ I^{3}[V], which only depends on |D(ξ)|. We first prove some preliminary results.

Lemma 1.9 Let ξ ∈ I^{2}[V] \ {0}, then GPf_{2}(ξ) ≤ (|D(ξ)| − 2)/2.

Proof : We use an induction on d := |D(ξ)|. If d = 4, then

ξ = X^{a} + X^{b} + X^{c} + X^{a+b+c} = X^{a}[[a + b, a + c]]

for some a, b, c ∈ V, so GPf_{2}(ξ) = 1. Now suppose that d > 4, then let a, b, c ∈ D(ξ) be distinct pairwise. Set ξ′ := X^{a} + X^{b} + X^{c} + X^{a+b+c} + ξ. Then ξ′ ∈ I^{2}[V] and |D(ξ′)| ≤ d − 2. By induction, GPf_{2}(ξ′) ≤ (d − 4)/2. It implies that

GPf_{2}(ξ) = GPf_{2}(X^{a}[[a + b, a + c]] + ξ′) ≤ 1 + GPf_{2}(ξ′) ≤ 1 + (d − 4)/2 ≤ (d − 2)/2

as wanted. □

Lemma 1.10 Let e = (e_{1}, …, e_{n}) be a family of linearly independent vectors of V where n ≥ 3. Then

GPf_{3}(η_{e}) = \{ \begin{matrix}(n − 2)/2 & if n is even; \\ (n − 1)/2 & otherwise. \end{matrix}

Proof : Set e_{0} := e_{1} +…+e_{n} and ξ := n+X^{e_{1}} +…+X^{e_{n−1}} +X^{e_{0}+e_{n}}. Then ξ ∈ I^{2}[V], |D(ξ)| ≤ n+1 and η_{e} = (1 + X^{e_{0}})ξ. Hence GPf_{3}(η_{e}) ≤ GPf_{2}(ξ) ≤ (|D(ξ)| − 2)/2 ≤ (n − 1)/2. Since Pf_{3}(η_{e}) = n − 2, we have n − 2 ≤ 2GPf_{3}(η_{e}). We obtain that

(n − 2)/2 ≤ GPf_{3}(η_{e}) ≤ (n − 1)/2.

Therefore GPf_{3}(η_{e}) = (n − 2)/2 if n is even and GPf_{3}(η_{e}) = (n − 1)/2 otherwise. □

Lemma 1.11 Let ξ ∈ I^{3}[V] \ {0}. Then there exist linearly independent e_{1}, …, e_{n} ∈ D(ξ) such that e_{1} + … + e_{n−1} ∈ D(ξ) and n ≥ 3.

Proof : Let m be the dimension of span_{𝔽_{2}}⟨D(ξ)⟩. Then m ≥ 3 because |D(ξ)| ≥ 8. Let e_{1}, …, e_{m} ∈ D(ξ) be linearly independent vectors, then there exist v_{1}, …, v_{r} ∈ span_{𝔽_{2}}⟨e_{1}, …, e_{m}⟩ such that

ξ = X^{e_{1}} + … + X^{e_{m}} + X^{v_{1}} + … + X^{v_{r}} .

Since ϵ_{1}(ξ) = 0, we have r ≥ 1. If r = 1, then v_{1} = e_{1} + … + e_{m} because ϵ_{1}(ξ) = 0; it implies that ϵ_{2}(ξ) = ∑_{1≤i<j≤m} e_{i} ∧ e_{j} ≠ 0 (because m ≥ 3) which is impossible. Hence r ≥ 2. Renumbering the vectors e_{1}, …, e_{m} and v_{1}, …, v_{r} if necessary, we may assume that v_{r} = e_{1} +…+e_{n−1} for some n ∈ ℕ, 3 ≤ n ≤ m. So e_{1}, …, e_{n}, e_{1} + … + e_{n−1} ∈ D(ξ) with n ≥ 3 and e_{1}, …, e_{n} linearly independent. □

In the next proposition we establish an upper bound for the dimension of the space spanned by the support of a ξ ∈ I^{3}[V]. (Originally, our proof of this proposition was given by induction on |D(ξ)|; the shorter proof included here was pointed out to us by Karim Becher.)

Proposition 1.12 Let ξ ∈ I^{3}[V]. Set d := |D(ξ)| and m := dim span_{𝔽_{2}}⟨D(ξ)⟩. Then m ≤ d/2; if moreover 0 ∈ D(ξ), then m ≤ (d/2) − 1.

Proof : Let ϵ: 𝔽_{2}[V] → ⋀^{∗} V be the map defined by

ϵ(X^{v_{1}} + … + X^{v_{d}}) = (1 + v_{1}) · . . . · (1 + v_{d}),

where · is the multiplication in the exterior algebra ⋀^{∗} V (Compare with the definition of the Stiefel-Whitney invariant in [Milnor, 1970]). It is easy to check that ϵ(ξ_{1} + ξ_{2}) = ϵ(ξ_{1}) · ϵ(ξ_{2}) for all ξ_{1}, ξ_{2} ∈ 𝔽_{2}[V] and ϵ(ξ) = 1 for all ξ ∈ I^{3}[V]. Let ξ ∈ I^{3}[V], set d := |D(ξ)| and m := dim span_{𝔽_{2}}⟨D(ξ)⟩. Suppose that m ≥ d/2. Write

ξ = X^{v_{1}} + … + X^{v_{n}} + X^{w_{1}} + … + X^{w_{n}}

where v_{1}, …, v_{n} ∈ V are linearly independent and d = 2n. Since ξ ∈ I^{3}[V], we have

ϵ(X^{v_{1}} + … + X^{v_{n}}) = ϵ(ξ)ϵ(X^{w_{1}} + … + X^{w_{n}}) = ϵ(X^{w_{1}} + … + X^{w_{n}}).

We obtain in particular that,

w_{1} ∧ … ∧ w_{n} = v_{1} ∧ … ∧ v_{n} ≠ 0 ∈ ^{⋀n}V.

This implies that span_{𝔽_{2}}⟨v_{1}, …, v_{n}⟩ = span_{𝔽_{2}}⟨w_{1}, …, w_{n}⟩ and m = d/2. Now assume that 0 ∈ D(ξ), then one of the w_{i}’s is equal to zero and w_{1} ∧ … ∧ w_{n} = 0 which leads to a contradiction. Hence m ≤ d/2 and m ≤ d/2 − 1 if 0 ∈ D(ξ). □

From the previous results we can now deduce our upper bound for the 3-Pfister number of elements in I^{3}[V].

Theorem 1.13 Let ξ ∈ I^{3}[V], then GPf_{3}(ξ) ≤ |D(ξ)|^{2}/16, hence Pf_{3}(ξ) ≤ |D(ξ)|^{2}/8.

Proof : We use an induction on d. If d = 0, then ξ = 0 and GPf_{3}(ξ) = 0. Suppose that d > 0. Observe that GPf_{3}(ξ) = GPf_{3}(X^{b}ξ) for all b ∈ V. Hence, replacing ξ by X^{b}ξ for some b ∈ D(ξ), we may assume that 0 ∈ D(ξ). Let e_{1}, …, e_{n} ∈ D(ξ) be linearly independent such that n ≥ 3 and e_{1} + … + e_{n−1} ∈ D(ξ). Set e := (e_{1}, …, e_{n}) and e_{0} := e_{1} + … + e_{n}. Then

ξ = X^{0} + X^{e_{1}} + … + X^{e_{n}} + X^{e_{0}+e_{n}} + X^{v_{1}} + … + X^{v_{r}}

for some v_{1}, …, v_{r} ∈ V and we have

ξ + η_{e} = \{ \begin{matrix}X^{0} + X^{e_{0}+e_{1}} + … + X^{e_{0}+e_{n−1}} + X^{v_{1}} + … + X^{v_{r}} & if n is even, \\ X^{e_{0}} + X^{e_{0}+e_{1}} + … + X^{e_{0}+e_{n−1}} + X^{v_{1}} + … + X^{v_{r}} & if n is odd. \end{matrix}

So ξ + η_{e} ∈ I^{3}[V] and |D(ξ + η_{e})| < |D(ξ)|. By induction, GPf_{3}(ξ + η_{e}) ≤ (d − 2)^{2}/16. Since n ≤ dim span_{𝔽_{2}}⟨D(ξ)⟩ and 0 ∈ D(ξ), by Proposition 1.12, we have n ≤ d/2 − 1. Therefore

GPf_{3}(ξ) ≤ GPf_{3}(η_{e}) + GPf_{3}(ξ + η_{e}) ≤ (n − 1)/2 + (d − 2)^{2}/16 ≤ d/4 − 1 + (d − 2)^{2}/16 ≤ d^{2}/16

which ends the proof. □

2. The 3-Pfister number of quadratic forms

In this section, we shall apply the results in the previous section to quadratic forms over a field F of characteristic different from 2 which contains a square root of −1.

Recall that V_{F} := F^{×}/F^{×2} and that

Ψ: 𝔽_{2}[V_{F}] → W (F): (a_{1}F^{×2}) + … + (a_{d}F^{×2}) ↦ [⟨a_{1}, …, a_{d}⟩]

is a surjective 𝔽_{2}-algebra homomorphism. For all m ∈ ℕ, m ≥ 1, the map Ψ carries m-fold Pfister elements in 𝔽_{2}[V_{F}] to m-fold Pfister forms in W (F). Hence Ψ(I^{m}[V_{F}]) = I^{m}(F) for all m ≥ 1 but, for [q] ∈ I^{m}(F), the number of elements in D(ξ) may be strictly greater than the dimension of q for all ξ ∈ I^{m}[V_{F}] such that Ψ(ξ) = [q].

For m ∈ {1, 2}, every field F satisfies the property

∀a_{1}, …, a_{d} ∈ F^{×}, [⟨a_{1}, …, a_{d}⟩] ∈ I^{m}(F) ⟹ (a_{1}F^{×2}) + … + (a_{d}F^{×2}) ∈ I^{m}[V_{F}],  (1)

meaning in particular that for m ∈ {1, 2} and for all [q] ∈ I^{m}(F), there exists ξ ∈ I^{m}[V_{F}] such that Ψ(ξ) = [q] and |D(ξ)| is less than or equal to the dimension of q. However, for m = 3 this is no longer true for every field F. Theorem 1.1 in [Brosnan et al., 2010] gives counterexamples: for any field k of characteristic different from 2 and for all even integer d ≥ 2, there exist a field extension F/k and [q] ∈ I^{3}(F) with q of dimension d, such that

Pf_{3}(q) ≥ \frac{2^{(d+4)/4} − d − 2}{7} .

If we take in particular a field k which contains a square root of −1 and a d ∈ ℕ such that

\frac{2^{(d+4)/4} − d − 2}{7} > (d^{2})/8,

then we have an example of a [q] ∈ I^{3}(F) with q of dimension d such that Pf_{3}(q) > (d^{2})/8. Write q = ⟨a_{1}, …, a_{d}⟩ and set ξ = (a_{1}F^{×2}) + … + (a_{d}F^{×2}) then ξ ∉ I^{3}[V_{F}] because otherwise

Pf_{3}(q) = Pf_{3}(Ψ(ξ)) ≤ Pf_{3}(ξ) ≤ (d^{2})/8

which leads to a contradiction. Thus, the combinatorial analogue of the Witt ring is less powerful to compute 3-Pfister numbers than it is to compute the 1- and 2-Pfister numbers.

Below we shall first characterize those fields for which property (1) holds for m = 3; for these particular fields, we thus find an upper bound for the 3-Pfister number using the combinatorial analogue. Thereafter we shall discuss the general case.

2.1. Case where we deduce an upper bound from the combinatorial analogue

The homomorphism Ψ induces a map

\overline{Ψ}: I^{2}[V_{F}]/I^{3}[V_{F}] → I^{2}(F)/I^{3}(F): ξ + I^{3}[V_{F}] ↦ Ψ(ξ) + I^{3}(F)

which is a surjective group homomorphism. Observe that the property (1) holds for m = 3 if and only if \overline{Ψ} is an isomorphism. In fact, we shall prove that this is further equivalent to Ψ being an isomorphism. Before we give other equivalent conditions to the assertion that \overline{Ψ} is an isomorphism, we need some definitions. In [1981], Ware says that a ∈ F^{×} is F^{×2}-rigid if F^{×2} +aF^{×2} ⊂ F^{×2} ∪aF^{×2} and that F is F^{×2}-rigid (or rigid) if a is F^{×2}-rigid for all a ∈ F^{×} \F^{×2} (here Ware’s original definition of rigidity for a field F is simplified by the assumption that −1 is a square in F). Following [Ware, 1981], we say that a valuation v on F is F^{×2}-compatible if 1 + 𝔪_{v} ⊂ F^{×2} where 𝔪_{v} is the maximal ideal of the valuation ring of F.

Proposition 2.1 The following conditions are equivalent:

1. \overline{Ψ} is an isomorphism;

2. for all a ∈ F^{×} \ F^{×2}, F^{×2} + aF^{×2} ⊂ F^{×2} ∪ aF^{×2} (i.e., F is F^{×2}-rigid);

3. There exists an F^{×2}-compatible valuation v on F such that (\overline{F} _{v}^{×} : \overline{F} _{v}^{×2}) ≤ 2, where \overline{F}_{v} is the residue field of F.

4. Ψ is an isomorphism;

Proof : (4)⇒(1) Trivial. (1)⇒(2) Let a ∈ F^{×} \ F^{×2} and x, y ∈ F^{×}. Set b := x^{2} + ay^{2}, then b ≠ 0 because a ∉ F^{×2} and the quadratic form ⟨1, a, b, ab⟩ is isotropic. Since it is a Pfister form, we have [⟨1, a, b, ab⟩] = 0 ∈ W (F). Hence (F^{×2})+(aF^{×2})+(bF^{×2})+(abF^{×2}) ∈ I^{3}[V_{F}] because \overline{Ψ} is an isomorphism. By Proposition 1.6, the element (F^{×2}) + (aF^{×2}) + (bF^{×2}) + (abF^{×2}) is equal to zero. Since a ∉ F^{×2}, we obtain that b ∈ F^{×2} ∪ aF^{×2}. (2)⇒(3) By [Ware, 1981, Theorem 3.3], there exists an F^{×2}-compatible valuation v on F such that (\overline{F} _{v}^{×} : \overline{F} _{v}^{×2}) ≤ 2. (3)⇒(2) Let v be an F^{×2}-compatible valuation on F such that (\overline{F} _{v}^{×} : \overline{F} _{v}^{×2}) ≤ 2. Let O_{v} denote the valuation ring of F and for a ∈ O_{v}, let \overline{a} denote the residue class of a in \overline{F}_{v}. By [Arason et al., 1987, Proposition 1.5 (1)], all elements a ∈ F^{×} such that a ∉ O_{v}^{×} ·F^{×2} are F^{×2}-rigid. Now let a ∈ O_{v}^{×} ·F^{×2} be such that a ∉ F^{×2} and let x, y ∈ F^{×}. Then there exist b ∈ O_{v}^{×} and z ∈ F^{×} such that a = bz^{2} and b ∉ F^{×2}. Since 1 + 𝔪_{v} ⊂ F^{×2} and b ∉ F^{×2}, we have \overline{b} ∉ \overline{F} _{v}^{×2}. Observe that \overline{F}_{v} is \overline{F} _{v}^{×2}-rigid because (\overline{F} _{v}^{×} : \overline{F} _{v}^{×2}) ≤ 2, so \overline{b} is \overline{F} _{v}^{×2}-rigid. By [Arason et al., 1987, Proposition 1.5 (2)], we deduce that b is F^{×2}-rigid. Hence

x^{2} + ay^{2} = x^{2} + b(zy)^{2} ∈ F^{×2} ∪ bF^{×2} = F^{×2} ∪ aF^{×2}.

Thus F is F^{×2}-rigid. (2)⇒(4) It is easy to prove that for all a_{1}F^{×2}, …, a_{d}F^{×2} ∈ F^{×}/F^{×2} distinct pairwise,

{a_{1}x_{1}^{2} + … + a_{d}x_{d}^{2} ∈ F^{×} | x_{1}, …, x_{d} ∈ F } = a_{1}F^{×2} ∪ … ∪ a_{d}F^{×2}.

Let ξ be in the kernel of Ψ. Let a_{1}F^{×2}, …, a_{d}F^{×2} ∈ F^{×}/F^{×2} be pairwise distinct elements such that ξ = (a_{1}F^{×2}) + … + (a_{d}F^{×2}) and assume that d ≥ 1. Since Ψ(ξ) = 0, the quadratic form ⟨a_{1}, …, a_{d}⟩ is isotropic. So a_{d} ∈ a_{1}F^{×2} ∪ … ∪ a_{d−1}F^{×2}, leading to a contradiction. Hence ξ = 0 and Ψ is an isomorphism. □

By [Wadsworth, 1983, Proposition 1.2]), if v is an F^{×2}-compatible valuation on F and the residue characteristic is different from 2, then (F, v) is a 2-Henselian valued field. Any valued field with residue characteristic different from 2 is contained in a Henselian (hence 2-Henselian) valued field (F, v) such that (\overline{F} _{v}^{×} : \overline{F} _{v}^{×2}) ≤ 2. Indeed, let (k, w) be a valued field with residue characteristic different from 2. Let (K, v) be a Henselization of (k, w) (see [Efrat, 2006, Theorem 15.3.5]). Assume that (\overline{K}_{v}^{×} : \overline{K}_{v}^{×2}) > 2 and choose α ∈ \overline{K}_{v}^{×} \\overline{K}_{v}^{×2}. Now we take a maximal field extension M of \overline{K}_{v} in a separable closure of \overline{K}_{v} such that α ∉ M^{×2}, then M^{×} = M^{×2}∪αM^{×2}, so (M^{×} : M^{×2}) = 2. By [Efrat, 2006, Corollary 16.1.4], there exists an algebraic field extension F over K such that \overline{F}_{v} = M. Then (\overline{F} _{v}^{×} : \overline{F} _{v}^{×2}) = 2 and (F, v) is Henselian since (K, v) is Henselian and F is an algebraic extension of K. Similarly, taking for M a quadratic closure of \overline{K}_{v} we can also find an F such that (\overline{F} _{v}^{×} : \overline{F} _{v}^{×2}) = 1.

Now we can give an upper bound for the 3-Pfister number:

Theorem 2.2 Let (F, v) be a valued field such that 1 + 𝔪_{v} ⊂ F^{×2} and (\overline{F} _{v}^{×} : \overline{F} _{v}^{×2}) ≤ 2, and let [q] ∈ I^{3}(F) be such that q is of dimension d. Then Pf_{3}(q) ≤ (d^{2})/8.

Proof : Write q = ⟨a_{1}, …, a_{d}⟩. By Proposition 2.1, Ψ is an isomorphism. Hence the element ξ := (a_{1}F^{×2}) + … + (a_{d}F^{×2}) is in I^{3}[V_{F}]. Since |D(ξ)| ≤ d, one has

Pf_{3}(q) = Pf_{3}(Ψ(ξ)) ≤ Pf_{3}(ξ) ≤ (d^{2})/8

and we are done. □

2.2. General case

Now we study the case where Ψ may not be an isomorphism. Let [q] ∈ I^{3}(F) with q = ⟨a_{1}, …, a_{d}⟩. Set ξ := (a_{1}F^{×2}) + … + (a_{d}F^{×2}). Then \overline{Ψ}(ξ + I^{3}[V_{F}]) = [q] + I^{3}(F) = 0, hence ξ + I^{3}(F) is in the kernel of \overline{Ψ}. That is why we shall investigate the structure of the kernel of \overline{Ψ}.

First we introduce some notations. For e = (x_{1}F^{×2}, …, x_{r}F^{×2}) a family of linearly independent vectors of V_{F}, we set ξ_{e} := (r + 1)(F^{×2}) + (x_{1}F^{×2}) + … + (x_{r}F^{×2}) + (x_{1} … x_{r}F^{×2}) ∈ I^{2}[V_{F}].

Lemma 2.3 Let ξ ∈ I^{2}[V_{F}] and set d := |D(ξ)|. Suppose that d > 0, then we have one of the following situations:

1. ξ = ξ_{e} for some family e of linearly independent vectors of V_{F};

2. there exist an integer n, 3 ≤ n ≤ d − 2, and a_{1}F^{×2}, …, a_{n}F^{×2} ∈ D(ξ) linearly independent such that a_{1} … a_{n−1}F^{×2} ∈ D(ξ).

Proof : Let a_{1}F^{×2}, …, a_{d}F^{×2} ∈ D(ξ) be distinct pairwise such that (a_{1}F^{×2}, …, a_{m}F^{×2}) is a basis of span_{𝔽_{2}}⟨D(ξ)⟩, where m ≤ d. Because d > 0 and ξ ∈ I^{2}[V_{F}], we have d ≥ 4. Suppose that ξ ≠ ξ_{e} for all e. Then m ≤ d − 2 and m ≥ 3 because otherwise ξ = (a_{1}F^{×2}) + (a_{2}F^{×2}) + (F^{×2}) + (a_{1}a_{2}F^{×2}) with a_{1}F^{×2}, a_{2}F^{×2} linearly independent. Since the elements a_{1}F^{×2}, …, a_{d}F^{×2} are distinct pairwise, there exists i ∈ {m + 1, …, d} such that a_{i}F^{×2} = a_{σ(1)} … a_{σ(n−1)}F^{×2} for some permutation σ of {1, …, m} and for some integer n, 3 ≤ n ≤ m. Hence the situation (2) occurs. □

We recall a notation introduced in Subsection 1.2: for e := (x_{1}F^{×2}, …, x_{r}F^{×2}) a family of linearly independent vectors of V_{F} with r ≥ 3, we write η_{e} for the element

r(F^{×2}) + (x_{1}F^{×2}) + … + (x_{r}F^{×2}) + r(x_{0}F^{×2}) + (x_{0}x_{1}F^{×2}) … + (x_{0}x_{r}F^{×2})

of 𝔽_{2}[V_{F}], where x_{0}F^{×2} = x_{1} … x_{r}F^{×2}. We have η_{e} ∈ I^{3}[V_{F}] and Pf_{3}(η_{e}) = r − 2.

Lemma 2.4 Let r be an odd integer, r ≥ 3 and let e = (x_{1}F^{×2}, …, x_{r}F^{×2}) be a family of linearly independent vectors of V_{F}. Then ξ_{e} = ξ_{e′} + η_{e}, where e′ := (x_{0}x_{1}F^{×2}, …, x_{0}x_{r−1}F^{×2}) and x_{0} = x_{1} … x_{r}.

Proof : It is an easy computation. □

The next proposition describes the elements in the kernel of \overline{Ψ}.

Proposition 2.5 Let ξ + I^{3}[V_{F}] be in the kernel of \overline{Ψ} and set d := |D(ξ)|. Then either ξ ∈ I^{3}[V_{F}] or there exist η ∈ I^{3}[V_{F}] and a family e of r linearly independent vectors of V_{F} such that Pf_{3}(η) ≤ d^{2}, r is even, 2 ≤ r ≤ d − 2 and ξ = ξ_{e} + η.

Proof : We use an induction on d. Suppose that d = 0, then ξ = 0 ∈ I^{3}[V_{F}]. From now on, we assume that d > 0 and ξ ∉ I^{3}[V_{F}]. Since ξ ∈ I^{2}[V_{F}] and ξ ≠ 0, we have d ≥ 4. By Lemma 2.3, either ξ = ξ_{e} for some family e = (x_{1}F^{×2}, …, x_{r}F^{×2}) of linearly independent vectors of V_{F}, where r ∈ {d − 2, d − 1}, or ξ = (a_{1}F^{×2}) + … + (a_{d}F^{×2}), with a_{n+1} = a_{1} … a_{n−1}, where the vectors a_{1}F^{×2}, …, a_{n}F^{×2} are linearly independent and 3 ≤ n ≤ d − 2. If ξ = ξ_{e} and r = d − 2, then we choose η := 0. If ξ = ξ_{e} and r = d − 1, then by Lemma 2.4, ξ = ξ_{e′} + η_{e}, where e′ := (x_{0}x_{1}F^{×2}, …, x_{0}x_{d−2}F^{×2}) and x_{0} := x_{1} … x_{d−1}. One can choose η := η_{e} since η_{e} ∈ I^{3}[V_{F}] and Pf_{3}(η_{e}) = d − 3 ≤ d^{2}. Hence, we may assume that ξ = (a_{1}F^{×2}) + … + (a_{d}F^{×2}), with a_{n+1} = a_{1} … a_{n−1}, where the vectors a_{1}F^{×2}, …, a_{n}F^{×2} are linearly independent and 3 ≤ n ≤ d − 2. Set e := (a_{1}F^{×2}, …, a_{n}F^{×2}) and a_{0} := a_{1} … a_{n}.

1. Suppose that n is even. Then

ξ + η_{e} = (a_{0}a_{1}F^{×2}) + … + (a_{0}a_{n−1}F^{×2}) + (a_{n+2}F^{×2}) + … + (a_{d}F^{×2}),

hence ξ+η_{e}+I^{3}[V_{F}] = ξ+I^{3}[V_{F}] ≠ 0 with |D(ξ+η_{e})| ≤ d−2. By induction, there exist a family f of r linearly independent vectors of V_{F} and η ∈ I^{3}[V_{F}] such that r is even, 2 ≤ r ≤ d − 4, Pf_{3}(η) ≤ (d − 2)^{2} and ξ + η_{e} = ξ_{f} + η. Then ξ = ξ_{f} + η + η_{e} with η + η_{e} ∈ I^{3}[V_{F}] and

Pf_{3}(η + η_{e}) ≤ (d − 2)^{2} + n − 2 ≤ (d − 2)^{2} + d − 2 ≤ d^{2}

as wanted.

2. Suppose that n is odd, then

ξ + η_{e} = (F^{×2}) + (a_{0}a_{1}F^{×2}) + … + (a_{0}a_{n−1}F^{×2}) + (a_{0}F^{×2}) + (a_{n+2}F^{×2}) + … + (a_{d}F^{×2}).

If |D(ξ + η_{e})| < d then we can proceed as in the case where n is even. Hence we may assume that |D(ξ + η_{e})| = d. Suppose that ξ + η_{e} = ξ_{f} where f = (y_{1}F^{×2}, …, y_{r}F^{×2}) is a family of linearly independent vectors of V_{F}; if r = d − 2 then ξ = ξ_{f} + η_{e} with Pf_{3}(η_{e}) ≤ n − 2 ≤ d^{2} and if r = d − 1, then ξ = ξ_{f′} + η_{f} + η_{e} where f′ := (y_{0}y_{1}F^{×2}, …, y_{0}y_{d−2}F^{×2}), y_{0} = y_{1} … y_{d−1} and Pf_{3}(η_{f} + η_{e}) ≤ d − 3 + n − 2 ≤ 2(d − 2) ≤ d^{2}. Now suppose that ξ + η_{e} ≠ ξ_{f} for all family f, then by Lemma 2.3,

ξ + η_{e} = (b_{1}F^{×2}) + … + (b_{d−1}F^{×2}) + (F^{×2})

with b_{m+1} = b_{1} … b_{m−1}, 3 ≤ m ≤ d − 2 and b_{1}F^{×2}, …, b_{m}F^{×2} linearly independent, because F^{×2} ∈ D(ξ + η_{e}). Set f := (b_{1}F^{×2}, …, b_{m}F^{×2}) and b_{0} := b_{1} … b_{m}. If m is even, then

ξ + η_{e} + η_{f} = (b_{0}b_{1}F^{×2}) + … + (b_{0}b_{m−1}F^{×2}) + (b_{m+2}F^{×2}) + … + (b_{d−1}F^{×2}) + (F^{×2})

and if m is odd then

ξ + η_{e} + η_{f} = (b_{0}b_{1}F^{×2}) + … + (b_{0}b_{m−1}F^{×2}) + (b_{0}F^{×2}) + (b_{m+2}F^{×2}) + … + (b_{d−1}F^{×2}).

In both cases, ξ + η_{e} + η_{f} + I^{3}[V_{F}] = ξ + I^{3}[V_{F}] ≠ 0 with |D(ξ + η_{e} + η_{f})| ≤ d − 2. By induction, ξ + η_{e} + η_{f} = ξ_{g} + η where g is a family of r linearly independent vectors of V_{F}, r is even, 2 ≤ r ≤ d − 4 and Pf_{3}(η) ≤ (d − 2)^{2}. Then ξ = ξ_{g} + η + η_{e} + η_{f} with

Pf_{3}(η + η_{e} + η_{f}) ≤ (d − 2)^{2} + n − 2 + m − 2 ≤ (d − 2)^{2} + 2(d − 2) ≤ d^{2}

and we are done. □

We deduce two corollaries. First the existence of quadratic forms q = ⟨1, x_{1}, …, x_{r}, x_{1} … x_{r}⟩ such that [q] ∈ I^{3}(F) and x_{1}F^{×2}, …, x_{r}F^{×2} are linearly independent is precisely the obstruction for \overline{Ψ} to be an isomorphism.

Corollary 2.6 The equivalent conditions in Proposition 2.1 are also equivalent to the following condition: there do not exist linearly independent vectors x_{1}F^{×2}, …, x_{r}F^{×2} ∈ V_{F} such that r ≥ 2 is even and [⟨1, x_{1}, …, x_{r}, x_{1} … x_{r}⟩] ∈ I^{3}(F).

Proof : If there exist linearly independent vectors x_{1}F^{×2}, …, x_{r}F^{×2} ∈ V_{F} such that r ≥ 2 is even and [⟨1, x_{1}, …, x_{r}, x_{1} … x_{r}⟩] ∈ I^{3}(F), then (F^{×2})+(x_{1}F^{×2})+…+(x_{r}F^{×2})+(x_{1} … x_{r}F^{×2})+I^{3}[V_{F}] is a non-trivial element in the kernel of \overline{Ψ}, hence \overline{Ψ} is not an isomorphism. Conversely, assume that \overline{Ψ} is not an isomorphism, then there exists a non-trivial ξ + I^{3}[V_{F}] in the kernel of \overline{Ψ}. By Proposition 2.5, there exists a family e = (x_{1}F^{×2}, …, x_{r}F^{×2}) of linearly independent vectors of V_{F} such that r ≥ 2 is even and ξ + I^{3}[V_{F}] = ξ_{e} + I^{3}[V_{F}]. Then [⟨1, x_{1}, …, x_{r}, x_{1} … x_{r}⟩] ∈ I^{3}(F). □

Next we relate the 3-Pfister number of a given quadratic form to the one of some particular quadratic form.

Corollary 2.7 Let [q] ∈ I^{3}(F) with q = ⟨a_{1}, …, a_{d}⟩. Then we have one of the following situations: 1. (a_{1}F^{×2}) + … + (a_{d}F^{×2}) ∈ I^{3}[V_{F}] and Pf_{3}(q) ≤ (d^{2})/8;

2. there exist an even integer 2 ≤ r ≤ d − 2 and linearly independent vectors x_{1}F^{×2}, …, x_{r}F^{×2} ∈ V_{F} such that [⟨1, x_{1}, …, x_{r}, x_{1} … x_{r}⟩] ∈ I^{3}(F) and

Pf_{3}(q) ≤ d^{2} + Pf_{3}(⟨1, x_{1}, …, x_{r}, x_{1} … x_{r}⟩).

Proof : Set ξ := (a_{1}F^{×2}) + … + (a_{d}F^{×2}), then \overline{Ψ}(ξ + I^{3}[V_{F}]) = [q] + I^{3}(F) = 0, hence ξ + I^{3}[V_{F}] is in the kernel of \overline{Ψ}. Suppose that ξ ∈ I^{3}[V_{F}], then |D(ξ)| ≤ d and Pf_{3}(q) = Pf_{3}(Ψ(ξ)) ≤ Pf_{3}(ξ) ≤ (d^{2})/8. Suppose that ξ ∉ I^{3}[V_{F}], then by Proposition 2.5, there exist a family e = (x_{1}F^{×2}, …, x_{r}F^{×2}) of linearly independent vectors of V_{F} and η ∈ I^{3}[V_{F}] such that ξ = ξ_{e} + η, r is even, 2 ≤ r ≤ d − 2 and Pf(η) ≤ d^{2}. Since ξ_{e} + I^{3}[V_{F}] = ξ + I^{3}[V_{F}] is in the kernel of \overline{Ψ} we have [⟨1, x_{1}, …, x_{r}, x_{1} … x_{r}⟩] ∈ I^{3}(F). Moreover [q] = Ψ(ξ) = Ψ(ξ_{e}) + Ψ(η), so

Pf_{3}(q) ≤ Pf_{3}(Ψ(ξ_{e})) + Pf_{3}(Ψ(η)) ≤ Pf_{3}(Ψ(ξ_{e})) + Pf_{3}(η) ≤ Pf_{3}(⟨1, x_{1}, …, x_{r}, x_{1} … x_{r}⟩) + d^{2}

and the proof is complete. □

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