%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\magnification 1200
\documentstyle{amsppt}
\expandafter\redefine\expandafter\enddemo\expandafter{\expandafter\hfill
\expandafter$\expandafter\square\expandafter$\enddemo}

\topmatter
\title
ON SOME ELEMENTS OF THE BRAUER GROUP OF A CONIC
\endtitle

\author
A.S. Sivatski
\endauthor

\affil
Department of Mathematics-2 \\
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
Quadratic form, Brauer group, Quaternion algebra, Conic
\endkeywords

\def\a{\alpha}
\def\phi{\varphi}
\def\io{I_0}
\def\vf{v(f)}
\def\Vf{V(f)}
\def\Vfi{V(f_i)}
\def\gio{g_{I_0}}
\def\aio{a_{I_0}}
\def\si{\sum\limits_{i=1}^n}
\def\o{\overline}
\def\capi{\bigcap\limits_{i\in I}}
\def\capni{\bigcap\limits_{i\notin I}}
\def\cupi{\bigcup\limits_{i\in I}}
\def\cupni{\bigcup\limits_{i\notin I}}
\def\prni{\prod\limits_{i\notin I}}
\def\pri{\prod\limits_{i\in I}}
\def\fg{{f_i}{(\pri\gi )}^{-1}}
\def\v0{v_0}
\def\sai{\sqrt{a_I}}
\def\p{\{1,\dots ,p\}}
\def\d{\dots}
\def\z{\Bbb Z/2\Bbb Z}
\def\pa{\partial}
\def\br#1{\,_2\!\operatorname{Br} #1}
\def\Br{\operatorname{Br}}
\def\ind{\operatorname{ind}}
\def\res{\operatorname{res}}
\def\deg{\operatorname{deg}}
\def\div{\operatorname{div}}
\def\mod{\operatorname{mod}}
\def\ker{\operatorname{ker}}
\def\exp{\operatorname{exp}}
\def\disc{\operatorname{disc}}
\def\dim{\operatorname{dim}}
\def\cd{\operatorname{cd}}
\def\ai{a_I}
\def\s{\sqrt}
\def\ui{u_I}
\def\gi{g_I}
\def\vi{v_I}
\def\wk{\widetilde k}
\def\e{\emptyset}
\def\n{\{1,\dots ,n\}}
\def\bac{b_1+c_1\s {a_3}}
\def\bacc{b_2+c_2\s {a_3}}
\def\bacs{b_1^2 -a_3c_1^2}
\def\baccs{b_2^2 -a_3c_2^2}
\def\sab{(\sqrt a, \sqrt b_1,\dots ,\sqrt b_n)}
\def\ov{\overline}
\def\dd{D^{{\otimes}2^{n-1}}}

\abstract
Let $K$ be a field of characteristic not $2$,
$X$ a nonsingular projective conic over $K$, $n$ is a positive
integer and $a_i\in K^*$ for $1\le i\le n$.
We investigate elements of $_2 Br (X)$, which are sums of
quaternion algebras $(a_i,f_i)$
for some $f_i\in K(X)^*$. An application to a construction of indecomposable
division algebras of exponent $2$ is given.
\endabstract

%\subjclass
%\endsubjclass

\endtopmatter


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 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 $u$-invariant (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
$$\br (X)=\ker (\br K(X)\buildrel\pa\over\to\coprod
\limits_{x\in X}{K(x)}^*/{{K(x)}^*}^2),$$
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,
$$\pa_x (f,g)={(-1)}^{v_x(f)v_x(g)}{{f^{v_x(g)}}\over{g^{v_x(f)}}}
\in k(x)^*/{k(x)^*}^2,$$
where $v_x$ is the discrete valuation related to the point $x$,
and the symbol $(f,g)$ denotes a quaternion algebra.
Notice also that $$\res_{K(X)/K}\br K\subset\br (X).$$
From now on we fix a field $K$ of characteristic not $2$,
and elements $a_1,\d a_n\in K^*$
such that
$\o{a_1},\d \o{a_n}\in K^*/{K^*}^2$ are linearly independent over $\z$.
For any set
$I\subset\n$, put $a_I=\prod\limits_{i\in I}a_i$ (if $I=\emptyset$, then
$a_I=1$). For any polynomial $f\in K[t]$ denote by $l(f)$
the leading coefficient of $f$.
\proclaim{Proposition 1}
Let $X$ be a nonsingular projective conic over $K$,
$Q$ the corresponding quaternion algebra, $\a\in\br (X)$.
Suppose that the following conditions hold:

$1)$ $\a =\si(a_i,f_i)$ for some $f_i\in {K(X)}^*$.

$2)$ There exist at most two sets
$I\subset\n$ such that $Q_{K(\s{\ai})}=0$.

Then there exist $b_1,\d b_n\in K^*$ such that
$\a =\res_{K(X)/K}\si (a_i,b_i)$.
\endproclaim
\demo{Proof}
If $Q=0$, then $K(X)=K(t)$ is the rational function field.
The exact sequence
$$0\to\br K\buildrel\res\over\to\br K(t)\buildrel\pa\over\to\coprod
\limits_{x\in{\Bbb P}_K^1}{K(x)}^*/{{K(x)}^*}^2$$
shows that $\br (X)=\res_{K(t)/K}\br K$. There is a splitting homomorphism
$$c:\br K(t)\to\br K$$ such that
$c((f,g))=(l(f), l(g))$ for any $f,g\in K[t]$.
Applying the map $c$ to the equality
$$\alpha =\si (a_i, f_i),$$
we get what we need.

So assume
that $Q=(a,b)\not=0$. 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 $K(v_0)=K(\s a)$. In particular, $\deg (v_0)=2$. Assume first that
$K(v_0)\not=K(\sai )$ for each $I\subset\n$, i.e. $aa_I\notin {K^*}^2$.

For any $f\in K(X)$ denote by $\Vf$ the finite set of all the points
$v\not=v_0$ of $X$ such that $\vf$ is odd (we identify a point of $X$
with the corresponding
discrete $K$-valuation). Let $V=\bigcup\limits_{i=1}^n V(f_i)$.
For any $I\subset\n$ put $V_I=\capi V(f_i)\setminus\cupni V(f_i)$.
Clearly, $V=\bigcup\limits_{I}V_I$, and, moreover, this union is disjoint and
$V(f_i)=\bigcup\limits_{i\in I}V_I$.
For each $I\subset\n$ consider the divisor
$$\eufm a_I=\sum\limits_{v\in V_I}v -{1\over 2}
(\sum\limits_{v\in V_I}\deg v )v_0.$$
Obviously, $\deg\eufm a_I =0$.
Since any divisor on $X$ of degree $0$ is principal, we get that
$\eufm a_I=\div (\gi )=\sum\limits_{v\in X}v(g_I)v$ for some
$\gi\in K(X)^*$.
Notice that
$$v(\gi )=\cases 1 ,&\text{if}\ v\in V_I\\ 0,&\text{if}\
v\notin V_I ,v\not=v_0\endcases,$$
and, therefore,
$v(\fg )$ is even for any $i$ and $v\not=v_0$. Moreover, since
$\deg (\div (\fg ))=0$, $\deg v_0=2$, and $2\vert\deg v$
for any $v\in X$, we conclude that $v_0(\fg )$ is also even, i.e.
$v(\fg )$ is even for any $i$ and $v\in X$.

Therefore, we have $\div (\fg )=2\div (h_i)$, i.e.
$\fg =u_ih_i^2$ for some $u_i\in K^*$ and $h_i\in K(X)^*$.
This implies
$$\a_{K(X)}=\si (a_i,f_i)=\si (a_i,\ u_i\pri g_Ih_i^2)=\si (a_i,u_i)+
\sum\limits_{I}(a_I,g_I).$$
Now, let $I_0$ be a subset of $\n$, $v\in V_{I_0}$. Then we have
$$1=\pa_v (\a_{K(X)})=\pa_v(\sum\limits_I (a_I,g_I))=
\pa(a_{I_0},g_{I_0}),$$
since $v(g_I)=0$ if $I\not=I_0$.

On the other hand, if $v\notin V_{I_0}$ and $v\not=v_0$, then, obviously,
$\pa_v (a_{I_0}, g_{I_0})=1$, since $v(g_{I_0})=0$.
Thus, we conclude that $\pa_v (a_{I_0},g_{I_0})=1$ for any $v\not=v_0$.

Our next step is to show that $\pa_{v_0}(a_{I_0},g_{I_0})=1$ 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 $v\not=v_0$ such that $v(g_{I_0})$
is odd and $4\not\vert\deg v$. We have
$$1=\pa_v (a_{I_0}, g_{I_0})=a_{I_0} (\mod {{K(v)}^*}^2).$$
Hence we obtain that
$K(\s{a_{I_0}})\subset K(v)$. Furthermore, since
$\deg v$ is not divided by $4$,
we get that the degree $[K(v):K(\s{a_{I_0}})]$ is odd.
Since $Q_{K(v)}=0$, we conclude that $Q_{K(\s{a_{I_0}})}=0$.
Therefore,
by condition $2)$ of the proposition there are at most two
sets $I_0$, which can satisfy the unequality
$\pa_{v_0}(\aio ,\gio )\not=1$. Let, for instance,
$\pa_{v_0}(a_{I_{01}}, g_{I_{01}})\not=1$ and
$\pa_{v_0}(a_{I_{02}}, g_{I_{02}})\not=1$.
So we have
$$\prod\limits_{i\in I_{01}}a_i\prod\limits_{i\in I_{02}}a_i =
\pa_{v_0}((a_{I_{01}}, g_{I_{01}})+(a_{I_{02}}, g_{I_{02}}))=
\pa_{v_0}(\sum\limits_{I\not= I_{01}, I_{02}}(\ai ,\gi ))=1.$$
Thus, $a_{I_{01}}=a_{I_{02}}(\mod {K^*}^2)$ (recall that
$a\not=a_I (\mod {K^*}^2)$ for any $I$), which is impossible in view of
linear independency of $\o{a_1},\dots ,\o{a_n}\in K^*/{K^*}^2$.

The case where there is an only $I_0$ such that
$\pa_{v_0}(\aio ,\gio )\not=1$ is treated similarly.

Therefore, we conclude that $\pa_v (a_I , g_I)=1$ for any $I\subset\n$
and $v\in X$.
\proclaim{Lemma 2}
Assume that $\a\in\br (X)\setminus\res_{K(X)/K}(\br K)$. Then
$\ind (\a )\ge 4$.
\endproclaim
\demo{Proof}
The Merkurjev theorem on the norm residue homomorphism of degree $2$ [4]
claims that for any field $F$ the natural map $K_2(F)/2K_2(F)\to\br F$
is an isomorphism. In view of this it follows from [10]
that, if $\br (X)\not=\res_{K(X)/K}(\br K)$, then
$$\br (X)/\res_{K(X)/K}(\br K)\simeq\z,\ Q=2D\ \text{for some}
\ D\in\Br K\
\text{and}\ D_{K(X)}$$ is the nonzero element in the factorgroup
$\br (X)/\res_{K(X)/K}(\br K)$.

Therefore, we get that $\a =(D+\beta )_{K(X)}$, where
$\beta\in \br K$. Obviously, we may assume that $\a$ is
a division algebra. Let $D_1$ be the central division algebra over $K$
similar to $D+\beta$.
Since $\exp D_1=4$ we have $\ind D_1\ge 4$. Suppose that $\ind (\a )=2$.
Then by the Merkurjev index reduction theorem [5]
$D_1\simeq Q\otimes_K Q'$ for some
central division $K$-algebra $Q'$. Obviously, ${Q'}_{K(X)}={D_1}_{K(X)}=\a$.
Since $\exp D_1=4$ and $\exp Q=2$, we get that $\exp Q'=4$.
Applying the index reduction theorem again we obtain that
$Q'\simeq Q\otimes_K Q''$ for some central division $K$-algebra $Q''$,
which implies
$$D_1\simeq Q''\otimes_K Q\otimes_K Q\simeq Q''\otimes_K M_4(K),$$
a contradiction, since
$D_1$ is a division algebra. The lemma is proved.
\enddemo
Now we are able to finish the proof of Proposition $1$
in the case where $aa_I\notin {K^*}^2$ for each $I\subset\n$.
In view of Lemma $2$
we get that $(\ai ,\gi )=\res_{K(X)/K}\beta_I$ for some $\beta_I\in\br K$.
Since ${(\beta_I)}_{K(\sai )(X)}=0$, we have
$(\beta_I )_{K(\sai )}$ is either zero or $Q_{K(\sai )}$, which implies that
$\beta_I$ equals either $(\ai ,c_I)$ or $(\ai ,c_I)+Q$ for some $c_I\in K^*$.
Therefore,
$$\a =\res_{K(X)/K}(\sum\limits_I (\ai ,c_I))=
\res_{K(X)/K}(\si (a_i,\pri c_I)),$$ so we can
put $b_i=\pri c_I$.

It remains to consider the case where $aa_I\in {K^*}^2$ for some
$I\in\n$. Passing to the rational function field $K(t)$ we have
$$Q=(a,b)=(at^2+b, -ab),$$
and, obviously $(at^2+b)a_I\notin {{K(t)}^*}^2$.
Hence $\a_{K(X)(t)}=\si (a_i, p_i)$ for some polynomials $p_i\in K[t]$.
This implies that $\a =\si (a_i, l(p_i))$.
The proposition is proved.
\enddemo
\proclaim{Corollary 3}
In the notation of Proposition $1$ assume that
$\a =\res_{K(X)/K}\gamma$ for some
$\gamma\in\Br K$. Then $\gamma$ equals either $\si (a_i,b_i)$ or
$Q+\si (a_i,b_i)$ for some $b_i\in K^*$.
\endproclaim
\demo{Proof}
Since $\res_{K(X)/K}(\gamma -\si (a_i,b_i))=0$, and
$\ker (\Br K\to\Br K(X)) =\langle Q\rangle$ is the cyclic group
generated by $Q$, the corollary follows.
\enddemo

We can not drop condition $2)$ in Proposition $1$ as the following
two examples show.
\proclaim{Example A}
\endproclaim
Let $K$ be a field, $a_1,a_2,a_3\in K^*$, $D\in\br K$,
$D_{K(\s{a_1},\s{a_2},\s{a_3})}=0$, 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
$$D_{K(\s{a_3})}=(a_1,\bac )+(a_2,\bacc ) \tag 1$$
for some $b_1,b_2,c_1,c_2\in K$, and, moreover,
$$(a_1,\bacs )+(a_2, \baccs )=N_{K(\s{a_3})/K}D=0,$$ i.e.
$(a_1, \bacs )=(a_2,\baccs )$.
Put $Q=(a_1,\bacs )$, and let $X$ be the conic corresponding to $Q$.
Then we get
$$N_{K(X)(\s{a_3})/K(X)}(a_1,\bac )=(a_1,\bacs )_{K(X)}=0,$$
$$N_{K(X)(\s{a_3})/K(X)}(a_2,\bacc )=(a_2, \baccs )_{K(X)}=0.$$
Hence by [3]
$$(a_1, \bac )_{K(X)(\s{a_3})}=(a_1,f_1)_{K(X)(\s{a_3})},\tag 2$$
$$(a_2,\bacc )_{K(X)(\s{a_3})}=(a_2,f_2)_{K(X)(\s{a_3})} \tag 3$$
for some $f_1,f_2\in K(X)^*$.
Combining $(1)$, $(2)$ and $(3)$ we obtain that
$$(D-(a_1,f_1)-(a_2,f_2))_{K(X)(\s{a_3})}=0,$$ i.e.
$$D_{K(X)}=(a_1,f_1)+(a_2,f_2)+(a_3,f_3)$$
for some $f_1,f_2,f_3\in K(X)^*$. 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 $e_1,e_2,e_3\in K^*$, which would contradict to
the hypothesis on $D$.

Notice that in this example
$$Q_{K(\s{a_1})}=Q_{K(\s{a_1a_3})}=Q_{K(\s{a_2})}=Q_{K(\s{a_2a_3})}=0,$$
so there are at least four elements $a_I$ such that
$Q_{K(\s{\ai})}=0$.

\proclaim{Example B}
\endproclaim
Let $K$ be a field, $\s{-1}\in K$, $Q=(a_1,a_2)$ a nontrivial
quaternion algebra,
$X$ the corresponding conic. Put $\a =(a_1,y)+(a_2,x)\in \br K(X)$, where
$a_1x^2+a_2y^2=1$ is the equation of an affine part of the conic.
It is easy to see that
$\a\in\br (X)$ and $N_{K(X)/K(x)}\a =Q$. In particular,
$\a\notin\res_{K(X)/K}(\br K)$. Notice that since $\s{-1}\in K$, we have
$$Q=(a_1,a_2)=(a_1+a_2, -a_1a_2)=(a_1+a_2, a_1a_2),$$ hence
$$Q_{K(\s{a_1})}=Q_{K(\s{a_2})}=Q_{K(\s{a_1a_2})}=0,$$
so there are three elements $a_I$ such that $Q_{K(\s{\ai})}=0$.
\smallskip
We keep the conditions of Proposition $1$. Let $L/K$ be any field extension.
Consider the following abelian groups:
$$A(L/K)=\{\gamma\in\br K\ \text{such that}\
\res_{L/K}\gamma =\si (a_i, f_i)\
\text{for some}\ f_i\in L^*\},$$
$$B(L/K)=\{\ \si (a_i,b_i)\in\br K\ \text{where}\ b_i\in K^*\},$$
$$H(L/K)=A(L/K)/B(L/K).$$
Corollary $3$ claims that if $L=K(X)$, where $X$ is the nonsingular projective
conic corresponding to the quaternion algebra $Q$,
then $H(L/K)=\langle Q\rangle$
(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
\proclaim{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 $H(L/K)=0$.
\endproclaim
\demo{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 $u\in\br K$ is such that
$u_{K(t)}=\si (a_i, f_i)$ for some $f_i\in K[t]$. Then
$u=\si (a_i ,l(f_i))$,
which proves the assertion.

$3)$ This follows at once from $1)$ and $2)$.
\enddemo
\proclaim{Proposition 5}
Let $X=X_{\phi}$ be the projective quadric over $K$ corresponding to
the anisotropic quadratic form $\phi$. Put $L=K(X)$. Then

$1)$\ If $\dim\phi =4$ and $\disc (\phi )a_I\in {K^*}^2$
for some $I\not=\emptyset$, then $H(L/K)=\langle (a,b)\rangle$, where $(a,b)$ is an
arbitrary quaternion $K$-algebra such that
$C_0(\phi )\simeq (a,b)_{K(\s{\disc\phi })}$.

$2)$\ If $\dim\phi =4$ and $\disc (\phi )a_I\notin {K^*}^2$
for every $I$, then $H(L/K)=0$.

$3)$\ If $\dim\phi\ge 5$, then $H(L/K)=0$.
\endproclaim
\demo{Proof} $1), 2)$. Let $d=\disc (\phi )$, and let
$z^2-ax^2-by^2+abd=0$ be the equation of an affine part of $X$.
We have $$K(X)=K(x,y,\s{ax^2+by^2-abd})=K(x)(C),$$ where $C$ is the conic
over $K(x)$ corresponding to the quaternion $K(x)$-algebra
$(b, ax^2-abd)$.

Let $u\in A(L/K)$. By Corollary $3$ we have
$$u_{K(x)}=\delta +\si (a_i,f_i), \tag 4$$ where $\delta$ is either
$0$ or $(b,ax^2-abd)$, and $f_i\in K[x]$. Furthermore,
$$u= c(u_{K(x)})=c(\delta )+\si (a_i,l(f_i)),$$
where $c(\delta )$ is either
zero, or $(a,b)$. This proves that $H(L/K)\subset\langle (a,b)\rangle$.

Assume first that $da_I\in {K^*}^2$ for some $I$. Then, since
$${(ab)}^2d=a{(by)}^2+b{(ax)}^2-abz^2$$ 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 $(a,b)\in H(L/K)$.

Now assume that $da_I\notin {K^*}^2$ for every $I$. Compare the residues
at $(x^2-bd)$ on the left-hand and the right-hand sides of $(4)$. We have
$$\pa_{x^2-bd}(u_{K(x)})=0,\ \ \
\pa_{x^2-bd}(\ \si (a_i, f_i))=a_I\ \text {for some}\ I .\tag 5$$
Suppose that $\delta\not=0$, i.e. $\delta =(b,ax^2-abd)$.
The equality $\pa_{x^2-bd} (\delta )=b$ and $(5)$ imply
$ba_I\in {{K(\s{bd})}^*}^2$. Since $da_I\notin {K^*}^2$, we obtain that
$ba_I\in {K^*}^2$.
Thus, we have proved part $2)$ if $ba_I\notin {K^*}^2$ for any $I$.
On the other hand,
$$\phi_{K(t)}\simeq\langle 1,-ab(at^2+b), -(at^2+b), abd\rangle ,$$
and $(at^2+b)a_I\notin {{K(t)}^*}^2$.
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 $\psi$
of the form $\phi_{K(t)}$ such that $\disc (\psi)a_I\notin {{K(t)}^*}^2$.
Since the form $\phi_{K(t)(X_\psi )}$ is isotropic,
the extension $K(t)(X_{\psi})(X_{\phi})/K(t)(X_{\psi})$ is purely
transcendental. Applying part $2)$ and Lemma $4$ we get
$H(K(X_{\phi})/K)=0$.
\enddemo
Now we are ready to strentghen the main results of [7] and [8]. Recall the
corresponding notation.

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)" $\ind D=2$.
\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
\proclaim{Corollary 6}
Suppose that the triple $(F,D,\{a,b_1,\d ,b_n\})$ is admissible,
$\phi$ is an anisotropic form over $F$, and $\dim\phi\ge 5$.
Put $X=X_{\phi}$, and let $F(X)_{odd}$ be
a maximal odd degree extension of $F(X)$. Then
the triple $(F(X)_{odd},D_{F(X)_{odd}},\{a,b_1,\d ,b_n\})$
is admissible as well.
\endproclaim
\demo{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
$F(X)\subset F_1(X)\subset F_2(X)\subset L(X)=F{\sab }(X)$
such that $$\s a\notin F^*_2,\ \ [L:F_1]=8,\ \ [F_2:F_1]=4$$ and
$$D_{F_1(X)}\in\br (F_1(X)(\s a)/F_1(X))+\br (F_2(X)/F_1(X)).$$
Notice that $F_2=F_1(\s{d_1},\s{d_2})$, for some $d_1,d_2\in F^*$,
hence by [3] we have
$$D_{F_1(X)}=(a,f)+(d_1,f_1)+(d_2,f_2)$$
for some $f,f_1,f_2\in F_1(X)^*$.
By part $3)$ of Proposition $5$ we conclude that
$$D_{F_1}=(a,u)+(d_1,u_1)+(d_2,u_2)$$ for some $u,u_1,u_2\in F_1^*$,
which contradicts to condition $d)$ for
the admissible triple $(F,D,\{a,b_1,\d ,b_n\})$.

On the other hand, it has been shown in [8] that if
the triple $(K,D,\{a,b_1,\d ,b_n\})$
is admissible, then the triple
$K_{odd},D_{K_{odd}},\{a,b_1,\d ,b_n\})$
is admissible as well (see also Lemma $9$ below).
This finishes the proof of Corollary $6$.
\enddemo
We recall now some additional results from the papers [7] and [8].

If the triple $(F,D,\{a,b_1,\d ,b_n\})$ is admissible, and
$E=F((t_0))((t_1))\dots ((t_n))$ is the Laurent series field
in variables $t_0,t_1,\dots ,t_n$,
then the division algebra $A$ similar to the algebra
$$D\otimes_E (a,t_0)\otimes_E (b_1,t_1)\otimes\dots\otimes_E (b_n,t_n)$$
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
$F(\sqrt {b_1},\dots ,\sqrt{b_n})/F$ is not $4$-excellent, and if $D=(u,v)$,
the form $\langle uv,-u,-v,a\rangle$ provides a corresponding
counterexample [7].
Finally, if $k$ is a field and
elements $\ov {a},\ov{b_1},\d ,\ov{b_n}\in k^*/{k^*}^2$
are linearly independent over $\z$, then there exists an extension
$F/k$ and a quaternion algebra $D$ over $F$ such that
the triple $(F,D,\{a,b_1,\d ,b_n\})$ is admissible [8].

On the other hand, Corollary $6$ and the argument in Lemma $9$ below
imply that if
the triple $(F,D,\{a,b_1,\d ,b_n\})$ is admissible, then
there exists an extension $K/F$ such that
the triple $(K,D_K,\{a,b_1,\d ,b_n\})$ is admissible as well,
$u(K)=4$, and $K$ has no proper odd degree extension.

Summarizing all these results we obtain the following
\proclaim{Theorem 7}
Let $k$ be a field, $n\ge 2$. Assume that elements
$\ov {a},\ov{b_1},\d ,\ov{b_n}\in k^*/{k^*}^2$ are linearly independent
over $\z$. Then there exists an extension $K/k$ such that

$1)$\ $u(K)=4$.

$2)$\ $K$ has no proper extension of odd degree.

$3)$\ $K(\s{b_1},\dots ,\s{b_n})/K$ 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
$E=K((t_0))((t_1))\d ((t_n))$ 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 $\cd_2 (E)=n+3$).
\endproclaim
\smallskip
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 $2$.
\proclaim{Corollary 8}
Let $D\in\br K$, and either $\ind D\ge 4$, or $\ind D=2$ and
$D_{K(\sqrt{a_I})}\not=0$ for every $I$. Denote by $K(D)$ the function
field of the Severi-Brauer variety $SB(D)$ corresponding to $D$.
Then $H(K(D)/K)=\langle D\rangle$.
\endproclaim
\demo{Proof}
Let $\ind D=2^n$, where
$n\ge 2$. Let $D_1$ be the division algebra Brauer-equivalent to $D$,
$F=SB(D_1)$, and $L=SB(D)$. Then, since $SB(D_F)\simeq\Bbb P_F^{\deg D-1}$,
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 $\phi\in I^2(K)$ be a quadratic form
corresponding to $D$ under the isomorphism
$I^2(K)/I^3(K)\simeq \br K$. It is easy to see that we can choose $\phi$
such that $\dim\phi =2n+2$, and, moreover, no form of dimension
$< 2n+2$ corresponds to $D$ under this isomorphism.
Let $X=X_{\phi}$ be the projective quadric determined by the form $\phi$.
Consider the following diagram:
$$\CD K@>>>                      K(D)\cr
       @VVV                      @VVV\cr
    K(X)@>>> K(X)(D)
\endCD ,
$$
where all the maps are natural field embeddings.

If $n=2$, then
$\ind D_{K(X)}=2$, and $D_{K(X)(\s{a_I})}\not=0$ for any $I$, since
$D_{K(\s{a_I})}\not=0$, and
$\ker (\br F\to \br F(Y))=0$ for any field $F$ and any quadric $Y$ over $F$ of
dimension $>2$.

If $n\ge 3$, then $4\le \ind D_{K(X)}<\ind D$. So in both cases ($n=2$ or
$n\ge 3$) we can apply the induction hypothesis to the extension
$K(X)(D)/K(X)$. Let $u\in H(K(D)/K)$. By the induction hypothesis
$$u_{K(X)}\in H(K(X)(D)/K(X))\subset\langle D_{K(X)}\rangle .$$
We conclude that either
$u\in H(K(X)/K)$, or
$u+D\in H(K(X)/K)$. 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, $\ind D_L\ge 4$, and $H(L/K)=0$.
We need the following
\proclaim{Lemma 9}

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

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

$3)$\ Let $F$ be a field, $D\in\br F$ 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.

\endproclaim
\demo{Proof}

$1)$ Assume that $u\in\br K$ and $u_L =\si (a_i, b_i)$, where $b_i\in L^*$.
Then $$u=N_{L/K}(u_L)=\si (a_i,N_{L/K}b_i ),$$
which proves $1)$.

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

$3)$ This part is proved in [9] and [2].
\enddemo
We return to the proof of Corollary $8$ in the general case.
For an arbitrary field $F$ denote by $\widehat F$
a maximal odd degree extension of $F$, and by $\widetilde F$
the composite of all extensions $F(X_{\pi})$, where $\pi$ runs over all
anisotropic $3$-fold Pfister forms over $F$.
Consider
the following infinite tower of fields
$K=K_1\subset K_2\subset\dots$, where
$$K_{i+1}=\cases \widehat{K_i}\ \text{if}\
i\ \text{is odd},\cr \widetilde{K_i}\
\text{if}\ i\ \text{is even}\endcases.$$
Set $L=\bigcup\limits_{i=1}^{\infty}K_i$. 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.
\enddemo
\documentstyle{amsppt}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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


\ref
\key 1
\by Amitsur S., Rowen L., Tignol J.-P.
\paper Division algebras of degree $4$ and $8$ with involution
\jour Israel J. Math.
\yr 1979
\vol 33
\pages 133-148
\endref

\ref
\key 2
\by Kahn B.
\paper Quelques remarques sur le U-invariant
\jour S\'em. Th. Nombres Bordeaux
\yr 1990
\issue 2
\pages 155-161
\endref


\ref
\key 3
\by Lam T.Y., Leep D.B., Tignol J.-P.
\paper Biquaternion algebras and quartic extensions
\jour Pub. Math. I.H.E.S.
\yr 1993
\vol 77
\pages 63-102
\endref


\ref
\key 4
\by Merkurjev A.S.
\paper On the norm residue symbol of degree two
\jour Sov. Math. Dokl.
\yr  1981
\vol 24
\pages 546-551
\endref

\ref
\key 5
\by Merkurjev A.S.
\paper Simple algebras and quadratic forms
\jour Soviet Math. Docl.
\yr 1992
\vol 38
\pages 215-221
\endref

\ref
\key 6
\by Rowen L.H.
\paper Central simple algebras
\jour Israel J. Math.
\yr 1978
\vol 29
\pages 285-301
\endref

\ref
\key 7
\by Sivatski A.S.
\paper Nonexcellence of multiquadratic field extensions
\jour Journal of Algebra
\yr 2004
\vol 275
\issue 2
\pages 859-866
\endref

\ref
\key 8
\by Sivatski A.S.
\paper On indecomposable algebras of exponent $2$
\jour to appear in Israel J. Math
\endref

\ref
\key 9
\by Sivatski A.S.
\paper Invariants in the algebraic theory of quadratic forms
\jour Thesis, St. Petersburg State University
\yr 1988
\endref


\ref
\key 10
\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



\endRefs



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bigskip
\bigskip
\eightpoint
\smc
$$
\align
\text {e-mail:} \quad &\text{sivatsky}\@\text{AS3476.spb.edu} \cr
\endalign
$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye
\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye\bye
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%









