\titlea{\S11}{Fair Reduction and Weak Obedience}
\leftrunningtitle{Chapter II. Generic Splitting Theory}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S11 Fair Reduction and Weak
Obedience}


Our specialization theory of quadratic forms, developed in \S6 --
\S8, gave a satisfying basis for  understanding  the
splitting behaviour of quadratic forms under field extensions
(if the forms were not regular, we had to limit ourselves to
separable field extensions). There is one important point however, where
the specialization theory is disappointing.

For instance, let $\lambda \colon K \to L \cup \infty$ be a place
from a field $K$ of characteristic~$0$ to a field $L$ of
characteristic~$2$. If $\vf$ is a quadratic form over $K$ which
has good reduction \wrtl, then $\lambda_* (\varphi)$ is
automatically a strictly regular form. Conversely, given a
strictly regular form $\psi$ over $L$, one can easily find a
strictly regular ($=$ \nd) form $\vf$ over $K$ which has good
reduction \wrtl\ and such that $\lambda_* (\varphi) \cong
\psi$. One could then try to deduce properties of $\psi$ from
properties of the ``lifting'' $\vf$ of $\psi$, in the hope that
$\vf$ is easier to deal with than $\psi$ because $\Char K=0$. Good
examples can be obtained from the generalization of \S4,
Scholium~3, at the end of \S9, but we will not carry this out.

It is furthermore desirable  to lift a \nd\ form $\psi$ over
$L$, with quasilinear part $QL(\psi)\not=0$, to a form $\vf$ over
$K$. This is not possible with our specialization theory as it stands.

Let us review  the foundations of the current theory! We return to
the use of geometric language. Let $\lambda \colon K \to L \cup
\infty$ be a place with associated valuation ring $\o$. Let ${\nam m}$
be the  maximal ideal of $\o$ and $k = {\nam o}/{\nam m}$ its
residue class field. Finally, let $\overline{\lambda} \colon k \hookrightarrow
L$ be the field extension determined by $\lambda$.

Given a suitable quadratic space $E$ over $K$,
the idea was to attach a space $F$ over $k$ to $E$ and then to
define the specialization $\lambda_* (E)$ of $E$ \wrtl\ as the
space $L \otimes_{\overline{\lambda}} F$. We required $E$ to have
good reduction, i.e. $E \cong K \otimes_{{\nam o}} M$ for a
\nd\ quadratic module $M$ over $\o$. Then we could choose our
space $F$ over $k$ to be $F = M/{\nam m} M$.

The preceding is analogous to the specialization theory for
varieties in algebraic and arithmetic geometry (and related areas,
such as rigid analysis): One defines a class of
``\nd'' objects ${\Cal X}$ over the valuation ring $\o$ and associates
to it, by means
of a base extension, the objects $K \otimes_{{\nam o}} {\Cal X}$
and $k \otimes_{{\nam o}} {\Cal X}$ over $K$ and $k$. Finally one
decrees that
$L \otimes_{\overline{\lambda}} (k
\otimes_{{\nam o}} {\Cal X}) = L \otimes_{\lambda} {\Cal X}$
is
the specialization of $K \otimes_{{\nam o}} {\Cal X}$ \wrt\
$\lambda$.

Without any doubt, we have found in the \nd\ quadratic spaces,
as defined in \S6, Definition~3, a
respectable class ${\cal X}$ of such objects which easily comes to mind.
Nevertheless we are a bit unfair to all the quadratic spaces over
$K$, waiting to be specialized. It would be fairer to only require
that $E \cong K \otimes_{{\nam o}} M$ with $M$ a free quadratic
$\o$-module such that $M/{\nam m} M$ is \nd, and then to define
$\lambda_* (E) \colon = L
\otimes_{\lambda} M = L \otimes_{\overline{\lambda}} (M/{\nam m} M)$.
But then one should show, that up to isometry,
$\lambda_* (E)$ only depends on
$\lambda$ and $E$ and not on the choice of $M$.

This is possible, as we shall show now. In the following a {\it
quadratic module} over a valuation ring (in particular a field)
will always be understood to be a {\it free quadratic module of
finite rank}.

Let $\o$ be a valuation ring with maximal ideal ${\nam
m}$, quotient field $K$ and residue class field $k = {\nam o}/{\nam
m}$. If $2 \not\in {\nam m}$, then we already know all what
follows. On formal grounds, we allow nevertheless the
uninteresting case that $\Char k\ne 2$.

\begdefinition{1.}
\smallskip
\setitemindent{(a)}
\item{(a)} A quadratic module $M$ over $\o$ is called
{\it reduced \nd}\Idx{reduced nondegenerate quadratic module}\Idx{quadratic module!reduced nondegenerate} 
if the quadratic module $M/{\nam m} M$ over $k$
is \nd. (Note: If $\Char k\ne2$, this implies that $M$ itself is
\nd.)
\medskip
\item{(b)} A quadratic module $E$ over $K$ has {\it fair reduction}\Idx{fair reduction}\Idx{111fr@FR} 
(or:
\FR\ for short) {\it \wrtl} (or: {\it \wrt\ $\o$}), if there exists a
reduced \nd\ quadratic module $M$ over $\o$ such that $E \cong
K \otimes_{{\nam o}} M$.\fonote{In [K$_4$] the words ``nearly good
reduction'' are used instead of ``fair reduction''. We avoid this
terminology here, because  we will talk about a different
concept,  ``almost good reduction'' later (Chapter~IV).}
\enddefinition

It is our task to prove that in the situation of
Definition~1(b), the space $M/{\nam m} M$ is independent of the
choice of reduced \nd\ module $M$ up to isometry. We will however proceed
in a more general direction than necessary, in order to develop at
the same time the equipment necessary to obtain a  generalization
of the important Theorem~2(b) (\S8) for fair reduction instead of  good
reduction. First a very general definition.

\begdefinition{2.} Let $M = (M, q)$ and $M' = (M', q')$ be
quadratic modules over a ring $A$. We say that $M$ {\it
represents}\Idx{representation (quadratic module)}\Idx{quadratic module!representation}   
the quadratic module $M'$, and write $M'<M$,\Idx{111mplessm@$M'<M$ (quadratic module)} 
if the $A$-module
$M$ has a direct sum decomposition $M = M_1 \oplus M_2$ with
$(M_1, q \vert M_1) \cong (M', q')$.
\enddefinition

Now, if $M$ is a \q\ $\o$-module, we always regard $M$ as an
$\o$-submodule of the $K$-vector space $K \otimes_{{\nam o}} M$.
So we have $K \otimes_{{\nam o}} M = KM$. In the following,
we will almost always denote the \qf\ on $M$
by $q$, and its associated bilinear form $B_q$ by $B$.
We will also use $q$ to denote the \qf\ on $K \otimes_{{\nam o}} M$
with values in $K$, obtained from $q$. We will often denote the
module $M/{\nam m} M$ over $k$ by $\overline{M}$, and the image of
a vector $x\in M$ in $\overline{M}$ with $\overline{x}$. Likewise
we denote the image of a scalar $a \in {\nam o}$ in $k$ by $\overline{a}$.
Finally, we denote by $\overline{q}$ the quadratic form on
$\overline{M}$ with values in $k$, induced by $q$, i.e.
$\overline{q} (\overline{x}) = \overline{q(x)}$ for $x \in M$.
Furthermore, $\overline{B}$ will stand for the associated bilinear
form $B_{\overline{q}}$, i.e.
$\overline{B} (\overline{x}, \overline{y}) = \overline{B(x, y)}$ for $x, y
\in M$.

As before, we  call a \nd\ \qm\ over a field or valuation ring
a {\it \qs}, or just a {\it space}. If $M$ is a reduced \nd\ \q\
$\o$-module, then $\overline{M} = M/{\nam m} M$ is a space over
$k$. It is easy to show that $K\otimes M$ is a space over $K$ too
(see the Remark below, after Definition~3), but that is not so
important at the moment.

\beglemma{1.} Let $M$ be a \rnd\ \qm\ over $\o$.\break There exists a
decomposition $M = M_1 \perp M_2$ with $M_1$ strictly regular and
$B(M_2 \times M_2) \subset {\nam m}$. If such a decomposition is
given and $x$ is a primitive vector of $M_2$, then $q(x) \in
{\nam o}^*$.
\endlemma

\begproof{} We have a decomposition $\overline{M} = U \perp
QL(\overline{M})$ with $U$ strictly regular.\fonote{Recall that
${QL(\overline{M})}$ denotes the quasilinear part of
$\overline{M}$, see the beginning of \S6.} Let $x_1, \dots, x_r$
be vectors of $M$, such that the images $\overline{x}_1, \dots, \overline{x}_1$ in
$\overline{M}$ form a basis of the $k$-vector space $U$. Then the
determinant of the value matrix $(B (x_i, x_j))$ with $1 \leq i$, $j \leq
r$ is a unit of $\o$. Therefore, $M_1 \colon = \sum\limits_{i = 1}^r
{\nam o} x_i$ is a strictly regular space. Let $M_2$ be the
orthogonal complement $M_1^{\perp}$ of $M_1$ in $M$. Then
$M = M_1\, \bot \, M_2$ and $M_2 /{\nam m} M_2 = QL
(\overline{M})$. Since $QL(\overline{M})$ is quasilinear, we have
$B(M_2 \times M_2) \subset {\nam m}$. Conversely, if such a
decomposition $M = M_1\, \bot \, M_2$ is given, with $M_1$
strictly regular and $B (M_2 \times M_2) \subset {\nam m}$, then
$\overline{M}_2 = QL(\overline{M})$. Since $QL(\overline{M})$ is
anisotropic, every primitive vector $x$ in $M_2$ has a value
$q(x) \in {\nam o}^*$.\qed
\endproof

\beglemma{2.} Suppose again that $M$ is a \rnd\ \q\ $\o$-module.
Then $M$ is maximal among all finitely generated $($thus free$)$ $\o$-modules
$N \subset K \otimes_{{\nam o}} M$ with $q(N)
\subset {\nam o}$.
\endlemma

\begproof{} We  work in the \q\ $K$-module $E \colon = K \otimes_{{\nam o}}
M$. According to Lemma~1, there is a decomposition $M = M_1 \perp M_2$ with $M_1$
strictly regular and $B(M_2 \times M_2) \subset {\nam m}$. Let $x \in E$ and $q(M +
{\nam o} x) \subset {\nam o}$. We have to show that $x\in M$.

We write $x = x_1 + x_2$ with $x_1 \in K \otimes M_1$, $x_2 \in K \otimes
M_2$. For every $y \in M_1$, we have $B (x_1, y) = B (x, y) = q(x+y) - q(x) -
q(y) \in {\nam o}$. Since the bilinear form $B = B_q$ is \nd\ on
$M_1$, it follows that $x_1 \in M_1$. Therefore, $M + {\nam o} x = M + {\nam o}
x_2$ and so $q(x_2) \in {\nam o}$. Now, $x_2 = a z$ for a
primitive vector $z \in M_2$ and $a \in K$. Since $q(z) \in {\nam
o}^*$, we get $a \in {\nam o}$, and so $x_2 \in M$. We conclude
that $x\in M$.\qed
\endproof

\beglemma{3.} {\rm (Extension of \S6, Lemma~4.)} Let $M$ be a
\rnd\ \q\ $\o$-module and let $e$ be a primitive isotropic vector
in $M$. Then $e$ can be completed to a hyperbolic vector pair
$e,f$ in $M$.
\endlemma

\begproof{} We work again in $E = K \otimes M$. The ideal $B(e,
M)$ of $\o$ is finitely generated. Therefore we have $B(e, M) = a {\nam
o}$ for an element $a$ of $\o$. If $a$ were equal to zero, then
the vector $\overline{e} \in \overline{M}$ would lie in the
quasilinear part $QL(\overline{M})$ of $\overline{M}$. However,
$\overline{e} \ne 0$ and $\overline{q}
(\overline{e}) = 0$, so that $\overline e \not \in QL(\overline M)$
due to the anisotropy of $QL(\overline M)$. Therefore, $a \ne 0$.
Since $B(e, M) = a
{\nam o}$ and $q(M) \subset {\nam o}$, we have $q(M + {\nam o} (a^{-1} e))
\subset {\nam o}$. Thus, by Lemma~2, $a^{-1} e \in M$. Since $e$
is primitive, we conclude that $a^{-1} \in {\nam o}$, i.e. $a \in {\nam o}^*$.
Hence, $B(e, M) = {\nam o}$. As in the proof of \S6, Lemma~4, we
choose  $z\in M$ with $B(e,z)=1$ and complete $e$ to a hyperbolic
pair with the vector $f \colon = z - q(z) e$.\qed
\endproof

\begtheorem{1.} Let $M$ be strictly regular, $N$ a \rnd\ \q\
$\o$-module and $K \otimes M < K \otimes N$. Then $M<N$.
Furthermore, $N \cong M \perp P$ where $P$ is a \rnd\ \q\
$\o$-module.
\endtheorem

\begproof{} If $Q$ is a strictly regular \q\ submodule of $N$,
then $N = Q \perp P$ with $P \colon = Q^{\perp} = \{ x \in N \; \vert \; B
(x, Q) = 0 \}$, since $Q$ is \nd\ \wrt\ the bilinear form $B =
B_q$ (\S 5, Lemma 1). We have $\overline{N} = \overline{Q} \perp
\overline{P}$ and conclude that $\overline{P}$ is \nd, i.e. $P$ is
\rnd.

Because of this preliminary remark, it suffices to show that
$M<N$. Suppose first of all that $M$ is hyperbolic, i.e.
$M \cong r \times H$ for some $r > 0$, where $H$ denotes the \qm\
$\left[ {0 \atop 1} \; {1 \atop 0} \right]$ over ${\nam o}$. We
proceed by induction on $r$.

Since $K \otimes M < K \otimes N$, $K \otimes N$ contains
isotropic vectors. We can then choose a {\it primitive} isotropic
vector $e$ in $N$. This vector can  be completed to a
hyperbolic vector pair in $N$, by Lemma~3. Hence, $H<N$ and so
$N \cong H \perp N'$ where $N'$ is a \rnd\ \q\ $\o$-module. From
$r \times  (K \otimes H) < K \otimes N$, we get $K \otimes N \cong r \times (K
\otimes H) \perp G$ for some space $G$ over $K$. On the other
hand,
$K \otimes N \cong (K \otimes H) \perp (K \otimes N')$. By the
Cancellation Theorem over $K$ (\S6, Theorem~2), we get
$(r-1) \times (K \otimes H) \perp G \cong K \otimes N'$.
Therefore, $(r-1) \times (K \otimes H) < K \otimes N'$.  The
induction hypothesis gives us $(r-1) \times H < N'$ and so
$r \times H < N$.

In the general case, $K \otimes M < K \otimes N$ implies that
$K \otimes (M \perp (-M)) < K \otimes (N \perp (-M))$. Since
$M \perp (-M)$ is hyperbolic, it follows from above that
$M\perp(-M)<N\perp(-M)$, and so,
$N \perp (-M) \cong M \perp (-M) \perp P$
by the remark at the beginning of the proof, where $P$ is another
\rnd\ \q\ $\o$-module. By the Cancellation Theorem over $\o$
(\S6, Theorem~2), we may conclude that $N \cong M \perp P$, and so $M <
N$.\qed
\endproof

\begtheorem{2}{\rm [K$_4$, Lemma 2.8]}{\bf.} Suppose that $M$ and
$N$ are \rnd\ \q\ $\o$-modules with $K \otimes M < K \otimes N$.
The spaces $\overline{M} = M/{\nam m} M$ and $\overline{N} =
N/{\nam m} N$ over $k$ have the following properties:
\medskip
\setitemindent{$(a)$}
\item{$(a)$} $\overline{M} < \overline{N}$.
\medskip
\item{$(b)$} More precisely: there exist quadratic subspaces $S$
and $T$ of
$\overline{N}$ with $\overline{M} < S$, $\overline{N} = S \perp T$,
$T$ strictly regular,
$\dim S \leq \dim QL(\overline{M}) + \dim QL(\overline{N}) + \dim M$.
\medskip
\item{$(c)$} If $\overline{N}$ is anisotropic, then $M < N$.
\endtheorem

\begremark{} We will not need properties $(b)$ and $(c)$ later on. They
can however be obtained in the following proof at no extra
cost.
\endremark

\begproof{of Theorem~2.} (i) By Lemma~1, we can choose an
orthogonal decomposition $M = M_1 \perp M_2$ with $M_1$ strictly
regular and $B(M_2 \times M_2) \subset {\nam m}$. We have
$K \otimes M_1 < K \otimes N$. By Theorem~1, we have
$N \cong M_1 \perp N_2$, where $N_2$ is a \rnd\ \q\ $\o$-module.
Then $K \otimes M_1 \perp K \otimes M_2 < K \otimes M_1 \perp K \otimes
N_2$ implies $K \otimes M_2 < K \otimes N_2$ by the Cancellation
Theorem over $K$ (\S6, Theorem~2) as usual. If we could prove the
claims of the theorem for $M_2$ and $N_2$ instead of $M$ and $N$,
then they would follow immediately for $M$ and $N$ as well.
Hence we assume now, without loss of generality, that $B(M \times
M) \subset {\nam m}$.

(ii) If $M = \{ 0 \}$, nothing has to be done. So suppose that $M \ne \{ 0
\}$. Surely $k$ has characteristic~$2$. We suppose, without loss of generality that $F
\colon = K \otimes M$ is a subspace of $E \colon = K \otimes N$
(instead of only: $F$ is isomorphic to a subspace of $E$). Since
$\overline{M}$ is anisotropic, we have $M = \{ x \in F \; \vert \; q (x) \in {\nam o}
\}$ (as already ascertained before). Therefore, $N_1 \colon = N \cap F \subset M$.
Moreover, $N_1$ is a direct summand of the $\o$-module $N$, since
$N/N_1$ is torsion free and finitely generated, and thus free.
Hence $N_1$ itself is also free.

Let $m \colon = \dim M$. By the Elementary Divisor Theorem
for valuation rings, there exist bases $x_1, \dots,
x_m$ and $y_1, \dots, y_m$ of the free $\o$-modules $M$ and $N_1$
with $y_i = c_i x_i$, $c_i \in {\nam o} \quad (1 \leq i \leq m)$.
We suppose without loss of generality that $c_i = 1$ for $1 \leq i \leq s$ and $c_i \in {\nam
m}$ for $s < i \leq m$. $\{$It is allowed that $s = 0$ or $s =
m$.$\}$

If $\overline{N}$ is anisotropic, then $q(x) \in {\nam o}^*$ for
every primitive vector $x$ of $N$, and thus for every vector in
$N_1$ which is primitive in $N_1$. Hence $s=m$, i.e. $N_1=M$ and
$M<N$. This establishes property $(c)$.

(iii) Let $V$ be the image of $N_1$ in $\overline{N} = N/{\nam m}
N$. Since $N_1$ is a direct summand of $N$, we may make the
identification $V = \overline{N_1}$. Since $B(M \times M) \subset {\nam
m}$, we have $B(N_1 \times N_1) \subset {\nam
m}$, thus $\overline{B} (V \times V) = 0$. Hence $V$ is
quasilinear. We have
$$\leqalignno{\overline{M} &= k \overline{x}_1 \oplus \cdots \oplus k \overline{x}_m
\cong [a_1, \dots, a_m], \hbox{ with } a_i \colon = {\overline q} (\overline{x}_i) \in
k^*.&(1)\cr
V &= k \overline{y}_1 \oplus \cdots \oplus k \overline{y}_m \cong
[a_1, \dots, a_s] \perp (m-s) \times [0].&(2)}$$
Let $R \colon = QL(\overline{N})$ be the quasilinear part of $\overline{N}$
and $V_0 \colon = V \cap R$. Since $R$ is anisotropic, $V_0$ is
also anisotropic. From~(2) it follows that $V_0 < [a_1, \dots,
a_s]$, and so
$$[a_1, \dots, a_s] \cong V_0 \perp [b_1, \dots, b_{s-t}]\leqno{(3)}$$
for elements $b_i \in k^*$ and $t \colon = \dim V_0$. $\{$If
$t=s$, the righthand side of (3) should be read as $V_0$.$\}$ From
(2) and (3) we get a decomposition $V = V_0 \perp U$,  with
$$U \cong [b_1, \dots, b_{s-t}] \perp (m-s) \times
[0].\leqno{(4)}$$
Since $V \cap R = V_0$, we have $U \cap R = \{ 0 \}$. We choose a
submodule $W$ of $\overline{N}$ such that $\overline{N} = (R + U) \oplus W = R \oplus U
\oplus W$. Since $R$ is the quasilinear part of $\overline{N}$, $P
\colon = U \oplus W$ is strictly regular and $\overline{N} = R \perp P$.
Let $u_1, \dots, u_{m-t}$ be a basis of $U$, associated with the
representation~(4). $P$ is a symplectic vector space with respect
to the bilinear form $\overline{B} = B_{\overline{q}}$. Hence we
can complete $u_1, \dots, u_{m-t}$ to a basis $u_1, \dots,
u_{m-t}$, $z_1, \dots, z_{m-t}$ of a subspace $P_1$ of $P$ with
$\overline{B} (u_i, z_j) = \delta_{ij} \quad (1 \leq i$, $j \leq m-t)$.
As a quadratic space, $P_1$ is of the form
$$P_1 = (k u_1 + k z_1) \perp \dots \perp (k u_{m-t} + k z_{m-t})$$
and $P = P_1 \perp T$ for a strictly regular space $T$.

For $s - t < i \leq m-t$, we have $k u_i + k z_i \cong \left[ {0 \atop 1} \; {1
\atop 0} \right]$. Therefore, $[a_{t+i}] < k u_i + k z_i$ for this
$i$. Furthermore we have
$V_0 < R$ and $[b_i] < k u_i + k z_i$ for $1 \leq i \leq s-t$,
since $b_i = \overline{q} (u_i)$. Putting everything
together,  (1) and (3) yield
$$\overline{M} \cong V_0 \perp [b_1, \dots, b_{s-t}] \perp [a_{s+1}, \dots,
a_m] < R \perp P_1,$$
and $R \perp P = R \perp P_1 \perp T = \overline{N}$.
Let $S \colon = R \perp P_1$, then
$\overline{M} < S$ and $\overline{N} = S \perp T$.
Since $\overline{M} = QL(\overline{M})$ and $R =
QL(\overline{N})$, we have furthermore
$$\eqalign{\dim S & = \dim QL(\overline{N}) + 2 (m-t)\cr
& = \dim QL(\overline{N})
+ \dim QL(\overline{M}) + \dim \overline{M} - 2t \cr
                  & \leq \dim QL(\overline{N}) + \dim QL(\overline{M}) + \dim
                  M.}$$
\qed
\endproof

As a special case of Theorem~2, we obtain

\begcorollary{1.} Let $M$ and $M'$ be two reduced \nd\ \q\
$\o$-modules with $K \otimes M \cong K \otimes M'$. Then
$M/{\nam m} M \cong M'/{\nam m} M'$. \qed
\endcorollary

Now let $\lambda \colon K \to L \cup \infty$ again be a place and
$\o$ its associated valuation ring. As before, we denote by
$\overline{\lambda}$ the field embedding $k
\hookrightarrow L$ determined by $\lambda$. By Corollary~1 it
makes sense to make the following definition:

\begdefinition{3.} Let $E$ be a space over $K$ which has \FR\
\wrtl. Let $M$ be a \rnd\ \qom\ with $E \cong K \otimes M$.
We call the \q\ $L$-module $L \otimes_{\lambda} M = L
\otimes_{\overline{\lambda}} (M/{\nam m} M)$ (which is uniquely
determined by $E$ up to isometry) the {\it specialization of $E$
\wrtl},\Idx{specialization (in the presence of FR)} 
and denote it by $\lambda_* (E)$.
\enddefinition

\begnote{} If $E$ has good reduction \wrtl, then $\lambda_* (E)$
has the old meaning.
\endnote

\noindent{\it Caution!\ } If $\Char L=2$, then  $\lambda_* (E)$
can be a degenerate \q\ $L$-module, even if $E$ is strictly
regular. However, this does not happen  when the field embedding
$\overline \lambda$ is separable, see \S10.

\begremark{} If $M$ is a \rnd\ \qom, then the \q\ $K$-module $K \otimes
M$ is definitely nondegenerate, for we know that
$QL(K \otimes M) = K \otimes QL(M)$, and if $QL(K \otimes M)$ were
isotropic, then $QL(M)$, and thus $QL(M/{\nam m}M)$, would also be
isotropic. Therefore our assumption in Definition~3, that the \q\
$K$-module $E$ is \nd, is a natural one and does not cause a loss
of generality.
\endremark

We illustrate the concept of fair reduction with a few examples.

\begexample{1.} Let $\o$ be a valuation ring with $\Char K=0$,
$\Char k=2$, and let $\lambda \colon K \to k \cup \infty$ be the
canonical place of $\o$. We want to lift an arbitrary space $S$
over $k$ to a space over $K$ by means of $\lambda$. We choose a
decomposition
$$S \cong \left[ {\hfill \alpha_1 \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill \beta_1 \hfill} \right] \perp \dots \perp
\left[ {\hfill \alpha_m \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill \beta_m \hfill} \right]  \perp [\gamma_1, \dots,
\gamma_r]$$
with $\alpha_i$, $\beta_i$, $\gamma_j \in k$.
$\{$Since the quasilinear part of $S$ is anisotropic, the elements
$\gamma_1, \dots, \gamma_r$ are linearly independent over
$k^2$.$\}$ Next we choose pre-images $a_i$, $b_i$, $c_j$ of the
elements $\alpha_i$, $\beta_i$, $\gamma_j$ in ${\nam o}$.
Suppose that $r=2s$ is even. We choose elements $t_1, \dots, t_s \in {\nam
m}$. Then the $K$-space
$$E \colon = \left[ {\hfill a_1 \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill b_1 \hfill} \right] \perp \dots \perp
\left[ {\hfill a_m \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill b_m \hfill} \right]
                      \perp \left[ {c_1 \atop t_1} \; {t_1
\atop c_2} \right] \perp \dots \perp \left[ {\hfill c_{2s-1}\hfill \atop
\hfill t_s \hfill} \; {\hfill t_s \hfill
\atop \hfill c_{2s} \hfill} \right]$$
clearly has \FR\ \wrtl\ and $\lambda_* (E) \cong S$. If $r = 2
s+1$ is odd, we again choose elements $t_1, \dots, t_s \in {\nam
m}$. This time the $K$-space
$$E \colon = \left[ {\hfill a_1 \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill b_1 \hfill} \right] \perp \dots \perp
\left[ {\hfill a_m \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill b_m \hfill} \right]
                      \perp \left[ {c_1 \atop t_1} \; {t_1
\atop c_2} \right] \perp \dots \perp \left[ {\hfill c_{2s-1}\hfill \atop
\hfill t_s \hfill} \; {\hfill t_s \hfill
\atop \hfill c_{2s} \hfill} \right] \perp [c_{2s+1} ]$$
has \FR\ \wrtl\ and $\lambda_* (E) \cong S$. In the special case
that all $t_i=0$, we obtain a $K$-space
$$F \colon =  \left[ {\hfill a_1 \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill b_1 \hfill} \right] \perp \dots \perp
\left[ {\hfill a_m \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill b_m \hfill} \right] \perp [c_1, \dots, c_r]$$
for every $r$. Since $\Char K=0$, we can interpret $F$ (and more
generally $E$) as a bilinear space,
$$F \cong
\left( {\hfill 2a_1 \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill 2b_1 \hfill} \right) \perp \dots \perp
\left( {\hfill 2a_m \hfill \atop \hfill 1 \hfill} \; {\hfill 1
\hfill \atop \hfill 2b_m \hfill} \right)
                 \perp \langle 2c_1, \dots, 2c_r \rangle.$$
The elements $a_1, \dots, a_m$ can always be chosen to be $\ne0$
and we obtain the diagonalization
$$F \cong \langle 2 a_1, d_1, \dots, 2 a_m, d_m, 2 c_1,
\dots, 2 c_r \rangle$$
with
$$d_i \colon = \frac{4 a_i b_i-1}{2 a_i} \quad (1
\leq i \leq r).$$
\endexample

\begexample{2.} Let $k$ be an imperfect field of
characteristic~$2$, $\o$ the power series ring $k[[t]]$ in one
variable $t$, and so $K=k((t))$. Choose $c \in k \setminus k^2$
and let $E$ be the space
$\left[ {1 \atop 1} \; {\hfill 1 \hfill \atop \hfill ct^{-2} \hfill} \right]$
over $K$.

\begclaim{} $E$ has \FR, but does not have \GR\ \wrt\ $\o$.
\endclaim

\begproof{} We have $E \cong K \otimes M$, where $M$ is the \qom\
$\left[ {1 \atop t} \; {t \atop c} \right]$. Its reduction $\overline{M} = M/{\nam m}
M$ is the space
$\left[ {1 \atop 0} \; {0 \atop c} \right] = [1, c]$ over $k$,
which is anisotropic. Therefore $M$ is \rnd\ and $E$ has \FR.

Suppose \ftsoc\ that $E \cong K \otimes N$ where $N$ is a \qs\
over $\o$. Since $E$ does not have a quasilinear part, the same is
true for $N$. Hence,
$N \cong \left[ {\alpha \atop 1} \; {1 \atop \beta} \right]$ with $\alpha,
\beta \in {\nam o}$. We then have
$\left[ {1 \atop 1} \; {\hfill 1 \hfill
\atop \hfill ct^{-2} \hfill} \right] \cong \left[ {\alpha \atop 1} \; {1
\atop \beta} \right]$ over $K$. An inspection of the
Arf-invariants shows that
$ct^{-2} = \alpha \beta + x^2 + x$ for some $x \in k ((t))$.
We must have $x\ne0$. We write $x = \sum\limits_{i \geq d} a_i
t^i$ for some $d\in {\Bbb Z}$ and $a_d\ne0$. Since $\alpha \beta \in {\nam
o}$, we have $d = -1$ and $c = a_{-1}^2$. This is a contradiction
since $c$ is not a square in $k$. We conclude that $E$ does not
have \GR.\qed
\endproof
\endexample

\begexample{3.} Again let $\o=k[[t]]$ and $k$ a field of
characteristic~$2$. In \S7 we determined that the space
$E \colon =
\left[ {1 \atop 1} \; {\hfill 1 \hfill
\atop \hfill t^{-1} \hfill} \right]$
does not have \GR\ \wrt\ $\o$. Could it be that  $E$ has at least
\FR\ \wrt\ $\o$?

Let us assume that this is so. Then there exists a \rnd\ \qom\ $M$
with  $E \cong K \otimes M$. We choose a decomposition $M = M_1 \perp M_2$
with $M_1$ strictly regular and $B(M_2 \times M_2) \subset {\nam m}$.
We already know that $M$ is degenerate, so $\dim M_2 >0$. Since
$\dim M=2$ and $\dim M_1$ is even, we must have $M_1=0$. Hence
$\overline{M}$ is quasilinear. But $M$ is not quasilinear, since
$E$ is not quasilinear. Therefore we have a representation
$$M \cong \left[ {\hfill a \hfill \atop \hfill t^nc \hfill} \quad {\hfill
t^n c \hfill
\atop \hfill b \hfill} \right]$$
with $n \in {\Bbb N}$ and units $a, b, c \in {\nam o}^*$.
The space $\overline{M} \cong [ \overline{a}, \overline{b} ]$ is
anisotropic and so the element $\overline{a} \overline{b}$ is not
a square in $k$. Over $K$ we have
$$\left[{1 \atop 1} \; {\hfill 1 \hfill \atop \hfill t^{-1} \hfill}
\right] \cong K \otimes M  \cong \langle a \rangle \otimes
\left[ {\hfill 1 \hfill \atop \hfill t^n ca^{-1} \hfill} \quad {\hfill
t^n ca^{-1} \hfill
\atop \hfill ba^{-1} \hfill} \right]
\cong \langle a \rangle \otimes \left[ {1 \atop 1} \quad {\hfill
1 \hfill
\atop \hfill abc^{-2} t^{-2n} \hfill} \right].$$
Comparing Arf-invariants shows
$$abc^{-2} t^{-2 n} = t^{-1} + x^2 + x \leqno(\ast)$$
with $x\in K$. Let $v \colon = K \to {\Bbb Z} \cup \infty$ be the
valuation associated to $\o$. From $(\ast)$ we get $v(x) < 0$ and
thus $v(x) = - n$. Therefore
$x = t^{-n} \sum\limits_{i = 0}^{\infty} x_i t^i$ with $x_i \in k$, $x_0 \ne
0$. Comparing the coefficients of $t^{-2n}$ on the left and
righthand side of ($\ast$) gives
$$\overline{a} \overline{b} \overline{c}^{-2} = x_0^2.$$
So $\overline{a} \overline{b}$ is a square in $k$ after all, a
contradiction. Therefore $E$ does not have~\FR.
\endexample

We continue with the general theory. So
$\lambda \colon K \to L \cup \infty$ is again an arbitrary place.
We immediately obtain a consequence of  Theorem~2 which is not
%even (???)
contained in the results of \S8 in the case of good
reduction.

\begcorollary{2.} Let $E$ and $F$ be spaces over $K$ which have \FR\
\wrtl\ and with $F<E$. Then $\lambda_* F < \lambda_* E$. \qed
\endcorollary

This corollary engenders a substitution principle for quadratic
forms, modeled on \S3, Theorem~5.\Idx{Substitution Principle (quadratic form)}\Idx{quadratic form!Substitution Principle}

\begtheorem{3.} Let $(g_{kl} (t))_{1 \leq k, l \leq m}$ and $(f_{ij}
(t))_{1^\leq i, j \leq n}$ be symmetric matrices whose
coefficients are polynomials in variables $t = (t_1, \dots, t_r)$
over a field $k$. Let $L \supset k$ be a field extension of $k$
and $c = (c_1, \dots, c_r)$ an $r$-tuple with coefficients in $L$.
Suppose that the quadratic forms $[g_{kl}
(t)]$ and $[f_{ij} (t)]$ over $k(t)$ satisfy $[g_{kl} (t)] < [f_{ij} (t)]$.
Suppose also that the quadratic forms $[g_{kl} (c)]$ and $[f_{ij} (c)]$ over
$L$ are \nd. Then  $[g_{kl} (c) ] < [f_{ij}
(c)]$ $($over $L$$)$.\qed
\endtheorem

We return to an arbitrary place $\lambda \colon K \to L \cup
\infty$.

\begtheorem{4.}{\rm (Extension of \S8, Lemma~2(b).)} Let $G$ be a
space and $F$ a strictly regular space over $K$. Suppose that $F$
has \GR\ \wrtl. Finally, let $E \colon = F \perp G$.

Claim: $E$ has \FR\ \wrtl\ if and only if $G$ has \FR\ \wrtl. In
this case we have $\lambda_* E \cong \lambda_* F \perp \lambda_* G$.
\endtheorem

\begproof{} We have $F = K \otimes M$ where $M$ is a strictly
regular \qom. If $G$ has \FR\ \wrtl, then $G = K
\otimes P$ where $P$ is a \rnd\ \qom. The \qom\ $N \colon = M
\perp P$ is then also \rnd\ and
$$L \otimes_{\lambda} N = (L
\otimes_{\lambda} M) \perp (L \otimes_{\lambda} P) = F \perp G =
E.$$
Hence $E$ has \FR\ \wrtl\ and $\lambda_* E \cong \lambda_* F \perp \lambda_*
G$.

Suppose now that $E$ has \FR\ \wrtl. We have $E \cong K
\otimes N$ where $N$ is a \rnd\ \qom. By Theorem~1, $K \otimes M< K \otimes
N$ implies that $N \cong M \perp P$ where $P$ is a \rnd\ \qom.
Hence
$$F \perp (K \otimes P) \cong E \cong F \perp G.$$
Now $G \cong K \otimes P$ by the Cancellation Theorem over $K$ (\S6, Theorem~2).
Therefore $G$ has \FR\ \wrtl.\qed
\endproof

To the currently developed theory of fair reduction,
we can associate a variation of the weak specialization theory of
\S7, which we
will briefly present in the following.

\begtheorem{5.}{\rm (Extension of \S6, Theorem~$5$.)} Let $\o$ be
quadratically henselian. Let $(M,q)$ be a \rnd\ and anisotropic
\qom. Furthermore, let $e$ be a primitive vector in $M$. Then
$q (e) \in {\nam o}^*$.
\endtheorem

\begproof{} We choose a decomposition $M = N \perp M'$ with $N$
strictly regular and $B(M' \times M') \subset {\nam m}$.
Traversing the proof of \S6, Theorem~5 and replacing $M^{\perp}$
by $M'$ everywhere will give proof since all arguments will faithfully
remain valid.\qed
\endproof

As before, ${\nam o}^h$ denotes the henselization of the valuation
ring $\o$.

\beglemma{4.} Let $M$ be a \qom. Then
$M^h \colon = {\nam o}^h \otimes_{{\nam o}} M$ is \rnd\ if and
only if $M$ is \rnd.
\endlemma

\begproof{} This is evident since ${\nam o}^h/{\nam m}^h = k$, so
that $M^h/{\nam m}^h M^h$ is canonically isomorphic to the
quadratic $k$-module $M/{\nam m} M$.\qed
\endproof

Now the road is clear to extend, in an appropriate way, the main result of~\S7
(Theorem~4), using fair reduction. As before,
let\break $\lambda \colon K \to L \cup \infty$ be a place with associated
valuation ring $\o$. Let $S$ be a system of representatives of
$Q(K)/Q({\nam o})$ in ${\nam o}$, as introduced in \S7.

\begdefinition{4.} Let $E = (E, q)$ be a \q\ $K$-module (always
free of finite rank). We say that $E$ is {\it weakly obedient
\wrtl}\Idx{weakly obedient quadratic module}\Idx{quadratic module!weakly obedient} 
(or: \wrt\ $\o$) when $E$ has a decomposition
$$E = \; \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; E_s \leqno(\ast)$$
such that the \q\ $K$-module $(E_s, s^{-1}
(q \vert E_s))$ has \FR\ \wrtl\ for every $s\in S$. Every
decomposition of the form $(\ast)$ is called a {\it weakly
$\lambda$-modular}\Idx{weakly l-modular decomposition (quadratic module)@weakly $\lambda$-modular decomposition (quadratic module)}
\Idx{quadratic module!weakly l-modular decomposition@weakly $\lambda$-modular decomposition}\Idx{weakly o-modular decomposition (quadratic module)@weakly $\o$-modular decomposition (quadratic module)}
\Idx{quadratic module!weakly o-modular decomposition@weakly $\o$-modular decomposition} 
(or: {\it weakly $\o$-modular}) {\it
decomposition} of $E$.
\enddefinition

\begremark{} Let $E$ be \wo\ \wrtl. Then
$$E \cong
\; \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; \langle s \rangle \otimes (K \otimes M_s)$$
with \rnd\ \qom s $M_s$. For every $s\in S$ we may choose a
decomposition $M_s = N_s \perp M'_s$ with  strictly regular $N_s$
and\break $B_s (M'_s \times M'_s) \subset {\nam m}$, where $B_s$ is the
bilinear form associated to $M_s$. Just as indicated in \S7, we
see that
$$G \colon = \; \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; \langle s \rangle \otimes (K \otimes M'_s)$$
is anisotropic. Furthermore,
$$F \colon = \; \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; \langle s \rangle \otimes (K \otimes N_s)$$
is strictly regular. We have $E
\cong F \perp G$. Therefore $E$ is definitely not degenerate and
is thus a space over $K$.\qed
\endremark

\begtheorem{6.} {\rm (Extension of \S7, Theorem~4.)} Let $E$ be a
space over $K$ which is \wo\ \wrtl\ and let
$$E = \; \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; E_s = \; \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; F_s$$
be two weakly $\lambda$-modular decompositions of $E$.
Then\fonote{See \S6, Definition~10, for the definition of Witt
equivalence $\sim$.}
$\lambda_*
(E_1) \sim \lambda_* (F_1)$.
\endtheorem

\begproof{} The arguments in the proof of \S7, Theorem~4, remain
valid in the current more general situation. One should use
Theorem~5 and Lemma~4 above.\qed
\endproof

Now the following definition makes sense:

\begdefinition{5.} Let $E$ be a \qs, \wo\ \wrtl\ and let
 $E = \; \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; E_s$ be a weakly $\lambda$-modular decomposition of $E$.
Then we call $\lambda_* (E_1)$ a {\it weak specialization of $E$
\wrtl}.\Idx{weak specialization (more general)}\Idx{111lw2@$\lambda_W$ (more general)} 
We denote the Witt class of $\lambda_* (E_1)$ by $\lambda_W (E)$.
In other words, $\lambda_W (E) \colon = \{ \lambda_* (E_1) \}
\in \widehat{W}\!q (L)$. By Theorem~6, $\lambda_W (E)$ is uniquely
determined by $E$ and $\lambda$.
\enddefinition

\begremark{} Our proof of Theorem~6 is independent of the main
result Theorem~2 of the specialization theory developed above. We
could also have deduced the important Corollary~1 of Theorem~2
(specialization by \FR\ is well-defined) from Theorem~6, analogous
to the proof of \S8, Theorem~1  in the quadratic case. In other
words, we could have established the weak specialization theory
first and then develop from this the basic idea of specialization by \FR\
(Definition~3 above), just as before in \S7 and \S8.
\endremark

In the theorems of \S8 about the weak specialization of tensor
products (Theorem~6(ii), Theorem~7) we may not simply replace the
word \GR\ by \FR\ and the word ``obedient'' by ``weakly
obedient''. This already shows us that in our theory, good reduction does not
become superfluous in any way after the introduction of fair
reduction.

More important even is the observation that in the theory of
generic splitting in \S9 and \S10 good reduction appears center
stage, and not fair reduction: For example, let $\vf$ be a regular
\qf\ over a field $k$ and let $L \supset k$ be a field extension.
Let $(K_r \; \vert \; 0 \leq r \leq h)$ be a generic splitting
tower of $\vf$. If $\lambda \colon K_r \to L \cup \infty$ is a
place over $k$ for some $r \in \{ 1, \dots, h \}$, then the kernel
form $\vf_r$ of $\varphi \otimes K_r$ automatically has {\it good}
reduction \wrtl.
