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\expandafter\redefine\expandafter\enddemo\expandafter{\expandafter\hfill
\expandafter$\expandafter\square\expandafter$\enddemo}
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\topmatter

\title
ON INDECOMPOSABLE ALGEBRAS OF EXPONENT $2$
\endtitle

\author
A.S. Sivatski
\endauthor

\affil
Department of Mathematics \\
St.Petersburg Electrotechnical University \\
197376, St.Petersburg, Russia\\
\endaffil

\thanks
The work under this publication was partially supported by
Royal society Joint Project{\it "Quadratic forms and central simple algebras
under field extensions"}
\endthanks

\keywords
Brauer group, division algebra, conic, Laurent series field
\endkeywords

\def\br#1{\,_2\!\operatorname{Br}#1}
\abstract
For any $n\ge 3$ 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 $n\ge 3$ 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 $L_2/F$
$$W(L/F)\not= W(L_1/F)+W(L_2/F),\ \ \ _2 
Br (L/F)\not=_2 Br (L_1/F)+ _2 Br (L_2/F).$$

\endabstract

\subjclass 16K20, 14H05
\endsubjclass

\endtopmatter

\document







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\def\phi{\varphi}
\def\z{\Bbb Z/2\Bbb Z}
\def\d{\dots}
\def\s{\simeq}
\def\a{\alpha}
\def\i{\operatorname{ind}}
\def\deg{\operatorname{deg}}
\def\ll{\langle\!\langle}
\def\rr{\rangle\!\rangle}
\def\op{\operatorname}
\def\ot{\otimes}
\def\uv{u_1+v_1\sqrt a}
\def\uuvv{u_2+v_2\sqrt a}
\def\xyi{x_i+y_i\sqrt a}
\def\xyn{x_n+y_n\sqrt a}
\def\xxyyi{x^2_i-ay^2_i}
\def\xxyyn{x^2_n-ay^2_n}
\def\aj{A_j}
\def\bj{B_j}
\def\s{\sqrt}
\def\h{\widehat k}
\def\su{\coprod\limits_{z\in C_{\h}}{\h (z)}^*/{{\h (z)}^*}^2}
\def\dd{\partial_p}
\def\sa{\sqrt a}
\def\l{\longrightarrow}
\def\br#1{\,_2\!\operatorname{Br} #1}
\def\ov{\overline}
\def\da{D+(a,t_0)+\d +(b_{n-1},t_{n-1})}
\def\ft{F(\s p)((t_0))\d ((t_{n-1}))}
\def\ftx{F(\s p)((t_0))\d ((t_{n-1}))((x))}
\def\res{\operatorname{res}}
\def\ind{\operatorname{ind}}
\def\disc{\operatorname{disc}}
\def\exp{\operatorname{exp}}
\def\li{\lim\limits_{\scriptstyle{\longrightarrow}\atop\scriptstyle{n\ odd}}}
\def\lii{\lim\limits_{\scriptstyle{\longrightarrow}\atop\scriptstyle{n}}}
\def\t{K_{odd}((t^{1\over n}))}
\def\xx{K_{alg}((t^{1\over n}))}
\def\k{K_{odd}}
\document
\heading
{1. Preliminaries}
\endheading

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 $\br F$ 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 $m\ge 1$, $n\ge 3$, $n\ge m$.
An algebra $A$ of such a type is obtained by means of
the "generic" splitting of another algebra of index and exponent $2^n$.
However,
if $\exp A=2$ and $[A]=\sum\limits_{i=1}^p [(a_i,b_i)]$ 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 $n\ge 3$ an indecomposable algebra
of index $2^n$ over some field $E/k_0$
which is a sum of $n+1$ quaternion algebras
in $\br E$. 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 $\br F$ 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 $u,v\in F^*$ 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 $f\in {k(C)}^*$, 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
$A_L=\res_{L/F}A=A\ot_F L$. We will write simply $A\ot L$ instead of
$A\ot_F L$, 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 $\sim$ means similarity of algebras,
i.e. their equality in the Brauer group of the field.
\bigskip

\heading
{2. Auxiliary and related results}
\endheading

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 $\ov a$, $\ov b_1,\d ,\ov b_n\in k^*_0/{k^*_0}^2$
are linearly independent over $\Bbb Z/2\Bbb Z$.

\proclaim{Lemma 1}
There exist a tower of fields $k_0\subset k_1\subset\d\subset k_n$
and elements $x_i, y_i\in k^*_i$ for every $1\le i\le n$
such that the following conditions
hold:
\roster
\item "1)" The elements $\ov a,\ov b_1,\d ,\ov b_n$ remain linearly independent
in $k^*_n/{k^*_n}^2$.

\item "2)" Put $B_j=\sum\limits_{i=1}^{j} (b_i,\xyi )\in \br k_j(\sa )$.
Then for any $1\le j\le n$ $\i B_j=2$.
% for any $1\le j\le n$.

\item "3)" Put $A_j=\sum\limits_{i=1}^{j} (b_i,\xxyyi )\in \br k_j$.
Then for any $1\le i\le j\le n$ and a finite field extension $l/k_0$ such that
$(A_j)_{lk_j}=0$,
either $\sqrt{b_i}\in l$ or $\sqrt{ab_i}\in l$.
In particular, $[l:k_0]\ge 2^j$.
\endroster
\endproclaim

\demo{Proof}
Let $x_1,y_1$ be indeterminates, $k_1=k_0(x_1,y_1)$.
Put $$A_1=(b_1,x^2_1-ay^2_1)\in \br k_1,\ \  B_1=(b_1, x_1+y_1\sa )
\in \br k_1(\sa ).$$
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 $l_1=l(\s a)(y_1)$
and consider the second residue map
$$\partial_{x_1-\s a y_1}: \br l_1(x_1)\longrightarrow {l_1^*}/{l_1^*}^2$$
with respect to the linear polynomial $x_1-\s a y_1$. Since
$\partial_{x_1-\s a y_1}((b_1, x_1^2 -ay_1^2))=b_1$, we get that $b_1$
is a square in $l_1$, hence in $l(\s a )$, which is just what we need.

Now let $n\ge 2$.
Assume that we have already constructed a tower of fields
$k_0\subset k_1\subset\d\subset k_{n-1}$ and the algebras
$A_1,\d ,A_{n-1},B_1,\d ,B_{n-1}$ with the required conditions.
Suppose $B_{n-1}=(\uv ,\uuvv )$, where $u_1,u_2,v_1,v_2\in k^*_{n-1}$.
Let $t_1,t_2,t_3$ be indeterminates, and
$k_n=k_{n-1}(t_1,t_2,t_3)$.
Let further
$x_n,y_n\in k_n$ be such elements that
$$\xyn =(\uv ) t^2_1 +(\uuvv ) t^2_2 -(\uv )(\uuvv ) t^2_3 ,$$ i.e.
$$x_n=u_1t^2_1+u_2t^2_2-(u_1u_2+av_1v_2)t^2_3 ,$$
$$y_n=v_1t^2_1+v_2t^2_2-(u_1v_2+u_2v_1)t^2_3 .$$
Put
$$A_n=A_{n-1}+(b_n,\xxyyn )\in\br k_n,$$
$$B_n=B_{n-1}+(b_n,\xyn )\in\br k_n(\sqrt a).$$
It is easy to see that $\xyn$ is a square in $(B_{n-1})_{k_{n-1}(\s a)}$,
which implies that
$$(B_n)_{k_n(\s a)(\sqrt{\xyn})}=0,$$ hence
$\i B_n\le 2$. Obviously, $(B_{n-1})_{k_n(\s a)}\not= (b_n,x_n +y_n\s a )$, so
$\ind B_n =2$.

Now suppose that a finite extension $l/{k_0}$ is such that
$$0={(A_n)}_{lk_n}=(A_{n-1}+(b_n,\xxyyn ))_{lk_n}.$$
Then
$$0={(A_n)}_{lk_n(\sa )}={(A_{n-1})}_{lk_n(\sa )}+
(b_n, (x_n-y_n\sa )(\xyn ))_{lk_n(\sa )}.$$
Put $K=lk_{n-1}(\sa ,t_1,t_2 )$. We can view the algebra
$${(A_{n-1})}_{lk_n(\sa )}=(b_n,(x_n-y_n\sa )(x_n +y_n\sa ))_{lk_n(\sa )}$$
as an algebra over $K(t_3)$.
It has no residues with respect to the second residue map
$$\partial =\oplus\partial_p :
\br K(t_3)\longrightarrow\coprod\limits_{p} {K(p)}^*/{{K(p)}^*}^2,$$
where $p$ runs over all monic irreducible polynomials from $K[t_3]$.
On the other hand, the monic polynomials
$${P_1(t_3)=t^2_3-{{(\uv ) t^2_1 +(\uuvv ) t^2_2}\over {(\uv )(\uuvv )}}}
={ -{(\xyn)}\over{(\uv)(\uuvv)}}$$ and
$$P_2(t_3)=t^2_3-{{(u_1-v_1\sa )t^2_1 +(u_2-v_2\sa )t^2_2}\over
{(u_1-v_1\sa )(u_2-v_2\sa )}}=
{{-(x_n-y_n\sa )}\over{(u_1-v_1\sa )(u_2-v_2\sa )}}$$
are distinct.
Hence
$$0=\partial_{P_1}({(A_{n-1})}_{lk_n(\sa )})=b_n,$$
and so,
$$\s{b_n}\in lk_n(\sa ,t_1,t_2,\s{{(\uv ) t^2_1 + (\uuvv )t^2_2}\over
{(\uv )(\uuvv )}}).$$
This obviously implies $\s{b_n}\in lk_n(\sa )$.
Therefore, either $\s{b_n}\in lk_n$, or
$\s{ab_n}\in lk_n$. Since the extension $k_n/k_0$ is purely transcendental,
we get that either $\s{b_n}\in l$, or $\s{ab_n}\in l$.
In both cases it follows that $(b_n,\xxyyn )_{lk_n}=0$.
Since
$$0={(A_n)}_{lk_n(\sa )}={(A_{n-1})}_{lk_n(\sa )}+(b_n,\xxyyn )_{lk_n(\sa )},$$
we get $${(A_{n-1})}_{lk_n(\sa )}=0.$$
Since $k_n=k_{n-1}(t_1,t_2,t_3)$,
it follows that
$${(A_{n-1})}_{lk_{n-1}(\sa )}=0,$$
and by the induction assumption we
conclude that for any $1\le i\le {n-1}$ either $\s{b_i}\in l$, or
$\s{ab_i}\in l$. The induction step is done, so the lemma is proved.

Since $\i B_n=2$, it follows that $\i A_n=\i N_{k_n(\sa )/k_n}(B_n)=2$ or $4$.
If $\i A_n=2$, put $k=k_n$. If $\i A_n=4$ put $k=k_n(\phi )$, where $\phi$
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 $1\le i\le n$ either
$\s{b_i}\in l$ or $\s{ab_i}\in l$.

Now put $A={(A_n)}_k$, $B={(B_n)}_{k(\sa )}$.
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])
$$\br F@>{\op{res}}>>\br F(\sa )@>{\op{N}}>>\br F.$$
Since
$$0=A_F=N_{F(\sa )/F}(B_{F(\sa )}),$$
we have $B_{F(\sa )}=
D_{F(\sa )}$ for some $D\in\br F$. Since $\i B=2$, we get $\i D\le 4$.
It is well known that
if $\i D=4$, then
$D=D_1+D_2$ for some $D_1,D_2\in\br F$, such that $\i D_1=\i D_2=2$ and
${D_2}_{F(\sa )}=0$. Hence we may change $D$ for $D_1$ and so assume that
$\i D\le 2$.

From now on we will assume that $n\ge 2$.
Put $L=F(\sa ,\s{b_1},\d ,\s{b_n})$.
Let $F\subset F_1\subset F_2\subset L$ be a tower of
fields such that
$$\sa\notin F_2,\ \ [F_2:F_1]=4,\ \ [L:F_1]=8.$$
Obviously, $L=F_2(\sa )=F_1(\sa )F_2$, 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.

\proclaim{Proposition 2}
\roster

\item "1)" The elements $\ov a,\ov{b_1},\d ,\ov{b_n}$ remain linearly independent
in $F^*/{F^*}^2$.

\item "2)" $\i D=2$.

\item "3)" $D_L=0$.

\item "4)" The algebra $D_{F_1}$ does not decompose into a sum
$D_{F_1}=D_1+D_2$, where
$D_1\in\br (F_1(\sa )/F_1)$ and $D_2\in\br (F_2/F_1)$.
\endroster
\endproclaim

\demo{Proof}
$1)$ Obvious.

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

$3)$ Obvious, since $D_{F(\sa )}=\sum\limits_{i=1}^n (b_i, x_i+y_i\sa )$.

$4)$ Assume the contrary, i.e. that $D=D_1+D_2$, where
$D_1\in\br (F_1(\sa )/F_1)$ and $D_2\in\br (F_2/F_1)$. Then
$$B_{F_1(\sa )}=D_{F_1(\sa )}={D_2}_{F_1(\sa )}.$$
Denote the field $kl_1$ by $\h$, so that $F_1=\h(C)$.
Consider the points $z_1,\d ,z_m\in C_{\h}$ at which the algebra
$D_2$ has nonzero residues under the second residue map
$$\br F_1@> >>\su .$$
Since $B$ is defined over $k(\sqrt a )$, the algebra
${D_2}_{F_1(\sa )}=B_{F_1(\sa )}$
has no residues at all. This implies that
the residues of $D_2$ at the points $z_1,\d ,z_m$ are equal to $a$.
Recall that $F_2=kl_2(C)$, and let $c_1,\dots ,c_n\in k_0^*$
be elements such that
$l_2=k_0(\s {c_1},\d ,\s {c_n})$.
Therefore, since ${D_2}_{F_2}=0$,
we have for any $j$
$$a_{\h(z_j)(\s{c_1},\d ,\s{c_n})}\in {\h(z_j)(\s{c_1},\d ,\s{c_n})^*}^2.$$
Denote by $c_I$ the product $\prod\limits_{i\in I}c_i$ (the similar
notation will be used also in the sequel).
Thus, given any $j$, we have
$ac_I\in {{\h(z_j)}^*}^2$
for some $I\subset\{1,\d ,n\}$ depending on $j$, i.e.
$$\h(\s{ac_I})\subset \h(z_j).$$
\enddemo

\proclaim{Lemma 3}
For any $j$ we have $4\vert\deg\ {z_j}$.
\endproclaim
\demo{Proof}
Assume this is not the case. Consider the tower
$\h\subset \h(\s c )\subset \h(z_j)$, where
$c=ac_I$ and $I$ corresponds to the point $z_j$.
Notice that $\s c\notin\h =kl_1$, for otherwise $\s c\in kl_2$, and so
$\sa\in F_2$, a contradiction.
Hence $[\h(z_j):\h(\s c )]$ is odd.
Since $A_{\h(z_j)}=0$ we conclude that $A_{\h(\s c )}=0$.
Therefore, ${(A_n)}_{k_nl_1(\sqrt c)}=0$,
a contradiction to Lemma $1$, since $[l_1(\s c ):k_0]= 2^{n-1}$.
The lemma is proved.
\enddemo




\def\sa{(\sqrt a)}
\def\sb{(\sqrt b_1,\dots , \sqrt b_n)}
\def\sab{(\sqrt a, \sqrt b_1,\dots ,\sqrt b_n)}
\def\o{\overline}
\def\br#1{\,_2\!\operatorname{Br} #1}
\def\bra{\,_2\! \operatorname{Br} F(\sqrt a )}
\def\l{\longrightarrow}
\def\w{\widetilde}
\def\chr{\operatorname {char}}
\def\cd{\operatorname {cd}}
\def\f{((t_0))((t_1))\d ((t_n))}
\def\at{(a,t_0)\otimes(b_1,t_1)\otimes\d\otimes (b_n,t_n)}
\def\ot{\otimes}
\def\b{\operatorname{Br}}

Now we use the fact that any divisor of degree zero on a projective conic
is principal. Choose any $y\in C_{\h}$ such that
$\operatorname{deg} y =2$. Let $s=\sum\limits_{i=1}^m \operatorname{deg} z_j$.
Consider the divisor $\eufm a = -{s\over 2}y +\sum\limits_{i=1}^m z_i$.
Obviously, $\operatorname{deg} \eufm a =0$, hence there is $f\in F^*_1$
such that
$\eufm a =(f)$. The algebra $(a,f)$ has nonzero residues just at
the points $z_1,\dots ,z_m$, since by Lemma $3$ the number $s\over 2$
is even. Moreover, all these residues are equal to $a$. Therefore,
the algebra $\widehat{D_2}=D_2 + (a,f)$ has no residues at all, i.e.
$\widehat {D_2}\in \br(C_{\h})$.
Merkurjev's theorem ([M]) claims that $K_2\h(C)/2 K_2\h(C)\simeq\br\h(C)$.
Hence by ([Su], Lemma $5$) we get that either
$$\widehat {D_2}\in \operatorname{res}_{F_1/\h}(\br\h),$$ or
$$\br(C_{\h})/\operatorname{res}_{F_1/\h}(\br\h)=\Bbb Z/ 2\Bbb Z,$$
and the element $\widehat {D_2}$
is nontrivial in this factor group.

In the first case let $\widehat {D_2} =\operatorname{res}_{F_1/\h}\w D$
for some $\w D\in\br\h$.
Then $${(\w D +B)}_{\h\sa (C)}={(\widehat {D_2} +B)}_{\h\sa (C)}=
{(D_2 +B)}_{\h\sa (C)}={(D+B)}_{\h\sa (C)}=0.$$
Hence ${(\w D +B)}_{\h\sa}$ is either zero or $A_{\h\sa}$. But then
$$0=N_{\h\sa /\h}(\w D+B)=N_{\h\sa /\h}B=A_{\h},$$
hence ${(A_n)}_{l_1k_n}=0$, a contradiction to Lemma $1$, since
$[l_1:k_0]=2^{n-2}$.

In the second case, when
$\widehat{D_2}\notin \operatorname{res}_{F_1/\h}{(\br\h})$,
let $A=(a_1, a_2)$, where $a_1,a_2\in k^*$.
Then $F_1$ is the quotient field of the ring
$\h[u_1, u_2]/(a_1 u_1^2 + a_2 u_2^2 -1)$, where $u_1, u_2$ are
indeterminates. It is easy to see that
$N_{F_1/\h(u_1 )}(\widehat D_2 )=A$. Consider the commutative diagram
$$\CD \br F_1@>\operatorname{res}>>                      \br F_1(\s a)\cr
       @VNVV                      @VNVV\cr
    \br \h(u_1)@>\operatorname{res} >> \br \h\sa (u_1)
\endCD
$$
So we have
$$\eqalign{\operatorname{res}_{\h\sa (u_1)/\h(u_1)}A=
\operatorname{res}_{\h\sa (u_1)/\h(u_1)}\circ\
N_{F_1/\h(u_1)}(\widehat {D_2})=\cr N_{F_1\sa /\h\sa (u_1)}
\circ\ \operatorname{res}_{F_1\sa /F_1}
(\widehat{D_2})=N_{F_1\sa /\h\sa (u_1)}(B)=0 ,}$$
hence $\res_{\h\sa /k}(A)=0$,
a contradiction to Lemma $1$, since
$[l_1\sa :k_0]=2^{n-1}$.
Thus Proposition $2$ is proved.
\enddemo

Recall that $L=F(\s a,\s{b_1},\d ,\s{b_n})$. 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].

\proclaim{Corollary 4}
Let $F\subset L_1\subset L$, $F\subset L_2\subset L$ be
proper field subextensions of $L/F$.
Then
\roster
\item "1)" $D\notin \br (L_1/F)+\br (L_2/F)$. In particular,
$$\br (L/F)\not=\br (L_1/F)+\br (L_2/F).$$
\item "2)" $f\notin W(L_1/F)+W(L_2/F)$, where $f\in W(L/F)$ is
the $2$-fold Pfister form corresponding to $D$.
\endroster
\endproclaim

\demo{Proof}
$1)$\ Obviously we may assume that $[L:L_1]=[L:L_2]=2$ and $L=L_1L_2$.
Then $[L:L_1\cap L_2]=4$.
Choose a field $F_1$ such that
$$F\subset F_1\subset L_1\cap L_2,\ \
\s a\notin F^*_1 ,\ \ [L_1\cap L_2 :F_1]=2.$$ Then
$$[L:F_1]=8,\
[L_1:F_1]=[L_2:F_1]=4,\ L_1=F_1(\s{d_1},\s{d_2}),\
L_2=F_1(\s{d_1},\s{d_3})$$
for some $d_1,d_2,d_3$ belonging to the multiplicative group generated by
the elements $a,b_1,\dots ,b_n$.
We will show that $D_{F_1}\notin \br (L_1/F_1)+\br (L_2/F_1)$.
Assume the contrary.
Then $$D_{F_1}=(d_1,e_1)+(d_2,e_2)+(d_3,e_3)$$ for some $e_1,e_2,e_3\in F^*_1$.
Obviously, we may suppose that $a$ is equal
either to $d_i$, or to $d_id_j$ $(i<j)$, or to $d_1d_2d_3$.
We have
$$D_{F_1}=(d_1d_2,e_1)+(d_2,e_1e_2)+(d_3,e_3)=
(d_1d_2d_3, e_1)+(d_2, e_1e_2)+(d_3,e_1e_3).$$
Put $F_2=F_1(\s{d_2},\s{d_3})$.
We conclude that
$$D_{F_1}\in\br (F_1(\s a)/F_1)+\br (F_2/F_1),$$
which contradicts Proposition $2$.

$2)$ This part easily follows from part $1)$. Indeed,
assume that
$[f]=[f_1]+[f_2]$, where $f_1\in W(L_1/F)$, $f_2\in W(L_2/F)$.
Obviously, $\disc (f_1)=\disc (f_2)$.
Let $g_1,g_2\in I^2(F)$ be forms such that
$\dim (f_i\perp -g_i)_{an}\le 2$ for $i=1,2$. It is clear that
$g_i\in W(L_i/F)$. If $D_i\in\br F$ correspond to $g_i$
under the map $I^2(F)/I^3(F)\simeq\br F$, then
$D=D_1+(D_2 +(\disc f_2,u))$ for some $u\in F^*$,
a contradiction to part $1)$ since $\ll\disc (f_2)\rr\in W(L_2/F)$.
\enddemo

\proclaim{Lemma 5}
Let $K$ be a field, $K((t))_{odd}$ any odd degree extension of $K((t))$.
Then the following holds:
\roster
\item "1)"
There exists a field inclusion
${\k}((t))\hookrightarrow K((t))_{odd}$ over $K((t))$ for
some maximal odd degree extension $\k$ of $K$.
\item "2)"
${{K((t))}_{odd}}^*/{{{K((t))}_{odd}{}^*}^2}\simeq
{{K_{odd}}^*}/{{{K_{odd}}^*{}}^2}\oplus{\Bbb Z/2\Bbb Z}$, where the element
$t\in{{K((t))}_{odd}}^*/{{{K((t))}_{odd}{}^*}^2}$
corresponds to the nontrivial element of
$\Bbb Z/2\Bbb Z$ and
${{K_{odd}}^*}/{{{K_{odd}}^*{}}^2}$ naturally includes into
${{K((t))}_{odd}}^*/{{{K((t))}_{odd}{}^*}^2}$
under this isomorphism.
\endroster
\endproclaim
\demo{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 $L=\overline L ((\pi ))$, where $\overline L$ is
a subfield of $L$ isomorphic to the
residue field and $\pi$ is a uniformizer
([Se], Ch.2, Th.2). It is easy to check that we can choose for $\k$
the union of all finite extensions
of $K$ contained in ${K((t))_{odd}}$.

$2)$ Assume $L/{\k}((t))$ is a finite odd degree extension. Then
$$L\simeq \k ((\pi )),\ \ \
{L^*}/{L^*}^2\simeq {\k}^*/{{\k}^*}^2\oplus\z ,$$ where
$\pi\in{L^*}/{L^*}^2$
corresponds to the nontrivial element of
$\Bbb Z/2\Bbb Z$ and ${\k}^*/{{\k}^*}^2$ naturally includes into
$L^*/{L^*}^2$.
Denote by $v$ the discrete valuation on $L$. Obviously, $v(t)$ is odd, hence
$\pi t\in {\k}^*/{{\k}^*}^2\hookrightarrow L^*/{L^*}^2$.
Therefore, there exists another isomorphism
$${L^*}/{L^*}^2\simeq {\k}^*/{{\k}^*}^2\oplus\z ,$$ where
$t\in{L^*}/{L^*}^2$
corresponds to the nontrivial element of
$\Bbb Z/2\Bbb Z$ and ${\k}^*/{{\k}^*}^2$ naturally includes into
$L^*/{L^*}^2$. Also it is clear that the last isomorphisms are compatible
for various $L$, which completes the proof.
\enddemo

\proclaim{Corollary 6}
Let $K$ be a field,
$E=K((t_0))\d ((t_n))$ the iterated Laurent series
field. Let $E_{odd}$ be any maximal odd degree extension of $E$.
Then
$$\k\hookrightarrow E_{odd},\ \ \
{E_{odd}}^*/{{E_{odd}}^*}^2\simeq {K_{odd}^*}/{K_{odd}^*}^2
\oplus{(\Bbb Z/2\Bbb Z)}^{n+1}$$
for some maximal odd degree extension $\k$ of $K$, and under this isomorphism
the element $t_i$ $(0\le i\le n)$ corresponds to the nontrivial
element of the $i+1$-th factor of ${(\Bbb Z/2\Bbb Z)}^{n+1}$ while
${K_{odd}^*}/{K_{odd}^*}^2$ naturally includes into
${E_{odd}^*}/{E_{odd}^*}^2$.
\endproclaim
\demo{Proof}
Taking into account Lemma $5$, the proof is immediate by induction on $n$.
\enddemo

\proclaim{Lemma 7}
Let $K$ be a field, let $\o{c_0} ,\d ,\o{c_n}\in K^*/{K^*}^2$
be linearly
independent elements. Suppose the algebra $\a\in\br K$ is such that
$\a_{K(\s{c_0},\s{c_1},\d ,\s{c_n})}\not=0$.
Let further $t_0,\d ,t_n,x$ be indeterminates.
Then for any $0\le i\le {n+1}$
$$\i [{\a + (c_0,t_0)+\d + (c_n,t_n)}]_E\ge 2^{n+2},$$
where
$$E= K((t_0))\d ((t_{i-1}))((x))((t_{i}))\d ((t_n)).$$
\endproclaim

\demo{Proof}
The argument is by induction on $n$.
Suppose first that $i\le n$.
Put $$E_1=K((t_0))\d ((t_{i-1}))((x))((t_i))\d ((t_{n-1})).$$
By induction hypothesis
$\i[{\a + (c_0,t_0)+\d + (c_{n-1},t_{n-1})}_{E_1(\s{c_n})}]\ge 2^{n+1}$.
Applying Tignol's theorem ([T2], Prop. 2.4) we finishes the proof.
If $i=n+1$, the argument is similar with a slight modification.
\enddemo

\bigskip
\heading
{3. Construction of indecomposable algebras}
\endheading

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, $n\ge 2$, $a,b_1,\d ,b_n\in F^*$,
$D\in\br F$. We call the triple $(F,D,\{a,b_1,\d ,b_n\})$ admissible
if the following conditions hold:
\roster
\item "a)" The elements $\ov {a},\ov{b_1},\d ,\ov{b_n}\in F^*/{F^*}^2$
are linearly independent.
\item "b)" $\i D=2$.
\item "c)" $D_{F\sab }=0$.
\item "d)" For any tower $F\subset F_1\subset F_2\subset L=F{\sab }$
such that $$\s a\notin F^*_2,\ \ [L:F_1]=8,\ \ [F_2:F_1]=4$$ we have
$$D_{F_1}\notin\br (F_1(\s a)/F_1)+\br (F_2/F_1).$$
\item "e)" The field $F$ has no proper extensions of odd degree.
\endroster

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

\proclaim{Proposition 8}
\roster
\item "1)" If the triple $(F,D,\{a,b_1,\d ,b_n\})$
satisfies condition $a)-d)$ above, then the triple
$(F_{odd}, D_{odd},\{a,b_1,\d ,b_n\})$ is admissible.
\item "2)" If the triple
$(F,D,\{a,b_1,\d ,b_n\})$ is admissible,
$n\ge 3$, $c\in F^*$,
and $$F\sab =F(\s c,\s a,\s{b_1} ,\d ,\s{b_{n-1}}),$$ then the triple
$(F(\s c),D_{F(\s c)},\{a,b_1,\d ,b_{n-1}\})$ is admissible as well.
\item "3)" If the triple $(F,D,\{a,b_1,\d ,b_n\})$ is
admissible, and $L=F\sab$, then Corollary $4$ holds for
the extension $L/F$ and the algebra $D$.
\endroster
\endproclaim
\demo{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.
\enddemo
\smallskip

Now let $(F,D,\{a,b_1,\d ,b_n\})$ be an admissible triple.
Put $E=F((t_0))\d ((t_n))$. Suppose
$C$ is the division algebra over $E$ such that
$$C\sim D\ot (a,t_0)\ot (b_1,t_1)\ot\d\ot (b_n,t_n).$$
The main purpose of the present article is the following

\proclaim{Proposition 9}
\roster
\item "1)" $\i C=2^{n+1}$.
\item "2)" The algebra $C_{odd}\buildrel def\over =C_{E_{odd}}$
does not decompose into a tensor product of
two nontrivial algebras over any maximal odd degree extension $E_{odd}$
of $E$.
\endroster
\endproclaim
\demo{Proof}
$1)$\ Since $C_L=0$ we have
$\i C\le 2^{n+1}$. On the other hand, applying Lemma $7$
to the field $F_0=F((t_0))$ and the algebra $C_0=D\ot (a,t_0)$ we get
$$\i C=\i [(D+(a,t_0))+(b_1,t_1)+\d +(b_n,t_n)]\ge 2^{n+1}.$$

$2)$\ The proof is by induction on $n$.
Assume that the algebra $C_{odd}$ is decomposable,
$C_{odd}\simeq D_1\otimes_{E_{odd}} D_2$
and $\i D_1,\i D_2\ge 2$. Consider two cases:

$a)$\  $\i D_1=2$ or $\i D_2=2$.

Suppose for instance $\i D_1=2$. By Lemma $5$ $D_1\simeq (pt_I, st_J)$, where
$$ I,J\subset\{0,1,\d ,n\},
\ \ p,s\in F^*.$$
Case $a)$ divides in turn into two subcases.

$a1)$\ $I\not=\emptyset$ or $J\not=\emptyset$.
Assume $I\not=\emptyset$.

Suppose $I=\{k+1, k+2,\d ,n\}$ for some $k$ (the general case can be treated
similarly, using Lemma $7$).
Then we have
$$\eqalign{{C}_{E_{odd}(\s{pt_I})}=D+ (a,t_0)+
(b_1,t_1)\d + (b_{n-1},t_{n-1})+
(b_n,pt_{k+1}\d t_{n-1})=\cr
 D + (b_n,p)+ (a,t_0)+ (b_1,t_1)+\d + (b_k,t_k)
+(b_{k+1}b_n,t_{k+1})+\d
+ (b_{n-1}b_n, t_{n-1}).\cr}$$
The field $E_{odd}(\s{pt_I})$ can be considered as an iterated Laurent series
field, namely $E_{odd}(\s{pt_I})=F((t_0))\d ((t_{n-1}))((x))$, where
$x=\s{pt_I}$.
Put
$$L_1=F(\s a,\s{b_1},\d ,\s{b_k},\s{b_{k+1}b_n},\d ,
\s{b_{n-1}b_n}),\ \ L_2=F(\s{b_n}).$$
Since $\i{C}_{E_{odd}(\s{pt_I})}={1\over 2}\i C = 2^n$, it follows by Lemma $7$
that $$(D+ (b_n,p))_{L_1}=0.$$
Therefore,
$$D=(D+(b_n,p)) +(b_n,p)\in \br (L_1/F)+\br (L_2/F),$$
which contradicts Corollary $4$.

$a2)$\ $I=J=\emptyset$, i.e.
$D_1=(p,s)$, where $p,s\in F^*$.

Then $$\i[D+(p,s)+ (a,t_0)+ (b_1,t_1)+\d + (b_n,t_n)]=\i D_2=2^n,$$
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 $n=2$.

$b)$\
$\i D_1\ge 4$ and $\i D_2\ge 4$.

In particular, $n\ge 3$.
Obviously, either
$\i{D_1}_{E_{odd}(\s a )}=\i D_1$, or $\i{D_2}_{E_{odd}(\s a)}=\i D_2$.
Assume that $\i{D_1}_{E_{odd}(\s a)}=\i{D_1}$.
Since any finite extension of $E_{odd}$ is a tower of quadratic extensions,
there exists $c\in {E_{odd}}^*$ such that
$\i{D_1}_{E_{odd}(\s c)}={1\over 2}\i{D_1}$. Moreover, since
$F$ has no proper odd degree extensions,
we may assume by Lemma $5$ that $c=pt_I$ for some $p\in F^*$ and
$I\subset \{0,1,\d ,n\}$. Let
${D_1}_{E_{odd}(\s{pt_I})}\sim D'_1$,
where $D'_1$ is some nontrivial division algebra over
$E_{odd}(\s{pt_I})$.
If $I\not=\emptyset$ we come to a contradiction just as in case $a)$.
So suppose that $I=\emptyset$.
Since $\i{D_1}_{E_{odd}(\s a)}=\i{D_1}$, we have
$pa\notin {F^*}^2$, and
$$\i{C}_{E_{odd}(\s p)}=
\i [D+(a,t_0)+(b_1,t_1)+\d +(b_n,t_n)] _{E_{odd}(\s p)}=2^n.$$
It follows from Lemma $6$ that the elements
$\o a,\o{b_1},\d ,\o{b_n}$ are linearly dependent in
${F(\s p)}^*/{{F(\s p)}^*}^2$. This in turn means
that we may assume that either
$p=b_I$, or $p=ab_I$ for some
$\emptyset\not=I\subset\{1,\d ,n\}$. Let for instance
$p=b_I$, where $I=\{ k+1,\d ,n\}$
(the general case can be treated with a slight
modification, and is left to the reader).
Put
$$K=F(\s p)((t_0))\dots ((t_{n-1})),\ \
L=F(\s p)((t_0))\dots ((t_{n-1}))((\s{t_n})).$$
 Then
$${D\ot (a,t_0)\ot (b_1,t_1)\ot\d\ot (b_{n-1},t_{n-1})}_{L_{odd}}
\sim C_{L_{odd}}\sim D'_1\ot _{L_{odd}}D_2$$
for any maximal odd degree extension $L_{odd}$ of $L$,
containing $E_{odd}(\s p)$.
On the other hand, by Proposition $8$
the triple $(F(\s p),D_{F(\s p)},\{a,b_1,\d ,b_{n-1}\})$ is admissible.
Hence by the induction hypothesis for any maximal odd degree extension
$\k$ of $K$ we have
$${D\ot (a,t_0)\ot (b_1,t_1)\ot\d\ot (b_{n-1},t_{n-1})}_{K_{odd}}
\sim {C_1}_{K_{odd}},$$
$C_1$ being a division algebra
over $K$ such that it is indecomposable over $K_{odd}$ and $\i{C_1}=2^n$.
By the first part of Lemma $5$
we can choose $\k$ in such a way that $\k ((\s{t_n}))\hookrightarrow L_{odd}$
over $K((\s{t_n}))$. This implies
$${C_1}_{L_{odd}}\sim D'_1\ot _{L_{odd}}D_2 .$$
Since, obviously, $\i (D'_1\ot _{L_{odd}} D_2)=\i {C_1}_{L_{odd}}=2^n$, we have
$${C_1}_{L_{odd}}\simeq D'_1\ot _{L_{odd}} D_2.$$
By ([Se], Ch.2, Th.2) the field
$L_{odd}$ is the direct limit of fields $\k ((u))$,
where $u$ is an indeterminate. Hence
for some $u\in L_{odd}$
the algebras $D'_1$ and $D_2$ are defined over $\k ((u))$ and
$${C_1}_{K_{odd}((u))}\simeq D'_1\ot _{K_{odd}((u))} D_2.$$
It is well known that for any field $k$ and positive integer $m$ not divided by
$\chr k$
$$H^2(k((x)),\mu_m)\simeq H^2(k,\mu_m)\oplus H^1(k, \Bbb Z/m\Bbb Z),$$
where $\mu_m$ is the group of $m$th roots of unity.
From this and the cohomological interpretation of the Brauer group
it follows that
$$D'_1=\res_{K_{odd}(({u^{1\over s}}))/K_{odd}}{{\widetilde D}_1},\ \ \ \
D_2=\res_{K_{odd}(({u^{1\over s}}))/K_{odd}}{{\widetilde D}_2}$$
for a sufficiently large $2$-power $s$
and some central division algebras ${\widetilde D}_1$
and ${\widetilde D}_2$ over $K_{odd}$.
Since the natural map $\b\k\rightarrow\b\k{((u^{1\over s}))}$ is injective,
$${C_1}_{K_{odd}}=\widetilde{D_1}+\widetilde{D_2},$$
and by dimension count we conclude that
$${C_1}_{K_{odd}}\simeq\widetilde{D_1}\ot _{K_{odd}}\widetilde{D_2},$$
a contradiction to indecomposability of the algebra $C_1$ over $K_{odd}$.
\enddemo

Summing up the obtained results and changing the notation a bit we get
the following
\proclaim{Corollary 10}
Let $k$ be a field, $n\ge 3$.
Suppose the elements $a_1,\dots ,a_n\in k^*/{k^*}^2$
are linearly independent over $\Bbb Z/2\Bbb Z$. Then
there exists a field extension
$F/k$,
a quaternion algebra $D\in\br F$ and
a division algebra $C\in\br F((t_1))\dots ((t_n))$
of index $2^n$ such that
\roster
\item "1)" $C$ is indecomposable over any odd degree
extension of $E=F((t_1))\dots ((t_n))$.
\item "2)" $M_2(C)\simeq (a_1,t_1)\ot _E\dots\ot _E (a_n,t_n)\ot _E D$.
\item "3)" $D\in\br (F(\s{a_1},\dots ,\s{a_n})/F)$

\endroster
\endproclaim

\demo{Proof}
This immediately follows from Propositions $2$, $8$ and $9$.
\enddemo

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\frenchspacing
\widestnumber\key{ELTW}
\refstyle{D}
\Refs\nofrills{REFERENCES}

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\ref
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\by Merkurjev A.S.
\paper On the norm residue symbol of degree two
\jour Sov. Math. Dokl.
\yr  1981
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\ref
\key S
\by Sivatski A.S.
\paper Nonexcellence of multiquadratic field extensions
\jour Journal of Algebra
\yr 2004
\vol 275
\issue 2
\pages 859-866
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\ref
\key Sch
\by Scharlau W.
\paper Quadratic and Hermitian forms
\jour Springer, Berlin Heidelberg New York
\yr 1985
\endref

\ref
\key Se
\by Serre J.-P.
\paper Corps Locaux
\jour Hermann, Paris
\yr 1962
\endref

\ref
\key Su
\by Suslin A.
\paper The quaternion homomorphism for the function field on a conic
\jour Soviet Math. Docl.
\yr 1982
\vol 26
\issue 1
\pages 72-77
\endref

\ref
\key T1
\by Tignol J.-P.
\paper Corps {\`a} involution neutralis{\'e}s par une extension
ab{\'e}lienne {\'e}l{\'e}mentaire
\jour Springer Lecture Notes in Math.
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\ref
\key T2
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\yr 1987
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\pages 205-228
\endref

\endRefs
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Britta Heidrich                                          Tel +49-521-106-5022
Uni Bielefeld / Mathematik                               FAX +49-521-106-4743
Universitaetsstr. 25
33615 Bielefeld                   e-Mail heidrich@mathematik.uni-bielefeld.de
