\titlea{\S6}{Quadratic Modules over Valuation Rings}
\leftrunningtitle{Chapter I. Fundamentals of Specialization Theory}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S6 Quadratic Modules over
Valuation Rings}

In this section we let ${\nam o}$ be a valuation ring with maximal
ideal ${\nam m}$, quotient field $K$ and residue class field
$k = {\nam o}/{\nam m}$. The case ${\nam m} = 0$, i.e. $K = k ={\nam
o}$ is explicitly allowed. We shall present the  theorems
about quadratic modules over ${\nam o}$ necessary for our
specialization theory, and prove most of them.

\beglemma{1.} Let $(M, B)$ be  a free bilinear module over ${\nam
o}$, and let $N$ be a submodule of $M$. The submodule $N^{\perp}$
of $M$ is free with finite basis and is a direct summand of $M$.
Every submodule $L$ of $M$ with $M = N^{\perp} \oplus L$ is also
free with finite basis.
\endlemma

\begproof{} $M/N^{\perp}$ is torsion free and finitely generated,
hence free with finite basis. This follows from the fact that every
finitely generated ideal of ${\nam o}$ is principal, cf.  [CE, VII, \S
4]. (We used this already in the proof of Lemma~2, \S3.) If
$\overline{x}_1, \dots, \overline{x}_r$ is a basis of
$M/N^{\perp}$ and $x_1, \dots, x_r$ are the pre-images of the
$\overline{x}_i$ in $M$, then $L_0 \colon = {\nam o} x_1 +
\dots + {\nam o} x_r$ is free with basis $x_1, \dots, x_r$ and
$M = N^{\perp}\oplus L_0$. If $M = N^{\perp} \oplus L$ as well,
then $L \; \cong \; M/N^{\perp} \; \cong \; L_0$, so that $L$ is
also free with finite basis. Finally, $N^{\perp}$ is torsion free
and finitely generated, hence free with finite basis.\qed
\endproof

In particular, if $(M, B)$ is a free bilinear module over ${\nam o}$,
there is an orthogonal decomposition $M = M_0 \, \perp \, M^{\perp}$
with $B \vert{M^{\perp} \times M^{\perp}} = 0$.

\begdefinition{1.} $M^{\perp}$ is called the {\it radical} of the
bilinear module $(M, B)$. If $(M, q)$ is a free quadratic ${\nam
o}$-module, then the radical $M^{\perp}$ of $(M, B_q)$, equipped
with the quadratic form $q \vert{M^{\perp}}$, is called the {\it
quasilinear part} of $M$, and is denoted by $QL(M)$. If $M = M^{\perp}$,
the quadratic module $(M, q)$ is called {\it quasilinear}.
\enddefinition

\begdefinition{2.} Let $M$ be a free ${\nam o}$-module with basis
$e_1, \dots, e_n$. We call a vector $x \in M$ {\it primitive
in} $M$, if for the decomposition
$x = \lambda_1 e_1 + \dots + \lambda_n e_n$ $(\lambda_i \in
{\nam o})$ the ideal
$\sum\limits_1^m \lambda_i {\nam o}$ is equal to ${\nam o}$, i.e. at least
one $\lambda_i$ is a unit.
\enddefinition

This property is independent of the choice of basis: $x$ is
primitive when $M/{\nam o} x$ is torsion free, hence when
${\nam o} x$ is a direct summand of the module $M$.

\begdefinition{3.} We call a quadratic module $(M, q)$ over ${\nam
o}$ {\it nondegenerate} if it satisfies the following conditions:
\smallskip\noi
{\parindent=1.2cm
\litem{(Q0)} $M$ is free of finite rank.

\noi
\litem{(Q1)} The bilinear form $\overline{B}_q$, induced by $B_q$
on $M/M^{\perp}$ in the obvious way, is nondegenerate.

\noi
\litem{(Q2)} $q(x) \in {\nam o}^*$ for every vector $x$ in
$M^{\perp}$, which is primitive in $M^{\perp}$ (and hence in $M$).
\par}

\medskip\noi
If instead of (Q2), the following condition is satisfied

\smallskip\noi
{\parindent=1.2cm
\litem{(Q2$'$)} \quad
      $QL(M) = 0$ or $QL(M) \; \cong
      \; [\varepsilon]$ with $\varepsilon \in {\nam o}^*$,
\par}
\smallskip
\noi then we call $(M, q)$ {\it regular}. In the special case
$M^{\perp} = 0$, we call $(M, q)$ {\it strictly regular}.
\enddefinition

\begremark{} The strictly regular quadratic modules are identical
with the ``qua\-drat\-ic spaces'' over ${\nam o}$ in [K$_4$]. In case ${\nam o} =
K$, the term ``nondegenerate'' has the same meaning as in [K$_4$].
``Strictly regular'' is a neologism, constructed with the purpose of
avoiding collision with the confusingly many overlapping
terms in the literature.

From a technical point of view, Definition~3 is the core definition
of this book. The idea behind it is, that the term ``nondegenerate''
captures a possibly large class of quadratic ${\nam o}$-modules,
which work well in the specialization theory (see \S7 and
following).
$\{$We will settle upon the agreement that a quadratic $K$-module
$E$ has ``good reduction with respect to ${\nam o}$'' if
$E \; \cong \; K \otimes_{{\nam o}} M$, where $M$ is a
nondegenerate quadratic ${\nam o}$-module, see \S 7, Def.1.$\}$
The requirements (Q0) and (Q1) are obvious, but (Q2) and
(Q2$'$) deserve some explanation.

If $\Char K \ne 2$, i.e. $2 \ne 0$ in ${\nam o}$, then $q \vert{M^{\perp}} =
0$ and the requirement (Q2) implies that $M^{\perp} = 0$, hence implies --
in conjunction with (Q0) and (Q1) -- strict regularity. If $2 \in {\nam
o}^*$, then the nondegenerate quadratic ${\nam o}$-modules are the
same objects as the nondegenerate bilinear ${\nam o}$-modules in
the sense of our earlier definition.

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

On the other hand, property (Q2) has an annoying defect: it is
not always preserved under a basis extension. If
${\nam o}' \supset {\nam o}$ is another valuation ring, whose
maximal ideal ${\nam m}'$ lies over ${\nam m}$, i.e.
${\nam m}' \cap {\nam o} = {\nam m}$, and if $M$ is nondegenerate,
then ${\nam o}' \otimes_{{\nam o}} M$ can be degenerate.

However, if $M$ satisfies (Q2$'$), this clearly cannot happen.
Therefore we will have to limit ourselves later (from \S9
onwards) in some important cases to regular modules.

The arguments will be clearer however, when we allow arbitrary nondegenerate
quadratic modules as long as possible, and a lot of results  in
this generality are definitely important.

Our use of the terms ``regular'' and ``nondegenerate'' finds its
justification in the requirements of the specialization theory
presented in this book. Regular quadratic modules over valuation
rings and fields (in the sense of Definition~3) are just those
quadratic modules, for which the generic splitting
theory  in particular functions in a ``regular way'', as we are
used to in the absence of characteristic $2$ (cf. \S4), see \S9
below. The nondegenerate quadratic modules are those ones, for
which the generic splitting theory still can get somewhere in a
sensible way, see \S10 and \S12 below.

The author is aware of the fact that in other areas (number theory in particular)
a different terminology is used. For example,
Martin Kneser calls our regular modules
``half regular'' and our strictly regular modules
``regular'' in his lecture notes [Kn$_3$], and has a good reason
to do so. In the literature, the word ``nondegenerate'' is
almost exclusively  used for the more restrictive class of quadratic
modules, which we call ``strictly regular'', see Bourbaki [Bo$_1$, \S 3, No.
4] in particular. In our context, however, it would be a little
bit silly to give the label ``degenerate'' to  every quadratic module, which
is not strictly regular.
\endremark

We need some formulas, related to quadratic modules of rank $2$.

\beglemma{2.} Let $\alpha \in {\nam o}$, $\beta \in {\nam o}$, $\lambda
\in {\nam o}^*$. Then
$$\left[ {\alpha \atop \lambda} \quad {\lambda \atop \beta} \right] \; \cong \;
\left[ {\alpha \atop 1} \quad {1 \atop \lambda^{-2} \beta} \right], \leqno(1)$$
and furthermore
$$\langle \lambda \rangle \otimes \left[ {\alpha \atop 1} \quad {1 \atop \beta} \right] \; \cong \;
\left[ {\lambda\alpha \atop 1} \quad {1 \atop \lambda^{-1} \beta} \right].
\leqno(2)$$
Finally, if $\alpha \in {\nam o}^*$, $\beta \in {\nam o}$ then
$$\left[ {\alpha \atop 1} \quad {1 \atop \beta} \right] \; \cong \;
\langle \alpha \rangle \otimes \left[ {1 \atop 1} \quad {1 \atop \alpha \beta} \right].
\leqno(3)$$
\endlemma

\begproof{} Let $(M, q)$ be a free quadratic module with basis
$e, f$ and associated value matrix
$\left[ {\alpha \atop \lambda} \; {\lambda
\atop \beta} \right]$. Then $e, \lambda^{-1} f$ is also a basis of
$M$, whose associated value matrix equals
$\left[ {\alpha \atop 1} \; {1
\atop \lambda^{-2} \beta} \right]$. This settles (1). Furthermore
$\langle
\lambda \rangle \otimes \left[ {\alpha \atop 1} \; {1
\atop \beta} \right]$ is determined by the quadratic module
$(M, \lambda q)$, and so
$\langle \lambda \rangle \otimes \left[ {\alpha \atop 1} \; {1
\atop \beta} \right] \; \cong \; \left[ {\lambda\alpha \atop \lambda} \;
{\lambda \atop \lambda\beta} \right]$. Applying (1) to this,
results in (2). Clearly (3) is a special case of (2).\qed
\endproof

\begtheorem{1.} Let $M = (M, q)$ be a nondegenerate quadratic
module over ${\nam o}$.
\smallskip
\setitemindent{$(a)$}
\item{$(a)$} Then $M$ is an orthogonal sum of modules
$\left[ {\alpha \atop 1} \; {1
\atop \beta} \right]$ with $1-4 \alpha \beta \in {\nam o}^*$ and
modules $[\varepsilon]$ with $\varepsilon \in {\nam o}^*$.
\smallskip
\item{$(b)$} If $M$ is regular and $\dim M$ even $($recall that $\dim
M$ denotes the rank of the free module $M$$)$, then $M$ is strictly
regular and equal to an orthogonal sum of modules
$\left[ {\alpha \atop
1}\; {1 \atop \beta} \right]$ with $1-4 \alpha \beta \in {\nam o}^*$.
\endtheorem

\begproof{} First suppose that $2 \in {\nam o}^*$. Then $(M, B_q)$
is a bilinear space. Hence $(M, B_q)$ has an orthogonal basis, as
is well-known (cf. [MH, I, Cor.3.4]),  and so
$$(M, B_q) \; \cong \; \langle \varepsilon_1, \dots, \varepsilon_n \rangle.$$
Therefore $(M, q) \; \cong \; \left[ \frac{\varepsilon_1}{2}, \dots,
\frac{\varepsilon_n}{2} \right]$. For a binary quadratic module
$N = [a, b]$ with units $a, b \in {\nam o}^*$, we have
$$[a, b] \; \cong \; \left[ {a \atop 2a} \quad {2a \atop a+b} \right],$$
since, if $e, f$ is a basis of the module $N$ with value matrix
$\left[ {a \atop 0} \; {0 \atop b} \right]$, then $e, e+f$ is a
basis with value matrix
$\left[ {a \atop 2a} \; {2a \atop a+b} \right]$.
Lemma~(2), formula~(1) then yields
$$[a, b] \, \cong \; \left[ {a \atop 1} \quad {1 \atop (2a)^{-1} (a+b)}
\right].$$
Hence all the assertions of the theorem are clear when
$2 \in {\nam o}^*$.

Next suppose that $2 \in {\nam m}$. The quasilinear part
$M^{\perp}$ is of the form
$[\varepsilon_1, \dots, \varepsilon_r]$ with $\varepsilon_i \in {\nam
o}^*$. Let $N$ be a module complement of $M^{\perp}$ in $M$. Then
$(N, q\vert N)$ is strictly regular. We will show, by induction on
$\dim N$, that $N$ is the orthogonal sum of binary modules
$\left[ {\alpha \atop 1} \; {1 \atop
\beta} \right]$.


If $N = 0$, nothing has to be done. So suppose that $N \ne 0$. We
choose a primitive vector $e$ in $N$. Since the bilinear form
$B_q \vert{N \times N}$ is nondegenerate, there exists a vector
$f \in N$ with $B_q (e, f) = 1$. Let $\alpha = q(e)$, $\beta =
q(f)$. The determinant of the matrix
$\left( {2 \alpha \atop 1} \; {1
\atop 2 \beta} \right)$ is the unit $1 - 4 \alpha\beta$. Hence the
module ${\nam o} e + {\nam o} f$ is free with basis $e, f$. Using
Lemma~1 from \S5, we get an orthogonal decomposition
$N = ({\nam o} e + {\nam o} f) \, \perp \, N_1$. Now
${\nam o} e+ {\nam o} f \; \cong \; \left[ {\alpha \atop 1} \; {1 \atop \beta} \right]$,
and by our induction hypothesis,  $N_1$ is also an orthogonal sum of
binary quadratic modules of this form.

This establishes part (a) of the proof. In particular, $N$ has
even dimension. If $M$ is regular, then $r \leq 1$. Is in addition
$\dim M$ even, then $r= 0$, in other words, $M$ has to be strictly
regular.\qed
\endproof

If $M$ is a free ${\nam o}$-module, we interpret $M$ as a subset
of the $K$-vecor space $K \otimes_{{\nam o}} M$, by identifying
$x \in M$ with $1 \otimes x$. Then $K \otimes_{{\nam o}} M = KM$.

\begdefinition{4.} A quadratic ${\nam o}$-module $(M, q)$ is
called {\it isotropic} if there exists a vector $x \ne 0$ in
$M$ with $q(x) = 0$. Otherwise $(M, q)$ is called  {\it anisotropic}.
\enddefinition

\begremark{} In this definition of ``isotropic'', we proceed in a
different way, compared to \S3, Definition~6, since the bilinear
form $B_q$ can be degenerate. This would cause trouble over a
local ring instead of over ${\nam o}$.
\endremark

\beglemma{3.} Suppose that $(M, q)$ is a quadratic ${\nam o}$-module, and
that $M$ is free of finite rank over ${\nam o}$. The following
assertions are equivalent:
\smallskip
\setitemindent{$(a)$}
\item{$(a)$} $(M, q)$ is isotropic.
\smallskip
\item{$(b)$} $(K \otimes_{{\nam o}} M, q_K)$ is isotropic.
\smallskip
\item{$(c)$} The module $M$ has a direct summand $V \ne 0$ with $q \vert V
= 0$.
\endlemma

\begproof{} The implications (a) $\Rightarrow$ (b) and (c) $\Rightarrow$
(a) are trivial. (b) $\Rightarrow$ (c): Let $E = K \otimes_{{\nam o}} M =
KM$. Now $q = q_K \vert M$. The $K$-vector space $E$ contains a
subspace $W \ne 0$ with $q_K \vert W = 0$. Let $V \colon = W \cap
M$. The ${\nam o}$-module $V$ is a direct summand of $M$, since
$M/V$ is torsion free and finitely generated, hence free. We have
$KV = W$. Therefore certainly $V \ne 0$ and $q \vert V = 0$.
\qed
\endproof

\begdefinition{5.} Let $(M, q)$ be a quadratic ${\nam o}$-module.
A pair of vectors $e, f$ in $M$ is called {\it hyperbolic} if
$q(e) = q(f) = 0$ and $B_q (e, f) = 1$.
\enddefinition

Note that according to \S5, Lemma~1, we then have that
$$M = ({\nam o} e +
{\nam o} f) \, \perp \, N \; \cong \; \left[ {0 \atop 1} \; {1 \atop 0}
\right] \, \perp \, N$$
where $N$ is a quadratic submodule of $M$,
namely $N = ({\nam o} e + {\nam o} f)^{\perp}$.

\beglemma{4.} Let $(M, q)$ be a  nondegenerate quadratic ${\nam
o}$-module, and let $e$ be a primitive vector in $M$ with $q(e) =
0$. Then $e$ can be completed to a hyperbolic vector pair $e, f$.
\endlemma

\begproof{} We choose a decomposition $M = N \, \perp \,
M^{\perp}$ and write $e = x + y$ with $x \in N$, $y \in
M^{\perp}$. Suppose for the sake of contradiction that the vector
$x$ is not primitive in $N$, hence not primitive in $M$. Then
$y$ is primitive in $M$. According to condition (Q2), we then
have $q(y) \in {\nam o}^*$. Hence also $q(x) = - q(y) \in {\nam
o}^*$. Therefore $x$ has to be primitive, contradiction.

Hence $x$ is primitive in $N$. Since $B_q$ is nondegenerate on
$N$, there exists an element $z \in N$ with $B_q(x, z) = 1$. We also have
$B_q(e, z) = 1$. Clearly $f \colon = z - q(z) e$ completes the
vector $e$ to a hyperbolic pair.\qed
\endproof

\begtheorem{2}{\rm(Cancellation Theorem).} Let $M$ and $N$ be free
quadratic ${\nam o}$-modules. Let $G$ be a {\rm strictly regular}
quadratic ${\nam o}$-module, and suppose that
$G \, \perp \, M \; \cong \; G \, \perp \, N$. Then $M\; \cong \;
N$.
\endtheorem

For the proof we refer to [K$_3$], in which such a cancellation
theorem is  proved over local rings. An even more general
theorem can be found in Kneser's works [Kn$_2$], [Kn$_3$, Erg\"anzung zu Kap.~I].
A very accessible source for many aspects of the theory of
quadratic forms over local rings is the book [Ba] of R.~Baeza.
Admittedly, Baeza only treats strictly regular quadratic modules,
which he calls ``quadratic spaces''.

In the field case ${\nam o} = K$, Theorem~2 was already
demonstrated for $2 = 0$ by Arf [A] and for $2 \ne 0$ -- as is well-known -- by
Witt [W]. A very nice discussion of the cancellation problem over
fields, with a view towards the theory over local rings, can
again be found in  Kneser's work [Kn$_1$].

\begtheorem{3.} Let $M$ be a nondegenerate quadratic module over ${\nam
o}$. There exists a decomposition
$$M \; \cong \; M_0 \, \perp \, r \times \left[{0 \atop 1} \quad {1 \atop 0}
\right] \leqno(\ast)$$
with $M_0$ nondegenerate and anisotropic, $r \geq 0$. The number $r \in {\Bbb
N}_0$ and the isometry class of $M_0$ are uniquely determined by
$M$.
\endtheorem

\begproof{} Lemma 3 and Lemma 4 immediately imply the existence of
a decomposition as in~$(\ast)$. Suppose now that
$$M \; \cong \; M'_0 \, \perp \, r' \times \left[{0 \atop 1} \quad {1 \atop 0}
\right]$$
is another decomposition. Suppose without loss of generality that $r \leq
r'$. Then Theorem~2 implies that
$$M_0 \; \cong \; M'_0 \, \perp \, (r'-r) \times \left[{0 \atop 1} \quad {1 \atop 0}
\right].$$
Since $M_0$ is anisotropic, we must have $r = r'$.\qed
\endproof

In the field case ${\nam o} = K$, this theorem can again be found
back in Arf [A] for $2=0$ and in Witt [W] for $2\neq0$.

\begdefinition{6.} We call a decomposition $(\ast)$, as in
Theorem~3, a {\it Witt decomposition} of $M$. We call $M_0$ the
{\it kernel module} of the nondegenerate quadratic module $M$ and $r$ the
{\it Witt index} of $M$ and write $M_0 = \ker (M)$, $r = \ind
(M)$.

Furthermore, we call two nondegenerate quadratic modules
$M$ and $N$ over~${\nam o}$ {\it Witt equivalent}, denoted by $M \sim
N$, when $\ker (M) \; \cong \; \ker (N)$. We denote the Witt class of $M$,
i.e. the equivalence class of $M$ with respect to $\sim$, by $\{ M
\}$.
\enddefinition

It is now tempting to define an addition of Witt classes $\{ M \}$, $\{ N
\}$ of two nondegenerate quadratic modules $M, N$ over ${\nam o}$
in the same way as we did this for nondegenerate bilinear modules
over fields in \S2, namely $\{ M \} + \{ N \}
\colon = \{M \, \perp \, N \}$.

Although  the orthogonal sum of two nondegenerate quadratic modules
could be degenerate, the addition makes sense if one of the
modules $M$, $N$ is strictly regular, as we will show now.

\begproposition{} Let $M, M'$ be strictly regular quadratic ${\nam
o}$-modules with\break $M \sim M'$, and let $N, N'$ be nondegenerate
quadratic ${\nam o}$-modules with $N \sim N'$.  Then the modules
$M \, \perp \, N$ and $M' \, \perp \,N'$ are nondegenerate and
$M \, \perp \, N \sim M' \, \perp \, N'$.
\endproposition

\begproof{} It follows immediately from Definition~3 that
$M \, \perp \, N$ and $M' \, \perp \, N'$ are nondegenerate.
Suppose without loss of generality that $\dim M' \geq \dim M$.
Then
$M' \; \cong \; M \, \perp \, s \times \left[ {0 \atop 1} \; {1 \atop 0}
\right]$ for an $s \geq 0$. Therefore
$M' \, \perp \, N \; \cong \; (M \, \perp \, N) \, \perp \, s \times \left[ {0 \atop 1} \; {1 \atop 0}
\right]$, and so
$M' \, \perp \, N \sim M \, \perp \, N$. A similar argument shows
that $M' \, \perp \, N \sim M' \, \perp \, N'$.\qed
\endproof

\begdefinition{7a.} We denote the set of Witt classes of
nondegenerate quadratic ${\nam o}$-modules by $\widetilde{W}\!q ({\nam
o})$. We denote the subset of Witt classes of regular, resp. strictly
regular quadratic ${\nam o}$-modules by $W\!qr ({\nam o})$, resp.
$W\!q ({\nam o})$.
\enddefinition

Because of the proposition above, we now have  a well-defined
``addition'' of classes $\{ M \} \in W\!q ({\nam o})$ with classes
$\{ N \} \in \widetilde{W}\!q ({\nam o})$,
$$\{ M \} + \{ N \} \colon = \{ M \, \perp \, N \}.$$
The zero module $M = 0$ gives rise to a neutral element for this
addition,
$\{ 0 \} + \{ N \} = \{ N \}$, $\{ M \} + \{ 0 \} = \{ M \}$.
If $M$ and $N$ are strictly regular, then
$\{ M \} + \{ N \} = \{ N \} + \{ M \} =$
$\{ M \, \perp \, N \} \in W\!q ({\nam o})$.
Therefore $W\!q ({\nam o})$ is an abelian semigroup with respect
to this sum. The theorem at the end of \S5 shows that
$W\!q ({\nam o})$ is in fact an abelian group, since if
$(M, q)$ is a strictly regular quadratic ${\nam o}$-module, then
$\{ (M, q)\} + \{( M, -q)\} = 0$.

\begdefinition{7b.} We call the abelian group $W\!q ({\nam o})$
the {\it quadratic Witt group} of~${\nam o}$, the set
$\widetilde{W}\!q ({\nam o})$ the {\it quadratic Witt set} of
${\nam o}$ and $W\!qr ({\nam o})$ the {\it regular quadratic Witt
set} of ${\nam o}$.
\enddefinition

The group $W\!q ({\nam o})$ acts on the set $\widetilde{W}\!q ({\nam
o})$ by means of the addition, introduced above. One can show
easily that the isomorphism classes of nondegenerate quasilinear
quadratic ${\nam o}$-modules (see Definition~1) form a system of
representatives of the orbits of this action. $W\!qr ({\nam o})$
is a union of orbits.

If ${\nam m} = 0$, hence ${\nam o} = K$, then Witt classes of {\it
degenerate} quadratic modules can be defined without difficulties.
We will explain this next.

\begdefinition{8.} Let $M = (M, q)$ be a finite dimensional
quadratic module over $K$. The {\it defect} $\delta (M)$ of
$M$ is the set of all $x\in M^{\perp }$ with $q(x) = 0$.
\enddefinition

The defect $\delta (M)$ is clearly a subspace of $M$, and $M$ is
nondegenerate if and only if $\delta (M) = 0$. The form $q$
induces a quadratic form $\overline{q} \colon M/\delta (M) \to K$
on the vector space $M/\delta (M)$ in the obvious way: $\overline{q}
(\overline{x}) \colon = q(x)$ for $x \in M$, where
$\overline{x}$ is the image of $x$ in $M/\delta (M)$.  The
quadratic $K$-module $(M/\delta (M), \overline{q})$ is clearly
nondegenerate.

\begdefinition{9.} If $\delta (M) = 0$, i.e. if $M$ is
nondegenerate, then we call $M = (M, q)$ a {\it quadratic space}
over $K$. In general we denote the quadratic space
$(M/\delta(M), \overline{q})$ over $K$ by $\raise4pt\hbox{$
{{\scriptstyle\wedge} \atop {\textstyle M}}$}$, and call
$\raise4pt\hbox{${{\scriptstyle\wedge} \atop {\textstyle M}}$}$
the {\it quadratic space associated to} $M$.
\enddefinition

We have
$$M \; \cong \; \raise4pt\hbox{$
{{\scriptstyle\wedge} \atop {\textstyle M}}$} \, \perp \, \delta (M) \cong
\raise4pt\hbox{$
{{\scriptstyle\wedge} \atop {\textstyle M}}$}\, \perp \, s
\times [0]$$
with $s \colon = \dim \delta(M)$. Furthermore, given a
quadratic space $E$ over $K$, we obviously have
$$\delta (E \, \perp \, M) =
\delta (M)\quad \hbox{ and } \quad(E \, \perp \, M)^{\wedge} \; \cong \; E \, \perp \,
\raise4pt\hbox{$
{{\scriptstyle\wedge} \atop {\textstyle M}}$}.$$

Now it is clear, that in the field case, Theorem~3 can be expanded
as follows:

\begtheorem{3$'$.} Let $M$ be a finite dimensional  quadratic
$K$-module. There\break
exists
a decomposition
$$M \; \cong \; M_0 \, \perp \, r \times \left[ {0 \atop 1} \quad {1 \atop 0}
\right]$$
with $\raise4pt\hbox{$
{{\scriptstyle\wedge} \atop {\textstyle M}}$}_0$ anisotropic and $r \geq
0$. The number $r \in {\Bbb N}_0$ and the isometry class of $M_0$
are uniquely determined by $M$.
\endtheorem

\begdefinition{10.} We call $M_0$ the {\it kernel module} of $M$
and $r$ the {\it Witt index} of $M$,  and write
$M_0 = \ker M$, $r = \ind (M)$. Just as before, we call two finite
dimensional  quadratic $K$-modules $M$ and $N$ {\it Witt
equivalent}, and write $M \sim N$, when
$\ker (M) \; \cong \; \ker (N)$. We denote the Witt equivalence
class of $M$ by $\{ M \}$.
\enddefinition

\begremark{} It may seem more natural to call $M$ and $N$
equivalent when their ``kernel {\it spaces\/}''
$(\ker M)^{\wedge}$, $(\ker N)^{\wedge}$ are isometric. Our
results in \S7 and \S11 are a bit stronger if we use the finer
equivalence defined above.
\endremark

We denote the set of Witt classes $\{ M \}$ of finite dimensional
quadratic $K$-modules $M$ by $\widehat{W}\!q (K)$ and call this
the {\it defective quadratic Witt set} of $K$. The abelian group $W\!q (K)$
acts on $\widehat{W}\!q (K)$ in the usual way:
$\{ E \} +\{ M \} \colon = \{ E \, \perp \, M \}$ for $\{ E \} \in W\!q (K)$, $\{ M \}
\in \widehat{W}\!q (K)$. It even makes sense to add two arbitrary
Witt classes together:
$$\{ M \} + \{ N \} \colon = \{ M \, \perp \, N \}.$$
The result is independent of the choice of representatives
$M$, $N$, as can be shown by an argument, analogous to the proof
of the proposition, preceding Definition~7a. Therefore
$\widehat{W}\!q (K)$ is an abelian semigroup, having
$W\!q(K)$ as a subgroup.

{\it Look out!} If $\Char K=2$, the semigroup $\widehat{W}\!q (K)$
does not satisfy a cancellation rule. Is for example
$a \in K^*$, then $[a] \, \perp \, [a] \;
\cong \; [0] \, \perp \, [a]$, but it is {\it not} so that $[a] \sim [0]$.

To a field homomorphism $\alpha \colon K \to K'$ (i.e. a field
extension), we can associate a well-defined map
$\alpha_* \colon\widehat{W}\!q (K) \to \widehat{W}\!q (K')$, which
sends the Witt class $\{ M \}$ to the class $\{ K' \otimes_{\alpha} M
\}$. This map is a semigroup homomorphism.  By restricting
$\alpha_*$, we can find maps from $W\!q (K)$ to $W\!q (K')$ and
from $W\!qr (K)$ to $W\!qr(K')$, but (if $\Char K = 2$) in
general {\it no} map from $\widetilde{W}\!q(K)$ to
$\widetilde{W}\!q (K')$. This is the reason why we sometimes
will extend the Witt set $\widetilde{W}\!q (K)$ to the
semigroup $\widehat{W}\!q (K)$.

Let us now return to quadratic modules over arbitrary valuation
rings.

At the end of \S5 we introduced free hyperbolic modules. From now
on, we call those modules simply ``hyperbolic''.
$\{$ In this way, we stay in harmony with the terminology, used in the
literature, for valuation rings (or local rings in general), cf.
[B$_1$, V \S 2], [B$_2$].$\}$ Hence


\begdefinition{11.} A  quadratic ${\nam o}$-module $M$ is called
{\it hyperbolic} if
$$M \; \cong \; r \times \left[ {0 \atop 1} \; {1 \atop 0}
\right]$$
for some  $r \in {\Bbb N}_0$.
\enddefinition

So, if $M$ is not the zero module, this means that $M$ contains a
basis $e_1, f_1, e_2, f_2, \dots, e_r, f_r$, where
$e_i, f_i$ are  hyperbolic vector pairs, such that
${\nam o} e_i + {\nam o} f_i$ is orthogonal with
${\nam o} e_j + {\nam o} f_j$ for $i \ne j$. Clearly a hyperbolic
module is strictly regular.

\beglemma{5.} Let $M = (M, q)$ be a strictly regular  quadratic ${\nam
o}$-module. $M$ is hyperbolic if and only if $M$ contains a
submodule $V$ with $q \vert V = 0$ and $V = V^{\perp}$.
\endlemma

\begproof{} If $M$ is hyperbolic and $e_1, f_1, \dots, e_r, f_r$
is a basis, consisting of pairwise orthogonal hyperbolic vector
pairs, then the module
$V \colon = {\nam o} e_1 + {\nam o} e_2 + \dots + {\nam o} e_r$
clearly has the properties
$q \vert V = 0$ and $V = V^{\perp}$. Suppose now that $M$ is
strictly regular and that $V$ is a submodule satisfying these
properties. Since $M/V^{\perp} = M/V$ is torsion free and finitely
generated, it is free. Therefore $V$ is a direct summand of the
${\nam o}$-module $M$. $V$ is torsion free and finitely generated
as well, and so also free. The form $B_q$ gives rise to an exact
sequence
$$0 \longrightarrow V^{\perp} \longrightarrow M \longrightarrow
\raise4pt\hbox{${{\scriptstyle\vee} \atop {\textstyle V}}$} \longrightarrow
0$$
(cf. the proof of \S5, Lemma~2), from which we deduce that
$\dim M = \dim V + \dim V^{\perp} = 2\dim V$.

According to the theorem at the end of \S5, there is an orthogonal
decomposition $M = N\, \perp \, N^{\perp}$ with $V \subset N$ and
$N$ hyperbolic. Using $V$ to form $V^{\perp}$ in $N$, we see that
$V = V^{\perp}$. We conclude that $\dim N = 2 \dim V$ as well, so
that $N = M$.\qed
\endproof

\begtheorem{4.} Let $M$ be a strictly regular  quadratic
${\nam o}$-module with $K \otimes_{{\nam o}} M$ hyperbolic. Then
$M$ itself is hyperbolic.
\endtheorem

\begproof{} We apply  the criterion, given by Lemma~5, two times. As
before we interpret $M$ as an ${\nam o}$-submodule of $E \colon = K\otimes_{{\nam o}}
M$, i.e. $M \subset E$, $E= KM$. Now $E$ contains a subspace $W$
with $q \vert W = 0$, $W^{\perp} = W$. The intersection
$V\colon = W \cap M$ is an ${\nam o}$-submodule of $M$ with $q \vert V = 0$ and
$V^{\perp} = V$. (Here $V^{\perp}$ is considered in  $M$.)\qed
\endproof

\begdefinition{12.} The valuation ring ${\nam o}$ is called {\it
quadratically henselian} if for every $\gamma \in {\nam m}$, there
exists a $\lambda \in {\nam o}$ such that $\lambda^2-\lambda = \gamma$.
\enddefinition

\begremark{} One can convince oneself that this means that
``Hensel's Lemma'' is satisfied by monic polynomials of
degree~$2$. Every henselian valuation ring is of course
quadratically henselian (cf. [Ri, Chap. F], [E, \S 16]). In
Definition~12 we  isolated that part of the property
``henselian'', which is important for the theory of quadratic
forms. Note that, if ${\nam o}$ is quadratically henselian, then every
$\varepsilon \in 1 + 4 {\nam m}$ is a square in ${\nam o}$, since
$\varepsilon = 1 + 4 \gamma$ and $\gamma = \lambda^2 -
\lambda$ imply that $\varepsilon = (1-2 \gamma)^2$.
\endremark


\beglemma{6.} Let ${\nam o}$ be quadratically henselian. Let
$\alpha, \beta \in K$ and $\alpha \beta \in {\nam m}$. Then we
have that over $K$:
$$\left[ {\alpha \atop 1} \quad {1 \atop \beta} \right] \; \cong \;
\left[ {0 \atop 1} \quad {1 \atop 0} \right].$$
\endlemma

\begproof{} According to Lemma~2, formula (3), we have
$$\left[ {\alpha \atop 1} \quad {1 \atop \beta} \right]  \; \cong \; \langle
\alpha \rangle \otimes \left[ {1 \atop 1} \quad {1 \atop \alpha\beta}
\right].$$
There exists a $\lambda \in {\nam o}$ with $\lambda^2 + \lambda + \alpha \beta =
0$. If $e, f$ is a basis of the module
$\left[ {1 \atop 1} \; {1 \atop \alpha \beta}\right]$,
having the indicated value matrix, then
$q(\lambda e + f) = \lambda^2+ \lambda + \alpha \beta = 0$.
Therefore
$\left[ {\alpha \atop 1} \; {1 \atop \beta}
\right]$ is isotropic. Applying Lemma~4, with $\o=K$, gives
$\left[ {\alpha \atop 1} \; {1 \atop \beta}
\right] \; \cong \; \left[ {0 \atop 1} \; {1 \atop 0}
\right]$.\qed
\endproof

\begtheorem{5.} Let ${\nam o}$ be quadratically henselian. Let $(M,
q)$ be a nondegenerate anisotropic quadratic module over ${\nam
o}$. Then $q(e)$ is a unit for every vector $e \in M$ which is
primitive in $M$.
\endtheorem

\begproof{} We choose a decomposition $M = N \, \perp \,
M^{\perp}$. Let $e\in M$ be primitive. Suppose for the sake of
contradiction that $q(e) \in {\nam m}$. We write
$e = x+ y$ with $x \in N$, $y \in M^{\perp}$, and have
$q(e) = q(x) + q(y)$.
\medskip

\noi{\it Case 1:} $x$ is primitive in $M$, hence also in $N$. Then
there exists an $f \in N$ with $B_q (x, f) = 1$, since $B_q$ is
nondegenerate on $N$. We have $B_q(e, f) = 1$. Therefore
$${\nam o} e + {\nam o} f \; \cong \; \left[ {\alpha \atop 1} \; {1 \atop
\beta} \right]$$
with
$\alpha = q(e) \in {\nam m}$, $\beta = q(f) \in {\nam o}$.
According to Lemma~6,  ${\nam o} e + {\nam o} f$ is isotropic. This
is a contradiction, since $M$ is anisotropic.
\medskip

\noi{\it Case 2:} $x$ is not primitive in $M$. Now $x = \lambda
x_0$ with $x_0$ primitive in $M$, $\lambda \in {\nam m}$. The
vector $y$ has to be primitive in $M$. Since $q$ is nondegenerate,
it follows that $q(y) \in {\nam o}^*$. We have
$$q(e) = \lambda^2 q (x_0) + q(y) \in {\nam o}^*,$$
contradicting the assumption that $q (e) \in {\nam m}$.
\medskip

We conclude that $q(e) \in {\nam o}^*$.\qed
\endproof

Later on,   we will use Lemma~6 and Theorem~5 for a nondegenerate
quadratic module $M$ over an arbitrary valuation ring ${\nam o}$,
by going from $M$ to the quadratic module $M^h \colon = {\nam o}^h
\otimes_{{\nam o}} M$ over the henselisation ${\nam o}^h$ of ${\nam
o}$ (see [Ri, Chap.F], [E, \S 17]). $\{$Remark: One can also form
a ``quadratic henselisation''  ${\nam o}^{qh}$ in the obvious way.
It would suffice to take ${\nam o}^{qh}$ instead of ${\nam
o}^h$.$\}$ In order to do so, the following lemma is important.

\beglemma{7.} Let $M$ be a nondegenerate quadratic module over ${\nam
o}$. Then the quadratic module $M^h$ over ${\nam o}^h$ is also
nondegenerate.
\endlemma

\begproof{} The properties (Q0) and (Q1) from Definition~3
stay valid for every basis extension. Hence we suppose that
$M$ satisfies (Q0) and (Q1), so that we only have to consider
(Q2). We have
$(M^h)^{\perp} = ({\nam o}^h
\otimes_{{\nam o}} M)^{\perp} = {\nam o}^h \otimes_{{\nam o}} M^{\perp} =
(M^{\perp})^h$.


The property (Q2) for $M$ says that the quasilinear space
$k\otimes_{{\nam o}} M^{\perp} = k \otimes_{{\nam o}} QL(M)$ is anisotropic
over $k$. To start with, we can identify $k \otimes_{{\nam o}} M$
with the quadratic module $M/{\nam m} M$ over $k = {\nam o}/{\nam
m}$, whose quadratic form $\overline{q} \colon M/{\nam m} M \to k$
is induced by the quadratic form $q \colon M \to {\nam o}$
in the obvious way.
In the same way we then have
$k \otimes_{{\nam o}} M^{\perp} = M^{\perp}/{\nam m}
M^{\perp}$. A vector $x \in M^{\perp}$ is primitive if and only if
its image $\overline{x} \in M^{\perp}/{\nam m} M^{\perp}$ is
nonzero. (Q2) implies then that
$\overline{q} (\overline{x}) \ne 0$.

Let ${\nam m}^h$ denote the maximal ideal of ${\nam
o}^h$. It is well-known that ${\nam m}^h \cap {\nam o} = {\nam m}$
and that the residue class field ${\nam o}^h/{\nam m}^h$ is
canonically isomorphic to ${\nam o}/{\nam m} = k$.
We identify ${\nam o}^h/{\nam m}^h$ with $k$. Then
$k \otimes_{{\nam o}^h} M^h = k \otimes_{{\nam o}} M$ and also
$k \otimes_{{\nam o}^h} QL(M^h) = k \otimes_{{\nam o}} QL(M)$.
Therefore (Q2) is satisfied for $M$ if and only if (Q2) is
satisfied for $M^h$.\qed
\endproof
