On the Nonexcellence of Field Extensions F (π)/F

O. T. Izhboldin^{1}

Received: December 30, 1995 Revised: February 21, 1996

Communicated by Alexander S. Merkurjev

Abstract. For any n ≥ 3, we construct a field F and an n-fold Pfister form φ such that the field extension F (φ)/F is not excellent. We prove that F (φ)/F is universally excellent if and only if φ is a Pfister neighbor of dimension ≤ 4.

Keywords and Phrases: Quadratic forms, Pfister forms, excellent field extensions.

1991 Mathematics Subject Classification: Primary 11E04; Secondary 11E81, 12F20.

Let F be a field of characteristic different from 2 and φ be a non-degenerate quadratic form on an F -vector space V, by which V gets the structure of a non-degenerate quadratic space. Choosing an orthogonal basis of V we can write φ in the form a_{1}x_{1}^{2} + ⋯ + a_{d}x_{d}^{2}. In this case we use the notation φ = ⟨a_{1}, …, a_{d}⟩.

A quadratic form or space φ is called isotropic if φ(v) = 0 for some nonzero vector v ∈ V. We say that φ is anisotropic otherwise. Up to isometry, there is exactly one non-degenerated isotropic 2-dimensional quadratic space, namely the hyperbolic plane ℍ equipped with the form ⟨1, −1⟩. A non-degenerate quadratic space is called hyperbolic if it is isometric to the orthogonal sum of hyperbolic planes mℍ = ℍ ⊥ ⋯ ⊥ ℍ.

According to Witt’s main theorem any non-degenerate quadratic space V can be decomposed in the orthogonal sum V = V_{an} ⊥ V_{h}, where V_{an} is anisotropic and V_{h} ≅ mℍ is a hyperbolic space. (We will use ≅ to denote isometry of quadratic forms or spaces.) Moreover the quadratic space V_{an} is uniquely determined up to isometry. The restriction φ|_{V_{an}} is called the anisotropic part (or anisotropic kernel) of φ and is denoted by φ_{an}. The number m = \frac{1}{2} dim V_{h} is called the Witt index of φ.

For any quadratic space V and any field extension L/F one can provide V_{L} = V ⊗_{F} L with a structure of a quadratic space. The corresponding quadratic form we shall denote by φ_{L}. We say that a quadratic form φ over L is defined over F if there is a quadratic form ξ over F such that φ ≅ ξ_{L}.

1This work was partially supported by grant GAP300 of The International Science Foundation.

It is an important problem to study the behavior of the anisotropic part of forms over F under a field extension L/F. It occurs sometimes that any anisotropic form over F is still anisotropic over L (for example if L/F is of odd degree). In this case for any quadratic form φ over F the anisotropic part (φ_{L})_{an} of φ over L coincides with (φ_{an})_{L} and hence is defined over F.

However, very often φ becomes isotropic over L. In this case we do not know if the anisotropic part of φ over L is defined over F.

A field extension L/F is called excellent if for any quadratic form φ over F the anisotropic part (φ_{L})_{an} of φ over L is defined over F (i.e., there is a form ξ over F such that (φ_{L})_{an} ≅ ξ_{L}).

It is well known that any quadratic extension is excellent. Since any anisotropic quadratic form ψ over F is still anisotropic over the field of rational functions F (t), every purely transcendental field extension is excellent.

Among all field extensions the fields F (φ) of rational functions on the quadric hyper-surface defined by the equation φ = 0 are of special interest in the theory of quadratic forms. One of the important problems is to find a condition on φ so that the field extension F (φ)/F is excellent.

We say that F (φ)/F is universally excellent if for any extension K/F the extension K(φ)/K is excellent.

If φ is isotropic then F (φ)/F is purely transcendental, and it follows from Springer’s theorem that F (φ)/F is excellent and moreover is universally excellent. Thus it is sufficient to consider only the case of anisotropic forms φ.

In [Kn] Knebusch has proved that if φ is an anisotropic form such that F (φ)/F is excellent then φ is a Pfister neighbor. This means that there is a quadratic form π = ⟨1, −a_{1}⟩ ⊗ ⋯ ⊗ ⟨1, −a_{n}⟩ (called n-fold Pfister form) such that φ is similar to a subform of π and dim(φ) > \frac{1}{2} dim(π). This result gives rise to the natural question whether the field extension F (φ)/F is excellent for any Pfister neighbor φ. This problem can be easily reduced to the case of an n-fold Pfister forms φ.

If n = 1 then F (φ)/F is obviously excellent since F (φ)/F is a quadratic extension. Arason [ELW1, Appendix II] has proved that, for n = 2, F (φ)/F is always excellent (see also [R], [LVG]). Thus the answer to our question is yes for n-fold Pfister forms with n ≤ 2. It was an open problem whether F (φ)/F is excellent for any field F and any n-fold Pfister form φ over F (with n ≥ 3).

In [ELW2] some special cases of this problem were considered: for an n-fold Pfister form φ with n ≥ 3, the excellence of the field extension F (φ)/F was proved for all fields with ũ(F) ≤ 4. In [H2] Hoffmann considered another special case of the problem. An extension L/F is called d-excellent if for any quadratic form ψ of dimension ≤ d the anisotropic part (ψ_{L})_{an} of ψ over L is defined over F. Hoffmann has proved that the extension F (φ)/F is 6-excellent for any Pfister neighbor φ.

In this paper we prove that for any n ≥ 3 there is a field F and an n-fold Pfister form φ such that the field extension F (φ)/F is not excellent. Moreover Theorem 1.1 of our paper says that F (φ)/F is universally excellent if and only if φ is a Pfister neighbor of an n-fold Pfister form with n ≤ 2, (i.e., either dim φ ≤ 3 or φ is a 4dimensional form with det(φ) = 1). In §3 we use the main construction of the paper to study “splitting pairs” φ, ψ of quadratic forms. More precisely, we construct a “non standard pair” φ, ψ such that φ is isotropic over the function field F (ψ) of the quadric ψ.

On the Nonexcellence of Field Extensions F (π)/F 129

Remark. Some results of this paper were developed further by D. Hoffmann in [H4].

1. Main Theorem

We will use the following notation throughout the paper: by φ ⊥ ψ, φ ≅ ψ, and [φ] we denote respectively orthogonal sum of forms, isometry of forms, and the class of φ in the Witt ring W (F) of the field F. The maximal ideal of W (F) generated by the classes of even dimensional forms is denoted by I(F). We write φ ∼ ψ if φ is similar to ψ, i.e., kφ = ψ for some k ∈ F^{∗}. The anisotropic part of φ is denoted by φ_{an} and i_{W} (φ) denotes the Witt index of φ. We denote by ⟨⟨a_{1}, …, a_{n}⟩⟩ the n-fold Pfister form

⟨1, −a_{1}⟩ ⊗ ⋯ ⊗ ⟨1, −a_{n}⟩

and by P_{n}(F) the set of all n-fold Pfister forms. The set of all forms similar to n-fold Pfister forms we denote by GP_{n}(F). For any field extension L/F we put φ_{L} = φ ⊗ L, W (L/F) = ker(W (F) → W (L)).

Main Theorem 1.1. Let φ be an anisotropic form over F. Then the following conditions are equivalent.

(i) The field extension F (φ)/F is universally excellent, i.e., for any field extension E/F the extension E(φ)/E is excellent.

(ii) Either dim(φ) ≤ 3 or φ ∈ GP_{2}(F).

Proof of (ii) ⇒ (i). The case dim(φ) = 2 is obvious. If dim(φ) = 3 or φ ∈ GP_{2}(F) the excellence of the extension E(φ)/E was proved by Arason (see the introduction).

Proof of (i) ⇒ (ii). Since E(φ)/E is excellent for any extension E/F, we see that F (φ)/F is excellent. It was shown in [Kn, 7.13] that for F (φ)/F to be excellent it is necessary that φ is a Pfister neighbor. Let φ be a Pfister neighbor of the n-fold Pfister form π. Since F (φ) and F (π) are F -equivalent, we can replace φ by π, i.e., we can suppose that φ = π is an n-fold Pfister form. Thus it is sufficient to prove the following proposition.

Proposition 1.2. Let π be anisotropic n-fold Pfister form over the field F. If n ≥ 3 then there is a field extension E/F such that E(π)/E is not excellent.

2. Proof of Proposition 1.2

Lemma 2.1. Let π and τ be anisotropic n-fold Pfister forms over the field F. Then there is a field extension K/F such that the following conditions hold.

a) π_{K} = τ_{K},

b) π_{K} and τ_{K} are anisotropic.

Proof. Let φ be a Pfister neighbor of τ of dimension 2^{n−1} + 1. It follows from [H3, Theorem 4] that there exists a field extension K/F such that π_{K} is anisotropic and φ_{K} ⊂ π_{K}. Hence φ_{K} is a Pfister neighbor of π_{K}. Since φ_{K} is a Pfister neighbor of τ_{K}, we have π_{K} = τ_{K}. □

Lemma 2.2. Let τ and π be anisotropic n-fold Pfister forms over F. Suppose that there is a ∈ F^{∗} such that τ_{F (\sqrt{a})} and π_{F (\sqrt{a})} are isotropic. Then there is an extension E/F and x ∈ E^{∗} such that the following conditions hold.

1) π_{E(\sqrt{x})} = τ_{E(\sqrt{x})},

2) π_{E(\sqrt{x})} and τ_{E(\sqrt{x})} are anisotropic,

3) E/F is unirational.

Remark: We say that E/F is unirational, if there is a purely transcendental finitely generated field extension K/F such that F ⊂ E ⊂ K.

Proof. Since τ is an n-fold Pfister form and τ_{F (\sqrt{a})} is isotropic, we can write τ in the form τ = ⟨⟨a, b_{1}, …, b_{n−1}⟩⟩. Similarly, we can write π in the form π = ⟨⟨a, c_{1}, …, c_{n−1}⟩⟩. Let F̃ = F (A, B_{1}, …, B_{n−1}, C_{1}, …, C_{n−1}) be the rational function field in 2n − 1 variables over F̃.

Put τ̃ = ⟨⟨A, B_{1}, …, B_{n−1}⟩⟩ and π̃ = ⟨⟨A, C_{1}, …, C_{n−1}⟩⟩. Let γ = τ ⊥ −π and γ̃ = τ̃ ⊥ −π̃. Let E/F̃ be the universal field extension such that γ_{E} = γ̃_{E}, i.e., E = F̃_{h}, where F̃ = F̃_{0}, F̃_{1}, …, F̃_{h} is a generic splitting tower of the quadratic form γ ⊥ −γ̃.

It is well known that the following universal property of E holds: For any field extension K/F̃ the condition γ_{K} = γ̃_{K} implies that EK/K is purely transcendental.

Now we prove that conditions 1)–3) of the lemma hold for x = A.

1) We have [τ_{E(\sqrt{A})}] − [π_{E(\sqrt{A})}] = [γ_{E(\sqrt{A})}] = [γ_{E(\sqrt{A})}] = [τ_{E(\sqrt{A})}] − [π_{E(\sqrt{A})}] = 0. Hence [τ_{E(\sqrt{A})}] = [π_{E(\sqrt{A})}].

2) Let K/F be as in Lemma 2.1, i.e., τ_{K}, π_{K} are anisotropic and τ_{K} = π_{K}. We have [γ_{K}] = [τ_{K}] − [π_{K}] = 0

Let K̃ = K(A, B_{1}, …, B_{n−1}, C_{1}, …, C_{n−1}) be the rational function field in 2n − 1 variables over K. We have [γ_{K̃(\sqrt{A})}] = [τ_{K̃(\sqrt{A})}] − [π_{K̃(\sqrt{A})}] = 0 and [γ_{K̃(\sqrt{A})}] = [τ_{K̃(\sqrt{A})}]−[π_{K̃(\sqrt{A})}] = 0. Therefore [γ_{K̃(\sqrt{A})}] = [γ_{K̃(\sqrt{A})}]. Using the universal property

of E/F̃ we see that EK̃(\sqrt{A})/K̃(\sqrt{A}) is purely transcendental.

It is clear that K̃(\sqrt{A})/K is purely transcendental. Therefore EK̃(\sqrt{A})/K is purely transcendental. Hence τ_{EK̃(\sqrt{A})} and π_{EK̃(\sqrt{A})} are anisotropic. Therefore τ_{E(\sqrt{A})} and π_{E(\sqrt{A})} are anisotropic.

3) Let L = F̃(\sqrt{A/a}, \sqrt{B_{1}/b_{1}}, …, \sqrt{B_{n−1}/b_{n−1}}, \sqrt{C_{1}/c_{1}}, …, \sqrt{C_{n−1}/c_{n−1}}). It is clear that π_{L} = π̃_{L} and τ_{L} = τ̃_{L}. Therefore γ_{L} = γ̃_{L}. Using the universal property of E/F̃ we see that EL/L is purely transcendental. It is clear that L/F is purely transcendental. Hence EL/F is purely transcendental. Since E ⊂ EL we see that E/F is unirational. □

Lemma 2.3. Let F be a field and π be anisotropic n-fold Pfister form over F. Then there are a unirational extension E/F, an n-fold Pfister form τ over E, and x ∈ E^{∗} such that the following conditions hold.

1) π_{E(\sqrt{x})} = τ_{E(\sqrt{x})},
2) π_{E(\sqrt{x})} and τ_{E(\sqrt{x})} are anisotropic,
3) dim(π_{E} ⊥ −τ_{E})_{an} = 2^{n+1} − 4.

Proof. Write π in the form π = ⟨⟨a, b_{1}, b_{2}, …, b_{n−1}⟩⟩. Let F̃ = F (T_{1}, …, T_{n−1}) be the rational function field in n−1 variables over F. Let τ = ⟨⟨a, T_{1}, …, T_{n−1}⟩⟩. Obviously

(πF̃ ⊥ −τ)_{an} = ⟨⟨a⟩⟩⟨⟨b_{1}, …, b_{n−1}⟩⟩′ ; F̃ ⊥ −⟨⟨a⟩⟩⟨⟨T_{1}, …, T_{n−1}⟩⟩′.

Therefore dim(πF̃ ⊥ −τ)_{an} = 2^{n+1} − 4.

The quadratic forms π_{F̃(\sqrt{a})} and τ_{F̃(\sqrt{a})} are hyperbolic, i.e., all the conditions of

Lemma 2.2 hold for F̃, π, τ. Hence there is a unirational extension E/F̃ such that 1) π_{E(\sqrt{x})} = τ_{E(\sqrt{x})},

2) π_{E(\sqrt{x})} and τ_{E(\sqrt{x})} are anisotropic,

Since E/F̃ is unirational, we have dim(π_{E} ⊥ −τ_{E})_{an} = dim(πF̃ ⊥ −τ)_{an} = 2^{n+1} − 4. Finally E/F is unirational since E/F̃ is unirational and F̃/F is purely transcendental. □

Lemma 2.4. Let E be a field, n ≥ 3, x ∈ E^{∗}. Let π, τ ∈ P_{n}(E) be such that 1) π_{E(\sqrt{x})} = τ_{E(\sqrt{x})}.

2) π_{E(\sqrt{x})} and τ_{E(\sqrt{x})} are anisotropic.
3) dim(π ⊥ −τ)_{an} = 2^{n+1} − 4.

Let ψ = τ′ ⊥ ⟨x⟩ where τ′ is such that τ = τ′ ⊥ ⟨1⟩.

Then

a) ψ is anisotropic.

b) ψ_{E(π)} is isotropic.

c) There is no quadratic form γ over E such that (ψ_{E(π)})_{an} = γ_{E(π)}.

d) For any subform ξ ⊊ ψ the form ξ_{F (π)} is anisotropic, i.e., ψ is a minimal F (π)-form.

Proof. a) Obviously ψ_{E(\sqrt{x})} = τ_{E(\sqrt{x})}. By assumption we see that τ_{E(\sqrt{x})} is anisotropic. Hence ψ_{E(\sqrt{x})} is anisotropic. Therefore ψ is anisotropic too.

b) Suppose that ψ_{E(π)} is anisotropic. Since ψ_{E(\sqrt{x})} = τ_{E(\sqrt{x})} = π_{E(\sqrt{x})} we have [ψ_{E(π)(\sqrt{x})}] = [π_{E(π)(\sqrt{x})}] = 0. Since ψ_{E(π)} is anisotropic and ψ_{E(π)(\sqrt{x})} is hyperbolic, we conclude that ψ_{E(π)} = ⟨⟨x⟩⟩ξ where ξ is a quadratic form over E(π). Since dim(ξ) = 2^{n−1} is even, we have ξ ∈ I(E(π)). Therefore ψ_{E(π)} = ⟨⟨x⟩⟩ξ ∈ I^{2}(E(π)). Hence ψ ∈ I^{2}(E). Therefore [⟨⟨x⟩⟩] = [τ] − [ψ] ∈ I^{2}(E), a contradiction.

c) Suppose that (ψ_{E(π)})_{an} = γ_{E(π)} where γ is a quadratic form over E. It is clear that dim(γ) ≤ 2^{n} − 2. We have (ψ ⊥ −γ)_{an} ∈ W (E(π)/E). Since π is a Pfister form we conclude that (ψ ⊥ −γ)_{an} = πμ, with μ a quadratic form over E.

Since 2 = 2^{n} − (2^{n} − 2) ≤ dim(ψ ⊥ −γ)_{an} = 2^{n} + (2^{n} − 2) = 2^{n+1} − 2 and dim(π) = 2^{n} divides dim(πμ) we conclude that dim(μ) = 1. Writing μ in the form μ = ⟨k⟩ we have (ψ ⊥ −γ)_{an} = kπ. Hence [kπ] = [ψ] − [γ]. Therefore

[τ ⊥ −kπ] = [τ] − [kπ] = ([ψ] + [⟨⟨x⟩⟩]) − ([ψ] − [γ]) = [⟨⟨x⟩⟩ ⊥ γ].

Hence τ and kπ contain a common subform of dimension

\frac{1}{2}(dim(τ) + dim(kπ) − dim(⟨⟨x⟩⟩ ⊥ γ)) ≥ \frac{1}{2}(2^{n} + 2^{n} − 2^{n}) = 2^{n−1} ≥ 2^{3−1} = 4 > 3.

Therefore there is a 3-dimensional form ρ such that ρ ⊂ τ, ρ ⊂ kπ. Let a, b ∈ E be such that ρ ∼ ⟨1, −a, −b⟩. Let ε = ⟨⟨a, b⟩⟩. Obviously τ_{E(ε)} and π_{E(ε)} are isotropic. Since τ, π, and ε are anisotropic Pfister forms, we conclude that ε ⊂ τ and ε ⊂ π. Therefore dim(π ⊥ −τ)_{an} ≤ dim(π) + dim(τ) − 2 dim(ε) = 2^{n} + 2^{n} − 2 · 4 = 2^{n+1} − 8, a contradiction.

d) We can suppose that ξ is a (2^{n} − 1)-dimensional subform of ψ. let k ∈ E^{∗} be such that ξ ⊥ ⟨−k⟩ = ψ. Set ξ̃ = ξ ⊥ ⟨−xk⟩. We have

[τ] − [ξ̃] = [τ] − ([ξ] − [⟨xk⟩]) = ([ψ] + [⟨⟨x⟩⟩]) − ([ψ] + [⟨k⟩] − [⟨xk⟩]) = [⟨⟨x, k⟩⟩].

Let ρ = ⟨⟨x, k⟩⟩. We have [τ_{E(ρ)}] = [ξ̃_{E(ρ)}]. Comparing dimensions we see that τ_{E(ρ)} = ξ̃_{E(ρ)}. Therefore τ_{E(ρ,π)} = ξ̃_{E(ρ,π)}.

Our goal is to prove that ξ_{E(π)} is anisotropic. Let us suppose that ξ_{E(π)} is isotropic. Then ξ̃_{E(ρ,π)} is isotropic too. Therefore τ_{E(ρ,π)} is isotropic. Hence the Pfister form τ_{E(ρ)} becomes isotropic over the function field of the Pfister form π_{E(ρ)}. Therefore either τ_{E(ρ)} or τ_{E(ρ)} = π_{E(ρ)} is hyperbolic.

Suppose first that τ_{E(ρ)} is hyperbolic. Since ρ_{E(\sqrt{x})} = ⟨⟨x, k⟩⟩_{E(\sqrt{x})} is isotropic we conclude that τ_{E(\sqrt{x})} is isotropic. This contradicts the assumption in this lemma.

Let now τ_{E(ρ)} = π_{E(ρ)}. Then (τ ⊥ −π)_{an} ∈ W (E(ρ)/E). Hence (τ ⊥ −π)_{an} = ρλ with λ a quadratic form over E ([S, Ch.4,5.6]). Since dim(τ ⊥ −π)_{an} = 2^{n} − 4 and dim(ρ) = 4 we conclude that dim(λ) = (2^{n} − 4)/4 = 2^{n−2} − 1. Since n ≥ 3 we see that dim(λ) is odd and hence [λ] ≡ [⟨1⟩] (mod I(E)). Since ρ ∈ I^{2}(E) we have [ρλ] ≡ [ρ] (mod I^{3}(E)). Since τ, π ∈ P_{n}(E) and n ≥ 3, we see that [(τ ⊥ −π)_{an}] ≡ 0 (mod I^{3}(E)). We have

[ρ] ≡ [ρλ] = [(τ ⊥ −π)_{an}] ≡ 0 (mod I^{3}(E)).

Since dim(ρ) = 4 < 8 we conclude that ρ is hyperbolic. Therefore (τ ⊥ −π)_{an} = ρλ is hyperbolic. However dim(τ ⊥ −π)_{an} = 2^{n} − 4 > 0, a contradiction. □

Corollary 2.5. Let π be an anisotropic n-fold Pfister form over the field F. If n ≥ 3 then there is a unirational extension E/F such that E(π)/E is not excellent. □

This corollary completes the proof of Proposition 1.2 and Theorem 1.1.

Corollary 2.6. Let n ≥ 3. Then there are a field E, an n-fold Pfister form π over E, and a 2^{n}-dimensional form ψ over E such that ψ is an E(π)-minimal form. □

Corollary 2.7. Let n ≥ 3. Then there are a field E and 2^{n}-dimensional forms ψ and π over E such that ψ is an E(π)-minimal form and ψ is not similar to π. □

3. Nonstandard Splitting

An important problem in the theory of quadratic forms is to determine when an anisotropic quadratic form φ over F becomes isotropic over the function field F (ψ) of another form ψ. There are some well-known situations when this occurs and we list some of them in the following two definitions.

Definition 3.1. Let φ and ψ be anisotropic quadratic forms. We say that the ordered pair φ, ψ is elementary splitting (or elementary) if one of the following conditions holds.

1) There is a k ∈ F^{∗} such that kψ ⊂ φ;

2) There is a k ∈ F^{∗}, such that kφ ⊂ ψ and dim(φ) > dim(ψ) − i_{1}(ψ);

3) There is a ρ ∈ W (F (ψ)/F) such that dim(ρ) < 2 dim(φ) and kφ ⊂ ρ for some k ∈ F^{∗}.

Definition 3.2. Let φ and ψ be anisotropic quadratic forms. We say that the ordered pair φ, ψ is standard if there is a collection

φ_{0} = φ, φ_{1}, …, φ_{n−1}, φ_{n} = ψ

such that the pair φ_{i−1}, φ_{i} is elementary for each i = 1, 2, …, n.

It is clear that if the pair (φ, ψ) is elementary splitting or standard, then φ_{F (ψ)} is isotropic.

Examples 3.3. Let φ and ψ be anisotropic quadratic forms such that φ_{F (ψ)} is isotropic. Suppose that at least one of the following conditions holds

a) φ is a Pfister neighbor;

b) dim(ψ) ≤ 3, or ψ ∈ GP_{2}(F);

c) dim(φ) ≤ 5; Then the pair φ, ψ is elementary.

Proof. a) Let φ be a Pfister neighbor of ρ. Then condition 3) of Definition 3.1 is fulfilled.

b) By the excellence property of the field extension F (ψ)/F there exists an anisotropic form ξ over F such that (φ_{F (ψ)})_{an} = ξ_{F (ψ)}. Setting ρ = φ ⊥ −ξ one can see that condition 3) of Definition 3.1 holds.

c) Let dim(φ) ≤ 5. We can suppose that φ is not a Pfister neighbor and ψ ∉ GP_{2}(F) (see a), b) ). Then φ_{F (ψ)} is isotropic if and only if φ contains a subform similar to ψ (see [H1, Th. 1, Main Theorem]). Therefore condition 1) of Definition 3.1 holds. □

Example 3.4. Let F = ℝ(T), φ = ⟨T, T, T, 1, 1, 1, 1, 1⟩, ψ = ⟨T, T, 1, 1, 1, 1, 1, 1⟩. Then the pair φ, ψ is standard but not elementary.

Proof. Let ρ = ⟨T, T, 1, 1, 1, 1, 1⟩. Since ρ ⊂ φ, the pair (φ, ρ) is elementary. Since ρ ⊂ ψ and dim(ρ) = 7 > 8 − 2 = dim(ψ) − i_{1}(ψ), we see that the pair (ρ, ψ) is elementary. Since the pairs (φ, ρ) and (ρ, ψ) are elementary, we see that the pair (φ, ψ) is standard. It follows from Lemma 3.7 below that the pair (φ, ψ) is not elementary. □

In this section we construct a pair of anisotropic forms φ and ψ with φ_{F (ψ)} isotropic which is not standard.

Lemma 3.5. Let F be a field, n ≥ 3, x ∈ F^{∗}. Let π, τ ∈ P_{n}(F) be such that 1) π ≠ τ,

2) π_{F (\sqrt{x})} = τ_{F (\sqrt{x})},
3) π_{F (\sqrt{x})} and τ_{F (\sqrt{x})} are anisotropic.

Let φ = π′ ⊥ ⟨x⟩ and ψ = τ′ ⊥ ⟨x⟩. Then a) ψ and φ are anisotropic,

b) φ_{F (ψ)} and ψ_{F (φ)} are isotropic,

c) φ ≁ ψ.

Proof. a) Obviously ψ_{F (\sqrt{x})} = π_{F (\sqrt{x})} and ψ_{F (\sqrt{x})} = τ_{F (\sqrt{x})}. It follows from condition 3) that φ and ψ are anisotropic.

b) Let us suppose that φ_{F (ψ)} is anisotropic. Since φ_{F (\sqrt{x})} = π_{F (\sqrt{x})} and ψ_{F (\sqrt{x})} = τ_{F (\sqrt{x})} = π_{F (\sqrt{x})} we see that φ_{F (ψ,\sqrt{x})} = π_{F (π,\sqrt{x})}. Since π ∈ P_{n}(F) we conclude that φ_{F (ψ,\sqrt{x})} is hyperbolic. Therefore φ_{F (ψ)} = ⟨⟨x⟩⟩ξ where ξ is a quadratic form over F (ψ). Since dim(ξ) = 2^{n−1} is even, we have ξ ∈ I(F (ψ)). Therefore ψ_{F (ψ)} = ⟨⟨x⟩⟩ξ ∈ I^{2}(F (ψ)). Hence ψ ∈ I^{2}(F). Therefore [⟨⟨x⟩⟩] = [τ] − [ψ] ∈ I^{2}(F), a contradiction.

c) Suppose that kφ = ψ. Then [kπ] − [k⟨⟨x⟩⟩] = [kφ] = [ψ] = [τ] − [⟨⟨x⟩⟩]. Therefore [⟨⟨x, k⟩⟩] = [τ]−[kπ] ∈ I^{n}(F) ⊂ I^{3}(F). Since dim(⟨⟨x, k⟩⟩) = 4 < 8, we have [τ]−[kπ] = [⟨⟨x, y⟩⟩] = 0. Hence τ ∼ π. Since τ, π ∈ P_{n}(F) we see that τ = π, a contradiction. □

Lemma 3.6. Let π ∈ P_{3}(F) and x ∈ F^{∗} (x ∉ F^{∗2}) be such that π_{F (\sqrt{x})} is anisotropic. Let φ = π′ ⊥ ⟨x⟩. Suppose that ψ is an anisotropic quadratic form such that ψ_{F (φ)} and φ_{F (ψ)} are isotropic. Then dim(ψ) = 8.

By C(φ) (resp. C_{0}(φ)) we will denote the Clifford algebra (resp. even Clifford algebra) of the quadratic form φ. If they are central simple we denote their classes in the Brauer group of the underlying field by [C(φ)] (resp. [C_{0}(φ)]).

Proof. Since dim(φ) = 8 and φ_{F (ψ)} is isotropic, it follows from Hoffmann’s theorem [H3, §1, Theorem 1] that dim(ψ) ≤ 8.

Suppose that dim(ψ) ≤ 6. Since dim(φ) = 8 and ψ_{F (φ)} is isotropic, it follows from Hoffmann’s theorems [H1], [H2] that φ ∈ GP_{3}(F). Therefore x = det(φ) = 1, a contradiction.

Consider now the case dim(ψ) = 7. Since π_{F (ψ,\sqrt{x})} = φ_{F (ψ,\sqrt{x})} is isotropic we see that ψ_{F (\sqrt{x})} is a Pfister neighbor of π_{F (\sqrt{x})}. Therefore [C_{0}(ψ)_{F (\sqrt{x})}] = 0. Hence there is y ∈ F^{∗} such that [C_{0}(ψ)] = [\binom{x,y}{F}]. Let ρ = ⟨⟨x, y⟩⟩.

We claim that ψ_{F (ρ)} is an anisotropic Pfister neighbor. To prove this we consider the quadratic form ψ̃ = ψ ⊥ ⟨det(ψ)⟩. Since dim(ψ̃) = 8 and [C(ψ̃_{F (ρ)})] = [\binom{x,y}{F (ρ)}] = 0

we have ψ̃_{F (ρ)} ∈ GP_{3}(F (ρ)). If ψ_{F (ρ)} is isotropic then ψ̃_{F (ρ)} is isotropic too and hence hyperbolic. Therefore, (ψ̃)_{an} = ρμ. Since dim(ψ̃) = 6 or 8 we must have dim μ = 2 which implies ψ̃_{an} = ψ̃ ∈ GP_{3}(F). Therefore [C(ρ)] = [C_{0}(ψ)] = [C(ψ̃)] = 0. Hence, ρ is hyperbolic and ψ stays anisotropic over F (ρ), a contradiction.

Since ψ_{F (φ)} is isotropic, ψ_{F (ρ)} becomes isotropic over the functional field of the form φ_{F (ρ)}. Since ψ_{F (ρ)} is an anisotropic Pfister neighbor and dim(φ_{F (ρ)}) = 8 we see that φ_{F (ρ)} ∈ GP_{3}(F (ρ)) ⊂ I^{2}(F (ρ)). Since W (F)/I^{2}(F) → W (F (ρ))/I^{2}(F (ρ)) is injective we have φ ∈ I^{2}(F). Hence x = det(φ) = 1, a contradiction. □

Lemma 3.7. Let φ and ψ be anisotropic 8-dimensional quadratic form such that ψ ∉ GP_{3}(F) and the pair φ, ψ is elementary. Then φ ∼ ψ.

Proof. Since the pair φ, ψ is elementary, one of conditions 1)–3) of Definition 3.1 holds. Since dim(φ) = dim(ψ), both the conditions 1), 2) imply that φ ∼ ψ. Now we suppose that condition 3) holds, i.e., there is ρ ∈ W (F (ψ)/F) such that dim(ρ) < 2 dim(φ) = 16 and kφ ⊂ ρ. Since dim(ψ) > 4, the homomorphism W (F)/I^{3}(F) → W (F (ψ))/I^{3}(F (ψ)) is injective. Hence ρ ∈ I^{3}(F). Let σ ∈ P_{2}(F) be such that ψ contains a Pfister neighbor of σ. Then ρ ∈ W (F (ψ)/F) ⊂ W (F (σ)/F) and thus ρ_{an} ≅ σμ for some μ. If dim μ is odd then σ ≡ σμ = ρ ≡ 0 (mod I^{3}(F)), a contradiction. Thus dim μ is even and 8| dim(ρ_{an}). Therefore dim(ρ_{an}) = 8. Hence ρ_{an} ∈ GP_{3}(F). Since ρ_{F (ψ)} is hyperbolic, ψ is a Pfister neighbor in ρ_{an}. Since dim(ψ) = dim(ρ_{an}) = 8 we have ψ ∼ ρ_{an} ∈ GP_{3}(F), a contradiction. □

Lemma 3.8. Let n = 3, and let φ, ψ be as in Lemma 3.5. Then the pair φ, ψ is not standard.

Proof. Assume that the pair φ, ψ is standard. Then there is a collection

φ_{0} = φ, φ_{1}, …, φ_{n−1}, φ_{n} = ψ

such that the pair φ_{i−1}, φ_{i} is elementary for each i = 1, 2, …, n. Obviously, the quadratic forms φ_{F (φ_{i})} and (φ_{i})_{F (ψ)} are isotropic. Since ψ_{F (φ)} is isotropic (see Lemma 3.5) and (φ_{i})_{F (ψ)} is isotropic, we see that (φ_{i})_{F (φ)} is isotropic too. Thus φ_{F (φ_{i})} and (φ_{i})_{F (φ)} are isotropic. It follows from Lemma 3.6 that dim(φ_{i}) = 8.

Consider first the case ψ_{i} ∈ GP_{3}(F). Since (φ_{i})_{F (φ)} and is isotropic, φ is a Pfister neighbor of ψ_{i}. Since dim(φ) = dim(ψ_{i}) = 8 we have φ ∼ ψ_{i}. Hence φ ∈ GP_{3}(F), a contradiction.

Thus we have proved that dim(φ_{i}) = 8 and ψ_{i} ∉ GP_{3}(F) for each i = 1, 2, …, n. It follows from Lemma 3.7 that φ_{i−1} ∼ φ_{i}. We have

φ = φ_{0} ∼ φ_{1} ∼ ⋯ ∼ φ_{n} = ψ.

On the other hand, it follows from Lemma 3.5 that φ ≁ ψ. The contradiction obtained proves the lemma. □

Theorem 3.9. For any field F there is a unirational field extension E/F and a pair of 8-dimensional anisotropic quadratic forms φ and ψ over E such that φ_{E(ψ)} is isotropic, but the pair φ, ψ is not standard.

Proof. Let n = 3. Let E, π and τ be such as in Lemma 2.3. Set φ = π′ ⊥ ⟨x⟩, ψ = τ′ ⊥ ⟨x⟩. It is clear that all the conditions of Lemma 3.5 hold. Now the desired result follows immediately from Lemma 3.5 and Lemma 3.8. □

References

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[ELW2] Elman, R.; Lam, T.Y.; Wadsworth, A.R., Function fields of Pfister forms, Invent.Math. 51 (1979), 61–75.

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[H3] Hoffmann, D. W., Isotropy of quadratic forms over the function field of a quadric., Math. Z. 220 (1995), 461–466.

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[LVG] Lewis, D.W.; Van Geel, J., Quadratic forms isotropic over the function field of a conic, Indag. Math. 5 (1994), 325–339.

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[S] Scharlau, W., Quadratic and Hermitian Forms, Springer, Berlin, Heidelberg, New York, Tokyo, 1985.

O. T. Izhboldin

Department of Mathematics and Mechanics St.-Petersburg State University Petrodvorets, 198904, Russia