\titlea{\S23}{Regular Forms of Height $1$}
\leftrunningtitle{Chapter III. Some Applications}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S23 Regular Forms of Height $1$}

In the following let $\vf$ be a {\it regular} \an\ \qf\ {\it of
height} $1$ over a field $k$. The purpose of this section is the
proof of the following theorem.

\begtheorem{1.} $\vf$ is excellent.
\endtheorem

More particularly: if $\vf$ is of even dimension, then $\varphi = a \tau$
with $a \in k^*$ and $\tau$ a \Pf. If $\vf$ is of odd dimension,
then $\varphi = a\, \Pol_{\tau}([1])$ with $a \in k^*$ and $\tau$
a \Pf\ of dimension $\geq 4$. Is furthermore $\Char k\ne2$, then $\Pol_{\tau}
([1])$ is the pure part of $\tau$, cf. \S16.
\medskip

For $\vf$ of even dimension we already proved the theorem in \S20,
Theorem~5. Thus, suppose from now on that the dimension of $\vf$
is odd.
\medskip

We start with the proof for $\Char k\ne 2$ (cf. [K$_5$,
p.81 ff.]). We interpret $\vf$ as a symmetric bilinear form. Upon
multiplying $\vf$ by a scalar factor we may assume without loss of generality that
$d(\vf)=1$. $\{$The signed determinant $d(\vf)$ was introduced in
\S2.$\}$ 

We first show that $\vf$ does not represent the element $1$. By
way of contradiction assume that $\vf$ does  represent $1$. Then
we have an orthogonal decomposition $\varphi = \langle 1 \rangle \perp
\chi$. Since $h(\varphi) = 1$ we have $\varphi \otimes k(\varphi) \sim
\langle 1 \rangle$, and so $\chi \otimes k (\varphi) \sim 0$. As
usual the Norm Theorem and the Subform Theorem tell us that $a \varphi <
\chi$ for a suitable $a \in k^*$. This is absurd however since the
dimension of $\chi$ is smaller than the dimension of $\vf$.
Therefore $1 \not\in D(\varphi)$. The form $\tau \colon =
\langle 1 \rangle \perp (-\varphi)$ is thus anisotropic.

Let $L/k$ be an arbitrary field extension of $k$. If $\varphi
\otimes L$ is anisotropic, then $h (\varphi \otimes L) = 1$ since
$h(\varphi \otimes L) \leq h(\varphi)$ and $h(\varphi \otimes L)
\ne 0$. If we apply what we just proved to $\varphi \otimes L$
instead of $\varphi$, we see that $\tau \otimes L$ is anisotropic.
On the other hand, if $\varphi \otimes L$ is isotropic, then
$\varphi \otimes L \sim \langle 1 \rangle$, and so $\tau \otimes L
\sim 0$. This shows that $\tau \otimes L$ is either anisotropic or
hyperbolic. Thus $\tau$ has height~$1$. Furthermore $1 \in
D(\tau)$. As already established before (\S20, Theorem~5), $\tau$
is a \Pf. It follows that $\varphi = -\tau'$.

Let us now continue by proving Theorem~1 for $\Char k=2$ and
$\dim\vf$ odd, $\dim \varphi = 2 m+1$ with $m > 0$. Without loss
of generality we may assume that $\vf$ has quasi-linear part $QL (\varphi) =
[1]$. We have to show that $\vf$ is a close neighbour (cf. \S16)
of a \Pf.

We deal with the case $m=1$ separately. In this case we have
$$\varphi
\cong [1] \; \perp \; c \left[ {1 \atop 1} \; {1 \atop b}
\right]$$
with $c \in k^*$, $b \in k$, and so $\varphi$ is a close neighbour of
the \Pf\ $\langle 1, c \rangle \otimes \left[ {1 \atop 1}
\; {1 \atop b} \right]$.

From now on let $m \geq 2$, and thus $\dim \varphi \geq 5$. We
require an easy lemma of a general nature that we will  prove
at a later stage.

\beglemma{} Let $\psi$ and $\chi$ be \qf s over $k$, and assume
that $\psi \perp \chi$ is \nd. Then
$$\ind\, \psi
\; + \; \dim \chi \; \geq \; \ind (\psi \perp \chi).\eqno\square$$
\endlemma

We have an orthogonal decomposition
$$\varphi \; \cong \; [1] \;
\perp \; \rho$$
with $\rho$ strictly regular, $\dim \rho = 2 m$.
Since the height of $\vf$ is $1$ we have $\ind (\varphi \otimes
k(\varphi)) =m$. The lemma implies that
$$\ind (\rho \otimes
k(\varphi)) \; \geq \; m - 1 \; > \; 0.$$
So,  $\rho \otimes k(\varphi)$ is in particular  isotropic. Since $\varphi \otimes k(\rho)$ is
trivially also isotropic it follows that $k(\varphi) \sim_k
k(\rho)$ and then
$$\ind (\rho \otimes k(\rho)) \; \geq \; m -1.$$
Let us assume \ftsoc\ that $\ind (\rho \otimes k(\rho)) > m-1$,
thus $\rho \otimes k(\rho) \sim 0$. By earlier work we have that
$\rho = a \sigma$ where $\sigma$ is a \Pf. Hence $\varphi =
[1] \perp a \sigma$ is a neighbour of the \Pf\ $\tau \colon = \langle
1, a \rangle \otimes \sigma$. It follows from \S22, Theorem~1 that $k(\varphi)
\sim_k k(\tau)$.  This implies $k(\rho) \sim_k k(\tau)$,
since $k(\sigma)$ is also specialization equivalent to $k(\varphi)$ over
$k$. By \S21, Corollary to Theorem~1, this means $\sigma \cong
\tau$, which is not possible since $\dim \tau = 2 \dim \sigma$.
Therefore
$$\ind (\rho \otimes k(\rho)) \; = \; m-1.$$
Thus $\rho \otimes k(\rho)$ is the orthogonal sum of $m-1$
hyperbolic planes and an \an\ binary form. Hence $\rho \otimes
k(\rho)$ has non-vanishing Arf invariant. Thus we certainly have $\Arf (\rho) \ne 0$. Let
$\Arf (\rho) = a + \wp k$
and 
$$\sigma \colon = \left[ {1 \atop 1} \; {1 \atop a} \right].$$
Consider the form $\tau \colon = \sigma \perp \rho$. Then $\varphi < \tau$ and $\Arf (\tau) = 0$.
 $\{$Note that we are now in a situation which is similar to the proof involving the form $\tau$ in $\Char k\ne 2$.$\}$ Assume that $\varphi \otimes k(\sigma)$ is anisotropic. This form then has height~$1$ and
$$\varphi \otimes k(\sigma) = [1] \; \perp \; \rho \otimes k(\sigma).$$
Thus, we should get $\Arf (\rho \otimes k (\sigma)) \ne 0$, which is false. Therefore $\varphi \otimes
k(\sigma)$ is isotropic and it follows that
$$\varphi \otimes k(\sigma) \;
\cong \; [1] \; \perp \; m \times H.$$
Hence,  $m \times H < \tau \otimes k(\sigma)$. Since $\dim
\tau = 2 m +2$ and $\Arf (\tau \otimes k(\sigma)) = 0$ we obtain  $\tau \otimes k(\sigma) \sim 0$ 
and so $\rho \otimes k(\sigma) \sim 0$. By \S21, Theorem~1, there exist $a_1, \dots, a_m \in
k^*$ with $$\rho \; \cong \; \langle a_1, \dots, a_m \rangle
\otimes \sigma.$$
Consequently,
$$\tau \; \cong \; \langle 1, a_1, \dots, a_m \rangle
\otimes \sigma.$$

We will now show that $\tau$ is anisotropic. Assume for the sake of contradiction that this is not so. Then there exist vectors  $x_0, x_1, \dots, x_m \in k^2$ with 
$$\sigma (x_0) + \sum\limits_{i = 1}^m a_i \sigma (x_i) \; = \; 0$$
and not all  $x_i = 0$. Since $\rho$ is anisotropic, we must have $x_0 \ne 0$ and thus also   $\sigma (x_0) \ne 0$. Thus there are vectors $y_1, \dots, y_m \in k^2$  with $\sigma (y_i) = \sigma (x_0)^{-1}
\sigma (x_i)$ for $1 \leq i \leq m$. 
Hence,
$$1 +\sum\limits_{i = 1}^m a_i \sigma (y_i) \; = \; 0.$$
This contradicts the anisotropy of $\varphi$. Thus $\tau$ is indeed anisotropic.

Now it follows easily that $\varphi \otimes k(\tau)$ is isotropic. Indeed, if  $\varphi \otimes k(\tau)$ were anisotropic, then we could apply what we just proved to the form  $\varphi \otimes k(\tau)$ of height~$1$ and we would conclude that  $\tau \otimes k(\tau)$ would be anisotropic, which is of course false. 

Hence there exists a place from $k(\varphi)$ to $k(\tau)$ over
$k$, and it follows that 
$$\varphi \otimes k(\tau) \; \cong \; [1] \; \perp \; m \times H.$$
Thus  $\tau \otimes k(\tau) > m \times H$ and, since $\Arf
(\tau) = 0$ and $\dim \tau = 2 m +2$, even $\tau \otimes k(\tau)
\sim 0$. Furthermore, $1 \in D (\tau)$ and $\tau$ is anisotropic.
By earlier work $\tau$ is a Pfister form. The form $\varphi$ is a close neighbour of this Pfister form.

Let us finally supply  the proof of the lemma above. Then our theorem will be completely proved.

The field $k$ may have arbitrary characteristic. Let $\psi$ and $\chi$ be quadratic forms over $k$. If $\ind (\psi \perp \chi) \leq \dim \chi$, there is nothing to prove. Thus, suppose $\ind (\psi
\perp \chi) > \dim \chi$.  Let $E$ be the quadratic space of $\psi$ and $G$ the quadratic space of $\chi$. 
The space  $E \perp G$, nondegenerate by assumption, contains a totally isotropic subspace $V$ with 
$\dim V = \ind (\psi \perp \chi)$. Let $\pi \colon V \to G$ be the restriction of the natural projection 
 $E \perp G \to G$ to $V$. The kernel of $\pi$ is then $V\cap E$. This space is totally isotropic and
$$\dim (V \cap E) + \dim G \; \geq \; \dim V.$$
Therefore,
$$\dim (V \cap E) \; \geq \; \ind (\psi \perp \chi) -\dim \chi,$$
which establishes the assertion of the lemma.

