\titlea{\S10}{Separable Splitting}
\leftrunningtitle{Chapter II. Generic Splitting Theory}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S10 Separable Splitting}

{\it All fields occurring in this section are supposed to have
characteristic~$2$.} If $\vf$ is a nondegenerate quadratic form
over such a field $k$, then its quasilinear part (cf. \S6,
Definition~1), which we denote by $QL(\vf)$, is anisotropic. Now,
if $K \supset k$ is a field extension, then it can happen that
$QL(\varphi \otimes K) = QL(\varphi) \otimes K$ is isotropic, and thus that
$\varphi \otimes K$ is degenerate. This possibility prompted us,
when dealing with the theory of generic splitting in the previous
sections, to only allow regular forms, i.e. quadratic forms $\vf$
with $\dim QL(\vf)\leq 1$.

We will now make a course change and allow arbitrary nondegenerate
quadratic forms. We will however only tolerate a restricted class
of field extensions,  the so-called separable field extensions.
We will see that
a reasonably satisfying theory of generic splitting is still
possible in this case.

A field extension $K \supset k$ is called {\it separable}\Idx{separable field extension} 
if every
finitely generated subextension $E \supset k$ has a {\it
separating transcendence basis}, i.e. a transcendence basis $t_1, \dots,
t_n$, such that $E$ is  separably algebraic over $k(t_1, \dots,
t_n)$, cf. [Bo$_3$], [Lg$_1$, X, \S 6], [J, IV, \S 5]. If $K$ is
already finitely generated over $k$, then it suffices to check if
$K$ itself contains a separating transcendence basis over $k$
(loc. cit.).

{\it In what follows, a ``form'' will always be understood to be a
quadratic form.} The foundation for the rest of this section is

\begtheorem{1.} Let $\vf$ be a \nd\ form over $k$, and let $K \supset
k$ be a separable field extension. Then $\varphi \otimes K$ is
also \nd.
\endtheorem

\begproof{} We work in the algebraic closure $\widetilde{K}$ of
$K$. We have $QL(\varphi) = [a_1, \dots, a_r]$ with $a_i \in K$.
This form is anisotropic. We have to show that $QL(\varphi) \otimes
K$ is also anisotropic. Each $a_i$ has exactly one square root
$\sqrt{a_i} \in \widetilde{K}$, which is already in the radical
closure $k^{1/2^{\infty}} \subset K^{1/2^{\infty}}
\subset \widetilde{K}$ of $k$. Now, since $QL(\varphi)$ is
anisotropic, the elements $a_1, \dots, a_n$ are linearly
independent over the subfield $k^2 = \{x^2 \; \vert \; x \in k \}$
of $k$, or equivalently, the elements $\sqrt{a_1}, \dots,
\sqrt{a_n}$ are linearly independent over $k$. By an important
theorem about separable field extensions (``MacLane's Criterion'', loc.
cit.), the fields $K$ and $k^{1/2^{\infty}}$ are linearly disjoint
over $k$. Therefore, the elements $\sqrt{a_1}, \dots, \sqrt{a_r}$
are linearly independent over $K$. This shows that $QL(\varphi) \otimes
K$ is anisotropic.\qed
\endproof

If $\o$ is a valuation ring with maximal ideal ${\nam m}$,  we
denote the residue class field ${\nam o}/{\nam m}$ from now on by
$\kappa ({\nam o})$.

\begtheorem{2.} Let $\lambda \colon K \to L \cup \infty$ be a
place and $\vf$ a form which has \GR\ \wrtl.\fonote{This assumption
presupposes that $\vf$ is \nd\ (cf. \S7, Definition~1).} Suppose
that the form $\lambda_* (\varphi)$ is \nd. Suppose further that $K'
\supset K$ is a field extension and that $\mu \colon K' \to L \cup \infty$
is an extension of the place $\lambda$. Then the form $\varphi \otimes
K'$ has \GR\ \wrt\ $\mu$ and $\mu_* (\varphi \otimes K') \; \cong \; \lambda_*
(\varphi)$.
\endtheorem

\begproof{} Let ${\nam o} \colon = {\nam o}_{\lambda}$, ${\nam o}' \colon =
{\nam o}_{\mu}$. The field extension $\overline{\lambda} \colon \kappa
({\nam o}) \hookrightarrow L$ is a combination of the extensions
$\kappa ({\nam o})
\hookrightarrow \kappa ({\nam o}')$ and $\overline{\mu} \colon \kappa ({\nam
o}') \hookrightarrow L$, where the first extension is induced by
the inclusion ${\nam o} \hookrightarrow {\nam o}'$.

Let $E$ be a quadratic space for $\vf$ and $M$ a \nd\
quadratic $\o$-module with $E \cong K \otimes_{{\nam o}} M$. Then
$K' \otimes E =
K' \otimes_{{\nam o}'} M'$ with $M' \colon = {\nam o}' \otimes_{{\nam o}} M$.
The quasilinear quadratic $\kappa ({\nam o})$-module $G \colon =
\kappa ({\nam o}) \otimes_{{\nam o}} QL(M)$ is anisotropic. By
assumption,
$L \otimes_{\overline{\lambda}} G = QL(L \otimes_{\lambda} M)$
is also anisotropic. Therefore
$$\kappa ({\nam o}') \otimes_{\kappa ({\nam o})} G = \kappa ({\nam
o}') \otimes_{{\nam o}'} QL(M')$$
is anisotropic. This proves that $M'$ is a \nd\ quadratic $\o'$-module. Hence
$\varphi \otimes K'$ is \nd\ and has \GR\ \wrt\ $\mu$.
Furthermore, $\mu_* (\varphi \otimes K')$ corresponds to the
quadratic space
$$L \otimes_{\mu} M' = L \otimes_{\mu} ({\nam o}'
\otimes_{{\nam o}} M) = L \otimes_{\lambda} M.$$
Hence $\mu_* (\varphi
\otimes K') \; \cong \; \lambda_* (\varphi)$.\qed
\endproof

Now we can faithfully repeat the observation, made at start of our treatment
of the theory of generic splitting in \S4.

So, let $\vf$ be a \nd\ form over a field $k$ (of
characteristic~$2$), and let $K \supset k$ and $L \supset k$ be
field extensions of $k$ with $L \supset k$ separable. Let
$\lambda \colon K \to L \cup \infty$ be a place over $k$. On the
basis of Theorem~1, we can apply Theorem~2 to the trivial place
$k \hookrightarrow L$ and its extension $\lambda$. We see that
$\varphi \otimes K$ is \nd\ and has \GR\ \wrtl, and that
$\lambda_* (\varphi \otimes K) \; \cong \; \varphi \otimes L$.

Let $\varphi \otimes K \; \cong \; \varphi_1 \, \perp \, r_1 \times
H$ be the Witt decomposition of $\vf$. By an established argument
(\S8, Theorem~2(b)), $\vf_1$ has \GR\ \wrtl, and $\varphi \otimes L \; \cong \;
\lambda_* (\varphi_1) \, \perp \, r_1 \times H$. Hence, $\ind (\varphi
\otimes L) \geq \ind (\varphi \otimes K)$. If $K$ is also
separable over $k$ and if $K$ and $L$ are specialization
equivalent over $k$, it follows again that $\ind (\varphi \otimes
L) = \ind (\varphi \otimes K)$ and $\ker (\varphi \otimes L) \; \cong \;
\lambda_* (\varphi_1)$.

\begdefinition{} A {\it generic separable zero field of}\Idx{generic separable zero field} 
$\vf$ is a
separable field extension $K \supset k$ which has the following
properties:
\medskip
\setitemindent{(b)}

\item{(a)} $\varphi \otimes K$ is isotropic.
\medskip

\item{(b)} If $L \supset k$ is a separable field extension with
$\varphi \otimes L$ isotropic, then there exists a place
$\lambda \colon K \to L \cup \infty$ over $k$.
\enddefinition

According to Theorem~1, $\vf$ can become isotropic over a
separable field extension $L$ of $k$ only if $\varphi \ne QL(\varphi)$,
i.e. if $\dim \varphi - \dim QL(\varphi) \geq 2$. If $\vf$ is
such a form with $\dim \vf\not=2$, then the polynomial
$\varphi (X_1, \dots, X_n)$ is irreducible over the algebraic
closure $\widetilde k$ of $k$. This can easily be seen
by writing
$\varphi (X_1, \dots, X_n) = X_1X_2 + \psi (X_3, \dots, X_n)$
over $\widetilde k$,
after a coordinate transformation.
Here $\psi$ is a quadratic polynomial which is not the zero
polynomial. Therefore, we can again construct the field
$$k(\varphi) = \Quot \frac{k [X_1, \dots, X_n]}{(\varphi (X_1, \dots,
X_n))}.$$
This extends our definition of $k(\vf)$ in \S9 from regular forms
$\vf$ to \nd\ forms $\vf$. $\{$Obviously $k(\vf)$ will have its
original meaning when $\dim \varphi = 2$, $QL(\varphi) = 0$.$\}$

It is now easy to see that $k(\vf)$ is separable over $k$: Let
$x_1,\dots, x_n$ be the images of $X_1, \dots, X_n$ in $k(\varphi)$,
i.e. $k(\varphi) = k (x_1, \dots, x_n)$. After a coordinate
transformation we may suppose, without loss of generality, that
$$\varphi (X_1, \dots, X_n) = a_1 X_1^2 + X_1X_2 + b_1 X_2^2 + \psi (X_3,
\dots, X_n).$$
The elements $x_2, \dots, x_n$ form a transcendence basis of
$k(\vf)$ over $k$. If $x_1 \not\in k (x_2, \dots, x_n)$, then
$k(\vf)$ is a separable quadratic extension of $k(x_2, \dots,
x_n)$. If the form $\vf$ is isotropic, we can make it so that
$a_1=0$. Then $k(\varphi) = k (x_2, \dots, x_n)$ is purely
transcendental over $k$.

\begtheorem{3.} Let $\vf$ be a nondegenerate form over a field $K$ with $\varphi
\ne QL(\varphi)$ and let $\lambda \colon K \to L \cup \infty$ be a
place such that $\vf$ has \GR\ \wrtl. Suppose that also the form $\lambda_*
(\varphi)$ is \nd. 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 there exists such a place $\mu$, then $\lambda_*
(\varphi)$ is isotropic by an established argument, using
Theorem~2 above.

Suppose now that $\overline{\varphi} \colon = \lambda_* (\varphi)$
is isotropic. By the theory in \S9, we may suppose that
$\dim QL(\varphi) \geq 2$, thus $\dim \varphi \geq 4$. Just as in
the proof of Theorem~3 in \S9, we  see that $\lambda$ can be
extended to a place $\mu
\colon K (\varphi) \to L (\overline{\varphi}) \cup \infty$.
$\{$Note that this is also true in case $\overline{\varphi}$ is
anisotropic.$\}$ Since $\overline{\varphi}$ is isotropic,
$L(\overline{\varphi})$ is a purely transcendental extension of
$L$, as established above. Hence there exists a place $\rho$ from
$L(\overline{\varphi})$ to $L$ over $L$. The place
$\rho \circ \mu \colon K(\varphi) \to L \cup \infty$ extends $\lambda$.
\qed
\endproof

\begcorollary{} Let $\vf$ be a \nd\ form over a field $k$ with $\varphi
\ne QL(\varphi)$. Then $k(\vf)$ is a generic separable zero
field of $\vf$.\qed
\endcorollary

Since we have secured the existence of a generic separable
zero field, we obtain for every form $\vf$ over $k$ a {\it
generic separable splitting tower}\Idx{generic splitting tower}
$$(K_r \; \vert \; 0 \leq r \leq
h)$$
with higher indices $i_r$ and higher kernel forms
$\varphi_r \; (0 \leq r \leq h)$, in complete analogy with the
construction of  generic splitting towers in \S4 (just before
\S4, Definition~5). Thus $K_0/k$ is a {\it separable} inessential
field extension,\break $\varphi_r  =
\ker (\varphi \otimes K_r)$, and $K_{r+1}$ is a generic
separable zero field of $\vf_r$ for $r < h$. The height $h$
satisfies $h \leq \frac{1}{2} (\dim \varphi - \dim QL(\varphi))$.

In analogy with \S4, Theorem~3, we have the following theorem,
with mutatis mutandis the same proof.

\begtheorem{4.} Let $\vf$ be a \nd\ form over $k$. Let $(K_r \; \vert \; 0 \leq
r \leq h)$ be a generic separable splitting tower of $\vf$
with associated higher kernel forms $\varphi_r$ and indices $i_r$.
Let $\gamma \colon k \to L \cup \infty$ be a place such that $\vf$
has \GR\ \wrt\ $\gamma$. Suppose that the form $\gamma_*
(\varphi)$ is \nd. Finally, for an $m$ with $0 \leq m \leq h$, let
a place $\lambda \colon K_m \to L \cup \infty$ be given, which
extends $\gamma$ and which {\it cannot} be extended to $K_{m+1}$
in case $m < h$. Then $\vf_m$ has \GR\ \wrt\ $\lambda$. The form $\gamma_*
(\varphi)$ has kernel form
$\lambda_* (\varphi_m)$ and Witt index
 $i_0 + \dots + i_m$.\qed
\endtheorem

Now we can faithfully repeat Scholium~1 to 4 from \S4, but this
time with generic separable splitting towers and separable
field extensions. We leave this to the reader. $\{$In Scholium~3,
on should of course assume -- as in Theorem~4 above --
that $\gamma_* (\varphi)$ is \nd.$\}$


In particular, the generic separable splitting tower $(K_r \;
\vert \; 0 \leq r \leq h)$ regulates the splitting behaviour of
$\vf$ \wrt\ {\it separable} field extensions $L \supset k$, as
described in \S4, Scholium~1.
\bigskip

%\noindent[Problem: Is an inessential field extension $K_0/k$ always
%separable ?]\hfill\eop
