NONEXCELLENCE OF CERTAIN FIELD EXTENSIONS

A.S. Sivatski

Department of Mathematics-2 St.Petersburg Electrotechnical University 197376, St.Petersburg, Russia

Abstract. Consider towers of fields F_{1} ⊂ F_{2} ⊂ F_{3}, where F_{3}/F_{2} is a quadratic extension and F_{2}/F_{1} is an extension, which is either quadratic, or of odd degree, or purely transcendental of degree 1. We construct numerous examples of the above types such that the extension F_{3}/F_{1} is not 4-excellent. Also we show that if k is a field, char k ≠ 2 and l/k is an arbitrary field extension of degree 4, then there exists a field extension F/k such that the extension lF/F is not 4-excellent.

1. Notation and terminology

The purpose of this paper is to construct nonexcellent field extensions of certain types. Recall that a field extension L/F is called excellent (resp. n-excellent) if for any quadratic form φ over F (resp. for any quadratic form φ over F of dimension at most n) the anisotropic part of the form φ_{L} is defined over F. It is well known that any quadratic extension is excellent (see for example [4]). Obviously, the same is true for odd degree and purely transcendental extensions, since they can not make an anisotropic form isotropic. We consider below towers of fields F_{1} ⊂ F_{2} ⊂ F_{3}, where the extension F_{3}/F_{2} is quadratic and the extension F_{2}/F_{1} is either quadratic or of odd degree or purely transcendental of degree 1. We give numerous examples of extensions F_{3}/F_{1} of such types, which are not 4-excellent.

The main source of reference concerning quadratic forms over fields is the Scharlau book [4]. We keep the standard notation. All the fields considered in the sequel are of characteristic different from 2. By a form we mean a nonsingular quadratic form. If F is a field, then W (F) is the Witt ring of F and I(F) is the ideal of evendimensional forms in W (F). The anisotropic part of a form φ is denoted by φ_{an}. Slightly abusing notation we write φ = 0 if φ is hyperbolic, i.e. its anisotropic part is zero. By D(φ) we denote the set of nonzero values of φ. The function field of the projective quadric related to the form φ is denoted by F (φ). If L/F is a field extension, then φ_{L} is the extension of φ to L, i.e. the form φ ⊗_{F} L. For a, b ∈ F^{∗} the symbol ⟨⟨a, b⟩⟩ means the 2-fold Pfisterform ⟨1, −a, −b, ab⟩. If k is a field and p ∈ k[x] is an irreducible polynomial, by k(p) we denote the residue field k[x]/p. If

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Key words and phrases. Quadratic form, Pfister form, field extension, residue map.

The work under this publication was partially supported by Royal society Joint Project”Quadratic forms and central simple algebras under field extensions”

p is fixed, f ∈ k[x] and p does not divide f, then unless specified otherwise, \overline{f} is the image of f in k(p)^{∗}/k(p)^{∗}^{2}.

The main tool in constructing of the examples in question is the notion of the second residue homomorphism ∂_{p} : I^{2}/I^{3}(k(x)) → k(p)^{∗}/k(p)^{∗}^{2} for an arbitrary irreducible polynomial p in x over a field k ([4]). Recall that the residue of the 2-fold Pfisterform ⟨⟨f, g⟩⟩ at the polynomial p is computed as follows:

∂_{p}(⟨⟨f, g⟩⟩) = \{ \begin{matrix}0 if v_{p}(f) = v_{p}(g) = 0, \\ \overline{f} if v_{p}(f) = 0, v_{p}(g) = 1 \end{matrix} .

(v_{p} is the corresponding discrete valuation on k(x)). In fact these equations determine the map ∂_{p} on the whole group I^{2}/I^{3}(k(x)).

2. Nonexcellence of field extensions of degree 4

We begin with the following

Theorem 1. Let k be a field, a, b, c ∈ k, a ∈ k^{∗} \ k^{∗}^{2}, b^{2} − ac^{2} ∈ k^{∗} \ k^{∗}^{2}. Put l = k(\sqrt{b + c\sqrt{a}}) and F = k(t, x, y), where t, x, y are indeterminates. Then there exists a 4-dimensional quadratic form φ over F such that dim (φ_{lF})_{an} = 2 and the form (φ_{lF})_{an} is not defined over F. (This means that the field extension lF/F is not 4-excellent).

Proof. Let f = ((t^{2} − a)x^{2} − (b^{2} − ac^{2}))((t^{2} − a)y^{2} − 1),

φ_{1} ≃ (t + \sqrt{a})⟨1, −(b + c\sqrt{a})⟩,
φ_{2} ≃ (t + \sqrt{a})⟨1, −f(b + c\sqrt{a})⟩.

Let us find q ∈ F^{∗} such that the forms (t + \sqrt{a})⟨⟨q⟩⟩ and q(t + \sqrt{a})(b + c\sqrt{a})⟨⟨qf⟩⟩ are defined over F. Provided q satisfies this condition we get that the form

φ ≃ φ_{2} ⊥ −qφ_{1} ≃ (t + \sqrt{a})⟨⟨q⟩⟩ ⊥ q(t + \sqrt{a})(b + c\sqrt{a})⟨⟨qf⟩⟩

is defined over F as well.

The following lemma is well known, but in view of the absence of a convenient reference and for the sake of completeness we give the proof.

Lemma 2. Suppose that q ∉ F^{∗}^{2} ∪ aF^{∗}^{2} and the elements α, β ∈ F are such that ⟨⟨q, α^{2} − aβ^{2}⟩⟩ = 0. Then the form (α + β\sqrt{a})⟨⟨q⟩⟩ is defined over F. Moreover, if elements X, Y ∈ F are such that X^{2} − qY^{2} = α^{2} − aβ^{2}, then (α + β\sqrt{a})⟨⟨q⟩⟩ ≃ (2X + 2α)⟨⟨q⟩⟩.

Proof. Since (X + Y \sqrt{q})(X − Y \sqrt{q}) = (α + β\sqrt{a})(α − β\sqrt{a}), we have

\begin{matrix}N_{F (\sqrt{a},\sqrt{q})/F (\sqrt{a})}(α + β\sqrt{a} + X + Y \sqrt{q}) = \\ = (α + β\sqrt{a} + X + Y \sqrt{q})(α + β\sqrt{a} + X − Y \sqrt{q}) = (α + β\sqrt{a})(2X + 2α). \end{matrix}

Therefore, since obviously, ⟨⟨q, N_{F (\sqrt{a},\sqrt{q})/F (\sqrt{a})}(α + β\sqrt{a} + X + Y \sqrt{q})⟩⟩ = 0, we get

⟨⟨q, α+β\sqrt{a}⟩⟩ ≃ ⟨⟨q, (α+β\sqrt{a})N_{F (\sqrt{a},\sqrt{q})/F (\sqrt{a})}(α+β\sqrt{a}+X+Y \sqrt{q})⟩⟩ ≃ ⟨⟨q, 2X+2α⟩⟩, or, equivalently (α + β\sqrt{a})⟨⟨q⟩⟩ ≃ (2X + 2α)⟨⟨q⟩⟩. The lemma is proved. □

Since

((t^{2} − a)y)^{2} − \frac{t^{2} − a}{(t^{2} − a)y^{2} − 1}((t^{2} − a)y^{2} − 1)^{2} = t^{2} − a,

and

((t^{2} − a)x)^{2} − \frac{t^{2} − a}{(t^{2} − a)y^{2} − 1}f = (t^{2} − a)(b^{2} − ac^{2}) = N_{F (\sqrt{a})/F} (b + c\sqrt{a})(t + \sqrt{a}),

we obtain that

⟨⟨\frac{t^{2} − a}{(t^{2} − a)y^{2} − 1}, t^{2} − a⟩⟩ = ⟨⟨\frac{t^{2} − a}{(t^{2} − a)y^{2} − 1}f, (t^{2} − a)(b^{2} − ac^{2})⟩⟩ = 0.

Therefore, constructing the form φ we can put q = \frac{t^{2}−a}{(t^{2}−a)y^{2}−1}. By Lemma 2 we get

(t + \sqrt{a})⟨⟨q⟩⟩ ≃ (2(t^{2} − a)y + 2t)⟨⟨q⟩⟩,
q(t + \sqrt{a})(b + c\sqrt{a})⟨⟨qf⟩⟩ ≃ (2(t^{2} − a)x + 2(bt + ac))⟨q, −f⟩,

hence

φ ≃ (2(t^{2} − a)y + 2t)⟨1, −q⟩ ⊥ (2(t^{2} − a)x + 2(bt + ac))⟨q, −f⟩.

We are going to prove that the form φ provides a counterexample to 4-excellency of the field F. However, we don’t need the explicit form of the coefficients of φ, which look somewhat complicated.

It is obvious that

(φ_{lF})_{an} ≃ (φ_{2})_{lF} ≃ (t + \sqrt{a})⟨1, −f⟩_{lF} .

In particular, dim (φ_{lF})_{an} = 2. If the form (φ_{lF})_{an} is defined over F, then there is p ∈ F^{∗} such that

(t + \sqrt{a})⟨1, −f⟩ ≃ p⟨1, −f⟩,

i.e. ⟨⟨f, (t + \sqrt{a})p⟩⟩_{lF} = 0, which is equivalent to

⟨⟨f, (t + \sqrt{a})p⟩⟩ ≃ ⟨⟨b + c\sqrt{a}, u + v\sqrt{a}⟩⟩

over F (\sqrt{a}) for some u, v ∈ F. We will show that in fact this is impossible. More precisely we will prove the following a bit more general

Lemma 3. For any u, v, p ∈ F, w ∈ k(\sqrt{a}) one has

⟨⟨f, (t + \sqrt{a})wp⟩⟩ ≄ ⟨⟨b + c\sqrt{a}, u + v\sqrt{a}⟩⟩.  (1)

Proof. Put e = b^{2} − ac^{2}. If e ∉ ak^{∗}^{2}, i.e. l/k is not a Galois extension, then the residues at either t + \sqrt{a} or t − \sqrt{a} of the left hand and on the right hand parts of (1) are different, which proves the lemma in this case.

If e ∈ ak^{∗}^{2}, the argument is more complicated. Assume that

⟨⟨f, (t + \sqrt{a})wp⟩⟩ ≃ ⟨⟨b + c\sqrt{a}, u + v\sqrt{a}⟩⟩  (1’)

for some u, v, p ∈ k(t, y)[x], and that under this condition m = deg p is minimal (in this section by deg we mean the degree with respect to the variable x). In particular, the polynomial p is squarefree. Moreover, we may assume that deg(u + v\sqrt{a}) is minimal provided that deg p = m. It follows that (t^{2} − a)x^{2} − e does not divide p, for otherwise we would change p for −p\frac{(t^{2}−a)y^{2}−1}{(t^{2}−a)x^{2}−e}, a contradiction to minimality of deg p. Also (t^{2} − a)x^{2} − e does not divide u + v\sqrt{a}, for in the opposite case the isomorphism

⟨⟨f, (t + \sqrt{a})w(b + c\sqrt{a})p⟩⟩ ≃ ⟨⟨b + c\sqrt{a}, \frac{u + v\sqrt{a}}{(t^{2} − a)x^{2} − e}((t^{2} − a)y^{2} − 1)⟩⟩,

obtained from (1′) by adding ⟨⟨b + c\sqrt{a}, f⟩⟩ to its both parts, would imply a contradiction to minimality of deg(u + v\sqrt{a}).

Let p = ∏ p_{i}, where p_{i} ∈ k(t, y)[x] are irreducible polynomials, deg p_{i} ≥ 1. We need the following

Lemma 4. For any p_{i} the elements f and af are not squares in k(t, y)[x]/p_{i}.

Proof. Suppose for instance that af is a square in k(t, y)[x]/p_{1}. Consider two cases:

a) deg p_{1} ≥ 2, b) deg p_{1} = 1.

Case a).

Since af is a square in k(t, y)[x]/p_{1} we get P^{2} − af = p_{1}s, for some P, s ∈ k(t, y)[x], deg P ≤ deg p_{1} − 1. Since deg p_{1} ≥ 2 and deg f = 2 it follows that deg(p_{1}s) ≤ 2(deg p_{1} − 1), hence deg s ≤ deg p_{1} − 2. Over the field k(\sqrt{a}, t, y, x) we have

⟨⟨f, p_{1}s⟩⟩ ≃ ⟨⟨f, P^{2} − af⟩⟩ ≃ ⟨⟨af, P^{2} − af⟩⟩ = 0,

so

\begin{matrix}⟨⟨b + c\sqrt{a}, u + v\sqrt{a}⟩⟩ ≃ ⟨⟨f, (t + \sqrt{a})wp⟩⟩ ≃ ⟨⟨af, (t + \sqrt{a})wp⟩⟩ ≃ \\ ⟨⟨af, (t + \sqrt{a})w \frac{p}{p_{1}} s⟩⟩ ≃ ⟨⟨f, (t + \sqrt{a})w \frac{p}{p_{1}} s⟩⟩, \end{matrix}

a contradiction to minimality of deg p, since deg \frac{p}{p_{1}} s ≤ deg p − 2.

Case b).

Let p_{1} = Qx + P. We may assume that Q, P ∈ k(t)[y]. Since af is a square in k(t, y)[x]/p_{1}, we conclude that a((t^{2} − a)\frac{P^{2}}{Q^{2}} − e)((t^{2} − a)y^{2} − 1) ∈ k(t, y)^{∗}^{2}. This implies that (t^{2} − a)y^{2} − 1 is a divisor of (t^{2} − a)P^{2} − eQ^{2} as a polynomial in y. It follows that there are nonzero polynomials P_{1}, Q_{1} ∈ k(t)[y] and a rationl function g ∈ k(t)^{∗} such that deg_{y} P_{1}, deg_{y} Q_{1} ≤ 1 and (t^{2} − a)P_{1}^{2} − eQ_{1}^{2} = g((t^{2} − a)y^{2} − 1) (here by deg_{y} we mean the degree with respect to y). It is easy to see that this implies g⟨(t^{2} − a), −1⟩ ≃ ⟨(t^{2} − a), −e⟩ over k(t), a contradiction, since the discriminants on the left-hand and on the right-hand sides of the last isomorphism are different. The case where f is a would-be square in k(t, y)[x]/p_{1} is treated quite similar. Lemma 4 is proved. □

We return to the proof of Lemma 3. By Lemma 4 we get that every p_{i} is a divisor of the polynomial u + v\sqrt{a} ∈ k(\sqrt{a}, t, y)[x], since otherwise the residues at p_{i} (more precisely at some irreducible divisor of p_{i} over k(\sqrt{a}, t, y)[x]) on the left-hand and the right- hand sides of the equality ⟨⟨f, (t + \sqrt{a})wp⟩⟩ ≃ ⟨⟨b + c\sqrt{a}, u + v\sqrt{a}⟩⟩ would be different. So we obtain that u + v\sqrt{a} = p(u_{1} + v_{1}\sqrt{a}), hence

⟨⟨f, (t + \sqrt{a})wp⟩⟩ ≃ ⟨⟨b + c\sqrt{a} , p(u_{1} + v_{1}\sqrt{a})⟩⟩  (2)

for some u_{1}, v_{1} ∈ k(t, y)[x], and, since p is squarefree, the polynomials u_{1} + v_{1}\sqrt{a} and p are relatively prime. Multiplying by some square in k(t)^{∗} we may assume that v_{t^{2}−a}(p) equals either 0 or 1. Suppose that v_{t^{2}−a}(p) = 1. Then applying the automorphism W (F (\sqrt{a})) → W (F (\sqrt{a})) determined by t → −t to the equality (2) written as

⟨⟨f, (t − \sqrt{a})\frac{p}{t^{2} − a}w⟩⟩ ≃ ⟨⟨b + c\sqrt{a}, \frac{p}{t^{2} − a}(u_{1} + v_{1}\sqrt{a})(t^{2} − a)⟩⟩

and changing p(t, x, y) for p(−t, x, y)(t^{2} − a)^{−1}, we reduce the problem to the case v_{t^{2}−a}(p) = 0.

We claim that deg(u_{1} + v_{1}\sqrt{a}) = 0. Indeed, suppose s ∈ k(\sqrt{a}, t, y)[x] is a prime divisor of u_{1} + v_{1}\sqrt{a}. Since s does not divide ((t^{2} − a)x^{2} − e)p, we get that

∂_{s}(⟨⟨b + c\sqrt{a}, p(u_{1} + v_{1}\sqrt{a})⟩⟩) = ∂_{s}(⟨⟨f, (t + \sqrt{a})wp⟩⟩) = 1.

By the same argument as in Lemma 4 we can diminish deg(u_{1} +v_{1}\sqrt{a}) in (2), which contradicts minimality of deg(u + v\sqrt{a}) in (1′). Thus, we conclude that there are no prime divisors of u_{1} + v_{1}\sqrt{a} at all, i.e. u_{1}, v_{1} ∈ k(t, y). The equality (2) implies

⟨⟨(b + c\sqrt{a})f, (t + \sqrt{a})wp⟩⟩ ≃ ⟨⟨b + c\sqrt{a}, (t + \sqrt{a})(u_{1} + v_{1}\sqrt{a})w⟩⟩,

and the right- hand side of the last isomorphism is defined over k(\sqrt{a}, t, y). Denote it by π. Since π splits by the square root of

(b + c\sqrt{a})((t^{2} − a)y^{2} − 1)((t^{2} − a)x^{2} − e),

we conclude that π_{k(\sqrt{a},t,y)(τ)} = 0, where

τ ≃ ⟨⟨((t^{2} − a)y^{2} − 1)(b + c\sqrt{a})(t^{2} − a), −((t^{2} − a)y^{2} − 1)(b + c\sqrt{a})e⟩⟩.

Therefore, either π = 0, or

π ≃ τ ≃ ⟨⟨((t^{2} − a)y^{2} − 1)(b + c\sqrt{a})(t^{2} − a), −((t^{2} − a)y^{2} − 1)(b + c\sqrt{a})e⟩⟩. (3)

On the one hand, since v_{t^{2}−a}(p) = 0, we have

∂_{(t+\sqrt{a})}(π) = ∂_{(t+\sqrt{a})}(⟨⟨f(b + c\sqrt{a}), (t + \sqrt{a})wp⟩⟩) = b − c\sqrt{a},

so π ≠ 0, hence (3) holds. On the other hand,

∂_{(t−\sqrt{a})}(π) = ∂_{(t−\sqrt{a})}(⟨⟨f(b + c\sqrt{a}), (t + \sqrt{a})wp⟩⟩) = 1,

and

\begin{matrix}∂_{(t−\sqrt{a})}(⟨⟨((t^{2} − a)y^{2} − 1)(b + c\sqrt{a})(t^{2} − a), −((t^{2} − a)y^{2} − 1)(b + c\sqrt{a})e⟩⟩) = \\ (b + c\sqrt{a})e = b − c\sqrt{a}, \end{matrix}

a contradiction in view of (3). This proves Lemma 3, hence also Theorem 1. □

□ Corollary 5. Let k be a field, let l/k be an arbitrary field extension of forth degree. Then there is a field extension F/k such that the extension lF/F is not 4-excellent.

Proof. There is a field extension k_{1}/k, which is either cubic or trivial, such that lk_{1}/k_{1} is the tower of quadratic extensions. If this tower is not the composite of two quadratic extensions of k_{1} we can apply Theorem 1. The opposite case is considered in [5]. □

If k(\sqrt{b + c\sqrt{a}})/k is not a Galois extension i.e. e ∉ k^{∗}^{2} ∪ ak^{∗}^{2}, then Theorem 1 can be strengthened and the proof becomes much simpler. Namely, we have the following

Proposition 6. Let k be a field, a, b, c ∈ k, a ∈ k^{∗} \ k^{∗}^{2}, e = b^{2} − ac^{2} ∈ k^{∗} \ (k^{∗}^{2} ∪ ak^{∗}^{2}), l = k(\sqrt{b + c\sqrt{a}}). Let t be an indeterminate, F = k(t). Then there exists a 4-dimensional form φ over F such that the following holds:

For any field l̃⊃ l such that b − c\sqrt{a} is not a square in l̃and t is transcendental over l̃the form (φ_{l̃F})_{an} is not defined over F. In particular, the field l itself satisfies these conditions.

Proof. Let

φ ≃ ⟨2t, 2t(t^{2} − a), −2(bt + ac), −2e(bt + ac)(t^{2} − a)⟩.

It is easy to see that

φ_{F (\sqrt{a})} ≃ (t − \sqrt{a})⟨1, −(b − c\sqrt{a})⟩ ⊥ (t + \sqrt{a})⟨1, −(b + c\sqrt{a})⟩,

hence dim(φ_{l̃F})_{an} = 2. Assume that the form (φ_{l̃F})_{an} is define over F. Then there is z ∈ F^{∗} such that

(t − \sqrt{a})⟨1, −(b − c\sqrt{a})⟩ ≃ z⟨1, −(b − c\sqrt{a})⟩,

i.e. ⟨⟨b − c\sqrt{a}, (t − \sqrt{a})z⟩⟩_{l̃F} = 0. On the other hand, since b − c\sqrt{a} is not a square in l̃and z ∈ F^{∗}, the residue of this 2-fold Pfisterform at either t − \sqrt{a} or t + \sqrt{a} is nontrivial, a contradiction. □

3. Nonexcellence of 2n-degree extensions for an odd n

In this section we consider towers of two field extensions one of which is a Galois extension of odd degree and the other is quadratic.

Proposition 7. Let k be a field, l = k(α, β) is a finite Galois field extension of odd degree n > 1. Assume that α ∉ k and β ∉ k(α). Let further t, x, y be indeterminates, a, b, d ∈ k^{∗} such that d, −ab, −abd ∉ k^{∗}^{2}. Set φ ≃ ⟨a, b, x, abdx⟩, K = k(t, x, y), L = lK, E = L(\sqrt{−(xt^{2} + aα^{2})(by^{2} + abdxβ^{2})}). Then

1) dim (φ_{E})_{an} = 2.

2) The form (φ_{E})_{an} is not defined over K.

Proof. The argument is somewhat similar to the one in Theorem 1.

1) Obviously,

φ_{L} ≃ ⟨xt^{2} + aα^{2}, ax(xt^{2} + aα^{2}), by^{2} + abdxβ^{2}, adx(by^{2} + abdxβ^{2})⟩,

hence φ_{E} is isotropic and (φ_{E})_{an} ≃ ax(xt^{2} + aα^{2})⟨1, −d⟩.

2) Assume that the form (φ_{E})_{an} is defined over K. Then there is p ∈ K^{∗} such that

ax(xt^{2} + aα^{2})⟨1, −d⟩_{E} ≃ p⟨1, −d⟩_{E},

i.e. ⟨⟨d, ax(xt^{2} + aα^{2})p⟩⟩_{E} = 0. This means that

⟨⟨d, ax(xt^{2} + aα^{2})p⟩⟩_{L} ≃ ⟨⟨−(xt^{2} + aα^{2})(by^{2} + abdxβ^{2}), q⟩⟩_{L}  (4)

for some q ∈ L^{∗}.

We may assume that p ∈ k(t, x)[y] and m = deg p is minimal (in this section we mean by deg the degree with respect to the variable y). Notice also that by^{2} + abdxβ^{2} |̸p, for otherwise by^{2} + abdx(σβ)^{2}|p for any σ ∈ G(l/k) and comparing the residues of both parts of (4) at by^{2} + abdx(σβ)^{2}, with σβ ≠ β would give us a contradiction.

Now let p = ∏ p_{i}, where p_{i} ∈ k(t, x)[y] are irreducible polynomials in y.

Lemma 8. d is not a square in k(t, x)[y]/p_{i} for any p_{i}.

Proof. Suppose for instance that d is a square in k(t, x)[y]/p_{1}. We get P^{2} −d = p_{1}s for some P, s ∈ k(t, x)[y], deg P ≤ deg p_{1} − 1. Hence

deg s ≤ 2(deg p_{1} − 1) − deg p_{1} = deg p_{1} − 2.

Since ⟨⟨d, p_{1}s⟩⟩ ≃ ⟨⟨d, P^{2} − d⟩⟩ = 0, we can change the divisor p_{1} of p for s in (4), a contradiction to minimality of deg p. The lemma is proved. □

Let p_{i} = ∏ p_{ij}, where p_{ij} ∈ l(t, x)[y] are irreducible polynomials. Since l/k is an odd degree Galois extension, we have l(t, x)[y]/p_{ij} ≃ l(k(t, x)[y]/p_{i}) and, moreover, deg[l(k(t, x)[y]/p_{i}) : k(t, x)[y]/p_{i}] is odd. Therefore, by Lemma 8 d is not a square in l(k(t, x)[y]/p_{i}), hence in l(t, x)[y]/p_{ij}. This means that the form (4) has nonzero residues at each p_{ij}. This implies that q = pf, where f ∈ l(t, x)[y] and the polynomials f and p are relatively prime, so we can represent (4) as

⟨⟨d, ax(xt^{2} + aα^{2})p⟩⟩_{L} ≃ ⟨⟨−(xt^{2} + aα^{2})(by^{2} + abdxβ^{2}), pf⟩⟩_{L}.  (5)

Just as in Lemma 2 we may assume that in equality (5), where deg p = m is fixed, deg f is minimal. Denote the Pfisterform in (5) by π. Notice that by^{2} + abdxβ^{2} does not divide f, since otherwise we could change f for \frac{f}{(xt^{2}+aα^{2})(by^{2}+abdxβ^{2})}, a contradiction to minimality of deg f.

Lemma 9. deg f = 0.

Proof. Let f = ∏ f_{i}, where f_{i} ∈ l(t, x)[y] are irreducible polynomials. Since f_{i} do not appear on the left- hand side of (5), we have ∂_{f_{i}}(π) = 1, for any f_{i}. Assume first that, say deg f_{1} ≥ 2. Since ∂_{f_{1}}(π) = 1, we get that

P^{2} + (xt^{2} + aα^{2})(by^{2} + abdxβ^{2}) = f_{1}s

for some P, s ∈ l(t, x)[y] such that deg P ≤ deg f_{1} − 1. Since deg f_{1} ≥ 2 we get that

deg(P^{2} + (xt^{2} + aα^{2})(by^{2} + abdxβ^{2})) ≤ 2(deg f_{1} − 1),

hence deg s ≤ deg f_{1} − 2, and we are done just as in Lemma 8. Now suppose that deg f_{1} = 1, i.e. f_{1} = R(y − \frac{P}{Q}), where P, Q ∈ l(x)[t], R ∈ l(x, t). Since ∂_{f_{1}}(π) = 1, we get that −(xt^{2} + aα^{2})(bP^{2} + abdxβ^{2}Q^{2}) is a square in l(x)[t]. In particular,

xt^{2} + aα^{2}|bP^{2} + abdxβ^{2}Q^{2}.

Therefore, there are some nonzero polynomials P_{1}, Q_{1} ∈ l(x)[t] such that deg_{t} P_{1}, deg_{t} Q_{1} ≤ 1 and xt^{2} + aα^{2}|bP_{1}^{2} + abdxβ^{2}Q_{1}^{2}, i.e.

bP_{1}^{2} + abdxβ^{2}Q_{1}^{2} = g(xt^{2} + aα^{2})

for some g ∈ l(x). It is easy to see that this implies ⟨b, abdx⟩_{l(x)} ≃ ⟨gx, ga⟩, a contradiction, since the discriminants on the left-hand and on the right-hand sides of the last isomorphism are different. The lemma is proved. □

Adding to the both parts of (5) the Pfisterform ⟨⟨d, pf⟩⟩ we can represent (5) as

⟨⟨d, ax(xt^{2} + aα^{2})f⟩⟩_{L} ≃ ⟨⟨−d(xt^{2} + aα^{2})(by^{2} + abdxβ^{2}), pf⟩⟩_{L}.  (6)

By Lemma 8 the left-hand side of the last isomorphism is defined over l(t, x). Moreover, the right-hand side splits over l(t, x)(\sqrt{−d(xt^{2} + aα^{2})(by^{2} + abdxβ^{2})}). So, just as for the form π from section 2, we have that either

a) ⟨⟨d, ax(xt^{2} + aα^{2})f⟩⟩_{l(t,x)} = 0, or

b) ⟨⟨d, ax(xt^{2} + aα^{2})f⟩⟩_{l(t,x)} ≃ ⟨⟨−bd(xt^{2} + aα^{2}), −abx(xt^{2} + aα^{2})⟩⟩_{l(t,x)}.

Assume that case a) holds. We may suppose that f ∈ l(x)[t], p ∈ k(x, y)[t] and that f and p are squarefree as polynomials in t. Obviously, xt^{2} + aα^{2}|f, for otherwise we would have

d = ∂_{xt^{2}+aα^{2}}(⟨⟨d, ax(xt^{2} + aα^{2})f⟩⟩) = ∂_{xt^{2}+aα^{2}}(0) = 1,

a contradiction. By the similar reason if σα ≠ α (such σ exists, since β ∉ k(α)), then xt^{2} + a(σα)^{2} |̸f. Hence from (6) it follows that xt^{2} + a(σα)^{2} |̸p, if σα ≠ α, or equivalently xt^{2} + aα^{2} |̸p.

Let f = f_{1}(xt^{2} + aα^{2}). Then

\begin{matrix}1 = ∂_{xt^{2}+aα^{2}}(⟨⟨−d(xt^{2} + aα^{2})(by^{2} + abdxβ^{2}), pf⟩⟩) = \\ \overline{df_{1}(by^{2} + abdxβ^{2})p mod(xt^{2} + aα^{2})} ∈ l(t)(u)^{∗}/l(t)(u)^{∗}^{2}. \end{matrix}  (7)

Since β ∉ k(α) and [k(α, β) : k] is odd we get that β^{2} ∉ k(α), and so there is σ ∈ G(l/k) such that σα = α and σ(β^{2}) ≠ β^{2}. Applying such σ to (7) we get

\overline{dσf_{1}(by^{2} + abdx(σβ)^{2})p mod(xt^{2} + aα^{2})} = 1.  (8)

Combining (7) and (8) we get that

\overline{f_{1}σf_{1}(by^{2} + abdxβ^{2})(by^{2} + abdx(σβ)^{2}) mod(xt^{2} + aα^{2})} = 1.

Since \overline{f_{1}σf_{1} mod(xt^{2} + aα^{2})} ∈ l(t, x, \sqrt{−ax}) does not depend on y, and σ(β^{2}) ≠ β^{2}, we come to a contradiction. This shows that case a) is impossible.

Now assume that case b) holds, i.e.

⟨⟨d, ax(xt^{2} + aα^{2})f⟩⟩_{l(t,x)} ≃ ⟨⟨−bd(xt^{2} + aα^{2}), −abx(xt^{2} + aα^{2})⟩⟩_{l(t,x)}.

Comparing residues at x on the left-hand and on the right-hand sides of the last isomorphism we obtain that either −ab ∈ k^{∗}^{2}, or −dab ∈ k^{∗}^{2}, which is not so by the hypothesis. This shows that case b) is also impossible, which finishes the proof of Proposition 7. □

Corollary 10. Let n ≥ 2 be an integer, which is not an odd prime. Then there exists a finite 4-nonexcellent field extension of degree 2n.

Proof. If n is even, then 2n = 2^{k}m, where k ≥ 2 and m is odd. Let L/F be a 4-nonexcellent multiquadratic extension of degree 2^{k} (the existence of such extensions has been proven in [5]). Let further φ be a 4-dimensional form providing a corresponding counterexample. Consider the extension L(t^{\frac{1}{m}})/F (t) of degree 2n. Applying the specialization t = 0 (the first residue map) it is easy to see that the extension L(t)/F (t) is 4-nonexcellent with the form φ_{F (t)} as a counterexample. Since the extension L(t^{\frac{1}{m}})/L(t) is of odd degree, we conclude that φ_{F (t)} is a counterexample for the extension L(t^{\frac{1}{m}})/F (t) as well.

Now assume that a number n is odd, a prime p ≠ n divides n, and l/k is a Galois extension of degree n. Let k ⊂ k_{1} ⊂ l, where the Galois group G(l/k_{1}) is cyclic of order p, l = k(β) and α ∈ k_{1} \ k. Then the extension l/k and the elements α and β satisfy the hypothesis of Proposition 7, which implies an existence of a 4nonexcellent extension of degree 2n. □

4. Nonexcellence of extensions determined by hyperelliptic curves

In the last section of the present work we give examples of 4-nonexcellent extensions L/F, where L is the function field of a hyperelliptic curve over F. In this case the extension L/F is a tower of a purely trancendental extension of degree 1 and a quadratic extension. This question is of some interest, since the function field of a conic always determines an excellent extension ([1], [3]). More precisely the following statement holds.

Proposition 11. Let k be a field, a, b ∈ k^{∗} and the Pfisterform ⟨⟨a, b⟩⟩ is anisotropic. Suppose n > m ≥ 1 are positive integers. Let further F = k((t)), t being an indeterminate, and let X be a hyperelliptic curve over F defined by the equation y^{2} = ax^{2n} + bx^{2m} − t. Then the form (⟨1, −a, −b, t⟩_{F (X)})_{an} is not defined over F.

Proof. Since the hyperbolic plane ⟨1, −(ax^{2n} + bx^{2m} − t)⟩_{F (X)} is a subform of ⟨1, −a, −b, t⟩_{F (X)}, we get that the form ⟨1, −a, −b, t⟩_{F (X)} is isotropic, hence

dim (⟨1, −a, −b, t⟩_{F (X)})_{an} = 2.

Assume for a moment that there are d, e ∈ F^{∗} such that

(⟨1, −a, −b, t⟩_{F (X)})_{an} ≃ ⟨d, e⟩_{F (X)},

or, equivalently that the form φ ≃ ⟨1, −a, −b, t, −d, −e⟩ becomes hyperbolic over F (X). Then we have

(φ_{an})_{F (x)} ≃ ⟨⟨ax^{2n} + bx^{2m} − t⟩⟩ ⊗ τ  (8)

for some form τ. Assume that dim τ is odd. Then disc(φ_{an}) = ax^{2n} + bx^{2m} − t. On the other hand,

disc(φ_{an}) = −abdet ≠ ax^{2n} + bx^{2m} − t (mod F (x)^{∗}^{2}),

a contradiction. Hence dim τ is even, and so dim φ_{an} = 4. Applying the first residue map (in fact the specialization x = 0) to (8) we conclude that φ_{an} is similar to the form ⟨⟨−t, −c⟩⟩ for some c ∈ F^{∗}. Since ⟨⟨−t, t⟩⟩ = 0, we can assume that c ∈ k^{∗} and −c ∉ k^{∗}^{2}. Furthermore, since φ_{F (X)} = 0, we have

ax^{2n} + bx^{2m} − t ∈ D(⟨−c, −t, −ct⟩_{F (x)}).

Let r be a minimal nonnegative integer such that

t^{2r}(ax^{2n} + bx^{2m} − t) = −(cp_{1}^{2} + tp_{2}^{2} + ctp_{3}^{2})

for some p_{1}, p_{2}, p_{3} ∈ k[[t]][x] (such r exists by the Cassels-Pfister theorem, (see, for example [4])). Suppose r ≥ 1. Then p_{1} = tq_{1}, q_{1} ∈ k[[t]][x], and so

t^{2r−1}(ax^{2n} + bx^{2m} − t) = −(p_{2}^{2} + cp_{3}^{2} + ctq_{1}^{2}).

Hence \overline{p_{2}}^{2} + c\overline{p_{3}}^{2} = 0, where \overline{p_{2}}, \overline{p_{3}} ∈ k[x] are the reductions of p_{2}, p_{3} modulo the ideal (t) ⊂ k[[t]][x]. Since −c ∉ k^{∗}^{2}, we have \overline{p_{2}} = \overline{p_{3}} = 0 and so

p_{2} = tq_{2}, p_{3} = tq_{3}, q_{2}, q_{3} ∈ k[[t]][x].

Therefore,

t^{2r−2}(ax^{2n} + bx^{2m} − t) = −(cq_{1}^{2} + tq_{2}^{2} + ctq_{3}^{2}),

a contradiction to minimality of r. Thus, we conclude that r = 0, hence

ax^{2n} + bx^{2m} − t = −(cp_{1}^{2} + tp_{2}^{2} + ctp_{3}^{2}),

which implies ax^{2n} + bx^{2m} = −c\overline{p_{1}^{2}}. On the other hand, obviously, \frac{−(ax^{2n}+bx^{2m})}{c} ∉ k[x]^{2}, a contradiction, which proves Proposition 9. □

REFERENCES

[1] Arason J. Kr., Excellence of F (φ)/F for 2-fold Pfister forms, Appendix II in [2] (1977).

[2] Elman R., Lam T.Y. and Wadsworth A.R., Amenable fields and Pfister extensions, Queen’s Papers Pure Appl. Math. 46 (1977), 445-491.

[3] Rost M., Quadratic forms isotropic over the function field of a conic, Math. Ann. 288 (1990), 511-513.

[4] Scharlau W., Quadratic and Hermitian forms, Springer, Berlin Heidelberg New York (1985).

[5] Sivatski A.S., Nonexcellence of multiquadratic field extensions, Journal of Algebra 275 (2004), no. 2, 859-866.

e-mail: sivatsky@AS3476.spb.edu