\titlea{\S3}{Specialization of Symmetric Bilinear Forms}
\leftrunningtitle{Chapter I. Fundamentals of Specialization Theory}

%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S3 Specialization of Symmetric
Bilinear Forms}


In this section, a ``form'' will again be understood to be a
nondegenerate symmetric bilinear form. Let $\lambda
\colon K \to L \cup \infty$ be a place from the field $K$ to a
field $L$. Let ${\nam o} = {\nam o}_{\lambda}$ be the valuation
ring associated to $\lambda$ and ${\nam m}$ its maximal ideal. As
usual for rings, ${\nam o}^*$ stands for the group of units of ${\nam
o}$, so that ${\nam o}^* = {\nam o}\setminus {\nam m}$. This is
the set of all $x \in K$ with $\lambda (x) \ne 0, \infty$.

We will now denote the Witt class of a one-dimensional form $\langle a
\rangle$ over $K$ (or $L$) by $\{ a \}$. The group of
square classes $Q({\nam o}) = {\nam o}^*/{\nam o}^{*2}$ can be
embedded in $Q(K) = K^*/K^{*2}$ in a natural way via
$a {\nam o}^{*2} \mapsto a K^{*2}$. We interpret $Q({\nam o})$ as
a subgroup of $Q(K)$, so $Q({\nam o}) = \{ \langle a
\rangle \; \vert \; a \in {\nam o}^* \} \subset Q(K)$. Our
specialization theory is based on the following

\begtheorem{1.} There exists a well-defined additive map
$\lambda_W \colon W (K) \to W(L)$, given by $\lambda_W (\{ a\}) = \{
\lambda (a) \}$ if $a \in {\nam o}^*$, and $\lambda_W (\{ a\}) =0$
if $\langle a \rangle \not\in Q({\nam o})$ $($i.e. $(a
K^{*2}) \cap {\nam o}^* = \emptyset$$)$.\fonote{The letter $W$ in the
notation ${\lambda_W}$ refers to ``Witt'' or ``weak'', see \S1 and
\S7.}
\endtheorem

\begproof{} (Copied from  [K$_4$, \S 3].) Our place $\lambda$ is
a combination of the canonical place $K \to ({\nam
o}/{\nam m}) \cup \infty$ with respect to ${\nam o}$, and
a field extension $\overline{\lambda} \colon {\nam o}/{\nam
m} \hookrightarrow L$. Thus it suffices to prove the theorem for the
canonical place. So let $L = {\nam o}/{\nam m}$ and
$\lambda (a) = \overline{a} \colon = a + {\nam m}$ for $a \in
{\nam o}$.

We have a well-defined additive map $\Lambda \colon
{\Bbb Z} [Q(K)] \to W(L)$ such that $\Lambda (\langle a \rangle) = \{
\overline{a} \}$ if $a \in {\nam o}^*$, and $\Lambda (\langle a
\rangle) = 0$ if $\langle a \rangle \not\in Q({\nam o})$. Clearly
$\Lambda$ vanishes on all elements $\langle
a \rangle + \langle - a \rangle$ with $a \in K^*$.  According to
\S2, Theorem~3 we will be  finished if we can show that $\Lambda$
also disappears on every element
$$Z = \langle a_1 \rangle + \langle a_2 \rangle -
\langle a_3 \rangle - \langle a_4 \rangle $$
with $a_i \in K^*$ and $\langle a_1, a_2 \rangle \cong \langle a_3,
a_4 \rangle$. 

This will be the case when the four square classes $\langle a_i
\rangle$ are not all in $Q({\nam o})$.
Suppose from now on, without loss of generality, that $a_1 \in {\nam
o}^*$. Then we have $Z = \langle a_1 \rangle y$, where
$$y = 1 + \langle c \rangle - \langle b \rangle -
\langle bc \rangle$$
is an element such that $\langle 1, c \rangle \cong \langle b, bc \rangle
$. So $b = u^2 + w^2 c$ for elements $u, w \in K$. Clearly the
equation $\Lambda (\langle a \rangle x) = \{ \overline{a} \} \Lambda
(x)$ is satisfied for any $a \in {\nam o}^*$, $x \in {\Bbb
Z} [Q(K)]$. Therefore it is enough to verify that $\Lambda (y) =
0$. We suppose without loss of generality that $u$ and $w$ are not
both zero, otherwise we already  have that $y=0$.

Let us first treat the case $\langle c \rangle \in
Q({\nam o})$, so without loss of generality $c \in {\nam o}^*$.
Then we have
$$\Lambda (y) = (1 + \{ \overline{c} \}) \Lambda (1-
\langle b \rangle ).$$
If $\{ \overline{c} \} = \{ - 1 \}$,  we are done. So suppose
from now on that $\{ \overline{c} \} \ne \{ -1 \}$. Then the form
$\langle 1, \overline{c} \rangle$ is anisotropic over $L$.
Since we are allowed to replace $u$ and $v$ by $gu$ and $gv$ for
some $g\in K^*$, we may additionally assume  that $u$ and $v$
are both in ${\nam o}$, but not both in ${\nam m}$. Since $\langle 1,
\overline{c}
\rangle$ is anisotropic, we have $\overline{b} = \overline{u}^2 +
\overline{c} \overline{w}^2 \ne 0$ and
$$\Lambda (y) = (1 + \{ \overline{c} \}) (1 - \{ \overline{u}^2
+ \overline{c} \overline{w}^2 \}) = 0.$$

The case which remains to be tackled is when the square class
$cK^{*2}$ doesn't contain a unit from ${\nam o}$. Then $u^{-2} w^2
c$ is definitely not a unit and either $b = u^2 (1+d)$ or $b = w^2 c
(1+d)$ with $d \in {\nam m}$. Hence $\Lambda
(1 - \langle \overline{b} \rangle )$ is $0$ or $ 1 - \{
\overline{c} \}$, and both times $\Lambda (y) = 0$.
\qed
\endproof

\begscholium{} The map $\lambda_W \colon W(K) \to W(L)$ can be described very
conveniently as follows: Let $\varphi$ be a form over
$K$. If $\varphi$ is hyperbolic (or, more generally metabolic),
then $\lambda_W (\{ \varphi \} ) = 0$. If $\varphi$ is not
hyperbolic, then consider a diagonalisation
$\varphi \cong \langle a_1, a_2, \dots, a_n \rangle$. Multiply
each coefficient $a_i$, for which it is possible, by a square so
that it becomes a unit in ${\nam o}$, and leave the other
coefficients as they are. Let for example
$a_i \in {\nam o}^*$ for $1 \leq i \leq r$ and
$\langle a_i \rangle \not\in Q({\nam o})$ for $r < i \leq n$
(possibly $r = 0$ or $r = n$). Then $\lambda_W ( \{ \varphi
\}) = \{ \langle \lambda (a_1), \dots, \lambda (a_r) \rangle
\}$.
\endscholium

Let us now recall a definition from the Introduction~\S1.

\begdefinition{1.} We say that a form $\varphi$ over $K$ has {\it
good reduction with respect to}\Idx{good reduction (bilinear form)}\Idx{bilinear form!good reduction} 
$\lambda$, or that $\varphi$ is
$\lambda$-{\it unimodular} if $\varphi$ is isometric to a form
$(a_{ij})$ with $a_{ij} \in {\nam o}$ and $\det
(a_{ij}) \in {\nam o}^*$. We call such a representation $\varphi \cong
(a_{ij})$ a $\lambda$-{\it unimodular represention of}\Idx{l-unimodular (bilinear form)@$\lambda$-unimodular (bilinear form)}\Idx{bilinear form!l-unimodular@$\lambda$-unimodular}  
$\varphi$
(or a unimodular representation with respect to the valuation ring
 ${\nam o}$).
\enddefinition

This definition can be interpreted geometrically as follows. We
associate to $\varphi$ a couple $(E, B)$, consisting of an
$n$-dimensional $K$-vector space $E$ $(n = \dim \varphi)$ and a
symmetric bilinear form $B \colon E \times E \to K$ such that $B$ represents
the form $\varphi$ after a choice of basis of $E$.
We denote this by $\varphi \,\widehat{=}\, (E, B)$. Since
$\varphi$ has good reduction with respect to $\lambda$, $E$
contains a free ${\nam o}$-submodule $M$ of rank $n$ with $E =
KM$, i.e. $E = K \otimes_{{\nam o}} M$, and with
$B (M \times M) \subset {\nam o}$, such that the restriction
$B \vert{M \times M} \colon M \times M \to {\nam o}$ is a {\it
nondegenerate bilinear form over}\Idx{nondegenerate (bilinear form)}\Idx{bilinear form!nondegenerate} 
${\nam o}$, i.e. gives rise
to an isomorphism $x \mapsto B(x,-)$ from the ${\nam o}$-module
$M$ to the dual ${\nam o}$-module $\raise4pt\hbox{$
{{\scriptstyle\vee} \atop {\textstyle M}}$} = \Hom_{{\nam o}} (M, {\nam
o})$.


By means of Theorem~1 we can now quite easily find a solution of
the first problem posed in \S1.

\begtheorem{2.} Suppose that the form $\varphi$ over $K$ has good
reduction with respect to $\lambda$. Let $\varphi \cong (a_{ij})$
be a unimodular representation of $\varphi$. Then the Witt class
$\lambda_W (\{ \varphi \})$\Idx{111lw (bilinear form)@$\lambda_W$ (bilinear form)} 
is represented by the
$($nondegenerate!\,$)$
form $(\lambda (a_{ij}))$ over $L$. Consequently the form $(\lambda
(a_{ij}))$ is up to stable isometry independent of the choice of
unimodular representation. {\rm (Recall that, if two forms $\psi$
and  $\psi'$ are Witt equivalent and $\dim\psi=\dim\psi'$, then
$\psi\approx \psi'$.)}
\endtheorem

To prove this theorem, we need the following easy lemma about
lifting  orthogonal bases.

\beglemma{} Let $M$ be a finitely generated free ${\nam
o}$-module, equipped with a nondegenerate symmetric bilinear form
$B \colon M \times M \to {\nam o}$. Let $k \colon ={\nam o}/{\nam
m}$ and let $\pi \colon M \to M/{\nam m} M$ be the natural
epimorphism from $M$ to the $k$-vector space $M/{\nam m} M$.
Further, let $\overline{B}$ be the $($again nondegenerate$)$
bilinear form induced by $B$ on $M/{\nam m} M$,
$\overline{B} (\pi (x), \pi (y)) \colon = B(x, y) + {\nam m}$.
Suppose that the vector space $M/{\nam m} M$ has a basis $\overline{e}_1, \dots,
\overline{e}_n$, orthogonal with respect to $\overline{B}$. Then
$M$ has a basis $e_1, \dots, e_n$, orthogonal with respect to $B$,
with $\pi (e_i) = \overline{e}_i \quad (1 \leq i \leq n)$.
\endlemma

\begproof{} By induction on $n$, which obviously is the rank of the
free ${\nam o}$-module $M$. For $n=1$ nothing has to be shown. So
suppose that $n>1$. We choose an element $e_1 \in M$ with
$\pi (e_1) = \overline{e}_1$. Then $B (e_1, e_1) \in {\nam o}^*$
since $\overline{B} (\overline{e}_1, \overline{e}_1) \ne 0$. Hence
the restriction of $B$ to the module ${\nam o} e_1$ is a
nondegenerate bilinear form on ${\nam o} e_1$. Invoking a very
simple theorem (e.g. [MH,~p.5,~Th.3.2], \S 5, Lemma 1 below)
yields $M = ({\nam o}
e_1) \perp N$ with $N = ({\nam o} e_1)^{\perp} = \{ x
\in M \; \vert \; B (x, e_1) = 0 \}$.
The restriction $\pi \vert N \colon N \to M/{\nam m} M$ is then a
homomorphism from $N$ to $(k \overline{e}_1)^{\perp} = \bigoplus\limits_{i =
2}^n k \overline{e}_i$ with kernel ${\nam m} N$. By our induction
hypothesis, $N$ contains an orthogonal basis $e_2,
\dots, e_n$ with $\pi (e_i) = \overline{e}_i \quad (2 \leq i \leq
n)$ which can be completed by $e_1$ to form an orthogonal basis of
$M$ which has the required property.
\qed
\endproof

\begremark{} Clearly the lemma and its proof remain valid when ${\nam
o}$ is an arbitrary local ring with maximal ideal ${\nam m}$,
instead of a valuation ring.
\endremark

We also need the following

\begdefinition{2.} A {\it bilinear ${\nam o}$-module}\Idx{bilinear o-module@ bilinear ${\nam o}$-module} 
is a couple
$(M,B)$ consisting of an ${\nam o}$-module $M$ and a symmetric
bilinear form $B \colon M\times M \to {\nam o}$. A bilinear module
is called {\it free} when the ${\nam o}$-module $M$ is free of
finite rank. If $e_1, \dots, e_n$ is a basis of $M$, we write
$(M, B) \cong (a_{ij})$ with $a_{ij} \colon = B(e_i, e_j)$. If
$e_1, \dots, e_n$ is an orthogonal basis ($B(e_i, e_j) = 0$ for $i \ne
j$), then we also write $(M, B) \cong \langle a_1, \dots, a_n \rangle$ with $a_i
\colon = B(e_i, e_i)$.
\enddefinition

\begnote{} The form $B$ is nondegenerate exactly when $\det
(a_{ij})$ is a unit in ${\nam o}$, respectively when all $a_i$ are units
in ${\nam o}$.
\endnote

All this makes sense and remains correct when ${\nam o}$ is an
arbitrary commutative ring (with $1$), instead of a valuation
ring. As before, ``$\cong$'' stands for ``isometric'', also for
bilinear modules.


\begproof{of Theorem~2.} For $a \in {\nam o}$, let
$\overline{a}$ denote the image of $a$ in ${\nam o}/{\nam m}$.
We suppose for the moment that the bilinear space
$(\overline{a}_{ij})$ over ${\nam o}/{\nam m}$ is not hyperbolic.
Then it has an orthogonal basis. By the lemma, the bilinear module
$(a_{ij})$ over ${\nam o}$ also has an orthogonal basis. Hence
over ${\nam o}$,
\vskip-0.3cm
$$(a_{ij}) \cong \langle a_1, \dots, a_n \rangle
\leqno(\ast)$$

\noi for certain $a_i \in {\nam o}^*$.  The isometry $(*)$ is then
also valid over $K$, and so we have in $W(L)$
$$\lambda_W ( \{ \varphi \}) = \{ \langle \lambda (a_1), \dots,
\lambda (a_n) \rangle \}.$$
On the other hand $(*)$ implies that
$$(\overline{a}_{ij}) \cong \langle \overline{a}_1, \dots,
\overline{a}_n \rangle \quad \hbox{over} \; {\nam o}/{\nam m}.$$
If we now apply the (injective) homomorphism $\overline{\lambda}
\colon {\nam o}/{\nam m} \to L$ induced by $\lambda$ (thus we
tensor with the field extension given by $\overline{\lambda}$), we
obtain
$$(\lambda (a_{ij})) \cong \langle \lambda (a_1), \dots, \lambda
(a_n) \rangle$$
over $L$. Consequently the Witt class $\lambda_W ( \{ \varphi \})$
is represented by the form $(\lambda (a_{ij}))$.

Let us now tackle the remaining case, where the form
$(\overline{a}_{ij})$ over ${\nam o}/{\nam m}$ is hyperbolic.
We can apply what we just have proved
to the form $\psi \colon = \varphi \perp  \langle 1
\rangle$. This gives us
$$\eqalign{\lambda_W ( \{ \psi \}) & = \{ (\lambda (a_{ij}))
\perp  \langle 1 \rangle \}  \cr
                                   & = \{ (\lambda (a_{ij})) \}
+ \{ \langle 1 \rangle \}}$$
in $W(L)$.
On the other hand we have $\lambda_W ( \{ \psi \}) = \lambda_W ( \{
\varphi \}) + \{ \langle 1 \rangle \}$, and we find again that
$\lambda_W (
\{ \varphi \}) = \{ ( \lambda (a_{ij})) \}$.
\qed
\endproof

\begremark{} If $\Char L\not =2$, a hyperbolic form over ${\nam o}/{\nam
m}$ also has an orthogonal basis, so that the distinction between
the two cases above is unnecessary.
\endremark

\begdefinition{3.} If $\varphi$ has good reduction with respect to
$\lambda$, $\varphi \cong (a_{ij})$ with $a_{ij} \in {\nam o}$, $\det
(a_{ij}) \in {\nam o}^*$,  we denote the form $(\lambda
(a_{ij}))$ over $L$ by $\lambda_* (\varphi)$ and call it ``the''
{\it specialization of $\varphi$ with respect to $\lambda$.}\Idx{specialization}\Idx{111lsfi@$\lambda_* (\varphi)$}
\enddefinition

If $\Char L = 2$ we run into trouble with this definition, since
$\lambda_* (\varphi)$ is only up
to {\it stable isometry} uniquely determined by $\varphi$ .
We nevertheless use it, since it is so convenient. If $\Char L \ne
2$, $\lambda_*(\varphi)$ is up to isometry uniquely determined by
$\varphi$.

\begexample{} Every metabolic form $\varphi$ over $K$ has good
reduction with respect to $\lambda$. Of course is $\lambda_*
(\varphi) \sim 0$.
\endexample

\begproof{} It suffices to prove this in the case $\dim \varphi = 2$, so
$\varphi = \left( {a \atop 1} {1 \atop 0} \right)$
with $a \in K$. Let $\varphi \widehat{=} (E, B)$ and let $e, f$ be
a basis of $E$ with value matrix $\left( {a \atop 1} {1 \atop 0}
\right)$. Choose an element $c \in K^*$ with $ac^2 \in {\nam o}$.
Then $ce, c^{-1} f$ is a basis of $E$ with value matrix
$\left( {ac^2 \atop \hfill 1\hfill} {1 \atop 0} \right)$.\qed
\endproof

\begtheorem{3.} Let $\varphi$ and $\psi$ be forms over $K$, having
good reduction with respect to $\lambda$. Then $\varphi \,
\perp \, \psi$ also has good reduction with respect to $\lambda$
and
$$\lambda_* (\varphi \, \perp \, \psi) \approx
\lambda_* (\varphi) \, \perp \, \lambda_* (\psi).$$
\endtheorem

\begproof{} This is clear.\qed
\endproof

Until now we got on with our specialization theory almost without
any knowledge of bilinear forms over ${\nam o}$. Except for the
lemma above about the existence of orthogonal bases, we
needed hardly anything from this area. We could even have avoided using
this little bit of information if we would only have considered
diagonalised forms over fields.

We are still missing one important
theorem of specialization theory (especially for applications later on),
Theorem~4 below. For a proof of this theorem we need the basics of
the theory of forms over valuation rings, which we will present
next using a ``geometric'' point of view. In other words, we
interpret a form $\varphi$ over a field as an ``inner product'' on
a vector space and use more generally ``inner products'' on
modules over rings, while until now a form was usually interpreted
as a polynomial in two sets of variables $x_1, \dots, x_n, y_1, \dots,
y_n$.

For the moment we allow local rings instead of the valuation ring
${\nam o}$ since this won't cost us anything extra. So let $A$
be a local ring.

\begdefinition{4.} A {\it bilinear space}\Idx{bilinear space} 
$M$ over $A$ is a free
$A$-module $M$ of finite rank, equipped with a symmetric bilinear
form $B \colon M \times M \to A$ which is {\it nondegenerate},\Idx{nondegenerate (bilinear space)}\Idx{bilinear space!nondegenerate}
i.e. which determines an isomorphism $x \mapsto B (x, - )$ from
$M$ on the dual module $\raise4pt\hbox{${{\scriptstyle\vee} \atop {\textstyle M}}$} =
\Hom_A (M, A)$.
\enddefinition

\begremark{} We usually denote a bilinear space by the letter $M$.
If confusion is possible, we write $(M,B)$ or even $(M,B_M)$.
\endremark

In what follows, $M$ denotes a bilinear space over $A$, with
associated bilinear form $B$.

\begdefinition{5.} A {\it subspace}\Idx{subspace (bilinear space)}\Idx{bilinear space!subspace} 
$V$ of $M$ is a submodule $V$
of $M$ which is a direct summand of $M$, i.e. for which there exists another
submodule $W$ of $M$ with $M = V \oplus W$.
\enddefinition

To a subspace $V$ we can associate the orthogonal submodule
$$V^{\perp}= \{ x \in M \; \vert \; B (x, V) = 0 \},$$
and we have an exact sequence
$$0 \longrightarrow V^{\perp} \longrightarrow M \buildrel
\varphi \over \longrightarrow
\raise4pt\hbox{${{\scriptstyle\vee} \atop {\textstyle V}}$}
\longrightarrow 0.$$
Here $\raise4pt\hbox{${{\scriptstyle\vee} \atop {\textstyle V}}$}
= \Hom_A (V, A)$ and $\varphi$ maps $x \in M$ to the linear form $y \mapsto B(x, y)$ on
$V$. The sequence splits since $\raise4pt\hbox{${{\scriptstyle\vee} \atop {\textstyle
V}}$}$ is free. Thus $V^\perp$ is again a subspace of $M$.

\begdefinition{6.} A subspace $V$ of $M$ is called {\it totally
isotropic}\Idx{totally isotropic subspace (bilinear space)}\Idx{bilinear space!totally isotropic subspace}  
when $B(V, V) = \{ 0 \}$, i.e. when $V \subset
V^{\perp}$. $V$ is called a {\it Lagrangian subspace}\Idx{Lagrangian subspace (bilinear space)}\Idx{bilinear space!Lagrangian subspace} 
of $M$
when $V = V^{\perp}$. If $M$ contains a Lagrangian subspace, $M$
is called {\it metabolic}.\Idx{metabolic (bilinear space)}\Idx{bilinear space!metabolic} 
$M$ is called {\it anisotropic} if it
{\it doesn't} contain any totally
isotropic subspace $V \ne \{ 0 \}$.
\enddefinition

\beglemma{1.}
\setitemindent{$(b)$}
\item{$(a)$} Every bilinear space $M$ over $A$ has a
decomposition
$$M \cong M_0 \perp  M_1$$
with $M_0$ anisotropic and $M_1$ metabolic.
\smallskip
\item{$(b)$} Every metabolic space $N$ over $A$ is the orthogonal sum of
spaces of the form $\left( {a \atop 1} {1 \atop 0}\right)$,
$$N \cong \left( {a_1 \atop \hfill 1\hfill} {1 \atop 0}
\right) \perp \dots \perp \left( {a_r \atop \hfill 1\hfill} {1 \atop 0}
\right)$$
with $a_1, \dots, a_r \in A$.
\endlemma

These statements can be inferred from more general theorems, which
can be found in e.g. [Ba, \S 1], [K$_1$, \S 3],
[K$_2$, I \S 3], [KRW, \S 1].

\begremark{} If $2$ is a unit in $A$,  Witt's Cancellation
Theorem ([K$_3$], [R]) holds for bilinear spaces over $A$ and
every metabolic space over $A$ is even hyperbolic, i.e. is an
orthogonal sum $r \times \left( {0 \atop 1} {1 \atop 0}
\right)$ of copies of the ``hyperbolic plane''
$\left( {0 \atop 1} {1 \atop 0}\right)$ over $A$. Now the
anisotropic space $M_0$ in Lemma~1(a) is up to isometry uniquely
determined by $M$. If $2 \not\in A^*$ this is false in general.
\endremark

If $\alpha \colon A \to C$ is a homomorphism from $A$ to another
local ring $C$, we can associate to a bilinear space $(M, B) = M$
over $A$ a bilinear space\break
$(C \otimes_A M, B') = C \otimes_A M$ over
$C$ as follows: the underlying free $C$-module is the tensor
product $C \otimes_A M$ determined by $\alpha$, and the
$C$-bilinear form $B'$ on this module is obtained from $B$ by
means of a basis extension, so
$$B' (c \otimes x, d \otimes y) = cd \,\alpha (B(x, y))$$
$(x, y \in M; c, d \in C)$. The form $B'$ is again nondegenerate.  If
$(a_{ij})$ is the value matrix of $B$ with respect to a basis
$e_1, \dots, e_n$ of $M$, then $(\alpha (a_{ij}))$ is the value
matrix of $B'$ with respect to the basis $1\otimes e_1, \dots, 1 \otimes e_n$
of $C \otimes_A M$.

If $A$ doesn't contain any zero divisors and if  $K$ is the quotient
field of $A$, we can in particular use the inclusion $A
\hookrightarrow K$ to associate a bilinear space $K \otimes_A M$
to the bilinear space $M$ over $A$. Now we can interpret $M$ as an
$A$-submodule of the $K$-vector space $K \otimes_A M$
$(x = 1 \otimes x$ for $x \in M)$ and reconstruct $B$ from $B'$ by
restriction, $B = B' \vert {M \times M} \colon\;
M \times M \to A$.

Let us return to our place $\lambda \colon K \to L \cup
\infty$ and the valuation ring ${\nam o}$.

\beglemma{2.} Let $M$ be a bilinear space over ${\nam
o}$.\hfill\break
$(a)$ If $K \otimes_{{\nam o}} M$ is isotropic, then $M$ is
isotropic.\hfill\break
$(b)$ If $K \otimes_{{\nam o}} M$ is metabolic, then $M$ is
metabolic.
\endlemma

\begproof{} Let $E \colon = K \otimes_{{\nam o}} M$. We interpret
$M$ as an ${\nam o}$-submodule of $E$ and have $E=KM$.

(a) If $E$ is isotropic, there exists a subspace $W
\ne \{ 0 \}$ in $E$ with $W \subset W^{\perp}$. The ${\nam
o}$-submodule $V \colon = W \cap M$ of $M$ satisfies $KV = W$ and
so $V \ne \{ 0 \}$. Furthermore $V \subset V^{\perp}$. The ${\nam
o}$-module $M/V$ is torsion free and finitely generated, hence
free. This is because every finitely generated ideal in ${\nam o}$
is principal, cf. [CE, VII, \S 4]. Therefore $V$ is a totally
isotropic subspace of $E$.

(b) If $W = W^{\perp}$, then $V = V^{\perp}$. Hence $M$ is
metabolic.\qed
\endproof

Now we are fully equipped to prove the following important theorem
[K$_4$, Prop.2.2].

\begtheorem{4.} Let $\varphi$ and $\psi$ be forms over $K$. If
$\varphi$ and $\varphi \perp \psi$ have good reduction with
respect to $\lambda$, then $\psi$ also has good reduction with
respect to $\lambda$.
\endtheorem

\begproof{} Adopting geometric language, the statement says: Let
$F$ and $G$ be bilinear spaces over $K$ and $E \colon = F  \perp
G$. If $F$ and $E$ have good reduction, i.e. $F
\cong K \otimes_{{\nam o}} N$, $E \cong K \otimes_{{\nam o}} M$
for bilinear spaces $N$ and $M$ over ${\nam o}$, then
$G$ has good reduction as well.

By \S2, Theorem~1 there is a decomposition $G = G_0 \perp
G_1$ with $G_0$ anisotropic and $G_1$ metabolic. From above (cf. the example following
Definition~3), $G_1$ has good reduction. Hence it suffices to show
that $G_0$ has good reduction.

Now $E \perp (- F) \cong F \perp (- F) \perp G_0 \perp
G_1$.\fonote{If $E=(E,B)$ is a bilinear space, then $-E$ denotes
the space $(E,-B)$.} Since $F \perp (- F) \perp G_0$ is metabolic,
but $G_1$ anisotropic, $G_1$ is the kernel space of $E \perp (-
F)$. (``Kernel space'' is the pendant of the word ``kernel form''
($=$ anisotropic part) in the geometric language.)
We  decompose $M \perp (- N)$ following Lemma~1(a),
$M \perp  (- N) \cong R \perp S$ where $R$ is anisotropic and
$S$ metabolic. Tensoring with $K$ gives $E \perp (- F) \cong  K
\otimes_{{\nam o}} R \perp K \otimes_{{\nam o}} S$. Now $K \otimes_{{\nam o}}
S$ is metabolic and, according to Lemma~2, $K \otimes_{{\nam o}} R$
is anisotropic. Hence $K\otimes_{{\nam o}} R$ is also a kernel
space of $E \perp (- F)$. Applying \S2, Theorem~1 gives $K \otimes_{{\nam o}} R \cong
G_0$, and we are finished.\qed
\endproof

\begcorollary{} Let $\varphi$ and $\psi$ be forms over $K$ with
$\varphi \sim \psi$. If $\varphi$ has good reduction with respect
to $\lambda$, $\psi$ also has good reduction with respect to
$\lambda$ and $\lambda_* (\varphi) \sim \lambda_* (\psi)$. If
furthermore $\varphi \approx \psi$, then
$\lambda_* (\varphi) \approx \lambda_* (\psi)$.
\endcorollary

\begproof{} There are Witt decompositions $\varphi \cong \varphi_0
\perp \mu$, $\psi \cong \psi_0 \perp \nu$ with
$\varphi_0$, $\psi_0$ anisotropic and $\mu, \nu$ metabolic. As
established above, $\mu$ and $\nu$ have good reduction with
respect to $\lambda$ and $\lambda_* (\mu)$, $\lambda_* (\nu)$ are
metabolic. By assumption $\varphi$ has good reduction with respect
to $\lambda$ and $\varphi_0$ is isometric to $\psi_0$. Theorem~4
implies that $\varphi_0$ has good reduction with respect to
$\lambda$. Therefore $\psi_0$, and hence $\psi$, has good reduction
with respect to $\lambda$, and (according to Theorem~3)
$$\eqalign{\lambda_* (\varphi) & \approx \lambda_* (\varphi_0)
\perp \lambda_* (\mu) \sim \lambda_* (\varphi_0), \cr
           \lambda_* (\psi) & \approx \lambda_* (\psi_0)
\perp \lambda_* (\nu) \sim \lambda_* (\psi_0).}$$
Naturally $\lambda_* (\varphi_0) \approx \lambda_*
(\psi_0)$, so $\lambda_* (\varphi) \sim \lambda_* (\psi)$. If
$\varphi \approx \psi$, then $\varphi$ and $\psi$ have the same
dimension and so $\lambda_* (\varphi) \approx \lambda_*
(\psi)$.\qed
\endproof

Let us give a small illustration of Theorem~4.

\begdefinition{7.} Let $\varphi$ and $\psi$ be forms over a field
$K$. If there exists a form $\chi$ over $K$ with $\varphi \cong \psi \; \perp \;
\chi$, we say that $\varphi$ {\it represents}\Idx{representation (bilinear form)}\Idx{bilinear form!representation} 
the form $\psi$ and
write $\psi < \varphi$.\Idx{111psilessfi@$\psi < \varphi$ (bilinear form)}
\enddefinition

For example, the one-dimensional forms represented by $\varphi$
are exactly the square classes $\langle \varphi (x, x) \rangle$,
where $x$ runs through the anisotropic vectors of the space
belonging to $\varphi$.

\begtheorem{5}{\rm (Substitution Principle).}\Idx{Substitution Principle (bilinear form)}\Idx{bilinear form!Substitution Principle}
Let $k$ be a field and $K=k(t)$, where $t = (t_1, \dots, t_r)$ is
a set of indeterminates. Let $(f_{ij} (t))$ be a symmetric
$(n \times n)$-matrix and $(g_{kl} (t))$ a symmetric $(m
\times m)$-matrix, for polynomials  $f_{ij} (t) \in k [t]$,
and $g_{kl} (t) \in k [t]$. Let further be given a field extension
$k \subset L$ and a point $c \in L^r$ with $\det (f_{ij}
(c)) \ne 0$ and $\det (g_{kl} (c)) \ne 0$. If $\Char k=2$, also
suppose that the form $(f_{ij} (c))$ is anisotropic over $L$.

Claim: if $(g_{kl} (t)) < (f_{ij} (t))$ $($as forms over $K$$)$, then
$(g_{kl} (c)) < (f_{ij} (c))$ $($as forms over $L$$)$.
\endtheorem

\begproof{} Going from $k[t]$ to $L[t]$, we suppose without
loss of generality\break that $L=k$. For every $s \in \{ 1, \dots, r
\}$  there is exactly one corresponding place
$\lambda_s \colon k (t_1, \dots, t_s) \to k(t_1, \dots, t_{s-1}) \cup \infty$
with $\lambda_s (u) = u$ for all $u \in k (t_1, \dots, t_{s-1})$ and
$\lambda_s (t_s) = c_s$.
$\{$Read $k(t_1, \dots, t_{s-1}) = k$ when $s = 1$.$\}$ The
composition $\lambda_1 \circ \lambda_2 \circ \cdots \circ
\lambda_s$ of these places is a place $\lambda \colon K \to k \cup \infty$
with $\lambda (a) = a$ for all $a \in k$ and $\lambda (t_i) = c_i$ for $i = 1,
\dots, r$. Let $\varphi$ denote the form $(f_{ij} (t))$ over $K$ and
$\psi$ the form $(g_{kl} (t))$ over $K$.
$\{$Note that obviously
$\det (f_{ij} (t)) \ne 0$, $\det (g_{kl} (t)) \ne 0$.$\}$
The forms $\varphi$ and $\psi$ both have good reduction with respect to $\lambda$ and
$\lambda_* (\varphi) \approx (f_{ij} (c))$, $\lambda_* (\psi) \approx (g_{kl}
(c))$.

Now let $\psi < \varphi$. Then there exists a form $\chi$ over $K$
with $\psi \perp \chi \, \cong \, \varphi$. According to
Theorem~4, $\chi$ has good reduction with respect to $\lambda$ and
according to Theorem~3, $\lambda_* (\psi)
\perp \lambda_* (\chi) \approx \lambda_* (\varphi)$. Hence
$\lambda_* (\psi) \perp \lambda_* (\chi) \cong
\lambda_* (\varphi)$ if
$\Char k\not=2$. If $\Char k=2$ and $\lambda_* (\varphi)$ is
anisotropic, this remains true, since $\lambda_* (\varphi)$ is
up to isometry the unique
anisotropic form in the Witt class $\lambda_W (\varphi)$,
and
$\lambda_*~(\psi)\perp\lambda_*~(\chi)$ has the same dimension
as $\lambda_* (\varphi)$.
\qed
\endproof

Let us now return to our arbitrary place $\lambda \colon K \to L \cup
\infty$ and to the conventions made at the beginning of the
paragraph. The Lemmas~1 and 2 allow us to give an easier proof of
Theorem~2, which is interesting in its own right.
\medskip

\noindent{\it Second proof of Theorem 2.} We adopt the geometric
language. Let $E$ be a bilinear space over $K$, having good
reduction with respect to $\lambda$ and let $M$ and $N$ be
bilinear spaces over ${\nam o}$ with $E \cong K \otimes_{{\nam o}} M \cong K
\otimes_{{\nam o}} N$. We have to show that $L \otimes_{\lambda} M
\approx L \otimes_{\lambda} N$,
where the tensor product is taken over $\o$, and $L$ is regarded
as an $\o$-algebra via the homomorphism $\lambda|{\nam o}:\o\to
L$. The space $K \otimes_{{\nam o}} (M \perp (- N))$ is
metabolic. According to Lemma~2, $M \perp (- N)$ is metabolic.
Hence $L \otimes_{\lambda} (M \perp (- N)) = L
\otimes_{\lambda} M \perp (- L \otimes_{\lambda} N)$ is also
metabolic. Therefore $L \otimes_{\lambda} M \approx L
\otimes_{\lambda} N$.\qed
\bigskip

We can now describe the property ``good reduction'' and the
specialization of a form by means of diagonal forms as
follows.

\begtheorem{6.} Let $\varphi$ be a form over $K$, $\dim \varphi = n$.
\setitemindent{$(b)$}
\item{$(a)$} The form $\varphi$ has good reduction with respect to
$\lambda$ if and only if $\varphi$ is Witt equivalent to a diagonal form
$\langle a_1, \dots, a_r \rangle$ with units $a_i \in {\nam o}^*$.
In this case $\dim \lambda_* (\varphi) =
n$ and $\lambda_* (\varphi) \sim
\langle \lambda (a_1), \dots, \lambda (a_r) \rangle$. Furthermore
one can choose $r = n+2$.
\smallskip
\item{$(b)$} Let $2 \in {\nam o}^*$, i.e. $\Char L \ne 2$.
The form $\varphi$ has good reduction with respect to
$\lambda$ if and only if $\varphi$ is isometric to a diagonal form
$\langle a_1, \dots, a_n \rangle$ with $a_i \in {\nam o}^*$.
In this case $\lambda_* (\varphi) \; \cong \; \langle \lambda (a_1), \dots,
\lambda (a_n) \rangle$.
\endtheorem

\begproof{of part $(a)$.} If $\varphi \sim \langle a_1, \dots,
a_r \rangle$ with units $a_i \in {\nam o}^*$, then by the
Corollary $\varphi$ has good reduction with respect to $\lambda$
and $\lambda_* (\varphi) \sim \langle \lambda (a_1), \dots,
\lambda (a_r) \rangle$. Suppose now that $\varphi$ has good reduction with respect to
$\lambda$. The form
$\varphi$ corresponds to a bilinear space $K \otimes_{{\nam o}} M$ over
$K$, which comes from a bilinear space $M$ over~${\nam o}$. If $M/{\nam m}
M$ is not hyperbolic, then $M$ has an orthogonal basis by the
Lemma following Theorem~2. Therefore $\varphi \; \cong \; \langle a_1, \dots,
a_n \rangle$ with units $a_i \in {\nam o}^*$. In general we
consider the space $M' \colon = M \, \perp \, \langle 1, -1 \rangle$ over ${\nam
o}$. Then $M'/{\nam m} M'$ is definitely {\it not} hyperbolic.
Hence $\varphi
\, \perp \, \langle 1, - 1 \rangle \; \cong \; \langle b_1, \dots, b_{n+2}
\rangle$ with\break
units~$b_i$.\qed
\endproof

The proof of part (b) is similar, but simpler since now $M/ {\nam
m} M$ always has an orthogonal basis. We don't need the Corollary
here.
\bigskip


Finally, we consider the specialization of tensor products of
forms.

\begtheorem{7.} Let $\varphi$ and $\psi$ be two forms over $K$,
which have good reduction with respect to $\lambda$. Then $\varphi \otimes
\psi$ also has good reduction with respect to $\lambda$, and\break
$\lambda_* (\varphi \otimes \psi)
\approx \lambda_* (\varphi) \otimes \lambda_* (\psi)$.
\endtheorem

\begproof{} According to Theorem~6 we have the following Witt
equivalences, $\varphi \sim \langle
a_1, \dots, a_m \rangle$, $\psi \sim \langle b_1, \dots, b_n \rangle$, with
units $a_i, b_j \in {\nam o}^*$. Then $\varphi \otimes \psi \sim
\langle a_1 b_1, a_1 b_2, \dots, a_1 b_n, \dots, a_m b_n \rangle$.
Again according to Theorem~6, $\varphi \otimes \psi$ has good
reduction and
$$\eqalign{\lambda_* (\varphi \otimes \psi) & \sim \langle \lambda (a_1)
\lambda (b_1), \dots, \lambda (a_m) \lambda (b_n) \rangle \cr
           & \cong \; \langle \lambda (a_1), \dots, \lambda (a_m) \rangle
           \otimes \langle \lambda (b_1), \dots, \lambda (b_n) \rangle \cr
           & \sim \lambda_* (\varphi) \otimes \lambda_* (\psi).}$$
Now the forms $\lambda_* (\varphi \otimes \psi)$ and $\lambda_*
(\varphi) \otimes \lambda_* (\psi)$ both have the same dimension
as $\varphi \otimes \psi$. Therefore $\lambda_* (\varphi \otimes \psi) \approx \lambda_*
(\varphi) \otimes \lambda_* (\psi)$.\qed
\endproof

If we use a bit more multilinear algebra, we can come up with the
following conceptually more pleasing proof of Theorem~7.
\bigskip

\noindent{\it Second proof of Theorem~7.} Adopting geometric language,
$\varphi$ corresponds to a bilinear space $K \otimes_{{\nam o}} M$
and $\psi$ corresponds to a space $K\otimes_{{\nam o}} N$ with
bilinear  spaces $M$ and $N$ over ${\nam o}$.
Hence $\varphi \otimes \psi$  corresponds to the space
$$(K
\otimes_{{\nam o}} M) \otimes_K (K
\otimes_{{\nam o}} N) \; \cong \; K
\otimes_{{\nam o}} (M \otimes_{{\nam o}} N)$$
over $K$.
Now the free bilinear module $M \otimes_{{\nam o}} N$ is again
nondegenerate (cf. e.g. [MH, I \S 5]). Consequently $\varphi \otimes
\psi$ has good reduction with respect to $\lambda$, and $\lambda_* (\varphi \otimes
\psi)$ can be represented by the space
$L \otimes_{\lambda} (M \otimes_{{\nam o}} N)$, obtained from the
space $M\otimes_\o N$ by base extension to $L$ using the
homomorphism $\lambda|\o:\o\to L$. Now $L
\otimes_{\lambda} (M \otimes_{{\nam o}} N) \; \cong \; (L \otimes_{\lambda}
M) \otimes_L (L \otimes_{\lambda} N)$. In other words, $\lambda_* (\varphi \otimes
\psi)  \cong  \lambda_* (\varphi) \otimes \lambda_* (\psi)$,
since $\lambda_* (\varphi) \;  \widehat{=} \;
L \otimes_{\lambda} M$,
$\lambda_* (\psi) \; \widehat{=} \;
L \otimes_{\lambda} N$.\qed
