%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\magnification 1200
\documentstyle{amsppt}
\topmatter
\title
ON PROPERTY $D(2)$ AND COMMON SPLITTING FIELD OF TWO BIQUATERNION ALGEBRAS
\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
Quadratic form, Biquaternion algebra, Second residue map
\endkeywords

\abstract
Let $F$ be a field of characteristic $\not= 2$.
We say that $F$ has property $D(2)$ if
for any quadratic extension
$L/F$ and any two binary quadratic forms over $F$
having a common nonzero value over $L$ this value can be chosen in $F$.
We show that if $k$ is a field of characteristic $\not=2$ having at least
two distinct quadratic extensions, then for the field $k(x)$
property $D(2)$ does not hold.
Using this we construct two biquaternion algebras over a field
$K=k(x)((t))((u))$
such that their sum is a quaternion algebra, but they do not have
a common biquadratic (i.e. a field of the kind $K(\sqrt a, \sqrt b )$, where
$a,b\in K^*$) splitting field.

\endabstract

\subjclass 13C12, 13F20 
\endsubjclass

\endtopmatter

\document

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

\def\phi{\varphi}
\def\d{d_1^2-4a}
\def\dd{d_2^2-4a}
\def\o{\otimes}
\def\sq{\sqrt{x^2-4a}}
\def\l{\langle}
\def\r{\rangle}
\def\ll{\langle\!\langle}
\def\rr{\rangle\!\rangle}
\def\a{(a_1,b_1)}
\def\aa{(a_2,b_2)}
\def\p{\partial}
\def\ind{\operatorname{ind}}
\def\s{\simeq}
\def\x{x-d_1}
\def\xx{x-d_2}
\def\t{t+{a\over t}}
\def\w{\widehat k}
\def\sqq{x^2-4a}
\def\m{\matrix}
\def\e{\endmatrix}
\def\ab{\left (\m\alpha &\beta\cr \gamma &\delta\cr\e\right )}
\def\mx{\left (\m 1&0\cr x&1\cr\e\right )}
\def\mxx{\left (\m 1&0\cr -x&1\cr\e\right )}

\document

Let $F$ be a field of characteristic different from $2$.
By definition, $F$ has property $D(2)$ if for any quadratic extension $L/F$
and any two binary quadratic forms $q_1$, $q_2$ over $F$ the existence
of a common value of the forms ${q_1}_L$, ${q_2}_L$ implies the existence
of a common value of the forms ${q_1}_L$, ${q_2}_L$, which lies in $F$.
Examples of fields of characteristic $0$ not satisfying this property
has been given in [5]. Later in [1] starting from such a field,
it has been shown that the answers
to the following questions are negative in general:

$1)$ {\it Let $\a$ and $\aa$ be quaternion algebras over a field $K$.
Suppose $c\in K^*$ is such that
${\a\o\aa}_{K({\sqrt c})}$ is not a division algebra.
Is it true that there exists $d\in K^*$ such that
$$\a_{K(\sqrt c,\sqrt d)} = \aa_{K(\sqrt c,\sqrt d)} =0\ ?$$}

$2)$ {\it Let $\phi$ be an $8$-dimensional form from $I^2(K)$ whose Clifford
algebra has index $4$. Is it true that $\phi$ is a direct sum of two
forms similar to $2$-fold Pfister forms ?}

$3)$ {\it Let $\phi$ be a $14$- dimensional form from $I^3(K)$. Is it true that
$\phi$ is similar to the difference of the pure parts of two
$3$-fold Pfister forms ?}

By means of quite a different and very
nonelementary technique the negative answers to
the same questions have been given in [3].

The above questions stipulate our interest to property $D(2)$.
As far as we know, examples of fields of positive characteristic
not satisfying property $D(2)$ are not known. In this note independently
of characteristic of the field we give simple
counterexamples to property $D(2)$ such that the $2$-cohomological
dimension of the ground field $F$ equals $2$.
Using these examples we show that the following question
has the negative answer in general:

{\it Suppose $D_1$ and $D_2$ are two biquaternion algebras over the field $K$
such that $\ind (D_1+D_2)=2$.
Does there exist a common splitting field of $D_1$ and $D_2$ of the type
$K(\sqrt a ,\sqrt b )$, where $a,b\in K^*$ ?}
\smallskip

Our notation is standard. All the fields in the sequel are of characteristic
different from $2$. By form we always mean a quadratic form over a field.
Slightly abusing notation we often identify a form over $F$
with the corresponding
class in the Witt group $W(F)$.
By $\phi_{an}$ we denote the anisotropic part of the form $\phi$.
If $L/F$ is a field extension, and $\phi$ is a form over $F$, then
$D_L(\phi)$ is the set of nonzero values
of $\phi_L$, and $G_L(\phi)$ is the group
of multipliers of $\phi_L$, i.e. $G_L(\phi)=\{a\in L^*:\phi\s a\phi\}$.
By the $n$-fold Pfister form $\ll a_1,\dots ,a_n\rr$ we mean the form
$\l1,-a_1\r\o\dots\o\l 1,-a_n\r$. By $(a,b)$ we denote the quaternion algebra
with generators $1,i,j,k$ and relations $i^2=a,\ j^2=b,\ ij=-ji$.
If $k$ is a field, $t$ is an indeterminate and $p\in k[t]$ is
a monic irreducible
polynomial, then $\p_p:W(k(t))\to W(k[t]/p)$ is the second residue map,
i.e. the group homomorphism
determined by the following rule:
for a squarefree polynomial $f\in k[t]$
$$\p_p(\l f\r )=\cases 0\ \text{if}\
p\ \text {does not divide}\ f,\cr
\langle{f\over p}\rangle\
\text{if}\ p\ \text{divides}\ f\endcases.$$
The tensor product of central simple $K$-algebras is always considered
over the field $K$.
\smallskip

We turn to the construction of the examples in question.
Let $k$ be a field, $a,d_1,d_2\in k^*$ such that
$\d,\ \dd,\  (\d)(\dd)\notin {k^*}^2$, $x$ an indeterminate, $F=k(x)$,
$L=F(\sq )$.
Set $$q_1\s(\x)\ll\d\rr\in W(F),$$
$$q_2\s(\xx)\ll\dd\rr\in W(F) .$$

\proclaim{Proposition 1}
The field $F$ does not satisfy property $D(2)$, and
the forms $q_1, q_2$ and the quadratic extension $L/F$
provide a counterexample.
\endproclaim
\demo{Proof}
It is trivial to check that
$$\eqalign {q_1 ((\x -\sq){(\x )}^{-1},{(\x )}^{-1})\cr=
q_2((\xx -\sq){(\xx )}^{-1}, {(\xx )}^{-1})\cr=
2(x-\sq )\in L.}$$
This shows that $D_L(q_1)\cap D_L(q_2)\not=\emptyset$.

Now let us prove that $D_L(q_1)\cap D_L(q_2)\cap F =\emptyset$.
We will follow an idea in [6].
Let $y_1, y_2$ be indeterminates, $\w$ a maximal odd degree
extension of $k(y_1,y_2)$.
Since $\d$ and $\dd$ are distinct
nontrivial elements in ${k^*}/{k^*}^2$, it is easy to see that
$(\d ,y_1)\o_{\w} (\dd ,y_2)$ is a division algebra.
Put
$$\phi =\ll\d ,y_1\rr -\ll\dd ,y_2\rr .$$
In particular, $\phi\not=0$ over any quadratic extension
of $\w$.
Put $y=\sq ,\ x-y=2t$. Then $x+y={{2a}\over t}$, hence $x=t+{a\over t}$.
So we have $L=F(\sq )=k(t)$,\ $F=k(t+{a\over t})$.
Suppose that $u\in D_{\w (t)}(q_1)\cap D_{\w (t)}(q_2)\cap \w (x)$.
Obviously, we may assume that $u\in \w [x]$ and $u=u(x)$ is squarefree.
Then
$$(\x)u\in D_{\w (t)}(\ll\d\rr )=G_{\w (t)}(\ll\d\rr )=G_{\w (t)}(q_1),$$
$$(\xx)u\in D_{\w (t)}(\ll\dd\rr )=G_{\w (t)}(\ll\dd\rr )=G_{\w (t)}(q_2).$$
Therefore, we have
$$\eqalign{(q_1\ll y_1\rr - q_2\ll y_2\rr )_{\w (t)} =
((\x)u q_1\ll y_1\rr -(\xx)u q_2\ll y_2\rr )_{\w (t)}\cr =
u(\ll\d,y_1\rr -\ll\dd,y_2\rr ) _{\w (t)}=u\phi_{\w (t)} .} \tag 1$$
Similarly, since $4t=2(x-\sq )\in D_L(q_1)\cap D_L(q_2)$, we have
$$(q_1\ll y_1\rr -q_2\ll y_2\rr )_{\w (t)} =t\phi_{\w (t)} .\tag 2$$
Combining $(1)$ and $(2)$ we conclude that
$$u\phi_{\w (t)} =t\phi_{\w (t)}\tag 3 $$
Substituting $x$ for $\t$ in the left part of $(3)$
and comparing the residues at $t$ of the both parts of this
equality we see that the degree of the polynomial $u$ in $x$ is odd.
Since the field  $\w$ has no proper extensions of odd degree,
we conclude that there is $c\in\w$ such that $x-c$ divides $u$.
Notice that
$x-c ={{p(t)}\over t}$, where $p(t)=t^2 -ct +a$. Comparing the residues at
$p(t)$ of the both parts of the equality $(3)$ we get a contradiction, since
$\p_p(t\phi )=0$, and $\p_p(u\phi )=\phi_{\w (\sqrt{c^2-4a})}\not=0$.
The proposition is proved.
\enddemo

\proclaim{Corollary 2}
Let $k$ be a field, $b_1,b_2\in k^*$.
Then the following conditions are equivalent.

$1)$ The elements $1,b_1,b_2\in k^*/{k^*}^2$ are pairwise distinct.

$2)$ The field $k(x)$ has not property $D(2)$ and there exist $a\in k^*$,
$s_1,s_2\in k(x)^*$ such that the extension $k(x,\sq )/k(x)$ and the forms
$s_1\ll b_1\rr$ and $s_2\ll b_2\rr$ provide a counterexample.
\endproclaim
\demo{Proof}
$1)\Rightarrow 2)$. By Proposition $1$ it suffices to find $a,d_1,d_2\in k^*$
such that $b_1=\d$, $b_2=\dd$. One can put, for instance,
$d_1={{b_1-b_2+1}\over 2}$, $d_2={{b_2-b_1+1}\over 2}$,
$a={1\over 4}(d_1^2 -b_1)$.

$2)\Rightarrow 1)$. If $b_1$ is a square, then
$D_L(\ll b_1\rr )=L^*$ for any extension $L/k(x)$, hence
for any $s_1,s_2\in k(x)^*$ the forms $s_1\ll b_1\rr$ and $s_2\ll b_2\rr$
do not provide a counterexample to property $D(2)$. Assume that $b_1b_2$
is a square, i.e. $b_1\equiv b_2 (\mod{{k^*}^2} )$.
Let $L/k(x)$ be an arbitrary field extension.
Assume that $c\in D_L(s_1\ll b_1\rr )\cap D_L(s_2\ll b_2\rr )$.
Then $\ll b_1, cs_1\rr =\ll b_1,cs_2\rr=0$, hence $\ll b_1, s_1s_2\rr=0$,
and so
$$s_1\in D_L(s_1\ll b_1\rr )\cap D_L(s_2\ll b_2\rr )\cap k(x).$$
\enddemo

{\bf Remark.} If $k_0$ is an algebraically closed field, and $k$ is
the rational function field over $k_0$,
then $cd_2 (k)=1$ and $cd_2 (k(x))=2$. By Corollary $2$ the field $k(x)$
does not have property $D(2)$. On the other hand,
if $I^2(K)=0$, then
the field $K$ has property $D(2)$,
since $1$ is a value of any binary form over $K$.
\smallskip
It has been established in [1] that if the field $F$ has not property
$D(2)$, then
the field $F_1=F((t))$ has not property $CS$. Recall that property $CS$
for the field $F_1$
means that for any quaternion algebras $Q_1$, $Q_2$ over $F_1$ and
$c\in F_1^*$ such that ${(Q_1\o Q_2)}_{F_1(\sqrt c )}$
is not a division algebra there exists $d\in F_1^*$ such that
${Q_1}_{F_1(\sqrt c ,\sqrt d)}=0,\ {Q_2}_{F_1(\sqrt c ,\sqrt d)}=0$.
In fact, if the binary forms $s_1\ll b_1\rr$, $s_2\ll b_2\rr$ and
the quadratic extension $F(\sqrt c)/F$ provide a counterexample to
property $D(2)$ for the field $F$, then the quaternion algebras
$Q_1\simeq (b_1,s_1t)$, $Q_2\simeq (b_2,s_2t)$
give a counterexample to property $CS$ for the field $F((t))$.

Let now $F$ be an arbitrary field, for which property $CS$ does not hold.
Assume that quaternion algebras $Q_1$, $Q_2$ and a quadratic extension
$F(\sqrt c)/F$ provide a counterexample. In particular,
$(Q_1\o Q_2)_{F(\sqrt c)}$ is not a division algebra, which implies that
$Q_1\o Q_2\simeq (c,d)\o (e,f)$ for some $d,e,f\in F^*$. Notice that
$(e,f)\not=0$, for otherwise $Q_1$and $Q_2$ would have
a common quadratic subfield.
Consider the biquaternion algebras $D_1\simeq Q_1\o (c,u)$ and
$D_2\simeq Q_2\o (c,du)$ over the field $F((u))$.
\proclaim{Proposition 3}
$1)$ $\ind (D_1\o D_2)=2$.

$2)$ The algebras $D_1$ and $D_2$ do not
have a common biquadratic splitting extension. In other words,
for any $p,q\in F((u))^*$
either ${D_1}_{F((u))(\sqrt p,\sqrt q)}\not=0$, or
${D_2}_{F((u))(\sqrt p,\sqrt q)}\not=0$.
\endproclaim
\demo{Proof}

$1)$ $D_1 +D_2 =Q_1+Q_2+(c,u)+(c,du)=Q_1+Q_2+(c,d)=(e,f)$, which proves
the first part of the proposition.

$2)$ Assume the contrary, i.e. that
${D_1}_{F((u))(\sqrt p,\sqrt q)}=0$, ${D_2}_{F((u))(\sqrt p,\sqrt q)}=0$.
Since ${F((u))}^*/{{F((u))}^*}^2 =F^*/{F^*}^2\oplus\Bbb Z/2\Bbb Z$,
we may assume that $p\in F^*$. If $q\in F^*$, it is easy to see that
$F(\sqrt c)\subset F(\sqrt p,\sqrt q)$, i.e.
$F(\sqrt p ,\sqrt q )=F(\sqrt c ,\sqrt {c'})$ for some $c'\in F^*$.
Hence $${Q_1}_{F(\sqrt c,\sqrt {c'})}= {Q_2}_{F(\sqrt c,\sqrt {c'})}=0,$$
which is impossible, since the algebras
$Q_1$, $Q_2$ and the extension
$F(\sqrt c)/F$ provide a counterexample to property $CS$. Therefore, we may
suppose that $q=au$, where $a\in F^*$. Then, obviously,
$$(Q_1 +(c,a))_{F((u))(\sqrt p,\sqrt{au})}=
(Q_1 +(c,u))_{F((u))(\sqrt p,\sqrt q)}=0,$$
$$(Q_2 +(c,ad))_{F((u))(\sqrt p,\sqrt{au})}=
(Q_2 +(c,du))_{F((u))(\sqrt p,\sqrt q)}=0.$$
Hence,
$$(Q_1 +(c,a))_{F(\sqrt p)}=0,\ (Q_2 +(c,ad))_{F(\sqrt p)}=0,$$
which implies that
$${Q_1}_{F(\sqrt c,\sqrt p)}={Q_2}_{F(\sqrt c,\sqrt p)}=0,$$
a contradiction to the fact that the algebras $Q_1$, $Q_2$ and the extension
$F(\sqrt c)/F$ provide a counterexample to property $D(2)$.
The proposition is proved.
\enddemo
\smallskip
{\bf Remark.}
To give a concrete example of biquaternion algebras $D_1$ and $D_2$
satisfying the conditions of Proposition $3$ we start from the binary forms
$q_1\simeq (x-d_1)\ll\d\rr$ and $q_2\simeq (x-d_2)\ll\dd\rr$ over the field
$k(x)$ from Proposition $1$. Then
$Q_1\simeq (\d,(x-d_1)t)$,\ $Q_2\simeq (\dd, (x-d_2)t)$.
It is easy to check the following equality comparing the residues on its both
sides:

$$\eqalign{Q_1 +Q_2 =(\sqq ,-(x-d_1)(x-d_2)((d_1+d_2)x-4a-d_1d_2))+\cr
((\d)(\dd), ((d_1+d_2)x-4a-d_1d_2){(d_1+d_2)}^{-1}t).}$$

Tracing back the construction of the algebras $D_1$, $D_2$ from
the algebras $Q_1$, $Q_2$, we get over the field $k(x)((t))((u))$
$$D_1\simeq (\d,(x-d_1)t)\o (\sqq ,u),$$
$$D_2\simeq (\dd, (x-d_2)t)\o (\sqq ,-(x-d_1)(x-d_2)((d_1+d_2)x -4a-d_1d_2)u).$$

Notice that if $cd_2 k=1$, $K=k(x)((t))((u))$, then $cd_2 K=4$,
so one can construct
a counterexample with a field $K$ of $2$-cohomological dimension $4$.
On the other hand, there is no such a counterexample for a field of
cohomological dimension $2$. More precisely we have the following
\proclaim{Proposition 4}
Let $K$ be a field such that $I^3(K)=0$. Let further $D_1,D_2$ be
biquaternion algebras such that $\ind (D_1+D_2)=2$. Then there exist
$p,q\in K^*$ such that
$${D_1}_{K(\sqrt p,\sqrt q)}={D_2}_{K(\sqrt p,\sqrt q)}=0.$$
\endproclaim
\demo{Proof}
By ([2], Lemma 14.2) we have $I^3(K(\sqrt a))=0$ for any $a\in K^*$.
Let $\phi_1$, $\phi_2$ be Albert forms corresponding
to $D_1$, $D_2$. Then by ([4], Ch.2, Th.14.4) we get
$${(\phi_1 -\phi_2)}_{K(\sqrt b )}\in I^3(K(\sqrt b))=0$$
for any $b\in K^*$ such that $(D_1+D_2)_{K(\sqrt b)}=0$. Hence
$\phi_1\perp -\phi_2 \simeq\ll b\rr\otimes\psi$, where $\dim\psi =6$.
Let $\psi_1\in I^2(K)$ be such a form that $\psi =\psi_1 +\tau$, where
$\dim\tau =2$. Then we have
$$\phi_1 -\phi_2 =\ll b\rr\otimes\psi_1 +\ll b\rr\otimes\tau =\ll b\rr\tau ,$$
since $\ll b\rr\otimes\psi_1\in I^3(K)=0$. Therefore,
$\dim (\phi_1\perp -\phi_2 )_{an}\le 4$. This means that
$\phi_1$ and $\phi_2$ have a common $4$-dimensional subform, say
$$\phi_1\s\l a,b,c,d,e_1, -abcde_1\r ,$$
$$\phi_2\s\l a,b,c,d,e_2, -abcde_2\r .$$
Then, obviously,
$${\phi_1}_{K(\sqrt{-ab},\sqrt{-cd})}={\phi_2}_{K(\sqrt{-ab},\sqrt{-cd})}=0,$$
which is equivalent to
$${D_1}_{K(\sqrt{-ab},\sqrt{-cd})}={D_2}_{K(\sqrt{-ab},\sqrt{-cd})}=0,$$
so we can put $p=-ab$, $q=-cd$.
The proposition is proved.
\enddemo
The following theorem [7] is the main motivation
of the present note, and for the sake of completeness
we give here its proof.
\proclaim{Theorem 5}
Let $D_1$ and $D_2$ be biquaternion algebras over a field $k$ such that
$\ind (D_1+D_2)=2$. Then there exists a field extension $l/k$
of degree $4$ such that ${D_1}_l={D_2}_l=0$.
\endproclaim
\demo{Proof}
Let $D_1+D_2=Q=(a,b)$. Then by the dimension count $D_1\otimes Q\s M_2(D_2)$.
In particular, $Q\hookrightarrow M_2(D_2)$. It follows that there exist
$I,J\in M_2 (D_2)$ such that
$I^2=\left (\m a&0\cr 0&a\cr\e\right )$,
$J^2=\left (\m b&0\cr 0&b\cr\e\right )$, and $IJ=-JI$.

Let $I=\ab$.
We may assume that ${\alpha}^2\not=a$, and $D_2$ is a division algebra
(in the opposite case the theorem is simple and left to the reader).
\proclaim{Lemma 6}
There is a matrix $S\in GL_2(D_2)$ such that $SIS^{-1}=
\left (\m 0&1\cr a&0\cr\e\right )$.
\endproclaim
\demo{Proof}
Since ${\alpha}^2\not=a$ and
$I^2=\left (\m a&0\cr 0&a\cr\e\right )$,
we have $\beta\not=0$.
Then for any $x\in D_2$
$$\mx I{\mx}^{-1}=\mx\ab\mxx =
\left (\m {\alpha -\beta x}&*\cr *&*\cr\e\right ).$$
Hence setting $x={\beta}^{-1}\alpha$ we may assume that
$\alpha =0$. Since ${\ab}^2=\left (\m a&0\cr 0&a\cr\e\right )$, we get
$\beta\gamma =a$, $\beta\delta =0$. Since
$D_2$ is a division algebra,
we conclude that $\delta =0$, i.e.
$\ab =\left (\m 0&a\gamma^{-1}\cr \gamma&0\cr\e\right )$.
Put $C=\left (\m 1&0\cr 0&a\gamma^{-1}\cr\e\right )$. Then $$C\ab C^{-1}=
\left (\m 0&1\cr a&0\cr\e\right ).$$ The lemma is proved.
\enddemo
In view of the above lemma, conjugating by a suitable matrix we may assume
that $I=\left (\m 0&1\cr a&0\cr\e\right )$. Let
$J=\left (\m x&y\cr z&t\cr\e\right )$. Since
$$\left (\m 0&1\cr a&0\cr\e\right )\left (\m x&y\cr z&t\cr\e\right )=
-\left (\m x&y\cr z&t\cr\e\right )\left (\m 0&1\cr a&0\cr\e\right ),$$
we get that $z=-ay$, $t=-x$. Therefore,
${\left (\m x&y\cr -ay&-x\cr\e\right )}^2=
\left (\m b&0\cr 0&b\cr\e\right )$, which means that
$x^2-ay^2=b$ and $xy=yx$. Hence the subalgebra $l_1=k[x,y]$ is a subfield of
$D_2$. Moreover, $Q_{l_1}=0$. If $l$ is a maximal subfield of $D_2$ containing
$l_1$, then $Q_l={(D_2)}_l=0$, which implies ${(D_1)}_l=0$,
and proves Theorem $5$.
\enddemo










\documentstyle{amsppt}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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

\frenchspacing

\refstyle{D}
\Refs\nofrills{REFERENCES}
\widestnumber\key{10}


\ref
\key 1
\by Hoffmann D.W., Tignol J.-P.
\paper On $14$-dimensional quadratic forms in $I^3$,
$8$-dimensional forms in $I^2$, and the common value property
\jour Doc. Math.
\yr 1998
\pages 189-214
\endref

\ref
\key 2
\by Izhboldin O.T.
\paper Fields of $u$-invariant $9$
\jour Ann. Math.
\yr 2001
\vol 154
\pages 529-587
\endref

\ref
\key 3
\by Izhboldin O.T., Karpenko N.A.
\paper Some new examples in the theory of quadratic forms
\jour Math. Z.
\yr 2000
\vol 234
\pages 647-695
\endref

\ref
\key 4
\by Scharlau W.
\paper Quadratic and Hermitian forms
\jour Springer, Berlin Heidelberg New York
\yr 1985
\endref

\ref
\key 5
\by Shapiro D.B., Tignol J.-P., Wardsworth A.R.
\paper Witt rings and Brauer groups under multiquadratic extensions, II
\jour J. Algebra
\yr 1982
\vol 78
\pages 58-90
\endref

\ref
\key 6
\by Sivatski A.S.
\paper Nonexcellence of the function field of the product of two conics
\jour $K$-theory
\yr 2005
\vol 34
\pages 209-218
\endref

\ref
\key 7
\by Risman L.J.
\paper Zero divisors in tensor products of division algebras
\jour  Proceedings of the AMS
\yr 1975
\vol 51
\pages 35-36
\endref
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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

%\enddocument
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%














