\part{Chapter I}{Fundamentals of Specialization Theory}
\leftrunningtitle{Chapter I. Fundamentals of Specialization Theory}

\titlea{\S1}{Introduction: on the Problem of Specialization of
Quadratic and Bilinear Forms}
%\leftrunningtitle{Specialization of Quadratic and Symmetric Bilinear
%Forms}
\leftrunningtitle{Chapter I. Fundamentals of Specialization Theory}
\rightrunningtitle{\S1 Introduction}

Let $\varphi$ be a nondegenerate symmetric bilinear form over a
field $K$, in other words
$$\varphi (x, y) = \sum\limits_{i, j = 1}^n a_{ij} x_i y_j,$$
where $x = (x_1, \dots, x_n) \in K^n$ and $y = (y_1, \dots,
y_n) \in K^n$ are vectors, $(a_{ij})$ is a symmetric
$(n \times n)$-matrix with coefficients $a_{ij} = a_{ji} \in K$ and $\det
(a_{ij}) \ne 0$.
We like to write $\varphi = (a_{ij})$. The number of variables $n$
is called the {\it dimension of} $\varphi$, $n = \dim \varphi$.

Let also $\lambda \colon K \to L \cup \infty$ be a place,
${\nam o} = {\nam o}_{\lambda}$ the valuation ring associated to
$K$ and ${\nam m}$ the maximal ideal of ${\nam o}$.
We denote the group of units of ${\nam o}$ by
${\nam o}^*$, ${\nam o}^* = {\nam o} \setminus {\nam m}$.

We would like $\lambda$ to ``specialize'' $\varphi$  to a
bilinear form $\lambda_* (\varphi)$ over $L$.
When is this possible in a reasonable way?
If all $a_{ij} \in {\nam o}$ and if $\det (a_{ij}) \in {\nam
o}^*$,
then one can associate the nondegenerate form $(\lambda (a_{ij}))$
over $L$ to $\varphi$. This naive idea leads us to the following

\begdefinition{} We say that $\varphi$ has {\it good reduction with
respect to}
$\lambda$ when $\varphi$ is isometric to a form
$(c_{ij})$ over $K$ with $c_{ij} \in {\nam o}$, $\det (c_{ij})
\in {\nam o}^*$. We then call the form $(\lambda (c_{ij}))$
``the'' {\it specialization of} $\varphi$ with respect to $\lambda$. We denote
this specialization by $\lambda_* (\varphi)$.
\enddefinition

\begnote{} $\varphi = (a_{ij})$ is isometric to $(c_{ij})$ if and only if there
exists a matrix  $S \in \GL (n, K)$ with $(c_{ij}) = \;
^tS(a_{ij}) S$. In this case we write $\varphi \cong (c_{ij})$.
\endnote

We also allow the case $\dim \varphi = 0$, standing for
the unique bilinear form on the zero vector space,
the form $\varphi = 0$. We agree that the form $\varphi = 0$ has
good reduction and set $\lambda_* (\varphi) = 0$.

\begproblem{1.} Is this definition meaningful? Up to isometry $\lambda_*
(\varphi)$ should  be independent of the choice of
the matrix $(c_{ij})$.
\endproblem

We will later see that this indeed the case, provided $2
\not\in {\nam m}$, so that $L$ has characteristic $\not=2$. If
$L$ has characteristic $2$, then $\lambda_* (\varphi)$ is
well-defined up to ``stable isometry'' (see \S3).

\begproblem{2.} Is there a meaningful way in which one can
associate  a symmetric bilinear form over $L$ to $\varphi$,
when $\varphi$ has bad reduction?
\endproblem{}

With regards to this problem we would like to recall a classical
result of T.A.~Springer, which lets us suspect that finding
a solution to the problem is not completely beyond hope.
Let $v \colon K \to {\Bbb Z} \cup
\infty$ be a discrete valuation of a field $K$ with associated
valuation ring ${\nam o}$. Let $\pi$ be a generator of the maximal
ideal ${\nam m}$ of ${\nam o}$, so that ${\nam m} = \pi {\nam
o}$. Finally, let $k = {\nam o}/{\nam m}$ be the residue class
field of ${\nam o}$ and $\lambda \colon K \to k \cup \infty$ the
canonical place with valuation ring ${\nam o}$. We suppose that $2 \not\in {\nam
m}$, so that $\Char k \ne 2$ is.

Let $\varphi$ be a nondegenerate symmetric bilinear form over
$K$. Then there exists a decomposition $\varphi
\cong \varphi_0 \, \bot \, \pi \varphi_1$, where $\varphi_0$
and $\varphi_1$ have good reduction with respect to
$\lambda$. Indeed, we can
choose a diagonalisation $\varphi \cong \langle
a_1, \dots, a_n \rangle$. $\{$As usual
$\langle a_1, \dots, a_n \rangle$ denotes the diagonal matrix
$\left( \vcenter{\hbox{$a_1 \quad \quad 0$}
                       \vskip-0.1cm
                                \hbox{$\quad \ddots$}
                                \vskip-0.2cm
                                \hbox{$0  \quad \quad a_n$}}
                                \right)$.$\}$
Then we can arrange that $v(a_i)=0$ or $1$ for each $i$, by multiplying the $a_i$ by squares and renumbering  indices to get $a_i \in {\nam o}^*$
for $1 \leq i \leq t$ and $a_i = \pi \varepsilon_i$, where
$\varepsilon_i \in {\nam o}^*$ for $t < i \leq n$.
$\{$Possibly  $t = 0$, so that $\varphi_0 = 0$, or $t = n$, so that $\varphi_1
= 0$.$\}$

\begtheorem{ }{\rm (Springer 1955 [Sp]).} Let $K$ be complete with
respect to the discrete valuation $v$. If $\varphi$ is anisotropic
$($i.e. there is no vector $x \ne 0$ in $K^n$ with $\varphi (x, x) =
0$$)$, then the forms $\lambda_* (\varphi_0)$ and $\lambda_*
(\varphi_1)$ are anisotropic and up to isometry independent of
the choice of decomposition $\varphi \cong \varphi_0 \, \bot \,
\pi \varphi_1$.

Conversely, if $\psi_0$ and $\psi_1$ are
anisotropic forms over $k$, then there exists up to isometry a
unique anisotropic form $\varphi$ over $K$ with
$\lambda_* (\varphi_0) \cong \psi_0$ and
$\lambda_* (\varphi_1) \cong \psi_1$.
\endtheorem

 Given any place $\lambda
\colon K \to L \cup \infty$ and any form $\varphi$ over $K$,
Springer's theorem suggests to
look for a ``weak specialization'' $\lambda_W (\varphi)$ by
orthogonally decomposing $\varphi$ in a form $\varphi_0$ with good
reduction and a form $\varphi_1$ with ``extremely bad'' reduction,
subsequently forgetting $\varphi_1$ and setting $\lambda_W (\varphi) =
\lambda_* (\varphi_0)$.

Given an arbitrary valuation ring ${\nam o}$, this sounds like a
daring idea. Nonetheless we will see in \S3 that a weak
specialization can be defined in a meaningful way. Admittedly
$\lambda_W (\varphi)$ is not uniquely determined by $\varphi$ and $\lambda$
up to isometry, but up to so-called Witt equivalence.
In the situation of Springer's theorem, $\lambda_W (\varphi)$ is
then the Witt class of $\varphi_0$ and $\lambda_W (\pi \varphi)$ the
Witt class of $\varphi_1$.

A {\it quadratic form} $q$ of dimension $n$ over $K$ is a function
$q \colon K^n \to K$, defined by a homogeneous polynomial of
degree $2$,
$$q(x) = \sum\limits_{1 \leq i \leq j \leq n} a_{ij} x_ix_j$$
\noi
$(x = (x_1, \dots, x_n) \in K^n)$. We can  associate (a
possibly degenerate) symmetric bilinear form
$$B_q (x, y) = q(x+y) - q(x) - q(y)
                      = \sum\limits_{i = 1}^n 2 a_{ii} x_i y_i +
                      \sum\limits_{i < j} a_{ij} (x_iy_j + x_j
                      y_i)$$
to $q$. It is clear that  $B_q (x, x) = 2 q(x)$ for all $x \in
K^n$.

If $\Char K\not=2$,
%$\{\Char K$ denotes the characteristic of
%$K\}$
 then any symmetric bilinear form $\varphi$ over $K$
corresponds to just one quadratic form $q$ over $K$ with
$B_q=\varphi$, namely $q(x) = \frac{1}{2}\, \varphi (x, x)$.
In this way we can interpret a quadratic form as a symmetric
bilinear form and vice versa. In characteristic~$2$ however,
quadratic forms and symmetric bilinear forms are very different
objects.

\begproblem{3.} Let $\lambda \colon K \to L \cup \infty$ be a
place.
\smallskip

\noindent (a) To which quadratic forms $q$ over $K$ can we
associate  ``specialized'' quadratic forms $\lambda_* (q)$ over
$L$ in a meaningful way?\hfill\break
\noindent (b) Let $\Char L=2$ and $\Char K\not=2$, hence $\Char
K=0$. Should one specialize a quadratic form $q$ over $K$ with
respect to $\lambda$ as a quadratic form, or rather as a symmetric bilinear
form?
\endproblem

In what follows we will present a specialization theory for
arbitrary nondegenerate symmetric bilinear forms (\S3), but only
for a rather small class of quadratic forms, the so-called
``obedient'' quadratic forms (\S7). Problem~3(b) will be answered
unequivocally. If $q$ is obedient, $B_q$ will determine a really
boring bilinear form $\lambda_* (B_q)$ (namely a hyperbolic form)
which gives almost no information about $q$. However, $\lambda_*
(q)$  can give important information about~$q$.
If possible, a specialization in the quadratic sense is thus to be
preferred over a
specialization in the bilinear sense.
