ON SOME ELEMENTS OF THE BRAUER GROUP OF A CONIC

A.S. Sivatski

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

Abstract. Let K be a field of characteristic not 2, X a nonsingular projective conic over K, n is a positive integer and a_{i} ∈ K^{∗} for 1 ≤ i ≤ n. We investigate elements of _{2}Br(X), which are sums of quaternion algebras (a_{i}, f_{i}) for some f_{i} ∈ K(X)^{∗}. An application to a construction of indecomposable division algebras of exponent 2 is given.

The main purpose of this paper is to construct indecomposable central division algebras of an arbitrary 2-primary index more than 4 over fields with finite uinvariant (recall that the u-invariant of the field K is the maximal dimension of anisotropic quadratic forms over K). To do this we begin with a result concerning the Brauer group of a conic. Let K be a field of characteristic different from 2, and let X be a nonsingular projective conic over K. Recall that

2Br(X) = ker(_{2}Br K(X) →^{∂} ∐_{x∈X} K(x)^{∗}/K(x)^{∗}^{2}),

where K(X) is the function field of X, x runs over all closed points of X, and K(x) is the residue field at x. Moreover,

∂_{x}(f, g) = (−1)^{v_{x}(f)v_{x}(g)} \frac{f^{v_{x}(g)}}{g^{v_{x}(f)}} ∈ k(x)^{∗}/k(x)^{∗}^{2},

where v_{x} is the discrete valuation related to the point x, and the symbol (f, g) denotes a quaternion algebra. Notice also that

res_{K(X)/K} _{2}Br K ⊂ _{2}Br(X).

From now on we fix a field K of characteristic not 2, and elements a_{1}, … a_{n} ∈ K^{∗} such that \overline{a_{1}}, … \overline{a_{n}} ∈ K^{∗}/K^{∗}^{2} are linearly independent over ℤ/2ℤ. For any set I ⊂ {1, …, n}, put a_{I} = ∏_{i∈I} a_{i} (if I = ∅, then a_{I} = 1). For any polynomial

f ∈ K[t] denote by l(f) the leading coefficient of f.

Typeset by AMS-TEX

Key words and phrases. Quadratic form, Brauer group, Quaternion algebra, Conic.

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

Proposition 1. Let X be a nonsingular projective conic over K, Q the corresponding quaternion algebra, α ∈ _{2}Br(X). Suppose that the following conditions hold:

1) α = ∑_{i=1}^{n}(a_{i}, f_{i}) for some f_{i} ∈ K(X)^{∗}.

2) There exist at most two sets I ⊂ \{1, …, n\} such that Q_{K(\sqrt{a_{I}})} = 0.

Then there exist b_{1}, … b_{n} ∈ K^{∗} such that α = res_{K(X)/K} ∑_{i=1}^{n}(a_{i}, b_{i}).

Proof. If Q = 0, then K(X) = K(t) is the rational function field. The exact sequence

0 → _{2}Br K →^{res} _{2}Br K(t) →^{∂} ∐_{x∈ℙ_{K}^{1}} K(x)^{∗}/K(x)^{∗}^{2}

shows that _{2}Br(X) = res_{K(t)/K} _{2}Br K. There is a splitting homomorphism

c : _{2}Br K(t) → _{2}Br K

such that c((f, g)) = (l(f), l(g)) for any f, g ∈ K[t]. Applying the map c to the equality

α = ∑_{i=1}^{n}(a_{i}, f_{i}),

we get what we need.

So assume that Q = (a, b) ≠ 0. Let z^{2} = ax^{2} + by^{2} be the equation determining the conic X, and let v_{0} be the point of X determined by y = 0. Obviously, we have K(v_{0}) = K(\sqrt{a}). In particular, deg(v_{0}) = 2. Assume first that K(v_{0}) ≠ K(\sqrt{a_{I}}) for each I ⊂ {1, …, n}, i.e. aa_{I} ∉ K^{∗}^{2}.

For any f ∈ K(X) denote by V (f) the finite set of all the points v ≠ v_{0} of X such that v(f) is odd (we identify a point of X with the corresponding discrete K-

valuation). Let V = ⋃_{i=1}^{n} V (f_{i}). For any I ⊂ {1, …, n} put V_{I} = ⋂_{i∈I} V (f_{i})\ ⋃_{i∉I} V (f_{i}).

Clearly, V = ⋃_{I} V_{I}, and, moreover, this union is disjoint and V (f_{i}) = ⋃_{i∈I} V_{I}. For

each I ⊂ {1, …, n} consider the divisor

𝔞_{I} = ∑_{v∈V_{I}} v − \frac{1}{2}( ∑_{v∈V_{I}} deg v)v_{0}.

Obviously, deg 𝔞_{I} = 0. Since any divisor on X of degree 0 is principal, we get that 𝔞_{I} = div(g_{I}) = ∑_{v∈X} v(g_{I})v for some g_{I} ∈ K(X)^{∗}. Notice that

v(g_{I}) = \{ \begin{matrix}1, & if v ∈ V_{I} \\ 0, & if v ∉ V_{I}, v ≠ v_{0} \end{matrix} ,

and, therefore, v(f_{i}( ∏_{i∈I} g_{I})^{−1}) is even for any i and v ≠ v_{0}. Moreover, since

deg(div(f_{i}( ∏_{i∈I} g_{I})^{−1})) = 0, deg v_{0} = 2, and 2| deg v for any v ∈ X, we conclude

that v_{0}(f_{i}( ∏_{i∈I} g_{I})^{−1}) is also even, i.e. v(f_{i}( ∏_{i∈I} g_{I})^{−1}) is even for any i and v ∈ X.

Therefore, we have div(f_{i}( ∏_{i∈I} g_{I})^{−1}) = 2 div(h_{i}), i.e. f_{i}( ∏_{i∈I} g_{I})^{−1} = u_{i}h_{i}^{2} for

some u_{i} ∈ K^{∗} and h_{i} ∈ K(X)^{∗}. This implies

α_{K(X)} = ∑_{i=1}^{n}(a_{i}, f_{i}) = ∑_{i=1}^{n}(a_{i}, u_{i} ∏_{i∈I} g_{I}h_{i}^{2}) = ∑_{i=1}^{n}(a_{i}, u_{i}) + ∑_{I}(a_{I}, g_{I}).

Now, let I_{0} be a subset of {1, …, n}, v ∈ V_{I_{0}}. Then we have

1 = ∂_{v}(α_{K(X)}) = ∂_{v}(∑_{I}(a_{I}, g_{I})) = ∂(a_{I_{0}}, g_{I_{0}}),

since v(g_{I}) = 0 if I ≠ I_{0}.

On the other hand, if v ∉ V_{I_{0}} and v ≠ v_{0}, then, obviously, ∂_{v}(a_{I_{0}}, g_{I_{0}}) = 1, since v(g_{I_{0}}) = 0. Thus, we conclude that ∂_{v}(a_{I_{0}}, g_{I_{0}}) = 1 for any v ≠ v_{0}.

Our next step is to show that ∂_{v_{0}}(a_{I_{0}}, g_{I_{0}}) = 1 as well. Assume the converse. Then v_{0}(g_{I_{0}}) is odd. Since deg div(g_{I_{0}}) = 0, and deg v_{0} = 2, there exists v ≠ v_{0} such that v(g_{I_{0}}) is odd and 4 |̸ deg v. We have

1 = ∂_{v}(a_{I_{0}}, g_{I_{0}}) = a_{I_{0}}(mod K(v)^{∗}^{2}).

Hence we obtain that K(\sqrt{a_{I_{0}}}) ⊂ K(v). Furthermore, since deg v is not divided by 4, we get that the degree [K(v) : K(\sqrt{a_{I_{0}}})] is odd. Since Q_{K(v)} = 0, we conclude that Q_{K(\sqrt{a_{I_{0}}} )} = 0. Therefore, by condition 2) of the proposition there are at most two sets I_{0}, which can satisfy the unequality ∂_{v_{0}}(a_{I_{0}}, g_{I_{0}}) ≠ 1. Let, for instance, ∂_{v_{0}}(a_{I_{01}}, g_{I_{01}}) ≠ 1 and ∂_{v_{0}}(a_{I_{02}}, g_{I_{02}}) ≠ 1. So we have

∏_{i∈I_{01}} a_{i} ∏_{i∈I_{02}} a_{i} = ∂_{v_{0}}((a_{I_{01}}, g_{I_{01}}) + (a_{I_{02}}, g_{I_{02}})) = ∂_{v_{0}}( ∑_{I≠I_{01},I_{02}}(a_{I}, g_{I})) = 1.

Thus, a_{I_{01}} = a_{I_{02}}(mod K^{∗}^{2}) (recall that a ≠ a_{I}(mod K^{∗}^{2}) for any I), which is impossible in view of linear independency of \overline{a_{1}}, …, \overline{a_{n}} ∈ K^{∗}/K^{∗}^{2}.

The case where there is an only I_{0} such that ∂_{v_{0}}(a_{I_{0}}, g_{I_{0}}) ≠ 1 is treated similarly. Therefore, we conclude that ∂_{v}(a_{I}, g_{I}) = 1 for any I ⊂ {1, …, n} and v ∈ X.

Lemma 2. Assume that α ∈ _{2}Br(X) \ res_{K(X)/K}(_{2}Br K). Then ind(α) ≥ 4.

Proof. The Merkurjev theorem on the norm residue homomorphism of degree 2 [4] claims that for any field F the natural map K_{2}(F)/2K_{2}(F) → _{2}Br F is an isomorphism. In view of this it follows from [10] that, if _{2}Br(X) ≠ res_{K(X)/K}(_{2}Br K), then

2Br(X)/ res_{K(X)/K}(_{2}Br K) ≃ ℤ/2ℤ, Q = 2D for some D ∈ Br K and D_{K(X)}

is the nonzero element in the factorgroup _{2}Br(X)/ res_{K(X)/K}(_{2}Br K).

Therefore, we get that α = (D + β)_{K(X)}, where β ∈ _{2}Br K. Obviously, we may assume that α is a division algebra. Let D_{1} be the central division algebra over K similar to D + β. Since exp D_{1} = 4 we have ind D_{1} ≥ 4. Suppose that ind(α) = 2.

Then by the Merkurjev index reduction theorem [5] D_{1} ≃ Q⊗_{K} Q′ for some central division K-algebra Q′. Obviously, Q′_{K(X)} = D_{1}_{K(X)} = α. Since exp D_{1} = 4 and exp Q = 2, we get that exp Q′ = 4. Applying the index reduction theorem again we obtain that Q′ ≃ Q ⊗_{K} Q′′ for some central division K-algebra Q′′, which implies

D_{1} ≃ Q′′ ⊗_{K} Q ⊗_{K} Q ≃ Q′′ ⊗_{K} M_{4}(K),

a contradiction, since D_{1} is a division algebra. The lemma is proved. □

Now we are able to finish the proof of Proposition 1 in the case where aa_{I} ∉ K^{∗}^{2} for each I ⊂ {1, …, n}. In view of Lemma 2 we get that (a_{I}, g_{I}) = res_{K(X)/K} β_{I} for some β_{I} ∈ _{2}Br K. Since (β_{I})_{K(\sqrt{a_{I}})(X)} = 0, we have (β_{I})_{K(\sqrt{a_{I}})} is either zero or Q_{K(\sqrt{a_{I}})}, which implies that β_{I} equals either (a_{I}, c_{I}) or (a_{I}, c_{I}) + Q for some c_{I} ∈ K^{∗}. Therefore,

α = res_{K(X)/K}(∑_{I}(a_{I}, c_{I})) = res_{K(X)/K}(∑_{i=1}^{n}(a_{i}, ∏_{i∈I} c_{I})),

so we can put b_{i} = ∏_{i∈I} c_{I}.

It remains to consider the case where aa_{I} ∈ K^{∗}^{2} for some I ∈ {1, …, n}. Passing to the rational function field K(t) we have

Q = (a, b) = (at^{2} + b, −ab),

and, obviously (at^{2} + b)a_{I} ∉ K(t)^{∗}^{2}. Hence α_{K(X)(t)} = ∑_{i=1}^{n}(a_{i}, p_{i}) for some poly-

nomials p_{i} ∈ K[t]. This implies that α = ∑_{i=1}^{n}(a_{i}, l(p_{i})). The proposition is proved.

□

Corollary 3. In the notation of Proposition 1 assume that α = res_{K(X)/K} γ for

some γ ∈ Br K. Then γ equals either ∑_{i=1}^{n}(a_{i}, b_{i}) or Q + ∑_{i=1}^{n}(a_{i}, b_{i}) for some b_{i} ∈ K^{∗}.

Proof. Since res_{K(X)/K}(γ − ∑_{i=1}^{n}(a_{i}, b_{i})) = 0, and ker(Br K → Br K(X)) = ⟨Q⟩ is

the cyclic group generated by Q, the corollary follows. □

We can not drop condition 2) in Proposition 1 as the following two examples show.

Example A.

Let K be a field, a_{1}, a_{2}, a_{3} ∈ K^{∗}, D ∈ _{2}Br K, D_{K(\sqrt{a_{1}},\sqrt{a_{2}},\sqrt{a_{3}})} = 0, and D is not a sum of 3 quaternion algebras (the existence of such D has been established in [6] and [1]). Hence by [3] we have

(1)  D_{K(\sqrt{a_{3}})} = (a_{1}, b_{1} + c_{1}\sqrt{a_{3}}) + (a_{2}, b_{2} + c_{2}\sqrt{a_{3}})

for some b_{1}, b_{2}, c_{1}, c_{2} ∈ K, and, moreover,

(a_{1}, b_{1}^{2} − a_{3}c_{1}^{2}) + (a_{2}, b_{2}^{2} − a_{3}c_{2}^{2}) = N_{K(\sqrt{a_{3}})/K}D = 0,

i.e. (a_{1}, b_{1}^{2} − a_{3}c_{1}^{2}) = (a_{2}, b_{2}^{2} − a_{3}c_{2}^{2}). Put Q = (a_{1}, b_{1}^{2} − a_{3}c_{1}^{2}), and let X be the conic corresponding to Q. Then we get

N_{K(X)(\sqrt{a_{3}})/K(X)}(a_{1}, b_{1} + c_{1}\sqrt{a_{3}}) = (a_{1}, b_{1}^{2} − a_{3}c_{1}^{2})_{K(X)} = 0,
N_{K(X)(\sqrt{a_{3}})/K(X)}(a_{2}, b_{2} + c_{2}\sqrt{a_{3}}) = (a_{2}, b_{2}^{2} − a_{3}c_{2}^{2})_{K(X)} = 0.

Hence by [3]

(2)  (a_{1}, b_{1} + c_{1}\sqrt{a_{3}})_{K(X)(\sqrt{a_{3}})} = (a_{1}, f_{1})_{K(X)(\sqrt{a_{3}})},

(3)  (a_{2}, b_{2} + c_{2}\sqrt{a_{3}})_{K(X)(\sqrt{a_{3}})} = (a_{2}, f_{2})_{K(X)(\sqrt{a_{3}})}

for some f_{1}, f_{2} ∈ K(X)^{∗}. Combining (1), (2) and (3) we obtain that

(D − (a_{1}, f_{1}) − (a_{2}, f_{2}))_{K(X)(\sqrt{a_{3}})} = 0,

i.e.

D_{K(X)} = (a_{1}, f_{1}) + (a_{2}, f_{2}) + (a_{3}, f_{3})

for some f_{1}, f_{2}, f_{3} ∈ K(X)^{∗}. On the other hand, if Corollary 3 of Proposition 1 were valid, then we would have

D = (a_{1}, e_{1}) + (a_{2}, e_{2}) + (a_{3}, e_{3})

for some e_{1}, e_{2}, e_{3} ∈ K^{∗}, which would contradict to the hypothesis on D. Notice that in this example

Q_{K(\sqrt{a_{1}})} = Q_{K(\sqrt{a_{1}a_{3}})} = Q_{K(\sqrt{a_{2}})} = Q_{K(\sqrt{a_{2}a_{3}})} = 0,

so there are at least four elements a_{I} such that Q_{K(\sqrt{a_{I}})} = 0.

Example B.

Let K be a field, \sqrt{−1} ∈ K, Q = (a_{1}, a_{2}) a nontrivial quaternion algebra, X the corresponding conic. Put α = (a_{1}, y) + (a_{2}, x) ∈ _{2}Br K(X), where a_{1}x^{2} + a_{2}y^{2} = 1 is the equation of an affine part of the conic. It is easy to see that α ∈ _{2}Br(X) and N_{K(X)/K(x)}α = Q. In particular, α ∉ res_{K(X)/K}(_{2}Br K). Notice that since \sqrt{−1} ∈ K, we have

Q = (a_{1}, a_{2}) = (a_{1} + a_{2}, −a_{1}a_{2}) = (a_{1} + a_{2}, a_{1}a_{2}),

hence

Q_{K(\sqrt{a_{1}})} = Q_{K(\sqrt{a_{2}})} = Q_{K(\sqrt{a_{1}a_{2}})} = 0,

so there are three elements a_{I} such that Q_{K(\sqrt{a_{I}})} = 0.

We keep the conditions of Proposition 1. Let L/K be any field extension. Consider the following abelian groups:

A(L/K) = {γ ∈ _{2}Br K such that res_{L/K} γ = ∑_{i=1}^{n}(a_{i}, f_{i}) for some f_{i} ∈ L^{∗}},

B(L/K) = { ∑_{i=1}^{n}(a_{i}, b_{i}) ∈ _{2}Br K where b_{i} ∈ K^{∗}},

H(L/K) = A(L/K)/B(L/K).

Corollary 3 claims that if L = K(X), where X is the nonsingular projective conic corresponding to the quaternion algebra Q, then H(L/K) = ⟨Q⟩ (of course, it may happen that the image of Q in H(L/K) is zero). Now we will consider the case of a quadric of dimension more than 1. We will need the following

Lemma 4. 1) If F/L/K is a tower of field extensions, and H(L/K) = H(F/L) = 0, then H(F/K) = 0 as well. If H(F/K) = 0, then H(L/K) = 0.

2) If L/K is a purely transcendental extension, then H(L/K) = 0.

3) If H(F/K) = 0, and the extension L/K is such that LF/F is a purely transcendental extension, then H(L/K) = 0.

Proof. 1) This follows immediately from the definition of the groups H(F/K), H(L/K) and H(F/L).

2) In view of 1) it suffices to treat the case where L = K(t). Assume that u ∈

2Br K is such that u_{K(t)} = ∑_{i=1}^{n}(a_{i}, f_{i}) for some f_{i} ∈ K[t]. Then u = ∑_{i=1}^{n}(a_{i}, l(f_{i})),

which proves the assertion.

3) This follows at once from 1) and 2). □

Proposition 5. Let X = X_{φ} be the projective quadric over K corresponding to the anisotropic quadratic form φ. Put L = K(X). Then

1) If dim φ = 4 and disc(φ)a_{I} ∈ K^{∗}^{2} for some I ≠ ∅, then H(L/K) = ⟨(a, b)⟩, where (a, b) is an arbitrary quaternion K-algebra such that C_{0}(φ) ≃ (a, b)_{K(\sqrt{disc φ})}.

2) If dim φ = 4 and disc(φ)a_{I} ∉ K^{∗}^{2} for every I, then H(L/K) = 0.

3) If dim φ ≥ 5, then H(L/K) = 0.

Proof. 1), 2). Let d = disc(φ), and let z^{2} − ax^{2} − by^{2} + abd = 0 be the equation of an affine part of X. We have

K(X) = K(x, y, \sqrt{ax^{2} + by^{2} − abd}) = K(x)(C),

where C is the conic over K(x) corresponding to the quaternion K(x)-algebra (b, ax^{2} − abd).

Let u ∈ A(L/K). By Corollary 3 we have

(4)  u_{K(x)} = δ + ∑_{i=1}^{n}(a_{i}, f_{i}),

where δ is either 0 or (b, ax^{2} − abd), and f_{i} ∈ K[x]. Furthermore,

u = c(u_{K(x)}) = c(δ) + ∑_{i=1}^{n}(a_{i}, l(f_{i})),

where c(δ) is either zero, or (a, b). This proves that H(L/K) ⊂ ⟨(a, b)⟩. Assume first that da_{I} ∈ K^{∗}^{2} for some I. Then, since

(ab)^{2}d = a(by)^{2} + b(ax)^{2} − abz^{2}

and

(a, b)_{K(X)} = (a(by)^{2} + b(ax)^{2}, −ab) = (a(by)^{2} + b(ax)^{2} − abz^{2}, ab(a(by)^{2} + b(ax)^{2}),

we get that (a, b) ∈ H(L/K).

Now assume that da_{I} ∉ K^{∗}^{2} for every I. Compare the residues at (x^{2} − bd) on the left-hand and the right-hand sides of (4). We have

(5)  ∂_{x^{2}−bd}(u_{K(x)}) = 0, ∂_{x^{2}−bd}( ∑_{i=1}^{n}(a_{i}, f_{i})) = a_{I} for some I.

Suppose that δ ≠ 0, i.e. δ = (b, ax^{2} − abd). The equality ∂_{x^{2}−bd}(δ) = b and (5)

imply ba_{I} ∈ K(\sqrt{bd})^{∗}^{2}. Since da_{I} ∉ K^{∗}^{2}, we obtain that ba_{I} ∈ K^{∗}^{2}. Thus, we have proved part 2) if ba_{I} ∉ K^{∗}^{2} for any I. On the other hand,

φ_{K(t)} ≃ ⟨1, −ab(at^{2} + b), −(at^{2} + b), abd⟩,

and (at^{2} + b)a_{I} ∉ K(t)^{∗}^{2}. The above argument show that H(LK(t)/K(t)) = 0, and so by Lemma 4 we have H(L/K) = 0.

3) Obviously, we can choose a 4-dimensional subform ψ of the form φ_{K(t)} such that disc(ψ)a_{I} ∉ K(t)^{∗}^{2}. Since the form φ_{K(t)(X_{ψ})} is isotropic, the extension K(t)(X_{ψ})(X_{φ})/K(t)(X_{ψ}) is purely transcendental. Applying part 2) and Lemma 4 we get H(K(X_{φ})/K) = 0. □

Now we are ready to strentghen the main results of [7] and [8]. Recall the corresponding notation.

Let F be an arbitrary field, n ≥ 2, a, b_{1}, …, b_{n} ∈ F^{∗}, D ∈ _{2}Br F. We call the triple (F, D, {a, b_{1}, …, b_{n}}) admissible, if the following conditions hold:

a) The elements \overline{a}, \overline{b_{1}}, …, \overline{b_{n}} ∈ F^{∗}/F^{∗}^{2} are linearly independent.

b) ind D = 2.

c) D_{F (\sqrt{a},\sqrt{b}_{1},...,\sqrt{b}_{n})} = 0.

d) For any tower F ⊂ F_{1} ⊂ F_{2} ⊂ L = F (\sqrt{a}, \sqrt{b}_{1}, …, \sqrt{b}_{n}) such that

\sqrt{a} ∉ F_{2}^{∗}, [L : F_{1}] = 8, [F_{2} : F_{1}] = 4

we have

D_{F_{1}} ∉ _{2}Br(F_{1}(\sqrt{a})/F_{1}) + _{2}Br(F_{2}/F_{1}).

e) The field F has no proper extensions of odd degree.

Corollary 6. Suppose that the triple (F, D, {a, b_{1}, …, b_{n}}) is admissible, φ is an anisotropic form over F, and dim φ ≥ 5. Put X = X_{φ}, and let F (X)_{odd} be a maximal odd degree extension of F (X). Then the triple (F (X)_{odd}, D_{F (X)_{odd}}, {a, b_{1}, …, b_{n}}) is admissible as well.

Proof. Obviously, all the above conditions except maybe d) hold. Let us check d). First we will do this for the field F (X). So, suppose that there is a tower of fields F (X) ⊂ F_{1}(X) ⊂ F_{2}(X) ⊂ L(X) = F (\sqrt{a}, \sqrt{b}_{1}, …, \sqrt{b}_{n})(X) such that

\sqrt{a} ∉ F_{2}^{∗}, [L : F_{1}] = 8, [F_{2} : F_{1}] = 4

and

D_{F_{1}(X)} ∈ _{2}Br(F_{1}(X)(\sqrt{a})/F_{1}(X)) + _{2}Br(F_{2}(X)/F_{1}(X)).

Notice that F_{2} = F_{1}(\sqrt{d_{1}}, \sqrt{d_{2}}), for some d_{1}, d_{2} ∈ F^{∗}, hence by [3] we have

D_{F_{1}(X)} = (a, f) + (d_{1}, f_{1}) + (d_{2}, f_{2})

for some f, f_{1}, f_{2} ∈ F_{1}(X)^{∗}. By part 3) of Proposition 5 we conclude that

D_{F_{1}} = (a, u) + (d_{1}, u_{1}) + (d_{2}, u_{2})

for some u, u_{1}, u_{2} ∈ F_{1}^{∗}, which contradicts to condition d) for the admissible triple (F, D, {a, b_{1}, …, b_{n}}).

On the other hand, it has been shown in [8] that if the triple (K, D, {a, b_{1}, …, b_{n}}) is admissible, then the triple K_{odd}, D_{K_{odd}}, {a, b_{1}, …, b_{n}}) is admissible as well (see also Lemma 9 below). This finishes the proof of Corollary 6. □

We recall now some additional results from the papers [7] and [8].

If the triple (F, D, {a, b_{1}, …, b_{n}}) is admissible, and E = F ((t_{0}))((t_{1})) … ((t_{n})) is the Laurent series field in variables t_{0}, t_{1}, …, t_{n}, then the division algebra A similar to the algebra

D ⊗_{E} (a, t_{0}) ⊗_{E} (b_{1}, t_{1}) ⊗ ⋯ ⊗_{E} (b_{n}, t_{n})

does not decompose into a tensor product of two nontrivial central simple algebras over any odd degree extension of E, and ind A = 2^{n+1} [8]. Moreover, the multiquadratic extension F (\sqrt{b_{1}}, …, \sqrt{b_{n}})/F is not 4-excellent, and if D = (u, v), the form ⟨uv, −u, −v, a⟩ provides a corresponding counterexample [7]. Finally, if k is a field and elements \overline{a}, \overline{b_{1}}, …, \overline{b_{n}} ∈ k^{∗}/k^{∗}^{2} are linearly independent over ℤ/2ℤ, then there exists an extension F/k and a quaternion algebra D over F such that the triple (F, D, {a, b_{1}, …, b_{n}}) is admissible [8].

On the other hand, Corollary 6 and the argument in Lemma 9 below imply that if the triple (F, D, {a, b_{1}, …, b_{n}}) is admissible, then there exists an extension K/F such that the triple (K, D_{K}, {a, b_{1}, …, b_{n}}) is admissible as well, u(K) = 4, and K has no proper odd degree extension.

Summarizing all these results we obtain the following Theorem 7. Let k be a field, n ≥ 2. Assume that elements \overline{a}, \overline{b_{1}}, …, \overline{b_{n}} ∈ k^{∗}/k^{∗}^{2} are linearly independent over ℤ/2ℤ. Then there exists an extension K/k such that

1) u(K) = 4.

2) K has no proper extension of odd degree.

3) K(\sqrt{b_{1}}, …, \sqrt{b_{n}})/K is not 4-excellent, and the counterexample is provided by some 4-dimensional form of discriminant a.

4) There exists a central division algebra A over E = K((t_{0}))((t_{1})) … ((t_{n})) such that ind A = 2^{n+1}, and A_{L} does not decompose into a tensor product of two nontrivial central simple algebras over any odd degree extension L/E (notice that u(E) = 2^{n+3} and cd_{2}(E) = n + 3).

Next we compute the group H for the extension determined by the function field of the Severi-Brauer variety related to a central simple algebra of exponent 2.

Corollary 8. Let D ∈ _{2}Br K, and either ind D ≥ 4, or ind D = 2 and D_{K(\sqrt{a_{I}})} ≠ 0 for every I. Denote by K(D) the function field of the Severi-Brauer variety SB(D) corresponding to D. Then H(K(D)/K) = ⟨D⟩.

Proof. Let ind D = 2^{n}, where n ≥ 2. Let D_{1} be the division algebra Brauerequivalent to D, F = SB(D_{1}), and L = SB(D). Then, since SB(D_{F}) ≃ ℙ_{F}^{deg D−1}, the extension LF/F is purely transcendental. Hence by Lemma 4 it suffices to prove Corollary 8 in the case where D is a division algebra.

First consider the case where D is a tensor product of quaternion algebras. We will go on by induction on n. Let φ ∈ I^{2}(K) be a quadratic form corresponding to D under the isomorphism I^{2}(K)/I^{3}(K) ≃ _{2}Br K. It is easy to see that we can choose φ such that dim φ = 2n + 2, and, moreover, no form of dimension < 2n + 2 corresponds to D under this isomorphism. Let X = X_{φ} be the projective quadric determined by the form φ. Consider the following diagram:

\begin{matrix}K & ⟶ & K(D) \\ ↓ & & ↓ \\ K(X) & ⟶ & K(X)(D) \end{matrix},

where all the maps are natural field embeddings.

If n = 2, then ind D_{K(X)} = 2, and D_{K(X)(\sqrt{a_{I}})} ≠ 0 for any I, since D_{K(\sqrt{a_{I}})} ≠ 0, and ker(_{2}Br F → _{2}Br F (Y)) = 0 for any field F and any quadric Y over F of dimension > 2.

If n ≥ 3, then 4 ≤ ind D_{K(X)} < ind D. So in both cases (n = 2 or n ≥ 3) we can apply the induction hypothesis to the extension K(X)(D)/K(X). Let u ∈ H(K(D)/K). By the induction hypothesis

u_{K(X)} ∈ H(K(X)(D)/K(X)) ⊂ ⟨D_{K(X)}⟩.

We conclude that either u ∈ H(K(X)/K), or u + D ∈ H(K(X)/K). Since by Proposition 5 we have H(K(X)/K) = 0, this completes the proof in the case where D is a tensor product of quaternion algebras.

Thus, to prove the proposition in the general case it suffices to construct an extension L/K such that D_{L} is a tensor product of quaternion algebras, ind D_{L} ≥ 4, and H(L/K) = 0. We need the following Lemma 9.

1) If L/K is a finite odd degree extension, then H(L/K) = 0.

2) If π is a 3-fold Pfister form over K, then H(K(X_{π})/K) = 0.

3) Let F be a field, D ∈ _{2}Br F a division algebra. Assume that F has no proper extension of odd degree, and I^{3}(F) = 0. Then D is a tensor product of quaternion algebras.

Proof.

1) Assume that u ∈ _{2}Br K and u_{L} = ∑_{i=1}^{n}(a_{i}, b_{i}), where b_{i} ∈ L^{∗}. Then

u = N_{L/K}(u_{L}) = ∑_{i=1}^{n}(a_{i}, N_{L/K}b_{i}),

which proves 1).

2) This is a particular case of Proposition 5, since dim π = 8.

3) This part is proved in [9] and [2]. □

We return to the proof of Corollary 8 in the general case. For an arbitrary field F denote by F̂ a maximal odd degree extension of F, and by F̃ the composite of all extensions F (X_{π}), where π runs over all anisotropic 3-fold Pfister forms over F. Consider the following infinite tower of fields K = K_{1} ⊂ K_{2} ⊂ …, where

K_{i+1} = \{ \begin{matrix}K̂_{i} if i is odd, \\ K̃_{i} if i is even \end{matrix} .

Set L = ⋃_{i=1}^{∞} K_{i}. It is easy to see that L has no proper odd degree extensions, and

I^{3}(L) = 0. Applying Lemma 9 we get that H(L/K) = 0 and D_{L} is a tensor product of quaternion algebras. Moreover, by the index reduction formula for central simple algebras [5] we have ind D_{L} = ind D. These properties of the field L prove Corollary 8 in the general case. □

REFERENCES

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[3] Lam T.Y., Leep D.B., Tignol J.-P., Biquaternion algebras and quartic extensions, Pub. Math. I.H.E.S. 77 (1993), 63-102.

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

[8] Sivatski A.S., On indecomposable algebras of exponent 2, to appear in Israel J. Math.

[9] Sivatski A.S., Invariants in the algebraic theory of quadratic forms, Thesis, St. Petersburg State University (1988).

[10] Suslin A., The quaternion homomorphism for the function field on a conic, Soviet Math. Docl. 26 (1982), no. 1, 72-77.

e-mail: sivatsky@AS3476.spb.edu