\part{Chapter II}{Generic Splitting Theory}
\leftrunningtitle{Chapter II. Generic Splitting Theory}

\titlea{\S9}{Generic Splitting of Regular Quadratic Forms}
\leftrunningtitle{Chapter II. Generic Splitting Theory}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S9 Generic Splitting of Regular Quadratic Forms}

From now on we leave the geometric arena behind, and mostly talk of
quadratic and bilinear {\it forms}, instead of {\it spaces}, over
fields. The importance of  the geometric point of view was to
bring quadratic and bilinear modules over valuation rings into the
game. For our specialization theory, these modules were merely an
aid however, and their r\^ole has now more or less ended.

We should indicate one problem though, which occurs when we make
the transition from quadratic spaces to quadratic forms:
If $\varphi (x_1, \dots, x_n)$ and $\psi (x_1, \dots, x_n)$ are
two quadratic forms over a field $K$, they are isometric, $\psi  \cong
\varphi$, if the polynomial $\psi$ emerges from $\varphi$ through
a linear coordinate transformation. Now, if $\varphi$ -- i.e. the
space $(E,q)$ associated to $\varphi$ -- is degenerate, then it is
possible that not all of the coordinates $x_1, \dots, x_n$  ($n \colon = \dim
E$) occur in the polynomial $\psi$ (and possibly neither in
$\varphi$). The dimension of $\psi$ in the naive sense, i.e. the
number of occurring variables, can be smaller than the dimension
of $E$. In order to recover the space $(E,q)$ from $\psi$, one
would have to attach a ``virtual dimension'' $n$ to $\psi$, pretty
horrible!

As soon as one allows degenerate forms, the geometric language is
more precise -- and thus more preferred -- than the ``algebraic''
language. But now we only consider {\it regular} quadratic forms
($\widehat{=}$ regular quadratic spaces) over fields, so that
the forms are guaranteed to remain nondegenerate
under base field extensions.

Many concept which we introduced for quadratic spaces over fields
(\S5 -- \S8),
will now be used for quadratic forms, usually without any further
comments. {\it In what follows, a ``form'' is always a regular
quadratic form over a field.} We  recall once more the concepts \GR\
($=$ good reduction) and specialization of forms and try to be
as down to earth  as possible.

So, let $\varphi$ be a form over a field $K$. If $\dim \vf$ is
even (resp. odd), then $\varphi = [a_{ij} ]$ (resp. $\varphi \; \cong \;
[a_{ij} ] \, \perp \, [c]$), where $(a_{ij})$ is a symmetric
$(2m) \times (2m)$-matrix, such that $(r \colon = 2m)$
$$\det\,\left(\matrix{
           2a_{11} & a_{12} & \dots & a_{1r} \cr
           a_{21} & 2 a_{22} & \dots & a_{2r} \cr
           \vdots &  \vdots  &       & \vdots \cr
           a_{r1} & a_{r2} & \dots & 2a_{rr} \cr} \right) \ne 0$$
and $c\ne0$. (See the beginning of \S5 for the notation used
here.)

Let $\lambda \colon K \to L \cup \infty$ be a place and ${\nam o} = {\nam
o}_{\lambda}$ its valuation ring. The form $\vf$ has \GR\ \wrtl\
when $\varphi \; \cong \; [a_{ij} ]$, resp. $\varphi \; \cong \;
[a_{ij} ] \; \perp \, [c]$, such that all $a_{ij}$ are  in ${\nam o}$
and $c$, as well as the determinant above, are  units in $\o$. We
then have $\lambda_* (\varphi) \; \cong \; [\lambda (a_{ij}) ]$
resp.
$\lambda_* (\varphi) \; \cong \; [\lambda (a_{ij})] \, \perp \, [\lambda (c)
]$. By \S6, Theorem~1(b), $[ a_{ij}]$ is the orthogonal sum of $m$
binary forms $\left[ {a_i \atop 1} \; {1 \atop b_i} \right] \quad (1 \leq
i \leq m)$ with $a_i \in {\nam o}$, $b_i \in {\nam o}$, $1 - 4
a_i b_i \in {\nam o}^*$. Thus, after a coordinate transformation,
we have
$$\varphi (x_1, \dots, x_n) = \sum\limits_{i = 1}^m (a_i
x_{2i}^2 + x_{2i} x_{2i+1} + b_i x_{2i+1}^2) \qquad
                                      (\quad + c x_{2m+1}^2). \leqno(\ast)$$
We obtain $(\lambda_* \varphi) (x_1, \dots, x_n)$ from this form,
upon replacing the coefficients $a_i$, $b_i$, $c$ by $\lambda (a_i)$,
$\lambda (b_i)$, $\lambda (c)$.

Let $K' \supset K$ be a field extension. If $\varphi = \varphi (x_1,
\dots, x_n)$ is a form over $K$, we write $\varphi \otimes K'$\Idx{111fitensork@$\varphi \otimes K'$ ($\varphi$ over $K$, $K\subset K'$)} 
or
$\varphi \otimes_K K'$ for this form considered over $K'$.
If $(E,q)$ is
the quadratic space associated to $\vf$, i.e. $\varphi \; \widehat{=} \; (E,
q)$, then the basis extension $(K' \otimes_K E,
\; q_{K'})$ of $(E, q)$ is the quadratic space associated to
$\varphi \otimes K'$.\fonote{We write $\varphi \otimes K'$ instead
of $K'\otimes \varphi$ in order to be in harmony with the notation
in \S3 and in the literature. $\{$Many authors more briefly write
$\varphi_{K'}$. Starting with \S16 we occasionally will do this too.$\}$}

Let $\mu \colon K' \to L \cup \infty$ be an extension of the place
$\lambda \colon K \to L \cup \infty$. If $\vf$ has \GR\ \wrtl,
then $\vf \otimes K'$ has \GR\ \wrt\ $\mu$ and $\mu_* (\vf
\otimes K') \; \cong \; \lambda_* (\varphi)$. This is clear
now and was already established in \S 8, Theorem~8 anyway.

We can extend the observation we made, at the start of our treatment
of generic splitting in \S4, to regular quadratic forms. Thus let
$k$ be a field and $\vf$ a form over $k$. Furthermore, let $K$ and
$L$ be fields, containing $k$, and let $\lambda \colon K \to L \cup
\infty$ be a place over $k$. Then $\varphi \otimes K$ has \GR\
\wrtl\ and
$\lambda_* (\varphi \otimes K) \; \cong \; \varphi \otimes L$.

We use $H$ to denote the quadratic form\Idx{111hiss@$H= \left[ {0 \atop 1} \; {1 \atop 0} \right]$}
$\left[ {0 \atop 1} \; {1 \atop
0} \right]$, regardless of the field we are working in. The
bilinear form $B_{\varphi}$,   associated to $\varphi =
\left[ {0 \atop 1} \; {1 \atop 0} \right]$, is $\left( {0 \atop 1} \; {1 \atop
0} \right)$. We denote this bilinear form henceforth by
$\widetilde{H}$. (This notation differs from \S4!) If
$$\varphi \otimes K \; \cong \; \varphi_1 \, \perp \, r_1 \times H$$
is the Witt decomposition of $\varphi \otimes K$, then the form
$\vf_1$ has \GR\ \wrtl\ by \S8, Theorem~2(b) and we have
$$\varphi \otimes L \, \cong \; \lambda_* (\varphi_1) \, \perp \, r_1 \times
H.$$
Therefore, $\ind (\varphi \otimes L) \geq \ind (\varphi \otimes
K)$. If $K$ and $L$ are specialization equivalent over $k$ (see
\S4, Definition~1), then
$$\ind
(\varphi \otimes L) = \ind (\varphi \otimes K)\quad\hbox{ and }\quad\ker (\varphi \otimes L)
\; \cong \; \lambda_* (\varphi_1).$$

We transfer the definition of generic zero field\Idx{generic zero field} 
(\S4,
Definition~2) literally to the current situation. Just as in \S4
we define, for $n \colon = \dim \varphi \geq 2$, a field extension
$k(\vf)$ of $k$ as follows:\Idx{111kfi@$k(\vf)$ (including for $\Char k=2$)} 
If $n>2$ or $n=2$ and $\varphi \; \not\cong \;
H$, let $k(\vf)$ be the quotient field of the integral domain $k [X_1, \dots,
X_n]/(\varphi (X_1, \dots, X_n))$. $\{$It is easy to see that the
polynomial $\varphi (X_1, \dots, X_n)$ is irreducible.$\}$ If $\varphi \;
\cong \; H$ however, i.e. $\varphi (X_1, X_2) \; \cong \; X_1
X_2$, let $k(\varphi) = k(t)$ for an indeterminate $t$.

We want to show that $k(\vf)$ is a generic zero field of $\vf$.
This is very simple when $\vf$ is isotropic: Obviously $k$
itself is a generic zero field of $\vf$ in this case. We thus have
to show that $k(\vf)$ is specialization equivalent to $k$ over
$k$, which now means that there exists a place from $k(\vf)$ to
$k$ over $k$.

\beglemma{1.} If $\vf$ is an isotropic form over $k$, then
$k(\vf)$ is a purely transcendental field extension of $k$.
\endlemma

\begproof{} This is by definition clear when $\dim\vf=2$, i.e.
when $\varphi \cong H$. Now let $n \colon = \dim \varphi > 2$. We
have a decomposition $\varphi \cong H  \perp \psi$ and
may
suppose without loss of generality that $\varphi = H \, \perp \, \psi$. So, if
$X_1, \dots, X_n$ are indeterminates, we have
$$\varphi (X_1, \dots, X_n) = X_1 X_2 + \psi (X_3, \dots, X_n).$$
Therefore,\fonote{If $A$ is an integral domain, $\Quot A$ denotes
the quotient field of $A$.}
$$k (\varphi) = \Quot (k[X_1, \dots, X_n]/(X_1 X_2 + \psi (X_3,
\dots, X_n)) = k(x_1, \dots, x_n),$$
where $x_i$ of course denotes the image of $X_i$ in $k(\vf)$.
The elements $x_2, x_3, \dots, x_n$ are algebraically independent
over $k$ and $x_1 = - x_2^{-1} \psi (x_3, \dots, x_n)$. Therefore,
$k(\varphi) = k(x_2, \dots, x_n)$ is purely transcendental over
$k$.\qed
\endproof

Since $k(\vf)/k$ is purely transcendental, there are many places
from $k(\vf)$ to $k$ over $k$, cf.
[Bo$_2$, \S 10, Prop.~1].

We will move on to prove that also for anisotropic $\vf$ with
$\dim\vf\geq2$, $k(\vf)$ is a generic zero field of $\vf$. De
facto we will obtain a stronger result (Theorem~4 below), and we
will need its full strength later on as well. We require a lemma
about the extension of places to quadratic field extensions.

\beglemma{2.} Let $E$ be a field, $K$ a quadratic extension of $E$
and $\alpha$ a generator of $K$ over $E$. Let $p(T) = T^2 - aT + b \in
E[T]$ be the minimal polynomial of $\alpha$ over $E$. Furthermore,
let $\rho \colon E \to L \cup \infty$ be a place with $\rho (b) \ne \infty$
and $\rho (a) \ne \infty$, $\rho (a) \ne 0$. Finally, let $\beta$
be an element of $L$ such that
$$\beta^2 - \rho (a) \beta + \rho (b) =0.$$
Then there exists a unique place $\lambda \colon K \to L \cup \infty$
with $\lambda (\alpha) = \beta$, which extends~$\rho$.
\endlemma

\begproof{} Let $\o$ denote the valuation ring of $\rho$, ${\nam
m}$ its maximal ideal and $k$ the residue class field ${\nam o}/{\nam
m}$. Let $\gamma \colon E \to k \cup \infty$ be the canonical
place associated to $\o$. Then $\rho = \overline{\rho} \circ
\gamma$, where $\overline{\rho} \colon k \hookrightarrow L$ is a
field embedding. We may assume without loss of generality that $k$ is a subfield of
$L$ and that $\overline{\rho}$ is the inclusion map $k
\hookrightarrow L$. Every place $\lambda \colon K \to L \cup
\infty$ which extends $\rho$, has as image a field which is
algebraic over $k$. Therefore we may replace $L$ by the algebraic
closure of $k$ in $L$ and thus assume without loss of generality that $L$ is algebraic
over~$k$. If $c$ is an element of $\o$, then $\overline{c}$ will
denote the image of $c$ in ${\nam o}/{\nam m} =
k$, i.e. $\overline{c} = \gamma (c)$. We have $\beta^2 - \overline{a} \beta +
\overline{b} = 0$.

By the general extension theorem for places [Bo$_2$, \S 2,
Prop.~3], there exists a place from $K$ to the algebraic closure $\tilde{k}$ of
$k$, which extends $\gamma$. We choose such a place
$\delta \colon K \to \tilde{k} \cup \infty$. Let ${\nam O}$ be the
valuation ring of $K$, associated to $\delta$, ${\nam M}$ its
maximal ideal and $F \colon = {\nam O}/{\nam M}$ its residue class
field. Finally, let $\sigma \colon K \to F \cup \infty$ be the
canonical place of ${\nam O}$. Then $\sigma$ extends the place
$\gamma \colon E \to k \cup \infty$. By general valuation theory
[Bo$_2$, \S 8, Th.~1], we have $[F \colon k] \leq 2$. We may
envisage $F$ and $L$ as subfields of $\tilde{k}$, which both
contain $k$. One of the things we will show, is that $F$ is equal
to the subfield $L' \colon = k(\beta)$ of $L$.

The field extension $K/L$ is separable since the coefficient $a$
in the minimal polynomial $p(T)$ is different from zero. Let $j$
be the involution of $K$ over $E$, i.e. the automorphism of $K$
with fixed field $E$.

{\it Case 1:} $L'\neq k$. So $\beta \not\in k$ and $T^2 -
\overline{a} T + \overline{b}$ is the minimal polynomial of $\beta$ over $k$.
Now $\alpha^2 - \alpha a + b = 0$ implies $\sigma (\alpha)^2 - \sigma (\alpha)
\overline{a} + \overline{b} = 0$. Thus $\sigma (\alpha) = \beta$
or $\sigma (\alpha) = \overline{a} - \beta$. In particular, $L' = k (\sigma
(\alpha)) \subset F$. Since $[L' \colon k] = 2$ and $[F \colon k] \leq
2$, we have $L' = F$ and $F \subset L$.

We have $j (\alpha) = a - \alpha$, and thus $\sigma j (\alpha) = \overline{a} -
\sigma (\alpha)$. Since $\overline{a} \ne 0$ this implies  $\sigma j (\alpha) \ne
\sigma (\alpha)$. Therefore the places $\sigma$ and $\sigma \circ j$
from $K$ to $L$ are different. They both extend the place $\rho$.
By general valuation theory [Bo$_2$, \S 8, Th.~1], there can be no
other places from $K$ to $\tilde k$ which extend $\rho$. Now take $\lambda =
\sigma$ in case $\sigma (\alpha) = \beta$, and $\lambda = \sigma \circ j$
in case $\sigma (\alpha) = \overline{a} - \beta$. Then
$\lambda \colon K \to L \cup \infty$ is the only place which extends
$\rho$ and
maps $\alpha$ to $\beta$.

{\it Case 2:} $L'=k$. This time $\beta \in k$. Since $\beta$ is a
root of $T^2 - \overline{a} T + \overline{b}$, we have
$T^2 - \overline{a} T +
\overline{b} = (T - \beta) (T - \overline{a} + \beta)$ and
$\alpha^2 -\alpha \overline{a} + \overline{b} = 0$ implies again that $\sigma (\alpha)^2 -
\sigma (\alpha) \overline{a} + \overline{b} = 0$. Therefore,
$$(\sigma (\alpha) - \beta) (\sigma (\alpha) - \overline{a} + \beta) = 0.$$
Hence $\sigma (\alpha) = \beta$ or $\sigma (\alpha) = \overline{a} -
\beta$. In particular we have $\sigma (\alpha) \in k$.

Again the places $\sigma$ and $\sigma \circ j$ are different. By
general valuation theory, they are exactly all the places from $K$
to $\tilde k$ which extend $\rho$ and $[F \colon k] = 1$, i.e.
$F=k$. As in Case~1, we take $\lambda = \sigma$ in case $\sigma (\alpha)
= \beta$, and $\lambda = \sigma \circ j$ in case $\sigma (\alpha) =
\overline{a} - \beta$. Again $\lambda$ is the only place from $K$
to $L$ (even the only place from $K$ to $\tilde k$) which extends
$\rho$ and maps $\alpha$ to $\beta$.\qed
\endproof

\begtheorem{3.} Let $\lambda \colon K \to L \cup \infty$ be a
place and $\vf$ a form over $K$ with \GR\ \wrtl\ and $\dim \varphi \geq
2$. Let $\overline{\varphi} \colon = \lambda_* (\varphi)$. Then
$\lambda$ can be extended to a place $\mu \colon K (\varphi) \to L
(\overline{\varphi}) \cup \infty$.
\endtheorem

\begproof{} If $\vf$ is isotropic, then $K(\varphi)/K$ is purely
transcendental by Theorem~1. In this case $\lambda$ can be
extended in many ways to a place from $K(\varphi)$ to $L$. So
suppose that $\vf$ is anisotropic.

Let $n \colon = \dim \varphi$ and let $\o$ be the valuation ring
of $\lambda$. If $n=2$ we assume in addition that $\overline{\varphi} \;
\not\cong \; H$, deferring the case  $\overline{\varphi} \;
\cong \; H$ to the end of the proof.
After a linear transformation of $K^n$ we may
assume without loss of generality that $\vf$ is of the form ($\ast$) as in the
beginning of this section (thus with $a_i$, $b_i \in {\nam
o}$, $c \in {\nam o}^*$ etc.). We denote by  $\overline{\varphi}$  the form over $L$,
obtained from $\vf$ by replacing the coefficients $a_i$, $b_i$ and
-- if $n$ is odd~--~$c$ by $\overline{a}_i \colon = \lambda (a_i)$, $\overline{b}_i \colon =
\lambda (b_i)$, $\overline{c} \colon = \lambda (c)$.
Let $X_1, \dots, X_n$, resp. $U_1, \dots, U_n$ be indeterminates,
then we also write
$$\varphi (X_1, \dots, X_n) = a_1 X_1^2  + X_1 X_2 + b_1 X_2^2 +
\psi (X_3, \dots, X_n),$$
and accordingly
$$\overline{\varphi} (U_1, \dots, U_n) = \overline{a}_1 U_1^2  +
U_1 U_2 + \overline{b}_1 U_2^2 +
\overline{\psi} (U_3, \dots, U_n).$$
Furthermore we write
$$\eqalign{& K (\varphi) = \Quot \frac{K [X_1, \dots, X_n]}{(\varphi (X_1,
\dots, X_n))} = K(x_1, \dots, x_n), \cr
           & L(\overline{\varphi}) = \Quot \frac{L (U_1, \dots,
           U_n)}{(\overline{\varphi} (U_1, \dots, U_n))} = L (u_1, \dots,
           u_n), \cr}$$
where $x_i$ denotes of course the image of $X_i$ in $K(\vf)$ and $u_i$
the image of $U_i$ in $L(\overline{\varphi})$. We then have the
relations
$$\eqalign{& a_1 x_1^2 + x_1 x_2 + b_1 x_2^2 + \psi (x_3, \dots, x_n) = 0,
\cr
           & \overline{a}_1 u_1^2 + u_1u_2 + \overline{b}_1 u_2^2 +
           \overline{\psi} (u_3, \dots, u_n) = 0.\cr}$$
$\{$If $n = 2$,  the last summands on the left should be read as zero.$\}$
It is easy to see that the space
$N \colon = \left[ {a_1 \atop 1} \; {1
\atop b_1} \right]$ over ${\nam o}$ is isometric to a space
$\left[ {a'_1 \atop 1} \; {1
\atop b'_1} \right]$ with $a'_1 \in {\nam o}^*$.
$\{$Lift a suitable basis of $N/{\nam m} N$ to a basis of $N.\}$
Therefore we additionally assume, without loss of generality, that $a_1 \in {\nam
o}^*$, i.e. $\overline{a}_1 \ne 0$.

The elements $x_2, \dots, x_n$ are algebraically independent over
$K$ and likewise the elements $u_2, \dots, u_n$ are algebraically
independent over $L$. Let
$$E \colon = K(x_2,
\dots, x_n) \subset K(\varphi)$$
and
$$F \colon = L(u_2, \dots, u_n) \subset
L(\overline{\varphi}).$$
Our place $\lambda$ has exactly one
extension $\tilde{\lambda} \colon E \to F \cup \infty$ with $\tilde{\lambda} (x_i) =
u_i$\break $(2 \leq i \leq n)$ (cf. [Bo$_2$, \S 10, Prop.~2]).

Let $\tilde{{\nam o}}$ be the valuation ring of $\tilde{\lambda}$.
We have $K(\varphi) = E(x_1)$, $L (\overline{\varphi}) = F(u_1)$
with relations
$$x_1^2 + a x_1 + b = 0, \quad u_1^2 + \overline{a} u_1 + \overline{b} = 0,$$
where $a \colon = a_1^{-1} x_2$, $b \colon = a_1^{-1} (b_1 x_2^2 + \psi (x_3,
\dots, x_n))$ are in $\tilde{{\nam o}}$ and $\overline{a} =
\tilde{\lambda} (a)$, $\overline{b} = \tilde{\lambda} (b)$.

Surely $K(\varphi) \ne E$, because $\varphi \otimes K(\varphi)$ is
isotropic, but $\varphi \otimes E$ is anisotropic since the
extension $E/K$ is purely transcendental. Therefore $K(\varphi)$
is a quadratic field extension  of $E$. Furthermore
$\overline{a} = \overline{a}_1^{-1} u_2 \ne 0$. Hence we can apply
Lemma~2 to the place $\tilde{\lambda} \colon E \to
F \cup \infty$ and the field extensions $K(\varphi)/E$,
$L(\overline{\varphi})/F$. According to this lemma, there exists a
(unique) place $\mu \colon K(\varphi) \to L (\overline{\varphi})$
which extends $\tilde{\lambda}$ and maps $x_1$ to $u_1$. This
proves the theorem in case $\overline{\varphi} \; \not\cong \; H$.

Finally, let $n = 2$, $\varphi$ anisotropic, but
$\overline{\varphi}$ isotropic. We can still use the description
of $K(\vf)$ given above (this time with $\psi=0$), i.e. $\varphi = \left[ {a_1 \atop 1} \; {1
\atop b_1} \right]$ with $a_1 \in {\nam o}^*$, $b_1 \in {\nam o}$, $K(\varphi)
= K(x_1, x_2)$ with $x_2$ transcendental over $K$ and
$$a_1 x_1^2 + x_1x_2 + b_1 x_2^2 = 0. \leqno(1)$$

We want to extend $\lambda$ to a place $\mu \colon K(\varphi) \to L \cup
\infty$. In order to do this, we choose a nontrivial zero $(c_1,
c_2) \in L^2$ of $\overline{\varphi}$. Then we choose an extension
$\tilde{\lambda} \colon K(x_2) \to L \cup \infty$ of $\lambda$ with
$\tilde{\lambda} (x_2) = c_2$, cf. [Bo$_2$, \S 10, Prop.~1].
$\{$Take the ``variable'' $x_2 - c_2$ there.$\}$ We have
$$\overline{a}_1 c_1^2 + c_1c_2 + \overline{b}_1 c_2^2 = 0. \leqno(2)$$
If $c_2 = 0$, then $c_1 = 0$ by (2). Our zero is
nontrivial however, so $c_2 \ne 0$.

Equation (1) shows that $x_1^2 + a x_1 + b = 0$ with $a \colon = a_1^{-1} x_2$, $b
\colon = a_1^{-1} b_1 x_2^2$. We have $\tilde{\lambda} (a) =
\overline{a}_1^{-1} c_2 \ne 0$, $\tilde{\lambda} (b) = \overline{a}_1^{-1}
\overline{b}_1 c_2^2$. Equation~(2) shows that $c_1^2 + \tilde{\lambda} (a) c_1 +
\tilde{\lambda} (b) = 0$. Since $\vf$ is anisotropic, $K(\varphi) \ne
K(x_1)$, thus $[K(\varphi) \colon K(x_1)] = 2$. Lemma~2 tells us
that there exists a (unique) place $\mu \colon K(\varphi) \to L \cup
\infty$ which extends $\tilde{\lambda}$.
\qed
\endproof

\begremark{} In the proof we did not need the uniqueness
statement of Lemma~2. Nonetheless, it deserves some attention. For
example, one can use it to deduce from our proof that, given an
anisotropic $\overline{\varphi}$, there is {\it exactly one} place
$\mu \colon K(\varphi) \to L(\overline{\varphi}) \cup \infty$
which extends $\lambda$ and maps $x_i$ to $u_i$ $(1 \leq i \leq n)$.
Now, this holds for the generators $x_1, \dots, x_n$ of
$K(\varphi)$ and $u_1, \dots, u_n$ of $L(\overline{\varphi})$,
associated to the special representation ($\ast$) of $\vf$ above
(still with $a_1 \in {\nam o}^*$) and can, by means of a
coordinate transformation (with coefficients in $\o$, etc.), be
transferred to the case where $\varphi = [a_{ij}]$ for an
arbitrary symmetric matrix
$(a_{ij})$ over ${\nam o}$ with
$$\det\,\left(
\matrix{
           2a_{11}  & \dots & a_{1n} \cr
           \vdots   &       & \vdots \cr
           a_{n1}  & \dots & 2a_{nn} \cr}\right) \in {\nam o}^*.$$
\endremark

\begtheorem{4.} Let $\lambda \colon K \to L \cup \infty$ be a
place. Let $\vf$ be a form over $k$ with $\dim \varphi \geq 2$,
which has \GR\ \wrtl. Then $\lambda_* (\varphi)$ is isotropic if
and only if $\lambda$ can be extended to a place $\mu \colon K (\varphi) \to L \cup
\infty$.
\endtheorem

\begproof{} If $\lambda$ can be extended to a place $\mu$ from
$K(\varphi)$ to $L$, then $\lambda_* (\varphi)$ has to be
isotropic by the standard argument, which we already used in \S4,
just after Theorem~2.

Conversely, suppose that $\overline{\varphi} \colon = \lambda_*
(\varphi)$ is isotropic. By Theorem~3, there exists a place
$$\mu \colon K(\varphi) \to L(\overline{\varphi}) \cup \infty$$
which extends $\lambda$. By Lemma~1, $L(\overline{\varphi})/L$ is
a purely transcendental extension. Therefore there exists a
place
$$\rho \colon L (\overline{\varphi}) \to L \cup \infty$$
over $L$.
Now $\rho \circ \mu \colon K(\varphi) \to L \cup \infty$ is a
place which extends $\lambda$.\qed
\endproof

\begremark{} We could have adapted the proof of Theorem~3 in such
a way, that we would have obtained Theorem~4 immediately (see in
particular our argument there in the case
$\overline{\varphi}  \cong  H$). On the other hand, we can
obtain Theorem~3 from Theorem~4 by applying the latter to the
place $$j \circ \lambda \colon K (\varphi)
\to L(\overline{\varphi}) \cup \infty,$$
where $j$ is the inclusion
$L \hookrightarrow L(\overline{\varphi})$. For our further
investigations (\S10, \S12, \S13) it is better however, to isolate
Theorem~3 and its proof as a stopover to Theorem~4.
\endremark

\begcorollary{} Let $\vf$ be a form over a field $k$ with
$\dim\vf\geq 2$. Then $k(\vf)$ is a generic zero field of $\vf$.
\endcorollary

\begproof{} Clearly $\varphi \otimes k (\varphi)$ is isotropic.
Now let $L \supset k$ be a field extension with $\varphi \otimes
L$ isotropic. Applying Theorem~4 to the trivial place  $k \hookrightarrow
L$, we see that there exists a place $\mu \colon k (\varphi) \to L \cup \infty$ over
$k$.\qed
\endproof

Since Theorems~1 and 2 of \S4 are subcases of our current
Theorem~4, they are now proved. The definitions and theorems
following these two theorems remain valid. Later on they will be
used without further comments in renewed generality, namely for
arbitrary characteristic instead of characteristic $\not=2$,
regular quadratic forms instead of the nondegenerate symmetric
bilinear forms which occur there. In particular, for every
regular quadratic form $\vf$ over $k$,  we have a generic splitting tower
$(K_i \; \vert \; 0 \leq i \leq h)$ with higher indices $i_r$ and
higher kernel forms $\varphi_r$.

\begcomments{}
It should be noted that I.~Kersten and U.~Rehmann have used a
different route in [Ke R, \S 6] to construct, for every form $\vf$
over $k$ and every $r \leq \left[ \frac{\dim \varphi}{2} \right]$,
a field extension of $k$ which is generic for the splitting off of
$r$ hyperbolic planes. In particular one can already find
a generic splitting tower of $\vf$ in their work. In [KR] the
generic splitting of regular quadratic forms in arbitrary
characteristic is based on these foundations.

The results about forms of height~1 and 2, cited at the end of our
\S4, have so far only been established for characteristic $\not=2$
in the literature.
\endcomments
