\titlea{\S5}{An Elementary Treatise on Quadratic Modules}
\leftrunningtitle{Chapter I. Fundamentals of Specialization Theory}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S5 An Elementary Treatise on
Quadratic Modules}

We want to construct a specialization theory for {\it quadratic}
forms, similar to the theory in \S3 for symmetric bilinear forms.
In order to do this we need some definitions and theorems about
quadratic modules over rings, and in particular over valuation
rings.

Let $A$ be any ring (always commutative, with $1$). We recall some
fundamental definitions and facts about quadratic forms over $A$,
cf. [MH, Appendix~1].

\begdefinition{1.} Let $M$ be an $A$-module. A {\it quadratic form
on} $M$ is a function $q \colon M \to A$, satisfying the
following conditions:
\smallskip
\setitemindent{(1)}
\item{(1)} $q(\lambda x) = \lambda^2 q(x)$ for $\lambda \in A$, $x \in M$.
\smallskip
\item{(2)} The function $B_q \colon M \times M \to A$, $B_q (x, y) \colon = q
(x + y) - q(x) - q(y)$ is a bilinear form on $M$.
\medskip
\noindent The pair $(M,q)$ is then called a {\it quadratic module} over $A$.
\enddefinition

Note that the bilinear form $B_q$ is symmetric. Furthermore
$B_q(x,x)=2q(x)$ for all $x\in M$. If $2$ is a unit in $A$, $2 \in
A^*$, we can retrieve $q$ from $B_q$. Also, every symmetric
bilinear form $B$ on $M$ comes from a quadratic form $q$ in this case, namely
$q(x) = \frac{1}{2} B(x, x)$. So, if $2 \in A^*$, bilinear modules (see \S 3, Def.~2)
and quadratic modules over $A$ are really the same objects.

If $2 \not\in A^*$, and $2$ is not a zero-divisor in $A$, we can
still identify quadratic forms on an $A$-module $M$ with special
symmetric bilinear forms, namely the forms $B$ with $B(x,x)\in 2A$
for all $x\in M$ (``even'' bilinear forms). However, if $2$ is a
zero-divisor in $A$, then quadratic and bilinear modules over $A$
are very different objects.

In what follows, primarily {\it free quadratic modules} will play
a r\^{o}le, i.e. quadratic modules $(M,q)$ for which the
$A$-module is free and always of finite rank. If $M$ is a free
$A$-module with basis $e_1, \dots, e_n$ and $(a_{ij})$ a symmetric
$(n \times n)$-matrix, then there exists exactly one quadratic
form $q$ on $M$ with $q(e_i) = a_{ii}$ and $B_q
(e_i, e_j) = a_{ij}$ for $i \ne j \quad (1 \leq i, j \leq n)$,
namely
$$q \left(\sum\limits_{i = 1}^n x_i e_i \right) = \sum\limits_{i = 1}^n
a_{ii} x_i^2 + \sum\limits_{1 \leq i < j \leq n} a_{ij} x_i x_j.$$

We denote this quadratic module $(M,q)$ by a symmetric matrix in
square brackets, $(M, q) = [a_{ij}]$, and call $[a_{ij}]$ the {\it
value matrix} of $q$ with respect to the basis $e_1, \dots, e_n$.
If $(a_{ij})$ is a diagonal matrix with coefficients $a_1, \dots,
a_n$, then we write $(M, q) = [a_1, \dots, a_n]$.

\begdefinition{2.} Let $(M_1, q_1)$ and $(M_2, q_2)$ be quadratic
$A$-modules. The {\it orthogonal sum} $(M_1, q_1) \, \perp \, (M_2,
q_2)$ is the quadratic $A$-module
$$(M_1 \oplus M_2,\break q_1 \perp q_2),$$
consisting of the direct sum $M_1 \oplus M_2$ and the form
$$(q_1 \perp q_2) (x_1 + x_2) \colon = q_1 (x_1) + q_2 (x_2)$$
for $x_1 \in M_1$, $x_2 \in M_2$.
\enddefinition

If $(M_1, q_1) = [A_1]$ and $(M_2, q_2) = [A_2]$ are free
quadratic modules with corresponding symmetric matrices $A_1,
A_2$, then
$$(M_1, q_1) \, \perp \, (M_2, q_2) = \left[ {A_1 \atop 0} \; {0 \atop A_2}
\right].$$

Now it is also clear how to construct a multiple orthogonal sum\break $(M_1, q_1) \,
\perp \dots \perp \, (M_r, q_r)$. In particular we have for
elements $a_1, \dots, a_r \in A$ that
$$[a_1] \, \perp \dots \perp \, [a_r] = [a_1, \dots, a_r ].$$

Let $(M,q)$ be a quadratic ${\nam o}$-module and suppose that
$M_1$ and $M_2$ are submodules of the ${\nam o}$-module $M$, then
we write $M = M_1 \, \perp \,M_2$, when $B_q (M_1, M_2) = 0$
(``internal'' orthogonal sum, in contrast with the ``external''
orthogonal sum of Definition~2). Clearly
$(M, q) \; \cong \; (M_1, q \vert{M_1}) \, \perp \, (M_2,
q\vert{M_2})$ in this case.

If $\beta \colon M \times M \to A$ is a -- {\it not necessarily
symmetric} -- bilinear form, then $q(x) \colon = \beta (x, x)$ is
a quadratic form on $M$. If $M$ is free, one can easily verify
that every quadratic form on $M$ is of this form. Furthermore, two
bilinear forms $\beta, \beta'$ give rise to the same quadratic
form $q$ exactly when $\beta -\beta' = \gamma$ is an {\it
alternating} bilinear form: $\gamma (x, x) =
0$ for all $x \in M$, and hence $\gamma (x, y) = - \gamma (y, x)$ for all $x, y
\in M$.

Suppose now that $(M_1, B_1)$ is a free bilinear module (cf. \S3)
and that $(M_2, q_2)$ is a free quadratic module. We equip the
free module $M \colon = M_1 \otimes_A M_2$ with a quadratic form
$q$ as follows: first we choose a bilinear form $\beta_2$ on $M_2$
with $q_2 (x) = \beta_2 (x,x)$ for all $x \in M_2$. Next we
form the tensor product $\beta \colon =
B_1 \otimes \beta_2 \colon M \times M \to A$ of the bilinear forms $B_1,
\beta_2$. This bilinear form $\beta$ is characterized by
$$\beta (x_1 \otimes x_2, y_1 \otimes y_2) = B_1 (x_1, y_1) \beta_2 (x_2,
y_2)$$
for $x_1, y_1 \in M_1$ and $x_2, y_2 \in M_2$ (cf. [Bo$_1$, \S 1, No.9]).
Finally we let $q(x) \colon = \beta (x, x)$ for $x \in M_1 \otimes
M_2$. This quadratic form $q$ is {\it independent} of the choice
of the bilinear form $\beta_2$, since if $\gamma_2$ is an
alternating bilinear form on $M_2$, then $B_1 \otimes \gamma_2$ is
an alternating bilinear form on $M$.

\begdefinition{3.} We call $q$ the {\it tensor product} of the
symmetric bilinear form $B_1$ and the quadratic form $q_2$,
denoted $q = B_1 \otimes q_2$, and call the quadratic module $(M,
q)$ the tensor product of the bilinear module $(M_1, B_1)$ and
the quadratic module $(M_2, q_2)$,
$(M, q) = (M_1, B_1) \otimes (M_2, q_2)$.
\enddefinition

The quadratic form $q = B_1 \otimes q_2$ is characterized by $B_q =
B_1 \otimes B_{q_2}$ and
$$q(x_1 \otimes x_2) = B_1 (x_1, x_1) \; q_2 (x_2)$$
for $x_1 \in M_1$, $x_2 \in M_2$. For a one-dimensional bilinear
module $\langle c \rangle$ and a quadratic free module $(M, q)$
there is a natural isometry $\langle c \rangle \otimes (M, q) \; \cong \; (M,
cq)$. In particular is for a symmetric $(n \times n)$-matrix $A$
$$\langle c \rangle \otimes [A] \; \cong \; [c A].$$

Later on we will often denote a quadratic module $(M, q)$ by the
single letter $M$ and an (always symmetric) bilinear module $(E,
B)$ by the single letter $E$. The tensor product $E \otimes M$ is
clearly additive in both arguments,
$$\eqalign{(E_1 \perp E_2) \otimes M \; & \cong \; (E_1 \otimes M) \,
\perp \, (E_2 \otimes M) \cr
           E \otimes (M_1 \, \perp \, M_2) \; & \cong \; (E \otimes M_1) \,
           \perp \, (E \otimes M_2).}$$
Consequently we have for example $(a_i, b_j \in A)$:
$$\langle a_1, \dots, a_r \rangle \otimes [b_1, \dots, b_s ] \; \cong \; [a_1
b_1, a_1 b_2, \dots , a_r b_s ].$$

Let $M = (M, q)$ be a free quadratic $A$-module and $\alpha \colon A \to
C$ a ring homomorphism. We associate to $M$ a quadratic
$C$-module $M' = (M', q')$ as follows: The $C$-module $M'$ is the
tensor product $C \otimes_A M$, formed by means of $\alpha$.
Choose a bilinear module $\beta \colon M \times M \to A$ with $q(x) = \beta (x, x)$
for all $x \in M$. Let $\beta' \colon M' \times M' \to C$ be the
bilinear form over $C$ associated to $\beta$, in other words
$$\beta' (c \otimes x, d \otimes y) = cd \; \alpha (\beta (x, y))
\leqno(\ast)$$
for $x,y \in M$, $c, d \in C$. Let $q' (u) \colon = \beta' (u, u)$ for $u
\in M'$. This quadratic form $q'$ is independent of the choice of
$\beta'$, since if $\gamma$ is an alternating bilinear form on
$M$, then the associated $C$-bilinear form $\gamma'$ on $M'$ is
again alternating. The form $q'$ can be characterized as follows:
$$q' (c \otimes x) = c^2 \alpha (q(x)),\quad
B_{q'} (c \otimes x, d \otimes y) = cd \; \alpha (B_q (x, y)),$$
for $x \in
M$, $y \in M$, $c \in C$, $d \in C$.

\begdefinition{4.} We say that the quadratic $C$-module $(M', q')$
arises from $M = (M, q)$ by means of a {\it base extension}
determined by $\alpha$, and denote $M'$ by $C \otimes_A M$
or by $C\otimes_\alpha M$
or, even more
precisely, by $C \otimes_{A, \alpha} M$. We also use the notation
$q' = q_C$.
\enddefinition

\noindent If $M$ is given by a symmetric matrix, $M = [a_{ij}]$ then
$C\otimes_A M = [\alpha (a_{ij})]$.

Given a bilinear $A$-module $M = (M,B)$, we similarly define
a bilinear $C$-module $C \otimes_A M = (C \otimes_A M, B_C)$,
where $B_C$ is determined in the obvious way by $B$, cf. ($\ast$)
above.

If $M = (M, q)$ is a quadratic module over $A$ and $N$ a submodule
of $M$, we also interpret $N$ as a quadratic module, $N = (N, q
\vert N)$ (quadratic submodule). Furthermore we denote by
$N^{\perp}$ the submodule of $M$ consisting of all elements $x \in M$
with $B_q (x, y) = 0$ for all $y \in N$. Of course, we also
interpret $N^{\perp}$ as a quadratic module. In particular we can
look at the quadratic submodule $M^{\perp}$ of $M$. If $M^{\perp}$
is free of finite rank $r$, then $M^{\perp}$ has the form $[a_1, \dots,
a_r]$ with elements $a_i \in A$.

Later on we will frequently use the following elementary lemma.

\beglemma{1}{\rm [MH, p.5]}{\bf.} Let $M = (M, B)$ be a bilinear $A$-module
and let $N$ be a submodule of $M$. Suppose that the bilinear form
$B$ is nondegenerate on $N$, i.e. the homomorphism
$x \mapsto B(x, -) \vert N$ from $N$ to the dual module
$\raise4pt\hbox{${{\scriptstyle\vee} \atop {\textstyle N}}$}
= \Hom_A (N, A)$ is an isomorphism. Then
$M = N \, \perp \, N^{\perp}$.
\endlemma

Let $M = (M, q)$ be a quadratic module and let $N$ be a submodule of
$M$ such that the bilinear form $B_q$ is
nondegenerate on $N$, then according to the lemma we also have
that $M~=~N~\!\perp\!~N^{\perp}$.

To finish this section, we  briefly examine {\it free
hyperbolic} modules, being quadratic modules of the form
$r\times \left[ {0 \atop 1} \; {1 \atop 0} \right]$ with
$r \in {\Bbb N}_0$, in other words direct sums of $r$ copies of
the quadratic module
$\left[ {0 \atop 1} \; {1 \atop 0} \right]$. $\{$When $r = 0$, the zero module is meant.$\}$

\beglemma{2.} Let $(M, q)$ be a quadratic $A$-module whose
associated bilinear form $B_q$ is nondegenerate. Let $U$ be a
submodule of $M$ with $q \vert U =0$.  Suppose that $U$ is free of
rank $r$ and that it is a direct summand of the $A$-module $M$. Then there
exists a submodule $N \supset U$ of $M$ with $N \; \cong \; r
\times \left[ {0 \atop 1} \; {1 \atop 0} \right]$ and
$M = N \,\perp \, N^{\perp}$.
\endlemma

\begproof{}(cf. [Ba, p.13 f]). We saw already in \S3 (just after
Definition~5) that there exists an exact sequence of $A$-modules
$$0 \longrightarrow U^{\perp} \longrightarrow M \buildrel
\alpha \over \longrightarrow
\raise4pt\hbox{${{\scriptstyle\vee} \atop {\textstyle U}}$}
\longrightarrow 0,$$
with
$\raise4pt\hbox{${{\scriptstyle\vee} \atop {\textstyle U}}$} \colon =
\Hom_A (U, A)$, where $\alpha$ maps an element $z \in M$  to the
linear form $B(z, -) \vert U$ on $U$. Since
$\raise4pt\hbox{${{\scriptstyle\vee} \atop {\textstyle U}}$}$
is free, this sequence splits. Choose a submodule $V$ of $M$ with
$M = U^{\perp} \oplus V$. Then
$\alpha \vert V \colon V \to \raise4pt\hbox{${{\scriptstyle\vee} \atop
{\textstyle U}}$}$ is an isomorphism. The modules $U$ and $V$ are
therefore in perfect duality with respect to the pairing
$U \times V \to A$, $(x, y) \mapsto B_q (x, y)$.

Now choose a -- not necessarily symmetric -- bilinear form
$\beta \colon V \times V \to A$ with $\beta (x, x) = q(x)$ for $x \in
V$. We then have an $A$-linear map $\varphi \colon V \to U$ such
that $\beta (v, x) = B_q (v, \varphi (x))$ for $v \in V$, $x \in
V$. Therefore
$$q(x- \varphi (x)) = q(x) - B_q (x, \varphi (x)) = q(x) - \beta (x,
x) = 0$$
for every $x \in V$. Let
$W \colon = \{ x - \varphi (x) \; \vert \; x \in V \}$. Since
$$B_q (u, x -\varphi (x)) = B_q (u, x)$$
for all $u \in U$ and $x \in
V$, the modules $W$ and $U$ are also in perfect duality with respect to $B_q$.
Furthermore is $q \vert W = 0$. Choose a basis $e_1, \dots, e_r$
for $U$ and let $f_1, \dots, f_r$ be the dual basis of $W$ with
respect to $B_q$, i.e. $B_q (e_i,f_j) = \delta_{ij}$. Then
$N \colon = U \oplus W$ can be written as
$$N = \raise5pt\hbox{$\newbot\limits_{{\scriptstyle i = 1}}^{\lower6pt\hbox{${\scriptstyle
r}$}}$} (A e_i + A f_i) \; \cong \; r \times \left[ {0
\atop 1} \quad {1 \atop 0} \right].$$
An appeal to Lemma~1 yields $M = N \, \perp \, N^{\perp}$.\qed
\endproof

Given a quadratic $A$-module $M = (M, q)$, we denote $(M, -q)$ by
$- M$, as is  already our practice  for bilinear spaces.

\begtheorem{} Let $M = (M, q)$ be a quadratic $A$-module. Suppose
that the bilinear form $B_q$ is nondegenerate and that $M$ is free
of rank $r$. Then
$$M \, \perp \, (-M) \; \cong \; r \times \left[ {0 \atop 1} \quad
{1 \atop 0} \right].$$
\endtheorem

\begproof{} Let $(E, \tilde{q}) \colon = (M, q) \, \perp \, (M, -
q)$. We interpret the module $E = M \oplus M$ as the
cartesian product $M \times M$. The diagonal
$D \colon = \{ (x, x) \; \vert \; x \in M \}$ is a free
$E$-submodule of rank $r$ and $D$ is a direct summand of $E$, for
$E = D \oplus (M \times \{ 0 \})$. Furthermore $q \vert D = 0$.
The statement now follows from Lemma~2.\qed
\endproof
