\titlea{\S2}{An Elementary Treatise on Symmetric Bilinear Forms}
%\leftrunningtitle{Specialization of Quadratic and Symmetric Bilinear
%Forms}
\leftrunningtitle{Chapter I. Fundamentals of Specialization Theory}
\rightrunningtitle{\S2 An Elementary Treatise on Symmetric Bilinear Forms}

In this section a ``form'' will always be understood to be a {\it
nondegenerate} symmetric bilinear form over a field. So let $K$ be
a field.

\begtheorem{1} {\rm (``Witt decomposition'')}.
\setitemindent{$(a)$}
\item{$(a)$} Any form $\varphi$ over $K$ has a decomposition
$$\varphi \cong \varphi_0  \perp  \left( {a_1 \atop \hfill 1
\hfill} {1 \atop 0} \right)  \perp  \dots  \perp
\left( {a_r \atop \hfill 1
\hfill} {1 \atop 0} \right)$$
with $\varphi_0$ anisotropic and $a_1, \dots, a_r
\in K \quad (r \geq 0)$.
\smallskip
\item{$(b)$} The isometry class of $\varphi_0$ is uniquely determined
by $\varphi$. $($Therefore $\dim \varphi_0$ and the number $r$ are
uniquely determined.$)$
\endtheorem
\bigskip

\noi To clarify these statements, let us recall the following:
\smallskip
\setitemindent{(5)}
\item{(1)} A form $\varphi_0$ over $K$ is called {\it anisotropic}
if $\varphi_0(x,x)\not=0$ for all vectors $x\not=0$.
\smallskip
\item{(2)} If $\Char K\not=2$, then we have for every $a\in K^*$
that
$$\left( {a \atop 1} {1 \atop 0} \right) \cong
\left( {0 \atop 1} {1 \atop 0} \right) \cong \langle 1, - 1
\rangle \cong \langle a, - a \rangle.$$
Is $\Char K=2$ however and $a\not=0$, then
$\left( {a \atop 1} {1 \atop 0} \right)  \not\cong
\left( {0 \atop 1} {1 \atop 0} \right)$.
Indeed if $\varphi =
\left( {0 \atop 1} {1 \atop 0} \right)$ we have $\varphi (x, x) = 0$
for every vector  $x \in K^2$, while this is not the case for
$\varphi =
\left( {a \atop 1} {1 \atop 0} \right)$. In characteristic $2$ we
still have
$\left( {a \atop 1} {1 \atop 0} \right) \cong
\langle a, - a \rangle \quad (a \in K^*)$,
but
$\left( {a \atop 1} {1 \atop 0} \right)$ {\it need not} be
isometric to
$\left( {1 \atop 1} {1 \atop 0} \right) \cong \langle 1, - 1
\rangle$.
\smallskip
\item{(3)} The form $\left( {0 \atop 1} {1 \atop 0} \right)$
is  given the name ``hyperbolic plane'' (even in
characteristic~2), and every form $\varphi$, isometric to an orthogonal sum
 $r \times \left( {0 \atop 1} {1 \atop 0} \right)$ of $r$
copies of $\left( {0 \atop 1} {1 \atop 0} \right)$,  is
called ``{\it hyperbolic}'' $(r \geq 0)$.
\smallskip
\item{(4)} Forms which are isometric to an orthogonal sum
$\left( {a_1 \atop \hfill 1\hfill} {1 \atop 0} \right)  \perp
\dots \perp \left( {a_r \atop \hfill 1\hfill} {1 \atop
0} \right)$ are called {\it metabolic} $(r \geq 0)$.
If $\Char K\not=2$, then every metabolic form is hyperbolic. This
is not the case if $\Char K=2$.
\smallskip
\item{(5)} If $\Char K=2$, then $\varphi$ is hyperbolic exactly
when {\it every} vector $x$ of the underlying vector space $K^n$
is isotropic, i.e. $\varphi (x, x) = 0$. If $\varphi$ is not
hyperbolic, we can always find an orthogonal basis such that
$\varphi \cong \langle a_1, \dots,
a_n \rangle$ for suitable $a_i \in K^*$.


One can find a proof of Theorem~1 in any book about quadratic
forms when $\Char K\not=2$ (see in particular  [Bo$_1$], [L],
[S]). Part~(b) of the theorem is then an immediate consequence of
Witt's cancellation theorem. There is no general cancellation
theorem in characteristic~$2$, as the following example shows:
$$\left( {a \atop  1} {1 \atop 0} \right)  \perp  \langle -
a \rangle \cong \left( {0 \atop  1} {1 \atop 0} \right)  \perp
\langle - a \rangle \leqno(\ast)$$
for all $a \in K^*$. If $e,f,g$ is a basis of $K^3$ which has the
left-hand side of $(*)$ as value matrix, then $e+g,f,g$ will be a
basis which has the right-hand side of $(*)$ as value matrix. For
characteristic~$2$ one can find proofs of Theorem~1 and the other
statements we made in [MH, Chap.I and Chap.III, \S 1],
[K$_1$, \S 8], [M, \S 4]. The following is clear from
formula~$(*)$:

\beglemma{1.} If a form $\varphi$ with $\dim
\varphi = 2r$ is metabolic, then there exists a form $\psi$ such that $\varphi  \perp
\psi \cong r \times \left( {0 \atop  1} {1 \atop 0} \right)
\perp \psi$.
\endlemma

\begdefinition{1.}
(a) In the situation of Theorem~1, we call the
form $\varphi_0$ the {\it kernel form} of $\varphi$  and $r$ the
{\it $(\!$Witt$)$ index} of $\varphi$. We write $\varphi_0 = \ker
(\varphi)$, $r = \ind (\varphi)$. $\{$In the literature one
frequently sees the notation $\varphi_0 = \varphi_{\an}$
(``anisotropic part'' of $\varphi$).$\}$

(b) Two forms $\varphi$, $\psi$ over $K$ are called {\it Witt
equivalent}, denoted by $\varphi \sim \psi$, if $\ker \varphi
\cong \ker \psi$. We write $\varphi \approx \psi$ when
$\ker \varphi \cong \ker \psi$ and $\dim \varphi = \dim \psi$. On
the basis of the next theorem, we then call $\varphi$ and $\psi$
{\it stably isometric}.
\enddefinition

\begtheorem{2.} $\varphi \approx \psi$ exactly when there exists a
form $\chi$ such that $\varphi \perp \chi \cong \psi
\perp \chi$.
\endtheorem

We omit the proof. It is easy when one uses
Theorem~1, Lemma~1 and the following lemma.

\beglemma{2.} The form $\chi \perp (- \chi)$ is  metabolic for
every form $\chi$.
\endlemma

\begproof{} From Theorem~1(a) we may suppose that $\chi$ is
anisotropic. If $\chi$ is different from the zero form, then
$\chi \cong \langle a_1, \dots, a_n \rangle$ with elements $a_i
\in K^*$ $(n \geq 1)$. Finally, $\langle a_i \rangle \perp
\langle - a_i \rangle \cong
\left( {a_i \atop  \hfill 1\hfill} {1 \atop 0} \right)$. \hfill
$\square$
\endproof

As is well-known, Witt's cancellation theorem (already mentioned
above) is valid if $\Char K \ne 2$. It says that two stably
isometric forms are already isometric: $\varphi \approx \psi \Rightarrow
\varphi \cong \psi$.

Let $\varphi$ be a form over $K$. We call the equivalence class
of $\varphi$ with respect to the relation $\sim$, introduced above,
the {\it Witt class} of $\varphi$ and denote it by $\{ \varphi
\}$. We can add Witt classes together as follows:
$$\{ \varphi \} + \{ \psi \} \colon = \{ \varphi \perp
\psi \}.$$

The class $\{ 0 \}$ of the zero form, whose members are exactly
the metabolic forms, is the neutral element of this addition. From
Lemma~2 it follows that $\{ \varphi \} + \{ - \varphi
\} = 0$. In this way, the Witt classes of forms over $K$ form an
abelian group, which we denote by $W(K)$.
We can also multiply Witt classes together:
$$\{ \varphi \} \cdot \{ \psi \} \colon = \{ \varphi \otimes
\psi \}.$$

\begremark{} The definition of the tensor product $\varphi \otimes
\psi$ of two forms $\varphi$, $\psi$ belongs to the domain of
linear algebra [Bo$_1$, \S 1, No. 9]. For diagonalizable forms we
have
$$\langle a_1, \dots, a_n \rangle \otimes \langle b_1, \dots,
b_m \rangle \cong \langle a_1 b_1, \dots, a_1 b_m, a_2 b_1,
\dots, a_n b_m \rangle.$$
We also have $\langle a_1, \dots, a_n \rangle \otimes
\left( {0 \atop 1} {1 \atop 0} \right) \cong n \times
\left( {0 \atop 1} {1 \atop 0} \right)$. Finally, for a form
$\left( {b \atop 1} {1 \atop 0} \right)$ with $b \ne 0$ we have
$$\langle a \rangle \otimes \left( {b \atop 1} {1 \atop 0}
\right) \cong \langle a \rangle \otimes \langle b, - b \rangle
\cong \langle ab, - ab \rangle \cong \left( {ab \atop \hfill
1\hfill} {1 \atop 0} \right).$$
Now it is clear that the tensor product of any given form and a
metabolic form  is again metabolic.  $\{$For a conceptual proof of
this see [K$_1$, \S 3], [MH, Chap.~I].$\}$ Therefore the Witt
class $\{ \varphi \otimes \psi \}$ is completely determined by the
classes $\{ \varphi \}$, $\{ \psi \}$, independent of the choice
of representatives $\varphi, \psi$.
\endremark

With this multiplication, $W(K)$ becomes a commutative ring. The
identity element is $\{ \langle 1 \rangle \}$. We call $W(K)$ the
{\it Witt ring} of $K$. For $\Char K \ne 2$ this ring was already
introduced by Ernst Witt in 1937 [W].

We would like to describe the ring $W(K)$ by generators and
relations. In characteristic $\not=2$ this was already known by
Witt [oral communication] and is implicitly contained in his
work~[W, Satz 7].

First we must recall the notion of determinant of a form. For $a\in K^*$,
the isometry class of a one-dimensional form $\langle a \rangle$ will again be
denoted by $\langle a \rangle$. The tensor product $\langle a \rangle \otimes
\langle b \rangle$ will be abbreviated by
$\langle a \rangle\langle b \rangle$. We have
$\langle a \rangle\langle b \rangle = \langle ab \rangle$ and
$\langle a \rangle \langle a \rangle = \langle 1 \rangle $.
In this way the isometry classes form  an abelian group of
exponent~$2$, which we denote by $Q(K)$. Given $a,b\in K^*$, it is
clear that $\langle a \rangle = \langle b \rangle $ exactly when
$b = ac^2$ for a $c\in K^*$. So $Q(K)$ is just the {\it group of
square classes} $K^*/K^{*2}$ in disguise. We identify $Q(K) = K^*/K^{*2}$.

It is well-known that for a given form $\varphi = (a_{ij})$ the
square class of the determinant of the symmetric matrix $(a_{ij})$
only depends on the isometry class of $\varphi$. We denote this
square class by $\det (\varphi)$, so $\det
(\varphi) = \langle \det (a_{ij}) \rangle$, and call it the {\it
determinant of} $\varphi$. A slight complication arises from the fact that
 the determinant is not
compatible with Witt equivalence. To remedy this, we introduce the {\it
signed determinant}
$$d(\varphi) \colon = \langle -1 \rangle^{\frac{n(n-1)}{2}}
\cdot \det (\varphi)$$
$(n \colon = \dim \varphi)$. One can easily check  that
$d (\varphi \perp \left( {a \atop 1} {1
\atop 0} \right) ) = d (\varphi)$, for any $a\in
K$.
Hence $d(\varphi)$ depends only on the Witt class $\{ \varphi \}$.
The signed determinant $d(\varphi)$ also has a disadvantage
though, in contrast with $\det(\varphi)$, $d(\varphi)$ does not
behave well with respect to the orthogonal sum. Let $\nu
(\varphi)$ denote the {\it dimension index} of $\varphi$,
$\nu(\varphi) = \dim
\varphi + 2 {\Bbb Z} \in {\Bbb Z}/2 {\Bbb Z}$. Then we have (cf. [S,
I~\S2])
$$d(\varphi \perp \psi) = \langle - 1 \rangle^{\nu(\varphi)
\nu (\psi)} d(\varphi) d (\psi).$$

Let us now describe $W(K)$ by means of generators and relations.
Every one-dimensional form $\langle a \rangle$ satisfies
$d(\langle a \rangle) = \langle a \rangle$. This innocent remark
shows that the map from $Q(K)$ to $W(K)$, which sends every
isometry class $\langle a \rangle$ to its Witt class $\{ \langle a \rangle
\}$, is injective. We can thus interpret $Q(K)$ as a subgroup of
the group of units of the ring $W(K)$, $Q(K) \subset W(K)^*$.

$W(K)$ is additively generated by the subset $Q(K)$, since every
non-hyperbolic form can be written as $\langle a_1, \dots, a_n
\rangle = \langle a_1 \rangle  \perp
\dots \perp \langle a_n \rangle$. Hence $Q(K)$ is a
system of generators of $W(K)$.  There is an
obviously surjective ring homomorphism
$$\Phi \colon \; {\Bbb Z} [Q(K)] \twoheadrightarrow W(K)$$
from the group ring ${\Bbb Z} [Q(K)]$ to $W(K)$.
Recall that ${\Bbb Z} [Q(K)]$ is the ring of formal sums $\sum\limits_g n_g
g$ with $g \in Q(K)$, $n_g \in {\Bbb Z}$, and almost all $n_g =
0$. $\Phi$ associates to such a sum an in $W(K)$ constructed
sum $\sum\limits_g n_g g$.

The elements of the kernel of $\Phi$ are  the relations on
$Q(K)$ we are looking for. We can write down some of those
relations immediately: for every $a \in K^*$ is
$\langle a \rangle + \langle -a \rangle $ clearly a relation. For
$a, b \in K^*$ and given $\lambda, \mu \in K^*$, the form
$\langle a,b \rangle$ represents the element
$c \colon = \lambda^2 a + \mu^2 b$. If $c \ne 0$, then we can find
another element $d \in K^*$ with $\langle a,b \rangle \cong \langle c, d \rangle $
Comparing determinants shows that $\langle d \rangle = \langle abc
\rangle $. Hence
$\langle a \rangle + \langle b \rangle
- \langle c \rangle - \langle abc \rangle = (\langle a \rangle
+ \langle b \rangle) (\langle 1 \rangle - \langle c \rangle)$
is also a relation. We have the technically important

\begtheorem{3.} The ideal $\Ker \Phi$ of the ring ${\Bbb Z}
[Q(K)]$ is additively generated $($i.e. as abelian group$)$  by the
elements $\langle a \rangle + \langle -a \rangle$, $a \in K^*$
and the elements $\langle a \rangle + \langle b \rangle -
\langle c \rangle - \langle abc \rangle$ with $a, b \in K^*$,
$\langle b \rangle \ne \langle -a \rangle$, $c = \lambda^2 a +
\mu^2 b$ with $\lambda, \mu \in K^*$.
\endtheorem

\begremark{} $\Ker \Phi$ is therefore generated  as an ideal  by the
element $\langle 1 \rangle + \langle -1 \rangle$ and the elements
$(\langle 1 \rangle + \langle a \rangle) (1 -
\langle c \rangle)$ with $\langle a \rangle \ne
\langle -1 \rangle $, $c = 1 + \lambda^2 a$ with $\lambda \in
K^*$. For application in the next section, the additive
description of $\Ker \Phi$ above  is more favourable though.
\endremark

A proof of Theorem~3, which also works in characteristic $2$, can
be found in [K$_1$, \S 5], [KRW, \S 1], [K$_2$ II, \S 4] (even
over semi-local rings instead of over fields\fonote{The case where
$K$  has only $2$ elements, $K = {\Bbb F}_2$, is not covered by the
more general theorems there. The statement of Theorem~3 for $K = {\Bbb
F}_2$ is trivial however, since $K$ has only  one square class $\langle 1
\rangle$ and $\langle 1,1 \rangle \sim 0$.}), [MH, p.85]. For
characteristic $\not=2$  the proof is a bit simpler, since every
form has an orthogonal basis in this case, see [S,~I~\S~9].
