
\titlea{\S27}{Definability of the Leading Form over the Base Field}

\leftrunningtitle{Chapter III. Some Applications}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S27 Definability of the Leading Form over the Base Field}

In the following  $\varphi$ denotes a regular (quadratic) form 
over an arbitrary field~$k$,  $F$ denotes a leading field for
$\varphi$ and $\sigma$ denotes the associated leading form of~$\varphi$.

\begdefinition{1.} We say that the leading form of $\varphi$ is {\it
defined over}\Idx{definable over a field (for leading forms)} 
$k$ if there exists a form $\tau$ over $k$ such that
$\sigma \cong \tau \otimes F$.
\enddefinition

Obviously this property is independent of the choice of the leading field
$F$. Moreover,  $\tau$ is to a large extent uniquely determined by 
$\sigma$. We namely have the following

\begtheorem{1}{\rm ([K$_6$, Prop.~9.2] for $\Char \ne 2$).}  Assume that the leading form of
$\varphi$ is defined over $k$.
\smallskip	
\setitemindent{$(iii)$}
\item{$(i)$}  If the dimension of $\varphi$ is  even, then there exists up to isometry precisely one
form $\tau$ over $k$ such that $\tau \otimes F
\cong \sigma$.
\smallskip	
\item{$(ii)$} If the dimension of  $\varphi$  is odd, then there exists up to isometry precisely one
form $\tau$ over $k$ such that $1 \in D(\tau)$ and
$\tau \otimes F \cong \sigma$.
\smallskip	
\item{$(iii)$} In both cases  $\tau$ is a   Pfister form.
\endtheorem

\begproof{} We choose a generic splitting tower $(K_r
\; \vert \; 0 \leq r \leq h)$ of $\varphi$ and assume without loss of generality that
$F = K_{h-1}$.  Let $\tau_1$ and $\tau_2$ be forms over
$k$ such that
$$\tau_1 \otimes K_{h-1}  \cong  \tau_2 \otimes K_{h-1}  \cong
 \sigma.$$
 
Assume that $\tau_1 \not\cong \tau_2$. We must have $h \geq 2$.
Let $s$ be the largest integer in  $[0, h-2]$ such that $\tau_1
\otimes K_s \not\cong \tau_2 \otimes K_s$, and let  $\zeta$ be the 
kernel form of $[\tau_1 \perp (- \tau_2)] \otimes K_s$. Then $\zeta
\ne 0$ and the dimension of $\zeta$ is even. For $0 \leq r \leq h$
let $\varphi_r$ denote the kernel form of $\varphi \otimes K_r$. The form 
$\zeta$ is split by $K_s (\varphi_s)$. By the usual argument (the Norm and Subform Theorems)
we thus have a decomposition
$$\varphi_s \perp \psi \cong  c  \zeta$$
for some $c \in K_s^*$ and some form $\psi$ over $K_s$. It follows that
$$\varphi_s \otimes K_{s+1}  \sim  (- \psi) \otimes K_{s+1},$$
and thus $\dim \psi \geq \dim \varphi_{s+1}$. We obtain
$$2 \dim \tau_1  \geq  \dim \zeta  \geq  \dim \varphi_s + \dim
\varphi_{s+1}  \geq  2 \dim \varphi_{s+1} + 2.
\leqno(\ast)$$

If $\dim \varphi$ is even, this gives a contradiction and so 
$\tau_1 \cong \tau_2$.

Assume now that  $\dim \varphi$ is odd and moreover that 
$[1] < \tau_1$ and $[1] < \tau_2$.  Then $\tau_1
\perp (- \tau_2)$ is isotropic and we obtain from
($\ast$) that
$$2 \dim \tau_1  \geq  \dim \zeta + 2  \geq  2 \dim \varphi_{s+1} +
4,$$
which is again a contradiction, so that   $\tau_1
\cong \tau_2$ as before.

In addition there always exists a form
$\tau$ over $k$ such that $1 \in D
(\tau)$ and $\tau \otimes K_{h-1} \cong \sigma$ (also when $\dim
\varphi$ is odd). Namely, if $\eta$ is a form over $k$ such that
$\eta \otimes K_{h-1} \cong \sigma$, and if an element  $c \in
D(\eta)$ is chosen, then  $c \eta$  is such a form  $\tau$.

Now let  $\tau$ be the uniquely determined form over  $k$ with  $1
\in D (\tau)$ and $\tau \otimes K_{h-1} \cong \sigma$. Further, let  $t
= (t_1, \dots, t_n)$ be an  $n$-tuple of unknowns over  $k$ with
$n \colon = \dim \tau$. We consider the forms $\tilde{\varphi}
\colon = \varphi \otimes k(t)$ and $\tilde{\tau} \colon = \tau
\otimes k(t)$. By \S 13, Theorem~3, $(K_r (t) \; \vert
\; 0 \leq r \leq h)$ is a generic splitting tower of
$\tilde{\varphi}$ and $\tilde{\tau} \otimes K_{h-1} (t)$  is the associated 
leading form of $\tilde{\varphi} \otimes k(t)$ (also in the event that
$\dim \varphi$ is odd). Since this leading form is a Pfister form we have
$$\tilde{\tau} \otimes K_{h-1} (t)  \cong  [\tau (t) \tilde{\tau}]
\otimes K_{h-1} (t).$$
The form $\tau (t) \tilde{\tau}$ also represents 1. It follows from what we already proved that 
$\tau (t) \tilde{\tau} \cong \tilde{\tau}$. This indicates that
$\tau$ is strongly multiplicative, and thus is a 
Pfister form  (cf. \S 20).\qed
\endproof

\begremark{} If  $\dim \varphi$ is odd, we cannot dispense with the assumption that
$1 \in D(\tau)$ in assertion  $(ii)$, as the following example shows.
Let $\rho$  be a  Pfister form, $a \in
k^*$ and $\langle 1, a \rangle \otimes \rho$ anisotropic. Then the form
$\varphi \colon = \rho \perp [a]$ is a neighbour of  $\langle 1, a
\rangle \otimes \rho$ with complementary form
$$\eta \colon = \Pol_{\rho \perp a \rho} (\rho \perp [a]) = \Pol_{a \rho}
([a]) = a \Pol_{\rho} ([1]) = a \rho'.$$
Thus $\varphi$ has height~2 and leading form $\rho \otimes
F$, for instance with $F = k (\varphi)$. However, we do have $\rho \otimes F \cong (-
a \rho) \otimes F$, which does {\it not} contradict our theorem since  
$1 \not\in D(- a \rho)$.
\endremark

\begdefinition{2.} Assume that $h(\varphi) > 0$ and that the leading form 
$\sigma$ of $\varphi$ is defined over  $k$. Let $\tau$ be the Pfister form
over $k$ with $\sigma \cong \tau \otimes F$. Then we say that the leading form
of $\varphi$ is {\it defined over $k$ by $\tau$}.\Idx{definable over a field by a Pfister form (for leading forms)}
\enddefinition

\begexamples{}
\smallskip	
\setitemindent{$(3)$}
\item{$(1)$} If $\varphi$ is excellent, then the kernel form of
$\varphi \otimes F$ is definable over $k$ by 
\S 22, Theorem~3  (for this terminology, see  \S 24(C)). This implies, also when  $\dim \varphi$
is odd, that the leading form of $\varphi$ is defined over   $k$.
\smallskip	
\item{$(2)$} Let $\varphi \sim a \tau \perp \psi$ with  $a \in k^*$, $\tau$
a Pfister form of degree $n \geq 1$, and $\psi$ strictly regular with
$\deg \psi \geq n +1$. By \S 25, Theorem~3  the leading form of
$\varphi$ is defined over  $k$ by $\tau$.
\smallskip	
\item{$(3)$} Let  $\dim \varphi$ be odd and  $\Delta (\varphi) \ne 1$. Let
$\tau$ be the anisotropic binary Pfister form with   $\Delta (\tau) =
\Delta (\varphi)$ (cf. \S 25, Lemma~1). The form $\psi \colon =
\varphi \perp (- \tau)$ has discriminant $\Delta (\psi) = 1$, and thus
degree $\geq 2$ (cf. \S 25, Satz~5). We have $\varphi \sim \tau
\perp \psi$ and $\deg \tau = 1$. Thus, by the previous example, the leading form
of $\varphi$ is defined over $k$ by $\tau$.
\endexamples

In the search for forms whose leading form is defined over the ground field, we may restrict ourselves to forms of even dimension by the following theorem.

\begtheorem{2}{\rm ([K$_6$, Prop. 9.4] for $\Char k \ne 2$).} Assume   that $\varphi$
is not split  and that $\dim \varphi$ is odd. Let $\tau$ be an anisotropic Pfister form
over $k$ $($of degree $\geq 2$$)$. The leading form of
$\varphi$  is defined over  $k$ by $\tau$ if and only if
this is the case for the companion $\Com (\varphi)$.
\endtheorem

\begproof{} Let  $\psi \colon = \Com (\varphi)$. If the leading form of
 $\psi$ is defined over   $k$ by $\tau$, then  \S 26,
Theorem~9 shows that the same is true for  $\varphi$  $\{$although
the leading fields of $\varphi$ and $\psi$ might not be equivalent, see
Theorem~4 below.$\}$

Assume now that the leading form of  $\varphi$ is defined over 
$k$ by $\tau$. We must show that the same is true for   $\psi$. Let
$F$ be a leading field  for  $\varphi$, and let $QL(\varphi)
= [a]$ for  $\Char k = 2$, $d(\varphi) = \langle - a \rangle$ for $\Char
k \ne 2$, where $a \in k^*$. In both cases we have $\varphi
\perp H \cong \psi \perp [-a]$ by \S 26, Theorem~3.

Furthermore,  $\ker (\varphi \otimes F) = c (\tau' \otimes F)$ for some 
$c \in F^*$. We see that $QL(\varphi \otimes F) = [c]$ for
$\Char k = 2$, $d(\varphi \otimes F) = \langle - c \rangle$ for $\Char
k \ne 2$.  Therefore  $c$ and $a$ represent the same square class of
 $F$, and we may write
$$\ker (\varphi \otimes F)  =  (a \tau') \otimes F.$$

Let $E$ be a leading field for  $\psi$ and $\sigma$ the leading form of
$\psi$ over $E$. By \S 26, Theorem~9,   $\sigma$ and $\tau$ have the same
dimension $\geq 4$. The kernel form of $\psi
\otimes E$ is $b \sigma$ for some $b \in E^*$.

Suppose for the sake of contradiction that $\sigma$ is not isomorphic to  $\tilde{\tau} \colon =
\tau \otimes E$. Then $\sigma \otimes E(\tilde{\tau})$ is
anisotropic. We introduce the form
$$\alpha \colon  =  b (\sigma \otimes E(\tilde{\tau}))  \perp  [-a]$$
over  $E(\tilde{\tau})$ and know from above that
$\varphi \otimes E(\tilde{\tau}) \sim \alpha$ and $\dim \alpha =
\dim \tau +1$.


And now for a little thought experiment. Let $T$ be a field extension
of $E(\tilde{\tau})$ for which
$\alpha \otimes T$ is isotropic. Then the kernel form of $\varphi
\otimes T$, which coincides with the kernel form of  $\alpha \otimes T$, has dimension
$\leq \dim \tau'$. Thus  there exists a place
from $F$ to $T$ over $k$. Since $\varphi \otimes F
\sim (a \tau') \otimes F$ and $\tau \otimes T$ splits, it follows that
$$\alpha \otimes T  \sim  \varphi \otimes T  \sim  a (\tau' \otimes
T)  \sim  [-a].$$
By the definition of  $\alpha$ and the Cancellation Theorem \S 26,
Theorem~6 we obtain  $\sigma \otimes T \sim 0$.

This thought experiment shows for starters that  $\alpha$ itself must be anisotrop\-ic
since $\sigma \otimes E(\tilde{\tau})$
is anisotropic. (Take $T = E(\tilde{\tau})$.) Secondly, it shows that
$\alpha$ splits over every field extension of
$E(\tilde{\tau})$, over which $\alpha$ is  isotropic.
Thus  $\alpha$ has height~1. By \S 23 $\alpha$ is up to a scalar factor
the unit polar of a Pfister form. Hence $\dim
\alpha +1$  is a 2-power. This is the contradiction we were looking for because
$\dim \alpha + 1 = \dim \tau + 2$ cannot be a 2-power since
$\dim \tau$ is a 2-power $\geq 4$. This proves $\sigma \cong
\tau \otimes E$.\qed
\endproof

We come to the main theorem of this section.

\begtheorem{3}{\rm ([K$_6$, Th. 9.6] for $\Char k \ne 2$).}  Let $n \in
{\Bbb N}$, let  $\tau$ be a Pfister form of degree $n$ over $k$ and let 
 $\varphi$ be a non-split $($strictly$)$ regular form of even dimension
over $k$. The following statements are equivalent:
\smallskip	
\setitemindent{$(ii)$}
\item{$(i)$} The leading form of   $\varphi$ is defined over  $k$ by
$\tau$.
\smallskip	
\item{$(ii)$}  $\varphi \equiv \tau\ \bmod J_{n+1} (k)$.
\smallskip	
\noi $\{$Recall: $J_r (k)$ is the module of Witt classes of strictly regular forms
over $k$ of degree $\geq r$.$\}$
\endtheorem

\begproof{} The implication $ (ii) \Rightarrow (i)$  was already established in
\S 25, Theorem~3. Assume now that $(i)$ holds. We want to show that the form
$\psi \colon = \varphi \perp (-\tau)$ has degree $> n$.  
In order to achieve this we will for the first time make serious use of the theory developed in \S13.

We assume without loss of generality that $\psi$ is not hyperbolic. Let
$(K_r\; \vert \; 0 \leq r \leq h)$ be a regular generic splitting tower of
$\varphi$ (cf. \S 13, Definition~5) and let
$\varphi_r \colon = \ker (\varphi \otimes K_r)$. For every $r \in [0,
h]$ we consider the field compositum $K_r\!\cdot\!k(\tau)$ over
$k$ in accordance with  \S 13, Definition~6. Let $J$ be the set of all $r \in [0, h]$
such that $\varphi_r \otimes K_r\!\cdot\!k(\tau)$ is anisotropic and let
$r_0 < r_1 < \dots < r_e$ be the elements of $J$. We know from \S
13, Theorem~3, that $(K_{r_i}\!\cdot\!k (\tau) \; \vert \; 0 \leq
i \leq e)$ is a generic splitting tower of  $\varphi \otimes
k(\tau)$ and, since $\varphi \otimes k(\tau) \sim \psi \otimes
k(\tau)$, also of  $\psi \otimes k(\tau)$. 
Since $\psi \not\sim0$ we have $e \geq 1$. Furthermore $\varphi \otimes (K_{h-1} \cdot
k(\tau)) \sim 0$, since $\varphi \otimes K_{h-1}$ has kernel form
$b(\tau \otimes K_{h-1})$ for some $b \in K_{h-1}^*$ and $\tau
\otimes k(\tau) \sim 0$. Thus $h-1 \not\in J$. Therefore
$$\ker (\psi \otimes K_{r_{e-1}} \cdot k(\tau))  =  \varphi_{r_{e-1}}
\otimes K_{r_{e-1}} \cdot k(\tau)$$
has dimension $> 2^n$. This means that $\psi \otimes k
(\tau)$ has degree  $> n$.

Let $E$ be a leading field for $\psi$ and $\rho$ the associated leading form of
$\psi$. By \S 13, Theorem~3,  $\rho$ splits over
$E \cdot k(\tau) = E (\tau \otimes E)$. This implies by
\S 21, Theorem~1, and since  $\dim \rho = \dim \tau$, that
$\rho \cong \tau \otimes E$, and so  $\psi \otimes E \sim a (\tau
\otimes E)$ for some $a \in E^*$. We therefore obtain
$$\varphi \otimes E  \sim  \tau \otimes E  \perp  a(\tau \otimes E)
 =  \langle 1, a \rangle \otimes (\tau \otimes E).$$
In particular we have $\deg (\varphi \otimes E) = n +1$ or $=
\infty$. We now make use of the lemma below. We apply it to 
$\varphi$ and the field extension $E$ of $k$ and obtain $\tau \otimes E \sim
0$. But this is nonsense because  $\tau \otimes E \cong \rho$. Thus we must have 
$\deg \psi > n$.\qed
\endproof

\beglemma{} Continuing with the assumptions of Theorem~$3$, let
$L$ be a field extension of  $k$ with $\deg
(\varphi \otimes L) > \deg \varphi$. Then $\tau \otimes
L$ splits.
\endlemma

\begproof{} Let  $(K_r \; \vert \; 0 \leq r \leq
h)$ be the affine standard tower  of $\varphi$ (\S 13, Definition~4). Thus,  $K_0 =
k$ and $K_{r+1} = K_r (\varphi_r)$ where $\varphi_r = \ker (\varphi
\otimes K_r)$. By \S 13, Theorem~3,  $\tau \otimes
K_{h-1} \cdot L$ splits.

Assume for the sake of contradiction that $\tau \otimes L$ is {\it not} split.
We certainly have $h \geq 2$. Let $s \in [0, h-2]$ be the largest integer with $\tau
\otimes K_s \cdot L$ anisotropic. Let $\tilde{\tau} \colon = \tau
\otimes K_s \cdot L$ and $\tilde{\varphi}_s \colon = \varphi \otimes
K_s \cdot L$.  Then $K_{s+1} \cdot L = (K_s \cdot L)
(\tilde{\varphi}_s)$ and $\tilde{\tau} \otimes (K_s \cdot L)
(\tilde{\varphi}_s) = 0$.  By the usual argument it follows that
$\tilde{\varphi}_s < c \tilde{\tau}$ for some $c \in (K_s \cdot
L)^*$. In particular, $\dim \varphi_s \leq \dim \tau = \dim
\varphi_{h-1}$, which is nonsense.  Therefore $\tau \otimes L$
must be split.\qed
\endproof

\begremark{} We proved the converse of statement $(i)$ in \S 25,
Theorem~3. For characteristic $\ne 2$ the converse of statement $(ii)$ of this theorem was also proved in [K$_6$, p.~23~ff.]: 
{\it If} $\varphi \equiv a \tau  \bmod J_{n+2} (k)$, {\it
then the $(h-1)$-st kernel form $\varphi_{h-1}$ is defined over $k$ by $a
\tau$}, $\varphi_{h-1} \cong (a \tau) \otimes K_{h-1}$.
So far it has not been possible to extend the proof in  [K$_6$] to 
characteristic~2. One runs into the problem formulated at the end of 
\S 24, Part~(C), which must be solved at least in a special case.
\endremark

By means of  Theorem~3 we can improve upon Theorem~2:

\begtheorem{4.} Let $\varphi$ again be a nondegenerate regular form of odd dimension over
$k$, and let  $\psi$ be the companion of 
$\varphi$. Let $a k^{*2}$ be the square class with  $[a] =
QL(\varphi)$, in case $\Char k = 2$, and $\langle a \rangle = -
d (\varphi)$, in case $\Char k \ne 2$. Let $(K_r \; \vert \; 0 \leq r
\leq h)$ be a generic splitting tower of  $\varphi$ with associated
kernel forms $\varphi_r \colon = \ker (\varphi \otimes
K_r)$ and  let $(L_s \; \vert \; 0 \leq s \leq e)$ be a generic splitting tower 
of $\psi$ with associated kernel forms $\psi_s
\colon = \ker (\psi \otimes L_s)$. Assume furthermore that the leading form of   $\varphi$ -- therefore also
the leading form of  $\psi$ -- is defined over
$k$ by $\tau$. Finally, let   $h > 1$.
$\{$Note that otherwise  $\varphi \cong a \tau'$, $\psi \cong a
\tau$, which is not of interest here.$\}$ Then the following statements are equivalent:
\smallskip	
\setitemindent{$(iii)$}
\item{$(i)$}  $\dim \varphi_{h-2} > \dim \tau + 1$.
\smallskip

\item{$(ii)$} $L_{e-1} \sim_k K_{h-1}$.
\smallskip	

\item{$(iii)$} $a \in D (\psi_{e-1})$, and so  $\psi_{e-1} \cong (a \tau)
\otimes L_{e-1}$.
\medskip	

\noindent If these statements do not hold, then  $L_{e-1} \sim_k K_{h-2}$.
\endtheorem

\begproof{} In  \S 26, Theorem~9, we already determined that there exists a place
from $L_{e-1}$ to $K_{h-1}$ over $k$. Furthermore we have
$\varphi \perp H \cong \psi \perp [-a]$, and so $\varphi
\otimes L_{e-1}  \sim  \psi_{e-1}  \perp  [-a]$. Hence there is certainly a place
from $K_{h-2}$ to $L_{e-1}$ over $k$. This place can be extended to a place
from $K_{h-1}$ to $L_{e-1}$ if and only if the form $\psi_{e-1} \perp [-a]$ is isotropic, so that
$a \in D (\psi_{e-1})$. This establishes the equivalence $(ii) \Leftrightarrow (iii)$.  

If the form  $\psi_{e-1}
\perp [-a]$ is anisotropic, then it is the kernel form of  $\varphi \otimes
L_{e-1}$. If  $\dim \varphi_{h-2} > \dim \tau +1$, this cannot happen for dimension reasons. 
Therefore  $\psi_{e-1} \perp [-a]$ is isotropic in this case. This proves
$(i) \Rightarrow (iii)$.

Now we assume that $(i)$ does {\it not} hold, in other words that
$\dim \varphi_{h-2} = \dim \tau +1$. We want to show that then
$L_{e-1} \sim_k K_{h-2}$. Since  $K_{h-1}$  is not equivalent to  $K_{h-2}$ over $k$,
this will imply that  $(iii)$ does {\it not} hold. This will prove the implication
$(iii) \Rightarrow (i)$,  in addition to the last statement of the theorem. 

We already established above that there is a place from $K_{h-2}$ to $L_{e-1}$ over $k$.  
We have to show that there is also a place over $k$ in the reverse direction. By the generic splitting theory this means that the kernel form of  $\psi \otimes K_{h-2}$ must have dimension  $\leq 2^n$.

Let $m \colon = \ind (\varphi \otimes K_{h-2})$. Then
$$2^n +1  =  \dim \varphi_{h-2}  =  \dim \varphi - 2m.$$
Furthermore we have  $\ind (\psi \otimes K_{h-2}) \geq m$. Hence,
$$\dim \ker (\psi \otimes K_{h-2})  \leq  \dim \psi - 2m  =  2^n
+2.$$
If the inequality is strict, then  $\dim \ker (\psi \otimes
K_{h-2}) \leq 2^n$, and we are done.

We assume then that the form  $\alpha \colon = \ker (\psi
\otimes K_{h-2})$  has dimension $2^n +2$, and want to obtain a contradiction. 
We have  $\deg \alpha \geq \deg \psi = n$, but
$\dim \alpha < 2^{n+1}$, since $n \geq 2$. This forces   $\deg
\alpha = n$. By \S 13, Theorem~3 it is now clear that  $\alpha$ has a leading form over
$K_{h-2}$, defined by $\tau \otimes K_{h-2}$.
(The form $\tau \otimes K_{h-2}$ is in particular anisotropic.) By Theorem~3 this means that
$$\alpha  \sim  (\tau \otimes K_{h-2})  \perp  \chi$$
for some form $\chi$ over $K_{h-2}$ of degree $>
n$. Let $\tilde{\tau} \colon = \tau \otimes K_{h-2}$. We have
$\chi \otimes K_{h-2} (\tilde{\tau}) \sim \alpha \otimes
K_{h-2} (\tilde{\tau})$. Therefore $\alpha \otimes K_{h-2}
(\tilde{\tau})$ has degree $> n$. However, $\dim \alpha <
2^{n+1}$. This forces $\alpha \otimes K_{h-2} (\tilde{\tau}) \sim
0$. The theory of divisibility by Pfister forms  (\S
21, Theorem~1) then tells us that  $\alpha \cong \gamma \otimes \tilde{\tau}$
for a suitable form $\gamma$ over $K_{h-2}$. Now  $\dim \alpha$ is in particular
divisible by  $\dim \tau = 2^n$. This is the contradiction we were looking for, since
$\dim \alpha = 2^n +2 < 2^{n+1}$. This establishes that $L_{e-1}$  is specialization equivalent to  $K_{h-2}$ over $k$.\qed
\endproof

In the proof of Theorem~4 we did not use Theorem~12 of \S26. (Would the proof have been simpler if we had?) If $(i)$ -- $(iii)$ in Theorem~4 are valid, this theorem does give us the additional information that  the tower $(L_s \; \vert \; 0 \leq s \leq e)$ is a generic splitting tower of  $\varphi$, and in particular that $h = e$. 

