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\topmatter

\title
NONEXCELLENCE OF CERTAIN FIELD EXTENSIONS
\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, Pfister form, field extension, residue map
\endkeywords

\def\mod{\operatorname{mod}}
\def\dim{\operatorname{dim}}
\def\deg{\operatorname{deg}}
\def\det{\operatorname{det}}
\def\disc{\operatorname{disc}}
\def\phi{\varphi}
\def\a{\sqrt a}
\def\ab{\alpha^2-a\beta^2}
\def\abs{\alpha +\beta\sqrt a}
\def\xy{X^2-qY^2}
\def\aa{t + \sqrt a}
\def\bca{b+c\sqrt a}
\def\bac{b^2-ac^2}
\def\ay{(t^2-a)y^2 -1}
\def\ax{(t^2-a)x^2 -(b^2-ac^2)}
\def\tx{(t^2-a)x^2 -e}
\def\ll{\langle\!\langle}
\def\rr{\rangle\!\rangle}
\def\k{k(t,y)[x]}
\def\l{\langle}
\def\r{\rangle}
\def\ala{t^2-a}
\def\uva{u+v\sqrt a }
\def\uuva{u_1+v_1\sqrt a}
\def\s{\simeq}
\def\o{\overline}
\def\pol{ax^{2n} +bx^{2m} -t}
\def\t{ xt^2 + a{\alpha}^2}
\def\u{by^2 + abdx{\beta}^2}
\def\kk{k(t,x)[y]}
\def\x{l(t,x)[y]}
\def\o{\overline}
\def\b{\beta}
\def\p{\partial}
\def\us{by^2 + abdx{(\sigma\beta )}^2}
\def\P{bP^2 + abdx{\beta}^2 Q^2}
\def\PP{b{P_1}^2 + abdx{\beta}^2 Q_1^2}
\def\ts{xt^2 + a{(\sigma\alpha )}^2}
\def\md{\mod (xt^2+a\alpha^2)}

\abstract
Consider towers of fields
$F_1\subset F_2\subset F_3$, where $F_3/F_2$ is a quadratic extension
and $F_2/F_1$ is an extension, which is either quadratic, or of odd degree,
or purely transcendental of degree $1$.
We construct numerous examples of the above types such
that the extension $F_3/F_1$ is not $4$-excellent. Also we show
that if $k$ is a field, $char\ k\not=2$ and
$l/k$ is an arbitrary field extension of degree $4$,
then there exists a field extension $F/k$
such that the extension $lF/F$ is not $4$-excellent.
\endabstract

%\subjclass
%\endsubjclass


\endtopmatter

\document

\heading
{1. Notation and terminology}
\endheading

The purpose of this paper is to construct nonexcellent field extensions
of certain types. Recall that a field extension $L/F$ is called excellent
(resp. n-excellent) if for any quadratic form $\phi$ over $F$ (resp. for
any quadratic form $\phi$ over $F$ of dimension at most $n$)
the anisotropic part of
the form $\phi_L$ is defined over $F$. It is well known that any quadratic
extension is excellent (see for example [4]). Obviously, the same is true
for odd degree and purely transcendental extensions, since they can not
make an anisotropic form isotropic. We consider below towers of fields
$F_1\subset F_2\subset F_3$, where the extension $F_3/F_2$ is quadratic
and the extension $F_2/F_1$ is either quadratic or of odd degree or
purely transcendental of degree $1$. We give numerous
examples of extensions $F_3/F_1$ of such types, which are not $4$-excellent.

The main source of reference concerning quadratic forms over fields is
the Scharlau book [4].
We keep the standard notation. All the fields considered in the sequel
are of characteristic different from $2$. By a form we mean a nonsingular
quadratic form.
If $F$ is a field, then
$W(F)$ is the Witt ring of $F$ and $I(F)$ is the ideal of evendimensional
forms in $W(F)$. The anisotropic part of a form $\phi$ is denoted
by $\phi_{an}$.
Slightly abusing notation we write $\phi =0$ if $\phi$ is
hyperbolic, i.e. its anisotropic part is zero.
By $D(\phi)$ we denote the set of nonzero values of $\phi$.
The function field of the projective quadric related to the form $\phi$
is denoted by $F(\phi )$.
If $L/F$ is a field extension, then $\phi_L$ is the extension
of $\phi$ to $L$, i.e. the form $\phi\otimes_F L$.
For $a,b\in F^*$ the symbol $\ll a,b\rr$ means
the $2$-fold Pfisterform $\l 1,-a,-b,ab\r$.
If $k$ is a field and $p\in k[x]$ is an irreducible polynomial, by $k(p)$
we denote the residue field $k[x]/p$. If $p$ is fixed, $f\in k[x]$ and
$p$ does not divide $f$, then unless specified otherwise,
$\overline f$ is the image of $f$ in $k(p)^*/{k(p)^*}^2$.

The main tool in constructing of the examples in question is the notion of
the second residue homomorphism
$\partial_p : I^2/I^3 (k(x))\rightarrow k(p)^*/{k(p)^*}^2$ for an arbitrary
irreducible polynomial $p$ in $x$ over a field $k$ ([4]). Recall that
the residue of the $2$-fold Pfisterform $\ll f,g\rr$ at the polynomial
$p$ is computed as follows:
$$\partial_p (\ll f,g\rr )
=\cases 0\ \text{if}\ v_p(f)=v_p(g)=0,\cr
\overline f\ \text{if}\ v_p(f)=0,\ v_p(g)=1
\endcases.$$
($v_p$ is the corresponding discrete valuation on $k(x)$).
In fact these equations determine the map $\partial_p$ on the whole group
$I^2/I^3 (k(x))$.

\heading
{2. Nonexcellence of field extensions of degree $4$}
\endheading

We begin with the following

\proclaim{Theorem 1}
Let $k$ be a field, $a,b,c\in k$,
$a\in k^*\setminus {k^*}^2$, $b^2-ac^2\in k^* \setminus {k^*}^2$.
Put $l=k(\sqrt{b+c\sqrt a})$ and $F=k(t,x,y)$, where $t,x,y$
are indeterminates.
Then there exists a $4$-dimensional quadratic form
$\phi$ over $F$ such that $\dim\ {({\phi}_{lF})}_{an}=2$ and the form
${({\phi}_{lF})}_{an}$ is not defined over $F$. (This means that
the field extension $lF/F$ is not $4$-excellent).
\endproclaim
\demo{Proof}
Let $f=(\ax )(\ay )$,
$$\phi_1 \s(\aa )\l 1,\ -(\bca )\r ,$$
$$\phi_2 \s(\aa )\l 1,\ -f(\bca )\r .$$
Let us find $q\in F^*$ such that the forms
$(\aa )\ll q\rr$ and $q(\aa )(\bca )\ll qf\rr$ are defined over $F$.
Provided $q$ satisfies this condition we get that the form
$$\phi\s\phi_2\perp -q\phi_1\s (\aa )\ll q\rr\perp q(\aa )(\bca )\ll qf\rr$$
is defined over $F$ as well.

The following lemma is well known, but in view of the absence of
a convenient reference and for the sake of completeness we give the proof.
\proclaim{Lemma 2}
Suppose that $q\notin {F^*}^2\cup a{F^*}^2$ and the
elements $\alpha ,\beta\in F$
are such that $\ll q,\ab\rr=0$. Then the form
$(\abs )\ll q\rr$ is defined over $F$. Moreover, if elements
$X,Y\in F$ are such that
$\xy=\ab$, then $(\abs )\ll q\rr\s (2X+2\alpha )\ll q\rr$.
\endproclaim
\demo{Proof}
Since $(X+Y\sqrt q)(X-Y\sqrt q)=(\abs )(\alpha -\beta\sqrt a)$, we have
$$\eqalign{N_{F(\sqrt a,\sqrt q)/F(\sqrt a)}(\abs +X+Y\sqrt q)=\cr
=(\abs +X+Y\sqrt q)(\abs +X-Y\sqrt q)= (\abs )(2X+2\alpha).}$$
Therefore, since obviously,
$\ll q,N_{F(\sqrt a,\sqrt q)/F(\sqrt a)}(\abs +X+Y\sqrt q)\rr =0$,
we get
$$\ll q,\abs\rr\s\ll q, (\abs )N_{F(\sqrt a,\sqrt q)/F(\sqrt a)}
(\abs +X+Y\sqrt q)\rr\s\ll q, 2X+2\alpha\rr ,$$
or, equivalently
$(\abs )\ll q\rr\s (2X+2\alpha )\ll q\rr$. The lemma is proved.
\enddemo

Since
$${((t^2-a)y)}^2-{{t^2-a}\over{(t^2-a)y^2-1}}{((t^2-a)y^2-1)}^2=t^2-a,$$ and
$${((t^2-a)x)}^2-{{t^2-a}\over {(t^2-a)y^2-1}}f =(t^2 -a)(b^2-ac^2)=
N_{F(\sqrt a)/F}(\bca )(\aa ),$$
we obtain that
$$\ll {t^2-a\over {(t^2-a)y^2-1}},\ t^2-a\rr =\ll {t^2-a\over {(t^2-a)y^2-1}}f,
\ (t^2-a)(b^2-ac^2)\rr =0.$$
Therefore,
constructing the form $\phi$
we can put $q={{t^2-a}\over{(t^2-a)y^2-1}}$. By Lemma $2$ we get
$$(\aa )\ll q\rr\s (2(\ala )y+2t)\ll q\rr ,$$
$$q(\aa )(\bca )\ll qf\rr\s (2(\ala )x +2(bt+ac))\l q,-f\r ,$$ hence
$$\phi\s (2(\ala )y+2t)\l 1,-q\r\perp (2(\ala )x+2(bt+ac))\l q,-f\r .$$
We are going to prove that the form $\phi$ provides a counterexample to
$4$-excellency of the field $F$. However, we don't need the explicit form
of the coefficients of $\phi$, which look somewhat complicated.

It is obvious that
$${(\phi_{lF})}_{an}\s{(\phi_2 )}_{lF}\s (\aa ){\l 1,\ -f\r}_{lF}.$$
In particular, $\dim{(\phi_{lF})}_{an}=2$. If the form
${(\phi_{lF})}_{an}$ is defined over $F$, then there is $p\in F^*$
such that $$(\aa )\l 1,\ -f\r\s p\l 1,\ -f\r ,$$ i.e.
${\ll f,\ (\aa )p\rr}_{lF}=0$, which is equivalent to
$$\ll f,\ (\aa )p\rr\s\ll \bca , \uva\rr$$ over $F(\a )$
for some $u,v\in F$. We will show that in fact this is impossible.
More precisely we will prove the following a bit more general
\proclaim{Lemma 3}
For any $u,v,p\in F$, $w\in k(\a )$ one has
$$\ll f,\ (\aa )wp\rr\not\simeq\ll\bca ,\ \uva\rr .\tag 1$$
\endproclaim
\demo{Proof}
Put $e=\bac$.
If $e\notin a{k^*}^2$, i.e. $l/k$ is not a Galois extension, then
the residues at either $t +\a$ or $t -\a$ of the
left hand and on the right hand parts of $(1)$ are different, which proves
the lemma in this case.

If $e\in a{k^*}^2$, the argument is more complicated.
Assume that
$$\ll f,\ (\aa )wp\rr\simeq\ll\bca ,\ \uva\rr \tag 1'$$
for some $u,v,p\in\k$, and that under this condition
$m=\deg p$ is minimal (in this section
by $\deg$ we mean the degree with respect to the
variable $x$). In particular, the polynomial $p$ is squarefree.
Moreover, we may assume that $\deg (u+v\a )$ is minimal provided that
$\deg p =m$.
It follows that $\tx $ does not divide $p$, for
otherwise we would change
$p$ for $-p{\ay\over\tx}$, a contradiction to minimality of $\deg p$.
Also $\tx $ does not divide $\uva $, for in the opposite case
the isomorphism
$$\ll f,\ (\aa )w(\bca )p\rr\simeq \ll \bca ,\ {{\uva}\over{\tx}}(\ay )\rr ,$$
obtained from $(1')$ by adding $\ll\bca ,f\rr$ to its both parts,
would imply a contradiction to minimality of $\deg (\uva )$.

Let $p=\prod p_i$, where $p_i\in\k $ are irreducible polynomials,
$\deg p_i\ge 1$. We need the following
\proclaim{Lemma 4}
For any $p_i$ the elements $f$ and $af$ are not squares in $\k /{p_i}$.
\endproclaim
\demo{Proof}
Suppose for instance that $af$ is a square in $\k /{p_1}$.
Consider two cases: a)\ $\deg p_1\ge 2$,\ \ b)\ $\deg p_1=1$.

Case a).

Since $af$ is a square in $\k /{p_1}$ we get
$P^2-af=p_1s$, for some $P,s\in\k$, $\deg P\le deg\ p_1 -1$.
Since $\deg p_1\ge 2$ and $\deg f =2$
it follows that $\deg (p_1 s)\le 2(\deg p_1 -1)$, hence
$\deg s\le \deg p_1 -2$. Over the field $k(\a,t, y,x)$ we have
$$\ll f,p_1s\rr\s\ll f,\  P^2-af\rr\simeq \ll af,\ P^2 -af\rr =0,$$ so
$$\eqalign{\ll\bca ,\ \uva\rr\simeq\ll f,\  (\aa )wp\rr\simeq
\ll af,\ (\aa ) wp\rr\simeq\cr
\ll af,\  (\aa ) w{p\over{p_1}}s\rr\simeq \ll f,\  (\aa )w{p\over{p_1}}s\rr ,}$$
a contradiction to minimality of $\deg p$, since
$\deg{{p\over{p_1}}s}\le\deg p -2$.


Case b).

Let $p_1 =Qx+P$. We may assume that $Q,P\in k(t)[y]$.
Since $af$ is a square in $\k /{p_1}$, we conclude that
$a((\ala ){{P^2}\over{Q^2}}-e)(\ay )\in {k(t, y)^*}^2$. This implies
that $\ay$ is a divisor of $(\ala )P^2 - eQ^2$ as a polynomial
in $y$. It follows that
there are nonzero polynomials $P_1, Q_1\in k(t )[y]$
and a rationl function $g\in k(t )^*$ such that
$\deg_y P_1 ,\ \deg_y Q_1\le 1$ and
$(\ala )P_1^2 -eQ_1^2 =g(\ay )$ (here by $\deg_y$ we mean the degree with
respect to $y$). It is easy to see that this implies
$g\l (\ala ), -1\r\s\l(\ala ), -e\r$ over $k(t )$,
a contradiction, since the discriminants
on the left-hand and on the right-hand
sides of the last isomorphism are different.
The case where $f$ is a would-be square in $\k /{p_1}$ is treated
quite similar. Lemma $4$ is proved.
\enddemo
We return to the proof of Lemma $3$. By Lemma $4$ we get that
every $p_i$ is a divisor of the
polynomial $\uva\in k(\a ,t ,y)[x]$, since otherwise
the residues at $p_i$ (more precisely at some irreducible divisor of $p_i$
over $k(\a ,t ,y)[x]$) on the left-hand and the right- hand sides
of the equality
$\ll f,\ (\aa )wp\rr\simeq\ll\bca ,\ \uva\rr$ would be different.
So we obtain that $\uva =p(\uuva )$, hence
$$\ll f,\ (\aa )wp\rr \simeq\ll\bca \  ,p(\uuva )\rr\tag 2 $$ for some
$u_1,v_1\in\k$, and, since $p$ is squarefree,
the polynomials $\uuva$ and $p$
are relatively prime. Multiplying
by some square in $k(t)^*$ we may assume
that $v_{\ala}(p)$ equals either $0$ or $1$.
Suppose that $v_{\ala}(p)=1$. Then applying the automorphism
$W(F(\sqrt a))\to W(F(\sqrt a))$ determined by $t\to -t$ to
the equality $(2)$ written as $$\ll f,\ (t -\a ){p\over{\ala}}w\rr
\simeq\ll\bca ,\ {p\over{\ala}}(\uuva )(\ala )\rr$$
and changing $p(t,x,y)$ for ${p(-t,x,y)}{(t^2-a)}^{-1}$,
we reduce the problem to the case $v_{\ala}(p)=0$.

We claim that $\deg (\uuva )=0$. Indeed, suppose $s\in k(\a ,t, y)[x]$
is a prime divisor of $\uuva$.
Since $s$ does not divide $(\tx )p$, we get that
$$\partial_s (\ll\bca ,\ p(\uuva )\rr )=\partial_s (\ll f,\ (\aa )wp\rr )
=1.$$
By the same argument as in Lemma $4$ we can diminish $\deg (\uuva )$ in $(2)$,
which contradicts minimality of $\deg (\uva )$ in $(1')$.
Thus, we conclude that there are no prime divisors of $\uuva$ at all, i.e.
$u_1, v_1\in k(t ,y)$.
The equality $(2)$ implies
$$\ll (\bca )f,\ (\aa )wp\rr\simeq\ll\bca ,\ (\aa )(\uuva )w\rr ,$$
and the right- hand side of the last isomorphism
is defined over $k(\a ,t ,y)$.
Denote it by $\pi$. Since $\pi$ splits by the square root of
$$(\bca )(\ay )(\tx ),$$ we conclude
that $\pi_{k(\sqrt a ,t,y)(\tau)}=0$, where
$$\tau\s\ll (\ay)(\bca )(\ala ), -(\ay )(\bca )e\rr .$$
Therefore, either $\pi =0$, or
$$\pi\simeq\tau\simeq\ll (\ay )(\bca )(\ala ),\ -(\ay )(\bca )e\rr .\tag 3 $$
On the one hand, since $v_{\ala}(p)=0$, we have
$$\partial_{(\aa )}(\pi)=\partial_{(\aa )}(\ll f(\bca ),\ (\aa )wp\rr )=
b-c\a ,$$ so $\pi\not=0$, hence $(3)$ holds.
On the other hand,
$$\partial_{(t -\a )}(\pi)=\partial_{(t -\a )}(\ll f(\bca ),\
(\aa )wp\rr )=1,$$ and
$$\eqalign{\partial_{(t -\a )}(\ll (\ay )(\bca )(\ala )
,\ -(\ay )(\bca )e\rr )=\cr
(\bca )e=b-c\a ,}$$ a contradiction in view of $(3)$.
This proves Lemma $3$, hence also Theorem $1$.
\enddemo
\enddemo

\proclaim{Corollary 5}
Let $k$ be a field, let $l/k$ be an arbitrary
field extension of forth degree. Then there is a field extension
$F/k$ such that the extension $lF/F$ is not
$4$-excellent.
\endproclaim
\demo{Proof}
There is a field extension $k_1/k$, which is either cubic or trivial,
such that $lk_1/k_1$ is the tower of quadratic extensions. If this
tower is not the composite of two quadratic extensions of $k_1$ we can
apply Theorem $1$. The opposite case is considered in [5].
\enddemo
\smallskip

If $k(\sqrt{\bca})/k$ is not a Galois extension i.e.
$e\notin {k^*}^2\cup a{k^*}^2$, then Theorem $1$ can be strengthened
and the proof becomes much simpler. Namely, we have the following

\proclaim{Proposition 6}
Let $k$ be a field, $a,b,c\in k$,
$a\in k^*\setminus {k^*}^2$,
$e=b^2-ac^2\in k^*\setminus ({k^*}^2 \cup a{k^*}^2)$,
$l=k(\sqrt{\bca})$.
Let $t$ be an indeterminate, $F=k(t )$.
Then there exists a $4$-dimensional form $\phi$ over $F$ such that
the following holds:

For any field $\widetilde l\supset l$
such that $b-c\a$ is not a square in $\widetilde l$
and $t$ is transcendental
over $\widetilde l$
the form ${(\phi_{\widetilde l F})}_{an}$ is not defined over
$F$.
In particular, the field $l$ itself satisfies these conditions.
\endproclaim
\demo{Proof}
Let $$\phi\simeq\langle 2t ,\ 2t (t^2 -a),\
-2(bt +ac),\ -2e(bt +ac)(t^2 -a)\rangle .$$
It is easy to see that
$$\phi_{F(\a )}\simeq (t -\a )\langle 1,\ -(b-c\a )\rangle\perp
(t +\a )\langle 1,\ -(b+c\a )\rangle ,$$
hence $\dim (\phi_{\widetilde l F})_{an}=2$.
Assume that the form $(\phi_{\widetilde l F})_{an}$ is define over $F$.
Then there is $z\in F^*$ such that
$$(t -\a )\langle 1,\ -(b-c\a )\rangle\simeq
z\langle 1,\ -(b-c\a )\rangle ,$$
i.e. ${\ll b-c\a ,\ (t -\a )z\rr}_{\widetilde l F} =0$.
On the other hand, since $b-c\a$ is not a square in $\widetilde l$
and $z\in F^*$,
the residue of this $2$-fold Pfisterform at either $t -\a$ or
$t +\a$ is nontrivial, a contradiction.
\enddemo
\heading
{3. Nonexcellence of $2n$-degree extensions for an odd $n$}
\endheading

In this section we consider towers of two field extensions one of which is
a Galois extension of odd degree and the other is quadratic.

\proclaim{Proposition 7}
Let $k$ be a field, $l=k(\alpha,\beta )$ is a finite Galois field extension
of odd degree $n>1$. Assume that $\alpha\notin k$
and $\beta\notin k(\alpha)$.
Let further $t,x,y$ be indeterminates, $a,b,d\in k^*$
such that $d,-ab,-abd\notin{k^*}^2$. Set
$\phi\simeq\langle a,b,x,abdx\rangle$, $K=k(t,x,y)$, $L=lK$,
$E=L(\sqrt{-(\t )(\u )})$. Then

$1)\ \dim {(\phi_E )}_{an}=2$.

$2)$ \ The form ${(\phi_E )}_{an}$ is not defined over $K$.
\endproclaim
\demo{Proof}
The argument is somewhat similar to the one in Theorem $1$.

$1)$ Obviously,
$${\phi}_{L}\simeq\langle \t , ax(\t ), \u ,adx(\u )\rangle ,$$
hence $\phi_E$ is isotropic and
${(\phi_E )}_{an}\simeq ax(\t )\langle 1,-d\rangle$.

$2)$
Assume that the form ${(\phi_E )}_{an}$ is defined over $K$. Then there is
$p\in K^*$ such that
$$ax(\t )\langle 1,-d\rangle_E\simeq p\langle 1,-d\rangle_E,$$
i.e. $\ll d, ax(\t )p\rr_E=0$.
This means that
$$\ll d, ax(\t )p\rr_{L}\simeq\ll-(\t )(\u ),q\rr_{L}\ \ \ \ \ \ \ \ \tag 4$$
for some $q\in {L^*}$.

We may assume that $p\in\kk$ and $m=\deg p$ is minimal (in this section
we mean by $\deg$ the degree with respect to the variable $y$).
Notice also that
$\u\not\vert p$, for otherwise $\us\vert p$ for any $\sigma\in G(l/k)$
and comparing the residues of both parts of $(4)$ at $\us$, with
$\sigma\beta\not=\beta$ would give us a contradiction.

Now let $p=\prod p_i$, where $p_i\in\kk$ are irreducible polynomials in $y$.

\proclaim{Lemma 8}
$d$ is not a square in $\kk /{p_i}$ for any $p_i$.
\endproclaim
\demo{Proof}
Suppose for instance that $d$ is a square in $\kk /{p_1}$.
We get $P^2 -d =p_1s$ for some $P,s\in\kk$, $\deg P\le \deg p_1 -1$.
Hence $$\deg s\le 2(\deg p_1 -1)-\deg p_1 =\deg p_1 -2.$$
Since $\ll d,\ p_1s\rr\simeq\ll d, P^2 -d\rr =0$,
we can change the divisor $p_1$ of $p$ for $s$ in $(4)$, a
contradiction to minimality of $\deg p$. The lemma is proved.
\enddemo

Let $p_i=\prod p_{ij}$, where $p_{ij}\in\x$ are irreducible polynomials.
Since $l/k$ is an odd degree Galois extension, we have
$\x /p_{ij}\simeq l({\kk /{p_i}})$ and,
moreover, $\deg [l({\kk/{p_i}}):{\kk }/{p_i}]$
is odd. Therefore, by Lemma $8$ $d$ is not a square in
$l({\kk /{p_i}})$, hence in ${\x }/{p_{ij}}$.
This means that the form $(4)$ has nonzero residues at each $p_{ij}$.
This implies that $q=pf$,
where $f\in\x$ and the polynomials $f$ and $p$ are relatively prime,
so we can represent $(4)$ as
$$
{\ll d, ax(\t )p\rr}_{L}\simeq {\ll -(\t )(\u ), pf\rr}_{L}. \tag 5
$$
Just as in Lemma $2$ we may assume that in equality $(5)$, where
$\deg p =m$ is fixed, $\deg f$ is minimal. Denote the
Pfisterform in $(5)$ by $\pi$. Notice that $\u$ does not
divide $f$, since
otherwise we could change $f$ for $f\over{(\t )(\u )}$,
a contradiction to minimality of $\deg f$.

\proclaim{Lemma 9}
$\deg f =0$.
\endproclaim
\demo{Proof}
Let $f=\prod f_i$, where $f_i\in\x$ are irreducible polynomials.
Since $f_i$ do not appear on the left- hand side of $(5)$,
we have $\p_{f_i}(\pi )=1$, for any $f_i$.
Assume first that, say $\deg f_1\ge 2$. Since ${\p}_{f_1}(\pi )=
1$, we get that
$$P^2 +(\t )(\u )=f_1s$$
for some $P, s\in\x$ such that
$\deg P\le \deg f_1 -1$. Since $\deg f_1\ge 2$ we get that
$$\deg (P^2 +(\t )(\u ))\le 2(\deg f_1 -1 ),$$ hence
$\deg s\le \deg f_1 -2$, and we are done just as in Lemma $8$.
Now suppose that $\deg f_1 =1$, i.e.
$f_1=R(y-{P\over Q})$, where $P,Q\in l(x)[t]$, $R\in l(x,t)$.
Since $\p_{f_1}(\pi )=1$, we get that
$-(\t )(\P )$ is a square in $l(x)[t]$.
In particular,
$$\t\vert\P.$$
Therefore,
there are some nonzero polynomials
$P_1, Q_1\in l(x)[t]$ such that $\deg_t P_1, \deg_t Q_1\le 1$ and
$\t\vert\PP$, i.e.
$$\PP = g(\t )$$
for some $g\in l(x)$.
It is easy to see that this implies
$\langle b, abdx\rangle _{l(x)} \simeq\langle gx,\ ga\rangle$,
a contradiction, since the discriminants on the left-hand
and on the right-hand sides of
the last isomorphism are different.
The lemma is proved.
\enddemo

Adding to the both parts of $(5)$ the Pfisterform $\ll d,\ pf\rr$
we can represent $(5)$ as
$${\ll d,\ ax(\t )f\rr}_L\simeq {\ll -d(\t )(\u ), pf\rr}_L .\tag 6 $$
By Lemma $8$ the left-hand side of the last isomorphism
is defined over $l(t,x)$. Moreover, the right-hand side splits over
$l(t,x)(\sqrt{-d(\t )(\u )})$.
So, just as for the form $\pi$ from section $2$, we have that either
\roster
\item"a)"  ${\ll d,\ ax(\t )f\rr}_{l(t,x)} =0$, or
\item"b)" ${\ll d,\ ax(\t )f\rr}_{l(t,x)}\simeq
{\ll -bd(\t ), -abx(\t )\rr}_{l(t,x)}$.
\endroster
Assume that case $a)$ holds. We may suppose that $f\in l(x)[t]$,
$p\in k(x,y)[t]$ and that $f$ and $p$ are squarefree as polynomials in $t$.
Obviously, $\t\vert f$, for otherwise
we would have
$$d=\p_{\t }(\ll d,\ ax(\t )f\rr )=\p_{\t }(0)=1,$$
a contradiction.
By the similar reason if $\sigma\alpha\not=\alpha$ (such $\sigma$ exists,
since $\beta\notin k(\alpha )$), then
$\ts\not\vert f$. Hence from $(6)$ it follows that $\ts\not\vert p$,
if $\sigma\alpha\not=\alpha$, or
equivalently $\t\not\vert p$.

Let $f=f_1 (\t )$. Then
$$
\eqalign{
& 1=\p_{\t }(\ll -d(\t )(\u ),\ pf\rr )=\cr
& \o{df_1(\u )p\ \md}\in l(t)(u)^*/{l(t)(u)^*}^2.}\tag 7
$$
Since $\beta\notin k(\alpha )$ and $[k(\alpha ,\beta ):k]$ is odd we get that
${\beta}^2\notin k(\alpha )$, and so there is $\sigma\in G(l/k)$ such that
$\sigma\alpha =\alpha$ and $\sigma ({\beta^2})\not=\beta^2$.
Applying such $\sigma$ to $(7)$ we get
$$\o{d{\sigma f}_1(\us )p\ \md}=1. \tag 8$$
Combining $(7)$ and $(8)$ we get that
$$\o{f_1\sigma f_1(\u )(\us )\ \md}=1.$$
Since $\o{f_1\sigma f_1\ \md}\in l(t,x,\sqrt{-ax})$
does not depend on $y$, and
$\sigma ({\beta}^2)\not={\beta}^2$, we come to a contradiction.
This shows that case $a)$ is impossible.

Now assume that case $b)$ holds, i.e.
$${\ll d,\ ax(\t )f\rr}_{l(t,x)}\simeq
 {\ll -bd(\t ),\ -abx(\t )\rr}_{l(t,x)} .$$
Comparing residues at $x$ on the left-hand and on the right-hand sides
of the last isomorphism we obtain
that either $-ab\in{k^*}^2$, or $-dab\in{k^*}^2$, which is not so by
the hypothesis. This shows that case $b)$ is also impossible, which
finishes the proof of Proposition $7$.
\enddemo
\proclaim{Corollary 10}
Let $n\ge 2$ be an integer, which is not an odd prime. Then there exists a
finite $4$-nonexcellent field extension of degree $2n$.
\endproclaim
\demo{Proof}
If $n$ is even, then $2n=2^km$, where $k\ge 2$ and $m$ is odd. Let $L/F$
be a $4$-nonexcellent multiquadratic extension of degree $2^k$ (the existence
of such extensions has been proven in [5]). Let further $\phi$ be
a $4$-dimensional form providing a corresponding counterexample.
Consider the extension $L(t^{1\over m})/F(t)$ of degree $2n$. Applying
the specialization $t=0$ (the first residue map) it is easy to see that
the extension $L(t)/F(t)$ is $4$-nonexcellent with the form $\phi_{F(t)}$ as a
counterexample. Since the extension $L(t^{1\over m})/L(t)$ is of odd degree,
we conclude that $\phi_{F(t)}$ is a counterexample for the extension
$L(t^{1\over m})/F(t)$ as well.

Now assume that a number $n$ is odd, a prime $p\not=n$ divides $n$,
and $l/k$ is a Galois extension of degree $n$. Let
$k\subset k_1\subset l$, where the Galois group $G(l/k_1)$ is cyclic
of order $p$, $l=k(\b )$ and $\alpha\in k_1\setminus k$. Then
the extension $l/k$ and the elements $\alpha$ and $\b$ satisfy the
hypothesis of Proposition $7$, which implies an existence of a $4$-
nonexcellent extension of degree $2n$.
\enddemo

\heading {4. Nonexcellence of extensions determined by hyperelliptic curves}
\endheading

In the last section of the present work we give examples of
$4$-nonexcellent extensions $L/F$, where $L$ is
the function field of a hyperelliptic curve over $F$.
In this case the extension $L/F$ is a tower of a purely
trancendental extension of degree $1$ and a quadratic extension.
This question is of some interest, since the function field of a conic
always determines an excellent extension ([1], [3]).
More precisely the following statement holds.

\proclaim{Proposition 11}
Let $k$ be a field,
$a,b\in k^*$ and the Pfisterform
$\ll a, b\rr$ is anisotropic. Suppose $n>m\ge 1$ are positive integers.
Let further $F=k((t))$,
$t$ being an indeterminate, and let $X$ be a hyperelliptic curve
over $F$ defined by the equation
$y^2 = \pol$. Then the form
${(\l 1,-a,-b,t\r _{F(X)})}_{an}$ is not defined over $F$.
\endproclaim

\demo{Proof}
Since the hyperbolic plane
$\l 1,\ -(\pol ) \r _{F(X)}$ is a subform of
$\l 1,-a,-b,t\r _{F(X)}$, we get that
the form $\l 1,-a,-b,t\r _{F(X)}$ is isotropic, hence
$$\dim {(\l 1,-a,-b,t\r _{F(X)})}_{an}=2.$$ Assume for a moment that
there are $d,e\in F^*$ such that
$${(\l 1,-a,-b,t\r _{F(X)})}_{an}\simeq {\l d,e\r_{F(X)}},$$ or,
equivalently
that the form $\phi\simeq\l 1,-a,-b,t,-d,-e\r$ becomes hyperbolic
over $F(X)$. Then we have
$${(\phi_{an})}_{F(x)}\simeq\ll \pol\rr\otimes\tau\tag 8$$
for some form $\tau$.
Assume that $\dim \tau$ is odd. Then $\disc (\phi_{an})=\pol $.
On the other hand,
$$\disc (\phi_{an})= -abdet\not= \pol\ (\mod {F(x)^*}^2),$$
a contradiction. Hence $\dim \tau$ is even, and so $\dim {\phi}_{an}=4$.
Applying the first residue map (in fact the specialization $x=0$) to $(8)$
we conclude that $\phi_{an}$ is similar to the form $\ll -t, -c\rr$
for some $c\in F^*$.
Since $\ll -t,t\rr=0$, we can assume that $c\in k^*$ and $-c\notin {k^*}^2$.
Furthermore, since $\phi_{F(X)}=0$, we have
$$\pol\in D(\l -c,-t,-ct\r_{F(x)}).$$ Let $r$ be a minimal nonnegative integer
such that
$$t^{2r}(\pol )=-(cp_1^2 +tp_2^2 +ctp_3^2)$$
for some $p_1, p_2, p_3\in k[[t]][x]$
(such $r$ exists by the Cassels-Pfister theorem, (see, for example [4])).
Suppose $r\ge 1$. Then
$p_1=tq_1$, $q_1\in k[[t]][x]$, and so
$$t^{2r-1}(\pol )=-(p_2^2+cp_3^2+ ctq_1^2).$$
Hence ${\o{p_2}}^2 +c{\o{p_3}}^2 =0$,
where $\o{p_2},\o{p_3}\in k[x]$ are the reductions of $p_2, p_3$
modulo the ideal $(t)\subset k[[t]][x]$.
Since $-c\notin {k^*}^2$, we have $\o{p_2}=\o{p_3}=0$ and so
$$p_2=tq_2,\ p_3=tq_3,\ q_2,\ q_3\in k[[t]][x].$$
Therefore,
$$t^{2r-2}(\pol )=-(cq_1^2 +tq_2^2 +ctq_3^2),$$ a contradiction
to minimality of $r$. Thus, we conclude that $r=0$, hence
$$\pol =-(cp_1^2 +tp_2^2 + ctp_3^2) ,$$
which implies $ax^{2n} +bx^{2m} =-c \o{p_1^2}$. On the other hand,
obviously,
${{-(ax^{2n} +bx^{2m})}\over c}\notin {k[x]}^2$, a contradiction, which
proves Proposition $9$.
\enddemo
\smallskip

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\refstyle{D}
\Refs\nofrills{REFERENCES}

\ref
\key 1
\by Arason J. Kr.
\paper Excellence of $F(\phi )/F$ for $2$-fold Pfister forms
\jour Appendix II in [2]
\yr 1977
\endref

\ref
\key 2
\by Elman R., Lam T.Y. and Wadsworth A.R.
\paper Amenable fields and Pfister extensions
\jour Queen's Papers Pure Appl. Math.
\yr 1977
\vol 46
\pages 445-491
\endref

\ref
\key 3
\by Rost M.
\paper Quadratic forms isotropic over the function field of a conic
\jour Math. Ann.
\yr 1990
\vol 288
\pages 511-513
\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 Sivatski A.S.
\paper Nonexcellence of multiquadratic field extensions
\jour Journal of Algebra
\yr 2004
\vol 275
\issue 2
\pages 859-866
\endref


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\ref
\key L
\by Leep D.
\paper Function field results
\jour Handwritten notes taken by T.Y. Lam
\yr 1989
\endref

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


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