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%%% Lieber Herr Rehmann,
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    \textup{2000} Mathematics Subject Classification}

\begin{document}

\title{Quadratic forms and Pfister neighbors in characteristic 2}


\author{Detlev W.\ Hoffmann}
\address{Laboratoire de Math\'ematiques, UMR 6623 du CNRS, Universit\'e de
Franche-Comt\'e, 16 Route de Gray, F-25030 Besan\c con Cedex, France}
\email{detlev@math.univ-fcomte.fr}

\author{Ahmed Laghribi}
\address{Fakult\"at f\"ur Mathematik, Universit\"at Bielefeld,
Postfach 100131, D-33501 Bielefeld, Germany}
\email{laghribi@mathematik.uni-bielefeld.de}
\thanks{Both authors have been supported in part by the European
research networks FMRX-CT97-0107 and HPRN-CT-2002-00287 ``Algebraic
$K$-Theory, Linear Algebraic Groups and Related Structures'',
and by the program INTAS 99-00817 ``Linear Algebraic Groups and Related
Linear and Homological Structures.''}


\subjclass{Primary 11E04; Secondary 11E81}

\date{}



\keywords{Quadratic form, Pfister form, Pfister neighbor, quasi-Pfister form,
quasi-Pfister neighbor, function
field of a quadratic form, degree of a quadratic form, splitting tower,
height of a quadratic form, singular quadratic form}

\begin{abstract}
We study Pfister neighbors and their characterization over fields
of characteristic $2$, where we include the case of singular forms.
We give a somewhat simplified proof of a theorem of Fitzgerald which
provides a criterion for when a nonsingular quadratic
form $q$ is similar to a Pfister form in terms of the hyperbolicity of
this form over the function field of a form $\phi$ which is dominated by
$q$.  From this,
we derive an analogue in characteristic $2$ of a result by Knebusch saying
that, in characteristic $\neq 2$, a form is
a Pfister neighbor if its anisotropic part over its own function field
is defined over the base field.  Our result includes certain cases
of singular forms, but we also give examples which show that Knebusch's
result generally fails in characteristic $2$ for singular forms.
As an application, we characterize certain forms of height $1$ in the
sense of Knebusch whose quasi-linear parts are of small dimension.
We also develop some of the basics of a  theory of totally singular
quadratic forms.  This is used to give a new interpretation of
the notion of the height of a standard splitting tower as introduced 
by the second author in \cite{La2}.
\end{abstract}

\maketitle

\section*{Table of contents}
\begin{itemize}
\item[1.] Introduction
\item[2.] Witt cancellation and Witt decomposition
\item[3.] Subforms and dominated forms
\item[4.] Further notions and results needed in proofs
\begin{itemize}
\item[4.1] Pfister forms and quasi-Pfister forms
\item[4.2] Function fields of quadratic forms and splitting towers
\item[4.3] The Arf-invariant and the Clifford invariant
\end{itemize}
\item[5.] Fitzgerald's theorem
\item[6.] Pfister neighbors and an analogue of a theorem by Knebusch in\\
 characteristic $2$
\item[7.] Forms of nondefective height $1$
\item[8.] Totally singular forms and quasi-Pfister neighbors
\end{itemize}

\section{Introduction}
An important aspect of the algebraic theory of quadratic forms is the study
of function fields of quadrics and the isotropy behaviour of quadratic
forms over such function fields.  Questions of this type have already,
sometimes implicitly, appeared in Pfister's seminal work on quadratic
forms in the 1960s, which then inspired a large number of researchers
in the early 1970s (e.g., Arason, Elman, Lam, Knebusch, Wadsworth, etc.)
to investigate further problems in the algebraic theory of quadratic forms
where the use  of function fields of quadratic forms became an indispensable 
tool or even a subject of study per se.

Knebusch's papers \cite{Kn2}, \cite{Kn3} constitute a major advance in
that they attempted for the first time to provide a self-contained
in-depth study of function fields of quadratic forms and the behaviour
of quadratic forms over such function fields, with particular emphasis laid 
upon the ``generic'' properties of such function fields.
These two papers lead to a wealth of new result, and opened many
possible pathways for further research by asking important questions,
some of which have only just recently been answered using new techniques
developed partly in the wake of Voevodsky's work on the Milnor conjecture.
Knebusch's papers were followed in the early 1980s by various articles on 
function fields
most notably by Fitzgerald, Elman-Lam, Elman-Lam-Wadsworth, etc.

Starting in the 1990s, function field questions once again became of interest
due to Merkurjev's construction of fields with $u$-invariant equal to
any given even number (the $u$-invariant of a field $F$ being the supremum of 
the dimensions of anisotropic forms over $F$ which are torsion in the
Witt ring of $F$).   Of particular importance became the question of
when an anisotropic form over $F$ becomes isotropic over the function field
$F(\psi )$ of another form $\psi$.  This question has been studied extensively
by the present authors, Izhboldin, Karpenko, Merkurjev, Vishik, etc. 

Most of the above mentioned studies have one thing in common :  they all
deal initially only with quadratic forms over fields of characteristic
$\neq 2$.  Several of the results obtained have been extended to quadratic forms over 
fields of characteristic $2$ (see, e.g., \cite{Ah}, \cite{AB1}, \cite{AB2}, 
\cite{AJM}, \cite{B1}, \cite{B2}, \cite{B3}, \cite{KnR}, \cite{MTW}).
However, in most of these articles the emphasis has been on nonsingular 
quadratic forms (forms without radical) or regular forms 
(nonsingular forms or those  with nonvanishing radical of dimension $\leq 1$).

Only recently have there been efforts to develop a theory
of function fields of quadratic forms in characteristic $2$ which
systematically includes the case of singular forms, following
the initial ideas developed in \cite{Kn2}, \cite{Kn3}.  In \cite{Kn4},
Knebusch embarks on a study of the generic aspects of such a theory
(places, specializations, etc.) which in its scope goes far beyond
previous attempts.  The aspect of the isotropy of quadratic forms
over function fields of quadratic forms (singular or not) and
the behaviour of forms over their own function field has been
treated systematically in \cite{La1}, \cite{La2}, \cite{LM}.
It is this latter aspect which will constitute the centerpiece of the 
present article.

The central topic of this paper is the extension to characteristic $2$
of two important theorems concerning Pfister forms and Pfister 
neighbors,  and whose
characteristic $\neq 2$ versions are due to Fitzgerald \cite{F}
and Knebusch \cite{Kn3}.  In characteristic $\neq 2$, Pfister
forms are tensor products of a finite number of binary
(regular) forms and they play a central role in the algebraic
theory of quadratic forms.  A Pfister neighbor of a Pfister form
is a form similar to an orthogonal summand of this Pfister form
and of dimension greater than half the dimension of this Pfister form.
It turns out that much of the information about a Pfister form
can already be retrieved from information about a 
neighbor of this Pfister form and vice versa.
Now anisotropic Pfister forms and  neighbors can intrinsically be 
characterized by how they behave over their own function
field.  In fact, an anisotropic form is similar to a Pfister form iff 
the form
is hyperbolic over its own function field (\cite{Kn2}, \cite{W}),
and an anisotropic form $\phi$ over $F$ is a Pfister neighbor if its 
anisotropic part over its own function field is defined over the ground
field $F$, i.e. there exists a form $\psi$ over $F$ such that
$\an{(\phi_{F(\phi )})}\cong\psi_{F(\phi)}$.   This
latter statement is due to Knebusch \cite[Th. 7.13]{Kn3} 
and also follows from a more general result due to Fitzgerald 
\cite[Th. 1.6]{F} which essentially can be stated as saying
that if an anisotropic form $\phi$ over $F$ becomes hyperbolic
over $F(q)$ for some form $q$ over $F$, and if $q$ is in a certain
sense ``big enough'' compared with $\phi$, then $\phi$ is similar
to a Pfister form.
 

The plan of the papers is as follows.  The next three sections provide
an account of some basic facts on quadratic forms in characteristic $2$
and list some of the deeper results which we will use in our proofs.
Since general references to facts about singular forms are relatively
sparse in the literature, we decided to include, sometimes in quite some
detail, rather basic and elementary results which might hitherto have gone
unnoticed or which might belong to mathematical folklore and for which
we did not find a suitable reference.   

Section \ref{section-witt}
deals with Witt cancellation and Witt decomposition which yield
a notion of Witt equivalence based on comparing anisotropic parts of
(possibly singular) forms.
 
Section \ref{section-domination} deals with the notions of
subform (i.e. an orthogonal summand of another form) and of
dominated form (i.e., the restriction of
a form to a sub-vector space of the underlying vector space of another form),
as well as the notion of a nonsingular completion of a singular form
(i.e. a nonsingular form of minimal dimension dominating a given
singular form).  These notions of domination and nonsingular completion
plus our notion of Witt equivalence provide very effective tools 
(in particular, the ``completion lemma'' \ref{regcompl})
in order to circumnavigate the use of Witt cancellation in the original
proofs of Knebusch's and Fitzgerald's theorems in characteristic $\neq 2$.
The main result of section \ref{section-domination} is Theorem
\ref{commondomform} which is the most general version in
characteristic $2$ of determining
the maximal dimension of a common dominated form of
two forms in terms of the dimensions of totally isotropic subspaces
of these two forms and of their orthogonal difference.

In section \ref{section-results}, we give definitions of various notions and
objects which will appear in the statements and proofs of our results 
(Pfister form, function field, splitting tower, degree, Arf-invariant and 
Clifford algebra, etc.), and we will state several deep results used in the 
proofs, some of them rather recent (e.g. \cite{AB1}, \cite{AB2}, \cite{HL}).

Section \ref{section-fitzgerald} is almost entirely devoted to the proof of 
Fitzgerald's theorem in characteristic $2$.  Our version of the
theorem was formulated so that it
also includes the case of singular forms (dominated by nonsingular forms) 
and their function fields.

Fitzgerald's theorem is then applied in section \ref{section-knebusch}
to prove a characteristic $2$
version of Knebusch's theorem.  It should be remarked that Knebusch's
theorem carries over to characteristic $2$ for nonsingular forms
without much difficulty, but we construct various types of
counterexamples showing that it fails in general for singular forms.
However, we conjecture that if the dimension $s$ of the radical of an
anisotropic form
$q$ over $F$  is less than $\dim q/2$, then Knebusch's theorem holds, and
we prove this conjecture for $s\leq 4$ and for a particular case when
$s=5$ (Theorem \ref{knebusch}). 

As an application, we classify in section \ref{section-height1} forms of 
nondefective height $1$ (i.e., forms which are not totally singular but 
whose anisotropic parts over their own function fields are totally singular) 
provided they satisfy the same hypotheses on the radical as the forms in our 
version of Knebusch's theorem (Theorem \ref{ndh1-s<6}).

The final section \ref{section-ts} stands somewhat apart as it deals 
almost  exclusively with totally singular forms.   This section can be 
read independently of the others (perhaps with the exception of Theorem 
\ref{heightrepeat}), but two or three of its results will be used in 
earlier sections (and might be taken for granted at that moment).  
It turns out that one can develop a meaningful theory of totally 
singular quadratic forms in characteristic $2$ by almost altogether 
ignoring the theory of nonsingular resp. non-totally singular forms.   
In particular, one has notions of forms which we call quasi-Pfister 
forms and quasi-Pfister neighbors with properties which
resemble in a surprising way many of the better known properties
of ``classical'' Pfister forms and Pfister neighbors.  Since
Pfister forms and neighbors are part of the main theme of the present
paper, it is for this reason too that we decided to include this section 
on totally singular forms. It should be noted that some of the results in 
this last section or variations of these results have first been shown by 
the second author in \cite{La3} where also other aspects of totally singular
forms have been investigated. Our approach here, however, is different
and based on the notion of what we call the norm field of a totally
singular form.  This norm field is a totally inseparable multiquadratic
extension of the field of squares of the base field which is in a natural 
way associated to a totally singular form and which encodes various 
properties of that form.  As an application of these techniques, we give in
Theorem \ref{heightrepeat} (resp. Theorem \ref{height}) an 
interpretation of the standard height (as defined in \cite{La2}) of an 
anisotropic form (not necessarily totally singular) as
the maximal height of a field tower such that the dimensions of the anisotropic 
parts of this form over the fields along this tower are strictly decreasing.


For all basic results on forms in characteristic $2$ which we mention 
without reference,
we refer  the reader to \cite{B1}, \cite{B2}, \cite{B3}, and sections 
1.4 and 2.4 in \cite{Pf}.


\section{Witt cancellation and Witt decomposition}\label{section-witt}
Let $F$ be a field of characteristic $2$.  A quadratic form over $F$
is a pair $(V,q)$ of an $F$-vector space $V$ and a function
$q\,:\,V\to F$ such that
\begin{enumerate}
\item $q(ax)=a^2q(x)$ for all $a\in F$ and all $x\in V$;
\item There exists a symmetric bilinear form $b_q\,:\,V\times V\to F$
such that $q(x+y)-q(x)-q(y)=b_q(x,y)$ for all $x,y\in V$.
\end{enumerate}
It is well-known that there exists a basis $e_i,f_i,g_j$ and
elements $a_i,b_i,c_j\in F$,  $1\leq i\leq r$,
$1\leq j\leq s$ of $V$ with $\dim V=2r+s$ such that
for all $v=\sum_{i=1}^r(x_ie_i+y_if_i)+\sum_{j=1}^sz_jg_j$ one
has 
$$q(v)=\sum_{i=1}^r(a_ix_i^2+x_iy_i+b_iy_i^2)+\sum_{j=1}^sc_jz_j^2\ .$$
For a quadratic form given by such a polynomial we will write
$$q=[a_1,b_1]\perp [a_2,b_2]\perp \cdots \perp 
[a_r,b_r]\perp\qf{c_1,\cdots ,c_s}$$
for short.  If $q$ is written in this way, we will call it a 
{\it normalized} form.  We will often and implicitly use 
certain isometries which 
are rather obvious or easy to prove.  We will refer to them as
{\it standard relations} and they are the following~:
$$\begin{array}{c}\qf{a}\cong\qf{x^2a},\ [a,b]\cong 
[ax^2,bx^{-2}] \quad \mbox{for all $a,b\in F$, $x\in F^*$};\\
\qf{a,b}\cong\qf{a,a+b}\cong\qf{b,a},\ [a,b]\cong [a,a+b+1]\cong
[b,a] \quad \mbox{for all $a,b\in F$};\\
\mbox{$[a,b]\perp\qf{c}\cong [a+c,b]\perp\qf{c}$} \quad  
\mbox{for all $a,b,c\in F$};\\
\mbox{$[a,b]\perp [c,d]\cong [a+c,b]\perp [c,b+d]$}\quad 
\mbox{for all $a,b,c,d\in F$}.
\end{array}$$
  

 

\begin{remark} In many books on forms in characteristic $2$,
the notation $\qf{c_1,\cdots ,c_s}$ is used 
for the {\it bilinear form} given by 
$b(X,Y)=\sum_{i=1}^sc_ix_iy_i$ for $X=(x_1,\cdots ,x_s)$,
$Y=(y_1,\cdots ,y_s)$.  Since we consider almost exclusively quadratic
forms and in order to simplify notations,  we use the notation
$\qf{c_1,\cdots ,c_s}$ for the diagonal quadratic form $q(X)=b(X,X)$,
and we denote the bilinear form by $\qf{c_1,\cdots ,c_s}_b$ instead.
\end{remark}

Let $K/F$ be a field extension.  If $\phi$ is a form over $F$, then
we will denote by $\phi_K$ the form obtained after scalar extension
from $F$ to $K$.  We write $D_K(\phi )$ for the elements represented
nontrivially by $\phi$ over $K$, $D^*_K(\phi )=D_K(\phi)\cap K^*$, 
and $G_K(\phi )$ for the group of similarity factors
of $\phi$ over $K$, i.e. $G_K(\phi )=\{ a\in K^*\,|\,a\phi_K\cong\phi_K\}$.

The subspace of $V$ spanned by the $g_i$ as above is nothing but the
{\it radical} $\rad (q)=V^{\perp}$ of $q$, i.e. $\spn_{V/F}\{g_1,\cdots ,g_s\}=
\{ x\in V\,|\,b_q(x,y)=0\ \forall y\in V\}$.  
If $s>0$, then $q_r=[a_1,b_1]\perp [a_2,b_2]\perp \cdots \perp [a_r,b_r]$ 
is generally not 
determined uniquely up to isometry.  However, $q_s=\qf{c_1,\cdots ,c_s}$ 
is always determined uniquely up to isometry as
the radical $V^{\perp}$ is determined uniquely, and $q_s$ in the above 
representation will be called the {\it quasi-linear part} of $q$ and
denoted by $\ql (q)$. 
In particular, $r$ and
$s$ are determined uniquely and we say that $q$ is of type $(r,s)$.
We note that the type of $q$ does not change after passing to an
extension field.

$q$ is said to be {\it nonsingular} if $s=0$.  If $s\neq 0$ (and $r=0$)
then $q$ is said to be {\it singular} ({\it totally singular}).  $q$ will be 
called {\it nondefective} if $q$ is nonsingular or if $\ql (q)$ 
is anisotropic.


Consider another totally singular form $q'_s=\qf{c'_1,\cdots ,c'_s}$.
Then it is an easy exercise to show that $q_s\cong q'_s$ (i.e. the
two quadratic forms are isometric) if and only if
$\spn_{F/F^2}\{c_1,\cdots ,c_s\}=\spn_{F/F^2}\{c'_1,\cdots ,c'_s\}$,
i.e. the elements $c_i$ generate the same sub-vector space of
the vector space $F$ over the field $F^2$ as the elements $c'_i$.
In particular, if $\{ d_1,\cdots ,d_t\}$, $t\leq s$,
is a basis of $\spn_{F/F^2}\{c_1,\cdots ,c_s\}$, then
$q_s\cong \qf{d_1,\cdots ,d_t,0,\cdots,0}$, with $\qf{d_1,\cdots ,d_t}$
anisotropic. We call $\qf{d_1,\cdots ,d_t}$ the {\it anisotropic part}
of $q_s$.  This anisotropic part is, by the above, uniquely determined
up to isometry, and in fact we can choose the elements $d_i$
among the $c_i$.  We thus get :

\begin{lem}\label{exc} Let $q$ be a quadratic form over $F$ of type $(r,s)$.
Let $E$ be any field extension of $F$.  Then the anisotropic part of 
$\ql (q_E)$ is defined over $F$.
\end{lem}

\begin{proof}  Let $\ql (q)=\qf{c_1,\cdots ,c_s}$, $c_i\in F$.  
After reindexation, let
$\{c_1,\cdots ,c_t\}$, $t\leq s$, be a basis of 
$\spn_{E/E^2}\{c_1,\cdots ,c_s\}$.  Then the anisotropic part of
$\ql (q_E)$ is isometric to $\qf{c_1,\cdots ,c_t}_E$ and thus defined
over $F$.
\end{proof}

\begin{remark} 
The above observation, as elementary as it is, seems to have gone
to some extent unnoticed.  This is evidenced for  example in
\cite[1.3]{AJM}, where a proof is found for a rather special case
of the above lemma, namely for $E$ an inseparable quadratic extension
and $q$ totally singular.
\end{remark}

A hyperbolic plane $\HH$ is a nonsingular binary form isometric to $[0,0]$,
and a nonsingular form is said to be hyperbolic if it is isometric
to an orthogonal sum of hyperbolic planes.  One verifies readily
that if $q=q_r\perp q_s$ is a quadratic form with $q_r$ nonsingular
and $q_s$ totally singular and anisotropic, and if $q$ is isotropic, then
$q\cong [0,0]\perp q'_r\perp q_s$ with $q'_r$ nonsingular.

We thus get the following result which is an analogue in 
characteristic $2$ to the usual Witt decomposition in
characteristic $\neq 2$.

\begin{prop}\label{wd}
Let $q$ be a quadratic form over $F$.  Then $q\cong i\times \HH\perp
\tilde{q}_r\perp \tilde{q}_s\perp j\times\qf{0}$ with $\tilde{q}_r$
nonsingular, $\tilde{q}_s$  totally singular and $\tilde{q}_r\perp \tilde{q}_s$
anisotropic. The form $\tilde{q}_r\perp \tilde{q}_s$ is uniquely determined 
up to isometry.  In particular, $i$ and $j$ are uniquely determined. 
\end{prop}

In view of this proposition, we call $\tilde{q}_r\perp \tilde{q}_s$
the {\it anisotropic part} of $q$ and we write $\an{q}$ for short.
$i$ resp. $j$ will be called the {\it Witt index} resp. {\it defect}
of $q$, denoted by $i_W(q)=i$ and $i_{\ql}(q)=j$. 
Furthermore, $i\times \HH\perp\an{q}$ will be called
the {\it nondefective part} of $q$.  
For nondefective forms (the case $j=0$), this result can be found
in \cite[page 160]{Arf} or in \cite[page 283]{Kn1}.
Note that $i_W(q)+i_{\ql}(q)$ is exactly the dimension of any 
maximal totally isotropic subspace of $q$ (a subspace $W$ of
the underlying vector space $V$ of $q$ is called totally isotropic
if $q(w)=0$ for all $w\in W$).  Therefore, we call
$i_{\ti}(q)=i_W(q)+i_{\ql}(q)$ the {\it total isotropy index} of $q$.

We call two forms $\phi$ and $\psi$ {\it Witt-equivalent}, denoted by
$\phi\sim\psi$, if $\phi_{an}\cong\psi_{an}$.  If $\phi$ and $\psi$ are
nonsingular, then $\phi\sim\psi$ iff $\phi\perp -\psi$ is hyperbolic, the 
usual definition of Witt equivalence.  Thus, the notion of
Witt equivalence for nonsingular forms extends in a natural way
to forms which are not necessarily nonsingular.

To prove the proposition, we need the following result on Witt cancellation
in characteristic $2$ (cf. \cite[Prop. 1.2]{Kn1}).


\begin{prop}\label{wc1}  Let $q$ and $q'$ be quadratic forms of the same
dimension and let $\phi$ be a nonsingular quadratic form.  
If $q\perp\phi\cong q'\perp\phi$ then $q\cong q'$.
\end{prop}

We also need the following cancellation result.

\begin{lem}\label{wc2} Let $q$ and $q'$ be nondefective quadratic forms 
of the same dimension.  If $q\perp j\times\qf{0}\cong q'\perp j\times\qf{0}$,
then $q\cong q'$.
\end{lem}

\begin{proof} Write $q\cong m\times\HH\perp q_0$ and $q'\cong m'\times\HH
\perp q'_0$ with $q_0$ and $q'_0$ anisotropic and, say,  $m'\geq m$.
Put $n=m'-m$ and $q'_1=n\times\HH\perp q'_0$.
 By Witt cancellation for common nonsingular orthogonal summands
(Proposition \ref{wc1}), it suffices to show that for 
$\phi =q_0\perp j\times\qf{0}$ and 
$\phi'= q'_1\perp j\times\qf{0}$, we have that 
$\phi\cong\phi'$ implies $q_0\cong q'_1$.

Let $V=W\oplus U$ and $V'=W'\oplus U'$ be the underlying vector spaces
of $\phi$ and $\phi'$, respectively, such that 
$\phi|_W=q_0$, $\phi'|_{W'}=q'_1$, and
$\phi|_U=j\times\qf{0}=\phi'|_{U'}$.  

Let now $\sigma\,:\,V\to V'$ be a vector space isomorphism which
is an isometry of $\phi$ and $\phi'$, i.e. $\phi'(\sigma(x))=\phi (x)$
for all $x\in V$.  Let $\pi\,:\,V'=W'\oplus U'\to W'$ be the projection
onto $W'$, and define $\tau\,:\,W\to W'$ by $\tau =\pi\circ\sigma|_W$.

If $w\in W$ and $\sigma (w)=w'+u'$, $w'\in W'$, $u'\in U'$,
then $\tau (w)=w'$ and thus 
$$q'_1(\tau (w))=\phi'(\tau(w))=\phi'(w')=\phi'(w'+u')=\phi'(\sigma (w))=
\phi(w)=q_0(w)\ ,$$
where the third equality follows from the fact that 
$u'\in U'\subset V^{\prime\perp}$ and $\phi'(u')=0$.
To show that $\tau$ is an isometry, it suffices to show that $\tau$
is bijective.  Now if $0\neq w\in W$ then $q_0(w)\neq 0$ as $q_0$ is 
anisotropic.  But then $0\neq q'_1(\tau (w))=q_0(w)$, hence 
$\tau (w)\neq 0$ and $\tau$ is therefore injective.  The bijectivity
follows as $\dim W=\dim W'$.
\end{proof}




\begin{proof}[Proof of Proposition \ref{wd}]
We only have to prove uniqueness of the decomposition.
Now $\tilde{q}_s\perp j\times\qf{0}\cong\ql (q)$, and it follows
immediately that $j$ is uniquely determined and
so is $\tilde{q}_s$ (up to isometry).  So it suffices to show
that if $i\times\HH\perp\tilde{q}_r\perp \tilde{q}_s\perp j\times\qf{0}
\cong i'\times\HH\perp\tilde{q}'_r\perp\tilde{q}_s\perp j\times\qf{0}$ with
$\tilde{q}_r$, $\tilde{q}'_r$ nonsingular and
$\tilde{q}_r\perp\tilde{q}_s$, $\tilde{q}'_r\perp\tilde{q}_s$ anisotropic, 
then $\tilde{q}_r\perp \tilde{q}_s\cong\tilde{q}'_r\perp\tilde{q}_s$.

By Lemma \ref{wc2}, we have 
$i\times\HH\perp\tilde{q}_r\perp \tilde{q}_s\cong 
i'\times\HH\perp\tilde{q}'_r\perp\tilde{q}_s$.  But then it follows
immediately from Proposition \ref{wc1} that $i=i'$ and
$\tilde{q}_r\perp \tilde{q}_s\cong \tilde{q}'_r\perp\tilde{q}_s$.
\end{proof}



\section{Subforms and dominated forms}\label{section-domination}

\begin{lem}\label{dominate}  
\begin{enumerate}
\item[(i)] Let $(q,V)$ and $(\phi ,W)$ be quadratic
spaces over $F$.  Then the following are equivalent.
\begin{enumerate}
\item[(1)]  There exists an injective isometry $t:(q,V)\to (\phi,W)$
(i.e. $t$ is an injective $F$-linear map $V\to W$ with $\phi (t(x))=q(x)$
for all $x\in V$).
\item[(2)]  There exist nonsingular forms $q_r$ and $\tau$, 
nonnegative integers $s'\leq s\leq s''$, 
$c_i\in F$ ($1\leq i\leq s''$) and $d_j\in F$ ($1\leq j\leq s'$)
such that $q\cong q_r\perp\qf{c_1,\cdots ,c_s}$ and
$$\phi\cong q_r\perp\tau\perp [c_1,d_1]\perp\cdots\perp [c_{s'},d_{s'}]\perp
\qf{c_{s'+1},\cdots ,c_{s''}}\ .$$
\end{enumerate}
\item[(ii)]  Suppose that the equivalent conditions in {\em (i)} hold.
With the same notations, let $U=t(V)^{\perp}=\{ x\in W\,|\,b_{\phi}(x,t(V))=0\}$.  
Then $\phi|_{U}\cong \tau\perp\qf{ c_1,\cdots ,c_{s''}}$.
\end{enumerate}
\end{lem}
\begin{proof} Part (ii) follows readily from part (i) and is left to
the reader.  As for part (i),  the implication $(2)\Rightarrow (1)$ is clear.

To prove the converse, we may assume that $V\subset W$ and that $t$ is
the canonical injection, so that $\phi|_V=q$.  Write $V=U\oplus\rad(q)$.
We then have that $q_r:=\phi|_U=q|_U$ is nonsingular.  Thus, with
$U^{\perp}=\{ x\in W\,|\,b_{\phi}(x,U)=0\}$, we get that $W=U\oplus U^{\perp}$.
Clearly, $\rad(q)\subset U^{\perp}$.  We also have 
$U^{\perp}=U'\oplus\rad{\phi}$ with $\phi|_{U'}$ nonsingular.
We consider two cases.

First case : Suppose that $\rad(q)\subset\rad(\phi )$.  We choose any basis
of $\rad(q)$ and extend it to a basis of $\rad(\phi)$ such that, with respect
to this basis, we get
$q|_{\rad(q)}\cong\qf{c_1,\cdots ,c_s}$ and $\phi|_{\rad(\phi)}\cong
\qf{c_1,\cdots ,c_{s''}}$, where $c_i\in F$ and 
$0\leq s=\dim(\rad(q))\leq s''=\dim(\rad(\phi))$.  (2) follows
readily by putting $\tau =\phi|_{U'}$ and by noting that $V$ (resp. $W$) 
decomposes
orthogonally into  $V=U\oplus\rad(q)$ (resp. $W=U\oplus U'\oplus\rad(\phi)$).

Second case : Suppose that $\rad(q)\not\subset\rad(\phi)$.  Let
$\{ e_1,\cdots ,e_s\}$ be a basis of $\rad(q)$ such that 
$e_1\notin\rad(\phi )$.  With the same notations as before, we have that
$\rad(q)\subset U^{\perp}=U'\oplus\rad(\phi)$.  Hence, there exists
an $f_1\in U'$ such that $b_{\phi}(e_1,f_1)=\lambda_1\neq 0$.  In particular,
$f_1$ is linearly independent of $\{ e_1,\cdots ,e_s\}$.
Put $e_1'=\frac{1}{\lambda_1}e_1$ and, for $2\leq i\leq s$, put
$e_i'=e_i-\frac{\lambda_i}{\lambda_1}e_1$, where $\lambda_i=b_{\phi}(e_i,f_1)$.
Then $\{ e_1',\cdots ,e_s'\}$ is again a basis of $\rad(q)$, and
furthermore $b_{\phi}(e_i',f_1)=\delta_{i1}$.

Let $\widetilde{U}=U\oplus Fe_1'\oplus Ff_1$, let $c_1=q(e_1')=\phi (e_1')$
and $d_1=\phi (f_1)$.  Then $\phi|_{\widetilde{U}}\cong
\phi|_U\perp [c_1,d_1]$.   Let $\widetilde{V}=V\oplus Ff_1$ and let
$\widetilde{q}=\phi|_{\widetilde{V}}$.  It follows readily that
$\rad(\widetilde{q})$ has basis $\{ e_2',\cdots ,e_s'\}$ and that 
$\widetilde{V}$ has an orthogonal decomposition $\widetilde{V}=
\widetilde{U}\oplus\rad(\widetilde{q})$, where 
$\widetilde{q}|_{\widetilde{U}}=\phi|_{\widetilde{U}}$ is nonsingular.
In particular, $\dim(\rad(\widetilde{q}))<\dim(\rad(q))=s$.  (2) follows now
readily by induction on $s$ together with the first case.
\end{proof}

\begin{remark}\label{dom-rem}  The proof actually shows
that $q_r$ can be any nonsingular form with the property that
$q\cong q_r\perp\ql (q)$.
\end{remark}

\begin{coro}\label{nsc}  Let $(q,V)$ be a quadratic form of type
$(r,s)$ and let $(\phi ,W)$ be a nonsingular quadratic form of dimension
$2r+2s$.  The following are equivalent.
\begin{enumerate}
\item[(1)] There exists an injective isometry $t:(q,V)\to (\phi,W)$.
\item[(2)] There exist a nonsingular form $q_r$ of dimension $2r$ and 
$c_j,d_j\in F$, $1\leq j\leq s$, such that for
$\sigma =\qf{c_1,\cdots ,c_s}$ and $\rho =[c_1,d_1]\perp\cdots\perp [c_s,d_s]$,
we have $q\cong q_r\perp\sigma$ and $\phi\cong q_r\perp\rho$.
\end{enumerate}
\end{coro}

\begin{defi}  Let $q$ and $\phi$ be forms over $F$.
\begin{enumerate}
\item[(i)] $q$ is said to be a {\it subform} of $\phi$ if there exists
a form $\tau$ such that $\phi\cong q\perp\tau$.  We then write
$q\subset\phi$.
\item[(ii)] $q$ is said to be {\it dominated} by $\phi$ if the
equivalent conditions in Lemma \ref{dominate} hold.  We then write
$q\prec\phi$ and call $\phi$ a dominant form of $q$.  With the same
notations as in Lemma \ref{dominate}, the form $\phi|_U$
is called the {\it complement}
of $q$ in $\phi$, and we write $\phi|_U=q^{C}_\phi$.
\item[(iii)]  $\phi$ is said to be a {\it nonsingular completion} ({\it n.s.c.})
of $q$ if the equivalent conditions in Corollary \ref{nsc} hold.
\end{enumerate}
\end{defi}

\begin{lem}\label{nsc-coefficients}  
Let $c_i,c_i',d_i\in F$, $1\leq i\leq n$, and suppose
that $\qf{c_1,\cdots ,c_n}\cong\qf{c_1',\cdots ,c_n'}$.  Then there
exist $d_1',\cdots ,d_n'\in F$ such that
$[c_1,d_1]\perp\cdots \perp [c_n,d_n]\cong [c_1',d_1']\perp\cdots\perp
[c_n',d_n']$.
\end{lem}

\begin{proof}  This follows readily from the standard relations
and the fact that by assumption, $\{c_1,\cdots ,c_n\}$ and
$\{c_1',\cdots ,c_n'\}$ span the same $F^2$-vector space.
\end{proof}

\begin{remark} \label{rem-dominate} (i) The notion of 
domination was introduced in \cite{La1}, \cite{LM} using condition (2) in Lemma 
\ref{dominate} as definition, but under the somewhat more restrictive
and less natural assumption that the $c_i$ are already given. If the dominating 
form is nonsingular (the most 
important case), then this notion of domination  
coincides with the one given here as follows immediately
from Lemma \ref{nsc-coefficients}.  In general, however, our notion
is less restrictive.  For example, let $F=K(X)$ be the rational function
field in one variable over the field $K$.  Then, according to our
definition, the form $\qf{1,x}$ is dominated by $\HH\perp\qf{1+x}$
as $\qf{1,x}\cong\qf{1,1+x}$, but not according to the definition
in \cite{La1}, \cite{LM} as $1+x\not\equiv 1,x\bmod F^{*2}$.     

Condition (2) is useful for practical purposes, e.g. 
when doing explicit computations with coefficients in a normalized form.
However, condition (1) is more natural and conceptual.

(ii) A subform of another form is obviously dominated by this other 
form. The notions of subform and dominated form coincide if either the
dominated form is nonsingular, or if both forms are totally singular.

(iii)  The notion of a complement has also been defined by
Knebusch \cite{Kn4}, but only in the case where the dominant form
is nonsingular.  In this situation and with above notations, 
Knebusch calls the complement $\phi|_U$ the polar of $\phi$
with respect to the subspace $U$.

(iv)  Assuming that $q\prec\phi$ and with the
same notations as in Lemma \ref{dominate}, we have that
$q^c_{\phi}\cong \tau\perp\qf{c_1,\cdots ,c_{s''}}$.
\end{remark}



The following easy lemma is quite useful and will be used frequently.
\begin{lem}\label{easy}
Let $q\prec\phi$. Then $q\perp\phi\sim q^c_{\phi}$.  In particular,
if $\sigma$ is a totally singular form over $F$ and  $\rho$ is a 
nonsingular completion of $\sigma$, then
$\rho\perp\sigma\sim\sigma$.
\end{lem}
\begin{proof}  This follows readily from the definitions of the complement
and  of a nonsingular 
completion, together with Remark \ref{rem-dominate}(iv) and the fact that 
$[c,d]\perp\qf{c}\sim\qf{c}\sim\qf{c,c}$.
\end{proof}

\begin{remark}  If $\phi$ is anisotropic and $q\prec\phi$, then for another
form $\gamma$ we have
$\gamma\cong q^c_{\phi}$ iff $q\perp\phi\sim\gamma$ and $\gamma$ 
is anisotropic.

If we assume neither anisotropy nor nondefectiveness for $\phi$, we need to 
invoke other hypotheses to get a similar result.
Suppose that $q\prec\phi'\subset\phi$ with $\phi'$
nonsingular.  Let $\gamma$ be another form.  Then $\gamma\cong q^c_{\phi}$
iff $q\perp\phi\sim\gamma$, $\dim\phi =\dim\gamma +\dim\ q$,
$\dim\rad(\gamma )=\dim\rad(\phi)+\dim\rad(q)$.  Indeed, we have
$q\perp\phi\sim q^c_{\phi}$ by the above Lemma, hence $\gamma\sim q^c_{\phi}$, 
and the conditions
on the dimensions plus the fact that $q$ is dominated by a nonsingular
form inside $\phi$ guarantee that $q^c_{\phi}$ and $\gamma$ are
of the same type. 
\end{remark} 





Suppose that $\phi$ and $\psi$ are nonsingular, $\sigma$ is totally singular,
and $\rho$ is a nonsingular completion of $\sigma$.  Then, in general,
$\phi\perp\sigma\cong\psi\perp\sigma$ does not imply $\phi\perp\rho
\cong\psi\perp\rho$.  For example, suppose that $[a,b]$ is anisotropic.
If we put $\phi =\rho =[a,b]$, $\psi =[0,0]$ and $\sigma =\qf{a}$,
then $\rho$ is a nonsingular completion of $\sigma$ and we clearly have 
$\phi\perp\sigma\cong\psi\perp\sigma$.  Now 
$\phi\perp\rho\cong\HH\perp\HH$ is hyperbolic, but $\psi\perp\rho\cong
\HH\perp [a,b]$ isn't.  However, we do have the following result which we will
use in section \ref{section-fitzgerald}.

\begin{lem}\label{regcompl}  Let $\phi$, $\psi$ be nonsingular, and let
$\sigma$  be totally singular.  
Let $\rho$ be a fixed nonsingular completion of
$\sigma$.   If $\phi\perp\sigma\cong\psi\perp\sigma$, then there exists
a nonsingular completion $\rho'$ of $\sigma$ such that 
$\phi\perp\rho\cong\psi\perp\rho'$.
\end{lem}
\begin{proof}  Let $W$ (resp. $V$) be the underlying vector space of
$\phi\perp\sigma$ (resp. $\psi\perp\sigma$) with corresponding orthogonal
decomposition $W=W_{\phi}\oplus W_{\sigma}$ 
(resp. $V=V_{\psi}\oplus V_{\sigma}$).  Let $W'\supset W$ be the underlying
vector space of $\phi\perp\rho$ such that $(\phi\perp\rho)|_W=\phi\perp\sigma$,
and let $W'=W_{\phi}\oplus W_{\rho}$ be the corresponding orthogonal
decomposition with $W_{\sigma}\subset W_{\rho}$.  The isometry
$\psi\perp\sigma\cong\phi\perp\sigma$ yields an injective isometry
$t:(\psi\perp\sigma ,V)\to (\phi\perp\rho ,W')$. The result now follows
readily by noting that there is an orthogonal decomposition
$W=t(V_{\psi})\oplus t(V_{\sigma})$ with $t(V_{\sigma})=W_{\sigma}$
(an isometry maps  radical onto radical) and  
$(\phi\perp\sigma )|_{t(V_{\psi})}\cong\psi$, and by applying 
Corollary \ref{nsc} together with Remark \ref{dom-rem}. 
\end{proof}

\begin{coro} Let $\phi$, $\psi$ be nonsingular, and let
$\sigma$, $\tau$  be totally singular.  
Let $\rho$ be a fixed nonsingular completion of
$\sigma$.   If $\phi\perp\sigma\perp\tau\cong\psi\perp\sigma\perp\tau$, 
then there exists
a nonsingular completion $\rho'$ of $\sigma$ such that 
$\phi\perp\rho\perp\tau\cong\psi\perp\rho'\perp\tau$.
\end{coro}

\begin{proof} 
Let $\eta$ be any n.s.c. of $\tau$. By Lemma \ref{regcompl} and
Lemma \ref{nsc-coefficients}, there exist n.s.c.'s
$\rho'$ of $\sigma$ and $\eta'$ of $\tau$ such that
$\phi\perp\rho\perp\eta\cong\psi\perp\rho'\perp\eta'$.
Adding $\tau$ on both sides and using $\eta\perp\tau\sim\eta'\perp\tau
\sim\tau$ (Lemma \ref{easy}), we get 
$\phi\perp\rho\perp\tau\sim\psi\perp\rho'\perp\tau$. Since
both forms are of the same type, we get
$\phi\perp\rho\perp\tau\cong\psi\perp\rho'\perp\tau$.
\end{proof}

It is known in characteristic $\neq 2$ that if the 
orthogonal difference of two nonsingular forms of dimension $\geq m$
has Witt index $\geq m$, then these two forms have isometric
subforms of dimension $m$.  This can be generalized to
characteristic $2$.

\begin{prop}\label{commondomform}  Let $\phi$ and $\psi$ be forms
over $F$ and let 
$n(\phi,\psi)=i_{\ti}(\phi\perp\psi )-\max\{ i_{\ql}(\phi ),i_{\ql}(\psi )\}$.
\begin{enumerate}
\item[(i)] Let $m\leq\min\{\dim\phi ,\dim\psi , n(\phi,\psi)\}$.
Then there exists a form $q$ of dimension $m$ with
$q\prec\phi$, $q\prec\psi$.
\item[(ii)]  Let $q$ be a form with $q\prec\phi$, $q\prec\psi$.  Then
$\dim q\leq n(\phi,\psi)$.
\item[(iii)]  $\min\{\dim\phi ,\dim\psi , n(\phi,\psi)\}$ is the maximal
dimension of forms $q$ with $q\prec\phi$, $q\prec\psi$.
\end{enumerate}
\end{prop}


\begin{proof}
(iii) follows readily from (i) and (ii).

\medskip

{\it Proof of} (i). First, we reduce the statement to the case where one 
of the two
forms is nondefective.  Write $\phi\cong\phi_{\nd}\perp r\times\qf{0}$
and $\psi\cong\psi_{\nd}\perp s\times\qf{0}$ with $\phi_{\nd}$, $\psi_{\nd}$
nondefective, and assume $r\leq s$.  Put $\phi'=\phi_{\nd}$ and
$\psi'=\psi_{\nd}\perp (s-r)\times\qf{0}$.  
We have $\max\{ i_{\ql}(\phi ),i_{\ql}(\psi )\}=s$,
$\max\{ i_{\ql}(\phi' ),i_{\ql}(\psi' )\}=s-r$ and
$i_{\ti}(\phi\perp\psi )=i_{\ti}(\phi'\perp\psi' )+2r$.  Hence,
$$m-r\leq\min\{\dim\phi' ,\dim\psi' , n(\phi',\psi')\}\ .$$
Assuming our result to be true in the case where one of the
forms is nondefective, we conclude that there exists a form
$q'$ with $\dim q'=m-r$ such that $q'\prec\phi'$ and $q'\prec\psi'$.
The form $q=q'\perp r\times\qf{0}$ is the desired form.

So let us from now on assume that $\phi$ is nondefective and that
$\psi\cong\psi_{\nd}\perp s\times\qf{0}$ with $\psi_{\nd}$ nondefective.
We then have 
$$i_{\ti}(\phi\perp\psi )-\max\{ i_{\ql}(\phi ),i_{\ql}(\psi )\}
=i_{\ti}(\phi\perp\psi )-s=i_{\ti}(\phi\perp\psi_{\nd} )\ .$$

We use induction on $m$.  
If $m=0$ there is nothing to show.
So suppose $m>0$.

\smallskip  

Assume that $\phi$ and $\psi$ have a
common nonsingular subform $\alpha$ with $\dim\alpha >0$. Then  
$\phi\cong\phi'\perp\alpha$,
$\psi\cong\psi'\perp\alpha$ and $i_{\ql}(\psi')=i_{\ql}(\psi)=s$.
 We now use the induction hypothesis
applied to $\phi'\perp\psi'$.  Indeed, Witt cancellation yields
$i_{\ti}(\phi'\perp\psi')-s\geq m-\dim\alpha$,
and $\min\{\dim\phi',\dim\psi',m-\dim\alpha\}=m-\dim\alpha$ then shows the
existence of a form $q'$ of dimension $m-\dim\alpha$ with
$q'\prec\phi'$, $q'\prec\psi'$.  Thus,  $q\cong q'\perp\alpha$
is the desired form.

Therefore, we may assume that $\phi$ and $\psi$ do not contain
a common nonsingular subform.  Applied to the form $\HH$, this implies
in particular that $\phi$ or $\psi_{\nd}$ is anisotropic.

\smallskip

Suppose that $\phi$ is isotropic.  Then, by the above,
we may assume that $\psi_{\nd}$ is anisotropic.
The nondefectiveness implies that  $\phi\cong\HH\perp\phi'$
and  we get  $i_{\ti}(\phi'\perp\psi)-s\geq m-1$

By induction, there
exists a totally singular form $q'$ such that  
$\dim q' =m'=\min\{\dim\phi',\dim\psi, m-1\}$ and 
$q'\prec\phi'$, $q'\prec\psi$.  Since by assumption
$m\leq \dim\phi$, $\dim\psi$ and $\dim\phi'=\dim\phi -2$,
there are two cases :
$m'=m-1$ (if $\dim\phi\geq m+1$) and $m'=m-2$ (if $\dim\phi =m$).

Suppose 
$m'=m-1$.  In this case, we can write
$q'\cong\beta\perp\gamma$ with $\beta$, $\gamma$ totally singular, 
$\psi\cong\beta_1\perp\gamma\perp\delta$ with
$\beta_1$ a nonsingular completion of $\beta$ and some form $\delta$.
Now $\dim\psi >m'$ implies that $\dim\beta_1+\dim\delta\geq 1$.
If $\dim\delta\geq 1$, then for any $a\in D_F(\delta)$
we have $\qf{a}\prec\HH$ as $D_F(\HH )=F$,
and hence $q'\perp\qf{a}\prec\phi'\perp\HH\cong\phi$ and
$q'\perp\qf{a}\prec\psi$.  If $\dim\beta_1\geq 1$, then
there exist $b,d\in F$ such that $\qf{b}\subset\beta$,
and $[b,d]\subset\beta_1$.  Using the
fact that $[b,d]\perp\qf{b}\cong\HH\perp \qf{b}\prec\phi$,
we now conclude that $[b,d]\subset\phi$ and
$[b,d]\subset\psi$, a contradiction to our
hypothesis of the nonexistence of common nonsingular
subforms.  

If $m'=\dim\phi'=m-2$, then $\dim\psi_{\nd}+m-1>\dim (\phi'\perp\psi_{\nd})$,
hence, for dimension reasons,  the underlying subspace of $\psi_{\nd}$
inside the underlying vector space of $\phi'\perp\psi_{\nd}$ 
intersects with the totally isotropic subspace of dimension
$i_{\ti}(\phi'\perp\psi_{\nd})=i_{\ti}(\phi'\perp\psi)-s\geq m-1$, which
in turn yields that $\psi_{\nd}$ is isotropic, which by nondefectiveness 
implies $\HH\subset\psi_{\nd}$, again a contradiction because
by assumption $\HH\subset\phi$ as well.

The same reasoning can be applied when
$\psi_{\nd}$ is isotropic.  In this case we write
$\psi\cong\HH\perp\psi'$, we simply replace $\phi'$ by $\psi'$,
$\psi$ and $\psi_{nd}$ by $\phi$ in the above argument and we note 
that $i_{\ti}(\psi'\perp\phi )\geq m-1+s\geq m-1$.

\smallskip

Thus, we may finally assume that $\phi$ and $\psi_{\nd}$ are anisotropic,
and that $\phi$ and $\psi$ do not contain common nonsingular subforms.

Let $V$ and $W$ be the underlying vector spaces of $\phi$ and $\psi$,
respectively.  By induction hypothesis, there exist subspaces
$V'\subset V$ and $W'\subset W$ with $\dim V'=\dim W'=m-1$
such that $q'\cong \phi|_{V'}\cong \psi|_{W'}$ with $q'$ totally
singular.  Thus, there exist orthogonal bases $\{ e_1,\cdots ,e_{m-1}\}$ and 
$\{ f_1,\cdots ,f_{m-1}\}$
of $V'$ and $W'$, respectively, with $\phi (e_i)=\psi (f_i)=c_i\in F^*$.
In particular, $\{ e_1+f_1,\cdots ,e_{m-1}+f_{m-1}\}$ is a 
basis of a totally isotropic subspace $U$ of $(\phi\perp\psi ,V\oplus W)$.

Let $W=W_{\nd}\oplus W_0$ with $\psi|_{W_{\nd}}\cong\psi_{\nd}$
and $W_0\subset\rad (\psi)$, $\psi|_{W_0}\cong s\times\qf{0}$.
We fix a basis $\{ g_1,\cdots ,g_s\}$ of $W_0$.
The anisotropy of $\phi$ implies that $q'$ is anisotropic, hence
$W'\cap W_0=\{ 0\}$.  In particular, we have a direct sum of 
sub-vector spaces $V'\oplus W'\oplus W_0$ of $V\oplus W$ and therefore
also a direct sum $U\oplus W_0$. 
Clearly,  $U\oplus W_0$ is a totally isotropic subspace
of $(\phi\perp\psi ,V\oplus W)$ of dimension $m-1+s$. 
This totally isotropic subspace is contained in a maximal totally
isotropic subspace of $(\phi\perp\psi ,V\oplus W)$ which, by assumption, 
is of dimension $\geq m+s$.

Hence, there exist $v\in V$ and $w\in W$ such that $v+w\notin U\oplus W_0$
and such that
$\{ e_1+f_1,\cdots ,e_{m-1}+f_{m-1}, g_1,\cdots ,g_s, v+w\}$ is a basis for
a totally isotropic subspace of dimension $m+s$.  In particular,
$\phi (v)+\psi (w)=0$.  Suppose that $\phi (v)=0$.  The anisotropy
of $\phi$ implies that $v=0$.  In particular, $w\neq 0$ and
$\psi (w)=0$, and we have that $\{ g_1,\cdots ,g_s,w\}$ is in
fact a totally isotropic subspace of $(\psi ,W)$ of dimension
$s+1$.  But we assumed that $\psi\cong\psi_{\nd}\perp s\times\qf{0}$
with $\psi_{\nd}$ anisotropic, which in particular yields that
$i_{\ti}(\psi)=s<s+1$, a contradiction.  Therefore,
$v\neq 0$ and the anisotropy of $\phi$, implies
$a=\phi (v)=\psi (w)\neq 0$.  Note also that 
$0=b_{\phi\perp\psi}(e_i+f_i,v+w)=b_\phi(e_i,v)+b_\psi(f_i,w)$
and therefore, $b_\phi(e_i,v)=b_\psi(f_i,w)$.

Suppose $v\in V'$.  Then there exists $w'\in W'$ with
$a=\phi (v)=\psi (w')=\psi (w)$.  Also, $b_\phi(e_i,v)=0$
for all $i$ and hence $b_\psi(f_i,w)=0$ for all $i$
and thus  $b_\psi (w',w)=0$.  We also clearly have $b_\psi (g_i,w)=0$ for
all $g_i$.  Suppose now that 
$w\notin W'\oplus W_0$. The above then shows that  the quadratic 
subspace of $(\psi ,W)$ spanned by $\{ w,w',g_1,\cdots ,g_s\}$ 
is isometric to $\qf{a,a}\perp s\times\qf{0}\cong\qf{a}\perp
(s+1)\times\qf{0}$, again a contradiction to the fact that
$i_{\ti}(\psi )=s$.  Hence, $w\in W'\oplus W_0$.  Let $v=\sum_{i=1}^{m-1}a_ie_i$
and $w=\sum_{i=1}^{m-1}b_if_i+\sum_{j=1}^sc_jg_j$, $a_i,b_i,c_i\in F$.
Then $\sum_{i=1}^{m-1}a_i^2\phi (e_i)=\phi (v)=\psi (w)=
\sum_{i=1}^{m-1}b_i^2\psi (f_i)$. Now $\phi (e_i)=\psi (f_i)=c_i$ 
and the fact that
$\phi|_{V'}$ and $\psi|_{W'}$ are anisotropic totally singular
implies that the $c_i$ are $F^2$-linearly independent, hence
$a_i^2=b_i^2$ for all $1\leq i\leq m-1$ and thus $a_i=b_i$.  
In particular, $v+w=\sum_{i=1}^{m-1}a_i(e_i+f_i)+\sum_{j=1}^sc_jg_j\in 
U\oplus W_0$, a contradiction.  

We conclude that $v\notin V'$.
A similar argument as above shows that $w\notin W'$.  
Let $\tilde{V}=
V'\oplus Fv$ and $\tilde{W}=W'\oplus Fw$.  One verifies
readily that $\phi|_{\tilde{V}}\cong\psi|_{\tilde{W}}$.
$q\cong\phi|_{\tilde{V}}$ is then the desired form of dimension $m$.

\medskip

{\it Proof of} (ii).  We may assume that $i_{\ql}(\psi)=s\geq r=i_{\ql}(\phi)$.
Suppose there exists a form $q$ with $q\prec\phi$, $q\prec\psi$ and
$\dim q>n(\phi,\psi)=i_{\ti}(\phi\perp\psi)-s$.

First, we reduce to the case where $q$ is totally singular.
If $q$ is not totally singular, then we can write $q\cong \rho\perp\sigma$
with $\rho$ nonsingular of dimension $\geq 2$ and $\sigma =\ql (q)$.
Now $q\prec\phi$, $q\prec\psi$ and the nonsingularity of $\rho$ readily 
imply that $\phi\cong\rho\perp\phi'$, $\psi\cong\rho\perp\psi'$ for 
certain forms $\phi'$, $\psi'$ with $\sigma\prec\phi'$, $\sigma\prec\psi'$.
We have $\rho\perp\rho\cong (\dim\rho)\times\HH$.  Thus,
$$\dim\sigma =\dim q-\dim\rho > i_{\ti}(\phi\perp\psi)-s-\dim\rho=
i_{\ti}(\phi'\perp\psi') -s\ .$$
We also have $i_{\ql}(\phi')=i_{\ql}(\phi)=r\leq s=i_{\ql}(\psi)=
i_{\ql}(\psi')$ as the quasi-linear parts of $\phi$ (resp. $\psi$) and 
$\phi'$ (resp. $\psi'$) are the same.

Hence, we get $\sigma\prec\phi'$, $\sigma\prec\psi'$, and 
$\dim\sigma >n(\phi',\psi')$ with $\sigma$ totally singular.  
Thus, to get a contradiction,
it suffices to consider the case where $q$ is totally singular.

\smallskip

So let us assume that $q$ is totally singular and, say, $\dim q=k$.
Then $q\prec\phi$ implies that there exist an integer $\ell$,
$0\leq \ell\leq k$, $c_1,\cdots ,c_k\in F$, $a_1,\cdots ,a_{\ell}\in F$,
and a form $\alpha$ such that
$q\cong\qf{c_1,\cdots ,c_k}$ and
$$\phi\cong \alpha\perp [a_1,c_1]\perp\cdots\perp [a_{\ell},c_{\ell}]\perp
\qf{c_{\ell +1},\cdots ,c_k}\ .$$
Using the standard relations, we can rewrite
$[a_1,c_1]\perp\cdots\perp [a_{\ell},c_{\ell}]\perp
\qf{c_{\ell +1},\cdots ,c_k}$ in such a way that we may assume
that there exist $r_1,r_2\geq 0$ with $\ell\geq r_1$, $k-\ell\geq r_2$
such that 
$$c_1=\cdots =c_{r_1}=c_{k-r_2+1}=\cdots = c_k=0$$
and such that $q'=\qf{c_{r_1+1},\cdots ,c_{k-r_2}}$ is anisotropic.
It follows that we can write $q\cong q'\perp (t\times\qf{0})$ with $q'$
anisotropic,
$$\phi\cong \phi'\perp (r_1\times\HH)\perp (r_2\times\qf{0})$$
with $q'\prec\phi'$ and $t=r_1+r_2$.

Using the fact that the anisotropic part $q'$ of $q$ is determined
uniquely, we obtain in a similar way that there exist $s_1,s_2\geq 0$
with $s_1+s_2=t$ such that 
$$\psi\cong\psi'\perp (s_1\times\HH)\perp (s_2\times\qf{0})$$
and $q'\prec\psi'$.  In fact, the anisotropy of $q'$
implies that $q'$ is dominated by the nondefective parts
of $\phi$ and $\psi$, i.e. we have $q'\prec\phi'_{\nd},\psi'_{\nd}$.

By assumption, $\dim q'+t=\dim q >i_{\ti}(\phi\perp\psi)-s$. We have
$$i_{\ti}(\phi\perp\psi)=i_{\ti}(\phi'\perp\psi')+r_1+r_2+s_1+s_2=
i_{\ti}(\phi'\perp\psi')+2t\ .$$
Also, $i_{\ql}(\phi')=r-r_2$, $i_{\ql}(\psi')=s-s_2$.  Therefore,
$$i_{\ti}(\phi'\perp\psi')=
i_{\ti}(\phi'_{\nd}\perp\psi'_{\nd})+(r-r_2)+(s-s_2)\ .$$
Thus,
$$\begin{array}{rcl}
\dim q' & > & i_{\ti}(\phi'_{\nd}\perp\psi'_{\nd}) +t + 
(r-r_2)+(s-s_2) -s\\
        &  &  \mbox{}\quad= i_{\ti}(\phi'_{\nd}\perp\psi'_{\nd})+
        s_1+s_2+(r-r_2)-s_2\\
        &  &  \mbox{}\quad= i_{\ti}(\phi'_{\nd}\perp\psi'_{\nd})+
        s_1+(r-r_2)\\
        &  &  \mbox{}\quad\geq i_{\ti}(\phi'_{\nd}\perp\psi'_{\nd})\ .
        \end{array}$$
Now $q'\perp q'\cong q'\perp (\dim q'\times\qf{0})\prec
\phi'_{\nd}\perp\psi'_{\nd}$.  Hence $i_{\ti}(\phi'_{\nd}\perp\psi'_{\nd})
\geq \dim q'$, a contradiction.
\end{proof}

\begin{coro}\label{commondomform1} Let $\phi$ and $\psi$ be nondefective 
forms and let
$q$ be a form of maximal dimension such that $q\prec\phi$, $q\prec\psi$.
Then $\dim q=\min\{\dim\phi ,\dim\psi,i_{\ti}(\phi\perp\psi )\}$ and
$q$ is nondefective.
\end{coro}

\begin{proof}  The statement about $\dim q$ follows immediately
from the previous proposition by noting that
nondefectiveness of $\phi$, $\psi$ implies 
$n(\phi,\psi)=i_{\ti}(\phi\perp\psi )$.

Suppose that $\eta\prec\phi$, $\eta\prec\psi$ and that $\eta$ is defective,
say, $\eta\cong \eta'\perp (t\times\qf{0})$ with $\eta'$ nondefective and 
$t>0$. Similar arguments as in the proof of part (ii) of the previous
proposition together with the fact that $\phi$, $\psi$ are 
nondefective imply that we can write $\phi\cong\phi'\perp (t\times\HH)$,
$\psi\cong\psi'\perp (t\times\HH)$ with $\eta'\prec\phi'$, $\eta'\prec\psi'$.
For $\tilde{\eta}=\eta'\perp (t\times\HH)$, we thus get 
$\tilde{\eta}\prec\phi$, $\tilde{\eta}\prec\psi$ and 
$\dim\tilde{\eta}=\dim\eta+t>\dim\eta$.
This implies the nondefectiveness of $q$.
\end{proof} 
 
 
\section{Further notions and results needed in proofs}\label{section-results}
\subsection{Pfister forms and quasi-Pfister forms}\label{subsection-pfister}
Similar to the case of characteristic $\neq 2$, one can construct
the Witt ring $WF$ of nondegenerate symmetric bilinear forms over $F$ and
the Witt group $W_qF$ of nonsingular quadratic forms over $F$.
$W_qF$ can be considered in a natural way as a $WF$-module
by the operation $b\otimes q(x\otimes y)=b(x,x)q(y)$.


An $n$-fold bilinear Pfister form is a form of type
$b=\qf{1,a_1}_b\otimes\cdots\otimes\qf{1,a_n}_b$ for some $a_i\in F^*$,
we write $\pff{a_1,\cdots ,a_n}_b$ for short.  If $q(X)=b(X,X)$
is the corresponding totally singular (hence quasi-linear) quadratic form,  
we
write $q=\pff{a_1,\cdots ,a_n}$ and call it an {\it $n$-fold
quasi-Pfister form}.  An {\it $(n+1)$-fold Pfister form} is then a
nonsingular quadratic form of type $\pff{a_1,\cdots ,a_n}_b\otimes [1,b]$
for some $a_i\in F^*$, $b\in F$.  This Pfister form is then said to be
of degree $n+1$ and we write
$\pfr{a_1,\cdots ,a_n;b}$ for short.  $\pfr{b}=[1,b]$ is thus a $1$-fold
Pfister form.  The set of forms isometric (resp. similar) to $n$-fold Pfister 
forms will be denoted by $P_nF$ (resp. $GP_nF$).  We denote 
the $WF$-submodule of $W_qF$
generated by $n$-fold Pfister forms by $I^nW_qF$, and we obtain a 
filtration $W_qF= IW_qF\supset I^2W_qF\supset\cdots $.

Note that this definition of degree has also been given in \cite[Sect. 6]{La2}.
It differs from the one given in
\cite{AB1}, \cite{AB2}, but ours is in closer analogy to the characteristic
$\neq 2$ case, where $n$-fold Pfister forms are of degree $n$ and
dimension $2^n$ just as according to our definition in characteristic $2$
(and not $2^{n+1}$ as in the definition in \cite{B2}).

As is the case in characteristic $\neq 2$,  $x\in F^*$ is represented 
by a Pfister form $\pi$ if and only if $x\pi\cong\pi$.  This property
is commonly referred to as $\pi$ being round.  The same
holds for quasi-Pfister forms (cf. also \cite[page 38]{Pf}).
In fact, anisotropic quasi-Pfister forms can be characterized in this
way as was noted by Knebusch \cite{Kn4}.  For completeness' sake we will sketch 
a proof of these facts in section \ref{section-ts} (see
Proposition \ref{qlpf}).  




  
  

\subsection{Function fields of quadratic forms and splitting towers}
\label{subsection-fctfld} 
Suppose the quadratic form $q$ (which is supposed to be not the 
zero form) is given by the polynomial
$q(X,Y,Z)=\sum_{i=1}^r(a_iX_i^2+X_iY_i+b_iY_i^2)+\sum_{j=1}^sc_jZ_j^2$.  
Then the corresponding polynomial
$q(X,Y,Z)$ is reducible iff the nondefective part $q_{\nd}$ of $q$ is either
of type
$(0,1)$ or of type $(1,0)$ and $q_{\nd}\cong\HH$.
It is absolutely irreducible iff $q$ is of type $(r,s)$ with $r\geq 1$ and
$\dim q_{\nd} \geq 3$ (see, e.g. \cite{Ah}).  
Suppose that $q(X,Y,Z)$ is irreducible.   Then the function field $F(q)$ of 
$q$ is defined to be the field of fractions of the quotient ring 
$F[X,Y,Z]/(q(X,Y,Z))$.  To avoid case distinctions, we put $F(q)=F$ if
$q(X,Y,Z)$ is reducible or $\dim q=0$.  We have that $F(q)/F$ is 
purely transcendental iff $q_{\nd}$ is isotropic.  If $q(X,Y,Z)$
is absolutely irreducible and $K/F$ is any field
extension, then the free compositum $K\cdot F(q)$ can be identified
with $K(q)$, the quotient field of the integral domain
$K[X,Y,Z]/(q(X,Y,Z))$.

Suppose $q$ is defective, say, $q\cong q_{\nd}\perp m\times\qf{0}$ with
$q_{\nd}$ the nondefective part of $q$, then  $F(q)$ is a purely
transcendental extension of $F(q_{\nd})$ of transcendence degree $m$.

Of particular importance in this paper will be the isotropy behaviour
of a quadratic form  over the function field of another quadratic form.
Since an anisotropic form stays anisotropic over purely transcendental
extensions and by the previous remarks, this isotropy question is only
of interest when both forms are anisotropic.   The following lemma
has been shown in \cite[Cor. 3.1]{La1} and will be used frequently.

\begin{lem}\label{ts-fctfield} Let $\phi$, $\psi$ be anisotropic forms
over $F$.
Suppose that $\psi$ is not
totally singular.  Then $\ql (\phi)_{F(\psi )}$ is anisotropic
and is the quasi-linear part of
$\an{(\phi_{F(\psi )})}$.  In particular,  if $\phi$ is of
type $(r,s)$, then $\an{(\phi_{F(\psi )})}$ is of type $(r',s)$
with $r'\leq r$.
\end{lem}

Of utmost importance are the following four results.  The first part of
it will be referred to later on as
the {\it domination theorem}, the last part is an analogue in
characteristic $2$ of the Arason-Pfister Hauptsatz and will be
referred to as such.

\begin{thm}\label{domination} \begin{enumerate}
\item[(i)] Let $\phi$, $\psi$ be forms over $F$ with 
$\phi$ anisotropic and $\psi$ nondefective.  
If $\phi_{F(\psi )}$ is hyperbolic,
then $ab\psi\prec\phi$ for each $a\in D_F(\phi )$, $b\in D_F(\psi )$.
\item[(ii)] Let $\phi$ be an anisotropic nonsingular form.
Then $\phi$ is similar (isometric) to a Pfister form if and only if
$\phi_{F(\phi)}$ is hyperbolic (and $1\in D_F(\phi )$).
\item[(iii)] Let $\psi$ be anisotropic nonsingular, and let $\phi$
be an anisotropic Pfister form.  Then $\psi_{F(\phi)}$ is hyperbolic
if and only if there exist $a_1,\cdots ,a_m\in F^*$ such that
$\psi\cong a_1\phi\perp\cdots\perp a_m\phi$.
\item[(iv)] Let $q$ be a nonzero anisotropic form
in $I^nW_qF$.  Then $\dim q\geq 2^n$.  If $\dim q=2^n$, then $q$
is similar to an $n$-fold Pfister form.
\end{enumerate} 
\end{thm}

\begin{remark}
The corresponding version of the domination theorem in
characteristic $\neq 2$ is an immediate consequence of the 
Cassels-Pfister subform theorem and has been first explicitly stated 
independently by Wadsworth \cite{W}, Arason \cite{Ar} and 
Knebusch \cite{Kn2}. The characteristic $2$ version of the Cassels-Pfister
subform theorem was first proved by Amer (cf. \cite[pp. 17--18]{Pf}).
This subform theorem plus the norm theorem in characteristic $2$
by Baeza \cite{B3} immediately yield the above domination theorem.
See also \cite{La1}, \cite{LM}, where the above domination theorem has been
stated perhaps for the first time explicitly. 

The characterization of forms
similar (isometric) to Pfister forms is also an immediate consequence
of the norm theorem \cite{B3} together with the characterization of
Pfister forms as being the anisotropic (strictly) multiplicative
forms, cf. \cite[p.38]{Pf}. 

Part (iii) is a direct consequence of (i) and (ii).

The Arason-Pfister Hauptsatz in characteristic $\neq 2$ was shown in
\cite{AP} and later in characteristic $2$ in \cite{B1}.  It is a 
straightforward consequence of the first two results.
\end{remark}

The next important theorem which we will use later on has been shown 
in characteristic $\neq 2$
in \cite{H1}.  The characteristic $2$ version below has been shown in
\cite{HL}.  Before that, a somewhat weaker version has been proved 
in \cite{LM}.

\begin{thm}\label{sep-2-power}  Let $\phi$ and $\psi$ be anisotropic
forms over $F$ such that, for some positive integer $n$, one has
$\dim\phi <2^n\leq\dim\psi$.  Then $\phi_{F(\psi)}$ is anisotropic.
\end{thm}



In characteristic
$\neq 2$, the foundations of a theory of generic splitting of quadratic 
forms, based on splitting towers as defined below, have been laid out by 
Knebusch
\cite{Kn1}, \cite{Kn2}.  Most of this generic theory goes through also
in characteristic $2$ for nonsingular forms, based on results in \cite{B1},
\cite{B3}.  The notion of {\it generic} splitting
has recently been extended to forms which are possibly singular in
a forthcoming book by Knebusch \cite{Kn4}.
A theory of {\it standard} splitting towers for arbitrary forms
in characteristic $2$ has been developed in \cite{La2}.
  
The standard splitting tower of $q$ is defined to be the following
sequence of field extensions of $F$.  We put $F_0=F$ and $q_0=\an{q}$,
the anisotropic part of $q$.  For $n\geq 1$, we put 
$F_n=F_{n-1}(q_{n-1})$ and $q_n=\an{((q_{n-1})_{F_n})}$.  The {\it (standard) 
height} $h(q)$ of $q$ is the smallest integer $r$ with $\dim q_r\leq 1$. 
Let $h=h(q)$ be the height of $q$. The form 
$q_i$, $0\leq i\leq h$ is called the $i$-th {\it kernel
form}. If $h\geq 1$, then we call
$F_{h-1}$ the {\it leading field} of $q$, and $q_{h-1}$ its {\it leading form}.
Suppose that $q$ is anisotropic of type $(r,s)$ with $r\geq 1$.
By the previous lemma, there exists an $1\leq i\leq h(q)$ such that,
for $1\leq j\leq h(q)$ one has that $q_j$ is of type $(r_j,s)$ with
$r_j<r_{j-1}$ if $j\leq i$,  and $r_j=0$ for $j\geq i$ (see also
\cite{La2}).  The value $i$ will be called the {\it nondefective
height} of $q$ as it is the largest $i$ such that the quasi-linear part
$\ql (q)$ of $q$ stays anisotropic over $F_i$, i.e. such that
$q_{F_i}$ stays nondefective, and we write $h_{\nd}(q)=i$.
In Knebusch's generic theory for quadratic forms \cite{Kn4}, it is shown
that the standard splitting tower up to $F_{h_{\nd}}$ is a generic
splitting tower according to his definition of genericity, and
his definition of height corresponds to what we call
nondefective height.  Some results on the standard splitting
tower and the height of singular forms will be shown in 
section \ref{section-ts}.  In particular, we will show

\begin{thm}\label{height}  Let $q$ be an anisotropic form of standard
height $h$.
Let $F=K_0\subset K_1\subset\cdots\subset K_m$ be any tower of fields
such that $\dim\an{(q_{K_{i-1}})}>\dim\an{(q_{K_i})}$ for each
$1\leq i\leq m$, then $m\leq h$.
\end{thm}

It should be noted that the dimensions or types, respectively,
of the forms
$\an{(q_{K_j})}$, $1\leq j\leq m$ above might not necessarily be found 
among the  dimensions or types, respectively, of the higher kernel forms
$q_i$, $1\leq i\leq h$ (see also Remark \ref{splitpatvar}).
However, we do have

\begin{prop}\label{splitpatwitt}  Let $q$ be an anisotropic form
over $F$ of type $(r_0,s_0)$, and let $(r_i,s_i)$ be the type of the
$i$-th kernel form $q_i$ in a standard splitting tower
$F=F_0\subset F_1\subset\cdots\subset F_h$ of $q$.  Let
$K/F$ be any field extension and let $(r,s)$ be the type
of $\an{(q_K)}$.  Then there exists $i\in\{ 0,\cdots ,h\}$
such that $r=r_i$.
\end{prop}

\begin{proof}  If $s_0=0,1$ then this has been
shown in \cite[Th. 1.3]{KnR}. If $r_0\leq 1$ or if $r=r_0$, the result is 
trivial.

So suppose that $r_0\geq 2$, $s_0\geq 1$ and let $i\in\{ 0,\cdots ,h-1\}$ be
such that $r<r_i$.  To get the desired result, it suffices
to show that $r\leq r_{i+1}$.
Now $r_i\geq 1$ and we have $F_{j+1}=F_{j}(q_{j})$ for $0\leq j\leq i$.
Define inductively $K_0=K$ and $K_{j+1}=K_{j}(q_{j})$ for 
$0\leq j\leq i$.  Clearly,
$F_j\subset K_j$ and the nondefective part of $(q_j)_{K_j}$ is of dimension
$\geq 2r_j+1\geq 3$ for all $0\leq j\leq i$, so that indeed all 
$K_j$, $0\leq j\leq i+1$ are well defined and can be identified
with the free compositum $K\cdot F_j$.
 
Note that 
$$\begin{array}{rcl} i_W((q_j)_{K_j}) & = & i_W(q_{K_j})-i_W(q_{F_j})\\
 & \geq  & i_W(q_{K})-i_W(q_{F_j})
\geq (r_0-r)-(r_0-r_j)=r_j-r>0\end{array}$$
for $0\leq j\leq i$.  Therefore, $K_j(q_j)/K_j$ is purely transcendental for
$0\leq j\leq i$, hence $K_{i+1}/K$ is purely transcendental.  Therefore, 
$r_0-r = i_W(q_K)=i_W(q_{K_{i+1}})\geq i_W(q_{F_{i+1}})=r_0-r_{i+1}$, and
thus $r\leq r_{i+1}$ as desired.
\end{proof}


Suppose now that $q$ is nonsingular.  Then the degree of $q$ is defined as 
follows. We put $\deg(q)=\infty$ if $q$ is hyperbolic.  Otherwise, 
$h(q)\geq 1$ and $q_{h-1}$ becomes hyperbolic over its own function field, 
which implies that $q_{h-1}$ is similar to an $n$-fold Pfister form over 
$F_{h-1}$ for some $n\geq 0$.  We then put $\deg(q)=n$.  The set 
$J_nF=\{ q\in W_qF\,|\,\deg (q)\geq n\}$
is a $WF$-submodule of $W_qF$ with $I^nW_qF\subset J_nF$.  Aravire and Baeza
have shown in \cite{AB1}, \cite{AB2} that in fact $J_nF= I^nW_qF$.
It should be noted that the corresponding result $I^nF=J_nF$ in characteristic
$\neq 2$ is known to be true for $n\leq 5$ due to the work on the
Milnor Conjecture in small degrees by Merkurjev-Suslin and Rost
(see, e.g. \cite[Th.~2.8, Remarque]{Ka}), and for all $n$ 
due to Orlov-Vishik-Voevodsky \cite{OVV} (see also \cite[Theorem 1.5]{AE}).

Another important theorem by Aravire and Baeza which we will use
is the following (cf. \cite[Theorem 5.3]{AB2}):

\begin{thm}\label{baeza}  Let $\pff{a_1,\cdots ,a_n}$ be an anisotropic
$n$-fold quasi-Pfister form.  Let $\phi\in I^{n+1}W_qF$
and $K=F(\pff{a_1,\cdots ,a_n})$.
If $\phi_K\in I^{n+2}W_qK$ then there exists $b\in F$
such that $\phi\equiv\pfr{a_1,\cdots a_n,b}\bmod I^{n+2}W_qF$.
\end{thm}


\subsection{The Arf-invariant and the Clifford algebra}\label{subsection-arf}
For basic facts on Clifford algebras and quaternion algebras which we state
without further comment,  we refer the reader to \cite{B2}, \cite{MTW},
\cite{D}.

Let $q$ be a quadratic form defined on the vector space $V$ and let
$T(V)=\bigoplus_{i=0}^\infty V^{\otimes n}$ denote the tensor algebra
of $V$.  Let $J(q)$ be the ideal in $T(V)$ generated by elements of type 
$x\otimes x-q(x)$, $x\in V$.  Then the Clifford algebra $C(q)$ of $q$
is defined to be the quotient $T(V)/J(q)$.  This algebra has a natural
$\ZZ/2\ZZ$-gradation, and the subalgebra $C_0(q)$ of degree $0$ elements is
called the even Clifford algebra.

If $q$ is nonsingular, then $C(q)$ is a central simple algebra over $F$
which, up to isomorphism, only depends on the isometry class of $q$,
and the center $Z(q)$ of $C_0(q)$ is a separable quadratic algebra
over $F$, i.e. there exists $\delta\in F$ with $Z(q)=F[X]/(X^2+X+\delta)$.
If we put $\wp (a)=a^2+a$, then $\wp (F)=\{ \wp (a)\,|\,a\in F\}$ is
an additive subgroup of $F$, and the class of $\delta$ modulo $\wp (F)$
is an invariant of the isometry class of $q$ and
will be called the Arf-invariant of $q$, denoted by $\Delta (q)$.
By abuse of notation, we will often identify $\Delta (q)$ with a
representative in $F$ of its class in $F/\wp (F)$.

More explicitly, if $q\cong a_1[1,b_1]\perp\cdots\perp a_n[1,b_n]$,
then $\Delta (q)=b_1+\cdots +b_n\in F/\wp (F)$, and
$C(q)=[b_1,a_1)_F\otimes \cdots \otimes [b_n,a_n)_F$, where $[b,a)_F$
($a\in F^*$, $b\in F$)
denotes the quaternion algebra generated by two elements $u,v$ over $F$
subject to the relations $u^2=a$, $v^2+v=b$, $uv=(v+1)u$.  This is
a $4$-dimensional central simple algebra which is trivial in the
Brauer group iff $a$ is a norm for the extension $F[X]/(X^2+X+b)$ over $F$.
For quaternion algebras, one has $[b,a)_F\otimes [b',a')_F
\cong [b,aa')_F\otimes [b+b',a')_F$.  In particular, considered as a
symbol in the Brauer group, it is additive in the first and multiplicative 
in the second
slot.  Two central simple algebras $A$, $B$ are said to be similar,
$A\sim B$, if they represent the same element in the Brauer group
$\Br (F)$.

We will need the following results.

\begin{prop}\label{albert}{\em (Albert \cite[Theorem 28, p. 108]{Al}.)}
Let $A$ be a central simple algebra over $F$ and let
$K=F(\sqrt{b_1},\cdots ,\sqrt{b_n})$, $[K:F]=2^n$.  If $A_K\sim 0$
then there exist $a_1,\cdots ,a_n\in F$ such that
$A\sim [a_1,b_1)_F\otimes \cdots \otimes [a_n,b_n)_F$.
\end{prop}

\begin{thm}\label{sah}{\em (Sah \cite[Theorem 2]{Sa}).}
Let $q$ be a nonsingular quadratic form over $F$.
\begin{enumerate}
\item[(i)] $q\in I^2W_qF$ iff $\Delta (q)=0\in F/\wp (F)$.
\item[(ii)]  If $q\in I^2W_qF$, then $q\in I^3W_qF$ iff
$C(q)\sim 0$.
\end{enumerate}
\end{thm}
  




\section{Fitzgerald's theorem}\label{section-fitzgerald}
Let $q$ be a nonsingular nonhyperbolic form over $F$.  We define 
$$N(q)=\dim q -2^{\deg q}\ .$$
With this definition, we now give our characteristic $2$ version of 
Fitzgerald's theorem \cite[Th. 1.6]{F}.  
\begin{thm}\label{fitz} Let $q$, $\eta_r$, $\phi_r$ be nonsingular forms 
over $F$, and let $\sigma$ be a totally singular form over $F$.  Let $\rho$ be
nonsingular with $\sigma\prec\rho$ and $\dim\rho =2\dim\sigma$, i.e.
$\rho$ is a nonsingular completion of $\sigma$.
Suppose that 
\begin{enumerate}
\item[(i)] $q$,
$\eta =\eta_r\perp\sigma$, $\phi =\phi_r\perp\sigma$ are anisotropic,
\item[(ii)] $\dim q,\ \dim\phi\geq 2$,
\item[(iii)] $q\sim\phi_r\perp\eta_r\perp\rho$, 
\item[(iv)] $\dim\eta <\dim\phi +2^{\deg q}$.
\end{enumerate}
Then $q$ is hyperbolic over $F(\phi )$ if and only if $q$ is similar to
a Pfister form and $q\cong\phi_r\perp\eta_r\perp\rho$.
\end{thm}

\begin{proof} ``if'':  Suppose that $q$ is similar to a Pfister form
(which is anisotropic by hypothesis), and that 
$q\cong\phi_r\perp\eta_r\perp\delta$.  Since $\phi\prec q$, 
$\dim\phi\geq 2$, we have that
$q$ becomes isotropic and hence hyperbolic over $F(\phi )$.

``only if'':  We will prove this in a sequence of claims.  
So from now on, we suppose that $q$ is hyperbolic over $F(\phi )$.


\begin{claim} $q\cong \phi_r\perp\eta_r\perp\rho$.\end{claim}
\begin{proof}  Since $\phi_r\perp\phi_r\sim 0$ and $\rho\perp\sigma\sim\sigma$
(see Lemma \ref{easy}),  we have that $q\perp\phi\sim \eta$.  Now
$\dim (q\perp\phi )\geq \dim\phi +2^{\deg q}>\dim\eta$ by assumption (iv).
Hence, $q\perp\phi$ is isotropic and there exists an $a\in F^*$ represented
both by $q$ and $\phi$ because both these forms are anisotropic.    
Since $q_{F(\phi)}$ is hyperbolic, it follows
from the domination theorem that $\phi\cong a^2\phi\prec q$.

Thus, there exists a nonsingular form $\tau$ and a nonsingular completion 
$\rho'$ of $\sigma$ with 
$q\cong\phi_r\perp\tau\perp\rho'\sim\phi_r\perp\eta_r\perp\rho$.
Witt cancellation for nonsingular forms implies 
$\tau\perp\rho'\sim\eta_r\perp\rho$.
Now $\tau\perp\rho'$ is anisotropic and $\tau\perp\sigma\prec\tau\perp\rho'$.
In particular, $\tau\perp\sigma$ is anisotropic.  Since $\rho'\perp\sigma\sim
\rho\perp\sigma\sim\sigma$, we get $\tau\perp\sigma\sim\eta_r\perp\sigma$.
Since the forms on both sides are anisotropic, we therefore obtain
$\tau\perp\sigma\cong\eta_r\perp\sigma$ and thus $\dim\tau =\dim\eta_r$.
Thus, the anisotropic form $q\cong\phi_r\perp\tau\perp\rho'$ and the form
$\phi_r\perp\eta_r\perp\rho$ have the same dimension, and since they are
Witt-equivalent, it follows that $q\cong \phi_r\perp\eta_r\perp\rho$.
\end{proof}

After scaling, we may therefore assume that $1\in D_F(q)\cap D_F(\phi)$.

Since an anisotropic nonsingular form becomes hyperbolic over its own function
field iff the form is similar to a Pfister form (Theorem \ref{domination}), 
we get that $q$ is similar to a Pfister form (and thus isometric to a Pfister 
form as $1\in D_F(q)$) iff $N(q)=0$.  Thus, it suffices to show
that under the above assumptions we cannot have $N(q)>0$.  

So suppose that $N(q)=m>0$.  We will show by induction on $m$ that this will
yield a contradiction.

\begin{claim} Let $F_1=F(q)$ and $q_1=\an{(q_{F_1})}$.  Then
$\phi_{F_1}$ is anisotropic, $q_1$ is a Pfister form of dimension
$2^{\deg q}$, and $\phi_{F_1}\prec q_1$.
\end{claim}

\begin{proof}
$N(q)>0$ implies that $h(q)\geq 2$. Then $q_1$ is not hyperbolic.  
Furthermore, if $\dim\sigma >0$, then $\sigma_{F_1}$ stays anisotropic 
(cf. \cite[Cor.3.1]{La1}).
In particular, $\phi_{F_1}$ will stay nondefective, so that as a polynomial
over $F_1$, $\phi_{F_1}$ will be irreducible.  It follows that the compositum
of $F_1$ and $F(\phi )$ is just $F_1(\phi)$.  Therefore,
$(q_1)_{F_1(\phi)}$ is hyperbolic.  Since the function field of a nondefective
isotropic form is purely transcendental over the ground field, and since the
non hyperbolic form $q_1$ becomes hyperbolic over $F_1(\phi )$, we have that
$\phi_{F_1}$ is anisotropic.  Clearly, $\deg q=\deg q_1$ and thus
$N(q_1)<N(q)$.

Put $\eta_1=\an{(\eta_{F_1})}$. As remarked above, the quasi-linear part 
$\sigma$ of $\eta$ will stay anisotropic (if $\dim\sigma >0$).  Hence,
there exists a nonsingular form $\eta_{1,r}$ over $F_1$ such that $\eta_1\cong
\eta_{1,r}\perp\sigma$. Now over $F_1$ we have
$$\phi_r\perp\eta_r\perp\sigma\sim_{F_1}\phi_r\perp\eta_{1,r}\perp\sigma\ .$$
By Lemma \ref{regcompl}, there exists a nonsingular completion $\rho_1$ of 
$\sigma$ over $F_1$ such that
$$q_1\sim_{F_1} \phi_r\perp\eta_r\perp\rho\sim_{F_1}
\phi_r\perp\eta_{1,r}\perp\rho_1\ .$$
Replacing $q$ by $q_1$, $\eta$ by $\eta_1$, $\eta_r$ by $\eta_{1,r}$, 
$\phi$ by $\phi_1=\phi_{F_1}$, $\phi_r$ by $\phi_{1,r}= (\phi_r)_{F_1}$,
$\sigma$ by $\sigma_1=\sigma_{F_1}$,
$\rho$ by $\rho_1$, it is easily verified that
these forms satisfy the hypotheses (i)--(iv).  The same reasoning as above
shows then that $q_1\cong \phi_{1,r}\perp \eta_{1,r}\perp\rho_1$.
Furthermore, since $N(q_1)<N(q)$, it follows
by induction that $q_1$ is similar to a Pfister form, and hence isometric
to a Pfister form since $1\in D_K(\phi_1)$ and $\phi_1\prec q_1$.
In particular, $\dim q_1=2^{\deg q_1}=2^{\deg q}$.\end{proof}

\begin{claim} $D_F(q)=G_F(q)$.\end{claim}
\begin{proof}  Since $1\in D_F(q)$ and $q$ anisotropic, we have 
$G_F(q)\subset D_F(q)=D^*_F(q)$.  Suppose
there exists an $x\in D_F(q)\setminus G_F(q)$.  Let $\beta =
\an{(\pff{x}_b\otimes q)}$.  By assumption, $\pff{x}_b\otimes q$ is isotropic
but not hyperbolic.  It follows readily that 
$D_F(q)\cap D_F(\beta)\neq\emptyset$.  Also, $\deg\beta\geq 1+ \deg q$
(see, e.g., \cite[Ch.4, Theorem 7.5]{S} in the case of characteristic $\neq 2$;
the proof for the case of characteristic $2$ is more or less identical). 

Now let $\beta_1=\an{(\beta_{F_1})}$.  Suppose that $\beta_1\sim 0$.  
The domination theorem and the fact that 
$D_F(q)\cap D_F(\beta)\neq\emptyset$ imply that $q\subset\beta$.
Then there exists a nonsingular form $q'$ with $\beta\cong q\perp q'
\sim q\perp xq$.  Now $\dim\beta <2\dim q$, hence  $\dim q'<\dim q$.
But we have $q'\sim xq$, hence $q$ is isotropic, a contradiction.
It follows that $\beta_1\not\sim 0$.  In particular, 
$\dim\beta_1\geq 2^{\deg \beta}\geq 2^{1+\deg q}$.

Since $x\in D_F(q)$, we have $1\in D_F(xq)$.  Now $1\in D_F(\phi )$, and
$xq$ becomes hyperbolic over $F_1=F(\phi )$.  The domination
theorem implies that there exists a nonsingular form $\eta_r'$ and a nonsingular
completion $\rho'$ of $\sigma$ such that $xq\cong\phi_r\perp\eta_r'\perp\rho'$
and $\dim\eta_r=\dim\eta_r'$.  

Over $F_1$ we have $\beta_1\sim q_1\perp xq_1$.  Now $\phi_{F_1}\prec q_1$
and, by the same reasoning as in the previous claim, $\phi_{F_1}\prec xq_1$.  
In particular,
$$2^{1+\deg q}\leq\dim\beta_1=\dim \an{(q_1\perp xq_1)}
\leq 2\dim q_1-2\dim\phi<2^{1+\deg q}\ ,$$
which is absurd.  Hence, $D_F(q)=G_F(q)$.
\end{proof}

Now let $K=F(X_1,\cdots ,X_n)$ be the rational function field in $n=\dim q$
variables over $F$.  Clearly, $\deg q_K=\deg q$, and working over $K$
instead of $F$, it is obvious that the forms $q_K$, $\phi_K$, $(\phi_r)_K$, 
$\eta_K$, $(\eta_r)_K$, $\sigma_K$ and $\rho_K$ satisfy the hypotheses (i)--(v).
By the previous claim, we therefore have $D_K(q)=G_K(q)$.  Hence
$q_K\cong q(X_1,\cdots ,X_n)q_K$, i.e., the anisotropic form $q$ is 
strictly multiplicative.  It follows that $q$ is a Pfister form, hence
$N(q)=0$, a contradiction to the assumption $N(q)>0$ and the proof of
the theorem is completed. 
\end{proof}

\begin{remark}  To get Fitzgerald's original version
in characteristic $\neq 2$ from our version above, one simply has
to omit $\sigma$, i.e. $\phi$ and $\eta$ are nonsingular,  
and one has to replace ``$+$'' by
``$-$'' in certain obvious places in the proof.  Furthermore,
in the original version, the condition on $q$ being anisotropic was
replaced by $q$ nonhyperbolic and $q\cong\phi\perp\eta$, and condition
(iv) by the then equivalent condition $N(q)<2\dim\phi$.  As a consequence,
the hyperbolicity of $q$ over $F(\phi )$  becomes equivalent to $q$ being 
similar to an {\it anisotropic} Pfister form.  

Our proof follows along the same lines as Fitzgerald's original proof.
Certain modifications became necessary because of the possibility of having
singular forms and dominated forms.  On the other hand, some of the
original arguments could be simplified by using the domination theorem more 
prominently in certain places.
\end{remark}

\begin{coro}\label{fitz-coro} Let $\phi$ and $q$ be forms such that $q$ is 
nonsingular and
anisotropic, $\phi\prec q$, and $2\dim\phi\geq \dim q$.  If $q_{F(\phi )}$
is hyperbolic, then $q$ is similar to a Pfister form.
\end{coro}





\section{Pfister neighbors and an analogue 
of a theorem by Knebusch in characterisic $2$}\label{section-knebusch} 

A form $q$ is said to be a {\it Pfister neighbor} if there exist 
$\pi\in P_nF$ for some $n$ and an $a\in F^*$ such that $q\prec a\pi$ and
$\dim q >2^{n-1}$.  We then say that $q$ is a Pfister neighbor of $\pi$.
In this case, the Pfister form $\pi$ is uniquely determined by $q$
up to isometry, and $q$ is isotropic iff $\pi$ is isotropic (and hence 
hyperbolic), cf. \cite[Prop. 3.1]{La1}.  Furthermore, it is rather
obvious that if $q$ is a Pfister neighbor of type $(r,s)$, 
then there exists an integer $n\geq 1$ such that $2r+s >2^{n-1}$ and 
$2r+2s\leq 2^n$.  In particular, $q$ is cannot be totally singular.
If $q$ is a Pfister neighbor of $\pi\in P_nF$, and if $x,y\in F^*$
such that $q\prec x\pi$ and $q\prec y\pi$, then the roundness of $\pi$
implies that $x\pi\cong y\pi$.  This together with the uniqueness of
$\pi$ implies that $q^c_{x\pi}$  only depends on the Pfister neighbor 
$q$ (see also \cite[Prop. 5.2]{La2}).  By abuse of language and notation, we will call it the  
{\it complement} of the Pfister neighbor $q$ and write $q^c$ for short.

Pfister forms can be characterized up to similarity by the property
that they become hyperbolic over their own function field, see
Theorem \ref{domination}(ii).  The behaviour of Pfister neighbors
over their own function field is equally well understood.

\begin{prop}\label{pfnb-fctfield}  Let $q$ be an anisotropic Pfister
neighbor. Then $\an{(q_{F(q)})}\cong q^c_{F(q)}$.
In particular, the anisotropic part of $q$ over $F(q)$ is
defined over $F$.
\end{prop}

\begin{proof}  Let $n$ and $\pi\in GP_nF$ such that over $F$,
$q\prec\pi$ and $\dim q>2^{n-1}$.  $\pi$ is anisotropic as $q$
is anisotropic, hence $q^c$ is anisotropic as well.   
Furthermore, $\pi\perp q\sim q^c$. Now
$\pi_{F(q)}$ is isotropic because $q_{F(q)}$ is isotropic, hence
$\pi_{F(q)}$ is hyperbolic, and we get $\an{(q_{F(q)})}\sim 
q^c_{F(q)}$.  But $\dim q>2^{n-1}>2^n-\dim q=\dim q^c$,
and by Theorem \ref{sep-2-power}, the anisotropic form
$q^c$ stays anisotropic over $F(q)$. Thus, 
$\an{(q_{F(q)})}\cong q^c_{F(q)}$.
\end{proof}

In characteristic
$\neq 2$, Knebusch \cite[Theorem 7.13]{Kn3} has shown that the converse 
of the last statement in the previous proposition also holds, 
more precisely, an anisotropic form $q$ with the
property that its anisotropic part over $F(q)$ is defined over $F$
must be a Pfister neighbor, so that together with the above result,
anisotropic Pfister neighbors can be characterized by this property.
This also holds in characteristic $2$ for nonsingular forms as follows 
readily from Theorem \ref{fitz} (see also Theorem \ref{knebusch} below
for more details).  However, for singular forms the converse
fails in general.  

\begin{exam}\label{counterex-sing}  
Let $\sigma$ be a totally singular anisotropic form.
Then $\sigma_{F(\sigma)}$ is defined over $F$.  This is clear by
Lemma \ref{exc}.  However, $\sigma$ is not a Pfister neighbor as 
we already mentioned above.
\end{exam}

\begin{exam}\label{counterex-r=1}  Here, we construct
counterexamples of forms of type $(1,s)$.

Let $q\cong q_{ns}\perp\sigma$ be 
an anisotropic form 
of type $(1,s)$ with $q_{ns}$ nonsingular and $\sigma$ totally
singular.  Lemma \ref{ts-fctfield} immediately implies that
$\an{(q_{F(q)})}\cong\sigma_{F(q)}$.  Now if $q$ is a Pfister
neighbor, then Proposition \ref{pfnb-fctfield} implies that
$q^c=\sigma$, hence there exists an $n$ with $\dim q =2+s>2^{n-1}$ 
and $2+2s =\dim q +\dim q^c=2^n$.  This implies $s=2^{n-1}-1$.
Thus, any anisotropic form of type $(1,s)$ with $s+1$ not a 
$2$-power will be a non-Pfister neighbor with the property that its 
anisotropic part over $F(q)$ is defined over $F$. (See also
\cite{La2}.)

Now nondefective forms of type $(1,1)$ are easily seen to be Pfister 
neighbors. However, there are  fields such that for any $n\geq 3$
there exist anisotropic forms of type $(1,2^{n-1}-1)$
which are not Pfister neighbors.  For one can construct fields
with anisotropic nonsingular forms of arbitrarily large dimension,
i.e. $u(F)=\infty$, but for which $I^3W_qF=0$.  This can be done
as in the characteristic $\neq 2$ case (where it was first done
by Merkurjev \cite{Me}) using the methods in \cite{MTW}. 
Over such a field, any anisotropic nonsingular form of dimension
$2^n$, $n\geq 3$, clearly contains a dominated form of type 
$(1,2^{n-1}-1)$.  This anisotropic form of dimension $2^{n-1}+1$ 
cannot be a Pfister neighbor
as $n$-fold Pfister forms are hyperbolic for $n\geq 3$ because
$I^3W_qF=0$. 
\end{exam}

\begin{exam}\label{counterex3} The construction we now describe
can be used to produce counterexamples of type $(r,s)$ subject to
the following conditions :  There exist integers $n,m,t\geq 0$ with
$n<m$, $2r+t>2^n$, $2r+2t\leq 2^{n+1}$, and $s=t+2^m$.
In particular, we get counterexamples of type
$(2^n,2^{n+1})$.

Consider a Pfister form
$\pi=\pfr{a_1,\cdots ,a_n;b}$ and an anisotropic totally singular 
Pfister form $\tau=\pff{a_1,\cdots ,a_m}$ with $m>n$. Let
$x\in F^*$ and put $\phi=\pi\perp x\tau$.
Let $E/F$ be any field extension such that  $\tau_E$
stays anisotropic but such that $\phi_E=\pi_E\perp x\tau_E$ is isotropic.
Then there exists $y\in D_E(\pi)\cap D_E(x\tau)$.  The roundness
of $\pi$ and $\tau$ implies that $\phi_E\cong y(\pi_E\perp\tau_E)$.

Now $\pi$ is an orthogonal sum of the forms $(\prod_{i\in I} a_i)[1,b]$
with $I\subset\{ 1,\cdots ,n\}$, and $\tau$ is an orthogonal
sum of the forms $\qf{\prod_{j\in J}a_j}$ with $J\subset\{ 1,\cdots ,m\}$.
Since $[u,v]\perp\qf{u}\cong\HH\perp\qf{u}$, it follows readily
that $\pi\perp\tau\sim\tau$, hence $\phi_E\sim y\tau_E\cong x\tau_E$.

Now let $\psi\prec\pi$ be a Pfister neighbor of $\pi$ with complement
$\psi^c$ and let $\eta =\psi\perp x\tau$ for some $x\in F^*$.   
Thus, if $E$ is a field extension with $\tau_E$
anisotropic, and if $\psi_E\perp x\tau_E$ is isotropic, then
the above shows that $\eta_E\sim\psi^c_E\perp x\tau_E$ 
because $\pi\perp\psi^c\sim\psi$. 

Suppose now that $\pi$ is anisotropic and that $x\in F^*$ is
such that $\pi\perp x\tau$ is anisotropic as well.
In particular, this implies that $\eta$ and $\psi^c\perp x\tau$
are anisotropic.  Note that $\eta$ is not totally
singular as the Pfister neighbor $\psi$ is not totally
singular.  Also, $\eta$ will not be a Pfister neighbor.
Indeed,  suppose $\eta$ is a Pfister neighbor.
Now $2^{m+1}\geq 2^m+2^{n+1}\geq\dim\eta >2^m+2^n$,  
thus $\eta$ is necessarily a Pfister neighbor of an
$(m+1)$-fold Pfister form.  If $(r,s)$ is the type
of $\eta$, this implies that $r+2s\leq 2^{m+1}$, 
a contradiction as $r\geq 1$ and $s\geq 2^m$.

Let $E=F(\eta )$.  Then, by the above, 
$\eta_E\sim (\psi^c\perp x\tau)_E$.  Thus, if
$(\psi^c\perp x\tau)_E$ is anisotropic, this gives
an example of an anisotropic non-Pfister neighbor $\eta$ 
whose anisotropic part over $F(\eta)$ is defined over the 
base field $F$.

This certainly holds if $\psi^c\perp x\tau$ is
totally singular (cf. Lemma \ref{ts-fctfield}), 
i.e. $\psi^c$ is totally singular.
This is the case if and only if the Pfister neighbor $\psi$
is of type $(r',s')$ with $2r'+2s'=2^{n+1}$.
(Knebusch \cite{Kn4} calls such Pfister neighbors
{\it close} Pfister neighbors.)  
This incidentally yields examples of non-Pfister neighbors
of nondefective height $1$.

To construct explicit situations with $(\psi^c\perp x\tau)_{F(\eta)}$ 
anisotropic, let $\pi$ and $\tau$ be as above, anisotropic,
and with $a_i,b\in F_0$, and let $\psi\prec\pi$
be a Pfister neighbor over $F_0$.  Let $F=F_0(x)$
with $x$ an indeterminate.  By the usual degree argument,
we see that $\pi\perp x\tau$ is anisotropic.
Let $K=F_0(\psi)$.  Then $\psi^c_K$ is anisotropic
(cf. Proposition \ref{pfnb-fctfield}), and so is
the totally singular form $\tau_K$.  Hence,
over $F(\psi )=F_0(\psi)(x)$, we have that
$(\psi^c\perp x\tau )_{F(\psi)}$ is anisotropic.
Now $\eta_{F(\psi)}$ is nondefective and isotropic, hence
$F(\psi)(\eta)/F(\psi)$ is purely transcendental,
thus $(\psi^c\perp x\tau )_{F(\psi)(\eta)}$ is
anisotropic and therefore,  $(\psi^c\perp x\tau)_{F(\eta)}$
is anisotropic as desired.
\end{exam}

A natural question is to ask under which conditions the converse
of the last statement in Proposition \ref{pfnb-fctfield} holds.
In all the examples above, the counterexamples are of type $(r,s)$
with $s\geq 2r$.  In view of this, we venture the following

\begin{conj}\label{pfnb-conj}  Let $q$ be an anisotropic form such
that the anisotropic part of $q$ over $F(q)$ is defined over $F$.
If $q$ is of type $(r,s)$ with $s<2r$, then $q$ is a Pfister neighbor.
\end{conj}

\begin{thm}\label{knebusch}  Conjecture \ref{pfnb-conj} is true
if $s\leq 4$ or if $s=5$ and there exist $a,b,c\in F$ such
that the quasi-linear part of $q$
is similar to $\pff{a,b}\perp\qf{c}$.
\end{thm}

To prove the theorem, we need some lemmas.

\begin{lem}\label{knebusch-lem1}
Let $\phi\cong\phi_r\perp\sigma$ be anisotropic of type
$(r,s)$, $r\geq 1$, with
$\sigma =\ql (\phi)$.
Suppose that the anisotropic part of $\phi_{F(\phi)}$ is defined over 
$F$. Then there exists a form $\eta\cong\eta_r\perp\sigma$ over $F$
with $\eta_r$ nonsingular such that
\begin{itemize} 
\item $\an{(\phi_{F(\phi)})}\cong \eta_{F(\phi )}$, and
\item $(\phi_r\perp\eta_r\perp\sigma)_{F(\phi )}\sim\sigma_{F(\phi)}$.
\end{itemize}
If, furthermore, there exists a nonsingular completion $\rho$
of $\sigma$ over $F$ such that 
$q\cong\phi_r\perp\eta_r\perp\rho\in I^nW_qF$
with $2^n>2s$, then $q$ is anisotropic and similar to a 
Pfister form.  In particular, $\phi$ is a Pfister neighbor.
\end{lem}

\begin{proof} Let $K=F(\phi )$.
By Lemma \ref{ts-fctfield}, $\sigma_K \cong\ql (q_K)$.  Hence, we may
assume that the anisotropic part over $K$ is defined over $F$
by a form $\eta$ whose quasi-linear part is $\sigma$, i.e.,
$\eta\cong\eta_r\perp\sigma$ with $\eta_r$ nonsingular.

Now $(\phi_r\perp\sigma)_K\sim (\eta_r\perp\sigma)_K$, hence
$(\phi_r\perp\eta_r\perp\sigma)_K\sim\sigma_K$

Let $\rho$ be an n.s.c. of $\sigma$ with 
$q\cong\phi_r\perp\eta_r\perp\rho\in I^nW_qF$ and $2^n>2s$.  By 
Lemma \ref{regcompl}, there exists over $K$ an n.s.c. $\tau$ of 
$\sigma_K$ such that $q_K\sim\tau_K\in I^nW_qF$.
Now $\dim\tau =2s<2^n$, and the Arason-Pfister Hauptsatz implies
$q_K\sim\tau\sim 0$.  It follows now readily from
Theorem \ref{fitz} that $q$ is
anisotropic and similar to a Pfister form.
Since $\dim\phi >\dim\eta$ and $\dim\phi+\dim\eta =\dim q$,
we have $\phi\prec q$ and $2\dim\phi >\dim q$.  Hence,
$\phi$ is a Pfister neighbor.
\end{proof}

\begin{lem}\label{knebusch-lem2}
Let $\phi\cong\phi_r\perp\qf{1,a_2,\cdots ,a_s}$ be nondefective of type 
$(r,s)$ with 
$s\geq 2$ and $\phi_r$ nonsingular.
Suppose that 
$2\leq t\leq s$ is such that $F(\sqrt{a_2},\cdots ,\sqrt{a_s})=
F(\sqrt{a_2},\cdots ,\sqrt{a_t})$.
Let $q$ be nondefective of type
$(r',s')$ with $\dim q\geq 5$ and $r'\geq 2$.  If $\an{(\phi_{F(q)})}$
is totally singular, then there exist $b_1,\cdots ,b_t\in F^*$
such that $\psi =\phi_r\perp [1,b_1]\perp a_2[1,b_2]\perp\cdots\perp
a_t[1,b_t]\perp (t-s)\times\HH$ is an n.s.c. of $\phi$
with  $\psi\in I^3W_qF$.
\end{lem}

\begin{proof}
Let $K=F(\sqrt{a_2},\cdots ,\sqrt{a_t})$.  By assumption,
$\qf{1,a_2,\cdots ,a_s}_K\sim\qf{1}_K$.  

Let $q'$ be
the nondefective part of $q$ over $K$.  Then $q'$ is of type
$(r',s'')$ and $\dim q'\geq 5$.  Indeed, if $r'\geq 3$, this
is clear, and if $r'=2$ then $s'\geq 1$ and hence $s'\geq s''\geq 1$.
Hence, $K(q)/K(q')$ is purely transcendental of
transcendence degree $s'-s''$.  

By assumption, $\phi_{F(q)}\sim\qf{1,a_2,\cdots ,a_s}_{F(q)}$, thus
$\phi_{K(q)}\sim\qf{1,a_2,\cdots ,a_s}_{K(q)}\sim\qf{1}_{K(q)}$, and since
anisotropic forms stay anisotropic over purely transcendental 
extensions, we obtain $(\phi_r\perp\qf{1})_{K(q')}\sim
\phi_{K(q')}\sim\qf{1}_{K(q')}$.
Let $\Delta =\Delta (\phi_r)$.  By Lemma \ref{regcompl} there
exists $a\in K(q')$ such that
$(\phi_r\perp [1,\Delta ])_{K(q')}\sim [1,a]$.  Now
$\phi_r\perp [1,\Delta ]\sim [1,a]\in I^2W_qK(q')$ by Theorem \ref{sah} and
therefore, by the Arason-Pfister Hauptsatz, $[1,a]\sim 0$.
In particular, $C(\phi_r\perp [1,\Delta ])$ splits over $K(q')$.

Since $K(q')/K$ can be obtained as a purely transcendental extension
followed by a separable quadratic extension, the index
of $C(\phi_r\perp [1,\Delta ])_K$ can only be $1$ or $2$.  If it
is $2$, then the class of $C(\phi_r\perp [1,\Delta ])$ in $\Br (K)$
is given by a quaternion division algebra which stays division
over $K(q')$ as $q'_K$ is nondefective and $\dim q'\geq 5$ 
(cf. \cite[Theorem 3]{MTW}), a contradiction.
Hence, $C(\phi_r\perp [1,\Delta ])_K$ is split, and by Albert's
result (Proposition \ref{albert}), it follows that there exist
$b_2,\cdots , b_{t}\in F$ with 
$C(\phi_r\perp [1,\Delta ])=[b_2,a_2)_F\otimes\cdots\otimes 
[b_{t},a_{t})_F$.  Let $b_1 =\Delta +b_2+\cdots +b_{t}$
and put 
$$\psi =\phi_r\perp [1,b_1]\perp a_2[1,b_2]\perp\cdots\perp
a_{t}[1,b_{t}]\perp (s-t)\times\HH\ .$$
Then one verifies rapidly that $\psi\in I^3W_qF$ and that $\psi$
is a nonsingular completion of $\phi$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{knebusch}]
Let $\phi\cong\phi_r\perp\sigma$ be anisotropic of type $(r,s)$ with 
quasi-linear part $\sigma$ and $2r>s$.  
Suppose that $\an{(\phi_{F(\phi )})}\cong\eta$ for some
$F$-form $\eta$.  Then, by Lemma \ref{knebusch-lem1},
we may assume that $\eta\cong\eta_r\perp\sigma$ with $\eta_r$
nonsingular, $\dim\eta_r<\dim\phi_r$.  

\smallskip

\noindent{\it The case $s=0$.}  In this case, $2>2s=0$.
Since $\phi\perp\eta\cong\phi_r\perp\eta_r\in IW_qF$, it follows
from Lemma \ref{knebusch-lem1} that $\phi\perp\eta$ is similar
to an anisotropic Pfister form of which $\phi$ is a Pfister neighbor.

\smallskip

\noindent{\it The case $s=1$.}  After scaling, we may assume that
$\sigma\cong \qf{1}$.  Let  $d =\Delta (\phi_r)+\Delta (\eta_r)$
Then $\rho\cong [1,d]$ is a regular completion of $\sigma$, and
by putting $q:=\phi_r\perp\eta_r\perp\rho$, it follows from
Theorem \ref{sah} that  
$q\in I^2W_qF$.  Now $2^2>2s=2$, hence
Lemma \ref{knebusch-lem1} implies that $q$ is a Pfister form
(note that $q$ represents $1$) and that $\phi$ is a Pfister
neighbor of $q$.

\smallskip

\noindent{\it The case $2\leq s\leq 3$.}  $2r>s$ implies that 
$r\geq 2$ and $\dim \phi\geq 6$.  Let $K=F(\phi )$.
By assumption $(\phi_r\perp\eta_r\perp\sigma )_K\sim\sigma_K$. 
By Lemma \ref{knebusch-lem2},
there exists an n.s.c. $\rho$ of $\sigma$ such that for
$q=\phi_r\perp\psi_r\perp\rho$ we get 
$q\in I^3W_qF$.  Since $2^3>6\geq 2s$,
Lemma \ref{knebusch-lem1} yields that $q$ is similar to
an anisotropic Pfister form of which $\phi\cong\phi_r\perp\sigma$
is a Pfister neighbor.

\smallskip

\noindent{\it The case $4\leq s\leq 5$.}  By assumption in the
case $s=5$ and after scaling, we may assume that 
$\sigma\cong\qf{1,a,b,c}$ or $\sigma\cong\qf{1,a,b,ab,c}$,
$a,b,c\in F^*$.

Let $L=F(\qf{1,a,b,ab})$.  By Theorem \ref{splitpat}(ii), 
$\qf{1,a,b,ab}_L\sim\qf{1,a}_L$.
Let $\sigma'=\an{(\sigma_L)}$.
The above shows that $\sigma'\cong\qf{1,a}$ or $\sigma'\cong\qf{1,a,c}$.

By Lemma \ref{knebusch-lem2}, there exists an n.s.c. $\rho$
of $\sigma$ such that $\psi:=\phi_r\perp\eta_r\perp\rho\in I^3W_qF$.

Now put $M=L(\phi )=K(\qf{1,a,b,ab})$.  Since 
$(\phi_r\perp\eta_r\perp\sigma)_K\sim\sigma_K$, there exists 
an n.s.c. $\rho'$ of $\sigma$ over $M$ such that 
$\psi_M\sim\rho'$.  We have
$i_{\ti}(\sigma_M)\geq s-3$, and $\sigma_M\prec\rho'$ with
$\rho'$ nonsingular.  Hence, $i_W(\rho')\geq s-3$ and
$\dim\an{\rho'}\leq 2s - 2(s-3)=6$.  The Arason-Pfister
Hauptsatz implies that $\rho'\sim 0$.  

Clearly, $\deg\psi_L\geq 3$ as $\psi_L\in I^3W_qL=J_3L$.
We claim that $\psi_L\in I^4W_qL=J_4L$.
Suppose that $\deg\psi_L=3$.  Let $N$ be the leading field
of $\psi_L$ and let $\pi\in GP_3N$ be the leading form.
We have $\psi_N\sim\pi$ and $0\sim\psi_{N(\phi)}\sim\pi_{N(\phi)}$.
Let $\phi'$ be the nondefective part of $\phi_L$ which,
by the above, is of type $(r,s')$ with $s'=\dim\sigma'\in\{ 2,3\}$
and $2r=\dim\phi_r\geq 6$.
Since $N(\phi)/N(\phi')$ is purely transcendental, we also have
$0\sim\psi_{N(\phi')}\sim\pi_{N(\phi')}$.  Now $N$
is obtained from $L$ by taking function fields of nonsingular
forms, hence the nondefective form $\phi'$ stays nondefective
over $N$ (Lemma \ref{ts-fctfield}). It follows from
the domination theorem (Theorem \ref{domination}) that there
exists $x\in N^*$ with $x\phi'\prec\pi$.  In particular,
$10\leq 2r+2s'\leq \dim\pi =8$, a contradiction.

Now $\psi\in I^3W_qF$ and $\psi_L\in I^4W_qL$.  Therefore,
Theorem \ref{baeza} yields that there exists $d\in F$ with
$\psi\equiv\pfr{a,b,d}\bmod I^4W_qF$.

Consider first the case $s=5$.  Now $\qf{1,a,b,ab}\prec\pfr{a,b,d}$
and $\qf{1,a,b,ab}\prec\rho$.  Hence, 
$i_{\ti}(\pfr{a,b,d}\perp\rho)=
i_W(\pfr{a,b,d}\perp\rho)\geq\dim\qf{1,a,b,ab}=4$.
Thus, $\pfr{a,b,d}\perp\rho\cong\rho'\perp (4\times\HH )$ with
$\rho'$ nonsingular and $\dim\rho'=10$.
$\rho'\perp\sigma\sim\pfr{a,b,d}\perp\rho\perp\sigma\sim
\pfr{a,b,d}\perp\sigma\sim\sigma$, which shows for 
dimension reasons that $\rho'$ is an n.s.c. of $\sigma$.
Furthermore, $q:=\phi_r\perp\eta_r\perp\rho'\in I^4W_qF$
and $2^4>2s=10$.  Lemma \ref{knebusch-lem1} implies that
$q$ is similar to a Pfister form of which $\phi\cong\phi_r\perp\sigma$
is a Pfister neighbor.

Suppose now that $s=4$.  Let $E=F(\sigma)$.  Similarly as
above, one shows that $\sigma''=\an{(\sigma_E)}$ is of dimension
$2$ or $3$, that the nondefective part $\phi''$ of $\phi_E$ is
of type $(r,s'')$ with $s''=\dim\sigma''$, and that
$\psi_{E(\phi )}\sim 0$ and therefore $\psi_{E(\phi'')}\sim 0$.
Since $\psi\equiv\pfr{a,b,d}\bmod I^4W_qF$, this implies
$\pfr{a,b,d}_{E(\phi'')}\equiv 0\bmod I^4W_qE(\phi'')$ and thus,
by the Arason Pfister Hauptsatz, $\pfr{a,b,d}_{E(\phi'')}\sim 0$.
If $\pfr{a,b,d}_E$ is anisotropic, then by the
domination theorem, there exists $y\in E^*$ with
$y\phi''\prec\pfr{a,b,d}$, hence $2r+2s''\leq \dim\pfr{a,b,d}=8$, 
a contradiction as
$10\leq 2r+2s''$.  Hence, $\pfr{a,b,d}_E\sim 0$.  
If $\pfr{a,b,d}$ is anisotropic, then by the domination
theorem and using the fact that both $\sigma$ and $\pfr{a,b,d}$
represent $1$, we get $\sigma\prec\pfr{a,b,d}$.  This is clearly
also true if $\pfr{a,b,d}$ is hyperbolic.  In any case,
similarly as above, there exists an n.s.c.  $\rho''$ of $\sigma$
with $\rho\perp\pfr{a,b,d}\sim\rho''$ and $\sigma\prec\rho''$.
By putting this time $q:=\phi_r\perp\eta_r\perp\rho''$,
we have $q\in I^4W_qF$ and we conclude as before.
\end{proof}

\section{Forms of nondefective height $1$}\label{section-height1}
Let $q$ be an anisotropic form of type $(r,s)$ which is not totally
singular, i.e. $r\geq 1$.  Recall that $q$ is of nondefective height $1$
if $\an{(q_{F(q)})}$ is totally singular, or in other words,
by Lemmas \ref{exc} and \ref{ts-fctfield}, $\an{(q_{F(q)})}\cong
\ql(q)_{F(q)}$.  Equivalently, $h_{\nd}(q)=1$ if and only
if $i_W(q_{F(q)})=r$.  (In Knebusch's terminology \cite{Kn4}, such a
form is said to be of height $1$.)  Note that anisotropic forms
of nondefective height $1$
have therefore the property that their anisotropic part over
their own function field is defined over the base field, thus we
can use our results from the previous section.

\begin{exam}\label{h1-r1}
Let $q$ be anisotropic of type $(1,s)$.  Then clearly, $h_{\nd}(q)=1$
(Lemma \ref{ts-fctfield}).
\end{exam}

\begin{exam}\label{close-pfnb} 
Let $q$ be an anisotropic Pfister neighbor of type $(r,s)$.  
Let $n$ be such that $2^{n-1}<\dim q =2r+s\leq 2^n$.  Then $q$ is a 
Pfister neighbor of some $n$-fold Pfister form and
necessarily $2r+2s\leq 2^n$.  The complement $q^c$ has dimension
$\dim q^c=2^n-2r-s$ and we have $\ql (q)\cong\ql (q^c)$
and $\an{(q_{F(q)})}\cong q^c_{F(q)}$ by Proposition \ref{pfnb-fctfield}.
It follows
immediately that $h_{\nd}(q)=1$ if and only if $2r+2s= 2^n$
if and only if $\ql (q)=q^c$, i.e. $q$ is a close Pfister neighbor
in  Knebusch's terminology. 
\end{exam}

\begin{exam}\label{h1-notclose}
We recall the construction in Example \ref{counterex3} of forms of 
nondefective height $1$  which are  not Pfister neighbors.
Let $\psi$ be a Pfister neighbor of type $(r',s')$ of
a Pfister form $\pi\cong\pfr{a_1,\cdots ,a_n;b}$, let
$\tau\cong\pff{a_1,\cdots ,a_m}$ with $m\geq n+1$, and suppose
that $\pi\perp x\tau$, $x\in F^*$, is anisotropic.  Suppose furthermore
that $2r'+2s'=2^{n+1}$ so that $\psi$ is a {\it close} Pfister
neighbor of $\pi$.  Let $q =\psi\perp x\tau$.  Then
$\ql (q)=\psi^c\perp x\tau$, $q$ is not a Pfister neighbor,
and $h_{\nd}(q)=1$ (see Example \ref{counterex3}).  
\end{exam}

\begin{lem}\label{ndh>1}  
Suppose that $q$ is anisotropic of type $(r,s)$, $r\geq 2$.
If there exists an integer $n\geq 1$ such that $r+s<2^n<2r+s$,
then $h_{\nd}(q)\geq 2$.
\end{lem}

\begin{proof}  Suppose that $h_{\nd}(q)=1$. 
Let $V$ be the underlying vector space of $q$,
and let $W\subset V$ be a subspace with $\dim W=r+s+1$ and
$q'=q|_W$.  By assumption, $i_W(q_{F(q)})=r$.  Hence, $V\otimes F(q)$ 
contains a totally
isotropic subspace (with respect to $q_{F(q)}$) of dimension $r$
which intersects with $W\otimes F(q)$ as $r+\dim W>\dim V$.
In particular, $q'_{F(q)}$ is isotropic.  But $q'\prec q$ and therefore,
$q'$ is anisotropic over $F$, and $\dim q'\leq 2^n<\dim q$,
a contradiction to Theorem \ref{sep-2-power}.
\end{proof}

We now come to the classification of forms of nondefective height $1$
of type $(r,s)$ with $r\geq 2$ and small values of $s$ (recall
that the case $r=1$ is trivial by Example \ref{h1-r1}).

\begin{thm}\label{ndh1-s<6}  Let $q$ be anisotropic of type $(r,s)$,
$r\geq 2$.  Suppose that $s\leq 4$ or $s=5$ and there exist $u,v,w\in F^*$
such that $\ql (q)$ is similar to $\pff{u,v}\perp\qf{w}$.  Then
$h_{\nd} (q)=1$ if and only if $q$ belongs to one of the following
(mutually exclusive) classes of forms~:
\begin{enumerate}
\item[(i)]  There exists an integer $n\geq 1$ such that $r+s=2^n$
and $q$ is a Pfister neighbor;
\item[(ii)] $(r,s)=(2,4)$ and there exist $a,b,c,d,e\in F^*$ such that
$q\cong d\pfr{a,b}\perp e\pff{a,c}$.
\end{enumerate}
\end{thm}

\begin{proof}
If $q$ is as in (i), then $q$ is in fact a close Pfister neighbor
of some $(n+1)$-fold Pfister form since $2r+2s=2^{n+1}$, and we have
that $h_{\nd} (q)=1$ by Example \ref{close-pfnb}.

If $q$ is as in (ii), then $h_{\nd}(q)=1$ by Example \ref{h1-notclose}.
Note that in this case, $q$ cannot be a Pfister neighbor as
$\dim q=8$ and $q$ is not similar to a $3$-fold Pfister form. 

Now assume that $h_{\nd}(q)=1$.  Then $\an{(q_{F(q)})}\cong
\ql (q)_{F(q)}$.  If $2r>s$, then by Theorem \ref{knebusch},
$q$ is a Pfister neighbor, say, of an $(n+1)$-fold Pfister form,
and we have $\an{(q_{F(q)})}\cong q^c_{F(q)}$
by Proposition \ref{pfnb-fctfield}.  Hence, $\dim q^c=\dim\ql (q)$.
Now $\ql (q)\subset q^c$ and thus
$q^c\cong\ql (q)$.  This readily yields that $q$ is a Pfister neighbor
of codimension $s$, i.e. $\dim q =2r+s=2^{n+1}-s$, and we get
$r+s=2^n$.

Since $s\leq 5$ and $r\geq 2$, the only cases which are not yet
covered are~:  $(r,s)\in \{ (2,4),\ (2,5)\}$.  The case $(r,s)=(2,5)$
cannot occur by Lemma \ref{ndh>1}.  

Let us finally assume that $(r,s)=(2,4)$ and write (after possibly scaling)
$q\cong \phi\perp\qf{1,u,v,w}$ with $\phi$ nonsingular of
dimension $4$ and $u,v,w\in F^*$.  By Proposition \ref{ts-ndeg}, the
anisotropy of $\qf{1,u,v,w}$ yields two possible cases~:
Either $[F^2(u,v,w):F^2]=8$, or $[F^2(u,v,w):F^2]=4$, and in this latter
case one has in fact $\qf{1,u,v,w}\cong\qf{1,u,v,uv}$.

Suppose first that $[F^2(u,v,w):F^2]=8$ and let $K=F(\pff{u,v})$.
It follows readily from Corollary \ref{qpf-isotr} that
$\pff{u,w}_K$ is anisotropic, hence also $\qf{1,u,w}_K$.  Now
$\qf{1,u,v}_K$ is isotropic (Proposition \ref{qpfnb}), hence
$\an{(\qf{1,u,v,w}_K)}\cong\qf{1,u,w}_K$.  Let  
$q'\cong\phi\perp\qf{1,u,w}$.  Then $q_K\sim q'_K$,
$q'_K$ is nondefective,     
and $K(q)$ is a purely transcendental
extension of $K(q')$.  Now $q_{F(q)}\sim\qf{1,u,v,w}$.
Hence, 
$$\qf{1,u,v,w}_{K(q)}\sim q_{K(q)}\sim q'_{K(q)}\sim\qf{1,u,w}_{K(q)}$$
and thus $q'_{K(q')}\sim\qf{1,u,w}_{K(q')}$.

If $q'_K$ is anisotropic, then $q'_K$ is of type $(r,s-1)=(2,3)$ and 
$h_{\nd}(q'_K)=1$ which, by the above, implies that $q'_K$ is a 
Pfister neighbor, necessarily of some $3$-fold, which is
impossible as $2r+2(s-1)=10>8$.  If $\an{(q'_K)}$ is of type
$(1,3)$, then $i_W(q'_K)=1$  and $K(q')/K$ is purely
transcendental.  This implies that the $5$-dimensional
form $\an{(q'_K)}$ stays anisotropic
over $K(q')$, impossible as $q'_{K(q')}\sim\qf{1,u,w}_{K(q')}$.
Hence, $q'_K$ is of type $(0,3)$, i.e. $q'_K\sim\qf{1,u,w}_K$.

Now by assumption, $q_{F(q)}\sim\qf{1,u,v,w}_{F(q)}$, and by 
Lemma \ref{knebusch-lem2}, there exists an n.s.c. $\psi$ of
$q$ with $\dim\psi =2r+2s=12$ and $\psi\in I^3W_qF$.  By 
Lemma \ref{regcompl}, there exists an n.s.c. $\tau$  over $K$ of
$\qf{1,u,v,w}_K\cong\qf{1,u,w,0}_K$ such that
$\psi_K\sim\tau\in I^3W_qK$.  Since $\dim\tau =8$ and since
$\tau$ is isotropic, it follows from the Arason-Pfister Hauptsatz that
$\psi_K\sim\tau\sim 0$.  By Theorem \ref{baeza}, there exists
$x\in F^*$ with $\psi\equiv\pfr{u,v,x}\bmod I^4W_qF$.
Now $i_W(\psi\perp\pfr{u,v,x})\geq 3$, hence
$\dim\an{(\psi\perp\pfr{u,v,x})}\leq 12+8-6=14<16$, and again by
the Arason-Pfister Hauptsatz, $\psi\perp\pfr{u,v,x}\sim 0$.
Hence,
$$\psi\perp\qf{1,u,v,w}\sim q\sim\pfr{u,v,x}\perp\qf{1,u,v,w}\sim
uv[1,x]\perp\qf{1,u,v,w}\ ,$$
and by comparing dimensions, we conclude that $q$ is isotropic,
a contradiction.

Thus, we have $[F^2(u,v,w):F^2]=4$ and $q\cong\phi\perp\qf{1,u,v,uv}$.
Again by Lemma \ref{knebusch-lem2}, there exist $x,y,z\in F$ such that
$\phi\perp [1,x]\perp [u,y]\perp [v,z]\perp [uv,0]\in I^3W_qF$.
Now there are no anisotropic forms of dimension $10$ in $I^3W_qF$
(\cite[pp. 129--130]{B2}), and thus, by the Arason-Pfister Hauptsatz,
there exists $\pi\in GP_3F$ such that 
$\phi\perp [1,x]\perp [u,y]\perp [v,z]\sim\pi$.  Hence,
$$\phi\perp [1,x]\perp [u,y]\perp [v,z]\perp\pi\perp\qf{1,u,v,uv}
\sim q\perp\pi\sim\qf{1,u,v,uv}\ .$$
This implies that $\an{(q\perp\pi )}\cong\qf{1,u,v,uv}$.  
Now $\dim (q\perp\pi )=16$,
hence $i_W(q\perp\pi )=6$,
and by Proposition \ref{commondomform}, there exists a form $\eta$
such that $\dim\eta =6$, $\eta\prec\pi$, $\eta\prec q$.
In particular, $\eta$ is a $6$-dimensional Pfister neighbor, therefore of
type $(3,0)$ or $(2,2)$. But $q$ is of type $(2,4)$ and $\eta\prec q$,
hence $\eta$ is of type $(2,2)$.  Then there exist
$a,b,d,e\in F^*$ such that $\eta\cong d\pfr{a,b}\perp e\qf{1,a}$
(see \cite[Prop. 3.2]{La1}).  Since $\eta\prec q$, it follows
that there exist $c,f\in F^*$ such that $q\cong d\pfr{a,b}\perp 
e\qf{1,a,c,f}$. Comparing quasi-linear parts, we obtain
$e\qf{1,a,c,f}\cong\qf{1,u,v,uv}$.  Propositions \ref{qlpf} and 
\ref{ts-ndeg}
readily imply that $\qf{1,u,v,uv}\cong\qf{1,a,c,f}\cong\qf{1,a,c,ac}$.
\end{proof}



  





\section{Totally singular forms and quasi-Pfister neighbors}\label{section-ts}
In Theorem \ref{knebusch} in the case $s=5$, we have 
considered only the case
of totally singular forms similar to forms of type
$\pff{a,b}\perp\qf{c}$.  In characteristic $\neq 2$,
forms similar to such diagonal forms are exactly the $5$-dimensional 
Pfister neighbors, so it seems natural to ask 
whether there is a meaningful notion of totally singular 
Pfister neighbors in characteristic $2$.  

The purpose of this section is to develop some basic notions
of a theory of totally singular forms in a way which is
as much independent of the theory of nonsingular forms
as possible and yet reflects some analogous properties,
including a notion analogous to that of Pfister neighbors.

Some basic properties have been mentioned before as they have been
used in the proofs of our results.  Let us recall some
of them to keep this section as self-contained as possible.
(See also  Lemma \ref{exc} and the remarks preceding it.)

Recall that the anisotropic part of a totally singular
form $\sigma =\qf{a_1,\cdots ,a_n}$ is uniquely determined by
the $F^2$-vector space spanned by $\{ a_1,\cdots ,a_n\}$ inside
$F$, and this $F^2$-vector space is nothing but $D_F(\sigma )\cup\{ 0\}$.
This immediately implies

\begin{prop}\label{ts-facts} \begin{enumerate}
\item[(i)]  There is a one-to-one correspondence
$$\begin{array}{rcl}
\left\{\begin{array}{l}\mbox{isometry classes of}\\ 
 \mbox{totally singular forms}\end{array}\right\} & \rightleftharpoons &
\left\{\begin{array}{l}\mbox{finite-dimensional}\\
\mbox{$F^2$-vector spaces inside $F$}\end{array}\right\}\times\NN_0 
\smallskip\\
\sigma & \rightarrow & (V,\dim\sigma -\dim_{F^2}V)\quad\mbox{for 
$V=D_F(\sigma )\cup\{ 0\}$}\smallskip\\
\qf{a_1,\cdots ,a_n}\perp m\times\qf{0} & \leftarrow & (V,m)\quad
\mbox{for an $F^2$-basis $\{ a_1,\cdots ,a_n\}$ of $V$}
\end{array}$$
\item[(ii)] $\phi =\qf{a_1,\cdots ,a_n}\subset\psi =\qf{b_1,\cdots ,b_m}$
if and only if $m\geq n$ and the sub-$F^2$-vector space inside $F$
spanned by $\{a_1,\cdots ,a_n\}$ is contained in
the sub-$F^2$-vector space inside $F$ spanned by  $\{ b_1,\cdots ,b_n\}$
if and only if $m\geq n$ and $D_F(\phi )\cup\{ 0\}
\subset D_F(\psi)\cup\{ 0\}$.
\item[(iii)]  {\em (Cf. Lemma \ref{exc}.)} Let
$\phi =\qf{a_1,\cdots a_n}$. If $E/F$ is any
field extension, then there exist (after possibly
reindexing) $0\leq m\leq n$ such that $\an{(\phi_E)}
\cong\qf{a_1,\cdots ,a_m}$. 
\end{enumerate}
\end{prop}





Recall that over a field $K$ of characteristic $\neq 2$, 
the norm group of
a quadric $Q$  associated to a (nonsingular) quadratic form $\phi$
is given by the subgroup of $K^*$ generated by elements $ab$ with
$a,b\in K^*$ represented by $\phi$ (see \cite[Lemme 2.2]{CT}).  
In analogy to this
we make the following definition :

\begin{defi}
Let $\sigma$ be a totally singular  form over
$F$.  The {\it norm field} of $\sigma$ over $F$,
denoted by $N_F(\sigma)$, is defined to be the field
$F^2(ab\,|\,a,b\in D_F(\sigma ))$. The
{\it norm degree} of $\sigma$ over $F$ is defined to be
$\ndeg_F (\sigma)=[N_F(\sigma ):F^2]$.
\end{defi}

The proof of the following lemma is easy and left to
the reader.

\begin{lem}
Let $\sigma\cong\qf{a_1,\cdots ,a_n}$ be a nonzero 
totally singular form with $a_1\neq 0$.  Then, for
each $a\in F^*$, we have
$N_F(a\sigma)=N_F(\sigma )=F^2(a_1a_2,\cdots ,a_1a_n)$.
\end{lem}

This implies in particular that the norm field and the norm degree
are invariants of the similarity class of a totally singular form.

This lemma also readily implies

\begin{coro}\label{nfld-ext}
Let $\sigma$ be a nonzero totally singular form and let $K/F$
be a field extension.  Let $b_1,\cdots b_m\in F$ be such that
$N_F(\sigma )=F^2(b_1,\cdots ,b_m)$.  Then $N_K(\sigma_K)=
K^2(b_1,\cdots ,b_m)$.
\end{coro}

We now come to the characterization of quasi-Pfister forms.


\begin{prop}\label{qlpf}  \begin{enumerate}
\item[(i)] Let $q=\pff{a_1,\cdots ,a_n}$ be a quasi-Pfister form over $F$.  
Then $N_F(q)=F^2(a_1,\cdots ,a_n)=D^*_F(q)\cup\{ 0\}$.  In particular, 
if $x\in D^*_F(q)$, then $q\cong xq$.
\item[(ii)] Conversely, let $q$ be a nonzero totally singular form such that
$q\cong xq$ for each $x\in D_F^*(q)$.  Then $\an{q}$ is a
quasi-Pfister form.
\item[(iii)] An anisotropic totally singular form 
$q$ is similar to a quasi-Pfister form if and only
if $\dim q=\ndeg_F(q)$.
\item[(iv)] There exists a natural bijection between anisotropic
$n$-fold quasi-Pfister forms and purely inseparable extensions
of $F^2$ of degree $2^n$ inside $F$ which is given by
$\pff{a_1,\cdots ,a_n}\leftrightarrow F^2(a_1,\cdots ,a_n)$.
\end{enumerate}
\end{prop}



\begin{proof}  
(i) Let $q=\pff{a_1,\cdots ,a_n}$ with $a_i\in F^*$.  
Clearly, $N_F(q)=F^2(a_1,\cdots ,a_n)$.
Also, we obviously have $F^2[a_1,\cdots ,a_n]= D_F(q)\cup\{ 0\}
\subset N_F(q)$.  Since the $a_i$ are algebraic over $F^2$,
we have $D_F(q)\cup\{ 0\}=N_F(q)$.  It follows
readily that the coefficients of the diagonal forms $q$ and $xq$,
$x\in N_F(q)^*=D_F^*(q)$ span the same sub-vector space of the 
$F^2$-vector space
$F$, namely $N_F(q)$.  Hence, $q\cong xq$. 

(ii) Since $\qf{0}\cong x\qf{0}$ for each $x\in F$, we clearly
may assume by Lemma \ref{wc2} that $q$ is anisotropic. 
$R=D_F(q)\cup\{ 0\}$ is an additive subgroup of $F$, and
the roundness of $q$ implies that $D_F(q)$ is closed under multiplication.
Hence, $R$ is a finite-dimensional $F^2$-algebra inside the field $F$
and one obviously has $\dim_{F^2}(R)=\dim q$.
In particular, $R$ is a field.  Let $q=\qf{a_1,\cdots ,a_m}$ and let 
$n\leq m$ be minimal such that, after possibly reindexing,
$F^2(a_1,\cdots ,a_m)=F^2(a_1,\cdots , a_n)$. Similarly as in part
(i), we get $R=F^2(a_1,\cdots , a_n)$ which, by the minimality of $n$,
yields $m=\dim_{F^2}R=2^n=\dim q$.
Now $\{ a_1,\cdots ,a_m\}$ is an $F^2$-basis of $R$, as is
$\{ \prod_{i\in I}a_i\,|\,I\subset\{ 1,\cdots ,n\}\,\}$.  It follows readily
that $q\cong\pff{a_1,\cdots ,a_n}$.

(iii)  Let $\ndeg_F(q)=2^n$
and $a_1,\cdots ,a_n$ with $N_F(q)=F^2(a_1,\cdots ,a_n)$.
Suppose that $\dim q=\ndeg (q)$.
We may assume that $q\cong\qf{1,\cdots }$ and thus
$V=D_F(q)\cup\{ 0\}\subset N_F(q)=D_F(\pff{a_1,\cdots ,a_n})\cup\{ 0\}$.
By Proposition \ref{ts-facts}(ii), we get $q\subset\pff{a_1,\cdots ,a_n}$
and hence $q\cong\pff{a_1,\cdots ,a_n}$ for dimension reasons.

The converse is clear.


(iv) is obvious and left to the reader. \end{proof}




\begin{prop}\label{ts-ndeg}  Let $\sigma$ be a nonzero totally
singular form.  
\begin{enumerate}
\item[(i)] There exists an $m\geq 0$ such that
$\ndeg_F (\sigma )=2^m$.
\item[(ii)] With $m$ as in (i), we have
$m+1\leq\dim\an{\sigma}\leq 2^m$.
\item[(iii)]  With $m$ as in (i) and if $a_1,\cdots ,a_m\in F^*$ 
are chosen such that
$N_F(\sigma )=F^2(a_1,\cdots ,a_m)$, then for each
$a\in D_F^*(\sigma)$ we have $a\an{\sigma }\subset
\pff{a_1,\cdots ,a_m}$.  Furthermore, if
$b_1,\cdots , b_m\in F^*$ are such that 
$a\an{\sigma }\subset\pff{b_1,\cdots , b_m}$, then
$\pff{a_1,\cdots ,a_m}\cong\pff{b_1,\cdots , b_m}$, and this
$m$-fold quasi-Pfister form is aniso\-tropic.
\end{enumerate}
\end{prop}
\begin{proof}  We may assume that $\sigma$ is
anisotropic and $\sigma\cong\qf{1,a_1,\cdots ,a_n}$.
Hence, $N_F(\sigma )=F^2(a_1,\cdots ,a_n)$.
Let $m\leq n$ be minimal such that  (after reindexation)
$N_F(\sigma )=F^2(a_1,\cdots ,a_m)$.  It is clear
that $2^m=\ndeg (\sigma)$, which shows (i).  
To prove (ii), we note that the anisotropy of
$\sigma$ implies that the $\{ 1,a_1,\cdots ,a_n\}\subset N_F(\sigma)$ 
are $F^2$-linearly independent and can therefore be extended
to an $F^2$-basis of the extension $N_F(\sigma)/F^2$, which
immediately yields that $\sigma\subset\pff{a_1,\cdots ,a_m}$.
Hence, $m+1\leq\dim\sigma\leq 2^m$. 
Part (iii) follows readily from
Proposition \ref{qlpf}(iv).
\end{proof}

\begin{prop}\label{ndeg-reduction}  Let $\sigma$ be anisotropic totally
singular, $\ndeg_F(\sigma )=2^m$ and let $a_1,\cdots , a_m\in F^*$ be
such that $N_F(\sigma )=F^2(a_1,\cdots , a_m)$.
 Let $K/F$ be any field extension such that $\sigma_K$ is isotropic.
Then $\ndeg_K(\sigma_K)=2^{\ell}<2^m$ and there exist
$\{ a_{i_1},\cdots ,a_{i_{\ell}}\}\subset \{ a_1,\cdots ,a_m\}$
such that $N_K(\sigma_K)=K^2(a_{i_1},\cdots ,a_{i_{\ell}})$.
In particular, for each $a\in D_K^*(\sigma )$, we have
$a\an{(\sigma_K)}\subset\pff{a_{i_1},\cdots ,a_{i_{\ell}}}$
\end{prop}

\begin{proof}  After scaling, we may assume that $1\in D_F(\sigma )$.
By Proposition \ref{ts-ndeg}(iii), $\sigma\subset\pff{a_1,\cdots ,a_m}$.
Now $\sigma_K$ is isotropic, hence also $\pff{a_1,\cdots ,a_m}_K$. 
By Proposition \ref{qlpf}(iv), this implies that
$[K^2(a_1,\cdots ,a_m):K^2]=2^{\ell}<2^m$.    Clearly, we may find
$\{ a_{i_1},\cdots ,a_{i_{\ell}}\}\subset \{ a_1,\cdots ,a_m\}$
with $K^2(a_1,\cdots ,a_m)= K^2(a_{i_1},\cdots ,a_{i_{\ell}})$.
Now $N_K(\sigma_K)=K^2(a_1,\cdots ,a_m)$.  All this, together with 
Proposition \ref{ts-ndeg}(iii) implies the desired result.
\end{proof}


\begin{defi}  A totally singular form $\sigma$
is called a {\it quasi-Pfister neighbor} if there exists
$a\in F^*$ and an anisotropic $m$-fold
quasi-Pfister form $\pi$ ($m\geq 0$) such that $a\sigma\subset\pi$ and
$2\dim\sigma>\dim\pi=2^m$.  In this situation, $\sigma$ is said
to be a quasi-Pfister neighbor of $\pi$.
\end{defi}



\begin{prop}\label{qpfnb}\begin{enumerate} 
\item[(i)]  If $\sigma$ is a quasi-Pfister neighbor of
the anisotropic $m$-fold quasi-Pfister form $\pi$, then 
$N_F(\sigma )=N_F(\pi)$.  In particular, if 
$\sigma$ is also a quasi-Pfister neighbor of 
another anisotropic $m$-fold quasi-Pfister form  $\tau$,
then $\pi\cong\tau$. 
\item[(ii)] A totally singular form
$\sigma$ is a quasi-Pfister neighbor if and only if 
$\sigma$ is anisotropic and $2\dim\sigma >\ndeg_F(\sigma )$.
\item[(iii)] Let $\sigma$ be a quasi-Pfister 
neighbor of the anisotropic $m$-fold quasi-Pfister form $\pi$, $m\geq 1$.
Let $E/F$ be a field extension.  Then $\sigma_E$ is
isotropic iff $\pi_E$ is isotropic.
\end{enumerate}
\end{prop}

\begin{proof} (i)  Clearly, $N_F(\sigma )\subset N_F(\pi)$.
Hence $\ndeg_F(\sigma )\leq\ndeg_F\pi =2^m$. Since $\sigma$ is
anisotropic of dimension $>2^{m-1}$, it follows from
Proposition \ref{ts-ndeg} that $\ndeg_F(\sigma )=2^m$ and thus
$N_F(\sigma )=N_F(\pi)$.  Now if $\sigma$ is also a 
quasi-Pfister neighbor of $\tau$, then $N_F(\pi )=N_F(\tau)$
and $\pi\cong\tau$ by Proposition \ref{qlpf}.

(ii) If $\sigma$ is a quasi-Pfister neighbor, say, of
the anisotropic $m$-fold quasi-Pfister form $\pi$,
then by (i) we have $2\dim\sigma >\dim\pi =2^m=
\ndeg_F(\pi)=\ndeg_F(\sigma )$.

Conversely, let $\sigma$ be anisotropic and
$\ndeg_F(\sigma )=2^m$.  Then there exist $a_1,\cdots ,a_m$
$\in F^*$ with $N_F(\sigma )= F^2(a_1,\cdots ,a_m)$ and 
$\pi=\pff{a_1,\cdots ,a_m}$ is anisotropic 
(Proposition \ref{qlpf}) and for each $a\in D_F^*(\sigma )$,
we have $a\sigma\subset\pi$ (Proposition \ref{ts-ndeg}).
Now $2\dim\sigma >\ndeg_F(\sigma )=2^m=\dim\pi$ implies
that $\sigma$ is a Pfister neighbor of $\pi$.

(iii) Clearly, the isotropy of
$\sigma_E$ implies that of $\pi_E$.
Now the definition of a quasi-Pfister neighbor
and its associated quasi-Pfister form imply by
Proposition \ref{ts-ndeg} that 
$N_E(\sigma_E )=N_E(\pi_E)$.
Hence $\ndeg_E(\sigma_E)=\ndeg_E(\pi_E)$.
Now $\pi_E$ being isotropic implies that
$\ndeg_E(\pi_E)\leq 2^{m-1}$, hence $\dim\an{(\sigma_E)}
\leq 2^{m-1}<\dim\sigma$.
\end{proof}

Recall that in characteristic $\neq 2$, the
splitting pattern of a quadratic form is the increasing
sequence of Witt indices of this form over the fields
in a generic splitting tower (see, e.g., \cite{HuR1}, \cite{HuR2}, 
\cite{H2}).
In characteristic $2$, it it more reasonable to define
a standard splitting pattern of a (possibly singular) form
to be the sequence of pairs $(r,s)$ denoting the types
of the anisotropic parts of this form over the fields in
the standard splitting tower (see, e.g. \cite{La2}). 
For totally singular forms, this just means the sequence of
the dimensions of the anisotropic parts over the fields
in the standard splitting tower (cf. subsection \ref{subsection-fctfld}).  
Hence, if $F=F_0\subset F_1\subset\cdots \subset F_{h-1}\subset F_h$
denotes the standard splitting tower of the nonzero totally singular
form $\sigma$ of height $h$, then the standard splitting pattern is given
by $(s_0,\cdots ,s_{h-1},s_h)$ with $s_i=\dim\an{(\sigma_{F_i})}$.

The following result is a more precise version of Proposition
\ref{ndeg-reduction} in the case where $K=F(\sigma)$.

\begin{lem}\label{firstkernel}
Let $\sigma\cong\qf{1}\perp\sigma'$ be an anisotropic totally
singular form of dimension $\geq 2$ and let $K=F(\sigma )$.  
Suppose that $\ndeg_F(\sigma )=2^m$ and let $b_1,\cdots ,b_m\in F^*$
be such that $N_F(\sigma )=F^2(b_1,\cdots ,b_m)$.  Then
$\ndeg_K(\sigma_K)=2^{m-1}$, $N_K(\sigma _K)=K^2(b_2,\cdots ,b_m)$.
In particular, $\an{(\sigma_K)}\subset\pff{b_2,\cdots ,b_m}_K$.
\end{lem}

\begin{proof}  Since $1\in D_F(\sigma)$, we have
$N_F(\sigma )=F^2(a\,|\,a\in D_F(\sigma'))$.  Hence, there
exists $c_1\in D_F(\sigma')$ with $c_1\notin F^2(b_2,\cdots ,b_m)$.
Thus, $F^2(b_1,\cdots ,b_m)=F^2(c_1,b_2,\cdots ,b_m)$.
Let $n=\dim\sigma'$ and let $c_2,\cdots c_n\in F^*$ such that
$\sigma'\cong\qf{c_1,\cdots ,c_n}$.  For $i\geq 2$, write
$c_i=u_i+c_1v_i$ with $u_i,v_i\in F^2(b_2,\cdots ,b_m)$.


Let $X_1,\cdots ,X_n$ be variables and put $L=F(X_1,\cdots ,X_n)$,
$d=\sum_{i=1}^nc_iX_i^2$.  Then $K=L(\sqrt{d})$ and $K^2=L^2(d)$.
If we put $r=X_1^2+\sum_{i=2}^nv_iX_i^2$ 
and $s=\sum_{i=2}^mu_iX_i^2$, then clearly 
$r,s\in L^2(b_2,\cdots ,b_m)$, $r\neq 0$ and $d=s+c_1r$.
Hence, $L^2(b_1,\cdots ,b_m)=L^2(c_1,b_2,\cdots ,b_m)=
L^2(d,b_2,\cdots ,b_m)=K^2(b_2,\cdots ,b_m)$.

By Corollary \ref{nfld-ext}, $N_K(\sigma_K)=K^2(b_1,\cdots b_m)=
L^2(d,b_1,\cdots ,b_m)$, and thus, by the above,
$N_K(\sigma_K)=K^2(b_2,\cdots ,b_m)$.

Since $L^2/F^2$ is purely transcendental, we clearly have
$2^m=[F^2(b_1,\cdots, b_m):F^2]=[L^2(b_1,\cdots ,b_m):L^2]=
[K^2(b_2,\cdots ,b_m):L^2]$.  Now $L^2(d)=K^2$ and
$[K^2:L^2]=2$.  Thus, $\ndeg_K(\sigma_K)=[N_K(\sigma_K):K^2]=
[K^2(b_2,\cdots ,b_m):K^2]=2^{m-1}$. 

Finally, $\an{(\sigma_K)}\subset\pff{b_2,\cdots ,b_m}_K$ follows
from Proposition \ref{ts-ndeg} and the fact that $1\in D_K(\sigma_K)$.
\end{proof}



It is now easy, to compute the standard splitting pattern
for quasi-Pfister neighbors.

\begin{thm}\label{splitpat} Let $\sigma$ be an anisotropic 
totally singular form.  Let 
$F=F_0\subset F_1\subset\cdots \subset F_{h}$ be the standard splitting
tower of $\sigma$ and let $\sigma_i=\an{(\sigma_{F_i})}$,
$s_i=\dim\sigma_i$.
Let $n_i$ and $m$ be such that $2^{n_i}<s_i\leq 2^{n_i+1}$ and
$\ndeg_F(\sigma )=2^m$.  Let $b_1,\cdots b_m\in F^*$ be such that
$N_F(\sigma )=F^2(b_1,\cdots b_m)$.  
\begin{enumerate}
\item[(i)] For $1\leq i\leq h$, we have
$\ndeg_{F_i}(\sigma_i)=2^{m-i}$,
$N_{F_i}(\sigma_i)=F_i^2(b_1,\cdots ,b_{m-i})$, and
$\max\{ m-i+1,2^{n_{i-1}}\}\leq\dim\sigma_{i}\leq 2^{m-i}$.
In particular, $h(\sigma )=m$.  Furthermore,
for every $a\in D_F(\sigma )$, we have
$a\sigma_i\subset\pff{b_1,\cdots ,b_{m-i}}_{F_i}$.
\item[(ii)]  $\sigma$ is a Pfister neighbor iff
$(s_0,s_1,\cdots ,s_h)=(\dim\sigma ,2^{m-1},2^{m-2},\cdots ,1)$.
In this situation, $a\sigma_i\cong\pff{b_1,\cdots ,b_{m-i}}_{F_i}$
for every $a\in D_F(\sigma )$.
\end{enumerate}
\end{thm}
\begin{proof}  (i) $\ndeg_{F_i}(\sigma_i)=2^{m-i}$,
$N_{F_i}(\sigma_i)=F_i^2(b_1,\cdots ,b_{m-i})$,
$m-i+1\leq\dim\sigma_{i}\leq 2^{m-i}$
and  $a\sigma_i\subset \pff{b_1,\cdots ,b_{m-i}}_{F_i}$
follow by induction from
Proposition \ref{ts-ndeg} and Lemma \ref{firstkernel}.

Now $2^{n_{i-1}}<\dim\sigma_{i-1}\leq 2^{n_{i-1}+1}$.
Theorem \ref{sep-2-power}  yields that any subform of
$\sigma_{i-1}$ of dimension $2^{n_{i-1}}$ stays anisotropic over 
$F_{i}$ and
will therefore be also a subform of
$\an{((\sigma_{i-1})_{F_i})}=\sigma_i$.  
Hence $\dim\sigma_i\geq 2^{n_{i-1}}$.

$h(\sigma )=m$ now follows readily by induction on $m$ 
and the definition of height.


(ii)  If $\sigma$ is a Pfister neighbor, then
$2^{m-1}<\dim\sigma\leq 2^m=\ndeg_F(\sigma)$ by Proposition
\ref{qpfnb}, and it follows
immediately from (i) that $\sigma$ has standard splitting pattern
$(\dim\sigma ,2^{m-1},2^{m-2},\cdots ,1)$.

Conversely, if the standard splitting pattern is of that type,
then $h(\sigma )=m$, hence $\ndeg_F(\sigma )=2^m$  and $\dim\sigma >2^{m-1}$.
Hence, $\sigma$ is a quasi-Pfister neighbor by Proposition \ref{qpfnb}.

The remaining statement follows immediately from (i) by
comparing dimensions.
\end{proof}



We now give  characterizations of quasi-Pfister
neighbors of small dimension.

\begin{prop} Let $\sigma$ be an anisotropic
totally singular form.
\begin{enumerate}
\item[(i)] If $\dim\sigma\leq 3$, then $\sigma$ is
a quasi-Pfister neighbor.
\item[(ii)] If $\dim\sigma =2^n$, then $\sigma$ is
a quasi-Pfister neighbor iff $\sigma$ is similar to
a quasi-Pfister form.
\item[(iii)]  If $\dim\sigma =5$, then $\sigma$ is
a quasi-Pfister neighbor iff there exist $a,b,c\in F^*$
such that $\sigma$ is similar to $\qf{1,a,b,ab,c}$.
\item[(iv)] If $\dim\sigma =6$, then $\sigma$ is
a quasi-Pfister neighbor iff there exist $a,b,c\in F^*$
such that $\sigma$ is similar to 
$\qf{1,a,b,ab,c,ac}$.
\item[(v)] If $\dim\sigma =7$, then $\sigma$ is
a quasi-Pfister neighbor iff there exist $a,b,c\in F^*$
such that $\sigma$ is similar to 
$\qf{a,b,c,ab,ac,bc,abc}$.
\end{enumerate}
\end{prop}

\begin{proof}
(i) and (ii) are easy and left to the reader.
It is also obvious that anisotropic forms similar to
forms of type $\qf{1,a,b,ab,c}$,  $\qf{1,a,b,ab,c,ac}$,
$\qf{a,b,c,ab,ac,bc,abc}$ are quasi-Pfister neighbors.
For the converse, let $\sigma$ be a quasi-Pfister neighbor
with $5\leq\dim\sigma\leq 7$

\smallskip

{\it The case $\dim\sigma =5$}.
After scaling, we may
assume that $\sigma\cong\qf{1,p,q,r,s}$, and by the
definition of a quasi-Pfister neighbor and
Proposition \ref{ts-ndeg}, we may furthermore
assume that $\sigma\subset\pff{p,q,r}$ with
$[F^2(p,q,r):F^2]=8$.  Then $0\neq s=u+v$ with 
$v\in D_F(\qf{1,p,q,r})\cup\{ 0\}$
and $u=pqx^2+pry^2+qrz^2+pqrt^2$ for suitable
$x,y,z,t\in F$.  If $u=0$ then $s\in D_F(\qf{1,p,q,r})$
and $\sigma$ is isotropic, a contradiction.  Hence,
$u\neq 0$ and we have $\qf{1,p,q,r,s}\cong\qf{1,p,q,r,u}$.

Suppose first that $t\neq 0$.  Then, after multiplying $u$
by a square, we may assume $t=1$.  Now
$\qf{1,p,q,r}\cong\qf{1,z^2+p,y^2+q,x^2+r}$ and
$(z^2+p)(y^2+q)(x^2+r)=u+w$ with $w\in D_F(\qf{1,p,q,r})$.
Put $f=z^2+p$, $g=y^2+q$, $h=x^2+r$.
Thus, we get  $\sigma\cong\qf{1,p,q,r,u}\cong
\qf{1,p,q,r,u+w}\cong\qf{1,f,g,h,fgh}$.
By putting $a=fg$, $b=fh$, $c=fgh$, we get
$c\sigma\cong\qf{1,a,b,ab,c}$.

Suppose now that $t=0$.  Since $u\neq 0$, it suffices
by symmetry to consider only the case $x\neq 0$.
After scaling by a square, we may assume $x=1$.  If
$y=z=0$ then $\sigma\cong\qf{1,p,q,r,u}\cong\qf{1,p,q,pq,r}$
and we are done.  Otherwise, again by symmetry, it suffices to
consider the case $y\neq 0$.  This time, we put $a=p+q(z/y)^2$,
$b=q+ry^2$.  Then $ab=u+w$ with $w\in F^2$ and we get
$\sigma\cong\qf{1,p,q,r,u}\cong\qf{1,a,b,r,u}\cong
\qf{1,a,b,r,u+w}\cong\qf{1,a,b,ab,r}$.

\smallskip

{\it The case $\dim\sigma =6$}.  By the $5$-dimensional
case, we may assume after scaling that 
$\sigma\cong\qf{1,p,q,pq,r,s}\subset\pff{p,q,r}$.
Then we can write $s=u+rv$ with $u,v\in D_F(\pff{p,q})\cup\{ 0\}$,
and the anisotropy of $\sigma$ implies $v\neq 0$.
Hence, $\sigma\cong\qf{1,p,q,pq,r,rv}$.  Also, we can
write $v=x^2+w$ with $w\in D_F(\qf{p,q,pq})$.  Since
$\qf{r,rx^2}$ is isotropic, we must have $w\neq 0$,
and thus $w\in F^2(p,q)\setminus F^2$.  But then there exists
$t\in\{ p,q\}$ with $F^2(p,q)=F^2(w,t)$ and we have
$\pff{p,q}\cong\pff{w,t}$.  Also, $\qf{r,rv}\cong\qf{r,rw}$
and thus $\sigma\cong\qf{1,p,q,pq,r,rw}\cong
\qf{1,w,t,tw,r,rw}$.

\smallskip

{\it The case $\dim\sigma =7$}.  By the $6$-dimensional
case, we may assume after scaling that 
$\sigma\cong\qf{1,p,q,pq,r,pr,s}\subset\pff{p,q,r}$.
We can write $s=u+v$ with $u\in D_F(\qf{qr,pqr})$ and
$v\in D_F(\qf{1,p,q,pq,r,pr})$.  The anisotropy implies
$u\neq 0$ and we get $\sigma\cong\qf{1,p,q,pq,r,pr,u}$.
Now $qru\in D_F(\qf{1,p})$ and thus, by Proposition \ref{qlpf}
(or Proposition \ref{ts-facts}), $up\qf{1,p}\cong u\qf{1,p}\cong qr\qf{1,p}$.
It follows that $up\sigma\cong\qf{p,q,r,pq,pr,qr,pqr}$.

\end{proof}

It should be noted that the description of quasi-Pfister
neighbors in dimensions $5,6,7$ bears a striking resemblance
to the well known characterization of Pfister neighbors
of the same dimensions in characteristic $\neq 2$
(see, e.g. \cite[p.10]{Kn3}).  However,
our methods to obtain the characterization are naturally very
different from those usually employed in characteristic $\neq 2$.

Let us now give a necessary condition
for the isotropy of an anisotropic totally singular form
over the function field of another totally singular form, which
then yields criteria for the isotropy of anisotropic quasi-Pfister
forms over function fields of totally singular forms.
Recall that anisotropic totally singular forms stay
anisotropic over function fields of forms which are not
totally singular, so there is no need to consider this case.

\begin{prop} Let $\phi$, $\psi$ be nonzero 
totally singular forms with $\phi$ anisotropic.
Let $K=F(\psi )$.  If $\phi_K$ is isotropic,
then $N_F(\psi )\subset N_F(\phi )$.
\end{prop}

\begin{proof} Without loss of generality, we may 
assume that $\psi$ is anisotropic, $\dim\psi\geq 2$,
and, after scaling, $\psi\cong\qf{1,b_1,\cdots ,b_n}$.

Let $\ndeg_F(\phi )=2^m$ and let $a_1,\cdots a_m\in F^*$ such that
$N_F(\phi )=F^2(a_1,\cdots a_m)$.  Let $\pi =\pff{a_1,\cdots a_m}$.
Then $\pi$ is anisotropic by Proposition \ref{qlpf}, and by
Proposition \ref{ts-ndeg}, $\phi$ is similar to a subform
of $\pi$.  Hence, $\phi_K$ isotropic  implies that $\pi_K$
is isotropic.

Let $i\in\{ 1,\cdots ,n\}$ and put $L'=F(\qf{1,b_i})=F(X)(\sqrt{b_iX^2})$
(where $X$ denotes a variable), $L=F(\sqrt{b_i})$.
By a  standard specialization argument (\cite[Lemma 4.5]{La1}), the
isotropy of $\pi_K$  implies
that of $\pi_{L'}$, and therefore also that of $\pi_L$.
Hence, necessarily $\ndeg_L(\pi_L)<\ndeg_F(\pi)=2^m$.
But $N_F(\pi )=N_F(\phi )=F^2(a_1,\cdots a_m)$,
and $N_L(\pi_L)=F^2(a_1,\cdots a_m,b_i)$.  If
$b_i\notin F^2(a_1,\cdots a_m)$, then 
$[F^2(a_1,\cdots a_m,b_i):F^2]=2^{m+1}$ and thus 
$[F^2(a_1,\cdots a_m,b_i):F^2(b_i)]=[L^2(a_1,\cdots a_m):L^2]
=\ndeg_L(\pi_L)=2^m$, a contradiction.  Thus,
$b_i\in F^2(a_1,\cdots a_m)$ for every $i$, hence we obtain
$N_F(\psi )=F^2(b_1,\cdots ,b_n)\subset N_F(\phi )$.
\end{proof}

\begin{coro}\label{qpf-isotr} Let $\pi$ be an anisotropic totally
singular $m$-fold Pfister form, $m\geq 1$, and
let $\sigma$ be a totally singular form.  Let $K=F(\sigma )$.
The following are equivalent~:
\begin{enumerate}
\item[(i)] $\pi_K$ is isotropic.
\item[(ii)] $\dim\an{\sigma}\geq 2$ and for any $a\in D_F^*(\sigma )$
one has $a\an{\sigma}\subset\pi$.
\item[(iii)] $\dim\an{\sigma}\geq 2$ and $N_F(\sigma )\subset
N_F(\pi)$.
\end{enumerate}
\end{coro}

\begin{proof}  Since $F(\sigma )/F(\an{\sigma})$ is purely
transcendental, we have that $\pi_K$ is isotropic iff 
$\pi_{F(\an{\sigma})}$ is isotropic.  Also,
$N_F(\sigma )=N_F(\an{\sigma})$, so we may assume without
loss of generality that $\sigma$ is anisotropic.

By the previous proposition, (i) implies (iii).
(iii) implies (ii) by Propositions \ref{qlpf} and
\ref{ts-ndeg}, and trivially (ii) implies (i).
\end{proof}
      
We conclude this section with some remarks on splitting towers and heights
of forms (singular or not). 
One of the important properties of generic splitting towers
in characteristic $\neq 2$ is the fact that if $\phi$ is an 
anisotropic form over the field $E$, and if $K$ is any
field extension of $E$, then there exists a field
$E_i$ in the generic splitting tower $E=E_0\subset E_1\subset
\cdots\subset E_h$ such that $\dim\an{(\phi_K)}=\dim\an{(\phi_{E_i})}$.

This property no longer holds in general for standard
splitting towers of totally singular forms as the following example
shows.

\begin{exam}\label{splitpatvar}
Consider a totally singular anisotropic
form  $q=\pff{a_1,\cdots ,a_n}\perp\qf{b}$, $n\geq 2$.  
Then $q$ is a quasi-Pfister 
neighbor of $\pff{a_1,\cdots ,a_n,b}$, and thus the standard splitting
pattern will be $(2^n+1,2^n,2^{n-1},\cdots , 2,1)$ by the preceding 
theorem.
Now for $K_m=F(\sqrt{a_1},\cdots ,\sqrt{a_m})$, $1\leq m\leq n-1$, 
it is not difficult to show that
$\an{(q_{K_m})}\cong\pff{a_{m+1},\cdots, a_n}\perp\qf{b}$, 
and it follows that $2^{n-m}+1=\dim\an{(q_{K_m})}$
does not show up in the standard splitting pattern.
\end{exam}

However, we do know by Proposition \ref{splitpatwitt} that the
dimension of the nonsingular part of the anisotropic part of a form
over a field extension will appear as dimension of the nonsingular part
of one of its higher kernel forms in the standard splitting tower.  
With our results 
on totally singular forms over field extensions, we are now able to prove
Theorem \ref{height} which shows that the maximal height of a tower of
fields over which the anisotropic parts of a form are of strictly
decreasing dimension is given by the standard height.
We restate the theorem for the readers convenience. 


\begin{thm}\label{heightrepeat}  Let $q$ be an anisotropic form of standard
height $h(q)=h$.
Let $F=K_0\subset K_1\subset\cdots\subset K_m$ be any tower of fields
such that $\dim\an{(q_{K_{i-1}})}>\dim\an{(q_{K_i})}$ for each
$i\in\{1,\cdots , m\}$. Then $m\leq h$.
\end{thm}

\begin{proof} Let $q_{(i)}=\an{(q_{K_i})}$ and denote by $(r_i,s_i)$
its type.  By assumption, $\dim q_{(i)}<\dim q_{(i-1)}$ 
and thus $r_i<r_{i-1}$ or $s_i<s_{i-1}$ for $1\leq i\leq m$.
Let $u=\card\{ i\,|\,1\leq i\leq m,\ r_i<r_{i-1}\}$ and
$v=\card\{ i\,|\,1\leq i\leq m,\ s_i<s_{i-1}\}$.    Then $u+v\geq m$.
On the other hand, by Proposition \ref{splitpatwitt}, we have
$u\leq h_{\nd}(q)$. 

Let $\sigma_i=\ql (q_{(i)})$ and $\sigma =\sigma_0=\ql (q)$.  
Now $\sigma_i=\an{(\sigma_{K_i})}$ and $\dim\sigma_i=s_i$.  
If $\ndeg_{K_i}(\sigma_i)=2^{m_i}$ and $s_i<s_{i-1}$, then by
Proposition \ref{ndeg-reduction}, $m_i<m_{i-1}$ because
the anisotropic form $\sigma_{i-1}$ becomes isotropic over $K_{i}$. Consequently,
$v\leq m_0$.  But $h(\sigma )=m_0$
by Theorem \ref{splitpat}(i), and therefore $v\leq h(\sigma )$.
We have $h(q)=h_{\nd}(q)+h(\sigma )$ (cf. \cite[Th. 4.6]{La2}), hence
$h(q)\geq u+v\geq m$.
\end{proof}
 




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\end{document}
