\titlea{\S26}{The Companion Form of an Odd-dimensional Regular Form}

\leftrunningtitle{Chapter III. Some Applications}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S26 The Companion Form of an Odd-dimensional Regular Form}


In this section $k$ is an arbitrary field and {\it $\varphi$ is always a regular form of odd dimension over $k$.}

If $\Char k \ne 2$, then  $\varphi$ is strictly regular. In characteristic~2 however, 
$\varphi$ has a one-dimensional quasilinear part, $QL(\varphi) = [a]$. In characteristic $\ne
2$ we also interpret  $\varphi$ as a bilinear form in the manner indicated at the beginning of \S
25.

\begtheorem{1.} Up to isometry there is a unique form $\psi$
over $k$ with the following four properties:
\smallskip	
\setitemindent{$(4)$}
\item{$(1)$}  $\psi$ is strictly regular.
\smallskip

\item{$(2)$} $\Delta (\psi) = 1$.
\smallskip

\item{$(3)$} $\varphi < \psi$.
\smallskip

\item{$(4)$} $\dim \psi = \dim \varphi +1$.
\endtheorem

\begproof{} If $\Char k\ne 2$, we need  $\psi
 = \varphi \perp \langle b \rangle$ for some $b \in k^*$ such that
the condition $d(\psi) = 1$ is satisfied.  So we take $b$ such that $\langle
b \rangle = - d (\varphi)$.

Now let $\Char k=2$. We choose an orthogonal decomposition 
$\varphi \colon = \chi \perp [a]$. Thus $\chi$ is strictly regular and $QL(\varphi) = [a]$. 
We need $\psi =\chi \perp \alpha$ for some binary strictly regular form
$\alpha$, to be determined.  The condition $\chi \perp [a] < \psi$ is equivalent with
$[a] < \alpha$ (cf. \S 17, Theorem~3). Thus, by  \S
25, Lemma 2$(a)$,  $\alpha$ should be chosen in such a way that $a \in
D(\alpha)$ and $\Delta (\alpha) = \Delta (\chi)$. By \S 25,
Lemma~1, there is exactly one binary Pfister form $\tau$ with $\Delta
(\tau) = \Delta (\chi)$. We must take $\alpha = a \tau$.\qed
\endproof

\begdefinition{1.} We call the form $\psi$, described in Theorem~1, the {\it companion}\Idx{companion}\Idx{111comfi@$\Com (\varphi)$} 
of $\varphi$. We denote this form, which is uniquely determined by $\varphi$, by  $\Com (\varphi)$.
\enddefinition

\begremark{} The mapping $\varphi \mapsto \Com (\varphi)$
satisfies the following rules:
\smallskip	
\setitemindent{$(4)$}
\item{$(1)$} If $\alpha$ is a strictly regular form with  $\Delta (\alpha) = 1$, then
$\Com (\varphi \perp \alpha) = \Com (\varphi) \perp \alpha$. In particular,
 $\Com (\gamma) \sim \Com (\varphi)$ for every regular form
$\gamma \sim \varphi$.
\smallskip

\item{$(2)$} $\Com (a \varphi) = a \Com (\varphi)$ for every $a \in k^*$.
\smallskip	

\item{$(3)$} $\Com (\varphi \otimes L) = \Com (\varphi) \otimes L$ for every field extension $L \supset k$.
\smallskip	

\item{$(4)$} If $\Char k \ne 2$, then
$\Com (\varphi) = \varphi \perp - d (\varphi)$.
\endremark

\begproof{}The rules (1) -- (3) follow immediately from the definition of a companion.
Rule (4) is clear from the proof of Theorem~1 above.\qed
\endproof

Conversely, every strictly regular form $\psi$ of even dimension and discriminant~1 generally acts  as the companion of many regular forms $\varphi$. More precisely the following theorem holds, which can easily be verified upon consulting the explanations about polars in \S 16.

\begtheorem{2.}
\smallskip	
\setitemindent{$(ii)$}
\item{$(i)$} Let  $\psi$ be a strictly regular form of even dimension 
over $k$ with $\Delta (\psi) = 1$. For every $a \in
D(\psi)$ the polar  $\varphi \colon = \Pol_{\psi} ([a])$ $($cf.
\S $16$$)$ is a strictly regular form of odd dimension with  $\Com (\varphi)
= \psi$. If $k$ is of characteristic~$2$, then  $[a] = QL(\varphi)$. If
$k$ is of characteristic $\ne 2$, then $\langle a \rangle = -
d(\varphi)$.
\smallskip	

\item{$(ii)$} The isometry classes $(\varphi)$ of strictly regular forms
$\varphi$ of odd dimension over  $k$  are in one-one correspondence with the pairs
$\bigl((\psi), [a]\bigr)$, where $(\psi)$ is the isometry class of a strictly regular form of even dimension and discriminant~$1$ $($thus $\{
\psi \} \in J_2 (k)$$)$ and $a \in D (\psi)$, via $\varphi =
\Pol_{\psi} [a]$ and $\psi = \Com (\varphi)$, $[a] = QL(\varphi)$
for $\Char k = 2$, $\langle a \rangle = - d(\varphi)$ for $\Char k \ne
2$.\newline \hbox{}\hfill$\square$
\endtheorem

The following theorem gives another method for retrieving $\varphi$ from $\psi$ and 
$d(\varphi)$ from $QL(\varphi)$, respectively, by means of the Cancellation Theorem (\S6, Theorem~2 -- for fields).

\begtheorem{3.} Let $\psi = \Com (\varphi)$.
\smallskip	
\setitemindent{$(a)$}
\item{$(a)$} If $\Char k \ne 2$, then $\varphi \perp H  \cong  \psi
\perp d(\varphi)$.
\smallskip	
\item{$(b)$} If $\Char k = 2$, then $\varphi \perp H  \cong  \psi
\perp QL(\varphi)$.
\endtheorem

\begproof{} $(a)$ This is clear since  $\psi =
\varphi \perp - d(\varphi)$ in this case.

$(b)$ Since  $QL(\varphi) = [a]$ we have
$$\varphi \perp H  \cong  \Pol_{\psi} ([a]) \perp H \cong 
\Pol_{\psi \perp H} ([a]).$$
Furthermore,  $\psi \perp H  \cong  \psi \perp \left[ {a \atop
1} \, {1 \atop 0} \right]$. Thus,
$$\Pol_{\psi \perp H} ([a])  \cong  \psi \perp [a].\eqno\square$$
\endproof

Next we present necessary and sufficient conditions for the companion $\Com (\varphi)$
to be isotropic or split.

\begtheorem{4.} The following statements are equivalent:
\smallskip	
\setitemindent{$(iii)$}
\item{$(i)$} $\Com (\varphi)$ is isotropic.
\smallskip	

\item{$(ii)$} There exists an element   $b \in k^*$ and a strictly regular form  $\chi$ of even dimension
with $\Delta (\chi) = 1$ and $\varphi \cong \chi \perp [b]$.
\smallskip	

\item{$(iii)$} There exists a strictly regular form 
$\chi$ of even discriminant with  $\Delta
(\chi) = 1$, $\chi < \varphi$ and $1 + \dim \chi = \dim \varphi$.
\smallskip	

\noi If $(i)$ -- $(iii)$ hold, then $\Com (\varphi) = \chi \perp H$ and, furthermore,
$[b] = QL(\varphi)$ when  $\Char k = 2$ and $\langle b
\rangle = d(\varphi)$ when  $\Char k \ne 2$.
\endtheorem

\begproof{} If  $\chi$ is a strictly regular form with $\chi <
\varphi$, $\dim \chi = \dim \varphi - 1$, then we have a decomposition
$\varphi \cong \chi \perp [b]$ for some $b \in k^*$. This establishes the equivalence
$(ii) \Leftrightarrow (iii)$. We also see that then
$QL(\varphi) = [b]$ in characteristic~2
and $d(\varphi) = \langle b \rangle$ in characteristic $\ne 2$.

Let $\psi \colon = \Com (\varphi)$. By Theorem~3 we have an isometry
$\varphi \perp H \cong \psi \perp [b]$. If $\psi$ is
isotropic, we have in addition a decomposition $\psi \cong \chi \perp
H$ and we have $\Delta (\chi) = 1$. It follows that $\varphi \cong \chi
\perp [b]$. Conversely, if $\varphi \cong \chi
\perp [b]$ with  $\Delta (\chi) = 1$,  then the form  $\chi
\perp H \cong \chi \perp \left[ {b \atop 1} \, {1 \atop 0} \right]$ satisfies the conditions of Theorem~1, and so  $\chi \perp H 
\cong  \psi$.\qed
\endproof

\begtheorem{5.}  $\Com (\varphi)$ splits if and only if $\varphi$ splits.
\endtheorem

\begproof{} If  $\varphi \cong r \times H \perp [a]$, then the form 
$\psi \colon = r \times H \perp \left[ {a \atop 1} \, {1
\atop 0} \right] \cong (r+1) \times H$ satisfies the four conditions of Theorem~1. Hence,
$\psi = \Com(\varphi)$. Conversely, if $\Com (\varphi)$ is hyperbolic, then $\varphi \perp H$ splits by Theorem~3.  This implies that $\varphi$ splits.\qed
\endproof

As a digression we remark that the unicity statement contained in Theorem~4 leads to a new kind of cancellation theorem:

\begtheorem{6.} Let  $a \in k$ and let  $\chi_1$, $\chi_2$ be two strictly regular
forms of even dimension with $\Delta (\chi_1) =
\Delta (\chi_2)$ and $\chi_1 \perp [a] \cong \chi_2 \perp [a]$. Then
 $\chi_1 \cong \chi_2$.
\endtheorem

\begproof{} This is trivial when  $a = 0$. Thus let $a \ne
0$. If $\Delta (\chi_1) = \Delta (\chi_2) = 1$, the statement follows immediately from Theorem~4.
If only $\Delta (\chi_1) = \Delta
(\chi_2)$, then $\Delta (\chi_1 \perp \chi_2) =
\Delta (\chi_2 \perp \chi_2) = 1$ by \S25, Lemma~2. We have $\chi_1 \perp \chi_2
\perp [a] \cong \chi_2 \perp \chi_2 \perp [a]$, and deduce that 
$\chi_1 \perp \chi_2 \cong \chi_2 \perp \chi_2$, and finally that  $\chi_1 \cong
\chi_2$ with the old Cancellation Theorem  \S 6, Theorem~2. \qed
\endproof

We make a general statement about the Witt indices of $\varphi$ and $\Com (\varphi)$:

\begtheorem{7.} We have
$$\ind (\varphi)  \leq \ind (\Com (\varphi))  \leq 1 + \ind (\varphi).$$
\endtheorem

\begproof{} Let  $\psi \colon = \Com (\varphi)$. Since  $\varphi <
\psi$ we have  $\ind (\varphi) \leq \ind (\psi)$. On the other hand we have an isometry
$\varphi \perp H \cong \psi \perp [a]$ by Theorem~3. Hence
$$\ind (\psi) \leq \ind (\psi \perp [a])  = 1 + \ind (\varphi).\eqno\square$$
\endproof

\begtheorem{8.}  $h (\varphi)  \leq  2 h (\Com (\varphi))$.
\endtheorem

\begproof{} Let  $\psi \colon = \Com (\varphi)$, $h \colon = h
(\varphi)$, $e \colon = h(\psi)$. Then $h+1$ is the number of possible
Witt indices of  $\varphi \otimes K$, where $K$ traverses all field extensions
of $k$, thus $h + 1 = 
\hbox{card}\, \SP (\varphi)$. Similarly,  $e + 1 =  \hbox{card} \,
\SP (\psi)$. Furthermore we have  $\Com (\varphi \otimes K) = \psi \otimes K$.
If  $\ind (\psi \otimes K) = j$, then  $\ind (\varphi
\otimes K)  = j$ or $j + 1$ by Theorem~7. However, if $\ind (\psi \otimes K)$
has the highest possible value $\frac{1}{2} \dim \psi$, then  $\ind
(\varphi \otimes K)$ also has the highest possible value
$\frac{1}{2} \dim \psi -1$ by Theorem~5. It follows that $h + 1 \leq 2 e +1$, and so
$h \leq 2e$. \qed
\endproof

In \S 25 we associated to every strictly regular form, which is not split, a {\it leading form}. We will now also introduce a leading form for regular forms of odd dimension, in characteristic~2. First some preliminaries.

\begdefinition{2.} Let $\tau$ be a (quadratic) Pfister form over
$k$. Since $1 \in D(\tau)$ we can form the polar
$\Pol_{\tau} [1]$, which we call the {\it unit polar of} 
$\tau$ and denote by $\tau'$.
\enddefinition

\begremarks{}
\smallskip	
\setitemindent{$(2)$}
\item{$(1)$} If  $\dim \tau = 2$ and $\Char k = 2$, then
$\tau' = [1]$, and forming the unit polar is not interesting.
On the other hand, if  $\dim \tau \geq 4$, then clearly  $\tau$ is the strictly regular companion
of $\tau'$ and $\tau'$ is the regular form of odd dimension 
associated to the pair $(\tau, [1])$ in the sense of Theorem~2.
\smallskip	
\item{$(2)$} For an arbitrary Pfister form  $\sigma$ over $k$ we introduced the {\it pure part} of $\sigma$, denoted by $\sigma'$, earlier  (\S 16). In $\Char k \ne 2$ this pure part is identical to the unit polar.
\endremarks


It stands to reason to look for a relationship between the generic spitting behaviour of the form 
$\varphi$ and its companion $\Com
(\varphi) = \psi$, and in particular to compare the leading forms of
$\varphi$ and $\psi$ (cf. \S23, Def.~2).  A result in this vein, presented below, can already be found in
[K$_5$, \S 5] for the characteristic $\ne 2$ case. A partial result in characteristic $\ne 2$ was anticipated in
\S 25, Theorem~2.

\begtheorem{9.} Let $\varphi$ be a regular form of odd dimension over
$k$ and $\psi \colon = \Com (\varphi)$ its companion. Let
$(K_r \; \vert \; 0 \leq r \leq h)$ be a generic splitting tower  of
$\varphi$ and $(L_s \; \vert \; 0 \leq s \leq
e)$ a generic splitting tower of $\psi$. We assume that 
$\varphi$, and so $($by Theorem~$5)$ also $\psi$,  does not split.
Let $\tau$ be the leading form of $\varphi$ over
$K_{h-1}$ and $\sigma$ the leading form of  $\psi$ over $L_{e-1}$. Then we claim:
\smallskip	
\setitemindent{$(iii)$}
\item{$(i)$}  $K_h \sim_k L_e$.
\smallskip	
\item{$(ii)$}  There is a place  $\lambda \colon L_{e-1} \to K_{h-1}
\cup \infty$ over $k$.
\smallskip	
\item{$(iii)$} $\sigma$ has good reduction with respect to every such place
$\lambda$,   and 
$\lambda_* (\sigma) \cong \tau$. In particular,
$\deg \tau = \deg \sigma = \deg \psi$.
\endtheorem

\begproof{} For every field extension $L$ of $k$ we have $\Com
(\varphi \otimes L) = \psi \otimes L$. Thus
$\psi \otimes L$ splits if and only if  $\varphi \otimes L$ splits by Theorem~5.
Hence  $K_h$ and $L_e$ are specialization equivalent  over $k$.

The kernel form $\varphi_{h-1}$ of $\varphi \otimes K_{h-1}$ is of the form
$a \tau'$ for some $a \in K_{h-1}^*$. $\{$Incidentally,  $a$ can be chosen in $k$, but we do not need that now.$\}$  We have
$$\psi \otimes K_{h-1}  =  \Com (\varphi \otimes K_{h-1})  \sim  \Com
(\varphi_{h-1})  =  a \tau. \leqno(\ast)$$
It follows that $\deg \tau \geq \deg \psi$. On the other hand, the kernel form
$\psi_{e-1}$ of $\psi \otimes L_{e-1}$ is of the form $b
\sigma$ for some $b \in L_{e-1}^*$. Hence,  $\varphi \otimes L_{e-1}
\sim b \sigma \perp [a]$, where  $[a] = QL(\varphi)$ in
characteristic~2 and $[a] = d(\varphi)$ in characteristic $\ne 2$. Since the form
$\sigma$ is anisotropic and of dimension at least~4, 
$b \sigma \perp [a]$ cannot split.

It follows from the general generic splitting theory that 
$\dim \sigma + 1 \geq \dim \tau'$, hence $\dim \sigma +2 \geq \dim
\tau$. Earlier we established that  $\dim \sigma \leq \dim \tau$. Since 
$\dim \sigma \geq 4$ and the integers  $\dim \sigma$ and  $\dim \tau$
are both 2-powers, it follows that  $\dim \sigma = \dim \tau$. Now 
($\ast$) tells us, by the generic splitting theory, that there exists a place
$\lambda$ from $L_{e-1}$ to $K_{h-1}$ over $k$, and furthermore that 
$b \sigma$ has good reduction with respect to every such place
$\lambda$ and that  $\lambda_* (b \sigma) = a \tau$.  Since $b
\sigma$ has good reduction with respect to    $\lambda$,  $b \sigma$ will certainly represent a unit
 $c$ in the valuation ring   ${\nam o}_{\lambda}$. Then we have
$b \sigma \cong c \sigma$, and it follows that 
$\sigma$ has good reduction with respect to  $\lambda$, and that  $\lambda_* (\sigma) = \lambda
(c) a \tau$. This forces $\lambda_* (\sigma) = \tau$.\qed 
\endproof

As an illustration of the results obtained so far, we classify the anisotropic regular forms of odd dimension whose strictly regular companion is of height~1.

\begtheorem{10.} Let $\varphi$ be  anisotropic. The companion
$\psi \colon = \Com (\varphi)$ is of height~$1$ if and only if one of the following two cases happens:
\smallskip	
\setitemindent{(B')}
\item{{\rm (A)}} $\varphi \cong a \tau'$ for some anisotropic Pfister form $\tau$ of degree 
$\geq 2$ and some $a \in k^*$.
\smallskip	
\item{{\rm (B)}} $\varphi \cong a \tau \perp [b]$ for some anisotropic Pfister form
$\tau$ of degree  $\geq 2$, some $a \in k^*$ and $b \in k^*$ with $- ab \not\in
D(\tau)$.
\medskip
\noi In case {\rm (A)} we have $\psi \cong a \tau$, and so  $\psi$ is anisotropic. In case 
{\rm (B)} we have $\psi \cong a \tau \perp H$, and so  $\psi$ is  isotropic. In case
{\rm (A)} $\varphi$ is of height~$1$, in case  {\rm (B)} of height~$2$. In both cases 
$\varphi$ is excellent.
\endtheorem

\begproof{} (1) Let  $\tau$ be a  Pfister form of degree $\geq 2$
and let $a$ and $b$ be elements of  $k^*$. If $\varphi \cong a \tau'$, 
then $\psi = a \tau$, and so $h(\psi) = 1$. Furthermore we have
$h(\varphi) = 1$ in this case. If $\varphi \cong a \tau \perp [b]$ and $- ab
\not\in D(\tau)$, then $\varphi$ is anisotropic, and $\psi \cong a \tau
\perp \left[ {b \atop 1} \, {1 \atop 0} \right] \cong a \tau \perp
H$. Therefore $\psi$ is also of height~1 in this case. Furthermore,
$$\varphi \perp b \tau'  \cong  a \langle 1, ab \rangle \otimes \tau.$$
Thus  $\varphi$ is a neighbour of the anisotropic Pfister form 
$\langle 1, ab \rangle \otimes \tau$ with complementary form $b
\tau'$. Hence $\varphi$ is excellent of height~2. 

(2) Assume now that $h(\psi) = 1$. By Theorem~7 we have  $\ind
(\psi) \leq 1$.

Case (A): $\psi$ is anisotropic. Then $\psi \cong c \tau$ for some
$c \in k^*$ and some Pfister form $\tau$ of degree $\geq 2$.
If $\Char k = 2$, and $QL(\varphi) = [a]$, then we may choose $c = a$
and have $\varphi = \Pol_{\psi} ([a]) \cong a \tau'$. If
$\Char k \ne 2$, then $\varphi$ is a strictly regular subform of
$\psi$, and so we obtain again that $\varphi \cong a \tau'$ for some $a \in k^*$.

Case (B): $\psi$ is isotropic.  Then $\psi \cong a \tau \perp
H$ for some $a \in k^*$ and some Pfister form $\tau$ of degree~2. By Theorem~3 we have
$\varphi \perp H \cong \psi \perp [b]$ for some $b \in k^*$. Hence,
$\varphi \cong a \tau \perp [b]$. Since $\varphi$ is anisotropic it follows that
$- ab \not\in D(\tau)$.\qed
\endproof

Next we will more closely consider and compare the generic splitting towers of 
$\varphi$  and $\psi \colon = \Com (\varphi)$. Assume as before that 
$\dim \varphi$ is odd. Let further  $ak^{*2}$ be the square class with 
$d (\varphi) = \langle - a \rangle$ in case
$\Char k \ne 2$,  and $QL(\varphi) = [a]$ in case $\Char k = 2$. Thus
$a \in D(\psi)$ and $\varphi = \Pol_{\psi} ([a])$. We need two lemmata.


\beglemma{1.} Let $\varphi_0 \colon = \ker (\varphi)$ and $\psi_0
\colon = \ker (\psi)$. The following statements are equivalent:
\smallskip	
\setitemindent{$(iii)$}
\item{$(i)$} $\Com (\varphi_0) = \psi_0$.
\smallskip	
\item{$(ii)$} $\Com (\varphi_0)$ is anisotropic.
\smallskip	
\item{$ (iii)$} $\ind (\psi) = \ind (\varphi)$.
\smallskip	
\item{$ (iv)$} $a \in D (\psi_0)$.
\endlemma

\begproof{} Let  $r \colon = \ind (\varphi)$, hence $\varphi 
\cong  \varphi_0 \perp r \times H$ and $\chi \colon = \Com
(\varphi_0)$. From Remark (1) above it follows that
$$\psi  \cong  \chi  \perp  r \times H.$$
This is the Witt decomposition of $\psi$ if and only if
$\ind (\psi) = r$ or, equivalently, if and only if  $\chi$ is anisotropic.
This already proves the equivalence of  $(i)$ -- $(iii)$.

Furthermore, $d(\varphi_0) = d(\varphi) = \langle - a \rangle$ if 
$\Char k \ne 2$, and  $QL(\varphi_0) = QL(\varphi) = [a]$ if
$\Char k = 2$. Thus $a \in D (\chi)$ and $\varphi_0 =
\Pol_{\chi} ([a])$. If $\chi = \psi_0$, then certainly  $a
\in D(\psi_0)$. This establishes $(i) \Rightarrow (iv)$.

Finally, assume that  $a \in D (\psi_0)$.
Of course $\psi_0 \sim \chi$. Hence,
$\Pol_{\psi_0} ([a]) \sim \Pol_{\chi} [a] = \varphi_0$, and so
$\varphi_0 \cong \Pol_{\psi_0} ([a])$, since the forms
$\varphi_0$ and $\Pol_{\psi_0} ([a])$ are both anisotropic. Therefore,
$\psi_0 = \Com (\varphi_0)$. This shows $(iv) \Rightarrow (i)$.\qed
\endproof

We assume from now on that  $\varphi$, and thus $\psi$,
is not split.

\beglemma{2.} Let $\psi = \Com (\varphi)$ be anisotropic. Let $\psi_1$
denote the kernel form of $\psi \otimes k(\psi)$.
\smallskip	
\setitemindent{$(iii)$}
\item{$(i)$} There exists a place from $k(\psi)$ to
$k(\varphi)$ over $k$.
\smallskip	
\item{$(ii)$}   $k(\varphi) \sim_k k(\psi)$ if and only if $a \in
D(\psi_1)$ or $\ind (\psi \otimes k(\psi)) \geq 2$.
\smallskip	
\item{$(iii)$}  If $k (\varphi)$ is {\rm not} specialization equivalent to  $k(\psi)$ over 
$k$, then $\varphi \otimes k(\psi)$ is anisotropic.
\endlemma

\begproof{} $(i)$ is clear since $\varphi$ is a subform of $\psi$.
Let  $K \colon = k (\psi)$. We have $\Com (\varphi \otimes K) =
\Com (\psi) \otimes K$. By the general theory of generic splitting it is clear that
$k(\varphi)$ is specialization equivalent to  $K$ over
$k$ if and only if $\varphi \otimes K$ is isotropic.

A priori we have  $\ind (\varphi \otimes K) = \ind (\psi \otimes K)$ or
$\ind (\varphi \otimes K) = \ind (\psi \otimes K) - 1$ (by Theorem~7). If
$\ind (\psi \otimes K) \geq 2$, then $\varphi \otimes K$ is definitely isotropic.
Assume now that $\ind (\psi \otimes K) = 1$, thus
$\psi_K \cong \psi_1 \perp H$. The form $\varphi \otimes K$ is of index~0
or~1. By Lemma~1,  $\varphi \otimes K$ is of
index 1 if and only if  $a \in D (\psi_1)$. This happens if and only if $\varphi
\otimes K$ is isotropic.\qed
\endproof

Let $h \colon = h(\varphi)$ and $e \colon = h(\psi)$. Further, let  $(L_s
\; \vert \; 0 \leq s \leq e)$ be a generic splitting tower of
$\psi = \Com (\varphi)$, and let $\psi_s \colon = \ker (\psi \otimes
L_s)$, $\eta_s \colon = \ker (\varphi \otimes L_s)$. We know that
$\varphi \otimes L_e$ splits, so that $\eta_e$ is one-dimensional and thus
$\eta_e = [-a]$.

We are looking for an integer $m \in [0, e]$, as large as possible, such that
$(L_s \; \vert \; 0 \leq s \leq m)$  is a {\it truncated generic splitting tower} of 
$\varphi$, which signifies  that there is a generic splitting tower
$(K_r \; \vert \; 0
\leq r \leq h)$ of $\varphi$ with $h \geq m$ and $K_r = L_r$
for $0 \leq r \leq m$.

\begtheorem{11.} Let $a \in D(\psi_m)$ for some $m \in [0, e-1]$.
\smallskip	
\setitemindent{$(b)$}
\item{$(a)$}  $(L_s \; \vert \; 0 \leq s \leq m)$ is a truncated generic splitting
tower of $\varphi$. Furthermore, $\ind
(\varphi \otimes L_s) = \ind (\psi \otimes L_s)$ and $\Com (\eta_s)
= \psi_s$ for $0 \leq s \leq m$.
\smallskip	

\item{$(b)$} If, in addition,  $a \not\in D(\psi_{m+1})$, then $\ind
(\varphi \otimes L_{m+1}) = \ind (\psi \otimes L_{m+1}) - 1$ and
$\Com (\eta_{m+1}) \cong \psi_{m+1} \perp H$.  The following alternative also holds:
if $\ind (\psi_m \otimes L_{m+1}) \geq 2$, then
$(L_s \; \vert \; 0 \leq s \leq m+1)$ is a truncated generic splitting tower
of $\varphi$. On the other hand, if $\ind (\psi_m \otimes
L_{m+1}) = 1$, then $\eta_m \otimes L_{m+1}$ is anisotropic, and so
$\eta_m \otimes L_{m+1} = \eta_{m+1}$ and
$$\ind (\varphi \otimes L_{m+1})  =  \ind (\varphi \otimes L_m)  =  \ind (\psi
\otimes L_m).$$
\endtheorem

\begproof{} By induction on  $m$. Lemma 1 and
Lemma~2 give us the case $m = 0$. Thus let $m > 0$.
From the induction hypothesis we know that $(L_s \; \vert \; 0
\leq s \leq m-1)$ is a truncated generic splitting tower of
$\varphi$ and that $\ind (\varphi \otimes L_s) = \ind (\psi \otimes
L_s)$ and $\Com (\eta_s) = \psi_s$ for $0 \leq s \leq m-1$. Let $K
\colon = L_{m-1}$. By assumption we have $L_m \sim_K K(\psi_{m-1})$
and $a \in D (\psi_m)$. It follows that the element $a$ is also represented by 
$\ker (\psi_{m-1} \otimes K (\psi_{m-1}))$. Furthermore,
$\psi_{m-1} = \Com (\eta_{m-1})$. From
Lemma~2 it follows that $K(\psi_{m-1}) \sim_K K(\eta_{m-1})$, and so $L_m
\sim_K K(\eta_{m-1})$. Hence, $(L_s \; \vert \; 0 \leq s
\leq m)$ is a truncated generic splitting tower of  $\varphi$.
Applying Lemma~1 to $\eta_{m-1} \otimes L_m$ further shows that
$\Com (\eta_m) = \psi_m$, and finally that
$\eta_{m-1} \otimes L_m$ and $\psi_{m-1} \otimes L_m$
have the same index. By the induction hypothesis
$\varphi \otimes L_{m-1}$ and $\psi \otimes L_{m-1}$ also have the same
index. Therefore $\varphi \otimes L_m$ and $\psi \otimes L_m$
have the same index. This proves part $(a)$ of the theorem. Applying 
Lemma~2 again, but this time to  $\eta_m$ and $\psi_m = \Com
(\eta_m)$, gives part $(b)$.\qed
\endproof

\begtheorem{12.} We have $h \geq e$. $\{$Recall that $h \colon =
h(\varphi)$, $e \colon = h(\psi)$. We assume that $h > 0$, $e > 0$.$\}$ 
If $a \not\in D(\psi_0)$, then we even have $h \geq e+1$.
The following statements are equivalent:
\smallskip	
\setitemindent{$(iii)$}
\item{$(i)$} $a \in D(\psi_{{e-1}})$.
\smallskip	
\item{$ (ii)$} $(L_s \; \vert \; 0 \leq s \leq e)$ is a generic splitting tower
of $\varphi$. $($In particular, $h = e$.$)$
\smallskip	
\item{$ (iii)$}  $\Com (\eta_s) = \psi_s$ for $0 \leq s \leq e-1$.
\smallskip	
\item{$ (iv)$} $\ind (\varphi \otimes L_s) = \ind (\psi \otimes L_s)$
for $0 \leq s \leq e-1$.
\smallskip	
\item{$ (v)$} $L_{e-1}$ is a leading field for $\varphi$.
\endtheorem

\begproof{} We distinguish three cases.

{\it Case $1$:}  $a \not\in D (\psi_0)$. For every $s \in [0, e]$
we establish the following. We have $a \not\in D(\psi_s)$, and so
$\psi_s \perp [-a]$ is anisotropic. By Theorem~3 we have $\psi \perp
[-a] \cong \varphi \perp H$. Thus $\psi_s \perp [-a] \sim
\eta_s$ by the anisotropy of both forms, hence $\eta_s \cong
\psi_s \perp [-a]$. For $s = e-1$ we obtain $\eta_{e-1} \cong b
\sigma \perp [-a]$ for some $b \in L^*_{e-1}$ and $\sigma$ the leading form of 
$\psi$ over $L_{e-1}$.

This form $\eta_{e-1}$ is a neighbour of the Pfister form $\langle 1, - ab
\rangle \otimes \sigma$ (which is thus anisotropic) and has
$\zeta \colon = - a \sigma'$ as complementary form, and thus 
$(- a \sigma') \otimes L_{e-1} (\eta_{e-1})$ as first higher kernel form. The
kernel form of $\varphi \otimes L$ has a different dimension for each of the fields  $L =
L_0, L_1, \dots, L_e$, $L_{e-1} (\eta_{e-1})$. Hence
$h > e$.

{\it Case $2$:}  $a \in D (\psi_{e-1})$. Since $\psi_e = 0$, and thus $a
\not\in D (\psi_e)$, we may apply Theorem~11. By part $(a)$ of said theorem, 
$(L_s \; \vert \; 0 \leq s \leq e-1)$ is a truncated generic splitting tower of
$\varphi$ and $\Com
(\eta_s) = \psi_s$ for every $s \in [0, e-1]$. In particular,
$\eta_{e-1}$ is not split. But $\eta_{e-1} \otimes L_e$ splits, since
$\varphi \otimes L_e$ splits. Thus, by part $(b)$ of Theorem~11, 
$(L_s \; \vert \; 0 \leq s \leq e)$ is a truncated generic splitting tower
of $\varphi$. Since
$\varphi \otimes L_e$ splits, this tower is actually a full generic splitting tower of
$\varphi$.

{\it Case $3$:} The rest. Now there exists a largest integer $m \in [0,
e-2]$ such that  $a \in D(\psi_m)$. By Theorem~11$(a)$,   $(L_s \; \vert \; 0
\leq s \leq m)$ is a truncated generic splitting tower of
$\varphi$ and $\Com (\eta_s) = \psi_s$ for $0 \leq s \leq m$. Furthermore, by
Theorem~11$(b)$,  $\Com (\eta_{m+1}) = \psi_{m+1} \perp H$,
and we have the following alternatives:

{\it Case $3(a)$:}  $\ind (\psi_m \otimes L_{m+1}) \geq 2$. Now
$(L_s \; \vert \; 0 \leq s \leq m+1)$ is a truncated generic splitting tower
of $\varphi$.

{\it Case $3(b)$:} $\ind (\psi_m \otimes L_m) = 1$. Now
$\eta_m \otimes L_{m+1} = \eta_{m+1}$.
\medskip

$(L_s \; \vert \; m+1 \leq s \leq e)$ is a generic splitting tower of
$\Com (\eta_{m+1}) \sim \psi_{m+1} \perp H$,
and we have $a \not\in D(\psi_{m+1})$. For this form $\psi_{m+1}$ we are in Case~1, discussed above,
and conclude that $h(\eta_{m+1}) > h(\psi_{m+1}) = e - (m+1)$,
and so $h(\eta_{m+1}) \geq e - m$. In Case 3$(a)$ we have  $h (\eta_{m+1})
= h-(m+1)$, and we obtain $h \geq e+1$. In case 3$(b)$ we have
$\eta_{m+1} \cong \eta_m \otimes L$ and we may conclude that
$h(\eta_{m+1}) \leq h (\eta_m) = h-m$. Therefore,
$e -m \leq h-m$ and so $e \leq h$.

Thus  $e \leq h$ in all cases, and in Case~2 we even have
$e < h$. Our analysis shows furthermore that all the conditions
$(i)$ -- $(iv)$ of the theorem hold in Case~2, and are thus equivalent. 
The implication $(ii) \Rightarrow (v)$ is trivial. Assume $(v)$, then
$\eta_{e-1} \cong a \tau'$ where $\tau$ is the leading form of
$\varphi$ over $L_{e-1}$. Thus $\psi_{e-1} \sim \Com
(a \tau') = a \tau$ and then $\psi_{e-1} \cong a \tau$, since the forms
$\psi_{e-1}$ and $\tau$ are both anisotropic. Hence $a \in D
(\psi_{e-1})$. This proves the implication  $(v) \Rightarrow (i)$.\qed
\endproof

Note that the analysis, carried out above,  also establishes the following:

\begcorollary{} If there exists an  $m \in [0, e-1]$ with $a \in D
(\psi_m)$, $a \not\in D(\psi_{m+1})$, and $\ind (\psi_m \otimes
L_m) \geq 2$, then $h > e$.
\endcorollary

