ON INDECOMPOSABLE ALGEBRAS OF EXPONENT 2

A.S. Sivatski

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

Abstract. For any n ≥ 3 we give numerous examples of central division algebras of exponent 2 and index 2^{n} over fields, which do not decompose into a tensor product of two nontrivial central division algebras, and which are sums of n + 1 quaternion algebras in the Brauer group of the field.

Also for any n ≥ 3 and any field k_{0} we construct an extension F/k_{0} and a multiquadratic extension L/F of degree 2^{n} such that for any proper subextensions L_{1}/F and L_{2}/F

W (L/F) ≠ W (L_{1}/F) + W (L_{2}/F), _{2}Br(L/F) ≠_{2} Br(L_{1}/F) +_{2} Br(L_{2}/F).

1. Preliminaries

Let F be a field of characteristic not 2. A well known result of Albert states that any central division algebra of index 4 and exponent 2 over F is a tensor product of two quaternion algebras ([A]). In [T1] Tignol proved that any algebra of index 8 and exponent 2 is similar to a tensor product of four quaternion algebras. Moreover, by Merkurjev’s theorem ([M]), any central simple algebra A of exponent 2 over a field is similar to a tensor product of quaternion algebras. In other words, any element [A] of _{2}Br F is a sum of classes of quaternion algebras. If the index of A is 2^{n}, then, obviously, the number of summands in this sum is not less than n. If this number equals n, then by dimension count A itself is isomorphic to a tensor product of quaternion algebras. We call a central simple algebra over a field decomposable if it is a tensor product of two nontrivial central simple algebras over the same field. Otherwise we say that the algebra is indecomposable. Obviously, if the algebra is indecomposable it is necessarily a division algebra, i.e. a skewfield. The first example of indecomposable algebra of index 8 and exponent 2 was given in [ART]. In ([K]) Karpenko gave examples of indecomposable algebras of exponent 2^{m} and index 2^{n} for any m ≥ 1, n ≥ 3, n ≥ m. An algebra A of such a type is obtained by means of the ”generic” splitting of another algebra of index and exponent 2^{n}.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 16K20, 14H05.

Key words and phrases. Brauer group, division algebra, conic, Laurent series field.

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

However, if exp A = 2 and [A] = ∑_{i=1}^{p}[(a_{i}, b_{i})] in these examples, it is unclear how

small one can choose the number of summands p. In this paper we construct for any field k_{0} and n ≥ 3 an indecomposable algebra of index 2^{n} over some field E/k_{0} which is a sum of n + 1 quaternion algebras in _{2}Br E. One can choose the field E having no proper odd degree extension. Moreover, all the summands are splitted by the same prescribed multiquadratic field extension of degree 2^{n} over E. The description of the field E and the quaternion algebras is quite transparent.

We refer the reader to [Sch] as the main source concerning central simple algebras and quadratic forms over fields. The notation used in the sequel is more or less standard and coincides with the notation in ([S]). All the fields in the paper are supposed to be of characteristic different from 2. If F is a field, then _{2}Br F stands for the 2-torsion of the Brauer group of F. Slightly abusing notation we will often identify a central simple algebra and its class in the Brauer group. W (F) and I(F) are respectively the Witt ring of F and the ideal of evendimensional quadratic forms in W (F). By K_{2}(F) we denote the Milnor’s K_{2} group of F. For u, v ∈ F^{∗} the symbol (u, v) denotes the quaternion algebra over F generated by the elements i and j with the relation

i^{2} = u, j^{2} = v ij = −ji.

If C is a projective conic over a field k, and p is a closed point of C, then k(C) is the function field of C, and k(p) is the residue field at p. If f ∈ k(C)^{∗}, then (f) is the divisor of the function f. The abbreviations res, N, ind, deg denote respectively restriction, norm, index and degree. If L/F is a field extension and A is a central simple algebra over F, then by definition A_{L} = res_{L/F} A = A ⊗_{F} L. We will write simply A⊗L instead of A⊗_{F} L, or even merely A, if the restriction is clear from the context. The tensor product of central simple algebras over F is always considered over F. The sign ∼ means similarity of algebras, i.e. their equality in the Brauer group of the field.

2. Auxiliary and related results

The crucial point in the examples that we are going to give is some quaternion algebra, which was constructed in ([S]) and which is the basic tool in the proof of nonexcellency of multiquadratic field extensions in general. For the convenience of the reader we recall the construction of this algebra, the more so that we need some additional properties of it.

Let k_{0} be a field. Suppose n is a positive integer and \overline{a}, \overline{b}_{1}, …, \overline{b}_{n} ∈ k_{0}^{∗}/k_{0}^{∗}^{2} are linearly independent over ℤ/2ℤ.

Lemma 1. There exist a tower of fields k_{0} ⊂ k_{1} ⊂ ⋯ ⊂ k_{n} and elements x_{i}, y_{i} ∈ k_{i}^{∗} for every 1 ≤ i ≤ n such that the following conditions hold:

1) The elements \overline{a}, \overline{b}_{1}, …, \overline{b}_{n} remain linearly independent in k_{n}^{∗}/k_{n}^{∗}^{2}.

2) Put B_{j} = ∑_{i=1}^{j}(b_{i}, x_{i} + y_{i}\sqrt{a}) ∈ _{2}Br k_{j}(\sqrt{a}). Then for any 1 ≤ j ≤ n

ind B_{j} = 2.

3) Put A_{j} = ∑_{i=1}^{j}(b_{i}, x_{i}^{2} − ay_{i}^{2}) ∈ _{2}Br k_{j}. Then for any 1 ≤ i ≤ j ≤ n and a

finite field extension l/k_{0} such that (A_{j})_{lk_{j}} = 0, either \sqrt{b_{i}} ∈ l or \sqrt{ab_{i}} ∈ l.

In particular, [l : k_{0}] ≥ 2^{j}.

Proof. Let x_{1}, y_{1} be indeterminates, k_{1} = k_{0}(x_{1}, y_{1}). Put

A_{1} = (b_{1}, x_{1}^{2} − ay_{1}^{2}) ∈ _{2}Br k_{1}, B_{1} = (b_{1}, x_{1} + y_{1}\sqrt{a}) ∈ _{2}Br k_{1}(\sqrt{a}).

We will prove the lemma by induction on n. If n = 1, conditions 1) and 2) are obvious. To check condition 3) in this case put l_{1} = l(\sqrt{a})(y_{1}) and consider the second residue map

∂_{x_{1}−\sqrt{a}y_{1}} : _{2}Br l_{1}(x_{1}) ⟶ l_{1}^{∗}/l_{1}^{∗}^{2}

with respect to the linear polynomial x_{1}−\sqrt{a}y_{1}. Since ∂_{x_{1}−\sqrt{a}y_{1}}((b_{1}, x_{1}^{2}−ay_{1}^{2})) = b_{1}, we get that b_{1} is a square in l_{1}, hence in l(\sqrt{a}), which is just what we need.

Now let n ≥ 2. Assume that we have already constructed a tower of fields k_{0} ⊂ k_{1} ⊂ ⋯ ⊂ k_{n−1} and the algebras A_{1}, …, A_{n−1}, B_{1}, …, B_{n−1} with the required conditions. Suppose B_{n−1} = (u_{1} + v_{1}\sqrt{a}, u_{2} + v_{2}\sqrt{a}), where u_{1}, u_{2}, v_{1}, v_{2} ∈ k_{n−1}^{∗}. Let t_{1}, t_{2}, t_{3} be indeterminates, and k_{n} = k_{n−1}(t_{1}, t_{2}, t_{3}). Let further x_{n}, y_{n} ∈ k_{n} be such elements that

x_{n} + y_{n}\sqrt{a} = (u_{1} + v_{1}\sqrt{a})t_{1}^{2} + (u_{2} + v_{2}\sqrt{a})t_{2}^{2} − (u_{1} + v_{1}\sqrt{a})(u_{2} + v_{2}\sqrt{a})t_{3}^{2},

i.e.

x_{n} = u_{1}t_{1}^{2} + u_{2}t_{2}^{2} − (u_{1}u_{2} + av_{1}v_{2})t_{3}^{2},
y_{n} = v_{1}t_{1}^{2} + v_{2}t_{2}^{2} − (u_{1}v_{2} + u_{2}v_{1})t_{3}^{2}.

Put

A_{n} = A_{n−1} + (b_{n}, x_{n}^{2} − ay_{n}^{2}) ∈ _{2}Br k_{n},
B_{n} = B_{n−1} + (b_{n}, x_{n} + y_{n}\sqrt{a}) ∈ _{2}Br k_{n}(\sqrt{a}).

It is easy to see that x_{n} + y_{n}\sqrt{a} is a square in (B_{n−1})_{k_{n−1}(\sqrt{a})}, which implies that

(B_{n})_{k_{n}(\sqrt{a})(\sqrt{x_{n}+y_{n}\sqrt{a}})} = 0,

hence ind B_{n} ≤ 2. Obviously, (B_{n−1})_{k_{n}(\sqrt{a})} ≠ (b_{n}, x_{n} + y_{n}\sqrt{a}), so ind B_{n} = 2. Now suppose that a finite extension l/k_{0} is such that

0 = (A_{n})_{lk_{n}} = (A_{n−1} + (b_{n}, x_{n}^{2} − ay_{n}^{2}))_{lk_{n}}.

Then

0 = (A_{n})_{lk_{n}(\sqrt{a})} = (A_{n−1})_{lk_{n}(\sqrt{a})} + (b_{n}, (x_{n} − y_{n}\sqrt{a})(x_{n} + y_{n}\sqrt{a}))_{lk_{n}(\sqrt{a})}.

Put K = lk_{n−1}(\sqrt{a}, t_{1}, t_{2}). We can view the algebra

(A_{n−1})_{lk_{n}(\sqrt{a})} = (b_{n}, (x_{n} − y_{n}\sqrt{a})(x_{n} + y_{n}\sqrt{a}))_{lk_{n}(\sqrt{a})}

as an algebra over K(t_{3}). It has no residues with respect to the second residue map

∂ = ⊕∂_{p} : _{2}Br K(t_{3}) ⟶ ∐_{p} K(p)^{∗}/K(p)^{∗}^{2},

where p runs over all monic irreducible polynomials from K[t_{3}]. On the other hand, the monic polynomials

P_{1}(t_{3}) = t_{3}^{2} − \frac{(u_{1} + v_{1}\sqrt{a})t_{1}^{2} + (u_{2} + v_{2}\sqrt{a})t_{2}^{2}}{(u_{1} + v_{1}\sqrt{a})(u_{2} + v_{2}\sqrt{a})} = \frac{−(x_{n} + y_{n}\sqrt{a})}{(u_{1} + v_{1}\sqrt{a})(u_{2} + v_{2}\sqrt{a})}

and

P_{2}(t_{3}) = t_{3}^{2} − \frac{(u_{1} − v_{1}\sqrt{a})t_{1}^{2} + (u_{2} − v_{2}\sqrt{a})t_{2}^{2}}{(u_{1} − v_{1}\sqrt{a})(u_{2} − v_{2}\sqrt{a})} = \frac{−(x_{n} − y_{n}\sqrt{a})}{(u_{1} − v_{1}\sqrt{a})(u_{2} − v_{2}\sqrt{a})}

are distinct. Hence

0 = ∂_{P_{1}}((A_{n−1})_{lk_{n}(\sqrt{a})}) = b_{n},

and so,

\sqrt{b_{n}} ∈ lk_{n}(\sqrt{a}, t_{1}, t_{2}, \sqrt{\frac{(u_{1} + v_{1}\sqrt{a})t_{1}^{2} + (u_{2} + v_{2}\sqrt{a})t_{2}^{2}}{(u_{1} + v_{1}\sqrt{a})(u_{2} + v_{2}\sqrt{a})}} ).

This obviously implies \sqrt{b_{n}} ∈ lk_{n}(\sqrt{a}). Therefore, either \sqrt{b_{n}} ∈ lk_{n}, or \sqrt{ab_{n}} ∈ lk_{n}. Since the extension k_{n}/k_{0} is purely transcendental, we get that either \sqrt{b_{n}} ∈ l, or \sqrt{ab_{n}} ∈ l. In both cases it follows that (b_{n}, x_{n}^{2} − ay_{n}^{2})_{lk_{n}} = 0. Since

0 = (A_{n})_{lk_{n}(\sqrt{a})} = (A_{n−1})_{lk_{n}(\sqrt{a})} + (b_{n}, x_{n}^{2} − ay_{n}^{2})_{lk_{n}(\sqrt{a})},

we get

(A_{n−1})_{lk_{n}(\sqrt{a})} = 0.

Since k_{n} = k_{n−1}(t_{1}, t_{2}, t_{3}), it follows that

(A_{n−1})_{lk_{n−1}(\sqrt{a})} = 0,

and by the induction assumption we conclude that for any 1 ≤ i ≤ n − 1 either \sqrt{b_{i}} ∈ l, or \sqrt{ab_{i}} ∈ l. The induction step is done, so the lemma is proved.

Since ind B_{n} = 2, it follows that ind A_{n} = ind N_{k_{n}(\sqrt{a})/k_{n}}(B_{n}) = 2 or 4. If ind A_{n} = 2, put k = k_{n}. If ind A_{n} = 4 put k = k_{n}(φ), where φ is an Albert form corresponding to A_{n}. It is easy to see that condition 3) remains valid for the field k and the algebra (A_{n})_{k}, i.e. if the extension l/k_{0} is finite and (A_{n})_{lk} = 0, then for any 1 ≤ i ≤ n either \sqrt{b_{i}} ∈ l or \sqrt{ab_{i}} ∈ l.

Now put A = (A_{n})_{k}, B = (B_{n})_{k(\sqrt{a})}. Let C be the projective conic over k corresponding to A. Set F = k(C). Then, obviously, A_{F} = 0. Consider the exact sequence ([M])

2Br F ⟶^{res} _{2}Br F (\sqrt{a}) ⟶^{N} _{2}Br F.

Since

0 = A_{F} = N_{F (\sqrt{a})/F} (B_{F (\sqrt{a})}),

we have B_{F (\sqrt{a})} = D_{F (\sqrt{a})} for some D ∈ _{2}Br F. Since ind B = 2, we get ind D ≤ 4. It is well known that if ind D = 4, then D = D_{1} + D_{2} for some D_{1}, D_{2} ∈ _{2}Br F, such that ind D_{1} = ind D_{2} = 2 and D_{2}_{F (\sqrt{a})} = 0. Hence we may change D for D_{1} and so assume that ind D ≤ 2.

From now on we will assume that n ≥ 2. Put L = F (\sqrt{a}, \sqrt{b_{1}}, …, \sqrt{b_{n}}). Let F ⊂ F_{1} ⊂ F_{2} ⊂ L be a tower of fields such that

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

Obviously, L = F_{2}(\sqrt{a}) = F_{1}(\sqrt{a})F_{2}, and F_{1} = kl_{1}(C), F_{2} = kl_{2}(C) for some multiquadratic extensions l_{1}/k_{0}, l_{2}/k_{0} of degree 2^{n−2} and 2^{n} respectively.

Proposition 2.

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

2) ind D = 2.
3) D_{L} = 0.

4) The algebra D_{F_{1}} does not decompose into a sum D_{F_{1}} = D_{1} + D_{2}, where

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

Proof. 1) Obvious.

2) Assume that ind D = 1, i.e. D = 0. Then B_{k(\sqrt{a})(C)} = 0, hence B_{k(\sqrt{a})} is either 0 or A_{k(\sqrt{a})}. In both cases A = N_{k(\sqrt{a})/k}B = 0, a contradiction.

3) Obvious, since D_{F (\sqrt{a})} = ∑_{i=1}^{n}(b_{i}, x_{i} + y_{i}\sqrt{a}).

4) Assume the contrary, i.e. that D = D_{1} + D_{2}, where D_{1} ∈ _{2}Br(F_{1}(\sqrt{a})/F_{1}) and D_{2} ∈ _{2}Br(F_{2}/F_{1}). Then

B_{F_{1}(\sqrt{a})} = D_{F_{1}(\sqrt{a})} = D_{2}_{F_{1}(\sqrt{a})}.

Denote the field kl_{1} by k̂, so that F_{1} = k̂(C). Consider the points z_{1}, …, z_{m} ∈ Ck̂ at which the algebra D_{2} has nonzero residues under the second residue map

2Br F_{1} ⟶ ∐_{z∈C_{k}} k̂(z)^{∗}/k̂(z)^{∗}^{2}.

Since B is defined over k(\sqrt{a}), the algebra D_{2}_{F_{1}(\sqrt{a})} = B_{F_{1}(\sqrt{a})} has no residues at all. This implies that the residues of D_{2} at the points z_{1}, …, z_{m} are equal to a. Recall that F_{2} = kl_{2}(C), and let c_{1}, …, c_{n} ∈ k_{0}^{∗} be elements such that l_{2} = k_{0}(\sqrt{c_{1}}, …, \sqrt{c_{n}}). Therefore, since D_{2}_{F_{2}} = 0, we have for any j

a_{k̂(z_{j})(\sqrt{c_{1}},...,\sqrt{c_{n}})} ∈ k̂(z_{j})(\sqrt{c_{1}}, …, \sqrt{c_{n}})^{∗}^{2}.

Denote by c_{I} the product ∏_{i∈I} c_{i} (the similar notation will be used also in the sequel).

Thus, given any j, we have ac_{I} ∈ k̂(z_{j})^{∗}^{2} for some I ⊂ {1, …, n} depending on j, i.e.

k̂(\sqrt{ac_{I}}) ⊂ k̂(z_{j}).
□

Lemma 3. For any j we have 4| deg z_{j}.

Proof. Assume this is not the case. Consider the tower k̂ ⊂ k̂(\sqrt{c}) ⊂ k̂(z_{j}), where c = ac_{I} and I corresponds to the point z_{j}. Notice that \sqrt{c} ∉ k̂ = kl_{1}, for otherwise \sqrt{c} ∈ kl_{2}, and so \sqrt{a} ∈ F_{2}, a contradiction. Hence [k̂(z_{j}) : k̂(\sqrt{c})] is odd. Since A_{k̂(z_{j})} = 0 we conclude that A_{k̂(\sqrt{c})} = 0. Therefore, (A_{n})_{k_{n}l_{1}(\sqrt{c})} = 0, a contradiction to Lemma 1, since [l_{1}(\sqrt{c}) : k_{0}] = 2^{n−1}. The lemma is proved. □

Now we use the fact that any divisor of degree zero on a projective conic is

principal. Choose any y ∈ Ck̂ such that deg y = 2. Let s = ∑_{i=1}^{m} deg z_{j}. Consider

the divisor 𝔞 = −\frac{s}{2}y + ∑_{i=1}^{m} z_{i}. Obviously, deg 𝔞 = 0, hence there is f ∈ F_{1}^{∗} such that

𝔞 = (f). The algebra (a, f) has nonzero residues just at the points z_{1}, …, z_{m}, since by Lemma 3 the number \frac{s}{2} is even. Moreover, all these residues are equal to a. Therefore, the algebra D̂_{2} = D_{2} + (a, f) has no residues at all, i.e. D̂_{2} ∈ _{2}Br(Ck̂). Merkurjev’s theorem ([M]) claims that K_{2}k̂(C)/2K_{2}k̂(C) ≃ _{2}Br k̂(C). Hence by ([Su], Lemma 5) we get that either

D̂_{2} ∈ res_{F_{1}/k̂}(_{2}Br k̂),

or

2Br(Ck̂)/ res_{F_{1}/k̂}(_{2}Br k̂) = ℤ/2ℤ,

and the element D̂_{2} is nontrivial in this factor group.

In the first case let D̂_{2} = res_{F_{1}/k̂} D̃ for some D̃ ∈ _{2}Br k̂. Then

(D̃ + B)_{k̂(\sqrt{a})(C)} = (D̂_{2} + B)_{k̂(\sqrt{a})(C)} = (D_{2} + B)_{k̂(\sqrt{a})(C)} = (D + B)_{k̂(\sqrt{a})(C)} = 0.

Hence (D̃ + B)_{k̂(\sqrt{a})} is either zero or A_{k̂(\sqrt{a})}. But then

0 = N_{k̂(\sqrt{a})/k̂}(D̃ + B) = N_{k̂(\sqrt{a})/k̂}B = Ak̂,

hence (A_{n})_{l_{1}k_{n}} = 0, a contradiction to Lemma 1, since [l_{1} : k_{0}] = 2^{n−2}. In the second case, when D̂_{2} ∉ res_{F_{1}/k̂} (_{2}Br k̂), let A = (a_{1}, a_{2}), where a_{1}, a_{2} ∈

k^{∗}. Then F_{1} is the quotient field of the ring k̂[u_{1}, u_{2}]/(a_{1}u_{1}^{2} + a_{2}u_{2}^{2} − 1), where u_{1}, u_{2} are indeterminates. It is easy to see that N_{F_{1}/k̂(u_{1})}(D̂_{2}) = A. Consider the commutative diagram

2Br F_{1} ⟶^{res} _{2}Br F_{1}(\sqrt{a})
N↓  ^{N}↓
2Br k̂(u_{1}) ⟶^{res} _{2}Br k̂(\sqrt{a})(u_{1})

So we have

res_{k̂(\sqrt{a})(u_{1})/k̂(u_{1})} A = res_{k̂(\sqrt{a})(u_{1})/k̂(u_{1})} ∘ N_{F_{1}/k̂(u_{1})}(D̂_{2}) =
N_{F_{1}(\sqrt{a})/k̂(\sqrt{a})(u_{1})} ∘ res_{F_{1}(\sqrt{a})/F_{1}}(D̂_{2}) = N_{F_{1}(\sqrt{a})/k̂(\sqrt{a})(u_{1})}(B) = 0,

hence res_{k̂(\sqrt{a})/k}(A) = 0, a contradiction to Lemma 1, since [l_{1}(\sqrt{a}) : k_{0}] = 2^{n−1}. Thus Proposition 2 is proved. □

Recall that L = F (\sqrt{a}, \sqrt{b_{1}}, …, \sqrt{b_{n}}). Proposition 2 implies the following corollary, which is of some independent interest, and which can be considered as a generalization of Th. 5.1 in [ELTW].

Corollary 4. Let F ⊂ L_{1} ⊂ L, F ⊂ L_{2} ⊂ L be proper field subextensions of L/F. Then

1) D ∉ _{2}Br(L_{1}/F) + _{2}Br(L_{2}/F). In particular,

2Br(L/F) ≠ _{2}Br(L_{1}/F) + _{2}Br(L_{2}/F).

2) f ∉ W (L_{1}/F) + W (L_{2}/F), where f ∈ W (L/F) is the 2-fold Pfister form corresponding to D.

Proof. 1) Obviously we may assume that [L : L_{1}] = [L : L_{2}] = 2 and L = L_{1}L_{2}. Then [L : L_{1} ∩ L_{2}] = 4. Choose a field F_{1} such that

F ⊂ F_{1} ⊂ L_{1} ∩ L_{2}, \sqrt{a} ∉ F_{1}^{∗}, [L_{1} ∩ L_{2} : F_{1}] = 2.

Then

[L : F_{1}] = 8, [L_{1} : F_{1}] = [L_{2} : F_{1}] = 4, L_{1} = F_{1}(\sqrt{d_{1}}, \sqrt{d_{2}}), L_{2} = F_{1}(\sqrt{d_{1}}, \sqrt{d_{3}})

for some d_{1}, d_{2}, d_{3} belonging to the multiplicative group generated by the elements a, b_{1}, …, b_{n}. We will show that D_{F_{1}} ∉ _{2}Br(L_{1}/F_{1}) + _{2}Br(L_{2}/F_{1}). Assume the contrary. Then

D_{F_{1}} = (d_{1}, e_{1}) + (d_{2}, e_{2}) + (d_{3}, e_{3})

for some e_{1}, e_{2}, e_{3} ∈ F_{1}^{∗}. Obviously, we may suppose that a is equal either to d_{i}, or to d_{i}d_{j} (i < j), or to d_{1}d_{2}d_{3}. We have

D_{F_{1}} = (d_{1}d_{2}, e_{1}) + (d_{2}, e_{1}e_{2}) + (d_{3}, e_{3}) = (d_{1}d_{2}d_{3}, e_{1}) + (d_{2}, e_{1}e_{2}) + (d_{3}, e_{1}e_{3}).

Put F_{2} = F_{1}(\sqrt{d_{2}}, \sqrt{d_{3}}). We conclude that

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

which contradicts Proposition 2.

2) This part easily follows from part 1). Indeed, assume that [f] = [f_{1}] + [f_{2}], where f_{1} ∈ W (L_{1}/F), f_{2} ∈ W (L_{2}/F). Obviously, disc(f_{1}) = disc(f_{2}). Let g_{1}, g_{2} ∈ I^{2}(F) be forms such that dim(f_{i} ⊥ −g_{i})_{an} ≤ 2 for i = 1, 2. It is clear that g_{i} ∈ W (L_{i}/F). If D_{i} ∈ _{2}Br F correspond to g_{i} under the map I^{2}(F)/I^{3}(F) ≃ _{2}Br F, then D = D_{1} + (D_{2} + (disc f_{2}, u)) for some u ∈ F^{∗}, a contradiction to part 1) since ⟨⟨disc(f_{2})⟩⟩ ∈ W (L_{2}/F). □

Lemma 5. Let K be a field, K((t))_{odd} any odd degree extension of K((t)). Then the following holds:

1) There exists a field inclusion K_{odd}((t)) ↪ K((t))_{odd} over K((t)) for some maximal odd degree extension K_{odd} of K.

2) K((t))_{odd}^{∗}/K((t))_{odd}^{∗}^{2} ≃ K_{odd}^{∗}/K_{odd}^{∗}^{2} ⊕ ℤ/2ℤ, where the element t ∈ K((t))_{odd}^{∗}/K((t))_{odd}^{∗}^{2} corresponds to the nontrivial element of ℤ/2ℤ and K_{odd}^{∗}/K_{odd}^{∗}^{2} naturally includes into K((t))_{odd}^{∗}/K((t))_{odd}^{∗}^{2} under this isomorphism.

Proof. 1) Any finite extension L of K((t)) is a complete valuation field of characteristic equal to the characteristic of its residue field. We have L = \overline{L}((π)), where \overline{L} is a subfield of L isomorphic to the residue field and π is a uniformizer ([Se], Ch.2, Th.2). It is easy to check that we can choose for K_{odd} the union of all finite extensions of K contained in K((t))_{odd}.

2) Assume L/K_{odd}((t)) is a finite odd degree extension. Then

L ≃ K_{odd}((π)), L^{∗}/L^{∗}^{2} ≃ K_{odd}^{∗}/K_{odd}^{∗}^{2} ⊕ ℤ/2ℤ,

where π ∈ L^{∗}/L^{∗}^{2} corresponds to the nontrivial element of ℤ/2ℤ and K_{odd}^{∗}/K_{odd}^{∗}^{2} naturally includes into L^{∗}/L^{∗}^{2}. Denote by v the discrete valuation on L. Obviously, v(t) is odd, hence πt ∈ K_{odd}^{∗}/K_{odd}^{∗}^{2} ↪ L^{∗}/L^{∗}^{2}. Therefore, there exists another isomorphism

L^{∗}/L^{∗}^{2} ≃ K_{odd}^{∗}/K_{odd}^{∗}^{2} ⊕ ℤ/2ℤ,

where t ∈ L^{∗}/L^{∗}^{2} corresponds to the nontrivial element of ℤ/2ℤ and K_{odd}^{∗}/K_{odd}^{∗}^{2} naturally includes into L^{∗}/L^{∗}^{2}. Also it is clear that the last isomorphisms are compatible for various L, which completes the proof. □

Corollary 6. Let K be a field, E = K((t_{0})) … ((t_{n})) the iterated Laurent series field. Let E_{odd} be any maximal odd degree extension of E. Then

K_{odd} ↪ E_{odd}, E_{odd}^{∗}/E_{odd}^{∗}^{2} ≃ K_{odd}^{∗}/K_{odd}^{∗}^{2} ⊕ (ℤ/2ℤ)^{n+1}

for some maximal odd degree extension K_{odd} of K, and under this isomorphism the element t_{i} (0 ≤ i ≤ n) corresponds to the nontrivial element of the i + 1-th factor of (ℤ/2ℤ)^{n+1} while K_{odd}^{∗}/K_{odd}^{∗}^{2} naturally includes into E_{odd}^{∗}/E_{odd}^{∗}^{2}.

Proof. Taking into account Lemma 5, the proof is immediate by induction on n. □

Lemma 7. Let K be a field, let \overline{c_{0}}, …, \overline{c_{n}} ∈ K^{∗}/K^{∗}^{2} be linearly independent elements. Suppose the algebra α ∈ _{2}Br K is such that α_{K(\sqrt{c_{0}},\sqrt{c_{1}},...,\sqrt{c_{n}})} ≠ 0. Let further t_{0}, …, t_{n}, x be indeterminates. Then for any 0 ≤ i ≤ n + 1

ind[α + (c_{0}, t_{0}) + ⋯ + (c_{n}, t_{n})]_{E} ≥ 2^{n+2},

where

E = K((t_{0})) … ((t_{i−1}))((x))((t_{i})) … ((t_{n})).

Proof. The argument is by induction on n. Suppose first that i ≤ n. Put

E_{1} = K((t_{0})) … ((t_{i−1}))((x))((t_{i})) … ((t_{n−1})).

By induction hypothesis ind[α + (c_{0}, t_{0}) + ⋯ + (c_{n−1}, t_{n−1})_{E_{1}(\sqrt{c_{n}})}] ≥ 2^{n+1}. Applying Tignol’s theorem ([T2], Prop. 2.4) we finishes the proof. If i = n + 1, the argument is similar with a slight modification. □

3. Construction of indecomposable algebras

Now we turn to the construction of indecomposable algebras of an arbitrary 2-primary index and exponent 2. From now on we change the notation a bit.

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.

Notice that Proposition 2 give us examples of triples satisfying conditions a)−d) of the above definition.

Proposition 8.

1) If the triple (F, D, {a, b_{1}, …, b_{n}}) satisfies condition a) − d) above, then the triple (F_{odd}, D_{odd}, {a, b_{1}, …, b_{n}}) is admissible.

2) If the triple (F, D, {a, b_{1}, …, b_{n}}) is admissible, n ≥ 3, c ∈ F^{∗}, and

F (\sqrt{a}, \sqrt{b}_{1}, …, \sqrt{b}_{n}) = F (\sqrt{c}, \sqrt{a}, \sqrt{b_{1}}, …, \sqrt{b_{n−1}}),

then the triple (F (\sqrt{c}), D_{F (\sqrt{c})}, \{a, b_{1}, …, b_{n−1}\}) is admissible as well.

3) If the triple (F, D, \{a, b_{1}, …, b_{n}\}) is admissible, and L = F (\sqrt{a}, \sqrt{b}_{1}, …, \sqrt{b}_{n}), then Corollary 4 holds for the extension L/F and the algebra D.

Proof. 1) Conditions a), b), c), e) are obvious, and applying the norm map we obtain condition d).

2) The proof is straightforward, and we leave it to the reader.

3) Obvious. □

Now let (F, D, {a, b_{1}, …, b_{n}}) be an admissible triple. Put E = F ((t_{0})) … ((t_{n})). Suppose C is the division algebra over E such that

C ∼ D ⊗ (a, t_{0}) ⊗ (b_{1}, t_{1}) ⊗ … ⊗ (b_{n}, t_{n}).

The main purpose of the present article is the following Proposition 9.

1) ind C = 2^{n+1}.

2) The algebra C_{odd} =^{def} C_{E_{odd}} does not decompose into a tensor product of two nontrivial algebras over any maximal odd degree extension E_{odd} of E.

Proof. 1) Since C_{L} = 0 we have ind C ≤ 2^{n+1}. On the other hand, applying Lemma 7 to the field F_{0} = F ((t_{0})) and the algebra C_{0} = D ⊗ (a, t_{0}) we get

ind C = ind[(D + (a, t_{0})) + (b_{1}, t_{1}) + ⋯ + (b_{n}, t_{n})] ≥ 2^{n+1}.

2) The proof is by induction on n. Assume that the algebra C_{odd} is decomposable, C_{odd} ≃ D_{1} ⊗_{E_{odd}} D_{2} and ind D_{1}, ind D_{2} ≥ 2. Consider two cases:

a) ind D_{1} = 2 or ind D_{2} = 2.

Suppose for instance ind D_{1} = 2. By Lemma 5 D_{1} ≃ (pt_{I}, st_{J}), where

I, J ⊂ {0, 1, …, n}, p, s ∈ F^{∗}.

Case a) divides in turn into two subcases.

a1) I ≠ ∅ or J ≠ ∅. Assume I ≠ ∅.

Suppose I = {k + 1, k + 2, …, n} for some k (the general case can be treated similarly, using Lemma 7). Then we have

\begin{matrix}C_{E_{odd}(\sqrt{pt_{I}})} = D + (a, t_{0}) + (b_{1}, t_{1}) ⋯ + (b_{n−1}, t_{n−1}) + (b_{n}, pt_{k+1} … t_{n−1}) = \\ D + (b_{n}, p) + (a, t_{0}) + (b_{1}, t_{1}) + ⋯ + (b_{k}, t_{k}) + (b_{k+1}b_{n}, t_{k+1}) + ⋯ + (b_{n−1}b_{n}, t_{n−1}). \end{matrix}

The field E_{odd}(\sqrt{pt_{I}}) can be considered as an iterated Laurent series field, namely E_{odd}(\sqrt{pt_{I}}) = F ((t_{0})) … ((t_{n−1}))((x)), where x = \sqrt{pt_{I}}. Put

L_{1} = F (\sqrt{a}, \sqrt{b_{1}}, …, \sqrt{b_{k}}, \sqrt{b_{k+1}b_{n}}, …, \sqrt{b_{n−1}b_{n}}), L_{2} = F (\sqrt{b_{n}}).

Since ind C_{E_{odd}(\sqrt{pt_{I}})} = \frac{1}{2} ind C = 2^{n}, it follows by Lemma 7 that

(D + (b_{n}, p))_{L_{1}} = 0.

Therefore,

D = (D + (b_{n}, p)) + (b_{n}, p) ∈ _{2}Br(L_{1}/F) + _{2}Br(L_{2}/F),

which contradicts Corollary 4.

a2) I = J = ∅, i.e. D_{1} = (p, s), where p, s ∈ F^{∗}.

Then

ind[D + (p, s) + (a, t_{0}) + (b_{1}, t_{1}) + ⋯ + (b_{n}, t_{n})] = ind D_{2} = 2^{n},

which contradicts Lemma 6.

Thus we have completed the proof in the case a). In particular, we have proved indecomposability of C in the case n = 2.

b) ind D_{1} ≥ 4 and ind D_{2} ≥ 4.

In particular, n ≥ 3. Obviously, either ind D_{1}_{E_{odd}(\sqrt{a})} = ind D_{1}, or ind D_{2}_{E_{odd}(\sqrt{a})} = ind D_{2}. Assume that ind D_{1}_{E_{odd}(\sqrt{a})} = ind D_{1}. Since any finite extension of E_{odd} is a tower of quadratic extensions, there exists c ∈ E_{odd}^{∗} such that ind D_{1}_{E_{odd}(\sqrt{c})} = \frac{1}{2} ind D_{1}. Moreover, since F has no proper odd degree extensions, we may assume by Lemma 5 that c = pt_{I} for some p ∈ F^{∗} and I ⊂ {0, 1, …, n}. Let D_{1}_{E_{odd}(\sqrt{pt_{I}})} ∼ D′_{1}, where D′_{1} is some nontrivial division algebra over E_{odd}(\sqrt{pt_{I}}). If I ≠ ∅ we come to a contradiction just as in case a). So suppose that I = ∅. Since ind D_{1}_{E_{odd}(\sqrt{a})} = ind D_{1}, we have pa ∉ F^{∗}^{2}, and

ind C_{E_{odd}(\sqrt{p})} = ind[D + (a, t_{0}) + (b_{1}, t_{1}) + ⋯ + (b_{n}, t_{n})]_{E_{odd}(\sqrt{p})} = 2^{n}.

It follows from Lemma 6 that the elements \overline{a}, \overline{b_{1}}, …, \overline{b_{n}} are linearly dependent in F (\sqrt{p})^{∗}/F (\sqrt{p})^{∗}^{2}. This in turn means that we may assume that either p = b_{I}, or p = ab_{I} for some ∅ ≠ I ⊂ {1, …, n}. Let for instance p = b_{I}, where I = {k + 1, …, n} (the general case can be treated with a slight modification, and is left to the reader). Put

K = F (\sqrt{p})((t_{0})) … ((t_{n−1})), L = F (\sqrt{p})((t_{0})) … ((t_{n−1}))((\sqrt{t_{n}})).

Then

D ⊗ (a, t_{0}) ⊗ (b_{1}, t_{1}) ⊗ … ⊗ (b_{n−1}, t_{n−1})_{L_{odd}} ∼ C_{L_{odd}} ∼ D′_{1} ⊗_{L_{odd}} D_{2}

for any maximal odd degree extension L_{odd} of L, containing E_{odd}(\sqrt{p}). On the other hand, by Proposition 8 the triple (F (\sqrt{p}), D_{F (\sqrt{p})}, \{a, b_{1}, …, b_{n−1}\}) is admissible. Hence by the induction hypothesis for any maximal odd degree extension K_{odd} of K we have

D ⊗ (a, t_{0}) ⊗ (b_{1}, t_{1}) ⊗ … ⊗ (b_{n−1}, t_{n−1})_{K_{odd}} ∼ C_{1}_{K_{odd}},

C_{1} being a division algebra over K such that it is indecomposable over K_{odd} and ind C_{1} = 2^{n}. By the first part of Lemma 5 we can choose K_{odd} in such a way that K_{odd}((\sqrt{t_{n}})) ↪ L_{odd} over K((\sqrt{t_{n}})). This implies

C_{1}_{L_{odd}} ∼ D′_{1} ⊗_{L_{odd}} D_{2}.

Since, obviously, ind(D′_{1} ⊗_{L_{odd}} D_{2}) = ind C_{1}_{L_{odd}} = 2^{n}, we have

C_{1}_{L_{odd}} ≃ D′_{1} ⊗_{L_{odd}} D_{2}.

By ([Se], Ch.2, Th.2) the field L_{odd} is the direct limit of fields K_{odd}((u)), where u is an indeterminate. Hence for some u ∈ L_{odd} the algebras D′_{1} and D_{2} are defined over K_{odd}((u)) and

C_{1}_{K_{odd}((u))} ≃ D′_{1} ⊗_{K_{odd}((u))} D_{2}.

It is well known that for any field k and positive integer m not divided by char k

H^{2}(k((x)), μ_{m}) ≃ H^{2}(k, μ_{m}) ⊕ H^{1}(k, ℤ/mℤ),

where μ_{m} is the group of mth roots of unity. From this and the cohomological interpretation of the Brauer group it follows that

D′_{1} = res_{K_{odd}((u^{\frac{1}{s}}))/K_{odd}} D̃_{1}, D_{2} = res_{K_{odd}((u^{\frac{1}{s}}))/K_{odd}} D̃_{2}

for a sufficiently large 2-power s and some central division algebras D̃_{1} and D̃_{2} over K_{odd}. Since the natural map Br K_{odd} → Br K_{odd}((u^{\frac{1}{s}})) is injective,

C_{1}_{K_{odd}} = D̃_{1} + D̃_{2},

and by dimension count we conclude that

C_{1}_{K_{odd}} ≃ D̃_{1} ⊗_{K_{odd}} D̃_{2},

a contradiction to indecomposability of the algebra C_{1} over K_{odd}. □

Summing up the obtained results and changing the notation a bit we get the following

Corollary 10. Let k be a field, n ≥ 3. Suppose the elements a_{1}, …, a_{n} ∈ k^{∗}/k^{∗}^{2} are linearly independent over ℤ/2ℤ. Then there exists a field extension F/k, a quaternion algebra D ∈ _{2}Br F and a division algebra C ∈ _{2}Br F ((t_{1})) … ((t_{n})) of index 2^{n} such that

1) C is indecomposable over any odd degree extension of E = F ((t_{1})) … ((t_{n})).

2) M_{2}(C) ≃ (a_{1}, t_{1}) ⊗_{E} … ⊗_{E} (a_{n}, t_{n}) ⊗_{E} D.

3) D ∈ _{2}Br(F (\sqrt{a_{1}}, …, \sqrt{a_{n}})/F)

Proof. This immediately follows from Propositions 2, 8 and 9. □

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e-mail: sivatsky@AS3476.spb.edu