APPLICATIONS OF CONICS TO CERTAIN QUADRATIC

FORMS OVER THE RATIONAL FUNCTION FIELDS

A.S. Sivatski

St.Petersburg Electrotechnical University 197376, St.Petersburg, Russia

Abstract. A few results on quadratic forms over fields are obtained. In particular, we show that for any forms φ_{1} and φ_{2} over a field k of characteristic different from 2 and a ∈ k^{∗}, the anisotropic part of the form φ_{1} ⊥ (t^{2} − a)φ_{2} over the rational function field k(t) is of the same type, i.e. there exist forms τ_{1} and τ_{2} over k such that (φ_{1} ⊥ (t^{2} − a)φ_{2})_{an} ≃ (τ_{1} ⊥ (t^{2} − a)τ_{2}). Also we determine the structure of certain Pfister forms over k(t), and describe the behavior of quadratic forms under biquadratic extensions of k in terms of some related forms over the function field of the product of two conics over k(x), or k(x, y).

1. Structure of certain quadratic forms over the rational function field

Let k be a field of characteristic different from 2, W (k) the Witt group of k. It is well known (see, for example, [Sch]) that the sequence of abelian groups

0 ⟶ W (k) ⟶^{res} W (k(t)) \underrightarrow{∐ ∂_{p}} ∐_{p∈𝔸_{k}^{1}} W (k_{p}) ⟶ 0,

is split exact. We consider here a point p ∈ 𝔸_{k}^{1} as a monic irreducible polynomial over k, k_{p} = k[t]/p is the corresponding residue field, and ∂_{p} : W (k(t)) → W (k_{p}) is the residue homomorphism well defined by the rule

∂_{p}(⟨f⟩) = \{ \begin{matrix}0 & if v_{p}(f) = 0 \\ ⟨\overline{fp^{−1}}⟩ & if v_{p}(f) = 1 \end{matrix}

The splitting map W (k(t)) → W (k) is defined by the rule ⟨f⟩ → ⟨l(f)⟩, where l(f) is the leading coefficient of the polynomial f ∈ k[t]. Let φ ∈ W (k(t)). Knowing the projections of φ to all direct summands of W (k(t)), it is easy to determine φ itself. However, only in some specific cases it is clear how to determine the anisotropic part φ_{an} of the form φ. One situation when it is possible is the case where the form φ has an only residue, and, moreover, at the linear polynomial t. Then, in view of the exact sequence above, φ = τ_{1} + tτ_{2} for some anisotropic forms τ_{1}, τ_{2} over k. It

Typeset by AMS-TEX

Key words and phrases. .

is well known that the form τ_{1} ⊥ tτ_{2} is anisotropic, hence φ_{an} ≃ τ_{1} ⊥ tτ_{2}. We start this paper by proving a similar result, where the linear polynomial t is replaced by the quadratic polynomial t^{2} − a, a being a nonsquare element of k^{∗}.

A few words about the notation. In the sequel all the fields are assumed to be of characteristic different from 2. The Pfister form ⟨⟨a_{1}, …, a_{n}⟩⟩ is the product ⟨1, −a_{1}⟩ ⊗ ⋯ ⊗ ⟨1, −a_{n}⟩ (take notice of the signs !). Usually, we will omit the sign ⊗ in products of quadratic forms. For the field extension L/F the kernel of the restriction map W (F) → W (L) is denoted by W (L/F). The extension L/F is called excellent if for any form φ over F the anisotropic part of the form φ_{L} is defined over F. The basic statement used throughout the paper is that the extension F (C)/F is excellent for any field F and a conic over F ([A], [R]).

Theorem 1.1. Let k be a field, φ_{1}, φ_{2} quadratic forms over k, m an odd positive integer. Then there exist forms τ_{1}, τ_{2} over k such that over k(t)

(φ_{1} ⊥ (t^{2m} − a)φ_{2})_{an} ≃ τ_{1} ⊥ (t^{2m} − a)τ_{2}.

Proof. First consider the case where m = 1. We may assume that the form (φ_{1} ⊥ (t^{2} − a)φ_{2}) is isotropic. Let C be the affine conic associated with the quaternion algebra (a, x) over the Laurent series field F = k((x)), with the equation t^{2} − a = xv^{2}. Notice that k(t) ⊂ F (C). The form

(φ_{1} ⊥ xφ_{2})_{F (C)} ≃ (φ_{1} ⊥ (t^{2} − a)φ_{2})_{F (C)}

is isotropic by the hypothesis. Since the extension F (C)/F is excellent, and W (F) = W (k) ⊕ xW (k), we get

(φ_{1} ⊥ xφ_{2})_{F (C)}_{an} ≃ (τ_{1} ⊥ xτ_{2})_{F (C)}

for some forms τ_{1} and τ_{2} over k. It follows, since W (F (C)/F) = ⟨⟨a, x⟩⟩W (F) ([Sch], Ch.4, Th. 5.4), that

φ_{1} + xφ_{2} = τ_{1} + xτ_{2} + ⟨⟨a, x⟩⟩q ∈ W (F) (∗)

for some form q over F. Since q ≃ q_{1} ⊥ xq_{2} for some forms q_{1}, q_{2} over k, and −x⟨⟨a, x⟩⟩ ≃ ⟨⟨a, x⟩⟩, we may assume that q is a form over k. Therefore,

φ_{1} − τ_{1} − ⟨⟨a⟩⟩q = x(τ_{2} − φ_{2} − ⟨⟨a⟩⟩q) ∈ W (F),

or, in other words,

φ_{1} − τ_{1} − ⟨⟨a⟩⟩q = τ_{2} − φ_{2} − ⟨⟨a⟩⟩q = 0 ∈ W (k).

So we have

φ_{1} − τ_{1} − ⟨⟨a⟩⟩q = (t^{2} − a)(τ_{2} − φ_{2} − ⟨⟨a⟩⟩q) = 0 ∈ W (k(t)),

and, consequently, we can replace x by t^{2} − a and k((x)) by k(t) in (∗), i.e.

φ_{1} + (t^{2} − a)φ_{2} = τ_{1} + (t^{2} − a)τ_{2} + ⟨⟨a, t^{2} − a⟩⟩q = τ_{1} + (t^{2} − a)τ_{2} ∈ W (k(t)), since ⟨⟨a, t^{2} − a⟩⟩ = 0. Notice that the form τ_{1} ⊥ (t^{2} − a)τ_{2} is anisotropic over k(t), since k(t) ⊂ F (C), and the form τ_{1} ⊥ xτ_{2} is anisotropic over F (C).

In the general case, where m is an arbitrary odd number, consider the odd degree field extension k(u)/k(t), where u = t^{\frac{1}{m}}. Assuming that the form φ_{1} ⊥ (u^{2m} −a)φ_{2} is isotropic over k(u), we see by the Springer theorem ([Sch], Ch.2, Th. 5.3) that the form φ_{1} ⊥ (t^{2} − a)φ_{2} is isotropic over k(t). By Theorem 1.1

(φ_{1} ⊥ (t^{2} − a)φ_{2})_{an} ≃ τ_{1} ⊥ (t^{2} − a)τ_{2}

for some forms τ_{1}, τ_{2} over k. Replacing t by t^{m}, and applying the Springer theorem again, we get

(φ_{1} ⊥ (t^{2m} − a)φ_{2})_{an} ≃ τ_{1} ⊥ (t^{2m} − a)τ_{2}.
□

Remark. If φ_{2} ≃ −φ_{1}, then the proof of Theorem 1.1 can be substantially simplified. Indeed, suppose the form φ_{1} ⊥ −(t^{2} − a)φ_{1} is isotropic, and φ_{1} is anisotropic. Then the form (φ_{1} ⊥ −(t − \sqrt{a})(t + \sqrt{a})φ_{1})_{k(\sqrt{a})((t−\sqrt{a}))} is isotropic, which implies that φ_{1}_{k(\sqrt{a})} is isotropic as well. Hence φ_{1} ≃ ⟨⟨a⟩⟩ψ ⊥ τ for some nonempty forms ψ and τ. We get

\begin{matrix}φ_{1} ⊥ −(t^{2} − a)φ_{1} = ⟨⟨a⟩⟩ψ + τ − (t^{2} − a)(⟨⟨a⟩⟩ψ + τ) = \\ (⟨⟨a⟩⟩ψ − (t^{2} − a)⟨⟨a⟩⟩ψ) + (τ − (t^{2} − a)τ) = τ − (t^{2} − a)τ, \end{matrix}

since (t^{2} − a)⟨⟨a⟩⟩ = ⟨⟨a⟩⟩, so we can finish the proof by induction on dim φ_{1}.

Corollary 1.2. Under the notation above the following two conditions are equivalent.

1) The form φ_{1} ⊥ (t^{2m} − a)φ_{2} is isotropic.

2) There exist forms τ_{1} and τ_{2} over k such that φ_{1} −τ_{1} = τ_{2} −φ_{2} ∈ W (k(\sqrt{a})/k), and dim τ_{1} + dim τ_{2} < dim φ_{1} + dim φ_{2}.

Proof. 1) ⟹ 2). By Theorem 1.1 we have

φ_{1} + (t^{2m} − a)φ_{2} = τ_{1} + (t^{2m} − a)τ_{2},

and dim τ_{1} + dim τ_{2} < dim φ_{1} + dim φ_{2}. Since φ_{1} − τ_{1} = (t^{2m} − a)(τ_{2} − φ_{2}), we get

φ_{1} − τ_{1} = τ_{2} − φ_{2} ∈ W (k(a^{\frac{1}{2m}})/k).

Since m is odd, by the Springer theorem we get

φ_{1} − τ_{1} = τ_{2} − φ_{2} ∈ W (k(\sqrt{a})/k).

2) ⟹ 1). The same argument, but in the opposite direction. □

If φ_{1}, φ_{2} ∈ I^{n}(k), then (φ_{1} ⊥ tφ_{2})_{an} ≃ τ_{1} ⊥ τ_{2}, where τ_{1}, τ_{2} ∈ I^{n}(k), since τ_{i} ≃ φ_{i}_{an} (i = 1, 2). The similar statement is not true even for n = 2 if one replaces the polynomial t by t^{2} − a, as the following counterexample shows.

Proposition 1.3. Let k_{0} be a field, k = k_{0}(z), a, b, c, u, v ∈ k_{0}^{∗} are such that ind((b, u) + (c, v))_{k_{0}(\sqrt{a})} = 4,

ψ ≃ ⟨1, −b, −u, abu⟩ ⊥ (t^{2} − a)z⟨1, −c, −v, acv⟩ ∈ W (k(t)).

Then ψ ≃ (φ_{1} ⊥ (t^{2}−a)φ_{2})_{an} for some forms φ_{1}, φ_{2} ∈ I^{2}(k), but ψ ≄ τ_{1} ⊥ (t^{2}−a)τ_{2} for any forms τ_{1}, τ_{2} ∈ I^{2}(k).

Proof. Clearly, ψ is anisotropic, ψ ∈ I^{2}(k(t)). By [T],Prop. 2.4 we have

ind C(ψ) = ind((b, u) + (c, v) + (a, −bcuvz)) = 8.

Moreover, comparing the residues, it is easy to see that ψ ≃ (φ_{1} ⊥ (t^{2} − a)φ_{2})_{an}, where φ_{1} ≃ ⟨1, −b, −u, abu, −zcv, zacv⟩, φ_{2} ≃ z⟨1, −c, −v, cv⟩. Obviously, φ_{1}, φ_{2} ∈ I^{2}(k). Suppose ψ ≃ τ_{1} ⊥ (t^{2} − a)τ_{2}, where τ_{1}, τ_{2} ∈ I^{2}(k). Then either dim τ_{1} = dim τ_{2} = 4, or dim τ_{1} = 8, dim τ_{2} = 0, or dim τ_{1} = 0, dim τ_{2} = 8. But the first case is impossible, since ind C(ψ) = 8. The second and the third cases are impossible, since ∂_{t^{2}−a}(ψ) ≠ 0 and ∂_{t^{2}−a}((t^{2} − a)ψ) ≠ 0. □

Given an even number 2n ≥ 4, Theorem 1.1 is not true in general for irreducible polynomials in t of degree 2n. To construct corresponding counterexamples we use the following statement, which immediately follows from [Si1], Prop.11.

Proposition 1.4. Let k_{0} be a field, n > m ≥ 1, k = k_{0}(x), a, b ∈ k_{0}^{∗}, ⟨⟨a, b⟩⟩ ≠ 0, p(t) = at^{2n} + bt^{2m} − x, X the affine curve over k determined by the equation y^{2} = at^{2n} + bt^{2m} − x. Then the form (⟨−a, −b, x, 1⟩_{k(X)})_{an} is not defined over k.

Corollary 1.5. Under the notation of Proposition 4

⟨−a, −b, x, p(t)⟩_{an} ≄ φ_{1} ⊥ p(t)φ_{2}

for any forms φ_{1}, φ_{2} over k.

Proof. Suppose ⟨−a, −b, x, p(t)⟩_{an} ≃ φ_{1} ⊥ p(t)φ_{2}. Then dim φ_{1} = dim φ_{2} = 1, and

(⟨−a, −b, x, 1⟩_{k(X)})_{an} ≃ (⟨−a, −b, x, p(t)⟩_{k(X)})_{an} ≃ (φ_{1} ⊥ p(t)φ_{2})_{k(X)} ≃ (φ_{1} ⊥ φ_{2})_{k(X)},

a contradiction to Proposition 1.4. □

The condition that m is odd is essential in Theorem 1.1, as the following example shows.

Proposition 1.6. For any even positive integer m = 2n ≥ 2 Theorem 1.1 does not remain true in general if one replaces t^{2} − a by t^{2m} − a.

Proof. Let k_{0} be a field, k = k_{0}(x, y, z). Consider the form

π ≃ ⟨⟨x(t^{2n} − y − zy^{−1}), (y + zy^{−1})(t^{4n} − 4z)⟩⟩

over k. Obviously, ∂_{p}(π) = 0 for any p ∈ 𝔸_{k}^{1} distinct from t^{2n} − y − zy^{−1} and t^{4n} − 4z. Moreover,

∂_{t^{2n}−y−zy^{−1}}(π) = −x⟨1, −(y+zy^{−1})((y+zy^{−1})^{2}−4z)⟩ = −x⟨1, −t^{2n}(y−zy^{−1})^{2}⟩ = 0,

\begin{matrix}∂_{t^{4n}−4z}(π) = −(y + zy^{−1})⟨1, −x(t^{2n} − y − zy^{−1})⟩ = \\ −(y + zy^{−1})⟨1, (t^{2n} − 2y)^{2}(4y)^{−1}x⟩ = −(y + zy^{−1})⟨1, xy⟩. \end{matrix}

Suppose π ≃ τ_{1} ⊥ (t^{4n} − 4z)τ_{2} for some forms τ_{1}, τ_{2} over k. Then, obviously, π ≃ ⟨⟨c_{1}, c_{2}(t^{4n} − 4z)⟩⟩, and, consequently, ⟨⟨x, y + zy^{−1}⟩⟩ = ⟨⟨c_{1}, c_{2}⟩⟩ for some c_{1}, c_{2} ∈ k^{∗}. Also

c_{1} = disc ∂_{t^{4n}−4z}(π) = −xy,

which implies that −xyc_{1} ∈ k_{t^{4n}−4z}^{∗}^{2}. It is easy to see that this implies −xyc_{1} ∈ k^{∗}^{2}∪zk^{∗}^{2}, or, equivalently, c_{1} ∈ −xyk^{∗}^{2}∪−xyzk^{∗}^{2}. Since ⟨⟨x, y+zy^{−1}⟩⟩ = ⟨⟨c_{1}, c_{2}⟩⟩, we conclude that either ⟨⟨x, y + zy^{−1}⟩⟩_{k(\sqrt{−xy})} = 0, or ⟨⟨x, y + zy^{−1}⟩⟩_{k(\sqrt{−xyz})} = 0, which is, clearly, impossible.

Summing up we see that

(φ_{1} ⊥ (t^{4n} − 4z)φ_{2})_{an} ≃ π ≄ τ_{1} ⊥ (t^{4n} − a)τ_{2},

where

φ_{1} ≃ (y + zy^{−1})⟨1, xy⟩ ⊥ ⟨⟨x, y + zy^{−1}⟩⟩, φ_{2} ≃ −(y + zy^{−1})⟨1, xy⟩,

and τ_{1}, τ_{2} are some forms over k. □

The following theorem determines the structure of certain Pfister forms over k(t).

Theorem 1.7. Let k be a field, π an n-fold Pfister form over k(t), a ∈ k^{∗} \ k^{∗}^{2}. Assume that ∂_{p}(π) = 0 for any p ∈ 𝔸_{k}^{1} different from t^{2} − a, and 0 ≠ ∂_{t^{2}−a}(π) ∈ res_{k(\sqrt{a})/k} W (k). Then π ≃ ⟨⟨c_{1}, …, c_{n−1}, c_{n}(t^{2} − a)⟩⟩ for some c_{1}, …, c_{n} ∈ k^{∗}.

Proof. Put F = k((x)). We have π = φ_{1} + (t^{2} − a)φ_{2} for some φ_{1}, φ_{2} ∈ W (k), hence

(φ_{1} + xφ_{2})_{F (C)} = (φ_{1} + (t^{2} − a)φ_{2})_{F (C)} = π_{F (C)}.

Therefore, there exists a Pfister form ρ over F = k((x)) such that

(φ_{1} + xφ_{2})_{F (C)} = ρ_{F (C)}

([ELW], Prop.2.10), or, in other words, φ_{1} + xφ_{2} − ρ = ⟨⟨a, x⟩⟩q for some form q ∈ W (k). Let ρ ≃ ⟨⟨c_{1}, …, c_{n−1}, c_{n}x^{m}⟩⟩, where c_{1}, …, c_{n} ∈ k^{∗}, m ∈ {0, 1}. Just as in the proof of Theorem 1.1 we can replace x by t^{2} − a. Thus, we get

φ_{1} + (t^{2} − a)φ_{2} − ⟨⟨c_{1}, …, c_{n−1}, c_{n}(t^{2} − a)^{m}⟩⟩ = ⟨⟨a, t^{2} − a⟩⟩q = 0 ∈ W (k(t)),

or, equivalently, π = φ_{1}+(t^{2}−a)φ_{2} = ⟨⟨c_{1}, …, c_{n−1}, c_{n}(t^{2}−a)⟩⟩, since ∂_{t^{2}−a}(π) ≠ 0. □

The example in Proposition 1.6 shows that the analog of Theorem 1.7 is false in general for polynomials t^{4n}−a. Also it cannot be generalized to arbitrary irreducible polynomial of degree 2m, where m ≥ 3 is odd, as the following example shows.

Proposition 1.8. Let m ≥ 3 be an odd number, k_{0} a field, k = k_{0}(x, y, z), p(t) = t^{2m} + xyt^{2} − x. Then

1) The form ⟨1, xy, −x, −p⟩ over k(t) is isotropic, or, in other words, ⟨1, xy, −x, −p⟩_{an} ≃ q⟨1, −yp⟩ for some q ∈ k[t]. Moreover, since −yp⟨1, −yp⟩ ≃ ⟨1, −yp⟩, we may assume that p does not divide q.

2) If f ∈ 𝔸_{k}^{1}, then

∂_{f}(⟨⟨yp, zq⟩⟩) = \{ \begin{matrix}0 & if f ≠ p \\ −y⟨⟨yz⟩⟩ & if f = p \end{matrix}

3) ⟨⟨yp, zq⟩⟩ ≄ ⟨⟨c_{1}, c_{2}p⟩⟩ for any c_{1}, c_{2} ∈ k^{∗}.

Proof. The first statement is obvious. In particular, yq ∈ k_{p}^{∗}^{2}, and yp ∈ k_{r}^{∗}^{2}, where r is any prime monic divisor of q, hence the second statement follows. Now suppose that ⟨⟨yp, zq⟩⟩ = ⟨⟨c_{1}, c_{2}p⟩⟩ for some c_{1}, c_{2} ∈ k^{∗}. Comparing the residues at p we get that c_{1}yz ∈ k_{p}^{∗}^{2} ∩ k^{∗}.

Lemma 1.9. k_{p}^{∗}^{2} ∩ k^{∗} = k^{∗}^{2}.

Proof. Let l = k(x, y, z)[u]/(u^{m} + xyu − x). Obviously, k_{p} = l(\sqrt{u}). Suppose there is a quadratic extension k(\sqrt{a}) of k containing in k_{p}. Then k_{p} = l(\sqrt{a}), i.e. au ∈ l^{∗}^{2}. Hence a^{m}x = N_{l/k}(au) ∈ k^{∗}^{2}, and so we may assume that a = x. On the other hand, we can consider k_{p} as the field k(u, y, z), where ux = u^{m+1}(1 − uy)^{−1}. Since this element is not a square in k(u, y, z), we come to a contradiction. □

Returning to the proof of Proposition 1.8 we get

⟨⟨c_{1}, c_{2}p⟩⟩ = ⟨⟨yp, zq⟩⟩ = ⟨1, −yp⟩−zq⟨1, −yp⟩ = ⟨1, −yp⟩−z⟨1, −x, xy, −p⟩ = −z⟨⟨x, y⟩⟩−yz⟨⟨yp, yz⟩⟩,

which implies, in view of c_{1}yz ∈ k^{∗}^{2}, that

0 = ⟨⟨c_{1}, c_{2}p⟩⟩_{k(\sqrt{yz})} = (−z⟨⟨x, y⟩⟩ − yz⟨⟨yp, yz⟩⟩)_{k(\sqrt{yz})} = −z⟨⟨x, y⟩⟩_{k(\sqrt{yz})},

which is, clearly, impossible. □

The main idea in Theorem 1.1 permits to obtain a short proof of Theorem 4.1 from ([RST]), which we formulate here in a bit different way.

Theorem 1.10. Let k be a field, a, b ∈ k^{∗}, a ∉ k^{∗}^{2}, ab ∉ k^{∗}^{2}. Let D be a finite-dimensional central division algebra over k. Then the following conditions are equivalent.

1) D ⊗_{k(t)} (a, t^{2} − b) is a division algebra over k(t).

2) D_{k(\sqrt{a})} and D_{k(\sqrt{ab})} are division algebras.

Proof. The implication 1) ⟹ 2) follows from the fact that

ind D ⊗_{k(t)} (a, t^{2} − b)_{k(\sqrt{a})} = ind D_{k(\sqrt{a})}

and

ind D ⊗_{k(t)} (a, t^{2} − b)_{k(\sqrt{ab})} = ind D_{k(\sqrt{ab})}.

As for the implication 2) ⟹ 1) assume that D_{k(\sqrt{a})} and D_{k(\sqrt{ab})} are division algebras. Then by [T], Prop. 2.4 the algebra D ⊗ (a, x) is a division algebra over k(x). Let C be the conic over k(x) with the equation t^{2} − b = xv^{2}. If the algebra D ⊗ (a, x)_{k(x)(C)} is a division algebra, we are done. If not, then by [M] D ⊗ (a, x) ≃ (b, x)⊗D′ for some division algebra D′ over k(x). Hence D⊗(ab, x) ≃ D′⊗M_{2}(k(x)) is not a division algebra. Therefore, again by [T], Prop. 2.4 D_{k(\sqrt{ab})} is not a division algebra, a contradiction to the hypothesis. □

2. Behavior of quadratic forms under biquadratic extensions

Our next purpose is to study the behavior of quadratic forms under a biquadratic extension. In particular, it turns out that a nonexcellent biquadratic extension over a field k gives rise to a nonexcellent extension of the field k(x, y) (resp. k((x))((y))) determined by the function field of the product of two conics over k(x, y) (resp. k((x))((y))). More precisely, let a, b ∈ k^{∗}, x, y indeterminates, C_{a}, C_{b} be the affine conics associated with the quaternion algebras (a, x) and (b, y) and determined by the equations t^{2} − a = xv^{2}, u^{2} − b = yw^{2}.

Lemma 2.1. Let φ a form over k. The following conditions are equivalent:

1) The form φ_{k(\sqrt{a},\sqrt{b})} is isotropic.

2) The form φ⟨⟨t^{2} − a, u^{2} − b⟩⟩_{k(t,u)} is isotropic.

3) The form φ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})} is isotropic.

4) The form φ⟨⟨x, y⟩⟩_{k((x))((y))(C_{a}×C_{b})} is isotropic.

Proof. 1) ⟹ 2). Since φ_{k(\sqrt{a},\sqrt{b})} is isotropic, either φ_{k(\sqrt{b})} is isotropic, or φ_{k(\sqrt{b})} ≃ α⟨⟨a⟩⟩ ⊥ ψ for some α ∈ k(\sqrt{b})^{∗} and a form ψ over k(\sqrt{b}). Since ⟨⟨a, t^{2} − a⟩⟩ = 0, in both cases the form φ⟨⟨t^{2} − a⟩⟩_{k(t)(\sqrt{b})} is isotropic. Hence, either φ⟨⟨t^{2} − a⟩⟩_{k(t)} is isotropic, or φ⟨⟨t^{2} − a⟩⟩ ≃ β⟨⟨b⟩⟩ ⊥ τ for some β ∈ k(t)^{∗} and a form τ over k(t). Again, in both cases the form φ⟨⟨t^{2} − a, u^{2} − b⟩⟩ is isotropic over k(t, u).

2) ⟹ 3). Obvious, in view of the equations t^{2} − a = xv^{2}, u^{2} − b = yw^{2}.

3) ⟹ 4). Obvious.

4) ⟹ 1). Since the extension k(\sqrt{a}, \sqrt{b})((x))((y))(C_{a}×C_{b})/k(\sqrt{a}, \sqrt{b})((x))((y)) is purely transcendental, and the form φ⟨⟨x, y⟩⟩_{k(\sqrt{a},\sqrt{b})((x))((y)))(C_{a}×C_{b})} is isotropic, the form φ⟨⟨x, y⟩⟩_{k(\sqrt{a},\sqrt{b})((x))((y))} is isotropic as well. Hence the form φ_{k(\sqrt{a},\sqrt{b})} is also isotropic. □

Under the notation of Lemma 2.1 we have the following

Theorem 2.2. The following conditions are equivalent:

1) The form φ_{k(\sqrt{a},\sqrt{b})}_{an} is defined over k.

2) The form φ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})}_{an} is defined over k(x, y).

3) The form φ⟨⟨x, y⟩⟩_{k((x))((y))(C_{a}×C_{b})}_{an} is defined over k((x))((y)).

Moreover, if these conditions are fulfilled, and (φ_{k(\sqrt{a},\sqrt{b})})_{an} ≃ τ, where τ is a form over k, then

φ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})}_{an} ≃ τ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})}

and

φ⟨⟨x, y⟩⟩_{k((x))((y))(C_{a}×C_{b})}_{an} ≃ τ⟨⟨x, y⟩⟩_{k((x))((y))(C_{a}×C_{b})}.

Proof. We induct on dim φ. If dim φ = 0, i.e. the form φ is empty, then all the conditions hold. 1) ⟹ 2). Suppose (φ_{k(\sqrt{a},\sqrt{b})})_{an} ≃ τ_{k(\sqrt{a},\sqrt{b})}, where τ is a form over k. Then φ − τ ∈ W (k(\sqrt{a}, \sqrt{b})/k), i.e. φ − τ = ⟨⟨a⟩⟩ψ_{1} + ⟨⟨b⟩⟩ψ_{2}, where ψ_{1}, ψ_{2} are forms over k ([ELW]). Then, obviously,

φ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})} = τ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})}.

By Lemma 2.1 the form τ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})} is anisotropic, which implies

φ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})}_{an} ≃ τ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})}.

1) ⟹ 3). The same argument with replacement of k(x, y) by k((x))((y)).

3) ⟹ 1). We may assume that the form φ_{k(\sqrt{a},\sqrt{b})} is isotropic. By Lemma 2.1 the form φ⟨⟨t^{2} − a, u^{2} − b⟩⟩ is isotropic, hence the form φ⟨⟨x, y⟩⟩_{k((x))((y))(C_{a}×C_{b})} is isotropic as well. Therefore, by the hypothesis

φ⟨⟨x, y⟩⟩_{k((x))((y))(C_{a}×C_{b})} = Φ_{k((x))((y))(C_{a}×C_{b})},

where Φ is a form over k((x))((y)), dim Φ < 4 dim φ. Since

W (k((x))((y))) = W (k) ⊕ xW (k) ⊕ yW (k) ⊕ xyW (k),

we have

Φ ≃ ψ_{1} ⊥ xψ_{2} ⊥ yψ_{3} ⊥ xyψ_{4},

where ψ_{i} are forms over k. On the other hand,

φ⟨⟨x, y⟩⟩ − Φ ∈ W (k((x))((y))(C_{a} × C_{b})/k((x))((y))).

Therefore, in view of [ELW] we get

φ⟨⟨x, y⟩⟩ − Φ = ⟨⟨a, x⟩⟩Φ_{a} + ⟨⟨b, y⟩⟩Φ_{b}, (∗)

where Φ_{a} and Φ_{b} are some forms over k((x))((y)). Since

Φ_{a} ≃ τ_{1} ⊥ xτ_{2} ⊥ yτ_{3} ⊥ xyτ_{4},
Φ_{b} ≃ ρ_{1} ⊥ xρ_{2} ⊥ yρ_{3} ⊥ xyρ_{4},

for some forms τ_{i} and ρ_{i} over k, it is easy to see that just as in Theorem 1.1 we can replace in (∗) x, y and k((x))((y)) by t^{2} − a, u^{2} − b and k(t, u) respectively. It follows that

φ⟨⟨t^{2} − a, u^{2} − b⟩⟩ = ψ_{1} ⊥ (t^{2} − a)ψ_{2} ⊥ (u^{2} − b)ψ_{3} ⊥ (t^{2} − a)(u^{2} − b)ψ_{4}. (∗∗)

Notice that ∑_{i=1}^{4} dim ψ_{i} < 4 dim φ. Multiplying if needed all the parts of equality (∗∗)

by −(t^{2} − a), or −(u^{2} − b), or (t^{2} − a)(u^{2} − b) we may assume that dim ψ_{4} < dim φ. We have

φ_{k(\sqrt{a},\sqrt{b})} = ∂_{t^{2}−a,u^{2}−b}(φ⟨⟨t^{2} − a, u^{2} − b⟩⟩) = ψ_{4},

where ∂_{t^{2}−a,u^{2}−b} : W (k(t, u)) → W (k(\sqrt{a}, \sqrt{b})) is the composition of the residues maps at t^{2} − a and u^{2} − b. It follows by [ELW] that

φ − ψ_{4} = ⟨⟨a⟩⟩f_{1} + ⟨⟨b⟩⟩f_{2}

for some forms f_{1}, f_{2} over k. Hence

ψ_{4}⟨⟨x, y⟩⟩_{k((x))((y))(C_{a}×C_{b})} = φ⟨⟨x, y⟩⟩_{k((x))((y))(C_{a}×C_{b})}.

By the induction hypothesis the form (ψ_{4}_{k(\sqrt{a},\sqrt{b})})_{an} is defined over k. Since (ψ_{4}_{k(\sqrt{a},\sqrt{b})})_{an} ≃ (φ_{k(\sqrt{a},\sqrt{b})})_{an}, we are done.

2) ⟹ 1). We may assume that the form φ_{k(\sqrt{a},\sqrt{b})} is isotropic. Then by Lemma 2.1 the form φ⟨⟨x, y⟩⟩_{k(x,y)(C_{a}×C_{b})} is isotropic, hence by the hypothesis of 2)

φ⟨⟨x, y⟩⟩_{k((x))((y))(C_{a}×C_{b})} = Φ_{k((x))((y))(C_{a}×C_{b})}

for some form Φ defined over k((x))((y)), dim Φ < 4 dim φ. The argument in the proof of implication 3) ⟹ 1) shows that the form (φ_{k(\sqrt{a},\sqrt{b})})_{an} is defined over k. □

Now let φ be a 4-dimensional anisotropic form over k, a, b ∈ k^{∗}, C_{1}, C_{2} the affine conics over k(x) associated with the quaternion algebras (a, x), (b, x) and determined by the equations t^{2} − a = xα^{−2}, u^{2} − b = xβ^{−2}.

Theorem 2.3. The following two conditions are equivalent:

1) The form φ_{k(\sqrt{a},\sqrt{b})} is isotropic and (φ_{k(\sqrt{a},\sqrt{b})})_{an} is defined over k.

2) The form φ⟨⟨x⟩⟩_{k(x)(C_{1}×C_{2})} is isotropic.

Proof. 1) ⟹ 2). Let ψ_{k(\sqrt{a},\sqrt{b})} ≃ (φ_{k(\sqrt{a},\sqrt{b})})_{an} for some form ψ over k. We have φ−ψ ∈ W (k(\sqrt{a}, \sqrt{b})/k), hence φ−ψ = ⟨⟨a⟩⟩φ_{1} +⟨⟨b⟩⟩φ_{2} for some forms φ_{1}, φ_{2} over k. Thus φ⟨⟨x⟩⟩_{k(x)(C_{1}×C_{2})} = ψ⟨⟨x⟩⟩_{k(x)(C_{1}×C_{2})}, and so the form φ⟨⟨x⟩⟩_{k(x)(C_{1}×C_{2})} is isotropic.

2) ⟹ 1). Let F = k(t). We may assume that φ ≃ ⟨1, −u, −v, uvd⟩. Let further C be a conic corresponding to the quaternion algebra (b, t^{2} − a). In view of the equations determining the conics C_{1}, C_{2} it is easy to see that the form φ⟨⟨t^{2}−a⟩⟩_{F (C)} is isotropic. Since the extension F (C)/F is excellent and the Clifford algebra of φ⟨⟨t^{2} − a⟩⟩ is similar to (d, t^{2} − a), we have

φ⟨⟨t^{2} − a⟩⟩ = f⟨⟨d, t^{2} − a⟩⟩ + ⟨⟨b, t^{2} − a⟩⟩τ,

where f ∈ k(t)^{∗} and τ is an even-dimensional form over k. It is easy to see that consequently

⟨⟨u, v, t^{2} − a⟩⟩ ≡ ⟨⟨d, t^{2} − a, Q⟩⟩ + ⟨⟨b, t^{2} − a, R⟩⟩ (mod I^{4}(F)) (∗)

for some squarefree nonzero polynomials Q, R ∈ k[t]. We may assume that deg Q + deg R is as small as possible. In particular, t^{2} − a divides neither Q, nor R. We are going to prove that there exists c ∈ k^{∗} such that R = cQ. Indeed, suppose that there exists a prime monic polynomial p such that p|Q, but p |̸R. Put π ≃ ⟨⟨d, t^{2} − a, p⟩⟩. Then

∂_{p}(π) = ∂_{p}(⟨⟨d, t^{2} − a, Q⟩⟩) ∈ ∂_{p}(⟨⟨u, v, t^{2} − a⟩⟩) + I^{3}(k_{p}) = I^{3}(k_{p}),

which implies ∂_{p}(π) = 0. Also ∂_{∞}(π) = 0, where ∂_{∞} is the residue map associated with the local parameter t^{−1} at the infinity point. By the Scharlau reciprocity law for W (F) ([Sch], Ch.6, Th. 3.5) we have s_{t^{2}−a}(∂_{t^{2}−a}(π)) = 0, where s_{t^{2}−a} is the transfer determined by the k-linear map k_{t^{2}−a} → k, taking 1 to 0, and t to

1. Thus, ⟨⟨d, p⟩⟩ ∈ res_{k_{t^{2}−a}/k} W (k). Hence ⟨⟨d, p⟩⟩ = ⟨⟨d, e⟩⟩ ∈ W (k_{t^{2}−a}) for some e ∈ k^{∗}, which implies that

⟨⟨d, t^{2} − a, p⟩⟩ = ⟨⟨d, t^{2} − a, e⟩⟩,

and we can replace Q by e\frac{Q}{p} in (∗). But this is a contradiction to minimality of deg Q + deg R. Quite similarly one can prove that the case where p |̸Q, but p|R is impossible as well. Therefore, we conclude that R = cQ for some c ∈ k^{∗}. It follows that

⟨⟨u, v, t^{2}−a⟩⟩ ≡ ⟨⟨d, t^{2}−a, Q⟩⟩+⟨⟨b, t^{2}−a, cQ⟩⟩ = ⟨⟨bd, t^{2}−a, Q⟩⟩+⟨⟨b, c, t^{2}−a⟩⟩ (mod (I^{4}(F)).

Hence ∂_{p}⟨⟨bd, t^{2} − a, Q⟩⟩ = 0 if p ≠ t^{2} − a. Comparing residues we get that

⟨⟨bd, t^{2} − a, Q⟩⟩ = ⟨⟨bd, t^{2} − a, e⟩⟩

for some e ∈ k^{∗}. Thus

⟨⟨u, v, t^{2} − a⟩⟩ ≡ ⟨⟨bd, e, t^{2} − a⟩⟩ + ⟨⟨b, c, t^{2} − a⟩⟩ mod(I^{4}(F)).

Taking the residue at t^{2} − a we get

⟨⟨u, v⟩⟩_{k(\sqrt{a})} ≡ (⟨⟨bd, e⟩⟩ + ⟨⟨b, c⟩⟩)_{k(\sqrt{a})} modI^{3}k(\sqrt{a})), (∗∗)

hence ⟨⟨u, v⟩⟩_{k(\sqrt{a},\sqrt{b})} = ⟨⟨d, e⟩⟩_{k(\sqrt{a},\sqrt{b})}. Finally, (∗∗) implies

\begin{matrix}⟨1, −u, −v, uvd⟩_{k(\sqrt{a},\sqrt{b})} = ⟨⟨u, v⟩⟩_{k(\sqrt{a},\sqrt{b})} + ⟨uvd, −uv⟩_{k(\sqrt{a},\sqrt{b})} ≡ (⟨⟨d, e⟩⟩ + ⟨uvd, −uv⟩)_{k(\sqrt{a},\sqrt{b})} \\ = ⟨⟨d⟩⟩⟨1, −e, −uv⟩_{k(\sqrt{a},\sqrt{b})} ≡ −euv⟨⟨d⟩⟩_{k(\sqrt{a},\sqrt{b})}(modI^{3}(k(\sqrt{a}, \sqrt{b}))), \end{matrix}

hence (φ_{k(\sqrt{a},\sqrt{b})})_{an} ≃ −euv⟨⟨d⟩⟩_{k(\sqrt{a},\sqrt{b})}. □

Theorem 2.3 can be reformulated without mentioning any conic at all. Namely, keeping the notation in this theorem we have the following

Corollary 2.4. Let α, β, z be indeterminates. Let further

p(α, β, z) = z^{4} − 2(aα^{2} + bβ^{2})z^{2} + (aα^{2} − bβ^{2})^{2} ∈ k(α, β)[z]

be the minimal polynomial of α\sqrt{a} + β\sqrt{b} over the field k(α, β). The following two conditions are equivalent.

1) The form φ_{k(\sqrt{a},\sqrt{b})} is isotropic and the form φ_{k(\sqrt{a},\sqrt{b})}_{an} is defined over k.

2) The form φ⟨⟨p(α, β, z)⟩⟩ is isotropic over k(α, β, z).

Proof. We have α^{2}(t^{2} − a) = β^{2}(u^{2} − b) = x, which implies (αt − βu)(αt + βu) = aα^{2} − bβ^{2}. Put z = αt − βu. Then it is easy to see by straightforward computation that k(x)(C_{1} × C_{2}) = k(α, β, z) and 4xz^{2} = p(α, β, z). Theorem 2.3 then yields the result. □

Corollary 2.5. Suppose the form φ⟨⟨p(α, β, z)⟩⟩ from Corollary 2.4 is anisotropic over k(α, β, z). Let ψ be a form over k, dim ψ ≥ 5. Then the form φ⟨⟨p(α, β, z)⟩⟩ remains anisotropic over the field k(ψ).

Proof. Recall that we may assume that φ ≃ ⟨1, −u, −v, uvd⟩. Suppose the form φ⟨⟨p(α, β, z)⟩⟩_{k(α,β,z)(ψ)} is isotropic. By Corollary 2.4 the form φ_{k(ψ)(\sqrt{a},\sqrt{b})} is isotropic and its anisotropic part is defined over k(ψ). It is easy to see that then (u, v)_{k(ψ)} = (a, x) + (b, y) + (d, z) for some x, y, z ∈ k(ψ)^{∗}. By [S2], Prop.5 (u, v) = (a, x′) + (b, y′) + (d, z′) for some x′, y′, z′ ∈ k^{∗}. Then the form φ_{k(\sqrt{a},\sqrt{b})} is isotropic, and the form (φ_{k(\sqrt{a},\sqrt{b})})_{an} is defined over k. By Corollary 2.4 the form φ⟨⟨p(α, β, z)⟩⟩ is isotropic. □

Remark. In Corollary 2.5 the condition dim ψ ≥ 5 is essential. For example, if ψ ≃ φ, then φ⟨⟨p(α, β, z)⟩⟩ is, obviously, isotropic.

Acknowledgement. I am very grateful to Professor A.S. Merkurjev, who posed the question considered in Theorem 1.1, which, in its turn inspired the further work on this paper.

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