\titlea{\S7}{Weak Specialization}
\leftrunningtitle{Chapter I. Fundamentals of Specialization Theory}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S7 Weak Specialization}

In this section $\lambda \colon K \to L \cup \infty$ is a place,
${\nam o} = {\nam o}_{\lambda}$ is the valuation ring associated
to $\lambda$, with maximal ideal ${\nam m}$  and residue class field
$k = {\nam o}/{\nam m}$. Given a quadratic module $E = (E, q)$ over
$K$, we want to associate a quadratic module
$\lambda_* (E)$ over $L$ to it, insofar this is possible in a
meaningful way.

In order to do this, we limit ourselves to the case where $E$ is
nondegenerate (see \S6, Definition~3, but with $K$ instead of ${\nam
o}$). This is not a real limitation, however. Since $K$ is a
field, every quadratic module $E =(E, q)$ of finite dimension over
$K$ has the form $E \; \cong \; F \, \perp \, [0, \dots, 0] = F \,
\perp \, s \times [0]$ with $F$ nondegenerate (see \S 6).
So, when we have associated a specialization $\lambda_*(F)$ to $F$ in
a satisfactory way, it is natural to put $\lambda_* (E) =
\lambda_* (F) \, \perp \, s \times [0]$.
We stay with nondegenerate quadratic $K$-modules, however, and
now call them {\bf quadratic spaces} over $K$, as agreed
in \S6, Definition~9.

\begdefinition{1.} Let $E$ be a quadratic space over $K$. We say
that $E$ has {\it good reduction with respect to} $\lambda$ when
$E \; \cong \; K \otimes_{{\nam o}} M$, where $M$ is a
$\underline{\hbox{nondegenerate}}$ quadratic ${\nam o}$-module.
\enddefinition

This is the ``good case''. We would like to put $\lambda_* (E) \colon = L
\otimes_{\lambda} M$, where the tensor product is formed by means
of the homomorphism $\lambda \vert{\nam o} \colon {\nam o} \to
M$, see \S5, Definition~4. This tensor product can more easily be
described as follows: First we go from the quadratic ${\nam
o}$-module $M = (M,q)$ to the quadratic space
$M/{\nam m} M = (M/{\nam m} M, \overline{q})$ over
$k = {\nam o}/{\nam m}$, where $\overline{q}$ is obtained from $q \colon M \to
{\nam o}$ in the obvious way. Next we extend scalars by
means of the field extension $\overline{\lambda} \colon k \hookrightarrow
L$, determined by $\lambda$, thus obtaining $\lambda_* (E) = L
\otimes_k (M/{\nam m} M)$.

Is this meaningful? With a similar argument as towards the end
of \S 3 (second proof of Theorem~2), we now show that the answer
is affirmative when $E$ is {\it strictly regular} (which means
that $E^{\perp} = 0$, cf. \S 6, Definition~3).

\begtheorem{1.} Let $E$ be strictly regular and let $M$, $M'$ be
nondegenerate quadratic ${\nam o}$-modules with $E \; \cong \; K \otimes_{{\nam o}} M \;
\cong \; K \otimes_{{\nam o}} M'$. Then $M \; \cong \; M'$, and hence $M/{\nam
m} M \; \cong \; M'/{\nam m} M'$.
\endtheorem

\begproof{} Clearly $M^{\perp} = 0$ and $(M')^{\perp} = 0$.
Hence $M$ and $M'$ are also strictly regular. By the theorem at
the end of \S5, the space
$K \otimes_{{\nam o}} [M' \, \perp \, (-M)] \; \cong \; E \, \perp \, (-
E)$ is hyperbolic. So $M' \, \perp \, (-M)$ is itself hyperbolic,
by \S6, Theorem~4. $M \, \perp \, (- M)$ is also hyperbolic, again
by the theorem at the end of \S5. The modules $M' \, \perp \, (-M)$ and $M \, \perp \,
(-M)$ have the same rank. Hence $M' \, \perp \, (- M) \; \cong \; M
\, \perp \, (- M)$, and so $M' \; \cong \; M$ by the
Cancellation Theorem (\S6, Theorem~2).\qed
\endproof

The problem remains to see that $\lambda_* (E)$ does not depend on
the choice of the module $M$, when $E$ has good reduction with
respect to $\lambda$, but only is nondegenerate. Furthermore,
in case of bad reduction (i.e. not good reduction), we would nevertheless
like to associate  a Witt class $\lambda_W (\{ E
\})$ over $L$  to $E$ in a meaningful way, just as for bilinear forms in
\S3 (``weak specialization''). Neither problem can be dealt with
as in \S3, in the first place because the Witt classes do not constitute an
additive group this time. Thus we have to seek out a new path.

\beglemma{1.}
Let ${\nam o}$ be quadratically henselian $($see $\S6$,
Definition~$12)$ and let $V= (V, q)$ be an {\it anisotropic} quadratic module
over $K$.
\smallskip
\setitemindent{(a)}
\item{$(a)$} The sets
$$\mu (V) \colon = \{ x \in V \; \vert \; q (x) \in {\nam o} \} \quad
\text{and} \quad \mu_+ (V) \colon = \{ x \in V \; \vert \; q (x) \in {\nam m}
\}$$
are ${\nam o}$-submodules of $V$.
\smallskip
\item{$(b)$} For any $x \in \mu (V)$ and $y \in \mu_+ (V)$, we have $q(x+y) - q(x) \in {\nam
m}$ and $B_q (x, y) \in {\nam m}$.
\endlemma

\begproof{}
(a) Let $x \in \mu (V)$ (resp. $x \in \mu_+ (V)$) and $\lambda
\in {\nam o}$, then clearly $\lambda x \in \mu (V)$ (resp. $\mu_+ (V)$).
So we only have to show that for any $x, y \in \mu (V)$ (resp. $\mu_+ (V)$),
we have $x + y \in \mu (V)$ (resp. $\mu_+ (V)$). Let $B \colon = B_q$.

Let $x \in \mu (V)$, $y \in \mu (V)$. Suppose for the sake of
contradiction that $x + y \not\in \mu (V)$, i.e.
$q(x+y) = q(x) + q(y) + B (x, y) \not\in {\nam o}$. Since
$q(x) \in {\nam o}$, $q(y) \in {\nam o}$, we must have that $B(x, y) \not\in {\nam
o}$. Hence $\lambda \colon = B(x, y)^{-1} \in {\nam m}$. The
vectors $x, \lambda y$ give rise to the value matrix
$\left[ {q(x) \atop 1} \; {1 \atop \lambda^{2} q(y)} \right]$.
According to \S 6, Lemma~6, the space $K x + Ky = Kx + K \lambda
y$ is isotropic. Contradiction, since $V$ is anisotropic.
Therefore $x + y \in \mu (V)$.

Next, let $x \in \mu_+ (V)$, $y \in \mu_+ (V)$. Suppose again for
the sake of contradiction that $x + y \not\in
\mu_+ (V)$. We just showed that $q(x+y) \in {\nam o}$. By our
assumption $q(x+y) \not\in {\nam m}$, so that $q(x+y) \in {\nam o}^*$.
Since $q(x) \in {\nam m}$, $q(y) \in {\nam m}$, we have $B(x, y) \in {\nam o}^*$.
Write $B(x, y) = \lambda^{-1}$ with $\lambda \in {\nam o}^*$.
Again, the vectors $x, \lambda y$ give rise to the value matrix
$\left[ {q(x) \atop 1} \; {1 \atop \lambda^{2} q(y)} \right]$.
Just as before, the space $Kx + Ky$ is isotropic, according to
\S6, Lemma~6. Contradiction! We conclude that
$x + y \in \mu_+ (V)$.

(b) Suppose for the sake of contradiction that there exist vectors
$x \in \mu (V)$, $y \in \mu_+ (V)$ with $q(x+y) - q(x) \not\in {\nam
m}$. We showed above that $q(x+y) \in {\nam o}$. Hence
$q(x+y) - q(x) = q(y) + B(x, y) \in {\nam o}$. By assumption,
this element doesn't live in ${\nam m}$ and is thus a unit. Since $q(y) \in
{\nam m}$, we also have $B(x, y) \in {\nam o}^*$. As before, we
write $\lambda =B(x, y)^{-1}$ and observe that $Kx + Ky$ is
isotropic, giving a contradiction. Therefore $q(x+y) - q(x) \in {\nam
m}$ and $B(x, y) = q(x+y) - q(x) - q(y) \in {\nam m}$ as well.\qed
\endproof


We remain in the situation of Lemma~1.
For $x \in \mu (V)$, $\lambda \in{\nam m}$, we have $\lambda x \in \mu_+ (V)$.
Therefore
$$\rho (V) \colon = \mu (V)/\mu_+ (V)$$
is a $k$-vector space in a natural sense $(k = {\nam o}/{\nam m})$.
We define a function $\overline{q} \colon \rho (V) \to k$ as
follows:
$$\overline{q} (\overline{x}) \colon = \overline{q(x)} \qquad (x \in \mu (V)),$$
where $\overline{x}$ denotes the image of $x \in \mu (V)$ in $\rho
(V)$ and $\overline{a}$ denotes the image of $a \in {\nam o}$ in
$k$. Lemma~1 tells us that the map $\overline{q}$ is well-defined.


For $x \in \mu (V)$, $a \in {\nam o}$, we have $\overline{q} (\overline{a}\, \overline{x}) =
\overline{q} (\overline{ax}) = \overline{q(ax)} = \overline{a}^2
q \overline{(x)} = \overline{a}^2 \overline{q} (\overline{x})$.
According to Lemma~1, the bilinear form
$B \colon = B_q$ has values in ${\nam o}$ on $\mu (V) \times \mu
(V)$ and values in ${\nam m}$ on $\mu (V) \times \mu_+ (V)$.
Therefore, it induces on $\rho (V)$ a symmetric bilinear form
$\overline{B}$ over $k$ with $\overline{B} (\overline{x}, \overline{y}) =
\overline{B(x,y)}$ for $x, y \in \mu (V)$. A very simple
calculation now shows that
$$\overline{q} (\overline{x} + \overline{y}) - \overline{q} (\overline{x}) -
\overline{q} (\overline{y}) = \overline{B} (\overline{x}, \overline{y})$$
for $x, y \in \mu (V)$. This furnishes the proof that
$\overline{q}$ is  a quadratic form on the $k$-vector space
$\rho (V)$  with
$B_{\overline{q}} = \overline{B}$.
The quadratic $k$-module $(\rho (V), \overline{q})$ is clearly
anisotropic.

\begdefinition{2.} (${\nam o}$ quadratically henselian.) We call
the quadratic $k$-module $\rho (V) = (\rho (V), \overline{q})$ the
{\it reduction} of the anisotropic quadratic $K$-module $V$ \wrt\ the
valuation ring ${\nam o}$.
\enddefinition

In order to associate to a quadratic space $E$ over $K$, by means
of the place $\lambda \colon K \to L \cup \infty$, a Witt class
$\lambda_W \{ E \}$ over $L$, the following path presents itself
now: Let ${\nam o}^h$ be the henselization of the valuation ring
${\nam o} = {\nam o}_{{\lambda}}$, ${\nam m}^h$ the maximal ideal of
${\nam o}^h$ and $K^h$ the quotient field
of ${\nam o}^h$. The residue class field
${\nam o}^h/{\nam m}^h$ is canonically isomorphic to
$k = {\nam o}/{\nam m}$ and will be identified with $k$.
Let
$$\lambda_W \{ E \} = \{ L \otimes_{\overline{\lambda}} \rho (\Ker (K^h
\otimes E))\},$$
where, as before,
$\overline{\lambda} \colon k \hookrightarrow L$ is the field
embedding, determined by $\lambda$. The space over $L$ on the
righthand side can then be considered to be a ``weak
specialization'' of $E$ with respect to
$\lambda$.

All good and well, if only we knew whether the vector space
$$\rho (\Ker (K^h \otimes E))$$
has finite dimension! To guarantee
this, we have to confine the class of allowed quadratic modules
$E$.

In the following, $v \colon K \to \Gamma \cup \infty$ is a
surjective valuation, associated to the valuation ring ${\nam o}$,
thus with $\Gamma$ the value group of $v$, $\Gamma \; \cong \; K^* /{\nam
o}^*$. We use additive notation for $\Gamma$ (so $v(xy) = v(x) + v(y)$
for $x, y \in K$).

Already in \S3, we agreed to view the square class group
$Q({\nam o})$ as a subgroup of $Q(K)$, and also to regard the elements of
$Q(K)$ as one-dimensional  bilinear spaces\fonote{Strictly speaking,
square classes are {\it isomorphism
classes} of one-dimensional spaces.} over $K$.

Now we choose a complement $\Sigma$ of $Q({\nam o})$ in
$Q(K)$, in other words, a subgroup $\Sigma$ of $Q(K)$ with $Q(K) = Q({\nam o})
\times \Sigma$. This is possible, since $Q(K)$ is elementary
abelian of exponent~$2$, i.e. a vector space over the field with
$2$ elements. Further, we choose, for every square class $\sigma \in
\Sigma$, an element $s \in {\nam o}$ with $\sigma = \langle s
\rangle$. For $\sigma = 1$ we choose the representative $s = 1$. Let
$S$ be the set of all elements $s$. For every $a \in K^*$ there
exists exactly one $s \in S$ and elements $\varepsilon \in {\nam o}^*$,
$b \in K^*$ with $a = s \varepsilon b^2$. Since $K^*/{\nam o}^* \; \cong \;
\Gamma$, it is clear that $S$ (resp. $\Sigma$) is a system of
representatives of $\Gamma /2 \Gamma$ in $K^*$ (resp. $Q(K)$) for the
homomorphism from $K^*$ to $\Gamma/2
\Gamma$ (resp. from $Q(K)$ to $\Gamma/2 \Gamma$), determined by
$v \colon K^* \to \Gamma$.

\begdefinition{3.} A quadratic space $E$ over $K$ is called {\it
obedient with respect to} $\lambda$  (or {\it obedient with respect
to} ${\nam o}$) if there exists a decomposition
$$E \; \cong \;
\raise6pt\hbox{$\newbot\limits_{s \in S}$}\;
\langle s \rangle \otimes (K \otimes_{{\nam o}} M_s),\leqno(\ast)$$
where $M_s$ is a nondegenerate quadratic ${\nam o}$-module.
(Of course  $M_s \ne 0$ for finitely many $s$ only.)
\enddefinition

Clearly this property does not depend on the choice of system of
representatives $S$.

\begremarks{}
\smallskip

\setitemindent{(5)}
\item{(1)} Obedience is much weaker than the property ``good
reduction'' (Definition~1).
\smallskip

\item{(2)} Let $E$ and $F$ be quadratic spaces over $K$, obedient
with respect to ${\nam o}$. If at least one of them is strictly
regular, then $E \, \perp \, F$ is obedient.
\smallskip

\item{(3)} If $\Char K\not=2$, then every quadratic space over $K$
is strictly regular. Hence the orthogonal sum of two obedient
quadratic spaces over $K$ is again obedient.
\smallskip

\item{(4)} If $2 \in {\nam o}^*$, i.e. $\Char L \ne 2$, then {\it
every} quadratic space $E$ over $K$ is obedient with respect to
$\lambda$, since $K$ also has characteristic $\not=2$ in this
case. $E$ has a decomposition in one-dimensional spaces, which are
clearly obedient.
\smallskip

\item{(5)} If $\Char K = 0$, but $\Char L = 2$, then every
quadratic space of odd dimension over $K$ is {\it disobedient}
($=$ not obedient) with respect to $\lambda$. To see this, let $E$
be an obedient quadratic space with orthogonal decomposition
($\ast$), as in Definition~3. Then the space $K \otimes_{{\nam o}}
M_i$ has quasilinear part $(K \otimes_{{\nam o}} M_i)^{\perp} = 0$
for every $i \in \{1, \dots, r \}$. Therefore $M_i$ also has
quasilinear part $M_i^{\perp} = 0$, and is thus strictly regular.
Hence the space $M_i/{\nam m} M_i$ over $k = {\nam o}/{\nam m}$ is
strictly regular, and so must have even dimension. We conclude
that $E$ must have even dimension.
\endremarks

Let us now give an example of a two-dimensional disobedient space
over a field $K$ of characteristic~$2$.

First, recall the definition of the {\it Arf-invariant} $\Arf(\varphi)$
of a strictly regular form $\varphi$ over a field $k$ of
characteristic~$2$. For $x\in k$, let $\wp(x):=x^2+x$.  Further,
let $\wp(k)$ denote the set of all $\wp(x)$, $x\in k$. This is a
subgroup of $k^+$, i.e. $k$ regarded as an additive group.
We choose a decomposition
$$\varphi \cong \left[ {a_1 \atop 1} \quad {1 \atop b_1}\right]
\perp \ldots \perp \left[ {a_m \atop 1} \quad {1 \atop b_m}\right]
\leqno(\ast)$$
and set
$$\Arf(\varphi):= a_1b_1 + \cdots +a_mb_m + \wp(k) \in
k^+/\wp(k).$$
It is well-known that $\Arf(\varphi)$ is independent of the choice
of the decomposition~$(*)$ ([A], [S, Chap.IX. \S4]).

\begexample{} Let $k$ be a field of characteristic~$2$ and
$K=k(t)$ the rational function field in one variable $t$ over $k$.
Further, let $\o$ be the discrete valuation ring of $K$ with
respect to the prime polynomial $t$ in $k[t]$, i.e.
$\o=k[t]_{(t)}$.

\begclaim{} The quadratic space $\left[ {1 \atop 1} \; {1 \atop t^{-1}}
\right]$ over $K$ is disobedient with respect to $\o$.
\endclaim

\begproof{} Suppose for the sake of contradiction that this space
is obedient. Then there exist a strictly regular space $M$ over
$\o$ and an element $u \in K^*$ with $\left[ {1 \atop 1} \; {1 \atop t^{-1}}
\right] \; \cong \; \langle u \rangle \otimes (K \otimes_{{\nam o}} M)$.
By \S6, Theorem~1, we have $M \; \cong \; \left[ {\alpha \atop 1} \; {1 \atop \beta}
\right]$ with $\alpha, \beta \in {\nam o}$. Hence, by \S6,
Lemma~2, (3), the following holds over $K$:
$$\left[ {1 \atop 1} \quad {1 \atop t^{-1}}
\right] \; \cong \; \langle \alpha u \rangle \left[ {1 \atop 1} \quad {1 \atop
\alpha\beta} \right].\leqno(1)$$
Comparing Arf-invariants of both sides, shows that there exists an
element $a \in K$  with
$$t^{-1} = \alpha \beta + a^2 + a. \leqno(2)$$
Hence there exists an element $a \in K$ with $t^{-1} + a^2 + a \in {\nam o}$.

If we now move to the formal power series field $k((t)) \supset
K$, we easily see that such an element $a$ does not exist: For if
$a = \sum\limits_{r \geq d} c_r t^r$ with
$d \in {\Bbb Z}$ and coefficients $c_r \in k$, $c_d \ne 0$, then
$${\textstyle \sum\limits_{r \geq d}}
c_r^2 t^{2r} + {\textstyle \sum\limits_{r \geq d}} c_r t^r + t^{-1} \in
k[[t]],$$
and so $d < 0$. The term of lowest degree, in the first sum on the
left, is $c_d^2 t^{2d}$. It cannot be compensated by other
summands on the left. Therefore $t^{-1} + a^2 + a \not\in {\nam
o}$, and the space $\left[ {1 \atop 1} \; {1 \atop t^{-1}}
\right]$ is disobedient.\qed
\endproof

In this proof we used the Arf-invariant. We remark that we can
use easier aids: Anisotropic quadratic forms like $\left[ {1 \atop 1} \; {1 \atop c}
\right]$ are norm forms of separable quadratic field extensions,
and are as such ``multiplicative''. Therefore, we can deduce
immediately from (1) that $\left[ {1 \atop 1} \; {1 \atop t^{-1}}
\right] \; \cong \; \left[ {1 \atop 1} \; {1 \atop \alpha \beta}
\right]$ and then, that there exists a relation~(2).
\endexample

\begremark{} Hitherto the quasilinear parts of free quadratic
modules played a predominantly negative role. This will continue to be the
case. Rather often they cause a lot of complications, compared to the theory of
nondegenerate bilinear modules. But sometimes quasilinear quadratic
modules can do good things. Suppose again that $K = k(t)$, ${\nam o} =
k[t]_{(t)}$ as in the example above. The space $\left[ {1 \atop 1} \; {1 \atop
t^{-1}} \right]$ is disobedient with respect to $\o$. However,
$\left[ {1 \atop 1} \; {1 \atop
t^{-1}} \right] \, \perp \, \left[ t^{-1} \right]$ is obedient,
since this space over $K$ is isometric to $\left[ {1 \atop 1} \; {1 \atop
0} \right] \, \perp \, \left[ t^{-1} \right]$, and hence to
$\left[ {0 \atop 1} \; {1 \atop
0} \right] \, \perp \, \left[ t \right]$.
\endremark

In connection with the definition of obedience (Definition~3), we agree
upon some more terminology. If $E = (E, q)$ is a quadratic space,
obedient with respect to $\o$, then we have an (internal)
orthogonal decomposition
$$E = \raise6pt\hbox{$\newbot\limits_{s \in S}$}\; E_s$$
of the following kind: Every vector space $E_s$ contains a free
$\o$-submodule $M_s$ with $E_s = KM_s$, such that $q \vert{E_s} = s
q_s$ for a quadratic form $q_s \colon E_s \to K$, which takes
values in $\o$ on $M_s$ and for which $(M_s, q_s \vert{M_s})$ is a
nondegenerate quadratic module. (Of course $E_s \ne 0$ for finitely
many $s\in S$ only.)

\begdefinition{4.} We call such a decomposition $E =
\raise6pt\hbox{$\newbot\limits_{s \in S}$}\; E_s$,
together with a choice of modules $M_s \subset E_s$, a
$\lambda$-{\it modular}, or also, an $\o$-{\it modular
representation} of the obedient quadratic space $E$.
\enddefinition

\beglemma{2.} Suppose again that $\o$ is quadratically henselian
and  $E=(E,q)$ is a quadratic space over $K$, obedient with
respect to $\o$. Let $E =
\raise6pt\hbox{$\newbot\limits_{s \in S}$}\; E_s$, $E_s = KM_s$ be
an $\o$-modular representation of $E$. Suppose furthermore that
$E$ is anisotropic. Then we have, with the notation of
Lemma~$1$,\fonote{In accordance with our earlier agreement, every
$E_s$ is considered as a quadratic subspace of $E$, $E_s = (E_s, q \vert{E_s})$.}
$$\eqalign{\mu (E) = M_1 \, & \perp \, \raise6pt\hbox{$\newbot\limits_{s \ne
1}$}\; \mu_+ (E_s), \cr
           \mu_+ (E) = {\nam m} M_1 \, & \perp \,
           \raise6pt\hbox{$\newbot\limits_{s \ne 1}$}\;
           \mu_+ (E_s).}$$
\endlemma

\begproof{} As above, we set $q \vert{E_s} = s q_s$. Let $x$ be a
vector of $E$ with $x\not=0$. Then there exist finitely many
pairwise distinct elements $s_1, \dots, s_r$ in  $S$, as well as
{\it primitive} vectors $x_i \in M_i \colon = M_{s_i}$ and scalars
$a_i \in K^*
\quad (1 \leq i \leq r)$, such that $x = \sum\limits_{i = 1}^r a_i
x_i$. We have
$$q(x) = \sum\limits_{i = 1}^r a_i^2 s_i q_i (x_i),$$
where we used the abbreviation $q_i \colon = q_{s_i}$. According
to \S6, Theorem~5, we have $q_i (x_i) \in {\nam o}^*$ for every
$i \in \{ 1, \dots, r \}$. Hence, for $i \ne j$, the values
$v(a_i^2 s_i q_i (x_i)) = v(a_i^2 s_i)$
and $v(a_j^2 s_j q_j (x_j)) = v(a_j^2 s_j)$ are different. Let $v(a_k^2
s_k)$ be the smallest value among the $v(a_i^2 s_i)$, $1 \leq i \leq
r$. Then $v(q(x)) = v(a^2_k s_k) = v (q(a_k x_k))$. This shows
that
$$\mu (E) =
\raise6pt\hbox{$\newbot\limits_{s \in
S}$}\; \mu (E_s), \quad \mu_+ (E) = \raise6pt\hbox{$\newbot\limits_{s \in
S}$}\;
\mu_+ (E_s).$$
For $s \ne 1$, $a \in K^*$ and primitive $y \in E_s$, we have
$q(ay) = sa^2 q_s (y) \not\in {\nam o}^*$.
Therefore, $\mu (E_s) = \mu_+ (E_s)$ for every $s \in S \setminus \{ 1 \}$.
We still have to determine the modules
$\mu (E_1)$ and $\mu_+ (E_1)$. So, let
$x \in E_1$, $x \ne 0$. We write
$x = ay$ with $a \in K^*$ and primitive $y \in M_1$. Then $q(x)
= a^2 q(y)$ and $q(y) \in {\nam o}^*$. Hence $q(x) \in {\nam o}$
exactly when $a \in {\nam o}$, and $q(x) \in {\nam m}$ exactly when $a \in
{\nam m}$. Therefore we conclude that
$\mu (E_1) = M_1$ and $\mu_+ (E_1) = {\nam m} M_1$.\qed
\endproof

An immediate consequence of the lemma is

\begtheorem{2.} Let $\o$ be quadratically henselian, and let $E = (E,
q)$ be an anisotropic quadratic space over $K$, obedient with
respect to $\o$. Let $E = \raise6pt\hbox{$\newbot\limits_{s \in
S}$}\; E_s$, $E_s = KM_s$ be an $\o$-modular representation of
$E$. Then we have for the reduction $\rho (E) =
(\rho (E), \overline{q})$ of $E$ \wrt\  ${\nam o}$:
$$(\rho (E), \overline{q}) \; \cong \; (M_1/{\nam m} M_1, \overline{q}_1),$$
where $\overline{q}_1 \colon M_1/{\nam m} M_1 \to k$ is the
quadratic form over $k$, determined by\break $q \vert{M_1}
\colon M_1 \to {\nam o}$ in the obvious way.\qed
\endtheorem

Furthermore, the lemma tells us that
$M_1 = \mu (E_1)$ and ${\nam m} M_1 = \mu_+ (E_1)$. In particular
we have

\begtheorem{3.} Let $\o$ be quadratically henselian and let $E = (E,
q)$ be an anisotropic quadratic space over $K$,  having good reduction
with respect to $\o$. Then $\mu (E) = (\mu (E), q \vert{\mu
(E)})$ is a nondegenerate quadratic $\o$-module with $E = K \otimes_{{\nam o}} \mu
(E)$, and $\mu_+ (E) = {\nam m} \mu (E)$. Note that $\mu (E)$ is
$\underline{\hbox{the only}}$  nondegenerate quadratic $\o$-module $M$ in $E$, with
$E = KM$.\qed
\endtheorem

We tone down Definition~4 as follows:

\begdefinition{5.} Let $E = (E, q)$ be a quadratic space over $K$,
obedient with respect to $\lambda$. A $\lambda$-{\it modular} (or
$\o$-{\it modular}) {\it decomposition of} $E$ is an orthogonal
decomposition $E = \raise6pt\hbox{$\newbot\limits_{s \in
S}$}\; E_s$, in which every space $\langle s \rangle \otimes E_s = (E_s, s
\cdot (q \vert{E_s}))$ has good reduction with respect to
$\lambda$.
\enddefinition

If $\o$ is \qh\ and $E$ is anisotropic then, according to
Theorem~3, every $\lambda$-modular decomposition corresponds to
{\it exactly one} $\lambda$-modular representation of $E$.

\beglemma{3.} Let $\o$ be \qh. Let $s_1, \dots, s_r$ be different
elements of $S$ and $M_1, \dots, M_r$ anisotropic nondegenerate
quadratic modules over $\o$. Then
$$E \colon = \raise5pt\hbox{$\newbot\limits_{{\scriptstyle i = 1}}^{\lower6pt\hbox{${\scriptstyle
r}$}}$} \langle s_i \rangle \otimes (K \otimes_{{\nam o}} M_i)$$
is an anisotropic quadratic space over $K$.
\endlemma

\begproof{} Let a primitive vector $x_i \in M_i$ be given for
every $i \in \{ 1, \dots, r \}$. Also, let $a_1, \dots, a_r \in K$
be scalars with
$$\sum\limits_{i = 1}^r s_i a_i^2 q_i (x_i) = 0, \leqno(\dagger)$$
where $q_i$ denotes the quadratic form on $M_i$. We must show that
all $a_i=0$.

Suppose \ftsoc\ that this is not so. After renumbering the $M_i$,
we can assume \wlog\ that for a certain
$t \in \{ 1, \dots, r \}$, $a_i \ne 0$ for $1 \leq i \leq t$
and $a_i = 0$ for $t < i \leq r$. By \S6, Theorem~5, $q(x_i) \in
{\nam o}^*$ for every $i \in \{ 1, \dots, t \}$. The values
$v (s_i a_i^2 q_i(x)) = v(s_i a_i^2)$ with $1 \leq i \leq t$ are pairwise
different. Since this contradicts equation~($\dagger$), all $a_i =
0$.\qed
\endproof

We arrive at the main theorem of this chapter.

\begtheorem{4.} Let $E$ be a quadratic
space over $K$, obedient \wrt\ $\o$. Let
$$E
= \raise6pt\hbox{$\newbot\limits_{s \in
S}$}\; E_s =
\raise6pt\hbox{$\newbot\limits_{s \in
S}$}\; F_s$$
be two $\o$-modular decompositions of $E$, and also let $M_1$,
$N_1$ be
nondegenerate quadratic modules with $E_1
\cong K \otimes_{{\nam o}} M_1$, $F_1 \cong K \otimes_{{\nam o}}
N_1$. Then the quadratic spaces $M_1/{\nam m} M_1$ and $N_1 /{\nam m}
N_1$ over $k = {\nam o}/{\nam m}$ are Witt equivalent.
\endtheorem

\begproof{} (a) For every $s \in S \setminus \{ 1 \}$ we choose
nondegenerate quadratic $\o$-modules $M_s$, $N_s$ with $E_s \; \cong \;
\langle s \rangle \otimes (K \otimes_{{\nam o}} M_s)$, $F_s \; \cong \;
\langle s \rangle \otimes (K \otimes_{{\nam o}} N_s)$.
Suppose first that $\o$ is quadratically henselian. In
accordance with \S6, Theorem~3, we choose Witt decompositions
$$M_s = M_s^{\circ} \, \perp \, i_s \times {\textstyle \left[ {0 \atop 1} \; {1 \atop
0} \right]}, \quad N_s = N_s^{\circ} \, \perp \, j_s \times {\textstyle \left[
{0 \atop 1} \; {1 \atop
0} \right]},$$
with $M_s^{\circ}$, $N_s^{\circ}$ anisotropic nondegenerate
quadratic $\o$-modules. By Lemma~3 above, the nondegenerate
quadratic $K$-modules
$$U \colon = \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \langle s \rangle \otimes (K \otimes_{{\nam o}} M^{\circ}_s), \quad V \colon =
\raise6pt\hbox{$\newbot\limits_{s \in
S}$} \langle s \rangle \otimes (K \otimes_{{\nam o}} N^{\circ}_s)$$
are anisotropic. Now,
$$E \; \cong \; U \, \perp \, \left(\raise2pt\hbox{${\textstyle
\sum\limits_{s \in S}}$} i_s \right) \times
{\textstyle \left[ {0 \atop 1} \; {1 \atop 0} \right]} \; \cong \; V \, \perp \, \left(
\raise2pt\hbox{${\textstyle \sum\limits_{s \in S}}$} j_s \right) \times {\textstyle \left[ {0
\atop 1} \; {1 \atop 0}
\right]}.$$
Hence $U \; \cong \; V$. According to Theorem~2, we have $\rho (U) \; \cong \;
M_1^{\circ}/{\nam m} M_1^{\circ}$, $\rho (V) \; \cong \; N_1^{\circ} /{\nam
m} N_1^{\circ}$. Therefore, $M_1^{\circ} /{\nam m} M_1^{\circ} \; \cong \;
N_1^{\circ} /{\nam m} N_1^{\circ}$. Furthermore,
$$M_1 /{\nam m} M_1 \; \cong \; M_1^{\circ} /{\nam m} M_1^{\circ} \; \perp \,
i_1 \times {\textstyle \left[ {0 \atop 1} \; {1 \atop 0}
\right]}, \; N_1 /{\nam m} N_1 \; \cong \; N_1^{\circ} /{\nam m} N_1^{\circ}
\, \perp \, j_1 \times {\textstyle \left[ {0 \atop 1} \; {1 \atop 0}
\right]}.$$
We conclude that
$M_1 /{\nam m} M_1 \sim N_1 /{\nam m} N_1$.

(b) Suppose next that $\o$ is arbitrary. We go from $(K, {\nam
o})$ to the henselization $(K^h, {\nam o}^h)$. (By definition,
$K^h$ is the quotient field of the henselian valuation ring ${\nam
o}^h$.) As is well-known, the valuation $v \colon K \to \Gamma \cup \infty$
extends uniquely to a valuation $v^h \colon K^h \to \Gamma \cup
\infty$, which therefore again has  $\Gamma$ as value group. Also,
${\nam o}^h$ is the valuation ring of $v^h$. Consequently, $S$ is
also a system of representatives of $Q(K^h)/Q({\nam o}^h) \;
\cong \; \Gamma/2 \Gamma$.

Given a free quadratic module $M$ over $\o$, we set -- as
before -- $M^h \colon = {\nam o}^h \otimes_{{\nam o}} M$ and for a
free quadratic module $U$ over $K$, we set \hbox{$U^h \colon = K^h \otimes_{K}
U$.} By \S6, Lemma~7, the quadratic ${\nam o}^h$-modules $M_s^h$,
$N_s^h$ are nondegenerate. Furthermore,
$E^h = \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; E_s^h = \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; F_s^h$ and $E_s^h \; \cong \; K^h \otimes_{{\nam o}^h} M_s^h, \; F_s^h \; \cong
\; K^h \otimes_{{\nam o}^h} N_s^h$ for every $s \in S$.

By above, the quadratic spaces $M_1^h/{\nam m}^h
M_1^h$ and $N_1^h/{\nam m}^h N_1^h$ over $k$ are equivalent. However, these
spaces can be canonically identified with $M_1/{\nam m} M_1$ and $N_1/{\nam m}
N_1$ (cf. the end of the proof of \S6, Lemma~7). We
conclude that $M_1 /{\nam m} M_1
\sim N_1/{\nam m} N_1$.\qed
\endproof

\begdefinition{6.} Let $E$ be a quadratic space over $K$, obedient
\wrt~$\lambda$. If $E = \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; E_s$ is a $\lambda$-modular decomposition of $E$ and $M_1$
a nondegenerate quadratic $\o$-module with $E_1 \; \cong \; K
\otimes_{{\nam o}} M_1$, then we call the quadratic space $L
\otimes_{{\lambda}} M_1$ a {\it weak specialization of $E$ \wrt\
$\lambda$}. (As before, $\otimes_{{\lambda}}$ denotes a
base extension  \wrt\ the homomorphism $\lambda \vert{\nam o} \colon {\nam
o} \to L$.)
By Theorem~4, 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.
$$\lambda_W (E) \colon = \{ L \otimes_{\lambda} M_1 \}.$$
(``W'' as in ``Witt'' or ``weak''.)
\enddefinition

If $\Char K=2$, then $M_1/{\nam m} M_1$ is a quadratic space over
$k$, to be sure. Nevertheless $L \otimes_{\lambda}
M_1 = L \otimes_{\overline{\lambda}} (M_1/{\nam m} M_1)$ can be
degenerate. In this case, we only have $\lambda_W (E) \in \widehat{W}\!q (L)$
(see \S 6, from Definition~10 onwards). If $\Char K\not=2$, this
cannot happen since $M_1/{\nam m} M_1$ is strictly regular in this
case. So now, $\lambda_W (E) \in W\!q (L)$.

\begtheorem{5.} If $E$ and $E'$ are quadratic spaces over $K$,
obedient \wrt\ $\lambda$, and if $E \sim E'$, then $\lambda_W (E) =
\lambda_W (E')$.
\endtheorem

\begproof{} Suppose \wlog\ that $\dim E \leq \dim E'$. Then $E' \; \cong \; E \,
\perp \, r \times \left[ {0 \atop 1} \; {1 \atop 0} \right]$ for a certain $r \in
{\Bbb N}_0$. If we choose a nondegenerate $\o$-module $M_1$ for
$E$, as in Definition~6, then $M'_1 \colon = M_1 \, \perp \, r \times
\left[ {0 \atop 1} \; {1 \atop 0} \right]$ is a possibe choice for
$E'$. (Here we consider $\left[ {0 \atop 1} \; {1 \atop 0} \right]$ over ${\nam
o}$ instead of over $K$.) Therefore, $L \otimes_{\lambda} M'_1 \sim L
\otimes_{\lambda} M_1$.\qed
\endproof

\begexample{} Let $K = k(t_1, \dots, t_n)$ be the rational
function field in $n$ variables $t_1, \dots, t_n$ over an
arbitrary field $k$. For every multi-index $\alpha = (\alpha_1, \dots, \alpha_n) \in {\Bbb
N}_0^n$, let $t^{\alpha}$ denote the monomial $t_1^{\alpha_1} \dots
t_n^{\alpha_n}$. We order the abelian group ${\Bbb Z}^n$
lexicographically and then have exactly one valuation
$v \colon K \to {\Bbb Z}^n \cup \infty$ with $v(t^{\alpha}) =
\alpha$ for every $\alpha \in {\Bbb N}_0^n$. Its valuation ring
contains the field $k$, and the residue class field ${\nam o}/{\nam
m}$ coincides with $k$. Let $\lambda \colon K \to k \cup \infty$
be the canonical place corresponding to $\o$. Furthermore, let $A$
be the set of all multi-indices $(\alpha_1,
\dots, \alpha_n)$ with $\alpha_i \in \{ 0,1 \}$ for every $i$, in other words
$A = \{0,1 \}^n \subset {\Bbb N}_0^n$. As system of
representatives $S$, in the above sense, we take the set of
$t^{\alpha}$ with $\alpha \in A$.

Suppose now that we are given a family
$(F_{\alpha} \; \vert \; \alpha \in A) = {\cal F}$ of $2^n$
quadratic spaces over $k$. We construct the space
$$E \colon = \raise6pt\hbox{$\newbot\limits_{\alpha \in
A}$} \; \langle t^{\alpha} \rangle \otimes (K \otimes_k F_{\alpha})$$
over $K$. For every $\alpha \in A$, $\langle t^{\alpha} \rangle \otimes
E$ is obedient \wrt\ $\lambda$ and
$$\lambda_W (\langle t^{\alpha} \rangle \otimes
E) = \{ F_{\alpha} \}.$$
Hence the family ${\cal F}$ can be
recovered from the space $E$, up to Witt equivalence. If all
$F_{\alpha}$ are anisotropic, we even get the $F_{\alpha}$ back from $E$ as
kernel spaces of the Witt classes $\lambda_W (\langle t^{\alpha} \rangle \otimes
E)$. In this case, $E$ is anisotropic as well. This can easily be
seen, using an argument similar to the one used in the proof of Lemma~3.
One could say that the family ${\cal F}$ is ``stored'' in the space
$E$ over $k((t_1, \dots, t_n))$.

This simple example reminisces of Springer's Theorem in \S1.
Without our theory of weak specialization, we can use
Springer's Theorem and  induction on $n$ to show that the $F_{\alpha}$ are uniquely
determined by $E$ (at least when $\Char k\not=2$). This is so,
because the valuation $v$ is ``$n$-fold discrete''. For more
complicated value groups, we cannot fall back on Springer's Theorem
for specialization arguments.

We return to the situation of arbitrary places $\lambda \colon K
\to k \cup \infty$.

\begdefinition{7.} (a) We denote the sets of Witt classes $\{ E \}$
of obedient quadratic spaces (\wrt\ $\o$)
and obedient strictly regular
quadratic spaces (\wrt\ $\o$) by $\widetilde{W}\!q (K, {\nam o})$ and $W\!q (K, {\nam o})$
respectively. These are thus subsets of the sets $\widetilde{W}\!q (K)$ and
$W\!q (K)$, introduced in \S6, Definition~7b.

(b) We define a map $\lambda_W \colon \widetilde{W}\!q (K, {\nam o}) \to
\widehat{W}\!q (L)$ by setting
$\lambda_W (\{ E \}) \colon = \lambda_W (E)$. This map is well-defined by Theorem~5.
\enddefinition


Clearly $W\!q (K, {\nam o})$ is a subgroup of the abelian group
$W\!q (K)$. The group $W\!q (K, {\nam o})$ acts on the set $\widetilde{W}\!q (K, {\nam
o})$ by restriction of the action of $W\!q (K)$ on $\widetilde{W}\!q
(K)$, explained in \S6. If $\Char K\not=2$, then naturally
$\widetilde{W}\!q (K, {\nam o}) = W\!q (K, {\nam o})$, and $\lambda_W$ maps $W\!q (K,
{\nam o})$ to $W\!q (L)$.

\begremark{} Let $E$ and $F$ be quadratic spaces over $K$,
obedient \wrt\ $\lambda$ and suppose that $E$ is strictly regular.
Obviously we then have
$$\lambda_W (E \, \perp \, F) = \lambda_W (E) + \lambda_W (F).$$

Therefore, restricting the map $\lambda_W \colon \widetilde{W}\!q (K, {\nam o}) \to
\widehat{W}\!q (L)$ gives rise to a homomorphism from $W\!q (K, {\nam o})$ to $W\!q
(L)$, which we also denote by $\lambda_W$. The map $\lambda_W
\colon \widetilde{W}\!q (K, {\nam o}) \to \widehat{W}\!q (L)$ is
equivariant \wrt\ the homomorphism
$\lambda_W \colon W\!q (K, {\nam o}) \to W\!q (L)$.

In \S3, we got a homomorphism $\lambda_W
\colon W(K) \to W(L)$, using a different method. Given a bilinear
space $E = (E, B)$ over $K$, does there exist a description of the
Witt class $\lambda_W (\{ E \})$, analogous to our current
Definition~5?

First of all, when  $2 \in {\nam o}^*$, i.e. $\Char L \ne 2$,
we find complete harmony between \S3 and \S6. In this case,
bilinear spaces over $K$ and $L$ are the same objects as quadratic
spaces, and every such space over $K$ is obedient \wrt\ $\lambda$.

\begtheorem{6.} If $\Char L\not=2$, the homomorphism $\lambda_W
\colon W(K) \to W(L)$ from $\S3$ coincides with the homomorphism
$\lambda_W \colon W\!q (K) \to W\!q (L)$, defined just now.
\endtheorem

\begproof{} It suffices to show that for a one-dimensional space
$[a] =\langle 2 a \rangle$, the element $\lambda_W (\{ \langle 2 a \rangle
\})$ from \S3 coincides with the currently defined Witt class $\lambda_W
([a])$. If $a \in {\nam o}^*$, then $\lambda_W (\{ \langle 2 a \rangle \}) = \{ \langle 2 \lambda
(a) \rangle \} = \{ [\lambda (a) ] \}$, according to \S3. However,
if $\langle a \rangle \not\in Q({\nam o})$, then $\langle 2 a \rangle \not\in Q({\nam o})$
and $\lambda_W (\{ \langle 2 a \rangle \}) = 0$. We obtain the
same values for $\lambda_W ([a])$, using Definition~6 above.\qed
\endproof

And so, when $\Char L\not=2$, the work performed hitherto gives us
a new~--~more conceptual -- proof of \S3, Theorem~1.

What is the situation when $\Char L=2$?  First, we encounter in all
generality definitions for bilinear spaces, analogous to
Definitions~1 and 5.

\begdefinition{8.} Let $E = (E, B)$ be a bilinear space over
$K$.
\smallskip
\setitemindent{(a)}
\item{(a)} $E$ has {\it good reduction \wrt\ $\lambda$} (or {\it
\wrt\ $\o$}) when $E \; \cong \; K \otimes_{{\nam o}} M$ for a
bilinear space $M$ over $\o$. $\{$Note: this is just a translation of
the original definition of ``good reduction'' of \S1 ($=$ \S3,
Definition~1) in geometric language.$\}$
\smallskip

\item{(b)} A {\it $\lambda$-modular} (or {\it $\o$-modular}) {\it
decomposition of} $E$ is an orthogonal decomposition $E =
\raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; E_s$, in which every space $\langle s \rangle \otimes E_s$
has good reduction \wrt\ $\lambda$.
\enddefinition

In contrast to the quadratic case, {\it every} bilinear space $E$
over $K$ has $\lambda$-modular decompositions. This is obvious,
since $E$ is the orthogonal sum of one-dimensional bilinear spaces
and copies of the space
$\left( {0 \atop 1} \; {1 \atop 0} \right)$, see \S 2.

Let $\lambda_W \colon W(K) \to W(L)$ be the map, introduced in
\S3. When $E$ is a bilinear space over $K$, we set $\lambda_W (E) \colon = \lambda_W
(\{ E \}) \in W(L)$.

\begtheorem{7.} Let $E$ be a bilinear space over $K$ and let $E =
\raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; E_s$ be a $\lambda$-modular decomposition of $E$. Also,
let $M_1$ be a bilinear space over $\o$ with
$E_1 \; \cong \; K \otimes_{{\nam o}} M_1$. Then
$$\lambda_W (E) = \{ L \otimes_{{\lambda}} M_1 \}.$$
\endtheorem

\begproof{} We have $\lambda_W (E) = \sum\limits_{s \in S} \lambda_W
(E_s)$. By \S3, Theorem~2, we know that $\lambda_W (E_1) = \{ L \otimes_{\lambda} M_1
\}$. It remains to show that $\lambda_W (E_s) = 0$ for every $s \in S$
with $s \ne 1$. Hence we have to show that if $\langle s \rangle \in
Q(K)$ is a square class, not in $Q({\nam o})$, and if $N$ is a
bilinear space over $\o$, then $\lambda_W (\langle s \rangle \otimes (K
\otimes_{{\nam o}} N)) = 0$.

By a lemma from \S3, $N$ has an orthogonal basis when the bilinear
space $N/{\nam m} N$ over $k$ is not hyperbolic. In this case we
have
$$\langle s \rangle \otimes (K \otimes_{{\nam o}} N) \; \cong \; \langle
s \varepsilon_1, \dots, s \varepsilon_r \rangle$$
with units $\varepsilon_i \in {\nam o}^*$. By definition of
$\lambda_W$, we have $\lambda_W ( \langle s \varepsilon_i \rangle ) = 0$ for all $i \in \{
1, \dots, r \}$, hence $\lambda_W ( \langle s \rangle \otimes (K
\otimes_{{\nam o}} N)) = 0$. If $N/{\nam m} N$ is hyperbolic, we
construct the space
$N' \colon = N \, \perp \, \langle 1 \rangle$ over ${\nam o}$.
Since $N'$ has an orthogonal basis, we get $\lambda_W ( \langle s \rangle \otimes (K
\otimes_{{\nam o}} N')) = 0$.
Since $\lambda_W ( \langle s \rangle
\otimes \langle 1 \rangle ) = 0$ as well, we conclude that $\lambda_W (\langle s \rangle
\otimes (K \otimes N)) = 0$.\qed
\endproof

Theorem~7 allows us to speak of  ``weak specialization'' of
bilinear spaces, in analogy with Definition~6 for quadratic
spaces.

\begdefinition{9.} Let $E$ be a bilinear space over $K$. Also, let
$E = \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; E_s$ be a $\lambda$-modular decomposition of $E$ and $M_1$
a bilinear space over $\o$ with
$E_1 \; \cong \; K \otimes_{{\nam o}} M_1$. Then we call the space
$L \otimes_{\lambda} M_1$ a {\it weak specialization} of the space
$E$ \wrt\ $\lambda$.
\enddefinition

According to Theorem~7, a weak specialization of $E$ \wrt\
$\lambda$ is completely determined by $E$ and $\lambda$ up to Witt equivalence.
One could ask if this can be shown in a direct --
geometric -- way, comparable to how we have done this for obedient
quadratic spaces above. Reversing direction, this would give a
new proof of \S3, Theorem~1, also when $\Char L=2$.

We leave this question open. After all, we have a proof of \S3,
Theorem~1, and with it a weak specialization theory for bilinear
spaces. This theory is more satisfying  than the
corresponding theory for quadratic spaces, in that we do not have
to demand obedience of the spaces.

We can now also make a statement about Problem~3b of \S1. Let
$\Char L=2$ and let $\Char K=0$. Let $(E, q)$ be a quadratic
space, obedient \wrt\ $\lambda$. Associated to it, we have the
bilinear space $(E, B_q)$. It may be that $(E, q)$ and $(E, B_q)$
are in principle  the same object, but it does make a difference
wether we weakly specialize $E$ as a quadratic or a bilinear
space. What is better?

Let $(M, q)$ be a regular ($=$ strictly regular) quadratic
$\o$-module. According to \S6, Theorem~1, we have a decomposition
$$(M, q) \; \cong \; {\textstyle \left[ {a_1 \atop 1} \quad {1 \atop b_1}
\right]} \, \perp \dots \perp \, {\textstyle \left[ {a_r \atop 1} \quad {1
\atop b_r}
\right]}.$$
(Note: $(M/{\nam m} M, \overline{q})$ is strictly regular, and
thus has even dimension.) Then
$$(M, B_q) \, \cong \; {\textstyle \left( {2a_1 \atop 1} \quad {1
\atop 2b_1}
\right)} \; \perp \dots \perp \, {\textstyle \left( {2a_r \atop 1} \quad {1
\atop 2b_r}
\right)}.$$
Hence,
$$L \otimes_{\lambda} (M, q) \; \cong \; {\textstyle \left[ {\lambda (a_1) \atop 1} \quad {1
\atop \lambda (b_1)}
\right]} \, \perp \dots \perp \, {\textstyle \left[ {\lambda (a_r) \atop 1} \quad {1
\atop \lambda (b_r)} \right]}.$$
However, $L \otimes_{\lambda} (M, B_q) \; \cong \; r \times \left( {0 \atop
1} \; {1 \atop 0} \right)$.

Therefore, a weak specialization of the space $(E, B_q)$ is always
hyperbolic and so gives hardly any information about $(E,q)$.
However, a weak specialization of $(E,q)$ \wrt\ $\lambda$ can give
an interesting result.
\bigskip

\noindent{\it Final Consideration.} A last word about the central
Definition~3 of an obedient quadratic space. It seems obvious to
formally weaken it, by considering quadratic {\it modules} $E$
instead of spaces $E$ over $K$, subject only to the requirements
that, as a
$K$-vector space,  $E$ should have  finite dimension and a decomposition
$(*)$ as presented there. But then $E$ is already nondegenerate.
We namely have
$$QL(E) \; \cong \; \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; \langle s \rangle \otimes (K \otimes_{{\nam o}} QL(M_s)),$$
and as in the proof of Lemma~3, we see that $QL(E)$ is anisotropic
(without the requirement that $\o$ is quadratically henselian).
Thus these ``obedient quadratic modules'' are the same objects as
those given by Definition~3.
