\titlea{\S16}{Some Forms of Height $1$}
\leftrunningtitle{Chapter III. Some Applications}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S16 Some Forms of Height $1$}

In this section $k$ can be any field. However, we are mostly
interested in fields of characteristic two. For any \nd\ \qf\
$\vf$ of height $h\geq 1$ (i.e. $\vf$ not split) over $k$, the
last but one kernel form $\varphi_{h-1}$ has height $1$. Thus, in
order to gain an understanding of \qf s, it is of prime importance
to have an overview of all anisotropic forms of height~$1$.

If $\Char k\not=2$, it is well-known (cf. the end of \S4) and it
will be shown in \S20 and \S23 that an anisotropic form $\vf$ over
$k$ has height~1 iff there exists an element $a \in k^*$ and a
Pfister form $\tau$ over $k$ with $\varphi \cong a \tau$,
$\dim \tau \geq 2$, in case  $\dim \varphi$ is even, $\varphi \cong a
\tau'$, $\dim \tau \geq 4$, in case $\dim \varphi$ is odd. Here
$\tau'$ denotes the pure part of $\tau$, cf. the  end of \S 4.
Moreover, in both cases $\tau$ is uniquely determined by $\vf$ up
to isomorphism.

Pfister forms can also be defined in case $k$ has characteristic~2
(see below). We will see that  diverse forms of height~1
can be obtained from quadratic Pfister forms in this case as well. Nonetheless, the
question remains open what other anisotropic quadratic forms of
height~1 there are. This is currently  perhaps the most important unsolved
problem about quadratic forms in characteristic~2.

\begdefinition{1.}
\smallskip

\setitemindent{(a)}
\item{(a)} Let $d \in {\Bbb N}_0$. A {\it bilinear $d$-fold
Pfister form}\Idx{bilinear Pfister form}\Idx{Pfister form!bilinear} 
over $k$ is a form
$$\tau = \langle 1, a_1 \rangle \otimes \cdots \otimes \langle 1, a_d \rangle$$
with
coefficients $a_i \in k^*$. $\{$We let $\tau = \langle 1 \rangle$, in case $d =
0$.$\}$
\medskip
\item{(b)} Let $d \in {\Bbb N}$. A {\it quadratic $d$-fold
Pfister form}\Idx{quadratic Pfister form}\Idx{Pfister form!quadratic}  
over $k$ is a form
$$\rho \otimes \left[ {1 \atop 1} \; {1\atop b} \right],$$
where $\rho$ is a bilinear Pfister form of degree $d-1$ and $b \in
k$, $1 - 4 b \ne 0$.\fonote{The form $[1]$ is thus {\it not}
considered to be a quadratic Pfister form, not even if $\Char
k\not=2$.}
\enddefinition

If $\tau$ is a bilinear $d$-fold Pfister form with $d \geq 1$,
then we have an orthogonal decomposition $\tau \cong \langle 1 \rangle \, \bot \, \tau
'$. It is well-known [Ba$_2$, p.257] that also in
characteristic~2,
the form $\tau'$ is uniquely determined by $\tau$ up to
isomorphism.\fonote{If $\tau$ is anisotropic, this follows easily
from \S20, Theorem~8 below.} We call $\tau'$ the {\it pure part of
the Pfister form $\tau$}.\Idx{pure part of a Pfister form} 
It seems that for a quadratic Pfister
form $\rho$ in characteristic~2, it does not make any sense to speak of a
pure part of $\rho$ (however, see \S26, Definition~2).

Next we will discuss some elementary properties of Pfister forms.

\begdefinition{2.} If $\vf$ is a quadratic or bilinear form over
$k$, then $D(\vf)$ denotes the set of all elements $a \in k^*$
which are values of the form $\vf$, $a = \varphi (c_1, \dots, c_n)$
for an  $n$-tuple $(c_1, \dots, c_n) \in k^n$.\fonote{If $\varphi$ is bilinear, and $c\in k^n$, we define $\varphi(c):=\varphi(c,c)$.} 
We denote by
$N(\varphi)$\Idx{represented element}\Idx{111dfi3@$D(\varphi)$}\Idx{similarity norm}\Idx{111nfi@$N(\varphi)$} 
the set of all $a \in k^*$ with $a \varphi \cong
\varphi$. We call the $a \in D (\varphi)$ {\it the elements
represented by $\vf$} and the $a \in N(\varphi)$ the {\it
similarity norms of $\vf$}.
\enddefinition

\begnote{} $D(\vf)$ is the set of all $a \in k^*$ with $\langle a
\rangle < \varphi$ in the bilinear case and $[a] < \varphi$ in the
quadratic case. Obviously, $N(\vf)$ is a subgroup of $k^*$ which
contains all squares of elements of  $k^*$. Furthermore, $D(\vf)$ is a union
of cosets of $N(\vf)$ in $k^*$.
\endnote

We call $N(\varphi)$ the {\it norm group}\Idx{norm group} 
of $\vf$. Analogously,
we define for a quadratic or bilinear module $E$ over $k$ the norm
group $N(E)$ and the set $D(E)$. For instance, if $E = (E, q)$ is
a quadratic module, then $D(E)$ is the set of all $a \in k^*$ for
which there exists an $x \in E$ with $q(x) = a$, and $N(E)$ is the
set of all $a \in k^*$ such that there exists a similarity
$\sigma\colon E \buildrel \sim \over \longrightarrow E$ with norm
$a$, i.e. a linear automorphisms $\sigma$ of $E$ with $q(\sigma (x)) = aq (x)$.

\begtheorem{1.} If $\tau$ is a bilinear or quadratic Pfister form
over $k$, then $D(\tau)=N(\tau)$.
\endtheorem

For this -- in characteristic $\not=2$ -- well-known theorem, due to
Pfister, Witt has given a particularly short proof which belongs
to the standard repertoire of all modern books on the algebraic
theory of quadratic forms (e.g. [S, p. 69 ff.], [P, p. 27]). The
proof can be extended to characteristic 2 without effort. We
recall it here for the convenience of the reader.

\begproof{of Theorem 1.} We use induction on the degree $d$ of
$\tau$. In the bilinear case  the start of the induction, $d=0$, is
trivial. If $d=1$ in the quadratic case, then $\tau$ is the
hyperbolic plane
$\left[ {0 \atop 1} \; {1
\atop 0} \right]$ or
the norm
form of a quadratic separable field extension of $k$.
In both cases the statement is clear.

Next we deduce $d$ from $d-1$. We write $\tau = \rho \, \bot \, a
\rho$, where $\rho$ is a (bilinear or quadratic) Pfister form of
degree $d-1$. Since $1 \in D (\tau)$, we clearly have
$N(\tau)~\subset~D(\tau)$. It is also clear that $N(\rho) \subset
N(\tau)$. Now, let $c \in D (\tau)$ be given. We write $c = \rho (x) + a \rho
(y)$ with vectors $x, y \in k^m$, $m = 2^{d-1}$. If $\rho (x) = 0$ or $\rho (y) =
0$, then $c$ is a similarity norm of $\tau$ since $a$ is one and
since $D (\rho) = N(\rho) \subset N(\tau)$ by our induction
hypothesis. Suppose now that $\rho (x)$ and $\rho (y)$ are both
nonzero. Then we have:
$$\eqalign{\tau & = \rho \, \bot \, a \rho \cong \rho (x) \rho \, \bot \, a \rho (y) \rho
\cong
\langle \rho (x), a \rho (y) \rangle \otimes \rho \cr
                & \cong \langle \rho (x) + a \rho (y), \; a \rho (x) \rho (y)
                (\rho (x) + a \rho (y)) \rangle \otimes \rho \cr
                & = (\rho (x) + a \rho (y)) \langle 1, a \rho (x) \rho (y)
                \rangle \otimes \rho \cr
                & \cong (\rho (x) + a \rho (y)) \langle 1, a \rangle \otimes
                \rho
                = (\rho (x) + a \rho (y)) \tau.}$$
Here we several times used the fact that $\rho (x)$ and $\rho (y)$ are
similarity norms of $\rho$.\qed
\endproof

\begtheorem{2.}
\smallskip
\setitemindent{$(ii)$}
\item{$(i)$} Let $\tau$ be an isotropic bilinear Pfister form over
$k$. Then there exists a bilinear Pfister form $\sigma$ over $k$
with $\tau \cong \langle 1, - 1 \rangle \otimes \sigma$. In
particular, $\tau$ is metabolic.
\medskip
\item{$(ii)$} Every isotropic quadratic Pfister form over $k$ is
hyperbolic.
\endtheorem

\begproof{} $(i):$ We use induction on the degree $d$ of $\tau$.
For $d=1$ the statement is clear. So suppose $d>1$. We choose a
factorization $\tau \cong \langle 1, a \rangle \otimes \rho$ where
$\rho$ is a Pfister form of degree $d-1$. If $\rho$ is isotropic
then, by our induction hypothesis, $\rho \cong \langle 1, - 1 \rangle \otimes \gamma$
for some Pfister form $\gamma$. Then $\tau \cong \langle 1, - 1 \rangle
\otimes \sigma$ with $\sigma \colon = \langle 1, a \rangle \otimes
\gamma$. Now suppose that $\rho$ is anisotropic. We have vectors $x, y \in k^m$, $m = 2^{d-1}$,
with $\rho (x) + a \rho (y) = 0$, but $x$ and $y$ not both zero.
Since $\rho$ is anisotropic, we must have $\rho (x) \ne 0$ and $\rho (y) \ne
0$. Theorem~1 then implies $-a = \rho (x) \rho (y)^{-1} \in N (\rho)$. Therefore,
$\tau \cong \langle 1, a \rangle \otimes \rho \cong \langle 1, - 1 \rangle \otimes \rho$.

$(ii):$ The argumentation is completely analogous.\qed
\endproof

\begcorollary{} If $\tau$ is an anisotropic quadratic  Pfister
form over $k$, then $\tau$ has height $1$.
\endcorollary

\begproof{} $\tau \otimes k (\tau)$ is an isotropic Pfister form
over $k(\tau)$. Thus, $\tau \otimes k (\tau) \sim 0$ by
Theorem~2.\qed
\endproof

If $\Char k\ne 2$, $\tau$ is a quadratic Pfister form over $k$ and
$\vf$ is a subform of $\tau$ with $\dim \varphi = \dim \tau - 1$,
then we have: If $\vf$ is isotropic, then $\vf$ splits. Indeed, we
have $\tau \cong \varphi \, \bot \, [a]$ for some $a \in k^*$. If
$\vf$ is isotropic, then $\tau$ is isotropic, thus $\tau \sim 0$
by Theorem~2 and hence $\varphi \sim [-a]$. As above, this
implies: if $\vf$ is anisotropic, then $\vf$ has height~$1$. In
characteristic~2, we cannot come to the same conclusion since
$\vf$ is not an orthogonal summand of $\tau$. As a replacement we
have the following theorem:

\begtheorem{3.} Let $\Char k=2$. Let $\rho$ be a bilinear Pfister
form of degree $d\geq 1$, let $\beta =
\left[ {1 \atop 1} \; {1
\atop b} \right]$
be a binary quadratic form and let $c \in D(\beta)$. Define
$\varphi \colon = \rho' \otimes \beta \, \bot \, [c]$, then:
\smallskip
\setitemindent{$(iii)$}
\item{$(i)$} If $\vf$ is isotropic, then $\varphi \cong (2^d-1) \times H \perp [c]$.
\medskip
\item{$(ii)$} If $\vf$ is anisotropic, then $\vf$ has height $1$.
\medskip
\item{$(iii)$} $k(\varphi)$ is specialization equivalent with $k(\rho \otimes
\beta)$ over $k$.
\endtheorem

\begproof{} Let $\vf$ be isotropic. Since $\vf$ is a subform of $\rho
\otimes \beta$, the form $\rho \otimes \beta$ is isotropic. Hence,
by Theorem~2,
$$\rho \otimes \beta = (\langle 1 \rangle \, \bot \, \rho') \otimes \beta \sim 0.$$
It follows that  $\rho' \otimes \beta \sim \beta$. $\{$Note that $\beta = -
\beta$.$\}$ Thus $\varphi  \sim \beta \, \bot \, [c] \sim [c]$,
since $\beta \, \bot \, [c]$ is isotropic and $\dim \beta = 2$.
This proves $(i)$.

Next, suppose that $\vf$ is anisotropic. Then $\varphi \otimes k
(\varphi)$ is isotropic and so $\varphi \otimes k(\varphi)$ splits
by what we already proved. Therefore $h (\varphi) = 1$.

Finally, let $K \supset k$ be a field extension with $( \rho \otimes
\beta)_K$ isotropic.\fonote{From now on, we sometimes write
$\psi_K$ instead of $\psi\otimes K$ for a form $\psi$ over $K$, as
announced in \S9.} Then $(\rho \otimes \beta)_K \sim 0$. Hence, $(\rho'
\otimes \beta)_K \sim \beta_K$ and $\varphi_K \sim (\beta \, \bot \, [c])_K \sim
[c]$. It is now clear that  $\rho \otimes \beta$ and $\varphi$
become isotropic over the same field extensions of $k$. Thus $k(\rho \otimes
\beta)$ is a generic zero field of $\vf$.\qed
\endproof

The class of examples of forms of height~$1$, given by this
theorem, can be expanded considerably, as we will show now.
But first some generalities. We need the following theorem from
the elementary theory of quadratic forms.

\begtheorem{4.} Let $E = (E, q)$ be a strictly regular space over
$k$ and let $F_1 = (F_1, q \vert F_1)$, $F_2 = (F_2, q \vert F_2)$
be quadratic submodules of $E$. Further, let $\alpha \colon F_1 \buildrel \sim \over
\longrightarrow F_2$ be an isometry. Then there exists an
automorphism $\sigma$ of $E$, i.e. an isometry $\sigma \colon E \buildrel \sim \over
\longrightarrow E$, which extends $\alpha$.
\endtheorem

For a proof, we refer to [Bo$_1$, \S 4, Th.1]. If $F_1$ (thus also
$F_2$) is strictly regular, then Theorem~4 can also be formulated
as a cancellation theorem and is then a special case of \S6,
Theorem~2. In characteristic $\ne 2$, this is just the Witt
Cancellation Theorem [W, Satz~4]. We dub Theorem~4 the ``{\it Witt
Extension Theorem}'' (although Theorem~4 was not formulated in
this generality by Witt).

Let $E = (E, q)$ again be a strictly regular space. Let $F$ be a
quadratic submodule of $E$. Then
$$F^{\perp} \colon = \{ x \in E \; \vert \; B_q (x, E) = 0 \}$$
is also a subspace of $E$. Since $B_q$ is \nd, we have
$(F^{\perp})^{\perp} = F$.
We call $F^{\perp}$ the {\it polar of $F$ in $E$} and write $F^{\perp} = \Pol_E (F)$.
If $F_1$ is a submodule of $E$ (w.r.t. $q$), isometric to $F$, then
$\Pol_{E}(F_1)$ is also isometric to $\Pol_E (F)$ by Theorem~4.
This fact supplies us with  very convenient terminology in the language
of forms (instead of spaces) as follows.

\begdefinition{3.} Let $\vf$ be a \qf\ over $k$ and $\tau$ a
strictly regular form over $k$. Suppose that $\varphi < \tau$. We
define a (possibly degenerate) form $\psi$ over $k$ as follows:
Let $(E, q) = E$ be a space, isometric to $\tau$ and $(F, q \vert F) =
F$ a submodule of $E$, isometric to $\vf$. Then we define  $\psi$
to be given by the submodule $\Pol_E (F)$ of $E$. $\psi$ is up to
isometry uniquely determined by $\vf$ and $\tau$. We write $\psi = \Pol_{\tau}
(\varphi)$ and call $\psi$ ``{\it the}'' {\it polar of  $\varphi$ in
$\tau$}.\Idx{polar}\Idx{111poltfi@$\Pol_{\tau}
(\varphi)$}
\enddefinition

We return to the geometric language. We suppose now that $k$ has
characteristic $2$. On a strictly regular space $(E, q)$ over $k$
we can then use symplectic geometry, since $B_q (x, x) = 0$ for every $x
\in E$.

\beglemma{1.} Let $F$ be a quadratic submodule of a strictly regular space
$E$ and let $P= QL(F)$ be the quasi-linear part of $F$. Then we
also have  $QL(\Pol_E (F))= P$. More precisely: let $G$ be a
complement of $P$ in $F$, i.e. $F = P \oplus
G = P \, \bot \, G$. Then there exists an orthogonal decomposition
$E = M \, \bot \, G \, \bot \, R$ with
$P \subset M$, $\dim M = 2 \dim P$, $F = P \, \bot \, G$. For
every such decomposition, we have $\Pol_E (F) = P \, \bot \, R$.\Idx{111polef@ $\Pol_E (F)$}
\endlemma

\begproof{} The space $G$ is strictly regular. Let $T = \Pol_E
(G)$, i.e. $E = T \, \bot \, G$. $T$ is also strictly regular and
$P \subset T$. Let $x_1, \dots, x_r$ be a basis of the vector
space $P$. Since $P$ is quasi-linear, we have
$B(x_i, x_j) = 0$ for all $i, j \in \{ 1, \dots, r \}$. By
elementary symplectic geometry, $T$ contains vectors $y_1, \dots, y_r$ with $B(x_i,
y_j) = \delta_{ij}$ for all $i, j \in \{ 1, \dots, r \}$. Let $M$
be the vector space spanned by $x_1, \dots, x_r$, $y_1, \dots,
y_r$. The bilinear form $B \vert M \times M$ is clearly \nd. Hence
the quadratic submodule $M$ of $T$ is \nd, i.e. is a quadratic
space,
and we have $T = M \, \bot \, R$
with a further quadratic space $R$. We have $E = M \, \bot \, R \, \bot \, G$.

Now let $E = M \, \bot \, R \, \bot \, G$ be a given
decomposition, with $P \subset M$,
$\dim M = 2 \dim P$, $F = P \, \bot \, G$. Then
$$\Pol_E (F) = \Pol_E (P) \cap \Pol_E (G) = \Pol_E
(P) \cap (M \, \bot \, R) = \Pol_M (P) \, \bot \, R.$$
By elementary symplectic geometry, we get $\Pol_M (P) = P$,
and so $\Pol_E (F) = P \, \bot \, R$.\qed
\endproof

In the language of forms, the lemma allows us to say the following.

\begscholium{} Let $\tau$ be a strictly regular \qf\ and $\vf$ an
arbitrary \qf\ with $\varphi < \tau$. Then there is an orthogonal
decomposition $\tau = \mu \, \bot \, \gamma \, \bot \, \rho$ with
$$QL (\varphi) <
\mu, \ \dim \mu = 2 \dim QL(\varphi), \ \varphi \cong QL(\varphi) \, \bot \, \gamma,
\ \Pol_{\tau} (\varphi) \cong QL (\varphi) \, \bot \, \rho.$$
If $QL (\varphi)\cong [a_1, \dots, a_r]$, then $\mu \cong \left[ {a_1 \atop 1} \; {1
\atop b_1} \right] \, \bot \, \dots \, \bot \, \left[ {a_r \atop 1} \; {1
\atop b_r} \right]$ with further elements $b_1, \dots, b_r \in k$.
We have $\Pol_{\tau} (\Pol_{\tau} (\varphi)) = \varphi$. If $\vf$
is \nd, then $\Pol_{\tau} (\varphi)$ is \nd,  since $\vf$ and $\Pol_{\tau} (\varphi)$ have the same
quasi-linear part.\qed
\endscholium

Later on, we will be particularly interested in the case where
$\vf$ itself is quasi-linear. Then $\gamma = 0$. We thus have a
decomposition $\tau = \mu \, \bot \, \rho$ with $\varphi < \mu$, $\dim \mu = 2 \dim
\varphi$, and so $\Pol_{\tau} (\varphi) = \varphi \, \bot \, \rho$.

\begremark{} The scholium remains valid in characteristic $\neq2$.
Then $QL(\varphi)$ is simply the radical of $\vf$, hence $a_1 = \dots =
a_r = 0$. Then we can also establish that $b_1 = \dots =b_r=0$. If
$\vf$ is \nd\ (i.e. $QL(\varphi) = 0$), then $\Pol_{\tau}
(\varphi)$ is \nd\ and $\tau = \varphi \, \bot \, \Pol_{\tau}
(\varphi)$. In characteristic $\ne 2$ the notion of polar is not
as interesting as in characteristic $2$.

\begtheorem{5.} Again let $\Char k=2$. Let $\tau$ be a quadratic
Pfister form over $k$ of degree $d \geq 2$ and let $\chi$ be an
{\rm anisotropic quasi-linear} form over $k$ with $\dim \chi <
2^{d-1}$. Suppose that $\chi < a \tau$ for some $a \in k^*$. Then,
for the polar $\varphi = \Pol_{a \tau}
(\chi)$,
we have the following:
\smallskip
\setitemindent{$(iii)$}
\item{$(i)$} $\vf$ is \nd, $QL(\varphi) = \chi$.
\medskip
\item{$(ii)$} If $\tau$ is isotropic, then $\vf$ splits.
\medskip
\item{$(iii)$} If $\vf$ is anisotropic, then $\vf$ has height $1$.
\medskip
\item{$(iv)$} The fields $k(\varphi)$ and $k(\tau)$ are
specialization equivalent over $k$.
\medskip
\item{$(v)$} If $\dim \chi < \frac{1}{4} \dim \tau$, then for
every form $\gamma$ with
$\varphi \cong \chi \, \bot \, \gamma$, $k(\gamma)$ is also
specialization equivalent to $k(\varphi)$ over $k$.
\endtheorem

\begproof{} 1) We assume without loss of generality that $a=1$, thus $\chi < \tau$. We
already know that $(i)$ holds. We choose a decomposition
$\varphi = \chi \, \bot \, \gamma$. Then $\Pol_{\tau} (\varphi) =
\chi$ and so, in particular, $QL(\varphi) =\chi$. Upon applying
the scholium to $\vf$, we see that $\rho = 0$ there. We thus have a
decomposition $\tau
= \mu \, \bot \, \gamma$ with $\chi < \mu$ and $2 \dim \chi = \dim
\mu$. Since $2 \dim \chi < \dim \tau$, we certainly have
$\gamma \ne 0$.

2) Now let $\tau$ be isotropic. By Theorem~2, we have $\tau \sim
0$. It follows that $\gamma
\sim \mu$ (note that $\mu = - \mu$) and then $\varphi \sim \chi \, \bot \, \mu$.
We have
$$\chi \; \cong \; [a_1, \dots, a_r], \quad \mu \; \cong \; \left[ {a_1 \atop 1} \; {1
\atop b_1} \right] \, \bot \, \dots \, \bot \, \left[ {a_r \atop 1} \; {1
\atop b_r} \right]$$
with elements $a_i, b_i \in k$. Now,
$$[a_i] \perp \left[ {a_i \atop 1} \; {1
\atop b_i} \right] \; \cong \; [a_i] \perp \left[ {0 \atop 1} \; {1
\atop b_i} \right] \; \cong \; [a_i] \perp H.$$
It follows that $\chi \, \bot \, \mu \cong \chi \, \bot \, r
\times H$, also $\varphi \sim \chi$. This proves $(ii)$.

3) If $\vf$ is anisotropic, then $\varphi \otimes k (\varphi)$ is
isotropic. Nevertheless, $\chi \otimes k (\varphi)$ remains
anisotropic, since $k(\varphi)$ is separable over $k$ (\S10,
Theorem~1). If we apply what we already proved to $\chi \otimes k(\varphi)$ and $\tau
\otimes k (\varphi)$, we see that $\varphi \otimes k(\varphi)$
splits. Hence $h(\varphi) = 1$.

4) If $\vf$ is isotropic, then the field extensions $k(\varphi)$ and
$k(\tau)$ of $k$ are both purely transcendental. Now suppose that
$\vf$ is anisotropic. $\varphi \otimes k (\tau)$ is isotropic
(even split). Of course, $\tau \otimes k (\varphi)$ is also
isotropic. Hence  $k(\varphi) \sim_k k (\tau)$ in this case also.

5) Finally, suppose that $\dim \chi < \frac{1}{4} \dim \tau =
2^{d-2}$. Now $\dim \mu < 2^{d-1}$, thus $\dim \gamma > 2^{d-1}$.
We have $\tau \otimes k
(\tau) \sim 0$, hence $\gamma \otimes k (\tau) \sim \mu \otimes
k(\tau)$. Since $\dim \mu < \dim \gamma$, it follows that $\gamma \otimes k
(\tau)$ is isotropic. Clearly $\tau \otimes k (\gamma)$ is also
isotropic. We conclude that $k(\gamma) \sim_k k (\tau)$.\qed
\endproof

We want to understand the connection between the forms $\varphi$ and $a \tau$
in Theorem~5 better. For this purpose we give another general
definition for quadratic forms over a field $k$ of
characteristic~$2$.

\begdefinition{4.} Let $\vf$ be an arbitrary form and $\tau$ a
strictly regular form over $k$. We say that $\tau$ is a {\it
strictly regular hull of $\vf$}, and write $\varphi <_s \tau$,\Idx{strictly regular hull}\Idx{111filesstau@$\varphi <_s \tau$}
when the following conditions hold:
\smallskip
\setitemindent{(a)}
\item{(a)} $\varphi < \tau$.
\medskip
\item{(b)} Let $(E, q) = E$ be a space, isometric to $\tau$ and
let $(F, q \vert F) = F$ be a quadratic submodule of $E$, isometric to $\vf$.
Then there {\it does not exist} a strictly regular subspace
$(E', q \vert E') = E'$ of $E$ with $F \,
\lower3pt\hbox{${\subset
\atop \ne}$}\, E' \, \lower3pt\hbox{${\subset \atop \ne}$}\,
E$.\qed
\enddefinition

By the Witt Extension Theorem (Theorem~4), the relation $\varphi <_s
\tau$ only depends on the isometry classes of $\vf$ and $\tau$,
and is independent of the choice of the spaces $E$ and $F$.

\beglemma{2.} Let $\vf$ be an arbitrary \qf\ and $\tau$ a strictly
regular \qf\ with $\varphi < \tau$. Then $\tau$ is a strictly
regular hull of $\vf$ \iff\ $\Pol_{\tau} (\varphi)$ is
quasi-linear.
\endlemma

\begproof{} Let $F \subset E$ be quadratic modules associated to
$\vf$ and $\tau$, as indicated in Definition~4. If $\tau$ is not a
strictly regular hull of $\vf$, then $E$ contains a strictly
regular subspace $E'$ with $F\subset E'$ and $E' \ne E$. We have
an orthogonal decomposition $E = E'
\perp E''$. Clearly, $E'' \subset \Pol_E (F)$. The space $E''$ is
$\ne 0$ and strictly regular. Hence $\Pol_E (F)$ is not
quasi-linear.

Now suppose that $\varphi <_s \tau$. We choose decompositions
$F = P \, \bot \, G$, $E = M \, \bot \, G \, \bot \, R$ which have
the properties exhibited in Lemma~1. We have $\Pol_E (F) = P \, \bot \,
R$. Since $M \, \bot \, G$ is strictly regular and contains the space
$F$, we must have $M \, \bot \, G = E$, i.e. $R = 0$. It follows
that $\Pol_E (F) = P$ is quasi-linear.\qed
\endproof

\begremark{} The definition of strictly regular hull is also
meaningful in characteristic $\ne2$ and Lemma~2 still holds (a
quasi-linear form then looks like $[0, \dots, 0]$), but again
this term is far less interesting now. If $\vf$ and $\tau$ are both
\nd, then $\varphi <_s \tau$ holds only if $\varphi \cong \tau$.
\endremark

\begdefinition{5.} Let $\vf$ be a \qf\ over $k$. We call $\vf$ a
{\it close Pfister neighbour}\fonote{In \S22 we will introduce the
more general notion of Pfister neighbour.}\Idx{close Pfister neighbour} 
if $\vf$ is \nd\ and
not quasi-linear and if there exist a Pfister form $\tau$  and an
$a \in k^*$ with $\varphi <_s a \tau$. More precisely, we then
call $\vf$ a {\it close neighbour of} $\tau$.\qed
\enddefinition

If $\vf$ is a close neighbour of $\tau$ and if $\vf$ is
anisotropic, then Theorem~5 tells us that $\vf$ has height $1$ and
that $k(\tau)$ is a generic zero field of $\vf$.

The relation ``strictly regular hull'' is compatible with taking
tensor products, as we will show now.

\beglemma{3.} Let $\vf$ be an arbitrary \qf\ and $\tau$ a strictly
regular \qf\ over $k$ with $\varphi < \tau$. Let $\sigma$ be a
bilinear \nd\ form over $k$. Then $\sigma \otimes \varphi < \sigma
\otimes \tau$ $($this is clear$)$ and $\Pol_{\sigma \otimes \tau} (\sigma \otimes
\varphi) \cong \sigma \otimes \Pol_{\tau} (\varphi)$.
\endlemma


\begproof{} Let $(E, q) = E$ be a \qs, isometric to $\tau$ and $(F, q \vert F) =
F$ a submodule, isometric to $\vf$. Furthermore, let $S$ be a
bilinear space, isometric to $\sigma$. By Lemma~1, we have
orthogonal decompositions $F = P \, \bot \, G$, $E = Q \, \bot \, G \, \bot \, R$
with $ P = QL(F) \subset Q$ and $\dim Q = 2 \dim P$. From those we
obtain analogous decompositions
$$S \otimes F = (S \otimes P) \, \bot \, (S \otimes G), \; S \otimes E = (S \otimes
Q) \, \bot \, (S \otimes G) \, \bot \, (S \otimes R).$$
Clearly, $S \otimes P$ is the quasi-linear part of $S \otimes F$ and
$\dim (S \otimes Q) = 2 \dim (S \otimes P)$. By Lemma~1, we get
$$\Pol_E (F) = P
\, \bot \, R\qquad \hbox{and}\qquad \Pol_{S \otimes E} (S \otimes F) = (S \otimes P) \, \bot
\, (S \otimes R).$$
Therefore,  $S \otimes \Pol_E (F) = \Pol_{S \otimes E} (S
\otimes F)$.\qed
\endproof

\begtheorem{6.} If $\varphi <_s \tau$ and if $\sigma$ is a \nd\
bilinear form, then $\sigma \otimes \varphi <_s \sigma \otimes \tau$.
\endtheorem

\begproof{} By Lemma~2, $\Pol_{\tau} (\varphi)$ is quasi-linear.
Using Lemma~3, it follows that $\Pol_{\sigma \otimes \tau} (\sigma \otimes \varphi) = \sigma
\otimes \Pol_{\tau} (\varphi)$ is quasi-linear. Again by Lemma~2,
we have
$\sigma \otimes \varphi <_s \sigma \otimes \tau$.\qed
\endproof

\begcorollary{} If $\vf$ is a close neighbour of a quadratic
Pfister form $\tau$, and if $\sigma$ is a bilinear Pfister form,
then $\sigma \otimes \varphi$ is a close neighbour of the Pfister
form $\sigma \otimes \tau$.\qed
\endcorollary

Thus it is  shown that for a large class of forms of height~1, if
$\vf$ is any such form and  $\sigma$ is any bilinear Pfister form,
then the tensor product $\sigma \otimes \varphi$  again has
height~1, insofar it is anisotropic. Such a phenomenon does not
happen in characteristic $\ne2$. If $\vf$ is an anisotropic form
of height 1 and not of the form $a \tau$ for some Pfister form
$\tau$, then (as mentioned above) we know  in characteristic $\ne2$
that $\vf$ has the form $a \tau'$, where $\tau'$ is the pure part
of some Pfister form.  If $\sigma$ is a further Pfister form
$\ne \langle 1 \rangle$, then $\sigma \otimes \varphi$ {\it does
not have} height~1, since neither $\dim
(\sigma \otimes \varphi)$ nor $1 + \dim (\sigma \otimes \varphi)$
is a 2-power in this case.

The following question suggests itself: Let $\vf$ be a close
neighbour of some Pfister form $\tau$. Is $\tau$
uniquely determined by $\vf$ (up to isometry)?

We will show that this is indeed so. We will also derive other
theorems about Pfister forms. However, for this purpose, we need
more tools from the general theory of \qf s, namely the so-called
Subform Theorem and a ``Norm Theorem''. Both theorems are
well-known in characteristic $\ne2$ and still do hold in
characteristic~2. We will derive them next, see \S17 and \S19.
