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\begin{document}
\title[Specialization of forms \dots]{Specialization of forms in the presence of characteristic 2: first steps}

\author[Manfred Knebusch]{Manfred Knebusch \\
University of Regensburg}

\address{Universit\"at Regensburg, NWF I - Mathematik,
D-93040 Regensburg, Germany}
\email{manfred.knebusch@mathematik.uni-regensburg.de}

\subjclass[2000]{Primary 11E08, 11E81. Secondary 11E39}
 \maketitle

\section{Introduction}
   We outline a specialization theory of quadratic and (symmetric)
   bilinear forms with respect to a place $ \lambda : K \rightarrow L
   \cup \infty $. Here $K, L$ denote fields of any characteristic.  We
   have to make a distinction between bilinear forms and quadratic
   forms and study them both over fields and valuation rings.

For bilinear forms this turns out to be essentially as easy as in
the case char $ L \neq 2$, albeit no general cancellation
law holds for nondegenerate  bilinear forms over a valuation
domain $ \O $, in which 2 is not a unit. For
quadratic forms things are more difficult mainly for two reasons.
1) Forms cannot be diagonalized. 2) The quasilinear part of an
anisotropic form over $ \O $ may become
isotropic over the residue class field of $ \O$.

Nevertheless a somewhat restricted specialization theory for
quadratic forms is possible which is good enough to establish a
fully fledged generic splitting theory. On the other hand it
seems, that for bilinear forms no generic splitting is possible.
(Most probably there does not exist a ``generic zero field'' for a
bilinear form over a field of characteristic 2.) But
specialization of bilinear forms is nevertheless important for
generic splitting of quadratic forms, since a bilinear form and a
quadratic form can be multiplied via tensor product to give
another quadratic form.

  All this is explicated in a recent book by the author [Spez]. The
   book contains more material than outlined here. In particular its
     last chapter IV gives a specialization theory of forms under
  ``quadratic places'', much more tricky than the theory for ordinary
    places. Miraculously this leads to a generic splitting theory
with respect to quadratic places which is as satisfactory as for ordinary
                               places.

If $ \varphi  $ is a quadratic form over a field $
K $ which has ``good reduction'' with respect to a place $
  \lambda : K \rightarrow L \cup \infty $ then our
specialization theory gives a quadratic form $
\lambda_{\ast} ( \varphi) $ over $ L$.  We also
develop a theory of ``weak specialization'', which associates to $
  \varphi $ only  a Witt class $ \lambda_{W} (
\varphi ) $ of forms over $ L$, but under a more general condition
on $ \varphi $ than just having good reduction. In the present
article weak specialization plays only an auxiliary role in order to
define specializations $ \lambda_{\ast} ( \varphi)$.  But weak
specialization is a key notion in establishing the specialization
theory for quadratic places (not described here, cf. [Spez, Chap.
IV]).\newpage
The book [Spez] is in German. It is now in the process
of translation into English by Thomas Unger. A preprint of the first
two chapters is available [Spez'].

Everything said in $ \S 2 - \S 6 $ of the present article
can be found with proofs and/or references in these two chapters.
I have freely borrowed from passages in Unger's translation. I
also give almost no references here to the work of others,
referring to the references in the book instead.

I thank Professors Ricardo Baeza and Ulf Rehmann for help in preparing
this manuscript.

\section{Specialization of symmetric bilinear forms}

We are given a place $ \lambda : K \rightarrow L \cup \infty
 $ and a symmetric bilinear form $ \varphi, $ i.e.,
a polynomial
\begin{equation}
\varphi (x, y) = \sum_{i,j=1}^{n} a_{ij} x_i y_i
\end{equation}
over $ K $ in two sets of variables $ x = (x_1,
\ldots , x_n ), y = (y_1, \ldots , y_n), $ with
coefficients $ a_{ij} = a_{ji} \in K$.  Under suitable
conditions (``good reduction'', see below) we want to associate to
$ \varphi $ a symmetric bilinear form $ \lambda_{
\ast} ( \varphi ) $ over $ L $ in a reasonable way.

We assume that $ \varphi $ is nondegenerate, i.e., $
  \det (a_{ij}) \in K^{\ast}, $ and we want that $
\lambda_{ \ast} ( \varphi ) $  is again a nondegenerate
form, of the same dimension $ n = \dim  \varphi $ as $
  \varphi$.

For the rest of this section a form always means a
{\it nondegenerate} {\it symmetric} {\it bilinear form.} We denote the
form $ \varphi $ above by the symmetric matrix $
(a_{ij})$.  Nondegeneracy  of  $ \varphi $ means
that $ \det (a_{ij}) \neq 0$.

We call two forms  $ \varphi = (a_{ij}), \psi = (b_{ij})
 $ {\it isometric} ( = isomorphic), and write $
\varphi  \cong \psi , $ if $ \dim \varphi = \dim \psi
 $ and $ \psi $ is obtained from $ \varphi
 $ by a linear change of coordinates, in matrix notation
\begin{equation}
(b_{ij}) = {}^{t} U (a_{ij})U
\end{equation}
with some $ U \in GL(n, K)$.

Let $ \O_{\lambda} $ denote the valuation ring
of $ \lambda, \O_{ \lambda} = \{ x \in K \;|\;
\lambda (x) \neq \infty \}$.

\begin{Def} We say that the form $ \varphi = (a_{ij}) $
has {\it good reduction} with respect to the place $
\lambda : K \rightarrow L \cup \infty , $ if there exists a
symmetric matrix $ (b_{ij}) $ with coefficients in $
  \O_{ \lambda} $ and $ \det (b_{ij}) $
a unit of $ \O_{ \lambda}, $ such that $
\varphi $ is isometric to the form $ (b_{ij}) $
over $ K$.  Alternatively we then say that $
\varphi $ is $ \lambda$-{\it unimodular}, and we
call an isometry $ \varphi \cong (b_{ij}) $ a $
\lambda$-{\it unimodular representation} of  $
\varphi$.

In this situation we are tempted to define
\begin{equation}
\lambda_{\ast} ( \varphi ) := ( \lambda (b_{ij})),
\end{equation}

hoping that - up to isometry - the form $ ( \lambda
(b_{ij})) $ does not depend on the choice of the $
\lambda$-unimodular representation of  $ \varphi$.

(N.B.: We do not care to identify a form with an isometric form,
thus abusively speaking of ``forms'' instead of isometry classes of
forms.). \\

In this hope justified? The answer will be
$$
\text{``Yes'', if } \ch\, L \neq 2, \mbox{and ``Nearly'', if } \, \ch\, L = 2.
$$

Our approach to the question will be via Witt rings. We briefly
recall the definition of the Witt ring $ {\rm W}(K)$.  We
call two forms $ \varphi $ and $ \psi $ over
$ K $ {\it stably isometric}, if there exists a
form $ \chi $ over $ K $ such that $
\varphi  \perp \chi \cong  \psi \perp  \chi $.  We then
write $ \varphi \approx \psi$.  If $\ch\,  K \neq 2
 $ then $ \varphi \approx \psi $ implies $
\varphi \cong \psi $ by Witt's cancellation theorem. For
$ \ch\,  K = 2 $, this is false.
\end{Def}

\begin{Def} We say that two forms $ \varphi $ and $
\psi $ over $ K $ are {\it Witt equivalent}, and then
write $ \varphi \sim \psi , $ if there exist numbers $
  r, s \in \mathbb{N}_{0} $ such that
$$
\varphi \perp r \times \left( \begin{array}{cc} 0 & 1 \\ 1 & 0
\end{array} \right) \approx \psi \perp s \times \left( \begin{array}{cc}
0 & 1 \\ 1 & 0 \end{array} \right).
$$

The Witt (equivalence) class of a form $ \varphi $ will
be denoted by $ \{ \varphi \}$.

Witt classes can be added and multiplied as follows:
\[
\begin{array}{rcl}
\{ \varphi \} + \{ \psi \} & : = & \{ \varphi \perp \psi \}, \\
&   & \\
\{ \varphi \} \cdot \{ \psi \} & : = & \{ \varphi \otimes \psi \},
\end{array}
\]
where $ \perp $ and $ \otimes $ denote the
usual orthogonal sum and tensor product of symmetric bilinear
forms. In this way the set of the Witt classes over $ K
$ becomes a well defined commutative ring with 1, the
{\it Witt ring} $ {\rm W}(K)$.  The zero element is
given by the class $ \left\{ \left( \begin{array}{cc} 0& 1\\
1 & 0 \end{array} \right) \right\} $ (or by the
zero-dimensional form $ \varphi = 0, $ which we admit),
and the unit element by the class $ \{ (1) \} $ of the
one-dimensional form (1). For any form $ \varphi $, we
have $ \{ \varphi \} + \{ - \varphi \} = 0$.  \\

A good insight into Witt equivalence is given by the following
Proposition 2.3. First a bit of notation. A form $ \varphi
 $ of dimension  $ n $ is called
{\it isotropic}, if there exists some $ x \in K^n,
  x \neq 0, $ with $ \varphi (x,x) = 0, $ and
{\it anisotropic} otherwise$.  \varphi $ is
called {\it metabolic} if
\[
\varphi \cong \left( \begin{array}{cc} a_1 & 1 \\ 1 & 0
\end{array} \right) \perp  \ldots \perp \left( \begin{array}{cc}
a_r & 1 \\ 1 & 0 \end{array} \right)
\]
for some $ i > 0 $ and $ a_1 , \ldots , a_r \in K.
$
\end{Def}

Every form $ \varphi $ has a decomposition
\[
\varphi \cong  \varphi_0 \perp \varphi_1
\]
with $ \varphi_0 $ anisotropic and $ \varphi_1
$ metabolic, called a {\it Witt decomposition} of $
\varphi$.

\begin{Prop} Let $ \varphi \cong  \varphi_0 \perp \varphi_1
 $ and $ \psi \cong  \psi_0 \perp \psi_1 $ be Witt
decompositions of two forms $ \varphi $ and $ \psi.
 $ Then $  \varphi \sim \psi $ iff $ \varphi_0
\cong \psi_0 $.  \bbox
\end{Prop}



 In particular, the anisotropic part $ \varphi_0 $ of $
 \varphi $ is uniquely determined by $ \varphi
 $ up to isometry. We call $ \varphi_0 $ the
 {\it kernel form} of  $ \varphi $ and write $
 \varphi_0 =\ker ( \varphi)$.

(Alternatively we may call $ \varphi_0 $ the
{\it anisotropic part} of $ \varphi $ and write $
  \varphi_0 = \varphi_{an}$.) \\

As a consequence of Proposition 2.3 we state

\begin{Coro} $ \varphi  \approx \psi $ iff $ \varphi
\sim \psi $ and $ \dim \varphi = \dim \psi$. \bbox
\end{Coro}


Given elements $ a_1 , \ldots , a_n \in K^{\ast } $, we
denote the diagonal form
$$ \left( \begin{array}{ccc} a_1 & &
0 \\ & \ddots & \\ 0 &  & a_n \end{array} \right) $$
 more
succinctly by $ \langle a_1, \ldots, a_n \rangle $.  We
have the rules
\[
\begin{array}{rcl}
\langle a_1, \ldots , a_n \rangle \perp \langle b_1, \ldots , b_m
\rangle & \cong & \langle a_1, \ldots, a_n, b_1, \ldots, b_m
\rangle, \\
&  & \\
\langle a_1, \ldots , a_n \rangle \otimes \langle b_1, \ldots ,
b_m \rangle & \cong & \langle a_1 b_1,  \ldots, a_1 b_m, a_2 b_1,
\ldots, a_n b_m \rangle, \\
&    & \\
\langle a, -a \rangle & \cong & \left( \begin{array}{cc} 0 & 1
\\ 1 & 0 \end{array} \right)
\end{array}
\]

If for a form $ \varphi $ there exists at least one
vector $ x $ with $ \varphi(x, x) \neq 0, $
then $ \varphi $ has an orthogonal basis, i.e$.
\varphi $ can be diagonalized, $ \varphi \cong \langle
a_1, \ldots, a_n \rangle $ for some $ a_i \in K^{\ast}.
 $ Otherwise $ \varphi $ is an orthogonal sum $
  m \times \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}
\right) $ of copies of the form $ \left(
\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right), $ hence
$ \varphi \sim 0$.   Thus the Witt ring $ {\rm W}(K)
$ is additively generated by the classes $ \{ \langle a
\rangle \} $ with $a$ running through $ K^{\ast}$.  \\

As a very special case of Proposition 1.1 we observe that two
classes $ \{ \langle a \rangle \}$,  $\{ \langle b \rangle \}
 $ are equal iff $ \langle a \rangle  \cong \langle b
\rangle $ iff the square  classes  $ aK^{\ast} $
and $ bK^{\ast 2} $ are equal. In the following we
identify the set of these Witt classes, and also the set of
isometry classes of one-dimensional forms over $ K $, with
the group $ Q (K) = K^{\ast }/ K^{\ast 2} $ of square
classes of $ K$.

We have $ \langle a \rangle \otimes \langle b \rangle =
\langle ab \rangle,  $ and thus may - and will - regard $
  Q(K) $ as a subgroup of the group $ {\rm W}(K)^{\ast }
 $ of units of the Witt ring $ {\rm W}(K)$.  \\

We return to the place $ \lambda : K \rightarrow L \cup
\infty $ with valuation ring  $ \O:=
\O_\lambda $.  Our specialization theory of
bilinear forms is based on the following theorem.

\begin{Theorem} There exists a well defined additive map $
\lambda_W : {\rm W}(K) \rightarrow {\rm W}(L), $ which can be
characterized as follows. If a is a unit of $ \O,
 $ then $ \lambda_W ( \langle a \rangle ) = \langle
\lambda (a) \rangle $.  If a square class $ \langle a \rangle = a
K^{\ast 2}  $ does not contain a unit of $ \O, $ then $ \lambda_W (
\langle a \rangle ) = 0 $.\footnote[1]{The letter $ W $ in the
notation $ \lambda_W $ refers to ``Witt'' or``weak'' (cf. also $ \S
5$).}

\end{Theorem}


This can be proved by using a description of the additive group
of $ {\rm W}(K) $ by generators and relations.  We have an
additive map $ \Lambda_W $ from the group ring $
\mathbb{Z}[Q(K)] $  to $ {\rm W}(L), $ which maps a
group element $ \langle a \rangle \in  Q(K) $ to $
\langle \lambda (a) \rangle $ if $ a \in
\O^{\ast}, $ and to $ 0 $ if $ a
K^{\ast 2} $ does not contain a unit of $ \O$.

The obvious surjection $ \mathbb{Z}[Q(K)] \rightarrow {\rm W}(K)
 $ has a kernel $ \frak{a} $ which can be described
explicitly (cf. [Spez, $ \S 2$]).

One then verifies that $ \Lambda_W ( \frak{a}) = 0$.
Thus $ \Lambda_W $ factors through an additive map $
  \lambda_W : {\rm W}(K) \rightarrow {\rm W}(L) $ with the properties
stated in the theorem.

\begin{Prop} Assume that the form $ \varphi $ has good
reduction under $ \lambda, $ and $ \varphi \cong
(b_{ij}) $ is a $ \lambda$-unimodular representation of
$ \varphi $.  Then
\begin{equation}
\lambda_W ( \{ \varphi \}) = \{ ( \lambda (b_{ij})) \}
\end{equation}

\end{Prop}
\bbox

This is obvious from Theorem 1.3 if $ (b_{ij}) $ is a
diagonal matrix. In the general case one has to argue that the
symmetric matrix can be ``diagonalized over $ \O$'',
i.e., there exists an equation
\begin{equation}
(b_{ij}) ={}^{t}U \left( \begin{array}{ccc} c_1 &  & 0 \\ & \ddots
& \\ 0 &  & c_n \end{array} \right)U
\end{equation}
with $ c_i \in \O^{\ast} $ and $ U \in
GL(n, \O)$.

This is not always true, but becomes true if we replace $ (b_{ij}),
$ say, by $ (b_{ij}) \perp \langle 1, -1 \rangle$.  The proof is
best understood in the geometric setting to be developed in $ \S 3$.

\begin{Coro} Assume that $ \varphi $ and $ \psi
 $ are forms over $ K  $ with good reduction and
that $ \varphi \cong (a_{ij}),  \psi \cong (b_{ij})
 $ are $ \lambda$-unimodular representations. If $
\varphi  \approx \psi, $ then $ ( \lambda (b_{ij}))
\approx ( \lambda (c_{ij}))$.
\end{Coro}

%\newpage

\noindent{\bf Proof.} We conclude by Theorem 2.5 that the forms $
  ( \lambda (b_{ij})) $ and $ ( \lambda (c_{ij}))
 $ are Witt equivalent, and then by Proposition 2.4 that they
are stably isometric, since they have the same dimension.
\bbox

In particular, if $ \varphi \cong (a_{ij}) $ and $ \varphi \cong
(b_{ij}) $ are two $ \lambda$-unimodular representations of a form $
\varphi $ over $ K $ then the forms $ ( \lambda (a_{ij})) $ and $ (
\lambda (b_{ij})) $ over $ L $ are stably isometric. Abusively we
call $ ( \lambda (a_{ij})) $ {\it ``the'' specialization of} $
\varphi $ {\it under} $ \lambda, $ and denote this form by $
\lambda_{\ast } ( \varphi), $ although $ \lambda_{\ast } ( \varphi)$
 is uniquely determined by $
  \varphi  $ and $ \lambda $ only up to
{\it stable} isometry.


\section{Bilinear Modules}

We now switch to the ``geometric language'' for bilinear and -
later ( $ \S 4$) - quadratic forms. Everything said in this
section is very well known.

We first fix the basic notation valid for the rest of the paper. $
  \O $ always denotes a valuation domain, $
\m $ its maximal ideal, $ k = \O/\m $ its
residue class field and $ K = {\rm Quot} ( \O) $ its
quotient field, $ \O^{\ast} $ denotes the
group of units of $ \O, $ hence $
\O^{\ast} = \O\backslash \m$.

The case $ \m = \{ 0 \}$, i.e., $ \O = K,
 $ is by no means excluded.

\vspace{8pt}

A {\it bilinear module} $ M = (M, B) $ over $
 \O $ consists of an $ \O$-module $
  M $ and a symmetric bilinear form $ B : M \times M
\rightarrow \O$.  If nothing else is said,
{\it we tacitly assume} {\it that the} $
 \O$-{\it module} $ M $ {\it is
free of finite rank} $ n$.  We write $ n = \dim M.
 $ If $ e_1, \ldots , e_n $ is a basis of $ M,
 $ then $ B $ is given by the symmetric $ n
\times n$-matrix $ (a_{ij}) $ with $ a_{ij} =
B(e_i, e_j)$.

Abusively we denote $ M = (M, B), $ or better, its
isometry class by this matrix $ (a_{ij})$.  If $
e_1, \ldots , e_n $ is an orthogonal basis, $ a_{ij} =
a_i \delta_{ij}, $ we denote the bilinear module $ M
 $ also by $ \langle a_1, \ldots , a_n \rangle$.

\vspace{8pt}


We call the bilinear module $ M $ (or the form $
B) $ {\it non degenerate} if $ B $ gives an
isomorphism of $ \O$-modules $ x \mapsto B(x,
-) $ from $ M $ to its dual module $ \check{M}
= {\rm Hom}_{\O} (M, \O)$.  This means that $
  \det (a_{ij}) \in \O^{\ast }$.  We then also
say that $ M $ is a {\it bilinear space} over $
  \O$.

\vspace{8pt}

It is well known that if $ M $ is a bilinear space
containing a vector $ x $ with $ B(x,x) \in
\O^{\ast}, $ then $ M $ has an orthogonal
basis, hence $ M \cong \langle a_1, \ldots , a_n \rangle
 $ with $ a_i \in \O^{\ast}$.  This fills
the gap in our sketch of proof of $ \S 2, $ Proposition
2.6.

\vspace{8pt}

We call a submodule $ N $ of a bilinear space $ M
 $ a {\it subspace} of $ M $, if $ N
$ is a direct summand of the module $ M$.  It will be
helpful to remember that every finitely generated  torsion free $
  \O$-module is free. Thus a submodule $ N
$ of $ M $ is a subspace iff $ M/N $ is
torsion free.

For any subset $ S $ of a bilinear module $ M
$ the module
\[
S^{\perp } = \{ x \in M \;|\; B(x, S) =  0 \}
\]
is a direct summand of $ M, $ since $ M/S^{\perp}
 $ is clearly torsion  free and finitely generated.

We call a bilinear space $ M $ {\it isotropic}, if
$ M $ contains a subspace $ U \neq 0 $, which
is ``totally isotropic'', i.e., $ B(U, U) = 0, $ in other
terms, $ U \subset U^{\perp}$.  Otherwise we call $
  M $ {\it anisotropic}.

\vspace{8pt}

Since $ \O $ has no zero divisors, and every
finitely generated ideal of $ \O $ is
principal, it is easily seen that $ M $ is isotropic
iff there exists a vector $ x \neq 0 $ in $ M
$ with $ B(x, x) = 0$.

\vspace{8pt}

Indeed, we may always write $ x = cz $ with $ c
\in \O $ and $ z $ a {\it primitive
vector} of $ M, $ i.e., a vector $ z $, such
that $ \O z $ is a direct summand of the
module $M$. \{ N.B.: If $ e_1, \ldots , e_n
 $ is a basis of $ M $ and $ z = a_1 e_1 +
\ldots + a_n e_n $, then $ z $ is primitive iff $
  a_1 \O + \ldots + a_n \O = \O$.\}

\vspace{8pt}

We call $ M $ {\it metabolic} if $ M $
contains a subspace $ U = U^{\perp}$.  Equivalently we
can say, that $ M $ is metabolic iff $ M $
contains a totally isotropic subspace $ U $ with  $
  2 \dim U = \dim M$.  Every metabolic space $ \,\ M $
has an orthogonal decomposition
\[
M \cong \left( \begin{array}{cc} a_1 & 1 \\ 1 & 0 \end{array}
\right) \perp \ldots \perp \left( \begin{array}{cc} a_m & 1 \\ 1
& 0 \end{array} \right)
\]
with some $ a_i \in  \O$.  Notice that in the
case $ \O = K $ our present terminology is in
complete harmony with $ \S 2, $ identifying isometry
classes of spaces and isometry classes of forms in the obvious
way.

\vspace{8pt}

Every bilinear space $ M $ is an orthogonal sum of an
anisotropic space $ M_0 $ and a metabolic space $
M_1$.  But now, in contrast to the case $ \O =
K $ (cf. Prop. 2.3), the isometry class of $ M_0 $
usually is not uniquely determined by $ M, $ if $
2 \not\in \O^{\ast}$.

\vspace{8pt}

For the rest of this section ``space'' always means ``bilinear
space''. Exactly as in $ \S 2 $ we define stable isometry
$ ( \approx ) $ and Witt equivalence $ ( \sim )
 $ of forms over $ \O, $and then proceed
as there to the Witt ring $ {\rm W}( \O) $
consisting of the Witt (equivalence) classes of spaces.

\vspace{8pt}

We denote the Witt class of a space $ M $ by $ \{ M \}$.  It turns
out that $ \{ M \} = 0,
 $ i.e., $ M \sim 0, $ iff $ M $ is
metabolic. Also, for every space $ M = (M, B) $ the
space $ (M, B) \perp (M, -B) $ is metabolic. Thus,
abbreviating the space $ (M, -B) $ by $ -M, $
we have $ \{ - M \} = - \{ M \} $ in $
{\rm W}(\O)$.  \\

The bilinear form $ B $ on $ M $ extends in a
unique way to a $ K$-bilinear form $ B' $ on the $
  K$-vector space $ E : = K \otimes_{ \O} M
   $ obeying the formula
\begin{equation}
B' (c \otimes x, d \otimes y) = cd\, B(x, y)
\end{equation}
for $ x, y \in M $ and $ c,d \in K$.
Identifying an element $ x $ of $ M $ with $
  1 \otimes x \in E, $ we regard the free module $ M
 $ as an $  \O$-submodule of $ E$.
We then have $ B'|_ {M \times M} = B$.  A basis $
e_1, \ldots , e_n $ of $ M $ over $
\O $ is also a basis of $ E $ over $
K, $ and the spaces $ M $ and $ E $ have
the same symmetric matrix $ (a_{ij}) $ with respect to $
  e_1, \ldots , e_n$.  We often write $ B $
instead of $ B'$.  \\

If $ U $ is a subspace of $ E, $ then $
U \cap M $ is a subspace of $ M, $ and $ K
\cdot ( U \cap M) = U$.  In this way the subspaces of $
E $ correspond uniquely  to the subspaces of $ M.
$ Clearly $ U $ is totally isotropic iff $ U \cap M
 $ is totally isotropic. Thus the following proposition is
pretty obvious.

\begin{Prop} Let $ M $ be a space over $
\O $ and $ E := K \otimes_{ \O} M$.

\begin{itemize}

\item[a)] $ E $ is isotropic iff $ M $ is
isotropic.

\item[b)] $ E $ is metabolic iff $ M $ is
metabolic.
\end{itemize}

\end{Prop}


It follows that the natural map $ \{ M \} \mapsto \{ K
\otimes_{ \O} M \} $ from $ W (\O)
 $ to $ {\rm W}(K), $ which is a ring homomorphism, is
injective. We will often regard $ {\rm W}( \O ) $ as
a subring of $ {\rm W}(K)$.  \\

The square class group $ Q( \O) :=
\O^{\ast}/\O^{\ast 2} $ of $
 \O $ injects into $ Q(K) =
K^{\ast}/K^{\ast 2} $ since clearly every unit of $
\O $ which is a square in $ K $ is a
square in $ \O$.  As previously in the case of
fields we identify a square class  $ a \O^{\ast 2},
  a \in \O^{\ast}, $ with the one-dimensional
space $ \langle a \rangle $ over $ \O
$ (more precisely, with its isometry class), and then observe
that the natural map $ Q( \O) \rightarrow {\rm W}(
\O) $ is injective, due to a natural commuting square

\begin{equation}
\xymatrix{
  Q(\O) \ar@{^{(}->}[d] \ar[r] & {\rm W}(\O) \ar[d] \\
  Q(K)  \ar@{^{(}->}[r]       & {\rm W}(K)          }
\end{equation}


%% \begin{equation}
%% \begin{array}{ccc}
%% Q( \O) & \rightarrow & {\rm W}( \O) \\
%% \downarrow &   &  \downarrow \\
%% Q(K) & \hookrightarrow & {\rm W}(K)
%% \end{array}
%% \end{equation}

Thus $ Q( \O) $ can -- and will -- be also
viewed as a subgroup of $ {\rm W}( \O)^{ \ast}$.  In
other terms, if $ a, b \in \O^{ \ast} $ then
$ \langle a \rangle \sim \langle b \rangle $ iff $
\langle a \rangle \cong \langle b \rangle$.  \\

Without invoking the commutative square (7) this can be also
proved by use of the signed determinant
\begin{equation}
d(M) := \langle (-1)^{\frac{n(n-1)}{2}} \det (a_{ij}) \rangle
\end{equation}
of an n-dimensional space $ M \cong (a_{ij}) $ over $
  \O$.  \\

We switch to a place $ \lambda : K \rightarrow L \cup \infty
 $ with valuation domain $ \O =
\O_{\lambda }$.  The notation from \S 2 ($\lambda_{W}$,
good reduction, $ \lambda_{\ast} $ etc.)
will be freely  used for spaces instead of forms. \\

Our place $ \lambda $ restricts to a ring homomorphism $
  \lambda |_\O $ from $ \O $
to $ L, $ and $ \lambda |_\O $
factors through  a field embedding $ \overline{\lambda }: k
\hookrightarrow L$.  The definition of good reduction
(Def. 2.1) and specialization under $ \lambda $
now reads  as follows.

\begin{Sch} A bilinear space $ E $ over $ K $
has good reduction under $ \lambda $ iff $ E \cong K
\otimes_{ \O} M $ for some bilinear space $ M
 $ over $ \O$.  In this case
\begin{equation}
\lambda_{\ast} (E) \approx L \otimes_{\lambda} M = L \otimes_{\overline
\lambda} \overline{M}.
\end{equation}

Here $ L \otimes_{ \lambda} M $ denotes the scalar extension of the
bilinear module $ M $ to $ L $  via $ \lambda \;|\; \O,
$\,\footnote[2]{The bilinear form of $ L \otimes_{ \lambda} M $ is
defined by a formula analogous to (6) above.} and $ \overline{M} $
is the bilinear space $ M /{\m M} $ over $ k $ obtained from $ M $
by reduction modulo $ \m$.
\end{Sch}
\bbox

\begin{Ex} Every metabolic space over $ K $ has good
reduction. This follows easily from the fact that, for any $
a,c \in K $ we have $ \left( \begin{array}{cc} a& 1 \\
1 & 0 \end{array} \right) \cong \left( \begin{array}{cc} ac^{2} &
1 \\ 1 & 0 \end{array} \right) $. \bbox
\end{Ex}

Corollary 2.7 tells  us that, if $ E  $
and $ F $ are spaces with good reduction and $ E
\approx F, $ then $ \lambda_{\ast} (E) \approx
\lambda_{\ast} (F). \{ $ In particular $ \lambda_{\ast}
(E) $ is well defined up to stable isometry. $\}$.  This
can now be proved in another, more transparent way as follows.


Let $ E \cong K \otimes_{ \O} M,\  F \cong K
\otimes_{ \O} N $ with spaces $ M, N $
over $ \O$.  Then  $ K \otimes_{
\O} (M \perp -N) \cong E \perp -F $ is metabolic,
hence $ M \perp -N $ is metabolic by Proposition 3.1,
and this implies that
\[
L \otimes_{\lambda} (M \perp -N) \cong L  \otimes_{\lambda} M
\perp (-L \otimes_{\lambda} N)
\]
is metabolic. Thus $ L \otimes_{\lambda} M $ ad $
L \otimes_{\lambda} N $ are Witt equivalent. Since these
spaces have the same dimension, they are stably isomorphic.
\\

We also want to describe the map $ \lambda_{W} $ from $
  \S 2 $ in geometric language.

Preparing for this we add more notation, which will be important
also for later sections. \\

We choose a surjective valuation $ v : K \rightarrow \Gamma
\cup \infty , $ essentially unique, associated with our
valuation domain $ \O$. So $ \Gamma \cong
K^{\ast} / \O^{\ast}. \{ $ We use additive notation
for $ \Gamma , $ so $ v (xy) = v(x)+ v(y).\} \,
\, $ We regard $ Q( \O) $ as a subgroup of $
  Q(K), $ and we choose a complement $ \Sigma $
of $ Q( \O ) $ in $ Q(K), $ i.e., a
subgroup $ \Sigma $ of $ Q(K) $ with $
Q(K) = Q( \O ) \times \Sigma $.  \\

This is possible, since the group $ Q(K) $ is
elementary abelian of exponent 2. Further, we choose, for every
square class $ \sigma \in \Sigma $ an element $ s
\in \O $ with $ \sigma =  \langle s \rangle .
 $ For $ \sigma = 1 $ we choose the representative
$ s = 1$.  Let $ S $ be the set of these
elements $ s$.  For every $ a \in K^{\ast} $,
 there exists exactly one $ s \in S $ and
elements $ \varepsilon \in  \O^{ \ast}, b \in
K^{\ast } $ with $ a = s \varepsilon b^{2}$.
Since $ K^{\ast } / \O^{\ast }  \cong \Gamma ,
$ it is clear that $ S $ (resp. $ \Sigma$) is a
system of representatives of $ \Gamma / 2 \Gamma $ in $
 K^{\ast } $ (resp. $ Q(K)$) for the homomorphism
from $ K^{\ast } $ (resp. $ Q(K)$) onto $
 \Gamma / 2 \Gamma $ determined by $ \upsilon :
K^{\ast } \rightarrow \Gamma $.
\\


\newcommand{\Perp}{\mathop{\mbox{\raisebox{-1pt}{\Large$\perp$}}}}

\begin{Def} A $ \lambda$-{\it modular decomposition} of
a bilinear space $ E $ over $ K $ is an
orthogonal decomposition
\[
E \cong \Perp_{s \in S}
\langle s \rangle \otimes (K
\otimes_{\O } M_s)
\]
with every $ M_s$ a space over $ \O $
and only finitely many $ M_s \neq 0$.  Here the
unadorned $ \otimes $ means tensor product over $
K$.  Instead of $  `` \lambda$-modular'' we also use
the word $ ``\O$-modular'', since not the place $
\lambda $ but only the valuation domain $ \O
 $ is involved. \bbox
\end{Def}


Every space $ E $ over $ K $ has a $
\lambda$-modular decomposition. Indeed, we may decompose $ E
 $ orthogonally in one-dimensional spaces and metabolic
planes, usually in many ways. One-dimensional spaces are products
$ \langle s \rangle \otimes \langle \varepsilon \rangle
$ with $ s \in S, \varepsilon \in  \O^{\ast },
 $ and metabolic spaces are orthogonal sums of forms $ \left(
\begin{array}{cc} a & 1 \\ 1 & 0 \end{array} \right) $ with $
  a \in \O$.  One then simply gathers summands
belonging to the same $ s \in S$.  \\

The following is now obvious from $ \S 2$.  \\

\begin{Sch} Assume that
\[
E \cong \Perp_{s \in S} \langle s \rangle \otimes (K \otimes_{
\O } M_s)
\]
is a $ \lambda$-modular decomposition of a space $ E
 $ over $ K$.  Then
\begin{equation}
\lambda_{W} ( \{ E \} ) = \{ L \otimes_{\lambda} M_1 \}.
\end{equation}
\end{Sch}
\bbox

In particular the space $ L \otimes_{ \O} M_1 $ over $ L $ is
uniquely determined by $ E $ up to Witt equivalence. In contrast  to
Scholium 3.2 we do not have a proof of this fact in simple geometric
terms. Thus we cannot assert that the present ``geometric language''
supersedes the ``algebraic language'' of $ \S 2$.  We call the space
$
 L \otimes_{ \O} M_1 $ a {\it weak
specialization of} $ E $ with respect to $
\lambda.$ \\

We add an important result about good reduction. Starting from
now we often abbreviate ``good reduction'' by ``GR''. \\

First notice the trivial fact, that, if $ E $ and $
F $ are spaces over $ K $ with GR under $
\lambda, $ then $ E \perp F $ has again GR under $
  \lambda, $ and
\begin{equation}
\lambda_{ \ast} (E \perp F) \approx \lambda_{ \ast} (E) \perp
\lambda_{ \ast} (F).
\end{equation}

\begin{Theorem} Let $ E $ and $ F $ bilinear
spaces over $ K$.  Assume that $ F $ and $
  E \perp F $ have GR under $ \lambda$.  Then $
  E $ has GR under $ \lambda.$
\end{Theorem}

\noindent{Proof.} Adding $ -F $ to the space $ F
 $  we retreat to the case that $ F $ is metabolic.
Let $ E \perp F \cong K \otimes_\O N $ with $
  N $ a space over $ \O $.  We choose
decomposition $ E \cong E_0 \perp E_1  $ and $ N
\cong N_0 \perp N_1 $ with $ E_0 $ and $ N_0
 $ anisotropic, $ E_1 $ and $ N_1 $
metabolic. \\


Then
\[
E_0 \perp E_1 \perp F \cong K \otimes_{\O} N_0 \perp K
\otimes_{\O } N_1.
\]

The spaces $ E_0 $ and $ K \otimes_{ \O }
N_0 $ are anisotropic, and the spaces $  E_1 \perp F
 $ and $ K \otimes_{\O } N_1 $ are
metabolic. We conclude by  Proposition 2.3 that $ E_0 \cong K
\otimes_{\O } N_0 $.  Thus $ E_0 $ has GR.
The space $ E_1 $ is metabolic, hence also has GR. Thus
$ E \cong E_0 \perp E_1 $ has GR.
\bbox


\section{Quadratic modules}

We retain the notation and conventions of  $ \S 3$.  In
particular, $ \O $ denotes a valuation domain,
and modules over $ \O $ will be free of finite
rank, if nothing else is said. \\

A {\it quadratic} module $ M = (M, q) $ over $
  \O $ is an $ \O$-module $ M
 $ equipped with a quadratic form $ q$.  This is a
function $ q: M \rightarrow \O $ such that $
  q(cx) = c^{2} q(x) $ for $ c \in  \O ,
  x \in M, $ and the map $ B_q : M \times M
\rightarrow \O $ given by
\begin{equation}
B_q (x, y) = q(x + y) - q(x) - q(y)
\end{equation}
is $ \O$-bilinear. If $ e_1, \ldots, e_n $
is a basis of $ M $ then $ q $ is determined
by the values $ a_i = q(e_i), a_{ij} = B(e_i, e_j)
$ for $ i \neq j$.  More precisely,
\begin{equation}
q(\sum_{1}^{n} x_i e_i ) = \sum_{i=1}^{n} a_i x_{1}^{2} + \sum_{i
< j} a_{ij} x_i x_j
\end{equation}
for $ x_1, \ldots , x_n \in \O$.  \\

We have an obvious notion of isometry (= isomorphism) between
quadratic module over $ \O, $ and most often will be only interested
in the isometry class of a quadratic module $ (M, q)$.  Slightly
abusively we abbreviate a quadratic module $ M $ with quadratic form
(13) by the symmetric matrix $ [ a_{ij}] $ in square brackets, with
$ a_{ij} \, \, $ as above ad $ a_{ii} := a_i $. The {\it associated
bilinear module} $\widetilde{M} := (M, B_{q}) $ is described by the
matrix $ (b_{ij}) $ with $ b_{ii} = 2 a_i, $ and $ b_{ij} = a_{ij}
 $ for  $ i \neq j$.  \\

The {\it orthogonal sum} of two quadratic modules $
(M_1 , q_1) $ and $ (M_2, q_2) $ over $
 \O $ is defined by
\[
(M_1, q_1 ) \perp (M_2, q_2) := (M_1 \oplus M_2, q_1 \perp q_2)
\]
with
\[
(q_1 \perp q_2)(x_1 + x_2) := q_1 (x_1) + q_2 (x_2)
\]
$ (x_1 \in M_1, x_2 \in M_2)$.  Notice that the associated
bilinear module of $ (M_1, q_1) \perp (M_2, q_2) $ is
the orthogonal sum $ \widetilde{M}_1  \perp \widetilde{M}_2$.  \\

Orthogonality in a quadratic module $ M = (M, q) $
refers to the bilinear form $ B_q$.  In particular, if $
  N_1 $ and $ N_2 $ are submodules of  $ M,
 $ then $ M = N_1 \perp N_2 $ means that $ M =
N_1 \oplus N_2 $ as an $ \O$-module and $
B_q (N_1, N_2) = 0$.  The following fact will be used
frequently.

%\vfill\eject

\begin{Lemma} Let $ M = (M, q) $ be a quadratic module.
Assume that $ N $ is a submodule of $ M $ and
the bilinear form $ B_q |_{N \times N} $ is
non-degenerate. Then
\[ M = N  \oplus N^{\perp}. \] \bbox
\end{Lemma}


Often we will denote a quadratic module by one letter, say $
M, $ without specifying the quadratic form on $ M.
$ We then usually denote this form by $ q $ and the
associated bilinear form $ B_q $ by $ B$.  \\

If $ 2 \neq 0 $ in $ \O $, then a
quadratic module $ M $ may be viewed  as a bilinear
module with $ B(x, x) \in 2 \O $ for every $ x \in M $
via the formula $ B(x,x) = 2q(x)$,
and if 2 is a unit of $ \O $ we may identify in
this way quadratic and bilinear modules over $ \O.
 $ But, if $ 2 = 0 $ in $ \O, $
bilinear and quadratic modules
over $ \O $ are rather different objects.

\begin{Def}\hfill
\begin{itemize}
\item[a)] A quadratic module $ N = (N, q) $ is called
{\it quasilinear} if $ B_q = 0$.

\item[b)] If $ M $ is any quadratic module over $
  \O, $ then the quadratic module
\[
M^{\perp} := \{ x \in M \;|\; B(x, M) = 0\}
\]
with the form $ q|_{M^{\perp}} $ is called
the {\it quasilinear part} of $ M$.  We denote it
by $ QL(M)$.
\end{itemize}
\end{Def}



\noindent$ M^{\perp} $ is a direct summand of the
$\O$-module $ M$.  Choosing any submodule $N$
of $ M $ with $ M = N \oplus M^{\perp}$ we have
$$M \cong N \perp QL(M)$$
as quadratic module. Moreover, $ QL(M) $ is an
orthogonal sum of quadratic modules of dimension (= rank) 1,
$$QL(M) = [a_1] \perp \ldots \perp [ a_n]$$
with $ a_i \in  \O$.

\bigskip

The following definition will be central for our theory of good
reduction and specialization of a quadratic form over a field
under a place.

\begin{Def} We call a quadratic module $(M, q)$ over
$\O$ {\it nondegenerate}, if it satisfies
the following conditions:

\begin{itemize}
\item[(Q0)] $ M $ is free of finite rank.
\item[(Q1)] The bilinear form $ \overline{B_q}, $
induced by $ B_q $ on $ M /M^{\perp} $ in the
obvious way, is nondegenerate.
\item[(Q2)] $ q(x) \in \O^{\ast} $ for every vector
$ x $ in $ M^{\perp}, $ which is primitive in
$ M^{\perp} $ (and hence in $ M)$.
\end{itemize}
\noindent
If instead of (Q2), the following condition is satisfied
\begin{itemize}
\item[(Q2${}^\prime$)] $ QL(M) = 0 $ or
$QL(M) \cong [\varepsilon]$
with $ \varepsilon \in \O^{\ast}$,
\end{itemize}

\noindent then we call $ (M, q) $ {\it regular}.\hfill\break
In the special case $ M^{\perp} = 0, $ we call $
(M, q) $ {\it strictly regular}.

\end{Def}


{\it Comment on conditions {\rm (Q2)} and {\rm (Q2')}.}

If $ 2 \neq 0 $ in $ \O, $ then $
  q|_{M^{\perp}} = 0 $ and the requirement (Q2) implies $
  M^{\perp } = 0, $ hence implies - in conjunction with
(Q0) and (Q1) - strict regularity. If $ 2 \in \O^{\ast} ,
 $ then the nondegenerate quadratic $
\O$-modules are the same objects as the nondegenerate
bilinear $ \O$-modules, as defined in $ \S 3$.  \\

Suppose now that $ 2 = 0 $ in $ \O$. The condition $ M^{\perp}  = 0,
$ in other words, strict regularity, is very natural but too limited
for applications. Indeed, if $ M^{\perp} = 0, $ then the bilinear
module $ (M, B_q) $ is nondegenerate and we have $ B_q(x,x) = 2q(x)
= 0 $ for every $ x \in M$. This implies that $ M $ has even
dimension, as is well-known. (To prove this, consider the bilinear
space $ K \otimes_{\O } \widetilde{M}. ) $ So, if we insist on using
strict regularity, we can only deal with quadratic forms of even
dimension. \\

On the other hand, property Q2 has an annoying effect: Q2 is not
always preserved under a base extension. If
$ \O'\supset \O $ is another valuation domain, whose
maximal ideal $ \m' $ lies over $ \m, $ i.e.,
$\m' \cap  \O= \m$, and if $ M $ is non
degenerate, then $ \O' \otimes_{\O }  M $
can be degenerate. However, if $ M $ satisfies
(Q2$^\prime$), this clearly cannot
happen. \bbox\\


In the case $ \m = 0, $ i.e., $ \O = K, $
we call a non degenerate quadratic $ \O$-module a
{\it quadratic space} over $ K$.  \\

We gather some facts about nondegenerate quadratic modules, all
to be found in [Spez, $ \S 6$]. In the following $ M =
(M, q) $ is a quadratic module over $ \O $.  \\

\begin{Fact} If $ M $ is nondegenerate then $ M
 $ is an orthogonal sum of quadratic modules $ \left[
\begin{array}{cc} \alpha & 1 \\ 1 & \beta \end{array} \right],
 $ with $ \alpha, \beta \in \O, 1 - 4
\alpha \beta \in \O^{\ast}, $ and modules $ [
\varepsilon] $ with $ \varepsilon \in \O^{\ast}$. \bbox
\end{Fact}


\begin{Fact} If $ M $ is regular and $ \dim M $
is even, then $ M $  is strictly regular and equal to an
orthogonal sum of modules $ \left[ \begin{array}{cc} \alpha
& 1 \\ 1 & \beta \end{array} \right], $ with $ \alpha,
\beta \in \O, \hbox{1 - 4} \alpha \beta \in \O^{\ast}$.
\bbox
\end{Fact}


\begin{Fact} Assume that  $ M $ is nondegenerate. Then
every primitive vector $ e \in M $ with $ q(e) = 0
 $ can be completed to a {\it hyperbolic vector pair},
i.e., a pair $ e, f $ with $ q(f) = 0 $ and $
  B(e, f) = 1$.  \bbox
\end{Fact}


As an illustration, how our conditions Q1 and Q2 can be put to work,
we give the proof of 4.6. We choose a decomposition $ M = N \perp
M^{\perp } $ and write $ e = x + y $ with $ x \in N, y \in M^{\perp
}$.  Suppose for the sake of contradiction that the vector $ x $ is
not primitive in $ N, $ hence not primitive in $ M.
 $ Then $ y $ is primitive in  $ M^{\perp},
 $ and thus $ q(y) \in \O^{\ast}$ by
condition Q2. Hence also $ q(x) = -q(y)\in \O^{\ast}
 $ and $ x $  has to be primitive, a
contradiction.\\
Thus $ x $ is primitive in $ N$.  Since $
  B_q $ is nondegenerate on $ N, $ there exists
some $ z \in N $ with $ B_q (x, z) = 1$.  We
also have $ B_q(e, z) = 1$.  Clearly $ f := z -
q(z)e $ completes the vector $ e $ to a hyperbolic
pair. \bbox

\begin{Fact} (Cancellation theorem) If $ M $ and $
N $ are quadratic $ \O$-modules and $ G $
is a {\it strictly regular} quadratic $
\O$-module with $ M \perp G \cong N \perp G, $
then $ M \cong N$.  \bbox
\end{Fact}


We call a quadratic $ \O$-module $ M $
{\it isotropic}, if $ M $ contains a vector $
  x \neq 0 $ with $ q(x) = 0, $
{\it anisotropic} otherwise. We call $ M $
{\it hyperbolic} if $ M $ is isometric to an
orthogonal sum $ r \times \left[ \begin{array}{cc} 0 & 1 \\
1 & 0 \end{array} \right] $ of $ r $ copies of the
``hyperbolic plane'' $ \left[ \begin{array}{cc} 0 & 1 \\
1 & 0 \end{array} \right] $ for some $ r \in
\mathbb{N}$.  As in the bilinear case one proves easily: \\

\begin{Fact} $ M $ is isotropic iff the quadratic module
$ K  \otimes_{\O} M $ over $ K $ is
isotropic, and $ M $ is hyperbolic iff $ K
\otimes_{\O} M $is hyperbolic. \bbox
\end{Fact}



\begin{Fact} (Witt decomposition). If $ M $ is
nondegenerate, then
\[
M \cong M_0 \perp r \times \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0
\end{array} \right]
\]
with $ M_0 $ anisotropic (and nondegenerate) and $
r \in \mathbb{N}_0$.  \bbox
\end{Fact}

It follows from 4.7 that the number $ r $ and the isometry class of
$ M_0 $ are uniquely determined by $
  M$.  We call $ r $ the (Witt-)
{\it index} of $ M  $ and $ M_0 $ the
{\it kernel module}, or the {\it anisotropic part} of
$ M, $ and we write $ r = \ind(M), M_0 = \ker
(M)$.  We often denote the hyperbolic plane $ \left[
\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] $ by $
  H $ (regardless which ring $ \O $ is
under consideration). \\

Notice also that, by 4.8, we have $ \ind (K
\otimes_{\O} M) = \ind (M) $ and
$ \ker (K \otimes_{\O} M) = K \otimes_{\O}  \ker (M)$.


\begin{Fact} If $ M = (M, q) $ is strictly regular then
\[
(M, q) \perp (M, -q) \cong (\dim M) \times H.
\]\bbox
\end{Fact}


If we write $ M $ for $ (M, q) $ we usually
write $ -M $ for $ (M, -q), $ following the
same practice as for bilinear modules. \\

\begin{Def} We call two nondegenerate quadratic modules $ M
 $ and $ N $ over $ \O $
{\it Witt-equivalent}, and write $ M \sim N $ if there exist natural
numbers $ s,t  $ such that
\[
M \perp s \times \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0
\end{array} \right] \cong N \sim t \times \left[
\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right].
\]

Due to 4.9  this happens iff $ \ker(M) \cong\ker (N)$.
The {\it Witt class} of $ M, $ i.e., the
equivalence of $ M $ under $ \sim , $ will be
denoted by $ \{ M \}$.  \bbox
\end{Def}


It is now easy to verify:

\begin{Fact} If $ M, M' $ are strictly regular and $
  N, N' $ are non degenerate quadratic modules over $
  \O $ with $ M \sim M' $ and $ N
\sim N', $ then $ M \perp N $ and $ M' \perp
N' $ are non degenerate and $ M \perp N \sim M' \perp
N'$. \bbox
\end{Fact}


\newcommand{\tWq}{\widetilde{\rm W\!q}}
\newcommand{\Wq}{{\rm W\!q}}
\newcommand{\Wqr}{{\rm W\!qr}}

\begin{Def} We  denote the set of Witt classes of nondegenerate
quadratic $ \O$-modules by $ \tWq (\O)$.
We denote the subsets of Witt classes of
regular, resp. strictly regular quadratic $
\O$-modules by $ \Wqr ( \O), $ resp$.
  \Wq ( \O)$. \bbox
\end{Def}


Due to 4.12 we have a well-defined ``addition'' of classes $
\{ M \} \in \Wq (\O) $ with classes $ \{ N \}
\in \tWq ( \O), $
\[
\{ M \} + \{ N \} := \{ M \perp  N \}.
\]


Restricting also $ \{ N \} $ to $ \Wq ( \O)
 $ we obtain on $ \Wq( \O) $ the structure
of an abelian group. This groups $ \Wq ( \O ) $
operates by addition on the set $ \tWq ( \O).
 $ The subset $\Wqr(\O ) $ is a union of
orbits.

\begin{Def} We call $ \Wq( \O) $ the
{\it quadratic Witt group} of $ \O$,\ \  $\tWq(\O ) $
the {\it quadratic Witt set}
of $ \O , $ and $\Wqr ( \O)
$ the {\it regular quadratic Witt set} of $ \O.
$ \bbox
\end{Def}


In the case $ \O = K $ we go further and define
Witt classes of arbitrary (finite dimensional) quadratic modules
over $ K $ as follows. \\

Starting with such a module $ M = (M, q) $ we define
the ``defect'' of $ M $  by
\[
\delta (M) = \{ x \in  M^{\perp} : q(x) = 0 \}.
\]

The form $ q $ gives us a quadratic form $ \overline{q}
 $ on $ M/ \delta (M) $ in the obvious way, hence a
quadratic space $ (M/ \delta (M), \overline{q}), $ which we
call the {\it quadratic space associated} to $ M
$ and denote by $ \widehat{M}$.  Clearly
\[
M \cong \widehat{M} \perp \delta (M) \cong \widehat{M} \perp s \times [0]
\]
with $ s := \dim \delta (M)$.  We have
\[
M \cong M_0 \perp r \times H
\]
with $ \widehat{M}_0 $ anisotropic and $ r \in
\mathbb{N}_0$.  \\

Again $ M_0 $ and  $ r $ are uniquely
determined by $ M, $ and again we call $ M_0
$ the {\it kernel module} $\ker (M) $ of $ M $ and
$ r $ the {\it index} $ \ind (M) $ of $ M.
$ Notice that now $ M_0 $ can be isotropic. Notice that
in contrast to strictly regular quadratic spaces we do not have
cancellation in general. For example $ [ a] \perp [a] \cong
[0] \perp [a] $ for any $ a \in K$.  \\

We call two quadratic modules $ M, N $ over $ K
 $ {\it Witt equivalent}, and write $ M \sim N $, if $
M \perp s \times H  \cong N \perp t \times H $ for some
number $  s,t $. This means the same as $\ker (M)
\cong\ker (N)$.  We denote the set of Witt equivalence classes $
  \{ M \} $ of quadratic  $ K$-modules $ M $
by $ \widehat{\rm Wq} (K)$, and we call $  \widehat{\rm Wq} (K) $
the {\it defective quadratic Witt set} of $ K$.
The value $ \{ M \} +  \{ N \} = \{M  \perp N \} $
makes $ \widehat{\rm Wq} (K) $ an abelian semigroup with
neutral element $ \{ 0 \} = \{ H \}$.  It contains
$ \tWq(K) $ as a subset and $ \Wq(K) $ as
a subgroup. \\

The reason why we need $ \widehat{\rm Wq}(K) $ instead of just $
  \tWq (K) $ is lack of functoriality of the latter
set. If $ K' \supset K $ is a field extension, we have
a well defined semigroup homomorphism $ \widehat{\rm Wq}(K)
\rightarrow \widehat{\rm Wq}(K') $ mapping a class $ \{ M \}
 $ to $ \{ K' \otimes_{K} M \}$.  This homomorphism
does not map $ \tWq (K) $ to $ \tWq
(K') $ in general (if $ \ch\; K = 2), $ since for a
space $ M $ over $ K $ the quasilinear part
of $ K' \otimes_{K} M  $ may be isotropic. \\

We return to an arbitrary valuation ring $ \O .
 $ If $ M_1 = (M_1, B_1) $ is a bilinear $
\O$-module and $ M_2 = (M_2, q_2) $ is a quadratic $ \O$-module, we
can install on the tensor product $ M_1 \otimes_{\O} M_2 $ a
quadratic form $ q := B_1 \otimes q_2 $ be choosing a (non
symmetric) bilinear form $ \beta_2  $ with $
 \beta_2 (z,z) = q_2 (z) $ for all $ z \in M_2,
$ taking the tensor product $ \beta := B_1 \otimes \beta_2
 $ on $ M_1 \otimes M_2, $ and putting $ q(x)
:= \beta (x,x)$.  The quadratic form $ q $ is
independent of the choice of $ \beta , $ and can be
characterized by the rules
\[
B_q = B_1 \otimes B_{q_{2}},\ \  q(x_1 \otimes x_2) = B_1 (x_1,
x_1) q_2 (x_2)
\]

$ (x_1 \in M_1, x_2 \in M_2)$.  \\

We denote the quadratic module $ (M_1 \otimes_{\O}
M_2, q) $ by $ M_1 \otimes M_2 $ for short. If $
  M_1 = \langle a_1 , \ldots , a_n \rangle , $ then
\[
M_1 \otimes M_2 \cong (M_2, a_1 q_2) \perp \ldots \perp (M_2, a_r
q_2).
\]
In particular, for any $ a \in K, $
\[
\langle a \rangle \otimes M_2 \cong (M_2, aq_2).
\]


If $ M_1 $ is non degenerate and $ M_2 $ is
strictly regular then $ M_1 \otimes M_2 $ is strictly
regular. It is now straightforward to verify that we have a well
defined product of Witt classes
\[
\{ M_1 \} \cdot \{M_2 \} = \{M_1 \otimes M_2  \},
\]
which turns $ \Wq(\O) $ into a module over the
ring $ {\rm W}( \O)$. \{ Notice in particular that,
if $ M_1 \approx M'_1 $ then $ M_1 \otimes M_2
\cong M'_1 \otimes M_2$. \}  Unfortunately there seems to be
no good way to let $ {\rm W}( \O) $ operate on $
 \tWq (\O )$.



\section{Weak specialization and good reduction}

As in previous sections $ \lambda : K \rightarrow L \cup
\infty $ is a place, $ \O =
\O_{\lambda} $ is the valuation domain of $
\lambda , \m $ its maximal ideal and $ k = \O /\m $
its residue class field. Let $ E = (E, q) $ be
a quadratic space over $ K$.

\begin{Def} We say that $ E $ has {\it good
reduction} (abbreviated: GR) with respect to $ \lambda $
if $ E \cong K \otimes_{\O} M $ with $ M$
a {\it non degenerate} quadratic $ \O$-module.\bbox
\end{Def}


In this situation we obtain from $ E $ a quadratic $
  L$-module
\begin{equation}
\lambda_{\ast } (E) := L \otimes_{\lambda} M = L
\otimes_{\overline{\lambda}} M/\m M
\end{equation}
(Notations analogous  to those in $ \S 3)$.  Notice that the
``reduced'' quadratic module $ M/\m M $ over $ k
$ is non degenerate, but $ L \otimes_{ \overline{\lambda}} M/\m M $
may be degenerate, since the quasilinear part of
$  M/\m M $ may become isotropic over $ L$.  \\

We would like to prove that the quadratic module $
\lambda_{\ast} (E) $ is independent of the choice of $
M$.  \\

Only then the notation $ \lambda_{\ast} (E) $ will be
justified. If $ E $ is strictly regular this can be
done by the same sort of geometric argument as used in $ \S 3
 $ in the bilinear case. To prove it in the general case we
would like to use an additive map $ \lambda_{W} : \widehat{\rm Wq}
(K) \rightarrow \widehat{\rm Wq} (L), $ similar to the map $
  \lambda_{W} : {\rm W}(K) \rightarrow {\rm W}(L) $ from $ \S 2,
 $ and then to proceed in a similar way as in $ \S 2 $
and $ \S 3 $ for bilinear spaces. But now a new path has to be
taken, since we do not have a presentation of $ \widehat{\rm Wq}(K)
$ by generators
and relations which fits well with the place $ \lambda$.  \\

Let $ \O^{h} $ denote the henselization of $
  \O, K^{h} $ its field of quotient (= the
henselization of $ K $ with respect to $
\O ). \, \lambda $ extends to a place $
\lambda^{h} : K^{h} \rightarrow L \cup \infty $ with
valuation ring $ \O^{h}, $ since $
 \O^{h} $ has the same residue class field $
\O^{h}/\m^{h} = \O/\m $ as $
\O$.  If $ M $ is a non degenerate
quadratic $ \O$-module, $ M^{h} :=
\O^{h} \otimes_{\O} M $ is again non
degenerate and
\[
L \otimes_{\lambda} M  = L \otimes_{\lambda^{h}} M^{h}.
\]

Thus we can retreat to the case that $ \O $ is
henselian. \\

Here the following lemma offers help.

%\newpage

\begin{Lemma} Assume that $ \O $ is henselian. Let $
  E = (E, q) $ be an anisotropic quadratic space over $
  K$.

\begin{itemize}

\item[(a)] The sets
\[
\mu (E) := \{ x \in E \;|\; q(x) \in \O \} \textrm{\ and\ }
 \, \mu_{+} (E) := \{ x \in E \;|\; q(x) \in \m \}
\]
are $ \O$-submodules of $ E$.

\item[(b)] For any $ x \in \mu (E) $ and $ y \in
\mu_{+} (E) $ we have $ q(x+y) - q(x) \in \m $ and
$ B_q (x, y) \in \m$.

\end{itemize}\bbox
\end{Lemma}



By this lemma
\[
\rho (E) := \mu (E) / \mu_{+} (E)
\]
is a $ k$-vector space in a natural sense $ (k =
\O/\m)$.  We define a function $ \overline{q}: \rho
(E) \rightarrow k $ as follows:
\[
\overline{q}(\overline{x}) := \overline{q(x)} \hspace{2cm} (x \in  \mu (E)),
\]

\vspace{12pt}

\noindent where $ \overline{x} $ denotes the image of $
x \in \mu(E) $ in $ \rho (E) $ and $ \overline{a}
 $ denotes the image of $ a \in  \O $ in $
  k$.  Lemma 4.1 tells us that the map $ \overline{q}
$ is well defined, and,  using the lemma further, one proves
easily that $ \overline{q} $ is a quadratic form on the $
k$-vector space $ \rho (E) $ with associated bilinear
form $ \overline{B} = B_{ \overline{q}} $ given by
\[
\overline{B}( \overline{x}, \overline{y}) = \overline{B(x,y)}.
\]

\vspace{12pt}

The quadratic $ k$-module $ (\rho (E), \overline{q}) $
is clearly anisotropic. \\

If $ \O $ is not necessarily henselian we are
motivated by this lemma to make the following Ansatz in order to
associate to a space $ E $ over $ K $ a Witt
class $ \lambda_{W}  \{ E \} $ over $ L: $
\begin{equation}
\lambda_{W}  \{ E \} := \{ L \otimes_{ \overline{\lambda}} \rho (\ker
(K^{h} \otimes E) \},\label{Ansatz}
\end{equation}

\vspace{12pt}

\noindent where, as before, $ \overline{\lambda} : k
\hookrightarrow L
$ is the field embedding determined by $ \lambda$.  \\

All good and well, if only we know whether the vector space
\[
\rho (\ker (K^{h} \otimes E))
\]

\vspace{12pt}

\noindent has finite dimension!  To guarantee this we have to
confine the class of allowed quadratic modules $ E$.  \\

As explicated in $ \S 3 $ we choose a system $ S
 $ of representatives of $ \Gamma / 2 \Gamma $ in $
  K $ (with $ 1 \in S), $ where $ \Gamma =
K^{\ast} / \O^{\ast} $ is the value group of the
natural valuation associated to $ \O $.  \\

\begin{Def} A quadratic space $ E $ over $ K $
is {\it obedient} with respect to $ \lambda $ if $
  E $ has an orthogonal decomposition
\begin{equation}
E =  \Perp_{s \in S} E_{s},\label{xxx}
\end{equation}
such that each space $ (E_s, s \cdot (q|_{E_s} )) $ has
GR under $ \lambda , $ hence
\begin{equation}
E_s = KM_s \cong \langle s \rangle \otimes (K \otimes_{\O
} M_s)\label{xxy}
\end{equation}
with $ M_s $ a non degenerate quadratic $ \O$-submodule of $ E_s
$.\footnote[3]{Of course, $ E_s \neq 0 $ only for finitely many $ s
\in S$. } Then (\ref{xxx}) is called a $ \lambda$-{\it modular
decomposition} of $ E, $ and (\ref{xxx}), (\ref{xxy}) is called a $
\lambda$-{\it modular representation} of $ E$.  Instead of $ ``
\lambda$-modular'' we also use the term $ `` \O$-modular''. \bbox
\end{Def}


It is no big deal to verify the following two lemmas.

\begin{Lemma} Assume that $ \O $ is henselian and
$E $ is an obedient anisotropic quadratic space over $K, $
with $ \O$-modular representation
{\rm (\ref{xxx}), (\ref{xxy})}. Then $ (M_1, q|_{M_1}) $ is the only non
degenerate quadratic $ \O$-submodule of $ E_1$, and
\[
\begin{array}{lcr}
\mu (E) & = & M_1 \perp \displaystyle\Perp_{s \neq 1} \mu_{+} (E_{s}), \\
&  & \\
\mu_{+} (E) & = & \m M_1 \perp \displaystyle\Perp_{s \neq 1} \mu_{+} (E_{s}).
\end{array}
\]

Thus
\[
( \rho (E), \overline{q}) \cong (M_1 /\m M_1, \overline{q_1})
\]
with $ q_1 := q|_{M_1}$.  \bbox
\end{Lemma}


\begin{Lemma} Let $ \O $ be henselian. Let
$ s_1, \ldots, s_r $ be different elements of $ S $
and $ M_1 , \ldots , M_r $ anisotropic nondegenerate
quadratic $ \O$-modules. Then

\[
E := \Perp^{n}_{i=1} \langle s_i \rangle \otimes (K
\otimes_{\O} M_i)
\]

\vspace{12pt}

is an anisotropic quadratic space over $ K$. \bbox
\end{Lemma}


We arrive at the main theorem of this section.

\begin{Theorem} Let $ E $ be a quadratic space over $
  K, $ obedient with respect to $ \O .
$ Let
\[
E  = \Perp_{s \in S} E_s  = \Perp_{s \in S} F_s
\]

\vspace{12pt}

\noindent be two $ \O$-modular decompositions of $ E,
 $ and also let $ M_1, N_1 $ be nondegenerate
quadratic $ \O$-modules with $ E_1 \cong K
\otimes_{\O} M_1, F_1 \cong K \otimes_{\O}
N_1$.  Then the quadratic spaces $ M_1 /\m M_1 $
and $ N_1 /\m N_1
 $ over $ k = \O/\m $ are Witt equivalent. \\
\end{Theorem}

For the proof one passes from $ K $ to $ K^{h},
 $ chooses Witt decompositions of the quadratic $
\O^{h}$-modules $ M_{i}^{h}, N_{j}^{h}, $ and
then computers the kernel space of $ E^{h} $ in two
different ways applying Lemma 5. Then Lemma 4 gives the result. \\

\begin{Def} Let $ E $ be a quadratic space over $
K, $ obedient with respect to $ \lambda$.  If $
  E = \displaystyle\Perp_{s \in S} E_s $ is a $ \lambda$-modular
decomposition of $ E $, and $ M_1 $ is a non
degenerate quadratic $ \O$-module with $ E_1
\cong K \otimes_{\O} M_1, $ then we call the
quadratic space
\[
L \otimes_{\lambda} M_1 = L \otimes_{ \overline{\lambda}} M_1/\m M_1
\]
a {\it weak specialization} of $ E $
with respect to $ \lambda$.  (As before, $
 \otimes_{ \lambda } $ denotes a base extension with
respect to the homomorphism $ \lambda |_\O :
\O \rightarrow L.) $
\end{Def}

\goodbreak
By Theorem 5.6, the space $ L \otimes_{\lambda} M_1 $ is uniquely
determined by $ E $ and $ \lambda, $ up to Witt equivalence. We
denote its Witt class  by $ \lambda_{W} (E), $ i.e.,
\[
\lambda_{W}  (E) := \{ L \otimes_{\lambda} M_1 \} \in \widehat{\rm
Wq} (L).
\]
(``W'' as in ``Witt'' or ``weak''.)
\bbox

\medskip

\noindent{\it Remark.} If $ E $ is strictly regular,
then $ E_1 $ is strictly regular, hence $ M_1
$ is strictly regular, and we conclude that $ \lambda_{W}
(E) \in \Wq (L)$.  In particular this happens if $ \ch
K \neq 2$.  If $ E $ is only regular then $
M_1 $ is still regular, hence $ \lambda_{W} (E) \in
\Wqr(L), $ since now the quasilinear part of $ L
\otimes_{\lambda} M_1 $ has at most dimension 1, hence is
anisotropic. If $ \ch L \neq 2 $ then quadratic spaces
over  $ K $ resp. $ L $ can be identified
with bilinear spaces over $ K $ resp. $ L, $
and the present  weak specialization coincides with the weak
specialization of $ \S 2 $ and $ \S 3$.  \bbox

\begin{Coro} If $ E $ and $ E' $ are quadratic
spaces over $ K, $ both obedient with respect to $
\lambda , $ and if $ E \sim E', $ then $
\lambda_{W} (E) = \lambda_{W}(E')$.
\end{Coro}


\noindent{Proof.} This can be quickly deduced from Theorem 5.6.
Suppose without loss of generality that $ \dim E \leq \dim E'.
 $ Then $ E' \cong E \perp r \times \left[
\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] $ for a
certain $ r \in \mathbb{N}_{0}$.  If we choose a non degenerate $ \O$-module $
M_1 $ for $ E_1, $ as in Definition 5.7, the $
M'_1 := M_1 \perp r \times \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0
\end{array} \right] $ is a possible choice for $ E'.
 $ Therefore, $ L \otimes_{\lambda} M'_1 \sim  L
\otimes_{\lambda} M_1$.\\\mbox{}  \hfill\bbox
\\

\begin{Rem} Let $ E $ and $ F $ be
quadratic spaces over $ K, $ obedient with respect to $
  \lambda , $ and suppose that $ E $ is
strictly regular.

Obviously we then have
\[
\lambda_{W} (E \perp F) = \lambda_{W} (E) + \lambda_{W} (F).
\]
\bbox
\end{Rem}

\{The addition of an element of $ \Wq (K)  $ and an
element of $ \tWq(K) $ has been explained in   \S 4.\}



We do not exploit here the full power of weak specializations but
use them only to justify the Ansatz (\ref{Ansatz}) from the beginning of the
section for specializing a space with good reduction.

\begin{Sch} Assume that $ E $ has GR under $
\lambda, E \cong K \otimes_{\O} M $ with $
M $ a non-degenerate quadratic $ \O$-module.
Then $ L \otimes_{\lambda} M $ is uniquely determined
by $ E $ and $ \lambda $ up to isometry.
Indeed, the Witt class of $ L \otimes_{\lambda} M $
does not depend on the choice of $ M $ by Theorem 5.6,
and $ \dim L \otimes_{\lambda} M = \dim E$.
\end{Sch}

\begin{Def} If $ E $ has GR under $ \lambda $
we define
\[
\lambda_{\ast} (E) :=  L \otimes_{\lambda} M,
\]
and we call $ \lambda_{\ast} (E)  $ the
{\it specialization} of $ E $ under $
\lambda$.  \\

If $ E $ and $ F $ are quadratic $
K$-spaces with GR under $ \lambda $ and $ E $
is strictly regular, then $ E \perp F $ has GR under $
  \lambda $ and clearly
\[
\lambda_{\ast} (E \perp F) = \lambda_{\ast} (E) \perp
\lambda_{\ast} (F).
\]
\end{Def}

By arguments analogous to the proof of Theorem 3.6 one now obtain
the following important fact.

\begin{Theorem} Let  $ F $ and $ G $ be
quadratic spaces over $ K$.  Suppose that $ F
$ is strictly regular. If $ F $ and $ F \perp G
 $ have GR with respect to $ \lambda, $ then $
G $ also has GR with respect to $ \lambda$.
\end{Theorem}

We mention that -- under some precaution -- weak specialization is
compatible with the tensor product of a bilinear and quadratic
space. For example the following holds.

\begin{Rem} Let $ F $ be a bilinear space and $ G
 $ a strictly regular quadratic space our $ K$.
Suppose that $ G $ has GR under $ \lambda$.
Then $ F  \otimes G $ is obedient with respect to $
  \lambda, $ and
\[
\lambda_{W} (F \otimes G) = \lambda_{W} (F) \lambda_{W}(G) =
\lambda_{W} (F) \{ \lambda_{\ast} (G) \}.
\] \bbox
\end{Rem}



%\newpage

\section{Generic splitting of quadratic forms}

In the following quadratic $ \O$-modules leave the
stage and will act only from the background. Openly we only deal
with quadratic spaces over fields. Thus we switch to the language
of quadratic forms (= homogeneous polynomials of degree 2) over
fields, freely using the  terminology of $ \S 4 $ and
$  \S 5 $ for forms instead of spaces. \\

If $ \varphi = \varphi (x_1 , \ldots, x_n) $ is a form%
\footnote[4]{\ ``Form'' will always mean``quadratic form''.} over a
field $ k $ and $ k \subset K $ is a field
extension then $ \varphi \otimes K $ denotes the
polynomial $ \varphi $ as an element of $ K[x_1,
\ldots , x_n] $ instead of $ k[x_1, \ldots , x_n]$.  \\

It turns out that we can extend the well known generic splitting
theory of forms over fields of characteristic $ \neq 2
$ to arbitrary fields, as long as we can guarantee that under the
relevant places $ \lambda : K \rightarrow L \cup \infty
$ a given  form $ \varphi  $ over $ K $ with
GR with respect to $ \lambda $\footnote[5]{\ This
assumption presupposes that $ \varphi $ is
nondegenerate (cf. $ \S 5, $ Def. 1).} has a
specialization $ \lambda_{\ast } (\varphi ) $ which is
again nondegenerate, i.e., the specialization $ \lambda_{\ast
} (QL(\varphi )) $ of the quasilinear part $ QL (
\varphi ) $ remains anisotropic $ \{$ Slogan: ``Do not
destroy the quasilinear part!''$ \}$ \\

\begin{Def} Let $ \varphi $ be a  nondegenerate form
over a field $ k$.  We call a field extensions $ k
\subset K$ \ $ \varphi$-{\it conservative} if $ K
\otimes \varphi $ is again nondegenerate, i.e., $ K
\otimes QL( \varphi ) $ is anisotropic. \bbox
\end{Def}

Notice that a separable field extension $k \subset K$  is $\varphi$-conservative
for every $\varphi$, since an anisotropic quasilinear form over $k$
remains anisotropic over $K$.

Notice also that, {\it if} $ \varphi $ {\it is
regular then every field extension} $ k \subset K $
{\it is} $ \varphi$-{\it conservative}, since
forms of dimension $ \leq 1 $ cannot become isotropic. \\

The generic splitting theory of a non degenerate form $
\varphi $ over $ k $ will deal with the Witt
decomposition of $  K  \otimes \varphi $ for $ K
 $ varying in the class of all $ \varphi$-conservative
field extension of $ k$.  \\

The following observation is crucial here.

\begin{Theorem} Let $ \lambda : K \rightarrow L \cup  \infty
 $ be a place and $ \varphi $ a form over $ K
 $ which has GR with respect to $ \lambda$.
Suppose that also $ \lambda_{\ast} (\varphi ) $ is
nondegenerate. Suppose further that $ K' \supset K $ is
a field extension and that $ \mu : K' \rightarrow L \cup
\infty $ is a place extending $ \lambda $.  Then
the form $ \varphi \otimes K' $ has GR with respect to
$ \mu $ and
\[
\mu_{\ast} ( \varphi \otimes K') = \lambda_{\ast} ( \varphi).
\]
\end{Theorem}

\noindent{\bf Proof.} Let $ \O :=
\O_{\lambda},\ \O' := \O_{\mu},
 $ and let $ k $ and $ k' $ denote the
residue class fields of $ \O $ and $ \O' $ respectively. The field
extension $ \overline{\lambda}: k \hookrightarrow L $ is a
composition of the extensions $ k \hookrightarrow k' $ and $
\overline{\mu} : k' \hookrightarrow L, $ where the first extension
is induced by
the inclusion $ \O \hookrightarrow \O'$.  \\

Let $ E $ be a quadratic space for $ \varphi $
and $ M $ a nondegenerate quadratic $
\O$-module with $ E \cong K \otimes_{\O}
M$.  Then $ K ' \otimes E = K' \otimes_{\O'} M'
 $ with $ M' := \O' \otimes_{\O} M.
 $ The quasilinear quadratic $ k$-module $ G := k
\otimes_{\O} QL(M) $ is anisotropic. By assumption,
$ L \otimes_{ \overline{\lambda}} G = QL ( L \otimes_{\lambda} M)
 $ is also anisotropic. Therefore
\[
k' \otimes_{k} G = k' \otimes_{\O\,'} QL(M')
\]
is anisotropic. This proves that $ M' $ is a
nondegenerate quadratic $ \O'$-module. Hence $
 \varphi \otimes K' $ is nondegenerate and has GR with
respect to $ \mu$.  Furthermore $ \mu_{\ast}
(\varphi \otimes K') $ corresponds to the quadratic space
\[
L \otimes_{\mu} M' = L \otimes_{\mu} ( \O'
\otimes_{\O} M) = L \otimes_{\lambda} M.
\]
Hence $ \mu_{\ast} ( \varphi \otimes K') = \lambda_{\ast}
(\varphi )$.  \bbox
\\


In the following $ \varphi $ is a nondegenerate form
over a field $ k$.

\begin{Sch} Let $ K \supset k, L \supset k $ be
field extensions of $ k, $ and let $ \lambda : K \rightarrow L \cup
\infty $ be a place over $ k, $ i.e., a place extending the trivial
place $ k \hookrightarrow L$.  Assume that $ L $ is $
\varphi$-conservative. Then Theorem 6.2 tells us that $ K $ is $
\varphi$-conservative, $ \varphi \otimes K
 $ has GR  with respect to $ \lambda $, and $
\lambda_{\ast} ( \varphi \otimes K) = \varphi \otimes L$.
\end{Sch}

Let $ \varphi \otimes K \cong \varphi_1 \perp r_{1} \times H
 $ be the Witt decomposition of $ \varphi$.  By
Theorem 5.12 it follows that $ \varphi_1 $ has GR with
respect to $ \lambda, $ and hence
\[
\varphi \otimes L = \lambda_{\ast} ( \varphi \otimes K) =
\lambda_{\ast} ( \varphi_1 ) \perp r \times H.
\]

\{We denote the hyperbolic plane $ \left[
\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] $ over
any field (or ring) by $H$.\}

Thus $ \ind ( \varphi \otimes L) \geq \ind ( \varphi \otimes
K), $ and, in case of equality,
$$\ker ( \varphi \otimes L)
= \lambda_{\ast} (\ker (\varphi \otimes K)).$$


It now follows that, if $ K $ and $ L $ are
specialization equivalent over $ k, $ i.e., there exists
also a place over $ k $ from $ L $ to $
K, $ then
\[
\ind ( \varphi \otimes L) = \ind ( \varphi \otimes K),
\]
and
\[
\ker ( \varphi \otimes L) = \lambda_{\ast} (\ker \varphi \otimes K)
\]
for any place $ \lambda $ from $ K $ to $
  L $ over $ k$.  \bbox

\begin{Def} We call a field extension $ K \supset k $ a
{\it generic zero field} of $ \varphi $, if $ K $ is $
\varphi$-conservative, and there exists a place from $ K $ to $ L $
over $
  \varphi$ for any $
 \varphi$-conservative field extension $ L \supset k
$ such that $ \varphi \otimes L $ is isotropic. \bbox
\end{Def}


Notice that $ \varphi \otimes L  $ is isotropic iff $
  \ind ( \varphi \otimes L ) \geq 1, $ since $
\varphi \otimes L  $ is nondegenerate. \\

Any two generic zero fields of $ \varphi $ are
specialization equivalent over $ k$.  \\

Fortunately generic zero fields of $ \varphi $ exist
whenever this  makes sense.

\begin{Theorem} Assume that $ \varphi $ is anisotropic
and not quasilinear, $ n := \dim  \varphi \geq 2$.

\begin{itemize}
\item[a)] The function field $ k( \varphi) $ of the
affine quadric $ \varphi(x_1, \ldots , x_n) = 0$,\
i.e., the quotient field of the integral domain
$$ k[x_1,\ldots , x_n]/( \varphi (x_1, \ldots, x_n)) $$
is a generic
zero field of $ \varphi$. \{N.B. $k(\varphi)$ is separable over $k$.\}

\item[b)] More generally the following holds. If $ \gamma : k
\rightarrow L \cup \infty $ is a place  such that $
 \varphi $ has GR under $ \gamma, $ and if $
  \gamma_{\ast} ( \varphi) $ is nondegenerate and
isotropic, there exists a place $ \lambda : k ( \varphi)
\rightarrow L \cup \infty $ extending $ \gamma $.
\end{itemize} \bbox
\end{Theorem}


We now can build a {\it generic splitting tower} $
(K_{r} \;|\; 0 \leq r \leq h) $ of $ \varphi $ in the
way well known from the case that $\ch\,  K \neq 2 $
(cf. [K], [KS], \hbox{[S], \dots)} and from the case that $\ch\,
k = 2, $ but $ \varphi $ regular (cf. [KR]).  \\

Take $ K_0 = k, $ or more generally, let
$ K_0 $ be field extension of $ k $ such that there exists a
place from $ K_0 $ to $ k  $ over $ k$,
with corresponding Witt decomposition
\[
\varphi \otimes K_0 \cong \varphi_0 \perp i_0 \times H
\]
(N.B.: $ i_0 = \ind ( \varphi))$.  If $ \varphi_0
 $ is quasilinear, we stop.

Otherwise we choose a generic zero field $ K_1 \supset K_0
 $ of $ \varphi_0, $ and then have a Witt
decomposition
\[
\varphi_0 \otimes K_1 \cong \varphi_1 \perp i_1 \times H
\]
etc. We could take $ K_0 = k, K_1 = k (\varphi_0), $ etc.
But for various problems it is useful
to allow other generic splitting towers. \\

We retain the terminology from the generic splitting theory in
characteristic $ \neq 2$.  In particular we call $
i_r $ the  {\it $ r$-th higher index} of $
 \varphi $ and $ h $ the {\it height} of
$ \varphi$.  The form $ \varphi_{h} $ is
quasilinear. \\

Precisely as in the characteristic $ \neq 2 $ case we obtain
from the above theorems immediately:

\begin{Theorem} Let $  \varphi $ be a non degenerate
form over $ k$.  Let $ (K_r \;|\; 0 \leq r \leq h)
$ be a generic splitting tower of $ \varphi $ with
associated higher indices $i_r$ and higher kernel forms $
\varphi_{r}$.  Let $ \gamma : k \rightarrow L \cup
\infty $ be a place such that $ \varphi $ has GR
with respect to $ \gamma $ and $ \gamma_{\ast}
(\varphi ) $ is non degenerate. We choose a place $
\lambda: K_m \rightarrow L \cup \infty $ extending  $
\gamma $ such that either $ m = r $ as $ m <
r, $ but $ \lambda $ cannot be extended to a place
from $ K_{m+1} $ to $ L$.  Then $
\varphi_m $ has GR with respect to $ \lambda$.
The form $ \lambda_{\ast} ( \varphi) $ has the kernel
form $ \lambda_{\ast} ( \varphi_m) $ and the Witt index
$ i_0 + \ldots + i_m$. \bbox
\end{Theorem}



If $ L \supset k $ is a $ \varphi$-conservative
field extension we may apply the theorem to the trivial place $
 \gamma : k \hookrightarrow L $ and obtain precise
information about the Witt decomposition of $ \varphi
\otimes L$.

%\newpage

\section{Epilogue}

\begin{itemize}

\item[A)] Perhaps the most urgent open problem in generic
splitting theory  is to determine all forms of height 1. Assume
that $ \varphi $ is anisotropic and $ h( \varphi) =
1$.  If $ \varphi $ is strictly regular, then it
turns out that $ \varphi $ is, up to a scalar factor, a
quadratic Pfister form. (cf. [Spez, $ \, \S 20]). $ If $
QL( \varphi) $ has dimension 1 then $ \varphi $ is,
up to scalar factor, a certain ``close neighbor'' of a quadratic
Pfister form (cf. [Spez, $ \, \S 22]), $ analogous to the pure
part of a Pfister form in the case of characteristic $ \neq
2$.  But, if $ \dim QL(\varphi) \geq 2 $ there
exist more forms of height 1 than  those which are Witt equivalent
to close Pfister
neighbors.

\item[B)] Let $ K $ and $ L $ be fields with $
  \ch\, K = 0,\  \ch\, L = 2$.  Given a place $
\lambda : K \rightarrow L \cup \infty $, it deserves interest
to ``lift'' a nondegenerate quadratic form $ \psi $ over
$ L $ to a form $ \varphi $ over $ K,
$ i.e., to exhibit a quadratic form $ \varphi $ over $
  K $ with GR with respect to $ \lambda $ and $
  \lambda_{\ast} ( \varphi ) \cong \psi$.  Then one can
hope to deduce properties of $ \psi $ from properties
of $ \varphi $.  \\

In the specialization theory outlined above such a lifting is
only possible if $ \psi $ is strictly regular. Indeed,
since a nondegenerate form $ \varphi $ over $ K
 $ as automatically strictly regular, also $ \,  \,
\lambda_{\ast} ( \varphi) $ has to be strictly regular. \\

\end{itemize}

Fortunately there exists a more general specialization theory
than the one explicated in $ \S 5$.  \\

Given a place $ \lambda : K \rightarrow L \cup \infty $
with valuation ring $ \O, $ we say that a quadratic
space $ E = (E, q) $ over $ K $ has
{\it fair reduction} with respect to $ \lambda, $
if $ E $ contains a free $ \O$-submodule
$ M $ with $ E = KM $ and $ q(M) \subset\O$,
such that $ (M/\m M, \overline{q|_M}) $ is
a quadratic space over $ \O/\m, $ while
$ (M,q|_M) $ may be degenerate. One can prove that then
\[
\lambda_{\ast} (E) := L \otimes_{\overline{\lambda}} (M/\m M) = L
\otimes_{\lambda} M
\]

\noindent is still well defined up to isometry by $ E $
and $ \lambda$.  This is the basis of a ``fair
specialization theory'' which parallels our theory in $ \S
5, $ cf. [Spez, II, $ \, \S 11$]. \\

It is now well possible to find for a non degenerate form $
\psi $ over $ L $ a form $ \varphi $
over $ K $ with fair reduction with respect to $
 \lambda $ and $ \lambda_{\ast}  ( \varphi) \cong
\psi $.  For fair specializations there also exists a theorem
completely analogous to the generic splitting theorem 6.6 above,
cf. [Spez, II, $ \, \S 12$].


%\newpage

\bigskip

%% \begin{center}
%% {\bf References}
%% \end{center}

\def\bib#1{\bibitem[#1]{#1}}
%%\def\SAuthor{References}

\begin{thebibliography}{00000}
\normalsize


\bib{Spez} {\it M.Knebusch}, ``Spezialisierung von
quadratischen und symmetrischen bilinearen Formen'' (2007),\hfill\break
http://www-nw.uni-regensburg.de/\char126.knm22087.mathematik.uni-regensburg.de/index.htm
(the author's homepage).
\\

\bib{Spez'} Translation of [Spez] by {\it Thomas Unger},
in preparation. Chapter I and II available as preprint. \\

\bib{K} {\it M.~Knebusch}, ``Generic splitting of quadratic
forms I'', Proc. London Math. Soc. (3) 33 (1976), 65-93. \\

\bib{KR} {\it M.~Knebusch, U.~Rehmann}, ``Generic splitting
towers and generic splitting preparation'', Contemporary Math. 272
(2000), 173-199. \\

\bib{KR} {\it M.~Knebusch, W.~Scharlau}, ``Algebraic
theory of quadratic forms, generic methods and Pfister forms'',
DMV-Seminar  N° 1 (Notes taken by Heisook Lee). Birkhauser
1980. \\

\bib{S} {\it W.~Scharlau}, ``Quadratic and hermitian
forms'' (Grundlehren  Math. Wiss. 270), Springer 1985. \\


\end{thebibliography}



\end{document}
