\titlea{\S13}{Regular Generic Splitting Towers and Base Extension}
\leftrunningtitle{Chapter II. Generic Splitting Theory}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S13 Regular Generic Splitting
Towers and Base Extension}

Before  turning  towards generic splitting towers, we will give
two general definitions which will also serve us well in later
sections.

As before, the word {\it form} over a field $k$ will always be
understood
to mean a \nd\ quadratic form over $k$. In the following let $\vf$
be a form over~$k$.

\begdefinition{1.} Let $\dim\vf$ be even, $\vf\neq 0$ and $QL
(\varphi) = 0$.
\medskip
\setitemindent{(b)}
\item{(a)} The {\it discriminant algebra} $\Delta (\varphi)$ is
defined as follows: If $\Char k\neq 2$, we let $\Delta (\varphi) \colon = k [X]
/ (X^2 - a)$, where $a$ is a representative of the signed
determinant $d(\tilde{\varphi}) = ak^{*2}$ of the bilinear form $\tilde{\varphi} =
B_{\varphi}$ associated to $\vf$ (cf. \S2). If $\Char k=2$, we let $\Delta (\varphi)
\colon = k [X]/(X^2 + X + c)$, where $c$ is a representative of
the Arf-invariant $\Arf(\varphi) \in k^+/\wp k$.
\medskip
\item{(b)} We define the {\it discriminant} of $\vf$ to be the
isomorphism class of $\Delta (\varphi)$ as $k$-algebra. We will
denote it sloppily by $\Delta (\varphi)$ as well. The discriminant
is independent of the choice of $a$ resp. $c$ above.
\medskip
\item{(c)} We say that $\Delta (\varphi)$ {\it splits} when  $\Delta
(\varphi)$ is not a field, i.e. when $\Delta (\varphi) \cong k \times
k$. We will symbolically write $\Delta (\varphi) = 1$ when $\vf$
splits and $\Delta (\varphi) \ne 1$ when $\vf$ doesn't split.
\enddefinition

\begremark{1.} This notation is not completely groundless, for the
isomorphism classes of quadratic separable $k$-algebras form a
group in a natural way with unit element $k \times k$. In this
group the equality $\Delta (\varphi \bot \psi) = \Delta (\varphi)
\Delta (\psi)$ holds. We will not discuss    the
group of quadratic separable $k$-algebras any deeper since we will not
make any serious use of it.
\endremark

\begremark{2.} If $K/k$ is a field extension,\fonote{From now on
we will often denote a field extension $F \hookrightarrow E$ by
$E/F$, as has been customary in algebra for a long time.}
then $\Delta (\varphi
\otimes K) = \Delta (\varphi) \otimes_k K$.
\endremark

\begremark{3.} If $\tau$ is the norm form of the $k$-algebra $\Delta
(\varphi)$, then $\Delta (\varphi) = \Delta (\tau)$. Is further
$\dim\vf=2$, then $\varphi \cong c \tau$ for some $c \in k^*$. If
we write $\tau = \left[ {1 \atop 1} {1 \atop \delta} \right]$,
then $\Delta (\varphi) = k [X]/(X^2 + X + \delta)$, also when
$\Char k\neq 2$.
\endremark

\begdefinition{2}{(cf. [KR, Def.1.1]).} Let $\dim\vf>1$. We say
that ``{\it $\vf$ is of outer type}'' when $\dim\vf$ is even,
$QL(\vf)=0$ and $\Delta(\vf)\not=1$. In all other cases (thus in
particular when $QL(\vf)\not=0$) we say that ``{\it $\vf$ is of inner
type}''.
\enddefinition

\begremark{4.} For those readers who are at home in the theory of
reductive  algebraic groups, we want to remark that Definition~2
leans on the concepts of inner/outer type in use there: is $\vf$
regular (i.e. $\dim QL(\vf)\leq 1$) and $\dim \vf\geq 3$, then the
group SO$(\vf)$ is almost simple. This group is of inner/outer
type if and only if this is the case for $\vf$ in the sense of
Definition~2.
\endremark

We want to construct ``regular'' generic splitting towers of
$\vf$, having
particularly convenient properties \wrt\
extensions of the ground field $k$, which are nonetheless
sufficiently general. Starting with such a regular generic
splitting tower of $\vf$ we will then construct a regular generic
splitting tower of $\varphi \otimes L$ for every
$\vf$-conservative field extension $L/k$.

For every form $\psi$ with $\dim\psi\geq 2+\dim QL(\psi)$ over a field
$K$ we introduced the field extension $K(\psi)$ of $K$ earlier
(\S 4; \S 9; \S 10, following Def.2). $K(\psi)$ was defined to be
the function field of the affine quadric $\psi(X)=0$ over $K$,
except when $\psi \cong \left[ {0 \atop 1} {1 \atop 0} \right]$,
in which case this quadric degenerates in two lines.  In this
situation we defined $K(\psi)=K(t)$ for some indeterminate $t$
over $K$. Sometimes it is more natural to use the {\it projective}
quadric $\psi(X)=0$ over $K$ instead of the affine quadric
$\psi(X)=0$. Thus we arrive at the following

\begdefinition{3.} Let $\psi$ be a (\nd\ quadratic) form over $K$
with $\dim \psi \geq 2 + \dim QL (\psi)$. We define a field
extension $K \{ \psi \}$ of $K$ as follows: If $\psi \cong
\left[ {0 \atop 1} {1 \atop 0} \right]$, we let $K \{ \psi \} =
K$. Otherwise we let $K \{ \psi \}$ be the subfield $K( \frac{x_1}{x_i}, \dots ,
\frac{x_n}{x_i})$ of $K(\psi) = \Quot K [X_1, \dots, X_n]/\psi (X_1, \dots,
X_n) = K(x_1, \dots, x_n)$ for some $i \in [1, n]$.\fonote{In
[K$_5$] this field is denoted by $K (\psi)_0$. $[1, n]$ denotes
the set $\{1,2,\ldots,n\}$.}
\enddefinition

In the main case $\psi \not\cong \left[ {0 \atop 1} {1 \atop 0}
\right]$ this field is obviously independent  of the choice of $i \in [1,
n]$. We have $K(\psi) = K \{ \psi \} (x_i)$ and $x_i$ is
transcendental over $K \{ \psi \}$. Also in the case
$\psi \cong \left[ {0 \atop 1} {1 \atop 0} \right]$ we have $K(\psi) = K \{ \psi
\} (t)$ where $t$ is an indeterminate.

$K(\psi)$ is an inessential extension of $K \{ \psi \}$.
Thus, all this time
we could have used $K \{ \psi \}$ instead of $K(\psi)$.

Back to our form $\vf$ over $k$! Let $h = h (\varphi)$ be the
height of $\vf$.

\begdefinition{4.} The {\it projective standard tower} of $\vf$ is the
field tower $(K_r \; \vert \; 0 \leq r \leq h)$ with $K_0 = k$, $K_{r+1} = K_r \{
\varphi_r \} \quad (0 \leq r \leq h - 1)$, where $\varphi_r$
denotes the kernel form of $\varphi \otimes K_r$. Similarly one
obtains the {\it affine standard tower} $(K'_r \; \vert \; 0 \leq r \leq h)$
of $\vf$ by replacing the $K_r \{ \psi_r \}$ with the function
fields $K_r (\varphi_r)$ of the affine quadrics $\varphi_r = 0$.
\enddefinition

\begremark{5.} From the definition of generic splitting tower (\S
12, just after Theorem~2) it follows immediately that $(K_r \; \vert \; 0 \leq r
\leq h)$ and $(K'_r \; \vert \; 0 \leq r \leq h)$ are both generic
splitting towers of $\vf$. For every $r \in \{ 0, \dots, h \}$ we
have that $K'_r$ is a purely transcendental field extension of
$K_r$ of transcendence degree $r$.
\endremark

Let us recall that a field extension $L/K$ is called {\it regular}
when it is separable and $K$ is algebraically closed in $L$. This
is synonymous with saying that $L$ and the algebraic closure
$\widetilde{K}$ of $K$ in $L$ are linearly disjoint, i.e. that $L \otimes_K
\widetilde{K}$ is a field, cf. [Lg$_2$, Chap.~3]. If $L/K$ is regular,
then the $K$-algebra $L \otimes_K E$ does not contain zero
divisors for any field extension $E/K$.

\begtheorem{1.} Let $(K_r \vert 0 \leq r \leq h)$ be the
projective standard tower of $\vf$. Then the field extensions
$K_r/K_{r-1}$ with $0 < r < h$ are all regular. The same is true
for $K_h/K_{h-1}$ in case $h>0$ and $\vf$ is of inner type.
Furthermore, the kernel form $\varphi_{h-1}$ of $\varphi \otimes
K_{h-1}$ has dimension $\geq 3$ in this situation. If $\vf$ is of
outer type, then $h\geq 1$ and $K_h =
K_{h-1} \otimes_k \Delta (\varphi)$. Furthermore, we then have
$\varphi_{h-1} \cong c (\tau \otimes K_{h-1})$, where $\tau$ is
the norm form of the field extension $\Delta (\varphi)$ of $k$ and
$c \in K^*_{h-1}$ is a constant. In particular we have $\dim
\varphi_{h-1} = 2$.
\endtheorem

\begproof{} If $\psi$ is a \nd\ \qf\ over a field $K$ and if
$\dim\psi>2$, then the extension $K \{ \psi \} /K$ is clearly
regular. This established the first statement of the theorem. If
$h>0$ and $\vf$ is of inner type, then $\varphi \otimes K_{h-1}$
is also of inner type and thus the kernel form $\varphi_{h-1}$ of
$\varphi \otimes K_{h-1}$ is of inner type. This shows that
$\dim \varphi_{h-1} \geq 3$ and so $K_h =
K_{h-1} \{ \varphi_{h-1} \}$ over $K_{h-1}$ is again regular.
Suppose now that $\vf$ is of outer type. $\vf$ does not split, so
that $h>0$. The extension $K_{h-1}/K_0$ is regular and $K_0/k$ is
inessential. Therefore $k$ is algebraically closed in $K_{h-1}$
and
$\Delta (\varphi_{h-1}) = \Delta (\varphi \otimes K_{h-1}) = \Delta
(\varphi) \otimes_k K_{h-1}$ is a field. If it would be true that $\dim \varphi_{h-1} >
2$, then $K_h = K_{h-1} \{\varphi_{h-1} \}$ over $K_{h-1}$ would
be regular. However, $\Delta (\varphi_{h-1}) \otimes_{K_{h-1}} K_h = \Delta (\varphi \otimes
K_h)$ is not a field since $\varphi \otimes K_h \sim 0$. Therefore
the dimension of $\varphi_{h-1}$ is $2$. By Remarks~2 and 3 above,
we have
$\Delta (\varphi_{h-1}) =
\Delta (\varphi \otimes K_{h-1}) = \Delta (\varphi) \otimes_k K_{h-1} =
\Delta (\tau) \otimes_k K_{h-1} = \Delta (\tau \otimes K_{h-1})$ (and $\tau
\otimes K_{h-1}$ is the norm form of  $K_h/K_{h-1}$). We conclude
that $\varphi_{h-1} \cong c (\tau \otimes K_{h-1})$ for some element $c$
of $K_{h-1}$.\qed
\endproof

\begtheorem{2.} Let $r \in [0, h]$ and let $E/k$ be a partially
generic splitting field of level $r$ of $\vf$.
\medskip
\setitemindent{(a)}
\item{$(a)$} If $r<h$, then $k$ is algebraically closed in $E$.
\medskip
\item{$(b)$} If $r=h$ and $\vf$ is of inner type, then $k$ is likewise
algebraically closed in $E$.
\medskip
\item{$(c)$} If $r=h$ and $\vf$ is of outer type, then $k$ has
algebraic closure $\Delta(\vf)$ in $E$.
\endtheorem

\begproof{} By Theorem~1 these statements hold when $E$ is the
$r$-field $K_r$ in the projective standard tower $(K_i \; \vert \; 0 \leq i \leq
h)$ of $\vf$. The theorem now follows in all generality from the
following simple lemma.
\endproof

\beglemma{1.} Let $K$ and $L$ be extensions of the field $k$ which
are specialization equivalent, $K \sim_k L$. Then every place
$\lambda \colon K \to L \cup \infty$ over $k$ maps the algebraic
closure of $k$ in $K$ isomorphically to the algebraic closure of
$k$ in $L$.
\endlemma

\begproof{of the lemma.} Let $K^{\circ}$ be the algebraic
closure of $k$ in $K$ and $L^{\circ}$ the algebraic closure of $k$
in $L$. Let $\lambda \colon K \to L \cup
\infty$ and $\mu \colon L \to K \cup \infty$ be places over $k$.
Since $K^{\circ}$ and $L^{\circ}$ are algebraic over $k$, the
places $\lambda$ and $\mu$ are finite on $K^{\circ}$ resp.
$L^{\circ}$. Their restrictions to $K^{\circ}$ and $L^{\circ}$ are
thus field homomorphisms $\lambda' \colon K^{\circ} \to L^{\circ}$,
$\mu' \colon L^{\circ} \to K^{\circ}$, which  are both the identity
on $k$. Now $\mu' \circ \lambda'$ is an endomorphism of $K^{\circ}
/k$ and thus automatically an automorphism of $K^{\circ}/k$.
Likewise, $\lambda' \circ \mu'$ is an automorphism of
$L^{\circ}/k$. The statement of the lemma is now clear.\qed
\endproof


Now we can precisely formulate
a desirable property of generic splitting towers.

\begdefinition{5.} We call a generic splitting tower $(K_r \;
\vert \; 0 \leq r \leq h)$ of $\vf$ {\it regular} when the
following holds:
\medskip
\setitemindent{(2)}
\item{(1)} For every $r$ with $0\leq r<h-1$ the extension
$K_{r+1}/K_r$ is regular.
\medskip
\item{(2)} If $\vf$ is of inner type and $h>0$, then $K_h/K_{h-1}$
is also regular. On the other hand, if $\vf$ is of outer type,
then $K_h$ is regular over the composite $K_{h-1} \cdot \Delta (\varphi) = K_{h-1} \otimes_k
\Delta (\varphi)$.
\enddefinition

\begexamples{} The projective standard tower of $\vf$ is regular
by Theorem~1. The affine standard tower of $\vf$ is likewise
regular.\qed
\endexamples

Now let $(K_r \; \vert \; 0 \leq r \leq h)$ be a  regular
generic splitting tower of $\vf$ and let $L/k$ be a
$\vf$-conservative field extension of $k$. Using $(K_r \;
\vert \; 0 \leq r \leq h)$ and $L$ we want to construct a generic
splitting tower of $\varphi \otimes L$.

\begdefinition{6.}
For every $r \in \{ 0, 1, \dots, h \}$ we construct a field
composite $K_r \cdot L$ of $K_r$ and $L$ over $k$ as follows: if $r <
h$, or if $r = h$ and $\varphi$ is of inner type, then $K_r \otimes_k L$
is free of zero divisors and we let $K_r \cdot L$ be the quotient
field of this ring, i.e. the uniquely determined free composite of
$K$ and $L$ over $k$.
If $r = h$ and $\varphi$ is of outer type, we distinguish the
cases where $\varphi \otimes L$ is of outer/inner type.

First, let $\varphi \otimes L$ be of outer type. Then
$\Delta (\varphi) \otimes_k L$ is a field. Let $K_h \cdot L$ be
the free composite of $K_h$ with $\Delta (\varphi) \otimes_k L$ over $\Delta (\varphi)$,
i.e. again the quotient field of $K_h \otimes_{\Delta (\varphi)} (\Delta (\varphi)
\otimes_k L) = K_h \otimes_k L$. Finally, let $\varphi \otimes L$
be of inner type.  Now $\Delta (\varphi)$ can be embedded over $k$
in $L$ in two ways. We choose one such embedding and set
$K_h \cdot L = \Quot
(K_h \otimes_{\Delta (\varphi)} L)$.\qed
\enddefinition

If necessary, we write more precisely $K_r \cdot _k L$ $(0 \leq r \leq
h)$ for the field composite $K_r \cdot L$. Later on we will call $K_r \cdot
L$ sloppily {\it ``the'' free composite of $K_r$ and $L$ over
$k$}, also in the case $r=h$, $\vf$ of outer type, $\vf\otimes L$
of inner type. It will never matter which of the two embeddings $\Delta (\varphi) \hookrightarrow
L$ we have chosen.

We use $\vf_r$ to denote -- as before -- the kernel form of $\varphi
\otimes K_r$ ($r$-th higher kernel form) and $i_r$ to denote the
Witt index of $\varphi \otimes K_r$ ($r$-th higher index, $0 \leq r \leq h$).

\begtheorem{3.} Let $J$ be the set of all $r \in [0, h]$ such that
$\varphi_r \otimes K_r \cdot L$ is anisotropic, i.e. such that
$\ind (\varphi \otimes K_r) = \ind (\varphi \otimes K_r \cdot L)$.
Let $r_0 < r_1 < \cdots < r_e$ be the elements of $J$.

\noindent Claim:

\setitemindent{(a)}
\item{$(a)$} $(K_{r_i} \cdot L \; \vert \; 0 \leq i \leq e)$ is a
regular generic splitting tower of $\varphi \otimes L$.
\medskip
\item{$(b)$} $K_{r+1} \cdot L/K_r \cdot
L$ is a regular inessential extension for every $r \in [0, h] \setminus
J$.
\endtheorem

For the proof of this technically very important theorem, we need
a general lemma about places.

\beglemma{2.} Let $K \supset k$ and $L \supset k$ be arbitrary
extensions of a field $k$ and let $\lambda \colon K \to L \cup
\infty$ be a place over $k$. Furthermore, let $E \supset k$ be a
field extension which is linearly disjoint from $K$ and $L$ over
$k$. Consider the free composites $K \cdot E$ and $L \cdot
E$ of $K$ resp. $L$ with $E$ over $k$.

Claim: $\lambda$ has a unique extension $\tilde{\lambda} \colon
\; K\cdot E \to L \cdot E \cup \infty$ to a place over $E$.
\endlemma

\begproof{} $K \cdot E$ is the quotient field of $K \otimes_k E$
and $L \cdot E$ is the quotient field of $L \otimes_k E$. Let
${\nam o} \colon =
{\nam o}_{\lambda}$ and let $\alpha \colon \; {\nam o} \otimes_k E \to L
\otimes_k E$ be the homomorphism of $E$-algebras induced by
$\lambda \vert {\nam o} \colon {\nam o} \to L$. Following the
general extension theorem for places [Bo$_2$, \S 2, Prop.3], we
choose a place
$\mu \colon \; K \cdot E \to (\widetilde{L \cdot E}) \cup \infty$
in the algebraic closure of $L \cdot E$ which extends the
homomorphism $\alpha$. We will now show that $\mu$ is the only
such extension of $\alpha$ and that $\mu (K \cdot E) \subset (L \cdot E) \cup
\infty$.

For this purpose we choose a basis $(\omega_i \; \vert \; i \in I)$ of $E$ over
$k$ (as $k$-vector space). Let $z \ne 0$ be an element in $K \cdot
E$. We write $z = \frac{x}{y}$, where  $x, y \in {\nam o} \otimes
E$ are both non-zero. We then have equations
$$x = u \cdot \sum\limits_{i \in I} a_i \otimes \omega_i \; , \quad y = v
\cdot \sum\limits_{i \in I} b_i \otimes \omega_i$$
with $u, v \in K$ and families
$(a_i \; \vert \; i \in I)$ and $(b_i \; \vert \; i \in I)$ in ${\nam
o}$, both of them not fully contained in the maximal ideal ${\nam m}$
of ${\nam o}$. Then
$$\mu \left( \sum\limits_{i \in I} a_i \otimes \omega_i \right) =
\sum\limits_{i \in I} \lambda (a_i) \omega_i \ne 0, \quad
           \mu \left( \sum\limits_{i \in I} b_i \otimes \omega_i \right) =
\sum\limits_{i \in I} \lambda (b_i) \omega_i \ne 0,$$
since $(\omega_i \; \vert \; i \in I)$ is also a family of
elements of $L \cdot E$, linearly independent over $L$. We thus
obtain
$$\mu (z) = \lambda \left( \frac{u}{v} \right) \cdot \left( \sum\limits_{i
\in I} \lambda (a_i) \omega_i \right) \cdot \left( \sum\limits_{i \in I}
\lambda (b_i) \omega_i \right)^{-1} \in (L \cdot E) \cup
\infty.$$\qed
\endproof

\begproof{of Theorem 3.} We assume \wlog\  that $\vf$ is
anisotropic and use induction on the height $h$. Remark that
$K_0 \sim_k k$  implies that $K_0 \cdot L \sim_L L$ by the lemma.
Therefore the extension $K_0 \cdot L/L$ is inessential.

For $h=0$ nothing more has to be done, so suppose that $h>0$.
First we assume that $\dim\vf>2$. We will deal with the (easier)
case $\dim\vf=2$ at the end.

The form $\varphi_1 = \ker (\varphi \otimes K_1)$ has height $h-1$
and regular generic splitting tower $(K_r \vert 1 \leq r \leq h)$.
We want to apply the induction hypothesis to $\vf_1$, this tower
and the field extension $K_1 \cdot L/K_1$.

Clearly the extension $K_1\!\cdot\!L/K_0\!\cdot\!L$ is regular
(cf. e.g. [Lg$_2$, p.58]). Furthermore, $K_0 \cdot L/L$ is
inessential and $L/k$ $\vf$-conservative. It follows that $K_1 \cdot
L/k$ is $\vf$-conservative and then that $K_1 \cdot L/K_1$ is
$\vf_1$-conservative. For every $r$ with $1 \leq r \leq h-1$ we
have
$$K_r \cdot_k L  = \Quot (K_r \otimes_k L) = \Quot (K_r
\otimes_{K_1} (K_1 \otimes_kL)) = K_r \cdot_{K_1} (K_1 \cdot_k L).$$
Also, $K_h \cdot_k L = K_h \cdot_{K_1} (K_1 \cdot_k L)$. This can
be seen using the same calculation in case $\vf$ is of inner type,
and also when $\vf$ is of outer type and $\Delta (\varphi)$ cannot
be embedded in $L$. If $\vf$ is of outer type and $\Delta (\varphi) \subset
L$, we perform the following calculation (bearing Definition~6 in
mind).
$$\eqalign{K_h \cdot_k L \colon =
K_h \cdot_{\Delta (\varphi)} L &= \Quot (K_h \otimes_{\Delta (\varphi)} L)\cr
&=
\Quot (K_h \otimes_{K_1 \cdot \Delta (\varphi)} (K_1 \cdot \Delta (\varphi)
\otimes_{\Delta (\varphi)} L)).}$$
We have $K_1 \cdot \Delta (\varphi) = K_1 \otimes_k \Delta (\varphi) = \Delta
(\varphi_1)$ and do indeed obtain again
$$K_h \cdot_k L  = \Quot (K_h \otimes_{\Delta (\varphi_1)} (K_1
\otimes_k L))  = K_h \cdot_{K_1} (K_1 \cdot_k L).$$
Thus we can apply the induction hypothesis, which tells us among
other things that $K_{r+1} \cdot L/K_r \cdot L$ is regular and
inessential for every $r \in [1, h] \setminus J$.
We now distinguish the cases $r_0\geq 1$ and $r_0=0$.

Suppose that $r_0 \geq 1$. By the induction hypothesis,
$(K_{r_i} \cdot L \vert 0 \leq i \leq e)$ is a regular
generic splitting tower of $\varphi \otimes K_1 \cdot L$. Let $F_0 \colon = K_0 \cdot
L$. The form $\varphi_0 \otimes F_0 = \varphi \otimes F_0$ is
isotropic. Therefore the extension $F_0 (\varphi_0 \otimes F_0) = K_0 (\varphi_0) \cdot_{K_0} F_0$
of $F_0$ is purely transcendental, and thus inessential. By the
lemma, $K_1 \cdot_{K_0} F_0 \sim_{F_0} K_0 (\varphi_0) \cdot_{K_0}
F_0$. Hence $K_1 \cdot_{K_0} F_0/ F_0$ is also inessential. In
analogy with the first calculation above, we find that
$$K_1 \cdot_{K_0} F_0 = K_1 \cdot_{K_0}
(K_0 \cdot_k L) = K_1 \cdot_k L,$$ so that $K_1 \cdot_k L/K_0 \cdot_k
L$ is inessential. This proves statement $(b)$ for the current case.

We saw that $K_0\cdot L/L$ is inessential. Thus
$K_1\cdot L/L$ is inessential. The generic splitting tower $(K_{r_i}\!\cdot\!L \vert 0 \leq i
\leq e)$  of $\varphi \otimes K_1\!\cdot\!L$ is thus also a
generic splitting tower of $\varphi \otimes L$.

We come to the case $\dim \varphi > 2$, $r_0 = 0$. Statement~$(b)$
is covered by the induction hypothesis. $\varphi \otimes L$ is
anisotropic. By the induction hypothesis, $\varphi \otimes K_1 \cdot
L$ has regular generic splitting tower $(K_{r_i} \cdot L \vert 1 \leq i \leq
e)$. The extension $K_{r_1}\cdot L/K_0\cdot L$ is regular. For the
proof of statement~$(a)$, it remains to be shown that
$K_{r_1}\cdot L$ is the generic zero field of $\varphi \otimes K_0 \cdot
L$. We already know that $K_{r_1}\!\cdot\!L/K_1\!\cdot\!L$ is
inessential. Thus it suffices to show that $K_1 \cdot L$ is the
generic zero field of $\varphi \otimes (K_0 \cdot L)$.

Let $F_0 \colon = K_0 \cdot L$. We have $\varphi_0 \otimes F_0 = \varphi
\otimes F_0$, and thus $F_0 (\varphi \otimes F_0) = K_0 (\varphi_0) \cdot_{K_0}
F_0$. Also $K_1 \sim_{K_0} K_0 (\varphi_0)$. By the lemma, this
shows that
$$K_1 \cdot_{K_0} F_0 \sim_{F_0} K_0 (\varphi_0) \cdot_{K_0}
F_0.$$
Furthermore, $K_1 \cdot_{K_0} F_0 = K_1 \cdot_k L$. Thus, $K_1 \cdot_k L \sim_{F_0} F_0
(\varphi_0 \otimes F_0)$, what we wanted to show.

Finally, assume that $\dim \varphi = 2$. Now we have $h=1$. The
field $\Delta (\varphi_0) = \Delta (\varphi) \otimes_k K_0$ is a
quadratic extension of $K_0$ and $K_1$ is a regular extension of $\Delta
(\varphi_0)$. We have a place $\lambda \colon K_1 \to \Delta
(\varphi_0) \cup \infty$ over $K_0$. After composing $\lambda$
with the non-trivial automorphism of $\Delta (\varphi_0)/K_0$ (if
necessary), we obtain a place $\mu \colon K_1 \to \Delta (\varphi_0) \cup \infty$
over $\Delta (\varphi_0)$. Therefore $K_1/ \Delta (\varphi_0)$ is
an inessential extension and regular.

Let $F_0 \colon = K_0 \cdot L$, $F_1 \colon = K_1 \cdot L$. We
again make a distinction between the cases $\varphi \otimes L$
isotropic, resp. anisotropic.

Suppose first that $\varphi \otimes L$ is isotropic. Now $J = \{ 1
\}$, and $\Delta (\varphi) \otimes_k L$ splits. We choose an
embedding $\Delta (\varphi) \hookrightarrow L$ and obtain from
this an embedding $K_0 \cdot
\Delta (\varphi) = \Delta (\varphi_0) \hookrightarrow K_0 \cdot L =
F_0$. By Definition~6, we now have
$$K_1 \cdot L \colon = K_1 \cdot_{\Delta (\varphi)} L = \Quot (K_1
\otimes_{\Delta (\varphi)} L) = \Quot (K_1 \otimes_{\Delta (\varphi_0)} (\Delta
(\varphi_0) \otimes_{\Delta (\varphi)} L)),$$
and furthermore $\Delta (\varphi_0) \otimes_{\Delta (\varphi)} L = K_0 \otimes_k L$.
It follows that
$$F_1 = K_1 \cdot L = K_1 \cdot_{\Delta (\varphi_0)} (K_0 \cdot L) = K_1
\cdot_{\Delta (\varphi_0)} F_0.$$

It is now clear that the extension $F_1/F_0$ is regular. Since $K_1/\Delta
(\varphi_0)$ is inessential, it follows furthermore from the lemma
that $F_1/F_0$ is inessential.  $F_0/L$ is also inessential. Thus
$F_1/L$ is inessential. This establishes statements~$(a)$ and $(b)$
in this case.

Finally, let $\dim \varphi = 2$ and assume that $\varphi \otimes
L$ is anisotropic. Now $J = \{ 0, 1 \}$. Statement~$(b)$ is
vacuous. We already know that $F_0/L$ is inessential. Furthermore,
$K_1/\Delta (\varphi_0)$ is regular and inessential. Therefore
$$F_1 = K_1 \cdot_{\Delta (\varphi_0)} (\Delta
(\varphi_0) \cdot_{K_0} F_0)$$
is regular and inessential over
$\Delta (\varphi_0) \cdot_{K_0} F_0
= \Delta (\varphi_0 \otimes F_0)$.
Furthermore, $\Delta (\varphi_0 \otimes F_0)
= F_0 \{ \varphi_0 \otimes F_0 \}$. Hence $F_1$ is the generic
zero field of the form $\varphi_0 \otimes F_0$. Therefore $(F_0,
F_1)$ is a regular generic splitting tower of $\varphi \otimes
L$.\qed
\endproof

\begcorollary{} In the situation of Theorem~3, $\varphi \otimes L$
has height $e$ and generic splitting pattern
$$\SP (\varphi \otimes L) = \{ i_0 + \dots + i_{r_j} \; \vert \; 0 \leq j
\leq e \}.$$
For every  $j \in [0, e]$ is $\varphi_{r_j} \otimes (K_{r_j} \cdot
L)$ furthermore the $j$-th higher kernel form of $\varphi \otimes
L$ with respect to the generic splitting tower constructed
here.\qed
\endcorollary
